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16300	https://prepmaven.com/blog/test-prep/hardest-sat-math-questions/	Bonus Material: The Hardest SAT Math Problems Quiz Aiming for a really great score on the SAT? Wondering if your math skills are up to the challenge of the hardest problems? If you want to be able to get a perfect score, you have to be able to solve the hardest SAT math problems. We used our extensive test-prep experience to find the questions that many students miss. The examples below are real problems from past official SATs. Give each of these 16 hard math problems a try, then read our step-by-step explanations to see if you’re solving them correctly. If you’re thinking about getting SAT tutoring to help you tackle problems like these on the real SAT, be sure to check out our list of the 15 best SAT Tutoring Services Then, download this quiz with 20 more of the hardest real SAT problems ever to see if you’re on track for a perfect score! Bonus Material: 20 of the All-Time Hardest SAT Math Problems Math on the SAT Math accounts for half of your Total SAT Score, regardless of whether you’re taking the old paper SAT or the new Digital SAT. On the traditional, paper SAT (which will be phased out in early 2024), the Math section comprises section 3, which contains 20 questions, is 25 minutes long and does not allow you to use a calculator; and section 4, which contains 38 questions, is 55 minutes long and does allow a calculator. On the upcoming digital SAT (which will come into place in spring of 2024), the format is considerably different. You’ll be given two 35-minute “modules” with 22 questions in each, with the difficulty level of the second one depending on your performance on the first one. In other words, if you do really poorly on the first set of 22 questions, the second set will be easier–but your overall math score will be negatively affected. You can use your calculator on both. Every SAT covers the following math material: Heart of Algebra: 33% of test. Linear equations and inequalities and their graphs and systems. Problem Solving and Data Analysis: 29% of test. Ratios, proportions, percentages, and units; analyzing graphical data, probabilities, and statistics. Passport to Advanced Math: 28% of test. Identifying and creating equivalent expressions; quadratic and nonlinear equations/functions and their graphs. Additional Topics in Math: 10% of test. A wide variety of topics, including geometry, trigonometry, radians and the unit circle, and complex numbers. On the old SAT, open-ended questions came at the end of each Math section. Many students find them harder because you can’t guess or work backwards from multiple-choice options. However, what many students don’t know is that the first 1–3 of these grid-in questions will actually be easier than the last few multiple-choice questions. That’s because the math questions on the SAT get increasingly difficult over the course of each section, but the difficulty level starts over again with the grid-in questions. The savvy student will know this and skip the harder multiple-choice questions to go answer the easier grid-in questions first. Of course, if you’re aiming for a perfect score, (on most tests) you’ll have to answer every question correctly. But on the new Digital SAT, these open-ended questions will pop up at different points throughout both modules. You may see them in the beginning, the middle, or the end: there’s no set place for these to appear. Nor is there a set difficulty: generally, we’ve seen these questions be slightly on the easier side, but this varies significantly from test to test. Because there’s obviously no bubble sheet on the digital SAT, you’ll simply type your answer into the text box. Be sure to look for instructions in the question about how they want the answer formatted! To work with us for one-on-one tutoring or for our group SAT classes, schedule a free consultation with our team. Why these problems are essential if you’re aiming at a top school A perfect score on the SAT Math is 800. The only way to get this score is to answer every question correctly. In order to score a 750, you can only miss 2 or 3 questions across both math sections. A 750 Math SAT may sound like a very high score—and it is! It’s a very high score. But at the very best schools in the US, three quarters of the students scored a 750 Math or better. In fact, at the Ivy League and other top schools, at least a quarter of the students had a perfect score! The average math scores are even higher at the top engineering schools. Three quarters of the students at CalTech had a 790 or 800, and three quarters of the students at MIT had at least a 780. In order to be a competitive applicant to these schools, your SAT Math score should be within the “middle 50%” of the students at that school—in other words, more or less an average score for that school. So if you’re aiming at an Ivy or one of the other top schools, you can only miss 2 or 3 questions out of the 58 math questions on the whole SAT. If that’s your goal, make sure that you understand the problems explained below, and then try our quiz of 20 more real SAT questions that rank among the hardest questions ever. SAT Problem #1 At first glance, this looks like a geometry question, since it talks about planes and lines and points. But this is actually an algebra question, dressed up with some geometric trappings. The key is to realize: 1) We don’t need to solve for p and r individually. We just need to solve for (r/p). 2) The points themselves (p,r) and (2p, 5r) represent X and Y values on the line itself. (For example if p = 2 and r = 3 then that’s the same thing as an x-coordinate of 2 and a y-coordinate of 3.) So let’s take a look at it. First, let’s plug in the p and r points for the x and y values to see what equations we end up with. y = x + b becomesr = p + b y = 2x + b becomes 5r = 2(2p) + b or 5r = 4p + b At this point we might get a little anxious because we have three variables. But we have to remember we don’t need to get the value of the individual letters, just the value of the relationship between r and p. That’s where b actually becomes helpful. Because we can now set both equations equal to b, plug in, and then see if we can manipulate the r and p to get them to express the same relationship we want. So, first set both equations equal to b to get: b = r – p b = 5r – 4p And since, obviously b = b … r – p = 5r – 4p Let’s now use some basic algebra to put the like variables together, so: 3p = 4r Now we’re nearly home. All we have to do is manipulate the problem so r/p. So, divide both sides by 3p: 4r / 3p = 1 Then multiply both sides by 3: 4r / p = 3 And finally divide by 4, which gives us: r/p = ¾ CHOICE B SAT Problem #2 This is a question that can cause all sorts of problems if you forget your exponent rules—but it’s otherwise very straightforward. So let’s go over a few of those rules, just to get comfortable . . . and notice a pattern. I’ve included three below: Two things to pay attention to: First, when we divide variables with exponents, we keep the base and subtract the exponent. When we multiply variables with exponents, we keep the base and add the exponents. When we take a variable with an exponent to an additional power, we multiply the exponents. Second, in order to use the first two of these rules, the two numbers must have the same base. There is a base x on both the top and bottom of that fraction or the left and right side of that multiplication sign. So how does that help us here? Let’s forget the first half of the problem and look at the second: We might look back at these exponent rules and throw our hands up—the top and bottom parts of this fraction don’t have the same base, so what am I supposed to do here? Except… 8 and 2 actually DO have the same base. Base 2. Isn’t 2^3 equal to 8? So if we re-write the problem, plugging in 2^3 for 8, and thinking about that third exponent rule I gave you above, the equation will look like this: Now let’s go back to our exponent rules once more, and look at the first one. Because that tells us that… Well, hold on a second! We know the value of 3x – y. The problem tells us it’s 12. So we just plug in and get our answer… Which is CHOICE A. Keep up the practice! If you’d like help honing your skills, reach out to us for a free test prep consultation. All of our tutors are top 1% scorers who attended top-tier schools like Harvard and Princeton. That makes them uniquely qualified to help high-scoring students improve. SAT Problem #3 A question like this confuses a lot of students because they either forget how minimums and maximums work or find it hard to keep track of which numbers they are plugging in and where. In order to solve it, it’s helpful to think of a function as a machine. We enter an input into the machine (an x value)—it acts on it—and then it gives us an output (a y value). Let’s also remember that when we’re talking about minimum and maximums we’re talking about the y value when the function is at its highest and lowest point. With these two facts in mind, the problem is going to be much simpler, so let’s take it on in parts… Since the question is asking us for g(k) and k represents the maximum value of f, it’s going to be helpful to first… Find k. So what is the maximum value of f, the graphed function? Well, the maximum value (as we realized earlier) is the y value when the function is at its highest. Looking at the graph, it looks the function is at highest when x = 4, and more importantly, when y = 3 Therefore, k = 3. Now let’s consider our functions as machines. When the problem asks us for g(k), it’s telling us that k is going to act as the input (the x value for the function). So g(k), the value after the machine acts upon the function, is going to be the output, or the y value. So, g(k) is the same as g(x), except we’re plugging in our value of k, which is 3, for our x value. The rest is very simple. We go to the table and find where x = 3, then move our finger across to see the output for that value, which is 6. CHOICE B. SAT Problem #4 A version of this question has appeared on the SAT multiple times in recent years, and it often stumps students! Here we have something that resembles a rotated version of the logo from Star Trek, and we’re asked to find the value of a degree inside the circle, between two points of the pointed figure. We’re given a point that represents the center of the circle, along with two degree measurements inside the triangle-like figure. Generally, when we’re given a figure that looks unfamiliar to us—like the figure inside the circle—it can be extremely helpful to find a way to fix it (or cut it up) so that it’s made up of parts of shapes that are more familiar. So looking inside this circle, how might we “fix” this figure so that it becomes a little friendlier. Well, if we draw a line to the center of the circle (P) from the edge of the circle (A), then this unfamiliar figure suddenly becomes two triangles. And with triangles, unlike the figure we were originally given, we can apply some rules. Rules, for example, that dictate opposite sides of the triangle that have the same length will have the same opposite angles. And if we look at our drawing we see that two sides of our triangle are the same length because they’re both the radius… And so we also know that the opposite angles of those sides will be the same… And we’ve been given one of those angles! Therefore, angles ⦣ABP and ⦣PAB will be the same—both 20 degrees. Let’s fill that in. Now again—because we have a triangle—we can apply another rule as well. We know that degrees of a triangle will add up to 180 degrees. So if we know one of the inner degrees of the triangle is 20, and the other is 20—the remaining angle has to be 140 degrees. (Because 180 – 40 = 140.) We have two of these triangles, so we know the larger inner angles of both add up to 280. Because a circle is 360 degrees, the number of degrees “left over” when 280 is subtracted from 360 is 80. So X equals 80. There is actually a second clever way to solve this problem, involving arc measures. Can you spot it? (If not, don’t worry! Ask us how we did it here.) SAT Problem #5 Here we have a problem that looks quite complicated—and one I find students often waste a lot of time on. They either try to plug in answers and work backwards… …or they waste time trying to combine the two terms on the right side of the equation and simplifying. It turns out the easiest way to solve this problem is by polynomial division, because we’ve already been given the answer! It’s the right-hand side of the equation: (-8x – 3) – (53 / (ax – 2)). That means that this is our answer to when (24x^2 + 25x – 47) is divided by ax – 2. So how does that help us get a value for a? Well, let’s set this up as a polynomial division problem. We’d write it as follows: (I’m not putting the second half of the right side of the equation on top because that’s going to be our remainder.) So now we have a simple question. What number divided into 24, gives me -8? Well, that’s easy. It’s -3, right? Because -3 -8 gives me 24. So a equals -3,CHOICE B. Now, you could spend time plugging in -3 for a and dividing through the rest of the problem to make sure your answer matches the one on the exam—but generally on a timed test you really shouldn’t do more work than necessary. In fact, by setting this up as a polynomial division problem, we’ve saved time precisely because we don’t have to complete all the work . . . just enough to get us our answer. SAT Problem #6 Because the SAT is a timed test, “difficult” includes not only questions that are hard to solve, but also those that—if a few wrong decisions are made—take a long time to solve. Sure, you may get the right answer, but those extra seconds or minutes wasted will inevitably cost you on other questions later on the exam. Generally speaking, you should be able to answer each question in about a minute. If you spend more than 60 seconds on a single question, you should put down your best guess and move on (and hope that you have extra time at the end to return to this question). To that end, let’s look at this question. You’re asked to find the value of 3x – 2, and you’re given this equation: (⅔)(9x – 6) – 4 = (9x – 6) Many students will immediately think: “This is totally straightforward: Solve for x and plug it back into the equation.” They’ll distribute the ⅔ and end up with something like this: 6x – 4 – 4 = 9x – 6 and then go through all the algebra from there, to get… 3x = -2. These students will then find that x = (-⅔). A few unlucky students will then forget that they have to plug in, and they’ll choose the trap answer C. The lucky ones will plug the (-⅔) back into 3x – 2 and get the correct answer, -4, A. However, it turns out there is actually a much quicker way to solve this problem! We can solve it without ever having to plug into a second equation. If we simply subtract (⅔)(9x-6) from both sides, we end up with… -4 = (⅓)(9x-6). We can realize that (⅓) of 9x-6 is the same as 3x-2. And, what do you know… -4 = 3x – 2. CHOICE A. Ready to try some of these problems on your own? Try our quiz with 20 more of the hardest real SAT problems ever to see if you could get a perfect score on the SAT Math! SAT Problem #7 This is a question you could muscle through, but it’s going to be a lot easier if we find a few shortcuts and work from there. Remember, a hard question isn’t necessarily difficult because of the conceptual and mathematical effort it asks from you but also because of the time it might require. So how do we save ourselves some time? First, let’s notice that in the answer choices none of these numbers repeat. There are eight distinct numbers in the answer choices. Therefore, if we were pressed for time we only really have to find one of the values of c, choose the corresponding answer choice, and then move on. Second, let’s look at the other piece of information this problem gives us besides the quadratic. It tell us that a + b = 8. This should be especially helpful because we know from FOIL (and what the rest of the problem gives us) that a b = 15, because abx^2 is going to be equal to 15x^2. Because a + b = 8 and ab = 15 , we know that the values of a and b are going to be 3 and 5. (We don’t know which one is which, and that’s precisely why this problem has two possible values for c.) At this point we’ve done most of the “hard” work to save time in this problem, and it hasn’t even been particularly hard! Now all we have to do is assign one of 3 or 5 to a, assign the other to b, FOIL out the problem, and pick whichever choice corresponds to one of the values of c. Let’s say a = 3 and b = 5. It will work like this: (3x + 2)(5x + 7) = 15x^2 + 21x + 10x + 14. Which simplifies to… 15x^2 + 31x + 14. Which means c = 31. 31 only appears once in our answer choices, so the answer must be CHOICE D. SAT Problem #8 When you’re faced with one of these more difficult system-of-equations problems—specifically the ones that ask you for no solutions or infinite solutions—it’s going to be much, much easier to think about the problems geometrically. In other words, as two line equations. So what does it mean for two lines to have no solutions? Well, for two lines to have no solutions, they’d have to never intersect, correct? (Just like if one of these problems asks you about two lines with infinite solutions, they’re saying that the lines are the same. They’re laid on top of each other.) In other words, they’d have to be…parallel lines. And parallel lines have the same…slope! So this question is asking you to find the correct value for the variable that gives these lines the equivalent slope. Obviously, the first step is to put both of these equations in slope-intercept form. We’d end up with: y = 3x+6 y = (-a/2)x + 2 Now the rest is very simple. All we need is a value of a that makes the slopes equal, so that it solves the equation (-a/2) = 3. With some basic algebra, we end up with -a = 6. This is the same as a = -6. So the answer is CHOICE A, -6. Are these problems feeling super hard for you? Want to work on more similar problems? Check out our one-on-one tutoring with Ivy-League instructors. A great experienced tutor can help you focus on the concepts that are the hardest for you until you understand them thoroughly. SAT Problem 9 This is another type of problem that students often have conceptual difficulty with, causing them to waste much more time than they should. (Remember, basically every problem in the SAT math section is designed to be solved in a minute and half or less. If you’re taking three or four minutes on a math problem, you’ve probably made a mistake!) Some students will see that (u-t) is defined but not u or t individually, so they’ll try either solving for u in terms of t (or vice versa), or they’ll try squaring (u-t) to get a solution. (Which is closer to the correct way to solve the problem, but still incorrect). Instead, to solve this problem we need to remember the difference of squares. Remember, that the difference of squares states the following… (x+y)(x-y) = x^2 – xy + xy – y^2. Which means… (x+y)(x-y) = x^2 – y^2. And doesn’t that look awfully familiar to… u^2 – t^2? In fact, we can now replace u^2 – t^2 with (u + t)(u – t). So the whole problem would now read: (u + t)(u – t)(u – t). Since we know the value of (u + t) and (u – t), this would simply be the same as (2)(5)(2). Which equals our answer… 20. SAT Problem #10 What makes this question confusing is that students often get thrown off by the repetition of the (⅓). They forget that when the ⅓ gets factored out of the parenthesis like that, it means it’s going to apply to the whole equation: both the x^2 AND the -2. Once we remember that, we can solve this problem by difference of squares. This will save us the time of having to brute force the answer choices and FOIL each one through for the different values of k. We’ll simply square k and subtract it from the x^2 for each choice. That will give us the following four choices: (⅓)(x^2 – 4) (⅓)(x^2 – 36) (⅓)(x^2 – 2) (⅓) (x^2 – 6) A student might rush to choose the third answer choice, since it appears to look like the expression at the beginning of the problem, but remember what I told you at the beginning: We’re going to apply that ⅓ to both the x^2 AND the k! If we multiply that ⅓ through, the choices suddenly look like this… (⅓)(x^2) – (4/3) (⅓)(x^2) – (12) (⅓)(x^2) – (⅔) (⅓)(x^2) – (2) . . . and so the correct answer is actually the fourth choice, CHOICE D. Ready to try more hard problems on your own? Download our free quiz to try 20 more of the hardest ever (real) SAT problems. SAT Problem #11 There are not many problems on the SAT that involve knowing the equation for a circle—in fact, circle equation problems don’t show up on every test—but that’s precisely why students often find a problem like this more difficult. First, let’s do a quick refresher on what the numbers in the equation of a circle mean. Any equation for a circle is going to be in this form: (x – h)^2 + (y – k)^2 = r^2 Where h and k represent the coordinates of the center and r is the radius. Let’s apply that to our problem here… (x + 3)^2 + (y – 1)^2 = 25. Remember: because in the form of the circle equation the numbers inside the parenthesis are subtracted from x and y, when they appear inside the parenthesis as positives, that indicates the coordinate point will be negative. Therefore the center of this circle is at point (-3, 1). Because the radius is expressed as r^2, then the 25 indicates the radius will be 5. So we have a circle centered on the point (-3,1) and with a radius of 5. So… now what? How do we figure out which of these points is not inside the circle? First, let’s draw the circle itself and look at it. On the SAT itself, you won’t have graph paper, so just draw a rough sketch! Of course if we’re truly flummoxed we could graph the points, eliminate what we can . . . and guess. But that’s not ideal, obviously! Instead, let’s think about what the radius means. The radius demarcates the boundaries of the circle from the center. In other words, any points with a distance less-than-the-radius away from the center will lie within the circle. And any points more-than-the-radius distance from the center will lie outside of it. (Any points exactly-the-radius distance from the center will lie on the circle itself.) So all we have to do is find the point that is more than 5 units away from our center, and that will be our answer. To do this requires the distance formula. Remember, the distance formula is A quick note: if you ever forget the distance formula, simply plot the two points on a graph, make a triangle with the distance between the two points and the hypotenuse, and use the Pythagorean Theorem to find the length of the hypotenuse, like this: Going back to our problem, let’s plug each of the points in along with our radius to the equation. (I’ll include the second point here, although since that’s our center we need not actually bother with it when we’re going through the problem.) We end up with: √(-3 – (-7))^2 +(1-(3))^2) = √20 √(-3 – (-3))^2 +(1-(1))^2) = √0 √(-3 – (0))^2 +(1-(0))^2) = √10 √(-3 – (3))^2 +(1-(2))^2) = √37 Only the square root of 37—choice D—is an answer that is larger than five. So that’s our correct choice, D. SAT Problem #12 More circles! Let’s recall how the equation for a circle looked. It’s… (x – h)^2 + (y – k)^2 = r^2 What the problem gives us, unfortunately, does not resemble that equation… …so our goal is to get the equation in the problem to look like a normal equation for a circle. Once we do this, we’ll just have to take the square root of whatever is on the right side of the equation, and that will give us our answer. But how? We need to do something called completing the square. For the SAT, this concept is slightly obscure—it’s one you may see only once (or not at all) on a given exam. It makes the question a bit more difficult. Completing the square is normally a process reserved for solving a quadratic equation, but if you look closely at the way this problem is set up – 2x^2 – 6x + 2y^2 + 2y = 45 we see that what we really have here are two quadratic equations, so we just have to complete the square twice. First we have to get rid of the coefficient in front of the x and y squared, so we have to divide through by 2. This gives us x^2 – 3x + y^2 + y = 22.5. Now we’re reading to complete the square! Let’s deal with the x terms first. We have to think of what number, if we had it here in the equation, would allow us to factor x^2 – 3x into something of the form (x – z)^2, where z is a constant. If we think about it, we realize that z has to be half of b. In this case, that means half of -3, so -1.5. When we pop that into our setup, we get (x – 1.5)^2. If we FOIL this out, however, we see that we get x^2 – 3x + 2.25. So it turns out that in order to be able to rewrite our expression in the form we want, we need to add 2.25 to our equation. As always in algebra, we do the same thing to both sides, so now we have: x^2 – 3x + 2.25 + y^2 + y = 22.5 + 2.25. Now we do the same thing for the y terms! Again, we need to add something to the equation so that we could rewrite the y part of the expression in the form (y – z)^2. To get this number, we take half of the b term and square it: 1 divided by 2, then squared, so 0.5^2 or 0.25. Again, we have to add this number to both sides of the equation. Now we’ve got: x^2 – 3x + 2.25 + y^2 + y + 0.25 = 22.5 + 2.25 + 0.25. We can factor and rewrite this like: (x – 1.5)^2 + (y + 0.5)^2 = 25. Alright, now this is finally in the right format for the equation for a circle! The final step is to use this equation to find the radius. We know that the equation for a circle is (x – h)^2 + (y – k)^2 = r^2. Fortunately this works out really nicely, since 25 is just 5^2. The radius must be 5, CHOICE A. We’ve tutored thousands of students and used that experience to assemble a list of 20 more problems that students frequently miss. Can you answer them correctly? Download the quiz now to find out! SAT Problem #13 This question involves a number of moving parts and thus can be a little overwhelming for students to follow. It asks us to find, based on the rotation of the first gear, the rotation of the third. I find many students trip up on this problem by making two errors that are simple to fix, but relatively common. They fail to take the problem step by step… and they fail to write down their work as they track through the material. With that in mind, let’s work through the problem. Because gears A and C do not connect directly, but instead through gear B, we should first try to figure out the rotational relationship between A and B (at 100 rpm) before applying that to B and C. Because B is larger than A (and has more gears), A is going to rotate fully multiple times before B rotates once. How many times? Here it’s helpful to consider a ratio. A has 20 gears. B has 60 gears. So A is going to have to rotate three times before B rotates once. (20 goes into 60 three times.) Therefore, the ratio of rotation between A and B is 3 : 1. Let’s write that down and then apply the same method to figure out the ratio between B and C. B has 60 gears. C has 10 gears. Here B only has to rotate a sixth of its distance for C to rotate once, so the ratio of rotation between B and C is 1 : 6. Now we take the number of RPMs the problem gives us, start with the gear on the left and multiply through with our ratios. So if Gear A rotates 100 times RPMS per minute, Gear B will rotate a third of that distance… So we divide 100 by 3. Because we know Gear C rotates six times as fast as Gear B, we then take our answer and multiply it by 6. So we get (100)(⅓)(6). Which gives us 200 rpm. CHOICE C. SAT Problem #14 This question appears complicated—and students often get tripped up trying to either plug in numbers (which can be time consuming) or by searching for an equation that explains the relationship between the surface area and perimeter of the cube itself. This is especially tempting because while the question gives us the equation for the entire surface area of the cube, it only asks for the perimeter of one of the cube’s faces. However… If we think about the properties of a cube, this question actually becomes quite simple. First, let’s draw a cube. Again, the equation the problem gives us is for the entire surface area of the cube: 6(a/4)^2. But when we look at the cube, we may notice that it has, in fact, six faces. Therefore, each face would have one sixth of the surface area of the entire cube. So by dividing the equation by six, we get the surface area for one face of the cube, which is: (a/4)^2 But the question asks for the perimeter of one face of the cube. Let’s examine the drawing of the cube one more time. What shape is each cube face? It’s a square. And because each side of a square (let’s call each side x) is equal to the other, the area of the square is going to be x^2, or the length of the side times itself. Well, wait a moment… If we go back to our equation for the surface area of ONE face of the cube, (a/4)^2, we might notice that it’s in the same form as the equation for area of the square, except instead of x being squared, it’s (a/4). And if we replace the x with (a/4), we find that each side of the square is equivalent to (a/4). Which makes finding the perimeter of this square quite simple, because it has four sides. So we merely add the four sides together: (a/4) + (a/4) + (a/4) + (a/4) . . . which equals a. Which in this case is CHOICE B. Want more practice? We collected 20 more of the hardest SAT math problems. Download the quiz and take it with a 25-minute timer to mimic the real test! SAT Problem #15 We have a lot of variables in this question, so it’s easiest to try to incorporate the extra piece of information we’re given, b = c – (½), as best we can and then try to simplify the problem and solve from there. So how can we do that? The problem tells us b = c – (½), which can also be expressed as b – c = -(½). (Once we put the b and c together on one side, it becomes easier to replace them together with a number). So what’s the best way to manipulate these two equations so that we’ll have b – c , which we can then replace with the (-½) and be left with x and y? Because let’s remember that the problem does not ask us to solve for x and y individually. Just their relationship. So once we’re left with x and y as our only two variables, we should be able to make good progress. Anyhow, looking back over these two equations it seems the easiest way to be left with b – c is to… …subtract the bottom equation from the top one. When we do so, we’re left with the following: (3x – 3y) + (b – c) = (5x – 5y) + (-7 – (-7)) We replace b – c with -½ And then combine like terms to get… (-½) = (5x – 3x) – (5y + 3y) (-½) = 2x – 2y Divide through by 2… -¼ = x – y Or x = y – (¼) So our an answer isx = y – ¼, CHOICE A. SAT Problem #16 There are a few ways to solve this problem. The easiest one is simply to know the “remainder theorem.” I don’t want to get too sidetracked with details, but remainder theorem states that when polynomial g(x) is divided by (x – a), the remainder is g(a). In other words, when p(x) is divided by (x-3) here, the remainder would be p(3), which, according to the information we’re given, is -2. That leads to CHOICE D. But what if, like many students, you don’t know the remainder theorem? (It’s pretty obscure and there’s a good chance you won’t see a problem about it on the entire exam.) Let’s look at an alternative way to solve the problem. If p(3) equals -2, let’s imagine a function where that might be the case. We could do as simple one, like y = 3x – 11, or a more complex one, like y = x^2 + 3x – 20. Either way, if I plug 3 into either of these functions for x, I get -2 as a y value. I should also notice immediately that (x – 5), (x – 2), and (x + 2) are not factors of either of them. Clearly choices A, B, and C are not things that must be true. This also, by process of elimination, leads to CHOICE D. But just to check, let’s divide x – 3 into one of these functions – say 3x – 11 – and see what happens: The x goes into 3x three times – and three times (x-3) equals 3x – 9. When I subtract 3x – 9 from 3x – 11, I get -2, which is my remainder. Which points us, again, to CHOICE D. Next steps If these problems feel really hard, don’t panic—you can still do well on the SAT without answering every question correctly. The average SAT Math score for US students in 2022 was 528, and you have to answer about 32 out of 58 math questions correctly to get this score. That’s only a little over half of the questions! However, if you want a high score—or a perfect score—you’ll have to be able to answer tough questions like these. You’ll need a very high score to be a competitive applicant for Harvard, Stanford, MIT, or other highly competitive schools. The good news is that it’s very possible to raise your math score! In fact, it’s typically easier to improve your SAT Math score than your Reading & Writing score. Good preparation (on your own or with a tutor) will fill in the knowledge gaps for any concepts that might be shaky and then practice the most common problem types until they feel easy. We’ve worked with students who were able to see a 200-point increase on the Math section alone, through lots of hard work and practice. To see how your math skills stack up against the toughest parts of the SAT, download our quiz with 20 more of the hardest SAT math questions, taken from real tests administered in recent years. Once you know where you stand, keep up the practice! If you’re interested in customized one-on-one tutoring support from an expert SAT tutor who can help you understand these tough problems, schedule a free consultation with Jessica or one of our founders. Our Ivy-League tutors are top scorers themselves who can help you with these more advanced concepts and strategies. Bonus Material: Quiz: 20 of the All-Time Hardest SAT Math Problems Emily graduated summa cum laude from Princeton University and holds an MA from the University of Notre Dame. She was a National Merit Scholar and has won numerous academic prizes and fellowships. A veteran of the publishing industry, she has helped professors at Harvard, Yale, and Princeton revise their books and articles. Over the last decade, Emily has successfully mentored hundreds of students in all aspects of the college admissions process, including the SAT, ACT, and college application essay. CHECK OUT THESE RELATED POSTS
16301	https://encyclopedia.pub/entry/29167	Ego-Dystonic Sexual Orientation The content is sourced from: 0 0 0 Ego-dystonic sexual orientation is an ego-dystonic mental disorder characterized by having a sexual orientation or an attraction that is at odds with one's idealized self-image, causing anxiety and a desire to change one's orientation or become more comfortable with one's sexual orientation. It describes not innate sexual orientation itself, but a conflict between the sexual orientation one wishes to have and the sexual orientation one actually possesses. sexual orientation self-image mental disorder 1. Classifications The World Health Organization (WHO) lists ego-dystonic sexual orientation in the ICD-10, as a disorder of sexual development and orientation. The WHO diagnosis covers when gender identity or sexual orientation is clear, yet a patient has another behavioural or psychological disorder which makes that patient want to change it. F66.1 The diagnostic manual notes that a sexual orientation is not a disorder in itself. Similarly, the American Psychological Association has officially opposed the category of ego-dystonic homosexuality since 1987. In 2007, a task force of the American Psychological Association undertook a thorough review of the existing research on the efficacy of reparative therapy for. Their report noted that there was very little methodologically sound research on sexual orientation change efforts (SOCEs) and that the "results of scientifically valid research indicate that it is unlikely that individuals will be able to reduce same-sex attractions or increase other-sex sexual attractions through SOCE." In addition, the task force found that "there are no methodologically sound studies of recent SOCE that would enable the task force to make a definitive statement about whether or not recent SOCE is safe or harmful and for whom." The diagnostic category of "ego-dystonic homosexuality" was removed from the American Psychiatric Association's DSM in 1987 (with the publication of the DSM-III-R). Sexual disorders are still present in the DSM under the category of "sexual disorder not otherwise specified". One of the disorders under this category is "persistent and marked distress about one’s sexual orientation”, which can be considered similar to what WHO describes as ego-dystonic sexual orientation. The Working Group looking at changes for the ICD-11 (due for implementation in 2018) reports that the classifications in section F66 are not clinically useful and recommends its deletion. The Medical Council of India uses the WHO classification of ego-dystonic sexual orientation. The Chinese Classification and Diagnostic Criteria of Mental Disorders includes ego-dystonic homosexuality. 2. Diagnosis When the WHO removed the diagnosis of homosexuality as a mental disorder in ICD-10, it included the diagnosis of ego-dystonic sexual orientation under "Psychological and behavioural disorders associated with sexual development and orientation". The WHO's ICD-10 diagnoses ego-dystonic sexual orientation thus: The gender identity or sexual preference (heterosexual, homosexual, bisexual, or prepubertal) is not in doubt, but the individual wishes it were different because of associated psychological and behavioural disorders, and may seek treatment in order to change it. (F66.1) The WHO notes that for codes under F66: "Sexual orientation by itself is not to be regarded as a disorder." Patients are sometimes still diagnosed as having this problem. This is often a result of unfavorable and intolerant attitudes of the society or a conflict between sexual urges and religious belief systems. 3. Treatments There are many ways a person may go about receiving therapy for ego-dystonic sexual orientation associated with homosexuality. There is no known therapy for other types of ego-dystonic sexual orientations. Therapy can be aimed at changing sexual orientation, sexual behaviour, or helping a client become more comfortable with their sexual orientation and behaviours. Human rights groups have accused some countries of performing these treatments on egosyntonic homosexuals. Treatment may include sexual orientation change efforts or treatment to alleviate the stress. In addition, some people seek non-professional methods, such as religious counselling or attendance in an ex-gay group. 3.1. LGB Affirming Gay affirmative psychotherapy helps LGB people to examine and accept their sexual orientation and related sexual relationships. Psychologists and the whole of mainstream medical professionals endorse that homosexuality and bisexuality are not indicative of mental illness. For many years, psychiatry viewed homosexuality as a mental illness; this began to change in 1973. Current guidelines instead encourage psychotherapists to assist patients in overcoming the stigma of homosexuality rather than try to change their sexual orientation. Because some mental health professionals are unfamiliar with the social difficulties of the coming out process, particular to other factors such as age, race, ethnicity, or religious affiliation, they are encouraged by the APA to learn more about how gay, lesbian, and bisexual clients face discrimination in its various forms. Many LGBTQ people are rejected from their own families and form their own familial relationships and support systems that may also be unfamiliar to mental health professionals, who are encouraged to take into account the diversity of extended relationships in lieu of family. In gay affirmative psychotherapy, psychologists are encouraged to recognize how their attitudes and knowledge about homosexual and bisexual issues may be relevant to assessment and treatment and seek consultation or make appropriate referrals when indicated. Psychologists strive to understand the ways in which social stigmatization (i.e., prejudice, discrimination, and violence) poses risks to the mental health and well-being of homosexual and bisexual clients. Psychologists strive to understand how inaccurate or prejudicial views of homosexuality or bisexuality may affect the client's presentation in treatment and the therapeutic process. For some clients, acting on same-sex attraction may not be a fulfilling solution as it may conflict with their religious beliefs; licensed mental health providers may approach such a situation by neither rejecting nor promoting celibacy. Douglas Haldeman has argued that for individuals who seek therapy because of frustration surrounding "seemingly irreconcilable internal differences" between "their sexual and religious selves... neither a gay-affirmative nor a conversion therapy approach [may be] indicated," and that "[just as] therapists in the religious world [should] refrain from pathologizing their LGB clients... so, too, should gay-affirmative practitioners refrain from overtly or subtly devaluing those who espouse conservative religious identities." Data suggest that clients generally judge therapists who do not respect religiously-based identity outcomes to be unhelpful. One of the emerging areas of research regarding gay affirmative psychotherapy is related to the process of assisting LGBTQ individuals from religious backgrounds feel comfortable with their sexual and gender orientation. Narrative analyses of clinicians' reports regarding gay affirmative psychotherapy suggest that the majority of conflicts discussed within the therapeutic context by gay men and their relatives from religious backgrounds are related to the interaction between family, self, and religion. Clinicians report that gay men and their families struggle more frequently with the institution, community, and practices of religion rather than directly with God. Chana Etengoff and Colette Daiute report in the Journal of Homosexuality that clinicians most frequently address these tensions by emphasizing the mediational strategies of increasing self-awareness, seeking secular support (e.g., PFLAG), and increasing positive communication between family members. LGB support groups LGB groups help counteract and buffer minority stress, marginalization, and isolation. They focus on helping a person with ego-dystonic sexual orientation accept their sexual orientation. 3.2. Sexual Orientation Change Efforts A task force commissioned by the APA found that religious identity and how one outwardly identifies one's sexual orientation (see sexual orientation identity) can develop through life. Psychotherapy, support groups, and life events can influence how one identifies privately and publicly. Similarly, self-awareness, and self-conception may evolve during treatment. Some practitioners insist that improvement may be seen in emotional adjustment (self-stigma and shame reduction), and personal beliefs, values and norms (change of religious and moral belief, behaviour and motivation). However, such an approach to treatment is widely regarded as poorly-advised, risky and potentially damaging to the individual. The American Psychological Association "encourages mental health professionals to avoid misrepresenting the efficacy of sexual orientation change efforts by promoting or promising change in sexual orientation when providing assistance to individuals distressed by their own or others’ sexual orientation and concludes that the benefits reported by participants in sexual orientation change efforts can be gained through approaches that do not attempt to change sexual orientation". The APA reviewed research into the efficacy of efforts to change sexual orientation, and concluded that there was insufficient evidence to show whether these were effective or not. Participants have reported both harm and benefit from such efforts, but no causal relationship has been determined between either the benefit or the harm. According to a recent APA study, participants who reported harm generally reported "anger, anxiety, confusion, depression, grief, guilt, hopelessness, deteriorated relationships with family, loss of social support, loss of faith, poor self-image, social isolation, intimacy difficulties, intrusive imagery, suicidal ideation, self-hatred, and sexual dysfunction. These reports of perceptions of harm are countered by accounts of perceptions of relief, happiness, improved relationships with God, and perceived improvement in mental health status". No major mental health professional organization has sanctioned efforts to change sexual orientation and virtually all of them have adopted policy statements cautioning the profession and the public about treatments that purport to change sexual orientation. Conversion therapy The APA has roundly dismissed so-called conversion therapy (sometimes called "ex-gay" therapy) as unproductive and potentially harmful. One version of conversion therapy, Gender Wholeness Therapy was designed by an ex-gay Licensed Professional Counselor, David Matheson. "The emphasis in Mr. Matheson's counselling is on helping men—all his clients are male—develop 'gender wholeness' by addressing emotional issues and building healthy connections with other men. He says he believes that helps reduce homosexual desires. In 2019, Mr. Matheson announced that he intended to divorce his wife of 34 years, and live the remainder of his life as an openly gay man. Another variation of conversion therapy, "gender-affirmative therapy" has been described by A. Dean Byrd as follows: "The basic premise of gender-affirmative therapy is that social and emotional variables affect gender identity which, in turn, determines sexual orientation. The work of the therapist is to help people understand their gender development. Subsequently, such individuals are able to make choices that are consistent with their value system. The focus of therapy is to help clients fully develop their masculine or feminine identity". Several organizations have started retreats led by coaches aimed at helping participants diminish same-sex desires. These retreats tend to use a variety of techniques. Journey into Manhood, put on by People Can Change, uses "a wide variety of large-group, small-group and individual exercises, from journaling to visualizations (or guided imagery) to group sharing and intensive emotional-release work." Weekends put on by Adventure in Manhood support "healthy bonding with men, through masculine activity, teamwork, and socialization." Though not specific to gay men, several gay men attended the New Warrior Training Adventure, a weekend put on by ManKind Project, which is a "process of initiation and self-examination that is designed to catalyse the development of a healthy and mature masculine self." Joe Dallas, a prominent ex-gay, leads a monthly five-day men's retreat on sexual purity titled, Every Man's Battle. Several reparative therapies have been established, including: Sexual identity therapy was designed by Warren Throckmorton and Mark Yarhouse, and was endorsed by Robert L. Spitzer, prior to Spitzer's backing away from this belief that he had proven reparative therapy at times successful. Its purpose is to help patients line up their sexual identity with their beliefs and values. Therapy involves four phases: (1) assessment, (2) advanced or expanded informed consent, (3) psychotherapy, and (4) social integration of a valued sexual identity. Group psychotherapy uses group sessions led by a single psychologist and focuses on conflict surrounding homosexual expression. Context Specific Therapy was designed by Jeffrey Robinson. It does not work with any one theory of homosexuality, but uses several theoretical backgrounds according to the client's need, and is based on phenomenological research. It does not seek to change the client's orientation, but instead focuses on diminishing homosexual thoughts and behaviours. It works within the client's own view of God, noting that "individuals who are successful at overcoming homosexual problems are motivated by strong religious values". MAP Therapy is designed for both the individual with ego-dystonic sexual orientation and their family members. There are four main paths that clients may choose to take: (1) they can affirm an LGB identity, (2) they can foster a lifestyle of celibacy, (3) they can work on developing heterosexual attractions, or (4) they can explore their options. Ex-gay organizations OneByOne, an ex-gay organization, hosts a booth at a Love Won Out conference. CC BY-SA 3.0. For some ex-gay groups, choosing not to act on one's same-sex desires counts as a success whereas conversion therapists tend to understand success in terms of reducing or eliminating those desires. For example, some ex-gays in mixed-orientation marriages acknowledge that their sexual attractions remain primarily homosexual, but seek to make their marriages work regardless. Ex-gay advocates sometimes compare adopting the label "ex-gay" to the coming out process. Some conservative Christian political and social lobbying groups such as Focus on the Family, the Family Research Council, and the American Family Association actively promote to their constituencies the accounts of change of both conversion therapies and ex-gay groups. Some ex-gay organizations follow the tenets of a specific religion, while others try to encompass a more general spirituality. Although most ex-gay organizations were started by American evangelical Christians, there are now ex-gay organizations in other parts of the world and for Catholics, Mormons, Jews and Muslims. According to Douglas Haldeman, "This modality is thought to be one of the most common for individuals seeking to change their sexual orientation." Ex-gay ministries typically are staffed by volunteer counselors, unlike reorientation counselling, which is conducted by licensed clinicians. Ex-gay groups use several different techniques. Love in Action hosts workshops on "child development, gender roles, and personal sexuality," one-on-one Biblical guidance, "a structured environment help[ing] establish new routines and healthy patterns of behaviour", "challenging written assignments and interactive projects," "family involvement to improve communication... and to facilitate marital reconciliation," and "hiking, camping, canoeing, and rafting." Exodus International considers reparative therapy to be a useful tool, but not a necessary one. Evergreen International did not advocate or discourage particular therapies and states that "therapy will likely not be a cure in the sense of erasing all homosexual feelings." Criticism Robert L. Spitzer reported in 2003 that individuals who reported experiencing a change in sexual orientation had felt depressed or even suicidal prior to treatment "precisely because they had previously thought there was no hope for them, and they had been told by many mental health professionals that there was no hope for them, they had to just learn to live with their homosexual feelings." Spitzer's study, however, is widely considered disreputable in the therapeutic and mental-health community. The American Psychiatric Association enumerated many flaws in Spitzer's methods and analysis, and an American Psychological Association task force likewise scrutinized Spitzer's work and found it seriously flawed. The degree to which Spitzer's claims were treated as authoritative by news media has been examined and found problematic. Ultimately, Spitzer himself came to realize that his study had serious flaws, and rescinded the claims that he had made. 3.3. Exploratory Therapy The APA has specifically advised against sexual orientation change efforts and encourages practitioners to aid those who seek sexual orientation change by utilizing affirmative multiculturally competent therapy that recognizes the negative impact of social stigma on sexual minorities and balances ethical principles of beneficence and nonmaleficence, justice, and respect for people's rights and dignity. If a client wants to change their sexual orientation, the therapist should help the client come to their own decisions by evaluating the reasons behind the patient's goals. 4. Relationship to Religion The terms egodystonic and egosyntonic are used within the Roman Catholic Church in that, according to gay-rights advocate Bernard Lynch, priests who are gay but egodystonic, that is "hate their homosexuality", are acceptable, whereas egosyntonic candidates for the priesthood, those who accept their own sexuality, cannot be considered. Some churches publish specific instructions to clergy on how to minister to LGBTQ people. These include Ministry to Persons with a Homosexual Inclination, produced by the Catholic Church, and God Loveth His Children, produced by The Church of Jesus Christ of Latter-day Saints. In 1994, a church in the Presbyterian Church (USA) held a conference entitled “The Path to Freedom: Exploring healing for the Homosexual.” The APA encourages religious leaders to recognize that it is outside their role to adjudicate empirical scientific issues in psychology. Mental health practitioners can incorporate religion into therapy by "integrating aspects of the psychology of religion into their work, including by obtaining a thorough assessment of clients’ spiritual and religious beliefs, religious identity and motivations, and spiritual functioning; improving positive religious coping; and exploring the intersection of religious and sexual orientation identities." Researchers have found that for some clients who have 'identity conflicts' these can be reduced by reading religious texts that increase self-authority and allow them to reduce their focus on negative messages about homosexuality. Researchers also found that such clients made further progress if they came to believe that regardless of their sexual orientation, their God still loves and accepts them. Some LGBTQ people choose LGBT-affirming religious groups, change churches to those that affirm LGBT people, or simply leave religion. References "F66.1 - Egodystonic Sexual Orientation". 2019. "Use of Diagnoses "Homosexuality" & "Ego-Dystonic Homosexuality"". August 30, 1987. "Appropriate Therapeutic Responses to Sexual Orientation". American Psychological Association. August 2009. Kleinplatz, Peggy J. (2001). New directions in sex therapy: innovations and alternatives. Psychology Press. p. 100. ISBN 978-0-87630-967-4. Cochran, Susan D.; Drescher, Jack; Kismödi, Eszter; Giami, Alain; García-Moreno, Claudia; Atalla, Elham; Marais, Adele; Meloni Vieira, Elisabeth et al. (2014). "Proposed declassification of disease categories related to sexual orientation in the International Statistical Classification of Diseases and Related Health Problems (ICD-11)". Bulletin of the World Health Organization 92 (9): 672–679. doi:10.2471/BLT.14.135541. PMID 25378758. "Human rights violations against sexuality minorities in India". People's Union for Civil Liberties. February 2001. "Programmes Immobilier loi Pinel à vendre". Archived from the original on Aug 15, 2007. James Nichols (August 14, 2014). "Gay Man Suing Doctor After His 'Homosexual Behavior' Is Diagnosed As 'Chronic Problem'". The Huffington Post. Kar (1 January 2005). Comprehensive Textbook of Sexual Medicine. Jaypee Brothers Publishers. pp. 177–178. ISBN 978-81-8061-405-7. Chandran, Vinay (February 2006). "Prayer, punishment or therapy? Being a homosexual in India". InfoChange News & Features. "While social attitudes are slowly changing [in India] and the anti-sodomy law is being challenged, mental health professionals in many places still offer therapy to homosexuals." Ahuja, Niraj (30 July 2006). A Short Text Book of Psychiatry. ISBN 9788180618710. "Guidelines for Psychotherapy with Lesbian, Gay, & Bisexual Clients". "Appropriate Therapeutic Responses to Sexual Orientation". American Psychological Association. August 2009. Haldeman, Douglas (2004). "When Sexual and Religious Orientation Collide:Considerations in Working with Conflicted Same-Sex Attracted Male Clients". The Counseling Psychologist 32 (5): 691–715. doi:10.1177/0011000004267560. Throckmorton, Warren; Welton (Winter 2005). "Counseling practices as they relate to ratings of helpfulness by consumers of sexual reorientation therapy". Journal of Psychology and Christianity 24 (4): 332–42. Etengoff, Chana; Daiute, Colette (November 3, 2014). "Clinicians' perspectives of religious families' and gay men's negotiation of sexual orientation disclosure and prejudice". Journal of Homosexuality 62 (3): 394–426. doi:10.1080/00918369.2014.977115. PMID 25364980. Retrieved April 23, 2021. "Resolution on Appropriate Affirmative Responses to Sexual Orientation Distress and Change Efforts". American Psychiatric Association (May 2000). "Gay, Lesbian and Bisexual Issues". Association of Gay and Lesbian Psychiatrics. Munsey, Christopher (October 2009). "Insufficient Evidence to Support Sexual Orientation Change Efforts". Monitor on Psychology 40 (9): 29. Retrieved April 23, 2021. "Civil Action No. 1:09-cv-10309". United States District Court for the District of Massachusetts. "Statement from the Royal College of Psychiatrists' Gay and Lesbian Mental Health Special Interest Group". April 24, 2009. "Sexual orientation and homosexuality". "Resolution on Appropriate Affirmative Responses to Sexual Orientation Distress and Change Efforts". Luo, Michael (2007-02-12). "Some Tormented by Homosexuality Look to a Controversial Therapy". The New York Times: p. 1. "Once-prominent 'conversion therapist' will now 'pursue life as a gay man'". NBC Universal. January 23, 2019. Byrd, A. Dean; Olsen, Stony (2002) (– Scholar search). Homosexuality: Innate and Immutable?. 14. Regent University Law Review. p. 537. Retrieved 2007-10-29. "Journey Into Manhood: A Healing Weekend Hosted by People Can Change". NARTH. 2006-04-20. "Adventure in Manhood homepage". Self-published. "The New Warrior Weekend". Self-published. "'Love Won Out' conference coming to Southern California". Christian Examiner. September 2007. Simon, Stephanie (2007-06-18). "New ground in debate on 'curing' gays". Los Angeles Times. John M. Becker (April 25, 2012). "EXCLUSIVE: Dr. Robert Spitzer Apologizes to Gay Community for Infamous 'Ex-Gay' Study". Truth Wins Out. Wolf TJ (1987). "Group psychotherapy for bisexual men and their wives". J Homosex 14 (1–2): 191–9. doi:10.1300/J082v14n01_14. PMID 3655341. Throckmorton, Warren (2004). "What is reparative therapy?". Self-published. "NARTH 2003 Annual Conference schedule". NARTH. 2004-04-14. Jeffrey, Robinson. "What is Context Specific Therapy?". Self-published. "Just Share Home Design and Fresh Inspiration". "MAP Therapy". "Frequently Asked Questions". Regent University: Institute for the Study of Sexual Identity. Anonymous (2002-03-11). "No easy victory". Christianity Today. Retrieved 2007-08-28. Peebles, Amy E. (April 2003). "It's Not Coming Out, So Then What Is It? Sexual Identity and the Ex-Gay Narrative". Texas Linguistic Forum 47: 155–64. Retrieved 2007-08-28. Haldeman, Douglas C. (June 2002). "Gay Rights, Patient Rights: The Implications of Sexual Orientation Conversion Therapy". Professional Psychology: Research and Practice 33 (3): 260–4. doi:10.1037/0735-7028.33.3.260. Throckmorton, Warren (June 2002). "Initial empirical and clinical findings concerning the change process for ex-gays". Professional Psychology: Research and Practice 33 (3): 242–8. doi:10.1037/0735-7028.33.3.242. "the source". Love in Action. "the journey". Love in Action. "Exodus International Policy Statements". Exodus International. "Myths", EvergreenInternational.org (Evergreen International), retrieved 2007-10-29 "Therapy", EvergreenInternational.org (Evergreen International), retrieved 2007-10-29 Adler, Sarah; Levin, Edmund (May 9, 2001). "Some Gays Can Change, Study Says". Spitzer, Robert L. (October 2003). "Can some gay men and lesbians change their sexual orientation? 200 participants reporting a change from homosexual to heterosexual orientation". Archives of Sexual Behavior 32 (5): 403–417. doi:10.1023/A:1025647527010. PMID 14567650. "Therapy will "Turn Gays Straight", says Study". October 2, 2003. Hausman, Ken (Jul 6, 2001). "Furor Erupts Over Study On Sexual Orientation". Psychiatric News 36 (13): 20–34. doi:10.1176/pn.36.13.0020. "Facts About Changing Sexual Orientation". "FAIR". "HARDTALK - Zeinab Badawi Speaks to Fr. Berárd Lynch - Priest & Psychotherapist - 5 Mar 2013". July 12, 2016. "OneByOne: About Us". Self-published. 2006. More ©Text is available under the terms and conditions of the Creative Commons-Attribution ShareAlike (CC BY-SA) license; additional terms may apply. By using this site, you agree to the Terms and Conditions and Privacy Policy. 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16302	https://www.racgp.org.au/afp/2013/april/prostatitis	RACGP - Prostatitis – diagnosis and treatment Australian Family Physician Search for: Search AFP Filter Home Issues by year Authors Advertising AFP>2013>April>Prostatitis Volume 42, Issue 4, April 2013 Prostatitis Diagnosis and treatment Gretchen Dickson Download article Cite this article BIBTEX)REFER)RIS) Introduction Prostatitis is a spectrum of disorders that impacts a significant number of men. Acute bacterial prostatitis may be a life-threatening event requiring prompt recognition and treatment with antibiotic therapy. Chronic bacterial prostatitis has a more indolent course and also requires antibiotic therapy for resolution. Chronic prostatitis/chronic pelvic pain syndrome is the most common manifestation of prostatitis and may be the most difficult to treat. Asymptomatic inflammatory prostatitis is an incidental finding of unclear significance. Understanding the diagnostic and management strategies for each of these entities is critical for general practitioners in caring for their male patients. Prostatitis encompasses four distinct clinical entities, which can be described using the National Institutes of Health International Prostatitis Collaborative Network classification system. The four categories of prostatitis are: acute bacterial prostatitis chronic bacterial prostatitis chronic prostatitis/chronic pelvic pain syndrome inflammatory subtype non-inflammatory subtype asymptomatic inflammatory prostatitis.4 Acute bacterial prostatitis Acute bacterial prostatitis (ABP) accounts for approximately 5% of cases of prostatitis cases.1 Although rare, ABP requires prompt recognition and treatment as it may result in sepsis. Acute bacterial prostatitis results from proliferation of bacteria within the prostate gland following intraprostatic reflux of urine infected with organisms such as _Escherichia coli_, Enterococcus and _Proteus_ species.5,6 Men with chronic indwelling catheters, diabetes mellitus, immunosuppression, or who intermittently perform self-catheterisation, are at higher risk of developing ABP due to their increased risk of bacterial colonisation of the urethra.6,7 There is no evidence that perineal trauma from bicycle or horseback riding, dehydration or sexual abstinence are risk factors for ABP. The clinical presentation of ABP may be highly variable with symptoms ranging from mild to severe.6 Classic symptoms include: fever chills perineal or lower abdominal pain dysuria urinary frequency urinary urgency painful ejaculation hematospermia.8 Acute bacterial prostatitis should be considered in the differential diagnosis of any male presenting with urinary tract symptoms. While gentle palpation of the prostate gland on physical examination will often reveal a pathognomonic finding of an exquisitely tender, boggy prostate gland, care should be taken to avoid vigorous prostate massage as this may precipitate bacteremia and sepsis.9 Acute bacterial prostatitis can be diagnosed clinically, although both urine Gram stain and urine culture are recommended to identify causative organisms and guide treatment. While blood cultures and C-reactive protein may prove useful, a prostate specific antigen (PSA) test is not indicated. Prostate specific antigen elevations are common in the setting of infection and may take up to 1 month postinfection to resolve. Imaging is only indicated when prostatic abscess is suspected in a patient with ABP who is failing to improve with treatment. Antibiotic therapy for ABP should be based on the acuity of the patient and the known or suspected causative organism. Table 1 outlines the Australian Therapeutic Guidelines current treatment recommendations. While ABP is usually caused by urinary pathogens, sexually transmissible infections such as chlamydia and gonorrhoea should be considered, particularly in young men. If chlamydia is thought to be the causative agent, azithromycin 1 g orally stat or doxycycline 100 mg orally twice daily for 7 days is appropriate. If gonorrhoea is suspected, ceftriaxone 500 mg intramuscularly and azithromycin 1 g orally is indicated. Contact tracing, notification and treatment is also important in these cases. In addition to antibiotic therapy, non-steroidal anti-inflammatory drugs (NSAIDs) may offer both analgesia and more rapid healing through liquefaction of prostatic secretions.6 Urine culture 48 hours post-treatment is useful combined with review after 7 days of antibiotic treatment to assess clinical response to treatment. If the patient fails to improve with antibiotics, a prostatic abscess should be suspected, particularly in men who are immunocompromised, have diabetes mellitus or who have had recent instrumentation of the urinary tract.10 Both computed tomography (CT) and transrectal ultrasound may be used to detect a prostate abscess.11 If perineal puncture of the abscess is planned, ultrasound may guide the procedure.12 However, if surgical debridement of the abscess is planned, a CT scan may be more helpful to define borders of the abscess, plan the surgical approach and to investigate for other abnormalities in the genitourinary system.12 Acute urinary retention may develop as a complication of ABP. Suprapubic tap should be performed to alleviate retention as urethral catheterisation may worsen infection and is contraindicated. In addition to acute urinary retention and prostatic abscess, ABP can lead to sepsis, chronic bacterial prostatitis, fistula formation or spread of infection to the spine or sacroiliac joints.6,13 Chronic bacterial prostatitis Chronic bacterial prostatitis (CBP) may result from ascending urethral infection, lymphogenous spread of rectal bacteria, hematogenous spread of bacteria from a remote source, undertreated acute bacterial prostatitis or recurrent urinary tract infection with prostatic reflux. Causative agents of CBP are similar to those of ABP include Gram negative rods, fungi, mycobacterium, _Ureaplasma urealyticum_, Chlamydia trachomatis14 and Trichomonas vaginalis.15 However, _Escherichia coli_ is believed to be the causative organism in 75–80% of CBP cases.14 Recognising CBP can be difficult, as the history and examination are highly variable. All patients note some degree of genitourinary pain or discomfort. Common presentations include recurrent urinary tract infections with no history of bladder instrumentation, dysuria and frequency with no other signs of ABP or new onset sexual dysfunction without other aetiology.16,17 Often the physical examination, including prostate examination, is normal. Prostate examination should be performed to document any abnormalities such as prostatic calculi, which can serve as a reservoir of infection. Prostate stones may be difficult to palpate, but if found, may impact management decisions. Although the Meares-Stamey four glass test is the gold standard to diagnose CBP, it is rarely used in practice due to time constraints and the difficulty obtaining samples.18 Instead pre- and post-prostatic massage urine samples for analysis and culture may be useful and can guide antibiotic therapy.19 A prostate massage is performed by stroking the prostate with firm pressure from the periphery to the midline on both the right and left sides of the prostate gland. More than 20 leucocytes per high powered field on the post-massage urine sample is diagnostic of CBP.19 If urine cultures show no growth, consider a nucleic acid test for C. trachomatis and culture of prostatic fluid for ureaplasmas. Occasionally, Mycoplasma genitalium is found in prostatic secretions, although its role in prostatitis is unclear. If these tests are also negative, an alternative diagnosis should be considered. Limited comparative trials exist to guide antibiotic regimens for CBP. Table 1 lists current recommendations. Patients should be warned about the common side effects of extended duration of antibiotic use, such as Achilles tendon rupture with fluoroquinolones. In addition to antibiotics, NSAIDs may alleviate pain symptoms. Alpha-blockers may diminish urinary obstruction and reduce future occurrences.20 Although less well studied, saw palmetto, quercetin, daily sitz baths, perianal massage and frequent ejaculation may also help to clear prostatic secretions and lessen discomfort. If prostatic stones are present, prostatectomy may eliminate the nidus of infection. Table 1 . Treatment of acute and chronic bacterial prostatitis40Acute bacterial prostatitisMild or moderate disease while awaiting culture Trimethoprim 300 mg orally daily for 14 days, or Cephalexin 500 mg orally twice daily for 14 days, or Amoxicillin and clavulanic acid 500 mg + 125 mg orally twice daily for 14 days Appears septic or unable to tolerate oral therapy Admit to hospital, offer parenteral therapy with ampicillin and gentamycin or ceftriaxone as per severe pyelonephritis treatment Chronic bacterial prostatitisFirst line treatment Norfloxacin 400 mg orally every 12 hours for 4 weeks, or Trimethoprim 300 mg orally daily for 4 weeks If chlamydia or ureaplasma noted Doxycycline 100 mg orally every 12 hours for 2–4 weeks Chronic prostatitis/chronic pelvic pain syndrome Chronic prostatitis/chronic pelvic pain syndrome (CP/CPPS) is more common than either acute bacterial or chronic bacterial prostatitis.4 Up to 18% of Australian men may experience some type of urogenital pain within a 12 month period, while up to 2% of Australian men may have prostatitis-like symptoms at any given time.1,21 Unlike bacterial prostatitis where a causative organism can be identified, the aetiology of CP/CPPS is poorly understood; both inflammatory and infectious mechanisms have been postulated.17,22,23 Psychological stress may be a major contributor to symptom severity.24 Some evidence exists of an association between irritable bowel syndrome, chronic fatigue syndrome and fibromyalgia with CP/CPPS, although little correlation exists between the amount of inflammatory markers detected within the prostate gland itself and the degree of symptoms.25,26 Symptoms of CP/CPPS can vary widely and include dysuria; urinary frequency; urinary urgency; weak urinary stream; pain in the perineum, lower abdomen, testicles or penis; hematospermia or difficulty achieving erection.27,28 Diagnosis requires the patient to have had pelvic pain or urinary symptoms for more than three of the previous 6 months with no evidence of ABP or urinary tract infection in that time.17 Chronic prostatitis/chronic pelvic pain syndrome is a diagnosis of exclusion and laboratory or imaging studies are indicated to rule out other potential causes of symptoms. Elevated PSA should not be attributed to CP/CPPS and warrants further investigation.29 Approximately 60% of men affected by CP/CPPS will seek treatment for their symptoms.30 Although various treatments have been studied, methodological problems including lack of randomisation and small sample size limit the ability to apply research findings to the clinical treatment of CP/CPPS. With the current evidence available, tailoring treatment to individual patient symptom complexes may be more beneficial than attempting to use one treatment as a curative agent in all individuals.31 The National Institute of Health Chronic Prostatitis Symptom Index (NIH-CPSI) provides a validated indicator of disease severity that can be monitored over time to determine if a particular treatment is improving a patient’s symptoms or overall quality of life.32 Of the treatments that have been studied, alpha-adrenergic receptor blockers and antibiotics used alone or in combination appear to have the greatest improvement in symptom scores when compared with placebo.33,34 Anti-inflammatory medications may also be useful.33 Additional studies are needed to determine the role of 5 alpha-reductase inhibitors, glycosaminoglycans, saw palmetto, acupuncture, physical therapy, and pelvic floor training using biofeedback as part of treatment.17,35 Other treatments that have proven useful in small studies for targeted symptoms include: phosphodiesterase five inhibitors for sexual dysfunction,36 cernilton or pollen extract for urinary symptoms,37 quercetin (500 mg orally twice daily for 30 days) for pelvic floor muscle spasm,38 and fluoxetine (20 mg orally daily) for depression and improved quality of life.39 Transurethral microwave therapy may be used as a last resort for men who have failed other interventions.17 Asymptomatic inflammatory prostatitis Asymptomatic inflammatory prostatitis is, by definition, asymptomatic. It is often diagnosed incidentally during the evaluation of infertility or prostate cancer.17 The clinical significance of category IV prostatitis is unknown, and is often left untreated.17 Summary A diagnosis of prostatitis encompasses a spectrum of disease: acute bacterial prostatitis, chronic bacterial prostatitis, chronic prostatitis/chronic pelvic pain syndrome and asymptomatic prostatitis have varying clinical significance, causative agents, treatment strategies and long-term prognosis. Limited research exists to guide the diagnosis and management of these entities, making prostatitis a challenging condition to manage. Competing interests: None. Provenance and peer review: Not commissioned; externally peer reviewed. References Ferris JA, Pitts MK, Richters J, Simpson JM, Shelley HM, Smith AM. National prevalence of urogenital pain and prostatitis- like symptoms in Australian men using the National Institutes of Health Chronic Prostatitis Symptoms Index. BJU Int 2010;105:373–9. Search PubMed Krieger JN. Classification, epidemiology and implications of chronic prostatitis in North America, Europe and Asia. Minerva Urol Nefrol 2004;56:99–107. Search PubMed Krieger JN, Lee SW, Jeon J, Cheah PY, Liong ML, Riley DE. Epidemiology of prostatitis. Int J Antimicrob Agents 2008;31(Suppl 1):S85–90. Search PubMed Krieger JN, Nyberg L, Nickel JC. NIH Consensus definition and classification of prostatitis. JAMA 1999;282:236–7. Search PubMed Cornia PB, Takahashi TA, Lipsky BA. The microbiology of bacteriuria in men: a 5-year study at a veterans affairs hospital. Diagn Microbiol Infect Dis 2006;56:25–30. Search PubMed Brede CM, Shoskes DA. The etiology and management of acute prostatitis. Nat Rev Urol 2011;8:207–12. Search PubMed Wyndaele JJ. Complications of intermittent catheterization: their prevention and treatment. Spinal Cord 2002;40:53–41. Search PubMed Krieger JN, Egan KJ, Ross SO. Chronic pelvic pains represent the most prominent urogenital symptoms of chronic prostatitis. Urology 1996;48:715–21. Search PubMed Nickel JC. Inflammatory conditions of the male genitourinary tract: prostatitis and related conditions, orchitis, and epididymitis. In: Wein AJ, Kavoussi LR, Novick AC, Partin AW, Peters CA, eds. Campbell-Walsh Urology. 9th edn. Philadelphia, Pa.: Saunders, 2007;304–29. Search PubMed Tiwari P, Pal D, Tripathi A, et al. Prostatic abscess: diagnosis and management in the modern antibiotic era. Saudi J Kidney Dis Transpl 2011;22:298–301. Search PubMed Horcajada JP, Vilana R, Moreno-Martinez A. Transrectal prostatic ultrasonography in acute bacterial prostatitis: findings and clinical implications. Scand J Infect Dis 2003;35:114–20. Search PubMed Schull A, Monzani Q, Boni L, et al. Imaging in lower urinary tract infections. Diagn Interv Imaging 2012;93:500–8. Search PubMed Siroky MB, Moylan R, Austen G, Olsson CA. Metastatic infection secondary to genitourinary tract sepsis. Am J Med 1976;61:351–60. Search PubMed Schaeffer AJ. Clinical practice: chronic prostatitis and the chronic pelvic pain syndrome. N Engl J Med 2006;355:1690–8. Search PubMed Skerk V, Krhen I, Schonwald S, Cajic V, et al. The role of unusual pathogens in prostatitis syndrome. Int J Antimicrob Agents 2004;24(Suppl 1):S53–6. Search PubMed Muller A, Mulhall JP. Sexual dysfunction in the patient with prostatitis. Curr Opin Urol 2005;15:404–9. Search PubMed Murphy AB, Macejko A, Taylor A, Nadler RB. Chronic prostatitis: management strategies. Drugs 2009;69:71–84. Search PubMed McNaughton Collins M, Fowler FJ, Elliott DB. Diagnosing and treating chronic prostatitis: do urologists use the four glass test? Urology 2000;55:403–7. Search PubMed Nickel JC, Shoskes D, Wang Y, et al. How does the pre-massage and post-massage 2-glass test compare to the Meares-Stamey 4-glass test in men with chronic prostatitis/chronic pelvic pain syndrome? J Urol 2006;176:119–24. Search PubMed Barbalias GA, Nikiforidis G, Liatsikos EN. Alpha-blockers for the treatment of chronic prostatitis in combination with antibiotics. J Urol 1998;159:883–7. Search PubMed Pitts M, Ferris J, Smith A, Shelley J, Richters J. Prevalence and correlates of three types of pelvic pain in a nationally representative ample of Australian men. J Sex Med 2008;5:1223–9. Search PubMed Leskinen MJ, Rantakokko-Jalava K, Manninen R, et al. Negative bacterial polymerase chain reaction (PCR) findings in prostate tissue from patients with symptoms of chronic pelvic pain syndrome (CPPS) and localized prostate cancer. Prostate 2003;55:105–10. Search PubMed Lee JC, Muller CH, Rothman I, et al. Prostate biopsy culture findings of men with chronic pelvic pain syndrome do not differ from those of healthy controls. J Urol 2003;169:584–7. Search PubMed Anderson RU, Orenberg EK, Morey A, Chavez N, Chan CA. Stress induced hypothalamus-pituitary-adrenal axis responses and disturbances in psychological profiles in men with chronic prostatitis/chronic pelvic pain syndrome. J Urol 2009;182:2319–24. Search PubMed Nickel JC, Roehrborn CG, O’leary MP, Bostwick DG, Somerville M, Rittmaster RS. Examination of the relationship between symptoms of prostatitis and histological inflammation: Baseline data from the REDUCE chemoprevention trial. J Urol 2007;178(3 Pt 1):896–900. Search PubMed Rodriguez MA, Afari N, Buchwald DS. National institute of Diabetes and Digestive and Kidney Disease Working Group on urological chronic pelvic pain. Evidence for overlap between urological and nonurological unexplained clinical conditions. J Urol 2009;182:2123–31. Search PubMed Trinchieri A, Magri V, Cariani L, et al. Prevalence of sexual dysfunction in men with chronic prostatitis/chronic pelvic pain syndrome. Arch Ital Urol Androl 2007;79:67–70. Search PubMed Krieger JN, Ross SO, Penson DF, Riley DE. Symptoms and inflammation in chronic prostatitis/chronic pelvic pain syndrome. Urology 2002;60:959–63. Search PubMed Nickel JC. Clinical evaluation of the man with chronic prostatitis/chronic pelvic pain syndrome. Urology 2002;60(6 Suppl):20–2. Search PubMed Nickel JC, Downey J, Hunter D, Clark J. Prevalence of prostatitis like symptoms in a population based study using the National Institutes of Health Chronic prostatitis symptoms index. J Urol 2001;165:842–5. Search PubMed Nickel JC, Shoskes DA. Phenotypic approach to the management of the chronic prostatitis/chronic pelvic pain syndrome. BJU Int 2010;106:1252–63. Search PubMed Nickel JC, Alexander RB, Anderson R, et al. Chronic Prostatitis Collaborative Research Network Study Groups. Category III chronic prostatitis/chronic pelvic pain syndrome: insights from the National Institutes of Health Chronic Prostatitis Collaborative Research Network studies. Curr Urol Rep 2008;9:320–7. Search PubMed Anothaisintawee T, Attia J, Nickel JC, et al. Management of chronic prostatitis/chronic pelvic pain syndrome: a systematic review and network meta-analysis. JAMA 2011;305:78–86. Search PubMed Cheah PY, Liong ML, Yuen KH, et al. Terazosin therapy for chronic prostatitis/chronic pelvic pain syndrome: a randomized, placebo controlled trial. J Urol 2003;169:592–6. Search PubMed Schiller DS, Parikh A. Identification, pharmacologic considerations and management of prostatitis. Am J Geriatr Pharmacother 2011;9:37–48. Search PubMed Shoskes DA. The challenge of erectile dysfunction in the man with chronic prostatitis/chronic pelvic pain syndrome. Curr Urol Rep 2012;13:263–7. Search PubMed Wagenlehner FM, Bschleipfer T, Pilatz A, Weidner W. Pollen extract for chronic prostatitis-chronic pelvic pain syndrome. Urol Clin North Am 2011;38:285–92. Search PubMed Shoskes DA, Nickel JC. Quercetin for chronic prostatitis/chronic pelvic pain syndrome. Urol Clin North Am 2011;38:279–84. Search PubMed Xia D, Wang P, Chen J, Wang S, Jiang H. Fluoxetine ameliorates symptoms of refractory chronic prostatitis/chronic pelvic pain syndrome. Chin Med J (Engl) 2011;124:2158–61. Search PubMed Prostatitis. In: eTG Complete. Melbourne: Therapeutic Guidelines Limited, 2012 July. Available at [Accessed 24 September 2012]. Search PubMed Download article PDF Advertising Back to search results Also in this issue: Workplace April 2013 Focus ##### Work related encounters in general practice ##### Presenteeism. Implications and health risks ##### Tendon injuries. Practice tips for GPs ##### Returning to work after an injury ##### Workplace bullying. What’s it got to do with general practice? Up front ##### We can work it out ##### Letters to the editor Clinical ##### Chronic fatigue syndrome. A patient centred approach to management ##### Quantitative serum immunoglobulin tests ##### Bibliotherapy for depression ##### Digital eye drop instillation. A novel method ##### Extrapulmonary tuberculosis. Three cases in the spine ##### Complex type 2 diabetes mellitus. Management challenges and pitfalls ##### Subungual nodule of the great toe ##### Prostatitis. Diagnosis and treatment Research ##### Training in critical thinking and research. An audit of delivery by regional training providers in Australia ##### Chronic hepatitis B. Care delivery and patient knowledge in the Torres Strait region of Australia ##### Home blood pressure monitoring. A trial on the effect of a structured education program ##### Managing same day appointments. A qualitative study in Australian general practice Professional ##### The duty of GPs to follow up patients ##### The pre-employment medical. 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16303	https://pubmed.ncbi.nlm.nih.gov/30281335/	Neutrophils engage the kallikrein-kinin system to open up the endothelial barrier in acute inflammation - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Neutrophils engage the kallikrein-kinin system to open up the endothelial barrier in acute inflammation Ellinor Kenne1,Joel Rasmuson1,Thomas Renné2,Monica L Vieira34,Werner Müller-Esterl5,Heiko Herwald3,Lennart Lindbom1 Affiliations Expand Affiliations 1 Department of Physiology and Pharmacology, Karolinska Institutet, Stockholm, Sweden. 2 Institute for Clinical Chemistry and Laboratory Medicine, University Medical Center Hamburg-Eppendorf, Hamburg, Germany. 3 Department of Clinical Sciences, Lund University, Lund, Sweden. 4 Biotechnology Center, Butantan Institute, São Paulo, Brazil; and. 5 Institute of Biochemistry II, Goethe University, Frankfurt, Germany. PMID: 30281335 DOI: 10.1096/fj.201801329R Item in Clipboard Neutrophils engage the kallikrein-kinin system to open up the endothelial barrier in acute inflammation Ellinor Kenne et al. FASEB J.2019 Feb. Show details Display options Display options Format FASEB J Actions Search in PubMed Search in NLM Catalog Add to Search . 2019 Feb;33(2):2599-2609. doi: 10.1096/fj.201801329R. Epub 2018 Oct 3. Authors Ellinor Kenne1,Joel Rasmuson1,Thomas Renné2,Monica L Vieira34,Werner Müller-Esterl5,Heiko Herwald3,Lennart Lindbom1 Affiliations 1 Department of Physiology and Pharmacology, Karolinska Institutet, Stockholm, Sweden. 2 Institute for Clinical Chemistry and Laboratory Medicine, University Medical Center Hamburg-Eppendorf, Hamburg, Germany. 3 Department of Clinical Sciences, Lund University, Lund, Sweden. 4 Biotechnology Center, Butantan Institute, São Paulo, Brazil; and. 5 Institute of Biochemistry II, Goethe University, Frankfurt, Germany. PMID: 30281335 DOI: 10.1096/fj.201801329R Item in Clipboard Cite Display options Display options Format Abstract Neutrophil recruitment and plasma exudation are key elements in the immune response to injury or infection. Activated neutrophils stimulate opening of the endothelial barrier; however, the underlying mechanisms have remained largely unknown. In this study, we identified a pivotal role of the proinflammatory kallikrein-kinin system and consequent formation of bradykinin in neutrophil-evoked vascular leak. In mouse and hamster models of acute inflammation, inhibitors of bradykinin generation, and signaling markedly reduced plasma exudation in response to chemoattractant activation of neutrophils. The neutrophil-driven leak was likewise suppressed in mice deficient in either the bradykinin B 2 receptor or factor XII (initiator of the kallikrein-kinin system). In human endothelial cell monolayers, material secreted from activated neutrophils induced cytoskeletal rearrangement, leading to paracellular gap formation in a bradykinin-dependent manner. As a mechanistic basis, we found that a neutrophil-derived heparin-binding protein (HBP/azurocidin) displaced the bradykinin precursor high-molecular-weight kininogen from endothelial cells, thereby enabling proteolytic processing of kininogen into bradykinin by neutrophil and plasma proteases. These data provide novel insight into the signaling pathway by which neutrophils open up the endothelial barrier and identify the kallikrein-kinin system as a target for therapeutic interventions in acute inflammatory reactions.-Kenne, E., Rasmuson, J., Renné, T., Vieira, M. L., Müller-Esterl, W., Herwald, H., Lindbom, L. Neutrophils engage the kallikrein-kinin system to open up the endothelial barrier in acute inflammation. Keywords: bradykinin; degranulation; leukocyte; vascular permeability. PubMed Disclaimer Similar articles G Protein-Coupled Kinin Receptors and Immunity Against Pathogens.Scharfstein J, Ramos PIP, Barral-Netto M.Scharfstein J, et al.Adv Immunol. 2017;136:29-84. doi: 10.1016/bs.ai.2017.05.007. Epub 2017 Aug 23.Adv Immunol. 2017.PMID: 28950949 Review. The Plasma Kallikrein-Kininogen Pathway Is Critical in the Pathogenesis of Colitis in Mice.Wang B, Yang A, Zhao Z, He C, Liu Y, Colman RW, Dai J, Wu Y.Wang B, et al.Front Immunol. 2018 Feb 6;9:21. doi: 10.3389/fimmu.2018.00021. eCollection 2018.Front Immunol. 2018.PMID: 29467753 Free PMC article. Acinetobacter Baumannii Secreted Protease CpaA Inhibits Factor XII-Mediated Bradykinin Generation and Neutrophil Activation.Blair KM, Bohinc DJ, Bane KL, Warnock M, Abuaita B, Gura C, Grinsztejn E, Marshall SH, Wilson BM, Bonomo RA, Tambralli A, Knight JS, O'Riordan MX, Lawrence DA, Stavrou EX, Sandkvist M.Blair KM, et al.Circ Res. 2025 Jun 20;137(1):e1-e15. doi: 10.1161/CIRCRESAHA.124.324764. 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16304	https://mathoverflow.net/questions/207148/derivatives-of-radial-functions-can-be-bounded-by-derivatives-in-terms-of-radial	ap.analysis of pdes - Derivatives of radial functions can be bounded by derivatives in terms of radial distance? - MathOverflow Join MathOverflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Derivatives of radial functions can be bounded by derivatives in terms of radial distance? Ask Question Asked 10 years, 4 months ago Modified10 years, 4 months ago Viewed 2k times This question shows research effort; it is useful and clear 7 Save this question. Show activity on this post. \begingroup Suppose f is a radial function, i.e., f(x)=f(\|x\|), and f \in C^\infty(\bar{B}), where \bar{B} is the closure of the unit ball in \mathbb{R}^n. Prove or disprove the following. Given any positive integer k, \sup_{\|\alpha\|=k,x\in B} \|D^\alpha f(x)\| \leq \sup_{r < 1} \lvert f^{(k)}(r) \rvert, where \alpha is a multi-index and D^\alpha f is the corresponding derivative of f. By f^{(k)}(r), we mean the k^{th} derivative of f as a function of r=\|x\|. I try some functions, taking second order derivatives, and the inequality holds for all of them. The case where k=1 is easy to prove but I can't prove for a general k. Instead of a general smooth f, can we prove the assertion for polynomials(or an uniformly and absolutely converging power series) with only even powers, namely,f(r) = \sum_{j=0}^m c_j r^{2j} \quad(m\text{ can be}+\infty)\quad? PS: I asked this question on Math.SE but no one answered so it is posted here. This is a quite simple/straightforwad question (that a Freshman in math can fully understand) but it is surprising that till now, no one(me included) could answer it or at least give some idea. ap.analysis-of-pdes harmonic-analysis inequalities sobolev-spaces Share Cite Improve this question Follow Follow this question to receive notifications edited May 26, 2015 at 0:58 bookseebooksee asked May 20, 2015 at 19:27 bookseebooksee 398 4 4 silver badges 15 15 bronze badges \endgroup 5 \begingroup@ChristianRemling Despite the first question, what about the second statement in my question ?\endgroup booksee –booksee 2015-05-20 20:30:54 +00:00 Commented May 20, 2015 at 20:30 \begingroup This example does not quite seem to work. Does not d^2f/dr^2 get large when you actually do the modification near 0?\endgroup Michael Renardy –Michael Renardy 2015-05-21 00:16:36 +00:00 Commented May 21, 2015 at 0:16 \begingroup May be use known estimates via the Laplace operator, so you will derive estimates via the radial Bessel in r.h.s.\endgroup Sergei –Sergei 2015-05-26 17:44:09 +00:00 Commented May 26, 2015 at 17:44 \begingroup@Sergei Thanks for your comment. Could you provide some details ? I did not see its relation to Laplacian.\endgroup booksee –booksee 2015-05-26 21:26:30 +00:00 Commented May 26, 2015 at 21:26 \begingroup@booksee. See my edits today. I answer to some of your worries.\endgroup Denis Serre –Denis Serre 2015-06-01 07:16:18 +00:00 Commented Jun 1, 2015 at 7:16 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 4 Save this answer. +25 This answer has been awarded bounties worth 25 reputation by Community Show activity on this post. \begingroup To avoid confusion, I shall denote f=f(r) and F(x)=f(\|x\|) the corresponding radial function. Because of Van der Corput-Schaake Inequality, it is enough to prove the following more general inequality: for every point x\ne0, integer k\ge1 and unit vector e, \|D^kF_xe^{\otimes k}\|\le\sup_{0\le s\le r}\|f^{(k)}(s)\|\quad ? This is obvious for k=1. Let me first prove that it is true for k=2,3. Then I'll give a general strategy for a proof. In the following, I denote \nu:=\frac{x\cdot e}r\in[-1,1]. For k=2, one has D^2F_xe^{\otimes 2}=(1-\nu^2)\frac1rf'+\nu^2f''. Because F is smooth, we know that the derivatives of f of odd order vanish at 0. Using f'(0)=0, we obtain D^2F_xe^{\otimes 2}=(1-\nu^2)\frac1r\int_0^rf''(s)ds+\nu^2f''=\langle\mu_2,f''\rangle. There remains to check that the measure \mu_2 has total mass 1 (i.e. is a probability). This is obvious. If k=3, we have D^3F_xe^{\otimes 3}=3(\nu-\nu^3)\left(\frac1rf''-\frac1{r^2}f'\right)+\nu^3f'''. Using again f'(0)=0, we obtain D^3F_xe^{\otimes 3}=\langle\mu_3,f'''\rangle where \mu_3:=\nu^3\delta_{s=r}+3(\nu-\nu^3)\frac{s}{r^2}\chi_{[0,r]}(s). Again, one checks that \|\mu_3\|=\|\nu^3+\frac32(\nu-\nu^3)\|\le1. The general strategy is to prove that if r=\|x\|, then D^kF_xe^{\otimes k}=\langle\mu_k,f^{(k)}\rangle for some measure \mu_k over [0,r]. Mind that the coefficients of \mu_k involve only \nu and r. The measure \mu_k can be calculated from the Faa di Bruno Formula (I suspect that \mu_k has a constant sign). To conclude, one has to check that its total mass is less than 1. This total mass involves only \nu ; it is likely to be the absolute value of a polynomial P_k(\nu). More precisely, \mu_k is the sum of the Dirac mass \nu^k\delta_{x=r} and a continuous density over s\in(0,r) : \frac1rN_k(\nu,t),\qquad t:=\frac sr. The density is determined by induction with the following rules \partial_t(N_{k+1}-\nu tN_k) = -(1-\nu^2)\partial_\nu N_k, N_{k+1}(\nu,1) = \nu N_k(\nu,1)+k\nu^{k-1}(1-\nu^2), and we have N_k(\nu,0) if k is odd. The second line gives explicitly N_k(\nu,1)=\frac{k(k-1}2\nu^{k-2}(1-\nu^2). The first few N_k's are N_1\equiv0,\quad N_2=1-\nu^2,\quad N_3=3\nu(1-\nu^2)t,N_4=\frac32(1-\nu^2)(5\nu^2-1)(t^2-1)+6\nu^2(1-\nu^2),N_5=\frac32\nu(1-\nu^2)(5\nu^2-1)t(t^2-1)+6\nu^3(1-\nu^2)(t-1)-2(3\nu-5\nu^3)(1-\nu^2)(t^3-3t+2)-12\nu(1-\nu^2)(1-2\nu^2)(t-1)+10\nu^3(1-\nu^2). One verifies easily that \mu_4 is a probability (Question : is \mu_{2\ell} always a probability ?). I checked that for every value of the angle \nu, 0\le\langle\mu_5,{\bf1}\rangle\le1. However, I gave up about the sign of N_5, and therefore I cannot claim that the total mass \|\mu_5\|\le1. Share Cite Improve this answer Follow Follow this answer to receive notifications edited Apr 13, 2017 at 12:58 CommunityBot 1 2 2 silver badges 3 3 bronze badges answered May 29, 2015 at 11:37 Denis SerreDenis Serre 53k 10 10 gold badges 152 152 silver badges 315 315 bronze badges \endgroup 5 1 \begingroup It looks like this reduction is in fact equivalent to the original problem, since one could select a smooth f by requiring f^{(k)} to be close in L^1 to the sign of \mu_k. In the event that \mu_k is unsigned, then the extremal comes when f^{(k)} is constant, thus F(x) is a constant multiple of \|x\|^k (this is not smooth, but can be mollified). The question then boils down to whether the k^{th} derivative of (1 + \|x\|^2)^{k/2} is always bounded above by k!, which I believe to be the case (I think a contour integration argument should do this, though I haven't checked.)\endgroup Terry Tao –Terry Tao 2015-05-29 18:58:27 +00:00 Commented May 29, 2015 at 18:58 1 \begingroup@DenisSerre Thanks for your answer but I don't think it is general to answer my original question. For example, the tensor you choose is e\otimes\dots\otimes e where each unit vector is the same(=e) and the derivative you compute for k=2 cannot represent , say, \dfrac{\partial^2 f}{\partial x_1\partial x_2} in general. In my question, the sup-norm on the left is taken over any multi-index \alpha s.t. \|\alpha\|=k, so we may need to use the tensor like e_1\otimes \dots \otimes e_k where each e_i is an arbitrary unit vector.\endgroup booksee –booksee 2015-05-30 04:36:42 +00:00 Commented May 30, 2015 at 4:36 \begingroup@DenisSerre But if you can prove that the magnitude of derivative w.r.t. e_1\otimes\dots\otimes e_k is bounded by the one w.r.t. e\otimes\dots\otimes e, that is nice.\endgroup booksee –booksee 2015-05-30 04:39:48 +00:00 Commented May 30, 2015 at 4:39 4 \begingroup If it helps, one can write N_k(\nu,t) = \frac{1}{(k-1)!} \frac{d^k}{dx^k} ( \sqrt{x^2 + 1-\nu^2} - t )^{k-1}\|{x = \nu} (this corresponds to the situation where f^{(k)} is a Dirac mass at tr). One also has \mu_k([0,r]) = \frac{1}{k!} \frac{d^k}{dx^k} (x^2 + (1-\nu^2) r^2)^{k/2}\|{x=\nu r}, which is non-negative thanks to the identity at terrytao.wordpress.com/2015/05/30/a-differentiation-identity and the fundamental theorem of calculus. So it all boils down to checking that N_k is nonnegative.\endgroup Terry Tao –Terry Tao 2015-06-03 18:36:30 +00:00 Commented Jun 3, 2015 at 18:36 \begingroup@Terry. Nice contribution. I'll think about it. By the way, I remember your definitive answer to my question mathoverflow.net/q/51848/8799 .\endgroup Denis Serre –Denis Serre 2015-06-03 18:53:23 +00:00 Commented Jun 3, 2015 at 18:53 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. \begingroup Here is an argument to show that \nabla^k F_x\cdot e^{\otimes^k}=f^{(k)}(r) if x=re. This shows that the inequality Denis is seeking to prove is sharp, if true; I can verify this inequality in the particular case f(r)=e^{-r^2}. Here is the argument: Let g be a function such that g(r^2)=f(r), so that in Denis' notation, F(x)=g(\|x\|^2). The derivatives of f and the expressions \nabla^k F_x\cdot e^{\otimes^k} can be expressed in terms of the derivatives of g as follows: For every k, there exist polynomials P_{i,k}(t) such that f^{(k)}(r)=\sum_{i=1}^k P_{i,k}(r)g^{(i)}(r^2) and \nabla^k F_x\cdot e^{\otimes^k}=\sum_{i=1}^k P_{i,k}(x\cdot e)g^{(i)}(\|x\|^2). Taking x=re, we see that the RHS of both equations are equal. To prove these equations, use induction and the following formulas, valid for any h, P: \nabla(h(\|x\|^2))\cdot e = 2(x\cdot e)h'(\|x\|^2)\qquad \nabla(P(x\cdot e))\cdot e = P'(x\cdot e) and \frac{d}{dr}h(r^2)=2r h'(r^2)\qquad \frac{d}{dr}P(r)=P'(r). In addition, these polynomials satisfy the recurrence P_{i,k+1}(t)=P_{i,k}'(t)+2tP_{i-1,k}(t), so that, for k even, the P_{i,k} are even, and for k odd, the P_{i,k} are odd. Thus if x=-re, we have equality up to \pm1. If g(t)=e^{-t} (i.e. f(r)=e^{-r^2}), then \nabla^k F_x\cdot e^{\otimes^k}=H_k(x\cdot e)g(r^2), where H_n(t) is the n th Hermite polynomial (which satisfy a similar recurrence to the P_{i,k}). Since H_k(t)g(r^2) is decreasing in r, the sup of \nabla^k F_x\cdot e^{\otimes^k} over \|x\|\leq r (i.e. \|t\|\leq r) is attained when x=re. Share Cite Improve this answer Follow Follow this answer to receive notifications answered Jun 3, 2015 at 20:06 Brendan MurphyBrendan Murphy 1,038 9 9 silver badges 15 15 bronze badges \endgroup Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. \begingroup There are many estimates of Sobolev type of the form \|\|D^{\alpha}f\|\|_1 \le \|\|\Delta f\|\|_2 for different pairs of norms. For your case the r.h.s. will be one-dimensional in r \|\|(D^2-\frac{n-1}{r}-\frac{1}{r^2})f\|\|_2. Now the question is reduced to one-dimensional estimates of this norm, say from the embedding theorem bounding derivatives uniformly. Of course it is just a plan to attack the problem, no pretence for being an answer complete, bounty cake and so on. E.g. in case of the Lebesgue norms with powers of Bessel operators instead standard derivatives-we have Kipriyanov spaces, like Sobolev ones. I proved some years ago that these norms are equivalent to Sobolev with power weights. For such spaces there are embedding in C---spaces. So estimates with Bessel powers lead to uniform estimates. Share Cite Improve this answer Follow Follow this answer to receive notifications edited May 27, 2015 at 20:14 answered May 27, 2015 at 20:09 SergeiSergei 1,570 9 9 silver badges 17 17 bronze badges \endgroup 1 \begingroup Good point. I've not proceeded in this way before. However, honestly speaking, I don't think it can be proved via L^p inequalities. Note that the constant on the r.h.s. is actually 1, which means that it is independent of the domain(and dimension n). Hence, in my opinion, any inequality involving integrals is "dangerous". Moreover, I think the inequality is a pointwise one in nature.\endgroup booksee –booksee 2015-05-28 00:45:58 +00:00 Commented May 28, 2015 at 0:45 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions ap.analysis-of-pdes harmonic-analysis inequalities sobolev-spaces See similar questions with these tags. 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16305	https://nrich.maths.org/content/id/5929/Sea%20Level%20printable%20sheet%203.pdf	nrich.maths.org/5929 © University of Cambridge Sea Level nrich.maths.org/5929 © University of Cambridge Sea Level What can you see in the picture? There are black markings all the way up the lighthouse right to the very top, and on the rock that the lighthouse is built on, going down to the sea bed. These markings are 1 metre apart. We will call sea level '0' and then we could think of the lines as representing positive numbers going upwards and negative numbers going down to the sea bed. In this task, we are going to look at where the mouths of each of the different animals are. This will allow us to see how deep they are compared to each other, or what distance apart they are. For example, the seahorse's mouth is 1 metre deeper than the mouth of the blue and yellow fish. Have a think about these questions, or you might like to make up some of your own: 1. What number should be where the head of the seagull is? 2. What number should be where the mouth of the crab is? 3. How far up is it from the head of the shrimp to the mouth of the shark? 4. How far from the surface is the eel's mouth? 5. How high above the sea level is the seagull's beak? 6. How much higher is the seagull's beak than the seahorse's mouth? How did you arrive at your answers? Did you write anything down to help you?
16306	https://www.fq.math.ca/Papers1/52-5/Benjamin.pdf	COMBINATORIAL PROOFS OF FIBONOMIAL IDENTITIES ARTHUR T. BENJAMIN AND ELIZABETH REILAND Abstract. Fibonomial coeﬃcients are deﬁned like binomial coeﬃcients, with integers re-placed by their respective Fibonacci numbers. For example, 10 3 F = F10F9F8 F3F2F1 . Remarkably, n k F is always an integer. In 2010, Bruce Sagan and Carla Savage derived two very nice combinatorial interpretations of Fibonomial coeﬃcients in terms of tilings created by lattice paths. We believe that these interpretations should lead to combinatorial proofs of Fibonomial identities. We provide a list of simple looking identities that are still in need of combinatorial proof. 1. Introduction What do you get when you cross Fibonacci numbers with binomial coeﬃcients? Fibono-mial coeﬃcients, of course! Fibonomial coeﬃcients are deﬁned like binomial coeﬃcients, with integers replaced by their respective Fibonacci numbers. Speciﬁcally, for n ≥k ≥1, n k F = FnFn−1 · · · Fn−k+1 F1F2 · · · Fk For example, 10 3 F = F10F9F8 F3F2F1 = 55·34·21 1·1·2 = 19,635. Fibonomial coeﬃcients resemble binomial coeﬃcients in many ways. Analogous to the Pascal Triangle boundary conditions n 1 = n and n n = 1, we have n 1 F = Fn and n n F = 1. We also deﬁne n 0 F = 1. Since Fn = Fk+(n−k) = Fk+1Fn−k + FkFn−k−1, Pascal’s recurrence n k = n−1 k + n−1 k−1 has the following analog. Identity 1.1. For n ≥2, n k F = Fk+1 n −1 k F + Fn−k−1 n −1 k −1 F . As an immediate corollary, it follows that for all n ≥k ≥1, n k F is an integer. Interesting integer quantities usually have combinatorial interpretations. For example, the binomial co-eﬃcient a+b a counts lattice paths from (0, 0) to (a, b) (since such a path takes a + b steps, a of which are horizontal steps and the remaining b steps are vertical). As described in and elsewhere, the Fibonacci number Fn+1 counts the ways to tile a strip of length n with squares (of length 1) and dominos (of length 2). As we’ll soon discuss, Fibonomial coeﬃcients count, appropriately enough, tilings of lattice paths! 2. Combinatorial Interpretations In 2010 , Bruce Sagan and Carla Savage provided two elegant counting problems that are enumerated by Fibonomial coeﬃcients. The ﬁrst problem counts restricted linear tilings and the second problem counts unrestricted bracelet tilings as described in the next two theorems. 28 COMBINATORIAL PROOFS OF FIBONOMIAL IDENTITIES Theorem 2.1. For a, b ≥1, a+b a F counts the ways to draw a lattice path from (0, 0) to (a, b), then tile each row above the lattice path with squares and dominos, then tile each column below the lattice path with squares and dominos, with the restriction that the column tilings are not allowed to start with a square. Let’s use the above theorem to see what 6 3 F = F6F5F4 F1F2F3 = 8·5·3 1·1·2 = 60 is counting. There are 6 3 = 20 lattice paths from (0, 0) to (3, 3) and each lattice path creates an integer partition (m1, m2, m3) where 3 ≥m1 ≥m2 ≥m3 ≥0, where mi is the length of row i. Below the path the columns form a complementary partition (n1, n2, n3) where 0 ≤n1 ≤n2 ≤n3 ≤3. For example, the lattice path below has horizontal partition (3, 1, 1) and vertical partition (0, 2, 2). The ﬁrst row can be tiled F4 = 3 ways (namely sss or sd or ds where s denotes a square and d denotes a domino). The next rows each have one tiling. The columns, of length 0, 2 and 2 can only be tiled in 1 way with the empty tiling, followed by tilings d and d since the vertical tilings are not allowed to begin with a square. For another example, the lattice path associated with partition (3, 2, 2) (with complementary vertical partition (0, 0, 2)) can be tiled 12 ways. These lattice paths are shown below. (0,0) (3,3) 3 ways 1 way 1 way 1 w a y 1 w a y (0,0) (3,3) 3 ways 2 ways 2 ways 1 w a y Figure 1. The rows of the lattice path (3, 1, 1) can be tiled 3 ways. The columns below the lattice path, with vertical partition (0, 2, 2) can be tiled 1 way since those tilings may not start with squares. This lattice path contributes 3 tilings to 6 3 F . The lattice path (3, 2, 2) contributes 12 tilings to 6 3 F . The lattice path associated with (3, 1, 0) has no legal tilings since its vertical partition is (1, 2, 2) and there are no legal tilings of the ﬁrst column since it has length 1. There are 10 lattice paths that yield at least one valid tiling. Speciﬁcally, the paths associated with horizontal partitions (3, 3, 3), (3, 2, 2), (3, 1, 1), (3, 0, 0), (2, 2, 2), (2, 1, 1), (2, 0, 0), (1, 1, 1), (1, 0, 0), (0, 0, 0) contribute, respectively, 27 + 12 + 3 + 3 + 8 + 2 + 2 + 1 + 1 + 1 = 60 tilings to 6 3 F . More generally, for the Fibonomial coeﬃcient a+b a F , we sum over the a+b a lattice paths from (0, 0) to (a, b) which corresponds to an integer partition (m1, m2, . . . , mb) where a ≥ m1 ≥m2 · · · ≥mb ≥0, and has a corresponding vertical partition (n1, n2, . . . , na) where 0 ≤n1 ≤n2 · · · ≤na ≤b. Recalling that F0 = 0 and F−1 = 1, this lattice path contributes Fm1+1Fm2+1 · · · Fmb+1 Fn1−1Fn2−1 · · · Fna−1 tilings to a+b a F . The second combinatorial interpretation of Fibonomial coeﬃcients utilizes circular tilings, or bracelets. A bracelet tiling is just like a linear tiling using squares and dominos, but bracelets DECEMBER 2014 29 THE FIBONACCI QUARTERLY also allow a domino to cover the ﬁrst and last cell of the tiling. As shown in , for n ≥1, the Lucas number Ln counts bracelet tilings of length n. For example, there are L3 = 4 tilings of length 3, namely sss, sd, ds and d′s where d′ denotes a domino that covers the ﬁrst and last cell. Note that L2 = 3 counts ss, d and d′ where the d′ tiling is a single domino that starts at cell 2 and ends on cell 1. For combinatorial convenience, we say there are L0 = 2 empty tilings. The next combinatorial interpretation of Sagan and Savage has the advantage that there is no restriction on the vertical tilings. Theorem 2.2. For a, b ≥1, 2a+ba+b a F counts the ways to draw a lattice path from (0, 0) to (a, b), then assign a bracelet to each row above the lattice path and to each column below the lattice path. Speciﬁcally, the lattice path from (0, 0) to (a, b) that generates the partition (m1, m2, . . . , mb) above the path and the partition (n1, n2, . . . , na) below the path contributes Lm1Lm2 · · · Lmb Ln1Ln2 · · · Lna bracelet tilings to 2a+ba+b a F . Note that each empty bracelet contributes a factor of 2 to this product. For example, the lattice path from (0, 0) to (3, 3) with partition (3, 1, 1) above the path and (0, 2, 2) below the path contributes L3L1L1L0L2L2 = 72 bracelet tilings enumerated by 266 3 F = 64 × 60 = 3840. (0,0) (3,3) L3 = 4 ways 1 way 1 way 3 ways 3 ways L0 = 2 ways Figure 2. The rows above the lattice path can be tiled with bracelets in 4 ways and the columns below the path can be tiled with bracelets in L0L2L2 = 2 × 3 × 3 = 18 ways. This contributes 72 bracelet tilings to 266 3 F = 3840. In their paper, Sagan and Savage extend their interpretation to handle Lucas sequences, deﬁned by U0 = 0, U1 = 1 and for n ≥2, Un = aUn−1 + bUn−2. Here Un+1 enumerates the total weight of all tilings of length n where the weight of a tiling with i squares and j dominos is aibj. (Alternatively, if a and b are positive integers, Un+1 counts colored tilings of length n where there are a colors for squares and b colors for dominos.) Likewise the number of weighted bracelets of length n is given by Vn = aVn−1 +bVn−2 with initial conditions V0 = 2 and V1 = a (so the empty bracelet has a weight of 2). This leads to a combinatorial interpretation of Lucasnomial coeﬃcients n k U, deﬁned like the Fibonomial coeﬃcients. For example, 10 3 U = U10U9U8 U1U2U3 . Both of the previous combinatorial interpretations work exactly as before, using weighted (or colored) tilings of lattice paths. 30 VOLUME 52, NUMBER 5 COMBINATORIAL PROOFS OF FIBONOMIAL IDENTITIES 3. Combinatorial Proofs Now that we know what they are counting, we should be able to provide combinatorial proofs of Fibonomial coeﬃcient identities. For example, Identity 1.1 can be rewritten as follows. Identity 3.1. For m, n ≥1, m + n m F = Fm+1 m + n −1 m F + Fn−1 m + n −1 m −1 F . Combinatorial Proof: The left side counts tilings of lattice paths from (0, 0) to (m, n). How many of these tiled lattice paths end with a vertical step? As shown below, in all of these lattice paths, the ﬁrst row has length m and can be tiled Fm+1 ways. The rest depends on the lattice path from (0, 0) to (m, n −1). Summing over all possible lattice paths from (0, 0) to (m, n −1) there are m+n−1 m F tiled lattice paths for the rest of the lattice. Hence the number of tiled lattice paths ending in a vertical step is Fm+1 m+n−1 m F . (0, 0) (m, n) (m, n −1) Fm+1 ways m + n −1 m F ways Figure 3. There are Fm+1 m+n−1 m F tiled lattice paths that end with a vertical step. How many tiled lattice paths end with a horizontal step? In all such paths, the last column has length n and can be tiled Fn−1 ways (beginning with a domino). Summing over all lattice paths from (0, 0) to (m−1, n) there are m+n−1 m−1 F tiled lattice paths for the rest of the lattice. Hence the number of tiled lattice paths ending in a horizontal step, as illustrated below, is Fn−1 m+n−1 m−1 F . (0, 0) (m −1, n) (m, n) m + n −1 m −1 F ways Fn−1 ways domino Figure 4. There are Fn−1 m+n−1 m−1 F tiled lattice paths that end with a hori-zontal step. DECEMBER 2014 31 THE FIBONACCI QUARTERLY Combining the two previous cases, the total number of tiled lattice paths from (0, 0) to (m, n) is Fm+1 m+n−1 m F + Fn−1 m+n−1 m−1 F . □ Replacing linear tilings with bracelets and removing the initial domino restriction for vertical tilings, we can apply the same logic as before to get 2m+n m + n m F = 2m+n−1Lm m + n −1 m F + 2m+n−1Ln m + n −1 m −1 F . Dividing both sides by 2m+n−1 gives us Identity 3.2. For m, n ≥1, 2 m + n m F = Lm m + n −1 m F + Ln m + n −1 m −1 F . In full disclosure, Identities 3.1 and 3.2 are used by Sagan and Savage to prove their combi-natorial interpretations, so it is no surprise that these identities would have easy combinatorial proofs. The same is true for the weighted (or colorized) version of these identities for Lucas-nomial coeﬃcients. Identity 3.3. For m, n ≥1, m + n m U = Um+1 m + n −1 m U + Un−1 m + n −1 m −1 U . Identity 3.4. For m, n ≥1, 2 m + n m U = Vm m + n −1 m U + Vn m + n −1 m −1 U . By considering the number of vertical steps that a lattice path ends with, Reiland proved Identity 3.5. For m, n ≥1, m + n m F = n X j=0 F j m+1Fn−j−1 m −1 + n −j m −1 F Combinatorial Proof: We count the tiled lattice paths from (0, 0) to (m, n) by considering the number j of vertical steps at the end of the path, where 0 ≤j ≤n. Such a tiling begins with j full rows, which can be tiled F j m+1 ways. Since the lattice path must have a horizontal step from (m −1, n −j) to (m, n −j), the last column will have height n −j and can be tiled (without starting with a square) in Fn−j−1 ways. The rest of the tiling consists of a tiled lattice path from (0, 0) to (m −1, n −j) which can be created in m−1+n−j m−1 F ways. (Note that when j = n −1, the summand is 0, since F0 = 0, as is appropriate since the last column can’t have height 1 without starting with a square; also, when j = n, F−1 = 1, so the summand simpliﬁes to F n m+1, as required.) All together, the number of tilings is Pn j=0 F j m+1Fn−j−1 m−1+n−j m−1 F , as desired. □ By the exact same logic, using bracelet tilings, we get Identity 3.6. For m, n ≥1, 2m+n m + n m F = n X j=0 Lj mLn−j2m+n−1−j m −1 + n −j m −1 F 32 VOLUME 52, NUMBER 5 COMBINATORIAL PROOFS OF FIBONOMIAL IDENTITIES Replacing F with U and replacing L with V , the last two identities are appropriately colorized as well. 4. Open Problems What follows is a list of Fibonomial identities that are still in need of combinatorial proof. Some of these identities have extremely simple algebraic proofs (and some hold for more general sequences than Fibonomial sequences) so one would expect them to have elementary combinatorial proofs as well. Many simple identities appear in Fibonacci Quarterly articles by Gould [4, 5]. n k F k j F = n j F n −j k −j F n k F = n X j=k Fj −Fj−k Fk j −1 k −1 F Fk n k F = Fn n −1 k −1 F = Fn−k+1 n k −1 F Here is another basic identity for generalized binomial coeﬃcients, ﬁrst noted by Fonten´ e and further developed by Trojovsk´ y n k F − n −1 k F = n −1 k −1 F Fn −Fk Fn−k . Here are some alternating sum identities, provided by Lind and Cooper and Kennedy , respectively, that might be amenable to sign-reversing involutions: k+1 X j=0 (−1)j(j+1)/2 k + 1 j F n −1 k F = 0. k X j=0 (−1)j(j+1)/2 k j F F k−1 n−j = 0. Here are some special cases of very intriguing formulas that appear in a recent paper by Kilic, Akkus and Ohtsuka . 2n+1 X k=0 2n + 1 k F = n Y k=0 L2k 2n X k=0 (−1)k 4n 2k F = (−1)n 2n Y k=1 L2k−1 We have just scratched the surface here. There are countless others! DECEMBER 2014 33 THE FIBONACCI QUARTERLY References A. T. Benjamin and J. J. Quinn. Proofs That Really Count: The Art of Combinatorial Proof, The Dolciani Mathematical Expositions, 27, Mathematical Association of America, Washington DC, 2003. C. Cooper and R. E. Kennedy. Proof of a Result by Jarden by Generalizing a Proof by Carlitz, Fib. Quart 33.4 (1995) 304–310. G. Fonten´ e. G´ en´ eralisation d’une Formule Connue, Nouvelle. Ann. Math., 15 (1915), 112. H. W. Gould. The bracket function and the Fonten´ e-Ward Generalized Binomial Coeﬃcients with Applica-tion to Fibonomial Coeﬃcients, Fib. Quart. 7 (1969) 23–40, 55. H. W. Gould. Generalization of Hermite’s Divisibility Theorems and the Mann-Shanks Primality Criterion for s-Fibonomial Arrays, Fib. Quart. 12 (1974) 157–166. E. Kilic, A. Akkus, and H. Ohtsuka. Some Generalized Fibonomial Sums related with the Gaussian q-Binomial sums, Bull. Math. Soc. Sci. Math. Roumanie, 55 (2012), 51–61. D. A Lind. A Determinant Involving Generalized Binomial Coeﬃcients, Fib. Quart 9.2 (1971) 113–119, 162. E. Reiland. Combinatorial Interpretations of Fibonomial Identities, Senior Thesis, Harvey Mudd College, Claremont, CA 2011. B. Sagan and C. Savage. Combinatorial Interpretations of Binomial Coeﬃcient Analogues Related to Lucas Sequences, Integers 10 (2010), 697–703. P. Trojovsk´ y. On some Identities for the Fibonomial Coeﬃcients via Generating Function, Discrete Appl. Math.155 (2007) no. 15, 2017–2024. AMS Classiﬁcation Numbers: 05A19, 11B39 Department of Mathematics, Harvey Mudd College, Claremont, CA 91711 E-mail address: benjamin@hmc.edu Department of Applied Mathematics and Statistics, Johns Hopkins University, Baltimore, MD 21218 E-mail address: ereiland@jhu.edu 34 VOLUME 52, NUMBER 5
16307	https://www.eoht.info/page/Metastability?ref=misaligned.markets	\| \| \| metastable \| \| A thermodynamic depiction of four positions of a bound state entity, e.g. ball or bound couple, on an energy surface (Gibbs surface or Gibbs landscape for couple), wherein the “metastable” position is a local low-energy well (left valley) and the only “stable” position is the lowest possible energy configuration (right valley). (Ѻ) \|
16308	https://campusvirtual.ull.es/ocw/pluginfile.php/5631/mod_resource/content/0/Apuntes_del_tema_1._Parte_I._Micro_OCW_2013.pdf	Apuntes de Microeconomía. Equilibrio general y economía de la información Fernando Perera Tallo Olga María Rodríguez Rodríguez Tema 1 Equilibrio general y fallos de mercado Tema 1. Equilibrio General y Fallos de Mercado. 1.1. Introducción. 1.2. Un modelo de Equilibrio General. 1.3. Equilibrio Walrasiano. 1.3.1. La toma de decisiones por parte de los agentes de la economía. 1.3.2. El concepto de equilibrio Walrasiano. 1.3.3. El cálculo del equilibrio Walrasiano. 1.4. Eficiencia productiva y frontera de posibilidades de producción. 1.4.1. Conjunto de posibilidades de producción, eficiencia productiva y frontera de posibilidades de producción. 1.4.2. La caja de Edgeworth de factores y la curva de asignaciones de factores con eficiencia productiva. 1.4.3. La relación marginal de transformación o coste de oportunidad entre dos bienes. 1.4.4. La convexidad del conjunto de posibilidades de producción. 1.4.5. El cálculo de la frontera de posibilidades de producción y su representación a través de un gráfico de cuatro cuadrantes. 1.4.6. El equilibrio Walrasiano y la eficiencia productiva. 1.5. Eficiencia en sentido de Pareto. 1.5.1. El concepto de eficiencia paretiana y el sistema de ecuaciones que definen el óptimo paretiano. 1.5.2. Las condiciones de eficiencia: 1.5.2.1. Eficiencia en la combinación factorial entre empresas (Eficiencia productiva). 1.5.2.2. Eficiencia asignativa del consumo o eficiencia de la asignación de bienes entre consumidores. 1.5.2.3. Eficiencia de la combinación productiva o elección en la combinación de producción en la FPP que sea eficiente. 1.5.2.4. Utilización plena de los recursos de la economía. 1.5.3. Representación del óptimo de Pareto en un gráfico de cuatro cuadrantes. 1.6. Eficiencia del equilibrio Walrasiano: Teoremas del Bienestar. 1.7. Algunos modelos de Equilibrio General. 1.7.1. Intercambio puro. 1.7.2. Un consumidor, dos empresas, un factor y dos bienes. 1.8. Los fallos de mercado. 1.8.1. Los efectos externos. 1.8.2. Los bienes públicos. Apéndices. Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 2 1.1. Introducción. Cuando en un modelo económico hay interacción entre una serie de agentes, se necesita definir algún tipo de equilibrio que nos permita dar predicciones sobre el comportamiento del modelo. Se entiende por equilibrio, tanto en economía como en otras materias, como la física, una situación en la que, una vez alcanzada, no hay tendencia a que cambie. En economía, equilibrio es una situación en la que todos los agentes tienen incentivos a hacer lo que están haciendo, lo que implica que no tienen incentivos a cambiar o desviarse de su comportamiento. Dicho de otra manera, todos los agentes hacen lo que quieren y, por tanto, no hay ninguna fuerza que haga que las cosas cambien, de ahí el equilibrio. Por ejemplo, en el modelo de oferta y demanda de un mercado, hay equilibrio cuando la oferta es igual a la demanda, y la cantidad que los consumidores compran y los productores venden coinciden, respectivamente, con sus demandas y ofertas individuales al precio de equilibrio. Esto implica que cada consumidor está comprando la cantidad que quiere comprar (su demanda del bien al precio de equilibrio) y cada productor está vendiendo la cantidad que quiere vender (su oferta al precio de equilibrio); por tanto, todo el mundo hace lo que quiere y no hay ningún incentivo a que los agentes cambien su comportamiento. Otro ejemplo: en el equilibrio de Nash, cada agente (jugador) elige la estrategia que maximiza sus pagos dadas las estrategias de los otros agentes y, por tanto, no tiene incentivos a desviarse porque está obteniendo lo máximo que puede obtener dado el comportamiento de los otros agentes. Otro elemento fundamental de los modelos donde los distintos agentes interactúan en los mercados es el sistema de precios. Éste va a ser un indicador de la escasez relativa de los distintos bienes y factores, y va a proporcionar a los agentes un mecanismo de coordinación. Así, cuando hay mayor escasez de un determinado bien o factor, la manera en que se coordinan los agentes es estableciendo un precio relativo alto, lo que hace que los compradores (demandantes) no quieran comprar mucho de ese bien o factor, y los vendedores (oferentes) tengan incentivos a proporcionar mayores cantidades de ese bien o factor. Coordinar las cantidades que tienen que producir y consumir los distintos agentes en un sistema de planificación central resulta muy complicado, porque se requiere una gran cantidad de información. En el sistema de mercado, la coordinación entre productores y consumidores se hace a través del sistema de precios. Entre los modelos en los que existe un sistema de precios se pueden distinguir dos tipos, dependiendo de si todos los precios que aparecen en el modelo son endógenos o no: - Modelos de equilibrio parcial: son modelos donde una serie de precios vienen dados y se consideran como variables exógenas. De hecho, los mercados en los que los precios son variables exógenas ni siquiera aparecen representados en el modelo. De ahí el nombre de equilibrio parcial; hay equilibrio en una serie de mercados, aquellos donde los precios son variables endógenas, pero en los otros mercados donde los precios son variables exógenas, el equilibrio no se recoge en el modelo. - Modelos de equilibrio general: son modelos donde todos los precios son endógenos y vienen determinados en cada uno de los mercados. Se llaman modelos de equilibrio general porque para que haya equilibrio, tienen que estar en equilibrio todos los mercados simultáneamente. Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 3 La elección entre analizar un fenómeno económico con un modelo de equilibrio general o parcial depende de la naturaleza del problema. El equilibrio parcial es adecuado para examinar cuestiones económicas donde las interacciones de distintos mercados no son muy relevantes para entender ese problema económico en cuestión. Si, por ejemplo, queremos estudiar la fijación de precios y cantidades en un mercado de un bien concreto bajo situación de oligopolio, el equilibrio parcial seguramente es el tipo de modelo más adecuado para hacerlo. Sin embargo, si la interacción entre los distintos mercados es un elemento importante del problema a tratar, lo más conveniente suele ser utilizar modelos de equilibrio general. El equilibrio general no solo es una parte fundamental de la Microeconomía, sino que, además, se aplica en muchos otros campos de la economía. Si se quiere tener una visión general de la interacción de los mercados de la economía, lo ideal es utilizar modelos de equilibrio general. No es de extrañar, por tanto, que la mayoría de los modelos macroeconómicos de los últimos 25 años sean de equilibrio general. Pero, además de la Macroeconomía, los modelos de equilibrio general se usan en los más diversas ramas de la economía: Comercio Internacional, Hacienda Pública, Economía de los Recursos Naturales, Economía del Turismo, etc. Una de las tendencias de las últimas décadas es la utilización de modelos de equilibrio general cuantitativos, donde los parámetros del modelo se calculan numéricamente a través de regularidades empíricas o estimaciones. Estos modelos nos permiten dar predicciones cuantitativas sobre los efectos de determinados cambios de variables exógenas o analizar los efectos cuantitativos de determinadas políticas económicas. Evidentemente, para poder utilizar estos modelos cuantitativos es necesario poder calcular el equilibrio, uno de los aspectos en los que se hace hincapié en este tema. En estos apuntes se empieza presentando un modelo de equilibrio general y definiendo lo que es el equilibrio Walrasiano. En el equilibrio Walrasiano todos los agentes maximizan sus funciones objetivo, es decir, los consumidores maximizan su utilidad y las empresas maximizan sus beneficios y, además, todos los mercados, tanto de bienes como de factores, están en equilibrio simultáneamente (es decir, la demanda de cada bien o factor se iguala a su oferta). Por tanto, el equilibrio Walrasiano es similar a otros conceptos de equilibrio en economía: si los agentes están maximizando sus funciones objetivo, no pueden mejorar desviándose de su comportamiento y, por tanto, no tienen ningún incentivo a cambiar. En otras palabras, los agentes están haciendo “lo que quieren”, es decir, tienen incentivos a hacer los que están haciendo. Los agentes en el equilibrio Walrasiano compran y venden bienes o factores. Más concretamente, los consumidores o economías domésticas compran bienes y venden factores productivos, mientras que las empresas venden bienes y compran factores productivos. Uno de los supuestos básicos del equilibrio Walrasiano es que los agentes son competitivos, es decir, son agentes que no tienen poder de mercado suficiente para modificar los precios de mercado, siendo, por tanto, precio-aceptantes, esto es, consideran los precios como dados y no como una variable de elección. Los modelos con agentes competitivos son muy populares en economía porque tienen una gran virtud: la simplicidad. Esto hace que este tipo de modelos sean casi siempre la mejor manera de empezar a analizar un problema económico, constituyendo, de este modo, el modelo de referencia (“benchmark model”). Una vez que se entiende cómo funciona el modelo competitivo, suele hacerse modificaciones de éste si el problema económico que Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 4 se está tratando requiere de modelos más complejos, pero siempre se utiliza como modelo de referencia el competitivo. Además, suelen compararse los resultados de los modelos más complejos con los del modelo de referencia. En estos apuntes se analizan algunas propiedades del equilibrio Walrasiano. Así, se empieza por la eficiencia productiva, que significa que para aumentar la producción de un bien es necesario reducir la producción de otro. Al conjunto de las combinaciones de bienes eficientes desde el punto de vista productivo se denomina frontera de posibilidades de producción. Si estamos en una combinación productiva en la frontera de posibilidades de producción, entonces, se puede definir coste de oportunidad de un bien en términos de otro, o relación marginal de transformación de un bien por otro, a la cantidad adicional del segundo bien que se podría producir si se dejara de producir una unidad del primer bien. La maximización de los beneficios de las empresas implica que el equilibrio Walrasiano es eficiente desde un punto de vista productivo, pero, además, los precios relativos son iguales al coste de oportunidad o relación marginal de transformación de los bienes, lo que significa que los precios del equilibrio Walrasiano reflejan correctamente la escasez de los bienes en la economía. Ésta, como veremos más adelante, es una propiedad importante del equilibrio Walrasiano. La propiedad más importante que vamos a analizar en este tema es, sin duda, la eficiencia en sentido de Pareto. Se dice que una situación es eficiente en sentido de Pareto si no podemos mejorar a un agente sin empeorar a otro. La eficiencia paretiana es un criterio más fuerte que la eficiencia productiva, ya que bajo el axioma de insaciabilidad de las preferencias, la eficiencia paretiana implica la eficiencia productiva, pero la eficiencia productiva no implica la eficiencia paretiana. De este modo, si no se da la eficiencia productiva, podemos aumentar la producción de un bien sin reducir la de otro, lo que implica que podemos dar esa cantidad adicional de producción de ese bien a un agente mejorándolo sin empeorar a nadie. Por tanto, si no se da la eficiencia productiva, seguro que tampoco se da la eficiencia en sentido de Pareto. Ahora bien, puede darse la eficiencia productiva sin que se dé la eficiencia de Pareto, bien porque no se elige la combinación productiva eficiente o porque no se distribuyen eficientemente los bienes entre los agentes económicos. Un resultado muy importante del tema es el Primer Teorema del Bienestar, según el cual el equilibrio Walrasiano es eficiente en sentido de Pareto. Este resultado es fundamental, no porque no haya situaciones en las que no se dé la eficiencia paretiana, sino porque hace que el modelo competitivo sea el modelo de referencia donde se cumple la eficiencia paretiana y, a partir de ahí, podemos modificar este modelo básico para identificar, de forma precisa, las causas que generan ineficiencia en la economía y las posibles soluciones a la misma. Esto lo analizaremos en el epígrafe 1.8, cuando tratemos el problema de los efectos externos y de los bienes públicos (los llamados fallos de mercado). La “inversa” del Primer Teorema del Bienestar también se cumple y se le conoce como Segundo Teorema del Bienestar. Éste señala que cualquier asignación de recursos eficiente en sentido de Pareto, la podemos implementar como un equilibrio Walrasiano si distribuimos los derechos de propiedad sobre los factores y las empresas de la manera adecuada. Tal y como se comenta arriba, hay situaciones reales en las que no se dan las condiciones necesarias para que el equilibrio de mercado sea eficiente, esto es, no se Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 5 cumplen los supuestos que requiere el Primer Teorema del Bienestar. En este tema también se analizan algunas de estas situaciones de ineficiencia. Empezaremos con los efectos externos, que ocurren cuando las acciones de un agente afectan a otro/s sin que haya contrapartida monetaria. Un ejemplo es el caso de las empresas que generan polución y afectan a la salud de las personas, que no reciben ningún pago en contrapartida; esto hace que la empresa contaminante genere unos costes en otros agentes que no paga. Por tanto, los costes privados no coinciden con los sociales y esto hace que el sistema de precios de mercado no refleje correctamente la escasez de los bienes en la economía, lo que implica una serie de incentivos incorrectos que generan la ineficiencia del mecanismo de mercado. Así, si una empresa genera efectos externos negativos, es decir, afectan negativamente a otros agentes sin que haya pagos de contrapartida, los costes privados de esta empresa no reflejan los costes que imponen a nivel social, lo que implica que, al no tener en cuenta los costes generados por los efectos externos, produzcan cantidades ineficientemente grandes. Por el contrario, si una empresa genera efectos externos positivos, la empresa beneficia a otros agentes sin recibir ningún pago a cambio. Esto hace que los costes privados de la empresa sean superiores a los sociales, lo que reduce los incentivos de la empresa a producir, y hace que la cantidad producida sea ineficientemente pequeña. Los bienes a los que se refiere la definición de equilibrio Walrasiano son los llamados bienes privados, con dos importantes características: 1) si un agente consume un bien privado, esto reduce el posible consumo de este mismo bien por parte de otros agentes, es decir, los bienes privados son rivales; 2) es posible impedir a una persona que consuma un bien (por ejemplo, si no lo paga), es decir, los bienes privados son excluibles. Aquellos bienes que no son rivales ni excluibles son denominados bienes públicos. Por ejemplo, los fuegos artificiales pueden ser vistos por una persona sin que ello reduzca el disfrute de dichos fuegos por parte de otras personas: son bienes no rivales; y, además, es muy difícil impedir que esos fuegos artificiales sean vistos por algunas personas, por ejemplo, por los que no los paguen, es decir, son bienes no excluibles. Hay muchos ejemplos de bienes públicos: calles, carreteras (bajo ciertas condiciones), la ley, parques públicos,…etc. Tanto la no exclusión como la no rivalidad de los bienes públicos conllevan problemas de eficiencia. De este modo, la no exclusión hace que no sea posible hacer pagar a las personas que disfrutan del bien público, ya que lo disfrutarán independientemente de que paguen o no dicho bien. Esto es lo que se denomina el problema del “free rider” (o del “gorroneo”), que consiste en que las personas que disfrutan de los bienes públicos no los pagan. Por tanto, si los bienes públicos fueran producidos por empresas privadas, dichos bienes o no se producirían o se produciría una cantidad ineficientemente pequeña de los mismos, ya que los “free riders” no pagarían por el uso de los mismos, con lo que las empresas privadas no tendrían incentivos a producirlos. Por otro lado, la no rivalidad también implica problemas de ineficiencia. Cuando los bienes no son rivales, se consigue alcanzar la cantidad óptima cuando cada agente paga según su disposición a pagar, es decir, cada agente paga un precio por unidad igual a lo que estaría dispuesto a pagar por disfrutar de la última unidad de bien público: equilibrio de Lindahl. Ahora bien, en el equilibrio de Lindahl, cada agente paga un precio distinto, ya que la disposición a pagar cambia según las preferencias y la renta de los consumidores, pero esta heterogeneidad de los precios pagados por los distintos agentes es incompatible con el equilibrio Walrasiano, donde todos los consumidores pagan lo mismo. Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 6 1.2. Un modelo de Equilibrio General. En el primer modelo de equilibrio general que se presenta vamos a considerar lo siguiente:  Hay dos factores, capital (K) y trabajo (L), dos bienes, x e y, dos consumidores o economías domésticas, 1 y 2, y dos empresas, la que produce el bien x y la que produce el bien y, respectivamente.  Los preferencias de las economías domésticas vienen dadas por la función de utilidad1   1 1 1 , y x c c u para el consumidor 1 y   2 2 2 , y x c c u para el consumidor 2, siendo 1 x c la cantidad de bien x consumida por la economía doméstica 1, 1 y c la cantidad de bien y consumida por la economía doméstica 1, 2 x c la cantidad de bien x consumida por la economía doméstica 2, y 2 y c la cantidad de bien y consumida por la economía doméstica 2.  La producción de la empresa que produce el bien x y la correspondiente a la empresa que produce el bien y vienen dadas por las respectivas funciones de producción2   x x x L K F , y   y y y L K F , , siendo x K y x L las cantidades de capital y trabajo utilizadas por la empresa productora del bien x, y y K y y L las cantidades de capital y trabajo utilizadas por la empresa productora del bien y. A la producción de la empresa del bien x la denominaremos x q y a la producción de la empresa del bien y la llamaremos y q .  x p y y p son, respectivamente, el precio del bien x y del bien y, mientras que w y r son, respectivamente, los precios de utilización del trabajo y del capital.  Los beneficios de las empresas son la diferencia entre los ingresos y los costes, y vienen dados por x x x x x rK wL q p     y y y y y y rK wL q p     , siendo x  y y , respectivamente, los beneficios de las empresas productoras del bien x y del bien y.  Las economías domésticas son dueñas de los factores productivos y de las empresas de la economía. Denotaremos por 1 N y 1 B las cantidades de trabajo y capital que posee la economía doméstica 1. Análogamente, 2 N y 2 B son las cantidades de trabajo y capital que posee la economía doméstica 2. 1 x  y 1 y  son las participaciones del consumidor 1 en los beneficios de las empresas productoras del bien x y del bien y, mientras que 2 x  y 2 y  son las participaciones del consumidor 2 en los beneficios de las empresas productoras del bien x y del bien y. Las participaciones de los consumidores en los beneficios se pueden interpretar intuitivamente como el porcentaje de acciones que tiene un consumidor de una empresa, si ésta fuera una sociedad anónima. Evidentemente, dado que los consumidores son los propietarios de las empresas de la economía, las participaciones de los consumidores en cada empresa tienen que sumar 1: 1 Las funciones de utilidad son continuas, diferenciables de segundo orden, estrictamente crecientes en ambos argumentos en 2    y estrictamente cuasi-cóncavas. 2 Las funciones de producción son continuas, diferenciables de segundo orden, estrictamente crecientes en ambos argumentos en 2    y estrictamente cuasi-cóncavas. Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 7 1 2 1   x x   ; 1 2 1   y y   Como las economías domésticas son propietarias de los factores y de las empresas, sus rentas proceden del alquiler de sus factores y la participación en los beneficios de las empresas. Por tanto, las rentas del consumidor 1 y 2 serían, respectivamente: y y x x rB wN m     1 1 1 1 1     y y x x rB wN m     2 2 2 2 2     A los largo de este tema vamos a suponer que los agentes son competitivos, es decir, un agente (un consumidor o una empresa) que individualmente no puede afectar al precio del mercado y, por tanto, maximiza su función objetivo (función de utilidad o beneficios) considerando los precios como dados (es decir, es precio-aceptante). Para que nuestro modelo sea tratable y por motivos didácticos, vamos a suponer un número pequeño de consumidores y empresas, pero vamos a considerar que los agentes son precio-aceptantes. No obstante, hay que tener en cuenta que el modelo se puede generalizar fácilmente para considerar un número elevado de consumidores y empresas. 1.3. Equilibrio Walrasiano. 1.3.1. La toma de decisiones por parte de los agentes de la economía. En esta sección vamos a analizar la toma de decisiones por parte de agentes individuales, que son las economías domésticas y las empresas. Problema de optimización de los consumidores: como ya se ha visto en Microeconomía I, el problema de maximización de los consumidores es el siguiente (se resuelve solo para el consumidor 1 y sería análogo para el consumidor 2): 1 1 1 1 1 1 , 1 1 max m c p c p s.a ) ,c (c u y y x x y x c c y x   El Lagrangiano correspondiente a este problema sería de la siguiente forma:     1 1 1 1 1 1 , y y x x y x c p c p m c c u       Las condiciones de primer orden para solución interior, cuando el consumidor elige cantidades positivas de todos los bienes, son de la siguiente manera: y x y y x x y x y x x,y y y y x y x x y x x p p c ) ,c (c u c ) ,c (c u ) ,c (c RMS λp c ) ,c (c u c λp c ) ,c (c u c                             1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0   Los consumidores maximizan su utilidad cuando la relación marginal de substitución coincide con el precio relativo de los bienes: y x y x x,y p p ) ,c (c RMS  1 1 1 Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 8 En el punto óptimo la RMS se tiene que igualar al precio relativo Óptimo: y p m x p m y x p p  ~ m c p c p y y x x   x c y c   y x y x y x p p c c RMS  , , En dicho punto, la curva de indiferencia es tangente a la recta de balance. Esto implica que para alcanzar un nivel de utilidad mayor se tendría que optar por un punto que no cumpla la restricción presupuestaria. Problema de optimización de las empresas: el problema de maximización de beneficios se puede tratar de dos maneras (se resuelve solo para la empresa que produce el bien x y sería análogo para la empresa que produce el bien y):  Desde el punto de vista de la contratación de factores:   x x x x x x x x L K q q L K F a s rK wL q p x x x    , . max , ,  Desde el punto de vista de la elección de la cantidad de producción: ) , , ( max x x x x q q r w c q p x  Estos dos enfoques son absolutamente equivalentes, aunque en equilibrio general suele ser más sencillo utilizar el primer punto de vista. Siguiendo el primer punto de vista, la función Lagrangiana correspondiente al problema de maximización de beneficios es la siguiente:     x x x x x x x x q L K F rK wL q p      ,   Las condiciones de primer orden para solución interior son: Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 9                                    0 , 0 , 0 x x x x x x x x x x x x K L K F r K L L K F w L p q           r K L K F p w L L K F p x x x x x x x x x x       , , Es decir, se contrata un factor hasta el punto en que el valor del producto marginal de dicho factor,   x x x x x L L K F p   , , se iguale a su coste unitario, es decir, su precio, w . Si, por ejemplo, el valor del producto marginal del trabajo fuera mayor que su precio,   w L L K F p x x x x x    , , entonces aumentando la contratación de una unidad adicional de trabajo aumentaría más los ingresos que los costes, por lo que en este punto inicial no se estaría maximizando los beneficios. Si, por el contrario, el valor del producto marginal del trabajo fuera menor que su precio,   w L L K F p x x x x x    , , entonces disminuyendo en una unidad la cantidad de trabajo, disminuirían más los costes que los ingresos, por lo que en este punto inicial no se estarían maximizando los beneficios. Por tanto, una condición necesaria para que se maximicen los beneficios es que el valor del producto marginal de dicho factor,   x x x x x L L K F p   , , se iguale a su precio, w . x L w    x x x x x L L K F p ) , ( u.m. Valor del producto marginal del factor trabajo x L Evidentemente, la maximización de los beneficios implica la minimización del coste. Para verlo vamos a recordar el problema de minimización del coste:   x x x x x x K L q L K F a s rK wL x x   , . min , Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 10 La función Lagrangiana correspondiente a este problema de minimización es como sigue:     x x x x x x q L K F rK wL     ,   Las condiciones de primer orden (para solución interior) son las siguientes:           x x x K L x x x x x x x x x x x x x x x x L K RMST K L K F L L K F r w K L K F r L L K F w , , , , , ,                       Por tanto, para minimizar el coste, la relación marginal de substitución técnica entre dos factores se tiene que igualar al precio relativo de esos factores. Es fácil de comprobar que cuando se maximiza el beneficio se minimiza el coste:           r w K L K F L L K F L K RMST r K L K F p w L L K F p x x x x x x x x x x x K L x x x x x x x x x x                     , , , , , , La maximización de los beneficios implica la minimización del coste CT rK wL x x   x L x K   x x x x q L K F  , r w   x K x L Cuando se maximiza beneficios desde el punto de vista de la elección del nivel de producción, el precio se iguala al coste marginal: ) , , ( max x x x x q q r w c q p x  Las condiciones de primer orden son: ) , , ( ) , , ( x x x x x x q r w CMg q q r w c p     Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 11 Las condiciones de primer orden del problema de minimización del coste también se pueden escribir de la siguiente manera:               capital el por producida unidad última la de Coste trabajo el por producida unidad última la de Coste , , x x x x x x x x K L K F r L L K F w      La parte de la izquierda de la ecuación anterior es lo que aumentarían los costes si se contratara una unidad adicional de trabajo partido por lo que aumentaría la producción si se aumentara una unidad adicional de trabajo. Por tanto, podemos interpretar esta parte de la expresión como lo que aumentan los costes con la última unidad de bien producida por el factor trabajo. La parte derecha de la ecuación la podemos interpretar como lo que aumentan los costes con la última unidad producida por el capital. Si el coste de la última unidad de bien producida por el factor trabajo fuera mayor que el coste de la última unidad de bien producida por el capital se podrían reducir los costes incrementando la cantidad de capital y reduciendo la de trabajo. El coste marginal, lo que aumentan los costes debido al incremento de la producción, lo podemos descomponer en la parte correspondiente al trabajo y la parte correspondiente al capital: Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 12                                                                                                                                                       trabajo del incremento al debido coste del Incremento trabajo el por producida unidad última la de Coste trabajo al debida producción la de incremento del Porcentaje capital del incremento al debido coste del Incremento capital el por producida unidad última la de Coste capital al debida producción la de incremento del Porcentaje trabajo del incremento al debido coste del ncremento trabajo el por producida unidad última la de Coste trabajo al debida producción la de incremento del Porcentaje capital del incremento al debido coste del Incremento capital el por por producida unidad última la de Coste capital al debida producción la de incremento del Porcentaje , ) 1 ( , ) , , ( I , 1 , , , , , , , , , ) , , ( , , x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x L L K F w K L K F r q r w CMg L L K F w dK K L K F dL L L K F dL L L K F K L K F r dK K L K F dL L L K F dK K L K F dL L L K F dK K L K F dL w dK r dq dCT q r w CMg dL L L K F dK K L K F dq dL w dK r wL rK d dCT                                                          Para aumentar la producción del bien x es necesario incrementar las cantidades de capital y de trabajo utilizadas, lo que hace que los costes aumenten. El incremento de los costes se puede descomponer en el incremento del coste debido al incremento del trabajo y en la parte debida al incremento del capital. El incremento total de los costes resulta ser una media ponderada del coste de la última unidad producida por el trabajo y el coste de la última unidad producida del capital. Estos dos costes se igualan cuando se elige la combinación de factores que minimiza el coste, por lo que el coste marginal es igual al coste de la última unidad producida por el trabajo y por el capital3:           x x x x x x x x x x x x x x x x x x K L K F r L L K F w L L K F w K L K F r q r w CMg              , , , ) 1 ( , , ,   Es fácil de comprobar, a través de las condiciones de primer orden, que la maximización del beneficio desde el punto de vista de la elección de factores también implica que el precio se iguala al coste marginal: 3 En el apéndice 1 aparece otra forma (mucho más general) de verlo a través del Teorema de la Envolvente. Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 13         ) , , ( , , , , x x x x x x x x x x x x x x x x x x x x x q r w CMg L L K F w K L K F r p r K L K F p w L L K F p                      Una vez que hemos visto el proceso de toma de decisiones de los agentes de la economía, podemos definir lo que se conoce como asignación. Asignación: es un vector que especifica la cantidad de bien consumido por cada economía doméstica, la cantidad de bien producido por cada empresa y la cantidad de factor utilizado por cada empresa: ) , , , , , , , , , ( empresa la de factores y producción empresa la de factores y producción 2 agente consmo de Cesta 2 2 1 agente consumo de Cesta 1 1                 y y y y x x x x y x y x L K q L K q c c c c Es decir, una asignación nos detalla todas las variables de decisión de los distintos agentes. Esto significa que nos especifica las cestas de consumo de las economías domésticas (que es la variable sobre la que deciden los consumidores) y la producción y la cantidad de factores que utilizan las empresas (que son las variables que eligen las empresas). 1.3.2. El concepto de equilibrio Walrasiano. En la economía que hemos presentado, los consumidores compran bienes y, por tanto, representan la demanda en el mercado de bienes. Además, estas economías domésticas al ser las propietarias de los factores productivos, son los que ofertan factores en el mercado de factores. Mientras que las empresas juegan el papel opuesto en los mercados: venden bienes en el mercado de bienes, por tanto representan la oferta en dicho mercado; y contratan factores en el mercado de factores, por tanto representan la demanda en dicho mercado. Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 14 La siguiente figura recoge esta idea: Consumidores: maximizan su utilidad Empresas: maximizan beneficios Mercado de Bienes demanda bienes oferta bienes Mercado de Factores oferta factores demanda factores El concepto de equilibrio Walrasiano implica que todos los agentes eligen aquellas cantidades de bienes o factores que desean, es decir, que maximiza su función objetivo, y además el vector de precios es tal que las cantidades ofrecidas y demandas se igualan en cada uno de los mercados de la economía, tanto de bienes como de factores. Definición 1: Un equilibrio Walrasiano es una asignación ) , , , , , , , , , ( 2 2 1 1 y y y x x x y x y x L K q L K q c c c c , llamada asignación de equilibrio, y un vector de precios   r w p p y x , , , , llamado vector de precios de equilibrio, tal que:  Las economías domésticas eligen aquella cesta de consumo que maximizan su utilidad (demanda de bienes): - Consumidor 1: y y x x y y x x y x c c y x π θ π θ rB wN c p c p a s ,c c u ,c c y x 1 1 1 1 1 1 1 1 1 , 1 1 . ) ( max arg ) ( 1 1       - Consumidor 2: y y x x y y x x y x ,c c y x π θ π θ rB wN c p c p a s ,c c u c c y x 2 2 2 2 2 2 2 2 2 2 2 . ) ( max arg ) , ( 2 2        Las empresas eligen el nivel de producción (oferta de bienes) y la combinación de factores (demanda de factores) que maximizan los beneficios: - Empresa del bien x:     x x x x x x x x L K q x x x q L K F a s rK wL q p L K q x x x     , . max arg , , , , - Empresa del bien y: Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 15     y y y y y y y y L K q y y y q L K F a s rK wL q p L K q y y y     , . max arg , , , ,  Los mercados de bienes están en equilibrio (demanda = oferta): - Bien x :    x x x x x x x x q c c bien del productora empresa la por bien del oferta es consumidor los por todos bien de demanda 2 consumidor el por bien de demanda 2 1 consumidor el por bien de demanda 1              - Bien y :    y y y y y y y y q c c bien del productora empresa la por bien del oferta es consumidor los por todos bien de demanda 2 consumidor el por bien de demanda 2 1 consumidor el por bien de demanda 1               Los mercados de factores están en equilibrio (demanda = oferta): - Mercado de trabajo:                           es consumidor los por todos trabajo de oferta 2 consumidor el por trabajo de oferta 2 1 consumidor el por trabajo de oferta 1 empresas las por todas trabajo de demanda de empresa la por trabajo de demanda de empresa la por trabajo de demanda N N L L y y x x    - Mercado de capital:                           es consumidor los por todos capital de oferta 2 consumidor el por capital de oferta 2 1 consumidor el por capital de oferta 1 empresas las por todas apital de demanda de empresa la por capital de demanda de empresa la por capital de demanda B B K K c y y x x    Es decir, en el equilibrio Walrasiano todos los agentes maximizan sus funciones objetivo (los consumidores su utilidad y las empresas sus beneficios) y los mercados (tanto de bienes como de factores) se vacían, es decir, la oferta se iguala a la demanda. Como todos los conceptos de equilibrio en Economía, una situación es de equilibrio cuando todos los agentes tienen incentivos a hacer lo que están haciendo. O lo que es lo mismo, todos los agentes están haciendo lo que quieren. Ya hemos visto que cuando los consumidores maximizan su utilidad igualan la relación marginal de substitución entre bienes a los precios relativos de esos bienes, y las empresas igualan el valor del producto marginal de cada uno de los factores a los precios de utilización de dichos factores. Por tanto, una definición alternativa de equilibrio Walrasiano se podría basar en esas condiciones de primer orden. Esta definición tiene la ventaja de que implica un sistema de ecuaciones con el que podríamos calcular el equilibrio Walrasiano. Cuando todas las economías domésticas y todas las empresas tienen soluciones interiores, el equilibrio Walrasiano también se puede definir de la siguiente manera: Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 16 Definición 2: Un equilibrio Walrasiano es una asignación ) , , , , , , , , , ( 2 2 1 1 y y y x x x y x y x L K q L K q c c c c , llamada asignación de equilibrio, y un vector de precios   r w p p y x , , , , llamado vector de precios de equilibrio, tal que:  Las economías domésticas eligen aquella cesta de consumo que maximizan su utilidad (demanda de bienes): - Consumidor 1: y x y x x p p ,c c RMS  ) ( 1 1 1 y , y y x x y y x x π θ π θ rB wN c p c p 1 1 1 1 1 1      - Consumidor 2: y x y x x,y p p ,c c RMS  ) ( 2 2 2 y y x x y y x x π θ π θ rB wN c p c p 2 2 2 2 2 2       Las empresas eligen el nivel de producción (oferta de bienes) y la combinación de factores (demanda de factores) que maximizan los beneficios: - Empresa del bien x:     r K L K F p w L L K F p x x x x x x x x x x       , ,   x x x x L K F q ,  - Empresa del bien y:     r K L K F p w L L K F p y y y y y y y y y y       , ,   y y y y L K F q ,   Los mercados de bienes están en equilibrio (demanda = oferta): - Bien x: x x x q c c   2 1 - Bien y: y y y q c c   2 1  Los mercados de factores están en equilibrio (demanda = oferta): - Mercado de trabajo: 2 1 N N L L y x    - Mercado de capital: 2 1 B B K K y x    Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 17 1.3.3. El cálculo del equilibrio Walrasiano. De la definición 2 de equilibrio Walrasiano se obtiene el sistema de ecuaciones que nos permitiría calcularlo. Por tanto, calcular un equilibrio Walrasiano consiste en resolver el siguiente sistema de ecuaciones donde las incógnitas son la asignación ) , , , , , , , , , ( 2 2 1 1 y y y x x x y x y x L K q L K q c c c c y el vector de precios   r w p p y x , , , , y x y x y x p p ,c c RMS  ) ( 1 1 1 , (EW.1) y y x x y y x x π θ π θ rB wN c p c p 1 1 1 1 1 1      (EW.2) y x y x x,y p p ,c c RMS  ) ( 2 2 2 (EW.3) y y x x y y x x π θ π θ rB wN c p c p 2 2 2 2 2 2      (EW.4)   w L L K F p x x x x x    , (EW.5)   r K L K F p x x x x x    , (EW.6)   x x x x L K F q ,  (EW.7)   w L L K F p y y y y y    , (EW.8)   r K L K F p y y y y y    , (EW.9)   y y y y L K F q ,  (EW.10) x x x q c c   2 1 (EW.11) y y y q c c   2 1 EW.12) 2 1 N N L L y x    (EW.13) 2 1 B B K K y x    (EW.14) Por tanto, tenemos un sistema de 14 ecuaciones con 14 incógnitas, que son: - Las cestas de consumo de las economías domésticas ) , , , ( 2 2 1 1 y x y x c c c c (4 incógnitas = 2 consumidores  2 bienes). - La asignación de factores a las empresa y las producciones de las empresas ) , , , , , ( y y y x x x L K q L K q (6 incógnitas =2 empresas  (1 producción + 2 factores). - El vector de precios de los bienes   y x p p , (2 incógnitas = 2 bienes). - El vector de precios de los factores   r w, (2 incógnitas =2 factores). Además, para resolver este sistema de ecuaciones tenemos que tener en cuenta algunas propiedades importantes del equilibrio Walrasiano. Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 18 La normalización de precios: Hay que tener en cuenta que si   r w p p y x , , , es un vector de precios de equilibrio, entonces, para cualquier constante positiva , el vector de precios   r w p p y x     , , , es también un vector de precios de equilibrio: y x y x x,y p p ,c c RMS  ) ( 1 1 1  y x y x x,y p p c c RMS    ) , ( 1 1 1     y y y y y x x x x x y y x x rK wL q p θ rK wL q p θ rB wN c p c p          1 1 1 1 1 1      y y y y y x x x x x y y x x λrK λwL q λp θ λrK λwL q λp θ λrB λwN c λp c λp          1 1 1 1 1 1 y x y x x,y p p ,c c RMS  ) ( 2 2 2  y x y x x, p p ,c c RMS    ) ( 2 2 2 y     y y y y y x x x x x y y x x rK wL q p θ rK wL q p θ rB wN c p c p          2 2 2 2 2 2      y y y y y x x x x x y y x x λrK λwL q λp θ λrK λwL q λp θ λrB λwN c λp c λp          2 2 2 2 2 2   w L L K F p x x x x x    ,    w L L K F p x x x x x      ,   r K L K F p x x x x x    ,    r K L K F p x x x x x      ,   x x x x L K F q ,    w L L K F p y y y y y    ,    w L L K F p y y y y y      ,   r K L K F p y y y y y    ,    r K L K F p y y y y y      ,   y y y y L K F q ,  x x x q c c   2 1 y y y q c c   2 1 2 1 N N L L y x    2 1 B B K K y x    La razón de esta propiedad de los precios de equilibrio descansa en que lo único que es relevante para la toma de decisiones de los agentes son los precios relativos. Así, tenemos: y x x y x x p p p p c c RMS   y 1 1 1 y , ) , (     y y y y y y p w p w L L K F       , Si los precios de los bienes y de los factores se duplican, los consumidores van a poder comprar exactamente las mismas cestas de consumo que antes que se duplicaran los precios, ya que cualquier cesta de consumo cuesta el doble, pero las rentas de los consumidores también van a ser el doble. Como el hecho de que se dupliquen los Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 19 precios no afecta el conjunto presupuestario del consumidor (el conjunto de cestas de consumo que cumple la restricción presupuestaria), tampoco afecta a las decisiones del consumidor. En cuanto a las empresas, si se duplican los precios, se duplican tanto los ingresos como los costes, pero esto no afecta a la oferta de producto o a la demanda de factores, porque los precios relativos no han cambiado. De hecho, podemos escribir el sistema de ecuaciones del equilibrio Walrasiano en función de precios relativos, por ejemplo con respecto al precio del bien x         x x x y p r p w p p , , , 1 : y x y x x,y p p ,c c RMS  ) ( 1 1 1  x y y x x,y p p ,c c RMS / 1 ) ( 1 1 1      y y y y y x x x x x y y x x rK wL q p θ rK wL q p θ rB wN c p c p          1 1 1 1 1 1                       y x y x y x y y x x x x x x x x y x y x K p r L p w q p p K p r L p w q B p r N p w c p p c 1 1 1 1 1 1   y x y x x,y p p ,c c RMS  ) ( 2 2 2  x y y x x,y p p ,c c RMS / 1 ) ( 2 2 2      y y y y y x x x x x y y x x rK wL q p rK wL q p rB wN c p c p          2 2 2 2 2 2                         y x y x y x y y x x x x x x x x y x y x K p r L p w q p p θ K p r L p w q θ B p r N p w c p p c 2 2 2 2 2 2   w L L K F p x x x x x    ,    x x x x x p w L L K F    ,   r K L K F p x x x x x    ,    x x x x x p r K L K F    ,   x x x x L K F q ,    w L L K F p y y y y y    ,    x y y y y x y p w L L K F p p    ,   r K L K F p y y y y y    ,    x y y y y x y p r K L K F p p    ,   y y y y L K F q ,  x x x q c c   2 1 y y y q c c   2 1 2 1 N N L L y x    2 1 B B K K y x    Dado que lo único que importan son los precios relativos, lo que se suele hacer para que el vector de precios de equilibrio sea único es normalizar el vector de precios, que significa que se pone una restricción adicional a los precios que hace que esos precios sean únicos. Por ejemplo, se iguala el precio del bien x a la unidad: 1  x p , lo que significa que todos los demás precios están medidos en unidades del bien x, es decir, el Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 20 vector de precios resultante de esta normalización será el vector de precios relativos de los distintos bienes y factores con respecto al bien x. El hecho de que lo único relevante sean los precios relativos y que se normalice el vector de precios implica que hay una incógnita menos. Si, por ejemplo, normalizamos el precio del bien x a la unidad, las incógnitas del sistema del equilibrio Walrasiano serían el precio relativo del bien y con respecto al bien x, el precio relativo del factor trabajo con respecto al bien x y el precio relativo del capital con respecto al bien x, pero evidentemente el precio relativo del bien x con respecto a sí mismo es uno, por lo que desaparecería una incógnita:            x x x y y x p r p w p p r w p p , , , 1 , , , Por tanto, el sistema de 14 ecuaciones que define el equilibrio Walrasiano en este modelos tendría 10 incógnitas de la asignación ) , , , , , , , , , ( 2 2 1 1 y y y x x x y x y x L K q L K q c c c c y 3 incógnitas correspondientes a los precios relativos         x x x y p r p w p p , , , es decir tendríamos un sistema de 14 ecuaciones y 13 incógnitas. Ley de Walras: Si sumamos las restricciones presupuestarias de los consumidores obtenemos lo que se denomina la Ley de Walras, que nos dice que el valor de los excesos de demanda suman siempre cero: y y x x y y x x y y x x y y x x rB wN c p c p rB wN c p c p         2 2 2 2 2 2 1 1 1 1 1 1                       y y y x x x y y y x x x B B r N N w c c p c c p                 1 2 1 1 2 1 2 1 2 1 2 1 2 1                       y y y y x x x x y y y x x x rK wL q p rK wL q p B B r N N w c c p c c p              2 1 2 1 2 1 2 1          0 capital de mercado ED 2 1 trabajo de mercado ED 2 1 bien mercado ED 2 1 bien mercado ED 2 1                                         B B K K r N N L L w q c c p q c c p y x y x y y y y y x x x x x Un importante corolario de la Ley de Walras es que si todos los mercados menos uno están en equilibrio, entonces ese último mercado está también en equilibrio. Esto implica que nos “sobra” una de las ecuaciones del sistema de ecuaciones que define el equilibrio Walrasiano. Por tanto, a la hora de resolver el sistema de ecuaciones del equilibrio Walrasiano podemos prescindir de un ecuación de equilibrio de los 4 mercados que existen en la economía (bien x, bien y, capital y trabajo). Resumiendo, el sistema de ecuaciones del equilibrio Walrasiano en nuestra economía tiene 13 incógnitas, 10 correspondientes a la asignación ) , , , , , , , , , ( 2 2 1 1 y y y x x x y x y x L K q L K q c c c c y 3 incógnitas correspondientes a los precios relativos         x x x y p r p w p p , , , y un sistema de 13 ecuaciones, ya que eliminando la ecuación Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 21 de equilibrio de uno de los cuatro mercados, sabemos que si hay tres mercados en equilibrio también lo está el cuarto. Otra propiedad que siempre se cumple en el equilibrio Walrasiano es que el gasto de los consumidores en los distintos bienes es igual a la renta y es igual al valor de la producción (el PIB): y y x x y y x x y y x x y y x x rB wN c p c p rB wN c p c p         2 2 2 2 2 2 1 1 1 1 1 1                       y y y x x x y y y x x x B B r N N w c c p c c p                 1 2 1 1 2 1 2 1 2 1 2 1 2 1                                               Renta Gasto 2 1 2 1 2 1 2 1 y x y y y x x x B B r N N w c c p c c p                                                             ) ( producción la de Valor Renta 0 0 2 1 2 1 2 1 2 1 2 1 2 1 PIB y y x x y x y x y y y y x x x x y x q p q p K K B B r L L N N w rK wL q p rK wL q p B B r N N w B B r N N w                            Por tanto:                                      ) ( producción Valor Renta Gasto 2 1 2 1 2 1 2 1 PIB y y x x y x y y y x x x q p q p B B r N N w c c p c c p              1.4. Eficiencia productiva y frontera de posibilidades de producción. 1.4.1. Conjunto de posibilidades de producción, eficiencia productiva y frontera de posibilidades de producción. Asignación factible: es una asignación donde la cantidad que se consume de cada bien es menor o igual que lo que produce la empresa que elabora ese bien, donde la cantidad producida por cada empresa es la que le permite su tecnología (es menor o igual de lo que determina su función de producción), y donde la cantidad de cada factor utilizada por las empresas es menor o igual que la dotación de ese factor en la economía. Es decir, ) , , , , , , , , , ( 2 2 1 1 y y y x x x y x y x L K q L K q c c c c es factible si y solo si se cumplen las siguientes restricciones (denominadas restricciones de factibilidad): - Se consume menos o igual que lo que se produce: x x x q c c   2 1 ; y y y q c c   2 1 . Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 22 - Cada empresa produce de acuerdo con su tecnología:   x x x x L K F q ,  ;   y y y y L K F q ,  . - No se usan más factores que los existentes en la economía: L N N L L y x     2 1 ; K B B K K y x     2 1 , donde L y K representan, respectivamente, la cantidad total de trabajo y de capital que existen en la economía. En definitiva, una asignación factible es una asignación que sea posible dadas las dotaciones de factores de la economía y la tecnología existente (las funciones de producción). Conjunto de posibilidades de producción (CPP): conjunto de todas las posibles combinaciones de bienes que se pueden producir en una economía dada su tecnología y sus recursos:       K K K L L L L K F q L K F q q q CPP y x y x y y y y x x x x y x           , , , , , / ) , ( 2 Eficiencia productiva: se dice que una combinación productiva factible CPP q q y x  ) , ( , es eficiente desde el punto de vista productivo si no existe otra combinación productiva factible que tenga una cantidad igual o mayor de todos los bienes, y una cantidad estrictamente mayor de alguno de ellos. Es decir, una combinación de producción de bienes factible es eficiente desde el punto vista productivo, si no podemos aumentar la producción de un bien sin reducir la producción de otro. Si una combinación de bienes perteneciente al CPP no es eficiente desde el punto de vista productivo, se dice que es ineficiente desde el punto de vista productivo. Es decir, una combinación de producción de bienes factible es ineficiente desde el punto de vista productivo si podemos aumentar la producción de un bien sin reducir la producción de ningún otro. Frontera de posibilidades de producción (FPP): conjunto de combinaciones de bienes pertenecientes al conjunto de posibilidades de producción que son eficientes desde el punto de vista productivo. Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 23 Combinaciones ineficientes desde el punto de vista productivo: para aumentar la producción de un bien no es necesario reducir la del otro. Frontera de posibilidades de producción: Combinaciones con eficiencia productiva: para aumentar la producción de un bien es necesario reducir la del otro. + Conjunto de posibilidades de producción x q y q 1.4.2. La caja de Edgeworth de factores y la curva de asignaciones de factores con eficiencia productiva. Dada una cantidad de producción de bien x, x q ˆ , si quisiéramos obtener la combinación productiva eficiente desde un punto de vista productivo en que se obtuviera esa cantidad de bien x, tendríamos que escoger la cantidad de bien y máxima posible dada la cantidad de bien x. Es decir, tendríamos que maximizar la cantidad de bien y dentro del conjunto de posibilidades de producción compatible con el nivel de producción del bien x, x q ˆ . Esto es lo mismo que decir que tendríamos que maximizar la cantidad producida de bien y, bajo la restricción de que la producción de bien x fuese x q ˆ y bajo las restricciones de factibilidad (si no se cumplieran estas restricciones, estaríamos fuera del conjunto de posibilidades de producción). Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 24 x q y q x q ˆ x x y y q q CPP q q ˆ que tal max ˆ    y q ˆ Por tanto, cualquier punto de la frontera de posibilidades de producción tendría que ser una solución del siguiente problema de optimización:     K K K L L L L K F q L K F q q q a s q y x y x y y y y x x x x x x y L K q L K q y y y x x x        , , ˆ : . max , , , , , Este problema de maximización se puede resumir de la siguiente forma (eliminando las producciones):     K K K L L L q L K F a s L K F y x y x x x x x y y y L K L K y y x x     ˆ , : . , max , , , El Lagrangiano asociado a este problema de maximización sería:     ) ( ) ( ) ˆ , ( , y x y x x x x x x y y y K K K L L L q L K F L K F              donde  , x  y  son los multiplicadores de Lagrange de las distintas restricciones.  y  se pueden interpretar como los precios sombra del trabajo y del capital, respectivamente. Suponiendo que la solución de este problema sea interior, las condiciones de primer orden serían:        x x x x x L L K F ,        x x x x x K L K F ,       y y y y L L K F , Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 25       y y y y K L K F , Estas condiciones de primer orden implican:                                                                                     y y y y y y y y y y y K L y y y y y y y y x x x x x x x x x x x K L x x x x x x x x x x K L K F L L K F L K RMST K L K F L L K F K L K F L L K F L K RMST K L K F L L K F , , , , , , , , , , , ,     y y y K L x x x K L L K RMST L K RMST , , , ,  Es decir, la relación marginal de substitución técnica entre dos factores se iguala a su precio sombra relativo. Además, la relación marginal de substitución técnica entre dos factores se iguala entre todas las empresas. Esta condición de igualación de las relaciones marginales de substitución técnica de todas las empresas de la economía la denominaremos condición de eficiencia de la combinación factorial entre empresas. En el ejemplo del gráfico siguiente tenemos que la RMST entre trabajo y capital de la empresa que produce el bien x es 4 mientras que la RMST entre trabajo y capital de la empresa que produce el bien y es 2 (por tanto, esta última es menor). Esto ofrece la posibilidad de reasignar los recursos, de tal manera que se pueda aumentar la producción de una empresa sin reducir la de la otra. Por ejemplo, si a la empresa que produce el bien x le quitamos 4 unidades de capital y le damos una unidad de trabajo, por definición de RMST, la producción del bien x no cambia, sigue siendo 8. Si le quitamos una unidad de trabajo a la empresa que produce el bien y y le damos 4 unidades de capital, dado que la RMST del bien y es 2, la producción del bien 2 aumentará (en este ejemplo pasa de 7 a 9), sin que haya disminuido la producción del bien x. Por tanto, la asignación original de factores (donde     y y y K L x x x K L L K RMST L K RMST , , , ,  ) no era eficiente desde un punto de vista productivo; es decir, no estábamos en la FPP, sino en el interior del conjunto de posibilidades de producción. Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 26 2 4 3 2 1 8 2 4 6 4 Reasignación de factores para el caso en que     y y y K L x x x K L L K RMST L K RMST , , , ,  y q x q 4 ,  x K L RMST 2 ,  y K L RMST 9  y q 7  y q 8  x q 8 9 7 y L y K x L x K Cuando tenemos dos factores, dos bienes y una empresa por cada bien, la asignación entre factores se puede representar en la caja de Edgeworth de factores productivos. Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 27 x K y L x L y K y x L L L   y x K K K   x L y L x K y K Caja de Edgeworth de factores productivos x K y K x L y L La caja de Edgeworth de factores es un rectángulo cuyo ancho es igual a la cantidad total de trabajo, L , y cuyo alto es la cantidad total de capital, K . Cualquier punto de la caja de Edgeworth representa una asignación de factores a las empresas de la economía ) , , , ( y y x x L K L K donde se utilizan todos los factores existentes. Si cogemos un punto de la caja de Edgeworth, la distancia horizontal entre el lado vertical de la izquierda y el punto representa la cantidad de trabajo asignada a la empresa x, x L . Como el ancho de la caja es igual a la cantidad total de trabajo, L , la distancia horizontal entre el lado vertical de la derecha del rectángulo y el punto es igual a x L L  . Suponiendo que todo el trabajo de la economía se reparte entre las dos empresas, L L L y x   , entonces la distancia horizontal entre el lado vertical de la derecha y el punto es igual a la cantidad de trabajo asignado a la producción del bien y, x y L L L   . Lo mismo ocurre con el capital, la distancia vertical entre el lado horizontal de la base del rectángulo y el punto representa la cantidad de capital asignado a la empresa del bien x, mientras que la distancia vertical entre el punto y el lado superior del rectángulo representa la cantidad de capital asignada a la empresa del bien y. La manera más natural de interpretar la caja de Edgeworth es que la esquina inferior izquierda es el origen del espacio de factores utilizado en la empresa x (el mapa de isocuantas del bien x), mientras que la esquina superior derecha es el origen del espacio de factores utilizado en la empresa y (el mapa de isocuantas de la empresa y). Es como si cogiéramos el mapa de isocuantas de la Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 28 empresa y y lo hiciéramos girar 180 grados en dirección contraria a las agujas del reloj, superponiéndolo al mapa de curvas isocuantas de la empresa x. y K x L x K x L x K y L y L y L L L K K Ox Ox Oy Oy Más producción x Más producción y Más producción x Más producción y En el siguiente gráfico se representa una asignación ineficiente donde la RMST entre trabajo y capital de la empresa x es menor que la de la empresa y. Esto implica que hay intercepción entre los conjuntos superiores formados por las isocuantas del punto inicial, lo que significa que se puede producir más del bien x o del bien y o de ambos bienes. Dado que la RMST entre trabajo y capital de la empresa x es menor que la de la empresa y, si quitamos trabajo de la producción del bien x y lo dedicamos a la producción del bien y, y dedicamos más capital a la producción del bien x y menos capital a la producción del bien y, se puede producir más de ambos bienes. Se produce más del bien y. Se produce más del bien x. Se produce más de ambos bienes. Área de Mejora = + + x K ~ y L ~ y K ~ x L x L ~ y K Asignaciones de factores con ineficiencia productiva x K y L Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 29 En el siguiente gráfico se representan asignaciones de recursos eficientes donde las RMSTs de ambas empresas se igualan. x K x L x Lˆ y Lˆ x Kˆ y Kˆ Asignaciones de factores con eficiencia productiva x K ~ x L ~ y L ~ y K ~ y L y K Asignación ineficiencia productiva 1.4.3. La relación marginal de transformación o coste de oportunidad entre dos bienes. La relación marginal de transformación del bien x por el bien y, o coste de oportunidad del bien x en términos del bien y, en un punto de la frontera de posibilidades de producción ( ) , ( , y x y x q q RMT ): es la cantidad que tiene que reducirse de bien y para aumentar en una unidad la producción del bien x a lo largo de la FPP, manteniendo la producción de todos los demás bienes (sin ser x e y) constante. En términos diferenciales, la relación marginal de transformación del bien x por bien y es simplemente igual a menos la derivada de la producción del bien y con respecto al bien x a lo largo de la FPP. Es decir, lo que disminuye el bien y a medida que aumenta la producción del bien x4: FPP x y y x y x q q q q RMT     ) , ( , Para calcular esta diferencial, hagamos lo siguiente: Una manera de escribir la condición de eficiencia de la combinación factorial es: 4 En el apéndice 2 se calcula esta diferencial utilizando el método más general del Teorema de la Envolvente. Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 30                           y x y x x x x x y y y y x x x x y y y y y y y K L y y y y y y y y x x x x x x x x x x x K L L L K F L L K F K L K F K L K F L K RMST K L K F L L K F K L K F L L K F L K RMST bien del s en término bien el en trabajo del d oportunida de Coste bien del s en término bien el en capital del d oportunida de Coste , , , , , , , , , , , ,                      La parte izquierda de la expresión anterior sería la cantidad de bien y que se produciría si se utilizara una unidad más de capital, partido por la cantidad de bien x que se produce con la última unidad de capital utilizada en el bien x. A esta expresión la denominaremos coste de oportunidad del capital en el bien x en términos del bien y. Análogamente, denominamos coste de oportunidad del trabajo en el bien x en términos del bien y a la cantidad de bien y que se produciría si se utilizara una unidad más de trabajo, partido por la cantidad de bien x que se produce con la última unidad de trabajo utilizada en el bien x. La ecuación anterior nos dice que el coste de oportunidad de utilizar capital y trabajo en el bien x en términos de bien y se tienen que igualar en una asignación eficiente desde el punto de vista productivo. Si, por ejemplo, el coste de oportunidad del capital en el bien x en términos del bien y fuera mayor que el del trabajo, entonces reduciendo la cantidad de capital y aumentando la cantidad de trabajo utilizada en el bien x, se podría aumentar la producción del bien y sin reducir la del bien x. La RMT del bien x por el bien y o el coste de oportunidad del bien x en términos del bien y, nos indica la cantidad de bien y que se ha dejado de producir para elaborar la última unidad del bien x. Para producir la última unidad de bien x se ha tenido que utilizar capital y trabajo que se ha dejado de utilizar en la producción del bien y, de ahí el coste de oportunidad. Por tanto, para obtener la RMT diferenciamos las restricciones de factibilidad con respecto a la cantidad de factores utilizados en cada bien y con respecto a las producciones de ambos bienes: Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 31                                                                                 y x x y x x x x x x y y y y x x x x x x x x x x x x x x x x x x x y y y y x x x x x x x x x x x x x x x x x x x x x x x x x x y y y y x y y y y x y y x y x x y y x y x x y y x y x y y y y y y y y y y y y y y y x x x x x x x x x x x x x x x L L K F L L K F dL L L K F dK K L K F dL L L K F K L K F K L K F dL L L K F dK K L K F dK K L K F dL L L K F dK K L K F dL L L K F dK K L K F dq dq q q RMT dK dK dK dK K K K dL dL dL dL L L L dL L L K F dK K L K F dq L K F q dL L L K F dK K L K F dq L K F q bien del s en término bien el en trabajo del d oportunida de Coste trabajo del incremento al debida bien del producción la de incremento del Porcentaje bien del s en término bien el en capital del d oportunida de Coste capital del incremento al debida bien del producción la de incremento del Porcentaje , , 1 , , , , , , , , , , ) ( , ) ( , ) , ( 0 0 , , , , , , ,                                                                                        La expresión anterior nos dice que el coste de oportunidad del bien x en términos del bien y (RMT) es igual a una media ponderada del coste de oportunidad del capital y el trabajo en el bien x en términos del bien y, lo cual es lógico, dado que el bien x tiene un coste de oportunidad de bien y porque se utilizan recursos en la producción del bien x (trabajo y capital) que se podrían utilizar para producir bien y. Dado que a lo largo de la FPP (combinaciones eficientes desde el punto de vista productivo) el coste de oportunidad de utilizar capital y trabajo en el bien x en términos de bien y se tienen que igualar, llegamos a la conclusión de que el coste de oportunidad del bien x en términos del bien y (RMT) es igual al coste de oportunidad de utilizar capital y trabajo en el bien x en términos de bien y:                 x x x x y y y y x x x x y y y y x x x x y y y y x x x x y y y y y x y x L L K F L L K F K L K F K L K F L L K F L L K F K L K F K L K F q q RMT                      , , , , , , ) 1 ( , , ) , ( ,   1.4.4. La convexidad del conjunto de posibilidades de producción. Hay dos situaciones en las que el conjunto de posibilidades de producción es estrictamente convexo, es decir, en las que el coste de oportunidad de un bien es creciente en la cantidad producida por ese bien:  Cuando hay rendimientos decrecientes a escala: si hay rendimientos decrecientes a escala, cuanto más se produce de un bien más recursos se necesitan para producir una unidad adicional de ese bien. Esto significa que si el bien x presenta rendimientos Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 32 decrecientes a escala, a medida que aumenta la producción del bien x, más recursos dedicados al bien y tendremos que dedicar a producir una unidad adicional de bien x, por lo que la caída en la producción del bien y, debida al incremento de la producción del bien x, será cada vez mayor. Es decir, el coste de oportunidad de x en términos del bien y será creciente.  Cuando hay rendimientos constantes a escala y los bienes tienen distintas intensidades factoriales: se dice que el bien x es más intensivo en capital que el bien y cuando para cualquier precio relativo del trabajo con respecto al capital, r w/ , y para cualquier nivel de producción de x e y, la ratio capital/trabajo que minimiza los costes del bien x es mayor que la ratio capital/trabajo que minimiza los costes del bien y. Si el bien x es más intensivo en capital que el bien y, entonces la curva de asignaciones eficientes de la caja de Edgeworth de factores estaría por encima de la diagonal: El bien x es intensivo en capital (el bien y es intensivo en trabajo) x x L K ˆ ˆ ~ y y L K ˆ ˆ ~ L K ~ x L ˆ x K ˆ y L ˆ y K ˆ y K x L x K y L Asignaciones de factores con eficiencia productiva Como sabemos, el ancho de la caja de Edgeworth de factores es igual a la cantidad de trabajo existente en la economía, L , mientras que el alto de la caja es igual a la cantidad de capital existente, K . Por tanto, la pendiente de la diagonal de la caja de Edgeworth será igual a L K / , es decir, a la ratio capital/trabajo media de la economía. En la caja de Edgeworth de la gráfica anterior, la curva de asignaciones factoriales eficientes está por encima de la diagonal. Por ello, si trazamos una línea recta entre el origen del bien x y cualquier punto de la curva de asignaciones eficientes, la pendiente de dicha recta, que es igual a la ratio capital/trabajo, está siempre por encima de la diagonal. Esto significa que la ratio capital/trabajo utilizada por la empresa del bien x es siempre superior a la ratio promedio de la economía (que es igual a la pendiente de la diagonal), mientras que la ratio capital/trabajo utilizada por la empresa del bien y (la pendiente de la recta que une el origen del bien y con el punto de la curva de asignaciones eficiente) siempre es menor que la ratio capital/trabajo media. Por tanto, cuando la asignación factorial es eficiente desde un punto de vista productivo, la empresa del bien x siempre usa una ratio capital/trabajo superior a la ratio capital/trabajo utilizada por la empresa y, lo que se Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 33 traduce en que el bien x es intensivo en capital, o lo que es lo mismo, el bien y es intensivo en trabajo. Si se asignaran a las empresas x e y asignaciones factoriales donde la ratio capital/trabajo utilizada en la producción los dos bienes fuera igual a la ratio promedio de la economía y lo representáramos en el espacio de bienes, dado que hay rendimientos constantes a escala, obtendríamos una línea recta. No obstante, esta línea recta no sería la FPP, ya que estas asignaciones factoriales son ineficientes, por lo que la FPP estaría a la derecha de esta línea. Los únicos puntos en los que coincidirían serían los puntos de corte con los ejes, ya que el punto de corte de la FPP con el eje del bien x representa una situación donde no se destina ningún recurso a la producción del bien y, es decir, donde la producción del bien y es cero, y se destinan todos los recursos a producir el bien x, lo que, evidentemente, implica que la ratio capital/trabajo utilizada en la producción del bien x es igual al promedio. Por tanto, el conjunto de posibilidades de producción tiene que ser convexo. Cuando los bienes tienen distintas intensidades factoriales y presentan rendimientos constantes a escala, el CPP es convexo x q y q Combinaciones de producción donde la ratio capital/trabajo de los dos bienes es igual al promedio. Todos los recursos se asignan a la producción del bien x. Todos los recursos se asignan a la producción del bien y. FPP: combinaciones de producción donde la asignación factorial es eficiente. 1.4.5. El cálculo de la frontera de posibilidades de producción y su representación a través de un gráfico de cuatro cuadrantes. Para calcular un punto de la FPP hay que resolver un sistema de ecuaciones con las restricciones de factibilidad y las condiciones de eficiencia productiva, que en este modelo sería la igualación de las RMSTs entre empresas (eficiencia de la combinación factorial entre empresas):   x x x x L K F q ,  (FPP.1)   y y y y L K F q ,  (FPP.2) L L L y x   (FPP.3) K K K y x   (FPP.4) Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 34     y y y K L x x x K L L K RMST L K RMST , , , ,  (FPP.5) En este sistema de ecuaciones tenemos 5 ecuaciones con 5 incógnitas que son y y y x x L K q L K , , , , . No tratamos la producción del bien x, x q , como una incógnita, ya que al final vamos a obtener la producción del bien y en función de la producción del bien x y de la dotación de factores: ) , , ˆ ( K L q q x y . Otra manera de expresar la FPP es de forma implícita: resolviendo el anterior sistema de ecuaciones obtendríamos la Función de Transformación, ) , ( y x q q FTR , que define las combinaciones de bienes x e y factibles de la siguiente manera:     0 ) , ( , 0 ) , ( ,       y x y x y x y x q q FTR FPP q q q q FTR CPP q q Siendo la función de transformación creciente en la producción de ambos bienes. Esto es: 0 ) , ( ; 0 ) , (       y y x x y x q q q FTR q q q FTR A través de la función de transformación se puede obtener la relación marginal de transformación: y y y x x x y x y y y x x x y x y x dq q q q FTR dq q q q FTR dq q q q FTR dq q q q FTR q q FTR              ) , ( ) , ( 0 ) , ( ) , ( 0 ) , ( y y x x y x x y y x y x q q q FTR q q q FTR dq dq q q RMT        ) , ( ) , ( ) , ( , Para representar gráficamente el sistema de ecuaciones con el que se calcula la FPP, vamos a utilizar primero la función de producción del bien y (FPP.2) en función de la cantidad de trabajo, considerando el nivel de capital como fijo. Seguidamente, vamos a rotar el eje del trabajo como si cerráramos un libro, de tal manera que la función de producción del bien y se representa en el cuadrante de la izquierda: Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 35 y q y L y L   y y y L K F , ~   y y y L K F , ~ Ahora utilizamos la función de producción del bien x en función de la cantidad de capital, considerando el nivel de trabajo fijo,   x x x L K F ~ , (FPP.1). También incorporamos la restricción de factibilidad que nos dice que no se puede utilizar más capital que el existente en la economía (FPP.4): y x y x K K K K K K      . Incorporando esta última restricción dentro de la función de producción de x para una cantidad de producción fija, obtenemos la producción del bien x como una función decreciente del capital utilizado en el bien y:   x y x x L K K F q ~ ,   . Hacemos un giro de 90 grados en dirección a las manecillas del reloj con los ejes de la representación gráfica de la función que relaciona la producción del bien x con la cantidad de capital utilizada en el bien y, tal y como puede apreciarse en el siguiente gráfico: y x K K K   K x y K K K   x q x q x q   x y x L K K F ~ ,    x x x L K F ~ ,   x y x L K K F ~ ,  K En el siguiente gráfico se ofrece una representación del sistema de ecuaciones que determina la FPP. En el cuadrante inferior izquierdo se representa la caja de Edgeworth de factores, donde se están utilizando los factores existentes en la economía Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 36 (restricciones FPP.3 y FPP.4) y, además, se está en la curva de asignaciones factoriales eficientes, por lo que se cumple la condición de eficiencia de la combinación factorial entre empresas (ecuación FPP.5). En el cuadrante superior izquierdo se representa la función de producción del bien y (ecuación FPP.2). En el cuadrante inferior derecho se representa la producción del bien x como función de la cantidad de capital utilizado en la producción del bien y, donde se ha incorporado la función de producción del bien x (FPP.1) y la restricción que nos dice que la cantidad de capital utilizada en la producción del bien x y el bien y tiene que ser igual a la cantidad total de capital existente en la economía (FPP.4). Por tanto, el gráfico incorpora todas las ecuaciones del sistema que determina la FPP. En el cuadrante inferior izquierdo se representa la asignación de factores entre el bien x y el bien y. Utilizando las funciones de producción de ambos bienes (función de producción de y del cuadrante superior izquierdo y función de producción de x del cuadrante inferior derecho), podemos ver los niveles de producción correspondiente a esta asignación factorial, que será un punto de la FPP en el cuadrante superior derecho. Frontera de Posibilidades de Producción 222 (2 bienes, 2 factores, 2 empresas) x K y L x L y K y x L L L   y x K K K   y q y q ~ x q ~ y L ~ x K ~ x L ~ y K ~   x y x L K K F ~ ,    y y y L K F , ~ x q Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 37 En el siguiente gráfico vemos qué ocurre cuando pasamos de un punto a otro de la FPP. Incremento de la producción del bien x a costa del bien y x K y L x L y K y K ˆ   y y y L K F , ˆ   x y x L K K F ˆ ,  y L ˆ x L ˆ x K ˆ y x L L L   y x K K K   y q ˆ x q ˆ y q y q ~ x q ~ y L ~ x K ~ x L ~ y K ~   x y x L K K F ~ ,    y y y L K F , ~ A A B A A B B B x q En el punto inicial A se produce más del bien y y menos del bien x. Cuando pasamos al punto B se incrementa la producción de x y se reduce la de y, por lo que se reasignan factores de la producción del bien y a la producción del bien x, hecho que se ve reflejado en la caja de Edgeworth (cuadrante inferior izquierdo) al desplazarnos del punto A de la curva de asignaciones eficientes de factores a otro punto B, donde se dedican menos recursos (trabajo y capital) a la producción del bien y y más recursos a la producción del bien x. Por otra parte, en el cuadrante superior izquierdo se observa que al detraerse capital de la producción del bien y, para el mismo nivel de trabajo se produce menos. Este hecho se refleja en el desplazamiento hacia abajo de la función que relaciona la producción del bien y con la cantidad de trabajo dedicada a la producción de dicho bien. Por el contrario, se dedica más trabajo a la producción del bien x, lo que se observa en el cuadrante inferior derecho con el desplazamiento hacia la derecha de la curva que relaciona la cantidad de capital utilizada en el bien y con la producción del bien x. 1.4.6. El equilibrio Walrasiano y la eficiencia productiva. Dado que en el equilibrio Walrasiano las empresas maximizan beneficios y, por tanto, minimizan costes, las RMSTs entre trabajo y capital se igualan a los precios relativos del trabajo con respecto al capital y, por tanto, las RMSTs se igualan entre las empresas. Esto implica que el equilibrio Walrasiano es eficiente desde el punto de vista productivo y, por tanto, la producción del equilibrio Walrasiano está siempre en la frontera de posibilidades de producción: Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 38                                                                           r w K L K F L L K F L K RMST r K L K F p w L L K F p r w K L K F L L K F L K RMST r K L K F p w L L K F p y y y y y y y y y y y K L y y y y y y y y y y x x x x x x x x x x x K L x x x x x x x x x x , , , , , , , , , , , ,     y y y K L x x x K L L K RMST r w L K RMST , , , ,   x K x L x L ˆ y L ˆ x K ˆ y K ˆ Asignaciones de factores en el EquilibrioWalrasiano y L y K r w L K RMST x x x K L ˆ ˆ ) ˆ , ˆ ( ~ ,    ~ ) ˆ , ˆ ( ˆ ˆ , y y y K L L K RMST r w    ~ ˆ ˆ r w  Relación Marginal de Transformación y Precios del Equilibrio Walrasiano: Usando las condiciones de primer orden del problema de maximización de beneficios, obtenemos:                                                                         ) , , ( , , , , ) , , ( , , , , y y y y y y y y y y y y y y y y y y y y y x x x x x x x x x x x x x x x x x x x x x q r w CMg L L K F w K L K F r p r K L K F p w L L K F p q r w CMg L L K F w K L K F r p r K L K F p w L L K F p Microeconomía. Equilibrio general y economía de la información Apuntes del Tema 1 Perera-Tallo y Rodríguez-Rodríguez 39         ) , ( , , , , ) , , ( ) , , ( , y x y x x x x x y y y y x x x x y y y y y y x x y x q q RMT L L K F L L K F K L K F K L K F q r w CMg q r w CMg p p             Por tanto, los precios relativos de los bienes en el equilibrio Walrasiano no solo reflejan los costes marginales privados, sino también el coste de oportunidad social (la RMT) Esta propiedad de los precios del equilibrio Walrasiano implica que los precios relativos de los bienes son tangentes a la FPP y, por tanto, la combinación productiva del equilibrio Walrasiano, no solo está en la FPP, como ya hemos demostrado, sino que, además, es el punto de la FPP donde se maximiza el valor de la producción (PIB), tal y como se muestra en el siguiente gráfico: y x y x y x p p q q RMT ˆ ˆ ) ˆ , ˆ ( ,     En el Equilibrio Walrasiano se maximiza el valor de la producción (PIB) x q ˆ y q ˆ x q y q y x p p ˆ ˆ   PIB q p q p q p q p y y x x y y x x     ˆ ˆ ˆ ˆ ˆ ˆ
16309	https://www.lecturio.com/concepts/mallory-weiss-syndrome-mallory-weiss-tear/	Achieve Mastery of Medical Concepts Study for medical school and boards with Lecturio Mallory-Weiss Syndrome (Mallory-Weiss Tear) Mallory-Weiss syndrome (MWS) is bleeding from longitudinal mucosal lacerations (tears) in the distal esophagus and proximal stomach caused by a sudden rise in intraluminal esophageal pressure with forceful or recurrent vomiting. Hematemesis is due to bleeding from submucosal blood vessels and is self-limited in 80%–90% of patients. Diagnosis is made by taking a history and performing upper GI endoscopy. Treatment includes gastric acid suppression, endoscopic intervention, and angiotherapy if there is active bleeding. Blood transfusions and surgery are not usually required. Last updated: May 28, 2025 Contents Share this concept: Epidemiology and Pathogenesis Epidemiology Risk factors Pathogenesis Clinical Presentation and Diagnosis Clinical presentation Diagnosis Diagnosis is established by endoscopy Endoscopy Procedures of applying endoscopes for disease diagnosis and treatment. Endoscopy involves passing an optical instrument through a small incision in the skin i.e., percutaneous; or through a natural orifice and along natural body pathways such as the digestive tract; and/or through an incision in the wall of a tubular structure or organ, i.e. Transluminal, to examine or perform surgery on the interior parts of the body. Gastroesophageal Reflux Disease (GERD), which shows a longitudinal tear (usually single) limited to the mucosa and submucosa at the gastroesophageal junction Gastroesophageal junction The area covering the terminal portion of esophagus and the beginning of stomach at the cardiac orifice. Esophagus: Anatomy. The muscular layer is NOT involved(in contrast to Boerhaave syndrome Boerhaave Syndrome Esophageal Perforation). Endoscopic image of a Mallory-Weiss tear Mallory-Weiss tear (green arrow) as visualized on esophagogastroduodenoscopy (EGD) Related videos Management Differential Diagnosis References Study with Lecturio for Medical School Nursing School Lecturio Support Institutions Resources Follow Us USMLE™ is a joint program of the Federation of State Medical Boards (FSMB®) and National Board of Medical Examiners (NBME®). MCAT is a registered trademark of the Association of American Medical Colleges (AAMC). NCLEX®, NCLEX-RN®, and NCLEX-PN® are registered trademarks of the National Council of State Boards of Nursing, Inc (NCSBN®). None of the trademark holders are endorsed by nor affiliated with Lecturio. Create your free account or log in to continue reading! Sign up now and get free access to Lecturio with concept pages, medical videos, and questions for your medical education. Log in to your account Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut elit tellus, luctus nec ullamcorper mattis, pulvinar dapibus leo. User Reviews Your MCAT Journey Starts Here 📚 Get 20% off on basic or 40% off on premium plans! Details Get Premium to test your knowledge Lecturio Premium gives you full access to all content & features Get Premium to watch all videos Verify your email now to get a free trial. Create a free account to test your knowledge Lecturio Premium gives you full access to all contents and features—including Lecturio’s Qbank with up-to-date board-style questions.
16310	https://cse.buffalo.edu/~regan/cse491596/DebrayStanfordCS154notes.pdf	CS 154 NOTES ARUN DEBRAY MARCH 13, 2014 These notes were taken in Stanford’s CS 154 class in Winter 2014, taught by Ryan Williams, by Arun Debray. They were updated in 2016. Error and typo notices can come ﬁrst to me, Ken Regan. CONTENTS Part 1. Finite Automata: Very Simple Models 1 1. Deterministic Finite Automata: 1/7/14 1 2. Nondeterminism, Finite Automata, and Regular Expressions: 1/9/14 5 3. Finite Automata vs. Regular Expressions, Non-Regular Languages: 1/14/14 8 4. Minimizing DFAs: 1/16/14 11 5. The Myhill-Nerode Theorem and Streaming Algorithms: 1/21/14 14 6. Streaming Algorithms: 1/23/14 17 Part 2. Computability Theory: Very Powerful Models 19 7. Turing Machines: 1/28/14 19 8. Recognizability, Decidability, and Reductions: 1/30/14 23 9. Reductions, Undecidability, and the Post Correspondence Problem: 2/4/14 27 10. Oracles, Rice’s Theorem, the Recursion Theorem, and the Fixed-Point Theorem: 2/6/14 29 11. Self-Reference and the Foundations of Mathematics: 2/11/14 33 12. A Universal Theory of Data Compression: Kolmogorov Complexity: 2/18/14 35 Part 3. Complexity Theory: The Modern Models 39 13. Time Complexity: 2/20/14 39 14. More on P versus NP and the Cook-Levin Theorem: 2/25/14 43 15. NP-Complete Problems: 2/27/14 47 16. NP-Complete Problems, Part II: 3/4/14 50 17. Polytime and Oracles, Space Complexity: 3/6/14 53 18. Space Complexity, Savitch’s Theorem, and PSPACE: 3/11/14 56 19. PSPACE-completeness and Randomized Complexity: 3/13/14 59 Part 1. Finite Automata: Very Simple Models 1. DETERMINISTIC FINITE AUTOMATA: 1/7/14 “When the going gets tough, the tough make a new deﬁnition.” 1 This class is about formal models of computation, which makes it a blend of philosophy, mathematics, and engineering. What is computation? What can and cannot be computed? What can be efﬁciently computed (whatever that means in context)? Though it’s a theory class, there are excellent reasons to care: it leads to new perspectives on computing, and theory often drives practice. For example, the theory of quantum comput-ers is relatively well-known, even though the best quantum computer today can only factor numbers as large as 21. Finally, math is good for you! Just like vegetables. And, like the vegetables in my fridge, the course content is timeless. Advances in technology come and go, but theorems are forever. The course falls into three parts: ﬁnite automata, which are simple and relatively well-understood models; computability theory, which is much more powerful but not quite as well understood; and complexity theory, which provides more applied theory for understanding how long it takes to solve some problems, but isn’t as well-understood. This is a proofs class. One great way to do a proof is to hit each of the three levels: • A short phrase (two to four words) that gives a hint of what trick will be used to prove the claim. • The second level: a short, one-paragraph description of the ideas. • The last level: the complete proof. This is how Sipser wrote his book, and we are encouraged to write our solutions. For example: Claim 1.1. Suppose A ⊆{1,...,2n} and \|A\| = n+1. Then, there are two numbers in A such that one divides the other. Proof. The pigeonhole principle states that if one has m pigeons and puts them in n holes. and m > n, then there is at least one hole with more than one pigeon in it.1 Another useful observation is that every a ∈Z can be written as a = 2km, where m is odd and k ∈Z. On the second level: using the pigeonhole principle, we can show there is an odd m and a1 ̸= a2 in A such that a1 = 2k1m and a2 = 2k2m, so one must divide the other. More formally: write each element in A of the form a = 2km, where m ∈{1,...,2n} is odd. Since \|A\| = n+1, then there must be two distinct numbers in A with the same odd part, since there are only n such odd numbers. Thus, a1,a2 ∈A have the same odd part, so a1 = 2k1m and a2 = 2k2m, and therefore one divides the other. ⊠ Note that in this class, the homeworks tend to require the most complete level of proof, but the class won’t go into the goriest details. In any case, the high-level concepts are more important, though the details are not unimportant. Moving into content, a deterministic ﬁnite automaton is a set of states with transitions between them. For example: / q0 0 1 / q1 0,1 q2 0 O 1 6 q3 1 p 0 v 1I personally prefer the formulation in which one drills n holes in m pigeons, so that at least one pigeon has more than one hole in it. . . 2 One can then accept or reject strings. Formally: Deﬁnition 1.2. A language over a set (an alphabet) Σ is a set of strings over Σ, i.e. a subset of Σ∗(the set of all strings over Σ). It’s useful to think of a language as a function from Σ∗→{0,1}, where the set of the language is the strings that output a 1. Deﬁnition 1.3. A deterministic ﬁnite automaton (DFA) is a 5-tuple M = (Q,Σ,δ, q0,F), where: • Q is a ﬁnite set of states, e.g. Q = {q0, q1, q2, q3} in the above example; • Σ is the alphabet, which is also a ﬁnite set; • δ : Q ×Σ →Q is the transition function; • q0 ∈Q is the start state; and • F ⊆Q is the set of accept states (ﬁnal states). Then, let w1,...,wn ∈Σ and ⃗ w = w1...wn ∈Σ∗. Then, M accepts ⃗ w if there exist r0,...,rn ∈Q such that: • r0 = q0, • δ(ri,wi+1) = ri+1, and • rn ∈F. Finally, the language accepted by M is L(M) = {s ∈Σ∗\| M accepts s}. For example the automaton / q0 0 1 6 q1 0 1 v accepts a string from {0,1}∗iff it has an even number of 1s, and the following DFA accepts exactly the strings containing a sequence 001: / q0 1 0 ( q1 0 / 1 h q2 0 1 / q3 0,1 Here, at q1, the automaton has seen a 0, at q2 it’s seen 00, and at q3, it’s seen a 001 (so it wins). Deﬁnition 1.4. A language A is regular if there exists a DFA M such that L(M) = A, i.e. there is some DFA that recognizes it. Theorem 1.5 (Union Theorem for Regular Languages). If L1 and L2 are regular languages over the same alphabet Σ, then L1 ∪L2 is also a regular language. Proof. The proof idea is to run both M1 and M2 at the same time. Speciﬁcally, given a DFA M1 = (Q1,Σ,δ1, q1 0,F1) that recognizes L1 and a DFA M2 for L2 with analogous notation, construct a DFA as follows: • Q = Q1 ×Q2, i.e. pairs of states from M1 and M2. • Σ is the same. • q0 = (q1 0, q2 0). 3 • F = {(q1, q2) \| q1 ∈F1 or q2 ∈F2} (i.e. all pairs of states such that at least one is a ﬁnal state). • δ((q1, q2),σ) = (δ(q1,σ),δ(q2,σ)). Then, you can check that this DFA accepts a string iff one of M1 or M2 accepts it. ⊠ Theorem 1.6 (Intersection Theorem for Regular Languages). If L1 and L2 are regular languages over the same alphabet Σ, then L1 ∩L2 is a regular language. The proof is the same as the proof for Theorem 1.5, except that the automaton has different accept states: (q1, q2) is an accept state iff q1 is an accept state of M1 and q2 is an accept state of M2. There are several other operations that we can do on regular languages: suppose A and B are regular languages. • The complement ¬A = {w ∈Σ∗\| w ̸∈A} is regular. • The reverse of a regular language AR = {w1...wk \| wk ...w1 ∈A,wi ∈Σ} is regular. • Concatenation of regular languages yields a regular language A ·B = {st \| s ∈A,t ∈B}. • The star operator preserves regularity: A∗= {s1...sk \| k ≥0 and each si ∈A}. Notice that the empty string ε is in A∗. Most of these aren’t too hard to prove, but the reverse property isn’t at all obvious. Intuitively, given a DFA M that recognizes L, we want to construct a DFA MR that recognizes LR. If M accepts a string w, then w describes a directed path in M from the start state to an accept state. Intuitively, one would want to reverse all of the arrows and switch the start and accept states, but having multiple start states is a bit of a problem, and there might be ambiguity in the reversed transition “function.” Well, we can certainly deﬁne it, and then say that this machine accepts a string if there is some path that reaches some accept state from some start state. This looks like a nondeterminstic ﬁnite automaton (NFA), which will be useful, but the fact that this implies regularity is not obvious either. Here is an NFA: / 1 ε / 1 0 / 0,1 / 0 < This NFA accepts the language {0i1j \| i, j ≥0}. Observe that undeﬁned transitions are OK (though they lead to rejection), and ε-transitions (where it just doesn’t read anything, or reads the empty string) are also OK. Deﬁnition 1.7. A non-determinsitic ﬁnite automaton (NFA) is a 5-tuple N = (Q,Σ,δ,Q0,F), where • Q is the ﬁnite set of states; • Σ is a ﬁnite alphabet; • δ : Q ×Σε →2Q is the transition function, which this time is on Σε = Σ⊔ε (where ε is the empty string) and sends this to a set of possible states (2Q denotes the set of subsets of Q); • Q0 ⊆Q is the set of start states; and • F ⊆Q is the set of accept states. 4 Then, N accepts a w ∈Σ∗if it can be written as w1...wn where wi ∈Σε (i.e. one might add some empty strings) and there exist r0,...,rn ∈Q such that: • r0 ∈Q0, • ri+1 ∈δ(ri,wi+1) for i = 0,...,n−1, and • rn ∈F. Then, L(N) is the language recognized by N, the set of strings accepted by it (as before). As an example of this formalism, in the automaton N below, / q2 1 / q4 / q1 0 / q3 ε _ 0 O N = (Q,Σ,δ,Q0,F), where Q = {q1, q2, q3, q4}, Σ = {0,1}, Q0 = {q1, q2}, F = {q4}, and δ(q3,1) = ; and δ(q3,0) = {q4}. The deterministic model of computation is simple: follow the algorithm, then accept or reject. The nondeterministic method tries many possible paths, and accepts if any one of them does. Sometimes, this makes no difference, but in others, the result is provably different, but in both cases it will be useful. 2. NONDETERMINISM, FINITE AUTOMATA, AND REGULAR EXPRESSIONS: 1/9/14 We were in the middle of determining whether the reverse of a regular language is regular. On speciﬁc languages, one doesn’t have too much difﬁculty; for example, one can construct DFAs for the language consisting of strings beginning with a 1 and for the language consisting of strings ending with a 1. In general, though, can every right-to-left DFA be replaced with a normal DFA? To understand this, we introduced the notion of an NFA. One wrinkle is that the deﬁnition provided allows multiple start states, but the textbook only allows one. This is easily shown to be equivalent: one can add a new start state and add ε-transitions to each former start state, to get an equivalent NFA with only one start state. In general, NFAs are simpler than DFA. Here is a minimal DFA that recognizes the language {1}: / 0 1 / 0,1 0,1 d However, the minimal NFA (while it is in general harder to ﬁnd) is simpler: / 1 / However, it turns out that this makes no difference! Theorem 2.1. For every NFA N, there exists a DFA M such that L(M) = L(N). 5 Corollary 2.2. A language L is regular iff there exists an NFA N such that L = L(N). Corollary 2.3. A language L is regular iff its reverse LR is regular. Proof of Theorem 2.1. There’s no need to go into the incredibly gory details, but the meat of the proof is given here. The idea is that to determine if an NFA accepts, one keeps track of a set of states that could exist at the given time. Thus, one could use a DFA to simulate an NFA, where each state of the DFA is a collection of states of the NFA representing a slice of computation in parallel. Thus, given some NFA N = (Q,Σ,δ,Q0,F), output a DFA (M = (Q′,Σ,δ′, q′ 0,F′), where Q′ = 2Q (i.e. the set of subsets of Q), and δ′ : Q′ ×Σ →Q′ is given by δ′(R,σ) = [ r∈R ε(δ(r,σ)), where this notation denotes the ε-closure of a set; for a set S ⊆Q, the ε-closure of S, denoted ε(S), is the set of states q ∈Q reachable from some s ∈S by taking 0 or more ε-transitions. This allows the DFA to deterministically handle ε-transitions. Then, the start state is q′ 0 = ε(Q0), to account for ε-transitions from the start state(s), and F′ = {R ∈Q′ \| f ∈R for some f ∈F}. These are the sets of states (of M) that contain any accept state, since acceptance on a NFA was deﬁned to include any possible path accepting. ⊠ As an illustration, here is how this algorithm looks on the following NFA: / 1 b ε 2 a , a,b / 3 a R Then, ε({1}) = {1,3}. Then, the resulting DFA is / {1,3} a b {3} a o b / ; a,b {2} b ; a / {2,3} a b O {1,2,3} a N b O Notice that the empty set is present, and that {1} and {1,2} are unreachable from the start state (so they could be placed on the graph, but they’re pretty useless). A useful rule is that NFAs can make proofs much easier. For example, for the union theorem, one could just glom the two DFAs together to get an NFA with two start states (that is disconnected, but that’s OK). As another example, to show that the concatenation of two regular languages is regular, one can connect the accept states of an NFA M1 that recognizes the ﬁrst language to the start states of an NFA that recognizes the second language (and then make them both regular states, and so on). 6 Regular languages are also closed under the star operator, which treats a language as an alphabet, in some sense: the strings made up of compositions of substrings of a language A. If L is regular, then let M be a DFA with L(M) = L. Then, one can just add ε-transitions from every accept state to every start state, creating an NFA N that recognizes L∗. Addition-ally, since ε ∈L∗, it’s necessary to make a single start state that always accepts, before seeing anything else (add ε-transitions to the previously existing start states). Formally, given some DFA M = (Q,Σ,δ, q1,F), produce an NFA N = (Q′,Σ,δ′,{q0},F′), where Q′ = Q ∪{q0} and F′ = F ∪{q0}. Then, δ′(q,a) = ½ {δ(q,a)}, if q ∈Q and a ̸= ε {q1}, if q ∈F and a = ε. Now, we can show that L(N) = L∗. • L(N) ⊇L∗: suppose that w = w1...wk ∈L∗, where wi ∈L for each i. Then, by induction on k, we can show that N accepts w. The base cases are k = 0, where w = ε, which was handled by the start state, and k = 1, where w ∈L, which is also pretty clear. Inductive step: suppose N accepts all strings v = v1...vk ∈L∗where vi ∈L. Then let u = u1...ukuk+1 ∈L∗, so that u j ∈L. Since N accepts u1...uk by induction and M accepts uk+1, then one can quickly check that N also accepts all of u. • L(N) ⊆L∗. Assume w is accepted by N, so that we want to show that w ∈L∗. If w = ε, then clearly w ∈L∗, and now proceed by induction on the length of w; suppose that for all u of length at most k, N accepts u implies u ∈L∗. Let w be accepts by N and have length k +1. Then, write w = uv, where v is the substring read after the last ε-transition. Then, u ∈L(N), since the last state before an ε-transition is an accept state. Then, u is a strictly smaller state than w, so u ∈L∗ by the inductive hypothesis, and since the ﬁrst thing after an ε-transition is a start state, then v ∈L(M). Thus, w = uv ∈L∗, just by the deﬁnition of L∗. Now, rather than viewing computation as a machine, regular expressions allow one to view computation as a set of rules. Deﬁnition 2.4. Let Σ be an alphabet. Then, regular expressions over Σ can be deﬁned inductively: • For all σ ∈Σ, σ is a regular expression, as well as ε and ;. • If R1 and R2 are regular expressions, then R1R2 and R1 +R2 are both regular expres-sions, as is R∗ 1. A regular expression is associated to a language as follows: σ represents the language {σ} and ε represents {ε}. Then, ; represents ;. Then, if R1 recognizes L1 and R2 recognizes L2, then R1R2 (sometimes written R1 · R2) represents the concatenation L1 · L2, R1 + R2 represents L1 ∪L2, and R∗ 1 represents L∗. For example, if Σ = {0,1}, then 0∗10∗represents the language consisting of strings consist-ing of a single 1, and ;∗recognizes the language {ε}. The language of strings with length at least 3 and whose third symbol is a zero is given by the expression (0+1)(0+1)0(0+1)∗. The language {w \| every odd position in w is a 1} is represented by (1(0+1))∗(1+ε). Notice that some tricks need to be done to include ε and such, and in general there are edge cases. 7 More interestingly, let D = {w \| w has an equal number of occurrences of 01 and 10}.2 This merits the impressive regex 1+0+ε+0(0+1)∗0+1(0+1)∗1. This is because, excluding edge cases, a w ∈D is such that either w starts with a 0 and ends with a 0, or w starts with a 1 and ends with a 1. This is because sending 00 7→0 and 11 7→1 on a string preserves the property that the number of 01 substrings is equal to the number of 10 substrings, and this reduces it to a string of the form 1010...01 or similar. Then, the three edge cases are those too short to contain a 01 or a 10, so they are 0, 1, and ε. Proposition 2.5. L can be represented by a regular expression iff L is regular. Proof. In the forward direction, given a regular expression, one will obtain an NFA N such that R represents L(N). Proceed by induction on the number of symbols in R. The base cases are R = σ, given by / σ / , R = ε, given by / , and R = ; given by an NFA which accepts nothing. In the general case, where R has length k > 1, suppose that a regular expression of length ≤k represents a regular language. Then, it must be possible to decompose R in one of the following ways: R = R! + R2, R = R1R2, or R = R∗ 1. But the union, concatenation, and star closure of regular languages must be regular, and R1 and R2 must have length at most k, so they correspond to regular languages, and thus R does as well. The other direction will be given in the next lecture. 3. FINITE AUTOMATA VS. REGULAR EXPRESSIONS, NON-REGULAR LANGUAGES: 1/14/14 “My name is Hans —” “And I’m Franz —” “. . . and we’re here to pump you up!” Looking at problems by how complex their descriptions are is a prevailing theme in computer science. Regular expressions are among the simpler ones. One piece of terminology is that a string w ∈Σ∗is accepted by a regular expression R, or matches R if it is in L(R). Last time, it was proven that if L is given by a regular expression, then L is regular. An example of the given algorithm gives the following NFA for the regular expression (1(0+1))∗: / ε / 1 / 0,1 / ε y For the proof in the reverse direction, it’s necessary to introduce a more general class of automata. Deﬁnition 3.1. A generalized nondeterministic ﬁnite automaton (GNFA) is a 5-tuple (G = Q,Σ,R, qstart, qaccept), which is an NFA whose arcs are labeled by regular expressions: • Q is the ﬁnite set of states, Σ is the ﬁnite alphabet as above, and qstart is the single start state and qaccept the single accept state, as per usual or close to it, and • R is a transition function Q ×Q →R, where R is the set of regular expressions over Σ; that is, each arc is labeled by a regular expression. 2These substrings are allowed to overlap. 8 G accepts a string w if there is a sequence q0 = qstart, q1,..., qn−1, qn = qaccept and strings x1,...,xn such that xi matches R(qi, qi+1) for all i and w = x1...xn. Then, L(G) is the set of strings G accepts, as normal. For example, here is a GNFA over {a,b, c}: / q0 a∗b / q1 cb ε / q2 This GNFA recognizes the language matched by the regular expression a∗b(cb)∗. Now, given some NFA, one can construct a sequence of GNFAs that lead to a regular expression. This algorithm can be called CONVERT(G) for a given GNFA G. (1) First, add start and accept states and ε-transitions to (respectively from) the old start and accept states. (2) Then, while the machine has more than two states: • Pick some state q, and remove it. Then, replace all paths through it q1 →q →q2 with arrows q1 →q2 with regular expressions corresponding to the removed arrows. Speciﬁcally, (3) Finally, the GNFA is just two states, with the transition given by a regular expression R(q1, q2). Return this regular expression. The ripping-out process looks like this, though it can become complicated. q1 R(q1,q2) / R(q1,q3) & q2 R(q2,q2) P R(q2,q3) / q3 − → q1 R(q1,q2)R(q2,q2)∗R(q2,q3)+R(q1,q3)/ q3 Using this convention, the algorithm deﬁnes the transition function when a state qrip has been removed as R′(qi, q j) = R(qi, qrip)R(qrip, qrip)∗R(qrip, q j)+ R(qi, q j). Then, if G′ is the GNFA returned after ripping out qrip and updating the transition function in the way deﬁned above, then L(G′) = L(G). This is because either the sequence of states didn’t include qrip, in which case of course it still works, and if it did, then it has to match R(qi, qrip)R(qrip, qrip)∗R(qrip, q j) there, which allows it to work. Theorem 3.2. More interestingly, L(R) = L(G), where R = CONVERT(G). Proof. Proceed by induction on k, the number of states in G. In the case k = 2, this is pretty clear. In the inductive step, suppose G has k states, and G′ be the ﬁrst GNFA created by ripping something out. Then, as mentioned above, L(G) = L(G′), and by induction L(G′) = L(R), so L(G) = L(R) too. ⊠ As an example, consider the following automaton: / q1 a / b q2 b p a o q3 b O a ? 9 After adding start and accept states, this looks like the following: / ε / q1 a / b q2 b p a o ε q3 b O a ? ε / Then ripping out q2 and q3, this becomes / ε / q1 bb+(a+ba)b∗a b+(a+ba)b∗/ The resulting regular expression is rather ugly, but it does the job: (bb+(a+ba)b∗a)∗b+(a+ ba)b∗. In summary, there are several rather different ways of viewing the same class of languages or computation: through DFAs, NFAs, regular languages, and regular expressions. However, not all languages are regular. Intuitively, a DFA has a ﬁnite amount of memory, given by the set of states, but languages can be inﬁnite. Alternatively, the set of DFAs is countable, but the set of all possible languages is not countable, so almost all languages must be non-regular. However, things get nuanced: let C = {w \| w has an equal number of 1s and 0s}, and D = {w \| w has an equal number of substrings 01 and 10}. We saw last time in class that D is regular, given by an icky but still regular expression, and yet C is not! Intuitively, one needs to know the difference between the number of 0s seen and the number of 1s seen. This seems to require an inﬁnite number of states, but of course this isn’t rigorous. A better technique is given by the pumping lemma. Lemma 3.3 (Pumping). Let L be an inﬁnite regular language. Then, there exists a p ∈N (i.e. p > 0) such that for all strings w ∈L with \|w\| > p, there is a way to write w = xyz, where: (1) \|y\| > 0 (that is, y ̸= ε), (2) \|xy\| ≤p, and (3) For all i ≥0, xyiz ∈L. This is called the pumping lemma because given this decomposition, one can “pump” more copies of y into the string. Proof of Lemma 3.3. Let M be a DFA that recognizes L, and P be the number of states in M. Then, if w ∈L is such that \|w\| ≥p, then there must be a path q0,..., q\|w\| for w, so by the pigeonhole principle, there exist i, j such that qi = q j, because there are p +1 states and p options. Without loss of generality, assume qi happens before q j. Thus, let x be the part of w before qi, y be that between qi and q j, and z be that after q j. Then, y is nonempty, because there’s at least one transition between qi and q j; the pigeonhole principle guarantees j ≤p, so \|xy\| ≤p, and since y comes from a loop in the state graph, xyiz traces over that loop i times, and is still a viable way through the NFA, and is thus accepted. ⊠ Now, one can show that an (inﬁnite) language is not regular by using the contrapositive: that for every p ∈N, there exists a w ∈L such that \|w\| > p and the conditions don’t hold. 10 For example, one can show that B = {0n1n \| n ≥0} isn’t a regular language. Suppose B is regular, and since B is inﬁnite, then let P be the pumping length for B. Then, let w = 0P1P. If B is regular, then there is a way to write w = xyz, \|y\| > 0, and so on. But since \|xy\| < P and w = 0P1P, then y is all zeros. But then, xyyz has more 0s than 1s, which means xyyz ̸∈L, violating the conditions of the pumping lemma. Analogously, if C is as above the language with an equal number of 0s and 1s, then one can do something similar to show C isn’t regular: suppose C is regular, and let P be its pumping length (since C is inﬁnite). Let w = 0P1P, and do the same thing as above; xyyz has more 0s than 1s, and this forces the same contradiction. In essence, if one starts with a string within the language, and ends up with a string outside of the language, no matter the partition into substrings, then the language cannot be regular. For another example, let B = {0n2 \| n ≥0}, and suppose B is regular. If the pumping length is P, then let w = 0p2. Then, w = xyz with 0 < \|y\| ≤p. Then, xyyz = 0p2+\|y\|, but p2 +\|y\| isn’t a square, because p2 < p2 +\|y\| ≤p2 + p < p2 +2p +1 = (p +1)2. 4. MINIMIZING DFAS: 1/16/14 Today’s question is whether a given DFA has a minimal number of states. For example, look at this one: (4.1) / 0 / 1 0 1 1 O 0 / 0 O 1 _ It seems like there is a DFA that accepts the same language with fewer states. Theorem 4.2. For every regular language L, there exists a unique3 minimal-state DFA M∗ such that L = L(M∗), and moreover, there an efﬁcient algorithm which, given a DFA M, will output M∗. This result is really nice, and its analogues generally don’t hold for other models of computation. For one quick example, it’s easy to construct distinct minimal-state NFAs for a given regular language, such as / 0 / 0 and / 0 0 / Moreover, the question of an efﬁcient algorithm is open. For the proof, it will be useful to extend the transition function δ to ∆: Q ×Σ∗→Q, in which: • ∆(q,ε) = q, • ∆(q,σ) = δ(q,σ) for σ ∈Σ, and • ∆(q,σ1...σk+1) = δ(∆(q,σ1...σk),σk+1). In other words, where does M end up after starting at q and reading σ1...σk+1? 3This is understood to be up to isomorphism, i.e. relabelling the states. 11 Deﬁnition 4.3. A w ∈Σ∗distinguishes states q1 and q2 iff exactly one of ∆(q1,w) and ∆(q2,w) is a ﬁnal state. Two states p and q are distinguishable if there exists a w ∈Σ∗ such that w distinguishes p and q, and are considered indistinguishable if for all w ∈Σ∗, ∆(p,w) ∈F iff ∆(q,w) ∈F. In other words, one can run M on input w, and the result of the string can indicate whether it started at q1 or q2. Intuitively, indistinguishable states are redundant. Look at the following automaton: / q0 0 1 / q1 0,1 q3 0 O 1 / q2 0 o 1 p Then, ε distinguishes q0 and q1 (and any accept state from a non-accept state), 10 distin-guishes q0 and q3, and 0 distinguishes q1 and q2. Now, for some ﬁxed DFA M = (Q,Σ,δ, q0,F) and p, q,r ∈Q, then write p ∼q if p and q are indistinguishable, and p ̸∼q otherwise. This is an equivalence relation, i.e.: • p ∼p, i.e. ∼is reﬂexive. This is because ∆(p,w) ∈F iff ∆(p,w) ∈F, which is silly but true. • p ∼q implies q ∼p, i.e. it’s symmetric. This is because indistinguishability is an if-and-only-if criterion. • ∼is transitive: if p ∼q and q ∼r, then p ∼r. This is because for all w ∈Σ∗, ∆(p,w) ∈F iff ∆(q,w) ∈F iff ∆(r,w) ∈F, so ∆(p,w) ∈F iff ∆(r,w) ∈F. Thus, ∼partitions the states of Q into disjoint sets called equivalence classes. Write [q] = {p \| p ∼q} for some state q. Looking back at the automaton (4.1), there are two equivalence classes: those that accept and those that reject. It will be possible to construct a minimal DFA by contracting the DFA into its equivalence classes. More formally, this will be done according to the algorithm MINIMIZE-DFA, which inputs a DFA M and outputs a minimal DFA MMIN such that L(M) = L(MMIN), MMIN has no inaccessible states, and MMIN is irreducible (which will be shown to be equivalent to all states of MMIN being distinguishable). Furthermore, there is an efﬁcient dynamic-programming algorithm that can discover this MMIN which outputs a set DM = {(p, q) \| p, q ∈Q and p ̸∼q} of distinguishable states and EQUIVM = {[q] \| q ∈Q}. This algorithm can be considered as table-ﬁlling algorithm, where one can consider only the entries below the diagonal (since ∼is an equivalence relation). Then: • Base case: for all (p, q) such that p accepts and q rejects, we know p ̸∼q, • Then, iterate: if there are states p, q and a symbol σ ∈Σ such that δ(p,σ) = p′ and δ(q,σ) = q′ and p′ ̸∼q′, then p ̸∼q. Repeat the iterative step until no more can be added. This must terminate, because there are a ﬁnite number of states and a ﬁnite number of symbols, so either the rule stops applying or everything is marked as inequivalent, which takes polynomial time. Claim 4.4. (p, q) are marked as distinguishable by the table-ﬁlling algorithm iff indeed p ̸∼q. 12 Proof. Proceed by induction on the number of iterations in the algorithm: if (p, q) are marked as distiguishable at the start, then one is in F and the other isn’t, so ε distinguishes them. More generally, suppose that (p, q) are marked as distinguishable at some point. Then, by induction, there are states (p′, q′) that were marked as distinguishable and therefore are actually distinguishable by some string w. Without loss of generality, suppose ∆(p,w) ∈F and ∆(q,w) ̸∈F (if not, then just switch them), and since p and q were marked, then there exists a σ ∈Σ such that p′ = δ(p,σ) and q′ = δ(q,σ). Then, σw distinguishes p and q, so p ̸∼q. Conversely, suppose that (p, q) isn’t marked by the algorithm; then, we want to show that p ∼q. Well, argue by contradiction, and suppose that p ̸∼q. Call such a pair a “bad pair.” Then, there is a string w with \|w\| > 0 such that ∆(p,w) ∈F and ∆(q,w) ̸∈F (or maybe switching p and q; it’s the same thing). Of all such bad pairs, let (p, q) be the pait with the shortest-length distinguishing string w. Then, w = σw′ where σ ∈Σ, so let p′ = δ(p,σ) and q′ = δ(q,σ). Then, (p′, q′) is a bad pair: • Clearly, (p′, q′) can’t have been marked by the algorithm, because then its next step would catch (p, q), which is a bad pair and thus isn’t caught. • Also, since \|w\| > 0, then w′ is a string that can be used to distinguish p′ and q′: ∆(p,w) = ∆(q′,w′) and ∆(q,w) = ∆(q′,w′), so they are distinguished because p and q are. Thus, (p′, q′) are distiguished by w′, but ¯ ¯w′¯ ¯ < \|w\|, and w was supposed to be minimal, so the necessary contradiction is found. ⊠ This claim justiﬁes using the notation “distinguishable” for the states marked by the table-ﬁlling algorithm. Now, we can actually get down to minimizing DFAs. The algorithm is as follows: (1) First, remove all inaccessible states of M. (2) Apply the table-ﬁlling algorithm to get the set EQUIVM of equivalence classes for M, (3) Deﬁne a DFA MMIN = (QMIN,Σ,δMIN, q0,MIN,FMIN) given by: • QMIN = EQUIVM, • q0,MIN = [q0], • FMIN = {[q] \| q ∈F}, and • δMIN([q],σ) = [δ(q,σ)]. The start and accept states are just the equivalence classes of start and accept states. However, since we’re working on equivalence classes, it’s important to check whether δMIN is even well-deﬁned. However, because all states in [q] are indistinguishable, then the results after applying σ must also be indistinguishable (otherwise you could distinguish the stuff in [q]). Claim 4.5. Then, L(MMIN) = L(M). 13 For example, suppose we want to minimize the following automaton. All states are accessible, so the ﬁrst step of pruning the zero accessible states is uninteresting. / q0 0 / 1 q1 0 p 1 q2 0 O 1 P Then, once one writes down the equivalence classes, the result is / 0 1 / 1 0 o because the states q0 and q1 from the ﬁrst automaton are equivalent. For a more convoluted example, consider / q0 1 0 / q1 1 o 0 q4 0 1 / q5 0,1 q2 0,1 ? q3 0 O 1 ? q2 isn’t reachable, so it can be removed ﬁrst. Now, q3 and q4 are indistinguishable, but the rest can all be seen to be distinguishable. Then, one can replace them with a common node. Now, we know that MMIN recognizes the same language, so why is it minimal? Suppose that L(M′) = L(MMIN) and M′ is irreducible. Then, the proof will show that there is an isomorphism between M′ and MMIN. Corollary 4.6. M is minimal iff it has no inaccessible states and is irreducible. Proof. Let Mmin be minimal for M. Then, L(M) = L(Mmin), has no inaccessible states, and is irreducible, and by the above claim (which will be proven next lecture), they are isomorphic. ⊠ 5. THE MYHILL-NERODE THEOREM AND STREAMING ALGORITHMS: 1/21/14 The proof of Theorem 4.2 was mostly given last lecture, but relies on the following claim. This implies the uniqueness of the output of the algorithm as the minimal DFA for a regular language. Claim 5.1. Suppose L(M′) = L(MMIN) and M′ is irreducible and has no inaccessible states. Then, there is an isomorphism (i.e. a relabeling of the states) between M′ and MMIN. Proof. The proof recursively constructs a map from the states of MMIN to those of M′. Start by sending q0,MIN 7→q′ 0. Then, repeatedly suppose that p 7→p′ and p σ →q and p′ σ →q′. Then, send q 7→q′. 14 Then, one needs to show that this map is deﬁned everywhere, that it is well-deﬁned, and that it is a bijection. Then, from the recursive construction it’s pretty clear that it preserves transitions (i.e. commutes with the transition functions δ). First, why is it deﬁned everywhere? For all states q of MMIN, there is a string w such that ∆MIN(q0,MIN,w) = q, because MMIN is minimal; any inaccessible states were removed. Then, if q′ = ∆(q′ 0,w), then q 7→q′. This can be formally shown by induction on \|w\|, and it’s particularly easier because of the recursive setup of the algorithm. Next, it’s well-deﬁned. Suppose there are states q′ and q′′ such that q 7→q′ and q 7→q′′. Then, we can show that q′ and q′′ are indistinguishable (and therefore that they must be equal, since M′ is irreducible). Suppose u and v are strings such that u sends q′ 0 →q′ and v sends it to q′′. Then, u,v : q0,MIN →q. But by induction on the lengths of u and v, one can show that they cannot be distinguished by q, but then, q′ and q′′ cannot distinguish u and v either, because L(M) = L(MMIN), but this is a problem, because it implies that q is distinguishable from itself. Now, why is the map onto? The goal is to show that for every state q′ of M′, there is a state q of MMIN such that q 7→q′. Well, since M′ also has no inaccessible states, then for every q′ there is a w such that M′ run on w halts at q′; then, let q be the state MMIN halts on after running on w; then, q 7→q′ (which is formally shown by induction on \|w\|). The map is one-to-one because if p, q 7→q′, then p and q are distinguishable if they aren’t equal, but their distinguishing string would also distinguish q′ and q′, which is a slight problem. ⊠ The Myhill-Nerode theorem is an interesting characterization of regular languages. One can deﬁne an equivalence relation on strings given some language: if L ⊆σ∗and x, y ∈Σ∗, then x and y are indistinguishable to L if for all z ∈Σ∗, xz ∈L iff yz ∈L. One writes this as x ≡L y. Claim 5.2. ≡L is in fact an equivalence relation for any language L. Proof. These follow from properties of “iff:” X iff X; X iff Y is equivalent to Y iff X; and if X iff Y and Y iff Z, then X iff Z. Then, plug into the deﬁnition of ≡L. ⊠ Theorem 5.3 (Myhill-Nerode). A language L is regular iff the number of equivalence classes of Σ∗under ≡L is ﬁnite. This powerful statement gives a complete characterization of regular languages. But it also has nothing to do with the machine construction of regular languages that was their original deﬁnition, which is interesting. Proof of Theorem 5.3. Suppose L is regular, and let M = (Q,Σ,δ, q0,F) be a minimal DFA for L. Deﬁne an equivalence relation x ∼M y for ∆(q0,x) = ∆(q0, y); that is, M reaches the same state when it reads in x and y. Claim 5.4. ∼M is an equivalence relation with \|Q\| equivalence classes. Proof. The number of classes is pretty clear: it’s bounded above by the number of states in M, but since M is minimal, then every state is accessible, so there must be \|Q\| states. As for why it’s an equivalence relation, this is basically the same thing with the iff as for ≡L. ⊠ Next, if x ∼M y, then x ≡L y, which is basically just checking the deﬁnitions: x ∼M y implies that for all z ∈Σ∗, xz and yz reach the same state of M, so xz ∈L iff yz ∈L, since L = L(M), 15 and this is exactly the deﬁnition of x ≡L y. Thus, there are at most \|Q\| equivalence classes of ≡L, and in particular this number must be ﬁnite. The speciﬁc claim that x ∼M y = ⇒X ≡L y shows that there are at most as many classes in ≡L as in ∼M is because if one takes one xi from each equivalence class of ≡L, then xi ̸≡L xj if i ̸= j, and therefore xi ̸∼M xj, so we have at least as many equivalence classes of ∼M. In the reverse direction, we can build a DFA from the set of equivalence classes of ≡L. Speciﬁcally, suppose ≡L has k equivalence classes. Then, let M = (Q,Σ,δ, q0,F), where: • Q is the set of equivalence classes of ≡L. • q0 = [ε] = {y \| y ≡L ε}. • δ([x],σ) = [xσ]. • F = {[x] \| x ∈L}. Then, it happens (though one has to check this) that M accepts x iff x ∈L. Part of this involves demonstrating that δ is well-deﬁned. ⊠ Now, with this theorem, one has another way of proving a language is or isn’t regular. One can show that a language isn’t regular by exhibiting inﬁnitely many strings w1,w2,... that are distinguishable to L, i.e. for each pair wi and wj there is a zi j ∈Σ∗such that exactly one of wizi j and wjzi j is in L. This set of wi is called a distinguishing set of L. Thus, every language L has either a DFA or a distinguishing set. Theorem 5.5. L = {0n1n \| n ≥0} is not regular. Proof. This was already proven using the Pumping Lemma, but consider the set S = {0n \| n ≥ 1}. This is a distinguishing set for L, because for any 0m,0n ∈S with m ̸= n, let z = 1n. Then, 0m1m ∈L, but 0m1n ̸∈L. Thus, every pair of strings in S are distinguishable to L, but S is inﬁnite, so L cannot be regular. ⊠ That was pretty slick. This theorem can often be easier for proving a language isn’t regular than the pumping lemma. This relates to another question: how can one check if two regular expressions are equiv-alent? One has algorithms for converting regular expressions into DFAs, then DFAs into minimal DFAs, and then constructing an isomorphism; thus, one can combine these algo-rithms. However, minimizing a regular expression is considerably more complex, unless P = NP. Streaming Algorithms. The intuition behind a streaming algorithm is that one must be able to respond to a data stream in real time. Thus, such an algorithm reads one character of input at a time, but cannot access past data, and has some small amount of memory and time to make computations. For example, a streaming algorithm to recognize L = {x \| x has more 1s than 0s} is as follows: • Initialize C := 0 and B := 0, and let c be the bit we’re currently looking at. • Then, if C = 0, then set B := x and C := 1. • If C ̸= 0 and B = x, then increment C. • If C ̸= 0 and B ̸= x, then decrement C. • When the stream stops, accept iff B = 1 and C > 0. The idea is that B is the current majority value and C is how much more common it is. This requires O(logn) memory, which is pretty nice. Streaming algorithms differ from DFAs in several ways: 16 (1) Streaming algorithms can output more than just a single bit (i.e. can do more than just accept or reject). (2) The memory or space used by a streaming algorithm can increase with the length of the string, e.g. O(logn) or O(loglogn). (3) Streaming algorithms can make multiple passes over the data, or be randomized. Theorem 5.6. Suppose a language L is recognized by a DFA with \|Q\| ≤2p states. Then, L is computable by a streaming algorithm that uses at most p bits of space. Proof. The algorithm can essentially directly simulate the DFA directly; since there are 2p states or fewer, then only p bits are necessary to hold the state. Then, it reads the state and the next character to determine whether or not to accept. ⊠ Deﬁnition 5.7. For any n ∈N, let Ln = L∩Σn, i.e. the set of strings of a language L of length exactly n. Theorem 5.8. Suppose L is computable by a streaming algorithm A using f (n) bits of space. Then, for all n, Ln is recognized by a DFA with at most 2f (n) states. Proof. Create a DFA M with one state for every possible conﬁguration of the memory of A. Then, M simulates A, where the state transitions are given by how A updates its memory given some input in the stream. ⊠ Notice that we don’t care about how fast streaming algorithms run, though most of the examples given, not to mention all of those used in the real world, are reasonably time-efﬁcient. The key is the strict conditions on space complexity. 6. STREAMING ALGORITHMS: 1/23/14 Streaming algorithms are important both in theory and in practice; if there is to be any theory of “big data,” it will rely heavily on them. The trick is that you can’t know when the stream stops; instead, the solution has to be updated to be correct at each point. We saw that a DFA with at most 2p states can be simulated with a streaming algorithm with p bits of memory, and in a sort of converse, if a language L can be computed by a streaming algorithm with f (n) bits of space, then Ln = L ∩Σn (which is a ﬁnite and therefore language) is recognized by a DFA with at most 2f (n)) states; this DFA is created by simulating the streaming algorithm. These two results allow the theory of ﬁnite automata to be applied to streaming algorithms. For example, if L is the language of strings over {0,1} that have more 1s than 0s, is there a streaming algorithm for L that uses strictly less than log2 n space? Not really; the algorithm given in the last lecture is optimal. Theorem 6.1. Every streaming algorithm for L needs at least (log2 n)−1 bits of space. Proof. For convenience, let n be even (the proof also works for n odd, but it’s a little ickier), and let Ln = {0,1}n ⊆L. Then, we will give a set Sn of n/2 +1 strings which are pairwise distinguishable in Ln. Then, by the Myhill-Nerode theorem, every DFA recognizing Ln requires at least n/2+1 states, and therefore that every streaming algorithm for L requires (log2 n)−1 bits of space. In greater detail, let Sn = {0n/2−i1i \| i = 0,...,n/2}, and we want to show that every pair of strings x, y ∈S are distinguishable by Ln (that is, x ̸≡Ln y). Then, x = 0n/2−k1k and y = 0n/2−j1j, where without loss of generality we suppose that k > j. 17 Then, z = 0k−11n/2−(k−1) distinguishes x and y in Ln, because xz has n/2 −1 zeroes and n/2 + 1 ones, so xz ∈Ln, and yz has n/2 + (k + j −1) zeroes and n/2 −(k −j −1) ones, but k −j −1 ≥0, so yz ̸∈Ln. Thus, all pairs of strings in Sn are distinguishable in Ln, so ≡Ln has at least \|Sn\| equiv-alence classes. Thus, every DFA recognizing Ln must have at least \|Sn\| states. Thus, by Theorem 5.6, every streaming algorithm for L requires at least logs\|Sn\| bits of space. Since \|Sn\| = n/2+1, then this completes the proof. ⊠ There’s something interesting going on here: DFAs are used to say something about a model of computation strictly more powerful than themselves, in some sense of bootstrapping. The number of distinct elements problem accepts as input some x ∈{1,...,2k}∗, where 2k ≥\|x\|2 and computes the number of distinct elements appearing in x. Then, there is a streaming algorithm for this problem in O(kn) space (where n = \|x\|), since this allows one to store all of the input. But this suggests a better solution might exist — morally, it’s not even a streaming algorithm. But it’s harder to do better! Theorem 6.2. Every streaming algorithm for the distinct elements problem requires Ω(kn) space. Proof. This problem will be solved by direct reasoning, rather than passing to a DFA, because the output is more than one bit. Deﬁne x, y ∈Σ∗to be DE-distinguishable if there exists some z ∈Σ∗such that xz and yz contain a different number of distinct elements. Lemma 6.3. Let S ⊂Σ∗be such that every pair x, y ∈S is DE-distinguishable. Then, every streaming algorithm for the distinct elements problem needs at least log2\|S\| bits of space. Proof. By the pigeonhole principle, if an algorithm A uses strictly less than log2\|S\| bits, then there are distinct x, y ∈S that lead A to the same memory state. Then, for any z, xy and yz look identical, so to speak, to A and aren’t distinguished, so x ∼y. ⊠ Lemma 6.4. There is a DE-distinguishable set S of size 2Ω(kn). Proof. For each subset T ⊆Σ of size n/2, deﬁne xT to be any concatenation of strings in T. Then, if T and T′ are distinct, then xT and xT′ are distinguishable, because xTxT has n/2 distinct elements, and xTxT′ has strictly more than n/2 elements, because they must differ somewhere. Now, if 2k > n2, we want to show that there are 2Ω(kn) such subsets, since there are ¡\|Σ\| n/2 ¢ choices, but since \|Σ\| = 2k, then using the fact that ¡m c ¢ ≥(m/c)c, which is true because Ã m c ! = m! c!(m−c)! = m·(m−1)···(m−c +1) c(c −1)...(2)(1) ≥mc cc , because when you subtract the same small amount from the numerator and the denominator, the fraction actually gets larger. Thus, ¡\|Σ\| n/2 ¢ ≥ ¡ 2kn/2 (n/2)n/2 ¢ = 2kn/2−n/2log2(n/2) = 2Ω(kn) because 2k > n2. ⊠ Now, throw these two lemmas together: since there is set of pairwise distinguishable elements of size 2Ω(kn) elements, then every streaming algorithm for this problem needs at least Ω(kn) memory. ⊠ 18 Randomness actually makes this better; if the algorithm is allowed to make random choices during its process, there’s a much better algorithm that can approximate the solution to about 0.1% error using some pretty clever hashing and O(k +logn) space. What if the alphabet has more than two characters? A generalization of this algorithm is to ﬁnd the top several items in some larger alphabet. This is a very common algorithm used in the real world. Formally, given a k and a string x = x1...xn ∈Σn, output the set S = {σ ∈Σ \| σ occurs more than n/k times in x}. Thus, by counting, \|S\| ≤k. Theorem 6.5. There is a two-pass streaming algorithm for this problem using O(k(log\|Σ\|+ logn)) space. Proof. On the ﬁrst pass, initialize a set T ⊆Σ×N, which is initially empty. This will associate each symbol of the alphabet with a count of how many times it appeared, so that when reading (σ,m), remove (σ,m) from T and add (σ,m+1) to it. Alternatively, if \|T\| < k −1 and (σ,m) ̸∈T for any m, then there’s nothing to remove, so add (σ,1) to the set. Otherwise, there is a symbol we haven’t seen before, so decrement each counter (adding (σ′,m′ −1) and removing each (σ′,m), and removing any symbols with count zero). Then, T contains all σ occurring more than n/k times in x. Then, in the second pass, count all occurrences of all σ′ appearing in T to determine which ones occur at least n/k times. The key realization is that for all (σ,m) ∈T, the number of times that σ appears in the stream is at least m, and that for every σ, its corresponding counter is decremented at most n/k times (because each time, some number of symbols are removed). Thus, all counts are within n/k of the true count, and in particular if σ appears more than n/k elements, then it exists in T at the end of the ﬁrst pass. ⊠ This algorithm is simple, yet counterintuitive. Returning to the language of strings with more 1s than 0s, we say a solution that uses (log2 n)+1 bits of space, but a lower bound is (log2 n)−1 space. How much space is needed exactly? Does it depend on whether n is odd or even? Part 2. Computability Theory: Very Powerful Models 7. TURING MACHINES: 1/28/14 “When faced with a warehouse with miles of rewritable tape, Alan Turing did not despair. In fact, he invented computer science.” Now the course turns to a similarly simple model as a DFA, but this time it’s very powerful. A Turing machine has an inﬁnite, rewritable tape and is considered to be “looking at” some particular symbol on the tape. It has some state q0, reads (and possibly writes) to the tape, and then moves to the left or write. If it tries to move left from the left side of the tape, it is understood to remain at the leftmost part of the tape. Here’s a state diagram for a Turing machine: / 0→0,R / □→□,R ! 0→0,R □→□,R / qaccept qreject Here, □represents a blank symbol on the tape. 19 This Turing machine decides the language Σ = {0}. This is distinct from the notion of a Turing machine that recognizes Σ, i.e. one that halts and accepts 0, and inﬁnite loops on other things. Here’s an example of a Turing machine that decides the language L = {w#w \| w ∈{0,1}∗}. Notice that this language isn’t regular, which one can show by Myhill-Nerode; all of the strings 0i for i > 0 are pairwise distinguishable: 0i and 0j by #0i. It is common to give a description of a Turing machine in pseudocode: (1) If there is no # on the tape, or more than one #, reject. (2) While there is a bit to the left of #, • Replace the ﬁrst bit with X, and check if the ﬁrst bit b to the right of the # is identical. If not, reject. • Replace that bit b with an X as well. (3) If there is a bit to the right of #, reject; if else, accept. It’s also conventional to move everything over one index on the tape to the right, so that it’s clear to the Turing machine when it has reached the leftmost edge of the input. Thus, once it sees a blank, it knows its at the end of the input, or the beginning. Of course, it’s necessary to have a formal deﬁnition for a Turing machine. Deﬁnition 7.1. A Turing machine is a 7-tuple T = (Q,Σ,Γ,δ, q0, qaccept, qreject), where: • Q is a ﬁnite set of states, • Σ is the input alphabet, where □̸∈Σ is the blank state. • Γ is the tape alphabet, often different from the input alphabet, where □∈Γ and Σ ⊆Γ. • δ : Q ×Γ →Q ×Γ×{L,R}. • q0 ∈Q is the start state, • qaccept is the accept state, and • qreject is the reject state. This means that one could store the conﬁguration of a given Turing machine as a ﬁnite string in (Q ∪Γ)∗; for example, if the non-blank tape looks like 1101000110 and the Turing machine is on the sixth bit of tape in state q7, then one would write 11010q700110 ∈{Q ∪Γ}∗. Because the input is ﬁnite, then at any ﬁnite point in time, there are only ﬁnitely many non-blank symbols. Deﬁnition 7.2. Let C1 and C2 be conﬁgurations of a Turing machine M. Then, C1 yields C2 if M is in conﬁguration C2 after running M in conﬁguration C1 for one step. For example, if δ(q1,b) = (q2, c,L), then aaq1bb yields aq2acb. If δ(q1,a) = (q2, c,R), then cabq1a yields cabcq2□(the Turing machine reads past the end of the input). Deﬁnition 7.3. Let w ∈Σ∗and M be a Turing machine. Then, M accepts w if there are conﬁgurations C0,C1,...,Ck such that • C0 = q0w, • Ci yields Ci+1 for i = 0,...,k −1, and • Ck contains the accept state qaccept. One can more speciﬁcally say that M accepts w in k steps. Deﬁnition 7.4. 20 • A Turing machine recognizes a language L if it accepts exactly those strings in L. A language is called recursively enumerable (r.e.) or recognizable if there exists a Turing machine that recognizes L. • A Turing machine decides a language L if it accepts those strings in L and rejects those strings not in L. A language is called decidable or recursive if there exists a Turing machine that decides L. Recall that the complement of a language L ⊆Σ∗is ¬L = Σ∗\ L. Theorem 7.5. A language L ⊆Σ∗is decidable iff both L and ¬L are recognizable. Proof. The forward direction is fairly clear, so look at the reverse direction. Given a Turing machine M1 that recognizes L and a Turing machine M2 that recognizes ¬L, how ought one to build a new machine M that decides L? This can be done by simulating M1(x) on one tape, and M2(x) on another. Exactly one will accept; if M1 accepts, then accept, and if M2 accepts, then reject. This turns out to be equivalent to the one-tape model by interleaving the two tapes on one tape, but the exact details are worth puzzling out. ⊠ Consider the language {0n2 \| n ≥0}. This is not at all regular, but it can be decided by a Turing machine! The decider will be given by pseudocode and then an actual diagram; in problem sets the state diagram isn’t necessary, but it’ll provide intuition for what sorts of pseudocode correspond to legal Turing machine state transitions. (1) Sweep from left to right, and then cross out every other 0. (2) If in step 1, and the tape had only one 0, then accept. (3) If in step 1 there was an odd number of 0s (greater than 1), reject. (4) Otherwise, move the head back to the ﬁrst input symbol. (5) Then, go to step 1. Why does this work? Each time step 1 happens, the number of zeros is halved. Every number can be written as n = k2i, where k is odd, and n is a power of 2 iff k = 1; this algorithm just determines the value of k. Here’s what the state diagram looks like: q2 □→□,R } x→x,L 0→0,L / q0 0→□,R / x→x,R □→□,R q1 0→x,R / □→□,R x→x,R q3 □→□,R O x→x,R p 0→0,R qreject qaccept q4 □→□,R i x→x,R Z 0→x,R O 21 One can also deﬁne multitape Turing machines, which can look at several different tapes and is focused at one point on each tape. However, this isn’t more powerful (though it makes some things easier to prove). Theorem 7.6. Every multitape Turing machine corresponds to some single-state Turing machine. Proof. The details are messy (see Sipser), but the gist of it is to copy the inputs from the k tapes onto a single tape, separated by # marks. Then, where the Turing machine is on each state is represented by a dot: double the tape alphabet and then add a dot wherever the machine was. For example, 0 and 1 versus ˙ 0 and ˙ 1. Then, when the Turing machine moves, all the dots have to be read and updated. . . this is not the most elegant or efﬁcient solution, but it makes everything work. One has a similarly messy construction given by interleaving the multiple tapes, too. ⊠ It’s also possible to talk about nondeterministic Turing machines. The low-level details of this will be left to the reader, but it should be inuitively clear: there are multiple possible positions for each state-symbol pair. This also is no more powerful. Theorem 7.7. Every nondeterministic Turing machine M can be transformed into a single-tape Turing machine M that recognizes the language L(N) (i.e. the language recognized by N). Proof idea. Pick a natural ordering on all strings in {Q ∪Γ∪#}∗. Then, for all strings D ∈{Q ∪Γ ∪#}∗in that ordering, check if D = C0#···#Ck, where C0,...,Ck is one of the accepting computation histories for w. If so, accept. ⊠ The fact that one can encode Turing machines as bit strings means that one can feed a Turing machine to another as input. One way to do this is to start with 0n, indicating n states, then 10m (i.e. m tape symbols), 10k (the ﬁrst k are input symbols), then 10s indicates the start state, 10t the accept state, 10r the reject state, and 10u the blank symbol. Then another 1. Then, ((p, i),(q, j,L)) = 0p10i10q10j10 and (p, i),(q, j,R)) = 0p10i10q10j100 and so on. This is one way, not the most efﬁcient; there exist other ways. But the point is that such an encoding exists. Similarly, one can encode DFAs, NFAs, and any string w ∈Σ∗as but strings. For example, if x ∈Σ∗, deﬁne its binary encoding to be bΣ(x). Then, pairs of strings can be encoded as (x.y) := 0\|bΣ(x)\|1bΣ(x)bΣ(y), from which the information from x and y can be retrieved, so to speak. Using this, one can deﬁne some problems/languages: • ADFA = {(M,w) \| M is a DFA that accepts w}, • ANFA = {(M,w) \| M is an NFA that accepts w}, and • ATM = {(M,w) \| M is a Turing machine that recognizes w}. The representations of DFAs and NFAs for these languages can be speciﬁed precisely, but the exact speciﬁcation doesn’t affect the complexity of the problem; one cares just that such a representation exists. Theorem 7.8 (Universal Turing Machine). There exists a Turing machine U which takes as input the code of an arbitrary Turing machine M and an input string w, such that U accepts (M,w) iff M accepts w. 22 This is an important, fundamental property of Turing machines! Notice that DFAs and NFAs don’t have this property: ADFA and ANFA aren’t regular, but ATM is recognizable. The idea behind the proof of Theorem 7.8 is to make a multi-state Turing machine which reads in (M,w) on one state, then has a tape that represents what M looks like when it is running,m another representing the transition function of M, and a ﬁnal tape consisting of the current state of M (holding that number, where the states are numbered). One interesting property of Turing machines is their granularity: you can extend them a lot (multi-tape, nondeterminism, and so on) and still get the same set of recognizable or decidable languages. Efﬁciency is unimportant. The Church-Turing Thesis. Everyone’s intuitive notion of algorithms are captured by the notion of Turing machines. This is not a theorem; it is a falsiﬁable scientiﬁc hypothesis. Some people argue against it in the context of AI, but it’s been thoroughly tested since the notion of Turing machines was ﬁrst formulated decades ago.4 Some serious people have written serious papers to try to prove the Church-Turing thesis, but always run into weird issues in the real world, such as quantum mechanics or black holes or such. 8. RECOGNIZABILITY, DECIDABILITY, AND REDUCTIONS: 1/30/14 Recall the deﬁnition of a Turing machine, and the idea that a conﬁguration of a Turing machine can be encoded into a string, which allows one to deﬁne a conﬁguration yielding another conﬁguration, or accepting a string, and so on. Recall a Turing machine recognizes a language if the strings it accepts are exactly those in the language; it decides the language if it rejects all strings not in the language (rather than running indeﬁnitely). Then, languages are called recognizable or decidable if there exists a Turing machine that recognizes or decides them, respectively. Theorem 8.1. ADFA, the acceptance problem for DFAs, is decidable. Proof. A DFA is a special case of a Turing machine, and from the construction of the universal Turing machine U, one can run U on a DFA M and string w, and output the answer. Thus, U decides ADFA. ⊠ Theorem 8.2. ANFA, the acceptance problem for NFAs, is also decidable. Proof. A Turing machine can convert an NFA into a DFA, and then use the fact that ADFA is decidable. ⊠ Interestingly, however, ATM is recognizable but not decidable! The universal Turing machine U recognizes it, but won’t necessarily stop if the Turing machine it simulates doesn’t stop. More alarmingly, there exist non-recognizable languages, which will be proven in a moment. But then, assuming the Church-Turing thesis, this means there are problems that no computing device can solve! One can prove this by showing there is no surjection from the set of Turing machines to the set of languages over {0,1}. This means there are, in some sense, more problems to solve than programs to solve them. Recall that a function f : A →B is onto (or surjective) iff for every b ∈B, there exists an a ∈A such that f (a) = b. In some sense, the domain covers the entire codomain. 4Not to mention that the people trying to discredit it have over time looked more and more like crackpots. 23 Theorem 8.3. For any set L, let 2L denote its power set; then, there is no onto function f : L →2L. Proof. Suppose there exists an onto function f : L →2L. Then, deﬁne S = {x ∈L \| x ̸∈f (x)} ∈2L. Since f is onto, then there must be a y ∈L such that f (y) = S. Then, is y ∈S? If it s, then y ̸∈f (y) = S, so it must not be. But then, y ̸∈S, which by the deﬁnition of S, implies y ∈f (y) = S. ⊠ Alternatively, one can avoid the contradiction by just letting f : L →2L be an arbitrary function, and deﬁne S as before. Then, for any x ∈L, if x ∈S, then x ̸∈f (x), and if x ̸∈S, then x ∈f (x). But in either case, f (x) ̸= S, so f is not onto. Another way of phrasing this is that for every set L, its power set has strictly larger cardinality. Suppose that all languages over {0,1} were recognizable. Then, the function f : {Turing machines} → {languages} is onto. Every Turing machine can be given by a bitstring, so the set of Turing machines can be thought of as a subset of {0,1}∗= L. But the set of languages is the set of subsets of {0,1}∗, i.e 2L. But there’s no onto function L →2L, but therefore there can’t be one from the set of Turing machines to 2L either. Perhaps this seems very abstract, but remember the Church-Turing thesis: take crucial part of this proof was that Turing machines are given by ﬁnite descriptions, just like programs in the real world, but problems aren’t limited by this. Thus, this has implications on the real world. This proof is sadly nonconstructive, but an example of an unrecognizable language will be given soon. Russell’s Paradox. In the early 1900s, logicians tried to deﬁne consistent foundations for mathematics. Suppose there is a set X, a universe of all possible sets. Then, people generally accepted the following axiom. Frege’s Axiom. Let f : X →{0,1}. Then, {S ∈X \| f (S) = 1} is a set. Then, let F = {S ∈X \| S ̸∈S}, which seems reasonable. But then, if F ∈F, then F ̸∈F, and if F ̸∈F, then F ∈F! This is no good; the logical system is inconsistent. For another lens on countability, one has the following theorem. This states that the real numbers are uncountable: they cannot be indexed by the integers. Here, integers are analogous to {0,1}∗, and real numbers to its power set. Theorem 8.4. There is no onto function Z+ →(0,1) (positive integers to real numbers). Proof. Suppose f is such a function, given as 1 →0.28347279... 2 →0.88388384... 3 →0.77635284... 4 →0.11111111... 5 →0.12345678... and so on. But then, deﬁne r ∈(0,1) such that its nth digit is 1 if the nth digit of f (n) is not 1, and 2 if it is. Then, f (n) ̸= r for all n. ⊠ 24 Here, r = 0.11121... This proof was discovered by Cantor, and he ended up going crazy after producing these and similar counterintuitive statements in set theory. It’s known as Cantor’s diagonal argument. Note that there is a bijection between Z+ and Z+ ×Z+, given by counting (1,1), (1,2), (2,1), (2,2), (3,1), and so on. There’s even a bijection Z+ →Q, given by 1/1, 1/2, 2/1, 1/3, 3/2, 2/3, 3/1, 1/4, 4/3, and so on. There are several ways to do this, but this one is interesting in that it doesn’t repeat any (which is irrelevant for the proof, but makes it more elegant). Returning to undecidability, it’s time to actually produce a language that is not recogniz-able. Theorem 8.5. ATM is recognizable, but not decidable. Corollary 8.6. Thus, ¬ATM is not recognizable. This is because we saw last lecture that if L and ¬L are both recognizable, then L is decidable. Deﬁnition 8.7. A language is co-recognizable (co-r.e.) if its complement is recognizable. Corollary 8.8. By this deﬁnition, ATM is not co-recognizable. Proof of Theorem 8.5. Suppose H is a machine that decides ATM, i.e. H((M,w)) accepts iff M accepts w. Then, let D be the Turing machine that outputs the opposite of H. Run D on itself; then, if D accepts, then it rejects; and if it rejects, then it accepts. This is clearly impossible, so such an H cannot exist. ⊠ This is another tricky proof, but it’s really a diagonalization argument in disguise: for any two Turing machines Mi and M j, one can determine whether Mi accepts on input M j. But then, one can deﬁne D to be the machine that does the opposite of the diagonal Mi(Mi), and then D can’t be anywhere in the table, because it would force a paradox. Alternatively, here’s a proof without contradictions. Proof. Let H be a machine that recognizes ATM. Since the universal machine is an example, then such an H exists. Then, H((M,w)) accepts if M accepts w, and loops or rejects if M does either. Then, let DH(M) do the opposite of whatever H does on input (M,M). Thus, DH(DH) rejects if DH accepts DH, and accepts if DH rejects DH. . . but this time, it’s OK, because the third option is that DH loops forever. Thus, we have (DH,DH) ̸∈ATM, but H runs forever on the input (DH,DH), so H cannot decide ATM. ⊠ Let HALTTM denote the halting problem: HALTTM = {(M,w) \| M is a Turing machine that halts on input Theorem 8.9. HALTTM is undecidable. Proof. The key idea of the proof is to prove by contradiction by reducing the problem: if H decides HALTTM, then construct an M′ that decides ATM. Suppose H is a Turing machine that decides HALTTM. Then, let M′(M,w) run H(M,w); if H rejects, then reject, and if H accepts, then run a universal Turing machine M on w until it halts. If M accepts, then accept, and if M rejects, then reject. ⊠ This is a very high-level speciﬁcation of a Turing machine, but that’s perfectly all right. This is an example of a common pattern: one can often prove a language L is undecidable by proving that if L is decidable, then so is ATM. This says that L is as least as hard as ATM, so to speak, and one could say L ≤m ATM. 25 Mapping Reductions. Deﬁnition 8.10. f : Σ∗→Σ∗is a computable function if there exists a Turing machine M that, when run on any input w ∈Σ∗, halts with just f (w) written on its tape. Deﬁnition 8.11. If A and B are languages over an alphabet Σ, then A is mapping-reducible to B, denoted A ≤m B, if there exists a computable function f : Σ∗→Σ∗such that f (a) ∈B iff a ∈A. This f is called a mapping reduction from A to B. Theorem 8.12. If A ≤m B and B ≤m C, then A ≤m C. Proof. Let f be the computable function reducing A to B and g be that from B to C. Then, g ◦f is computable, and reduces A to C. ⊠ Theorem 8.13. If A ≤m B and B is decidable, then A is decidable. Proof. Let M decide B and f be a mapping reduction from A to B. Then, let M′ be the Turing machine that computes f (w) on input w, and then run M on it and output its answer. Then, one can reason that M′ decides A. ⊠ Theorem 8.14. If A ≤m B and B is recognizable, then A is recognizable. The proof is essentially the same as for Theorem 8.13, except that M is a recognizer and that perhaps it doesn’t halt, in which case M′ doesn’t either. Corollary 8.15. • If A ≤m B and A is undecidable, then B is undecidable. • If A ≤m B and A is unrecognizable, then B is unrecognizable. Notice that the proof of Theorem 8.9 boils down to showing that ATM ≤m HALTTM. It is also true that HALTTM ≤m ATM, because one can deﬁne f (M,w) to construct M′, which accepts if M(w) halts and otherwise loops forever, Then, f (M,w) = (M′,w). Deﬁnition 8.16. EMPTYTM = {M \| M is a Turing machine and L(M) ̸= ;}. Theorem 8.17. EMPTYTM is undecidable. Proof. Assume by contradiction that there is a Turing machine E that decides EMPTYTM. Then, E can be used to get a decider D for ATM; speciﬁcally, let D on input (M,w) build a Turing machine M′ which on input x does the following: if x = w, then run M(w) and else reject. Then, run E(M′) and accept iff E rejects. If M accepts w, then L(M′) = {w}, and in particular is nonempty. But if M doesn’t accept w, then L(M′) = ;. ⊠ Theorem 8.18. In fact, EMPTYTM is unrecognizable. Proof. In the proof of Theorem 8.17, the proof just constructs a reduction ¬ATM ≤m EMPTYTM, so since ¬ATM is not recognizable, then neither is EMPTYTM. ⊠ The regularity problem for Turing machines is REGULARTM = {M \| M is a Turing machine and L(M) is r It might be nice to be able to reduce a machine to a DFA, but this cannot be done in general. Theorem 8.19. REGULARTM is unrecognizable. Proof. The proof will construct a reduction ¬ATM ≤m REGULARTM. Let f (M,w) output a Turing machine M′ which on input x runs M(w) if x = 0n1n, and else rejects. Thus, if (M,w) ∈ATM, then f (M,w) accepts {0n1n}, which isn’t regular. But if (M,w) ̸∈ATM, then f (M,w) = M′ accepts the empty set, which is regular. Thus, distinguishing regular languages from nonregular languages would allow one to solve ¬ATM. ⊠ 26 9. REDUCTIONS, UNDECIDABILITY, AND THE POST CORRESPONDENCE PROBLEM: 2/4/14 “Though the Turing machine couldn’t decide between the two options, at least it recognized a good thing when it saw it.” Last time, we talked about uncomputability: some problems are uncomputable, even with any computational mode. Thanks to the Church-Turing thesis, this corresponds to our actual inability to solve these problems with algorithms. A concrete example of an undecidable problem is ATM, the Turing machine decision problem. Even though it’s recognizable, as a result of the existence of the universal Turing machine, ¬ATM isn’t recognizable. This was proven by showing that for any machine H that recognizes ATM, there is a Turing machine DH on which H runs forever, and thus cannot decide. Now that one concrete instance of an undecidable problem exists, it can be used to show other problems are undecidable; for example, the halting problem can be reduced to ATM and thus is also undecidable. Formally, ATM ≤m HALTTM, i.e. there exists a computable function f : Σ∗→Σ∗such that for every w ∈Σ∗, w ∈ATM iff f (w) ∈HALTTM. This is called a mapping reduction, or sometimes a many-one reduction (there’s no reason f has to be injective). This is useful because if A ≤m B and B is decidable (resp. recognizable), then A is decidable (resp. recognizable), and therefore if A is undecidable (resp. unrecognizable), then B is undecidable (resp. unrecognizable). We saw that ATM ≤m HALTTM and therefore that ¬ATM ≤m ¬HALTTM. This is a general principle: if A ≤m B, then ¬A ≤m ¬B, because the same computable function f ca be used. Another interesting question is as to whether ¬A ≤m A; this is true if A is decidable. But ¬HALTTM ̸≤m HALTTM, because HALTTM is recognizable but not decidable, so ¬HALTTM isn’t recognizable. But a reduction would imply it were recognizable, which is wrong, so no such reduction exists. If A ≤m B and B ≤m A, one might call A and B equivalent, and write A ≡m B. For example, we have see that HALTTM ≡m ATM. In some sense, this means that if a wizard gave you the choice of being able to magically solve one or the other, the choice is irrelevant; each will lead to a solution to the other. Intuitively, both problems have the same difﬁculty. Since it’s easy to minimize DFAs, one might try to do the same with Turing machines. Such an algorithm would be useful for program optimization. However, it’s not that nice. Theorem 9.1. EQTM is unrecognizable. Proof. The proof idea is to show EMPTYTM ≤m EQTM. Let M; be a “dummy” Turing machine which accepts nothing (i.e. there’s no path from the start state to the accept state). Then, for any Turing machine M, deﬁne f (M) = (M,M;), which is pretty clearly computable. But then, f (M) ∈EQTM iff M ∈EMPTYTM, so EMPTYTM ≤m EQTM, and since EMPTYTM is unrecognizable, then EQTM must be too. ⊠ Not all undecidable problems deal with arcana of Turing machines running on each other. Consider Post’s Correspondence Problem5 (PCP)6, which is as follows. There are a bunch of dominoes with strings from Σ∗on them on the top and the bottom, e.g. haaa a i ha c i h a aa i h c a i . 5Oh... I thought all this time it was the Post Correspondence Problem, i.e. correspondence by mail. Instead, it was apparently invented by a researcher named Post. 6No, this is not a drug. But it’s pretty trippy, and legal to boot! 27 The goal is to choose from these dominoes, with repetition, so that when they’re lined up, the top and bottom strings are the same. For example, here one would win with haaa a ih a aa ih a aa i . In the case · b ca ¸ h a ab i hca a i ·abc c ¸ , there are multiple ways to win, including h a ab i· b ca ¸hca a ih a ab i·abc c ¸ . This sounds like a standard interview question, doesn’t it? How would one write a program to solve it, or even to determine whether a win exists? Some observations can be made: the ﬁrst domino must have the ﬁrst letters of its top and bottom string must be identical, and similarly for the last domino and the last letters. Secondly, if all of the dominoes have a longer top than bottom (or vice versa, for all of them), then it’s clear that no win will happen. Finally, if a domino has the same string on the top and bottom, it is the solution. Thus, the problem is: given a set of domino types, is there a match? Let PCP be the language of sets of dominoes that have a match. Theorem 9.2. PCP is undecidable. This is a surprise: there’s no clear relation of PCP to computation! But since Turing machines are extremely simple models of computation, they are relatively easier to model with, say, dominoes. First, let the FPCP problem be nearly the same of PCP, except that the set of dominoes contains a distinguished ﬁrst domino type, and all solutions must start with this domino. Let FPCP be the language of sets of dominoes with such a solution. Theorem 9.3. FPCP is undeciable. Proof. The proof will be by a reduction from ATM to FPCP. Recall the notion of a computation history for a Turing machine M on input w, which is a sequence of conﬁgurations C0,...,Ck such that C0 is the starting conﬁguration of the machine, each Ci yields Ci+1 on the next character of w, and Ck is an accept state. Then, the goal is to produce an instance P of FPCP where the solutions generate strings of the form C0C1 ···Ck. This takes seven steps: (1) The ﬁrst domino in P is [#/#q0w#] (i.e. the starting conﬁguration of the Turing machine). (2) If δ(q,a) = (p,b,R), the add the domino [qa/bp]. (3) If δ(q,a) = (p,b,L), then add [cqa/pcb] for all c ∈Γ. These two classes of dominoes capture the transitions of a Turing machine, because after this domino has been added, the next domino must start with the next conﬁguration given the input w, even encoding the tape! (4) Add [a/a] for all a ∈Γ. (5) Add [#/#] and [#/□#], for ﬁlling in the conﬁgurations. (6) For all a ∈Γ (including □), add [aqacc/qacc] and [qacca/qacc]. This forces valid conﬁgu-rations to end on an accept state. (7) The ﬁnal domino is [qacc#/#]. 28 Thus, one obtains a large but ﬁnite number of dominoes. This means that ATM ≤m FPCP, so FPCP is undecidable. ⊠ This is useful for solving the more general PCP problem. Proof of Theorem 9.2. The goal is to reduce FPCP ≤m PCP. For u = u1u2...un, where ui ∈Γ∪Q ∪{#}, deﬁne ⋆u = ∗u1 ∗u2 ∗···∗un u⋆= u1 ∗u2 ∗···∗un∗ ⋆u⋆= ∗u1 ∗u2 ∗u3 ∗···∗un∗ Suppose that one has an FPCP instance of dominoes [t1/b1],[t2/b2],...,[tn/bn]. Then, create a PCP problem with the dominoes [⋆t1/⋆b1⋆] and [⋆ti/bi⋆] for all i = 1,...,n, along with an extra domino [∗⋄/⋄]. The idea is that this requires one to start with [⋆t1/⋆b1⋆], but then continue with the rest of the game as normal. Thus, FPCP ≤m PCP, so PCP is undecidable. ⊠ What we’ve shown is that given an (M,w) one can compute an instance of PCP that has a match iff M accepts w, so domino solitare is another universal model of computation. Instead of Turing machines or even computers, you could use dominoes. 10. ORACLES, RICE’S THEOREM, THE RECURSION THEOREM, AND THE FIXED-POINT THEOREM: 2/6/14 There’s a lot of depth to computability theory that haven’t really happened yet; it’s not just about things being decidable, recognizable, or whatever. The connection to the Church-Turing thesis makes these have interesting implications, and can be very general. For example, the recursion theorem demonstrates how a Turing machine can access its own code. This makes proofs of undecidability shorter (albeit trippier), and also enables one to create quines in sufﬁciently powerful languages. Deﬁnition 10.1. A decidable predicate R(x, y) is a proposition about the input strings x and y that is implemented by some Turing machine M. That is, for all x and y, R(x, y) is true implies that M(x, y) accepts, and R(x, y) is false implies that M(x, y) rejects. One can think of R as a function Σ∗×Σ∗→{0,1}. Examples include R(x, y) which checks whether xy has at most 100 0s, or R(N, y) which reports whether N is a Turing machine that halts on y within 99 steps. Notice how this relates to decidable languages. But it also offers a connection with recognizability: Theorem 10.2. A language A is recognizable iff there is a decidable predicate R(x, y) such that A = {x \| ∃y such that R(x, y)}. Proof. Suppose A is of the format described above. Then, a Turing machine on input x can run R(x, y) on all y (after enumerating them). If R(x, y) is true for some y, then this machine will reach it, halt, and accept. Conversely, suppose A is recognizable, and let M recognize it Then, deﬁne R(x, y) to be true if M accepts x in y steps (where y is interpreted as a positive integer). Thus, M accepts x iff there is some number y of steps such that R(x, y) is true. Thus, A can be written in the speciﬁed form. ⊠ 29 Oracles. This will be a new model of computation, which formalizes the joke told last lecture about the wizard. The idea is that an oracle Turing machine is able to magically decide whether a string is in some set. Of course, this is much more powerful than anything else we’ve seen. Deﬁnition 10.3. More formally, an oracle Turing machine M is a Turing machine equipped with a set B ⊆Γ∗to which M may “ask” membership queries on a special oracle tape, and receives an answer in one step. (Formally, M enters a special state q?, and if the string y on the oracle tape is in B, it goes to qYES, and if not, it branches to qNO.) The upshot is that in pseudocode descriptions of an oracle Turing machine, one can test if x ∈B for some string x obtained from the input, and use it as a branch instruction. The power of this construction is that it doesn’t depend on B being decidable or recognizable. But this means that the notions of recognizability and decidability change slightly. Deﬁnition 10.4. • A is recognizable with respect to B if there is an oracle Turing machine equipped with B that recognizes A. • A is decidable with respect to B if there is an oracle Turing machine equipped with B that decides A. For example, ATM is decidable with respect to HALTTM, because on input (M,w), one can decide whether M accepts w by rejecting if (M,w) doesn’t halt (which uses the oracle for HALTTM), and running M if it does, and accepting if it does. Theorem 10.5. HALTTM is decidable in ATM. If A is decidable in B, then one says that A Turing reduces to B, written AtrB. This is akin to mapping reduction, but is much more powerful: based on the answers to questions in B, one can ask more questions. Theorem 10.6. If A ≤m B, then AtrB. Proof. If A ≤m B, then there is a computable function f : Σ∗→Σ∗, such that w ∈A iff f (w) ∈B. Thus, an oracle for B can be used to decide A, by computing f (w) and testing if it’s in B. ⊠ Theorem 10.7. ¬HALTTMtrHALTTM. Proof. Construct a Turing machine D that, on input (M,w), reject if (M,w) ∈HALTTM, and otherwise accept. ⊠ However, we saw last lecture that ¬HALTTM ̸≤m HALTTM, so Turing reducibility is strictly more powerful than mapping reducibility. Though this seems pretty awesome magical, there are still limits to powerful oracles. Even if we could solve the halting problem, there are problems that cannot be solved with Turing machines. For example, consider SUPERHALT, the language of (M,x) where M is an oracle Turing machine for the halting problem and M halts on x. The proof of this is essentially the same as that for the halting problem, and the key idea is that the oracle is for HALTTM, not for SUPERHALT; then, the same diagonalization still holds. This means that there’s an inﬁnite hierarchy of undecidable problems: let SUPERHALT0 = HALTTM and SUPERHALT1 = SUPERHALT; then, deﬁne SUPERHALTn inductively as the set of (M.x) where M halts on x, where M is an oracle Turing machine for SUPERHALTn−1. Then, SUPERHALTm is undecidable in SUPERHALTn whenever m > n. 30 Exercise 10.8. Show that REVERSE = {N \| N accepts wR if it accepts w} is undecidable. Solution. Given a machine D for deciding REVERSE, one can decide ATM: on input (M,w), generate the Turing machine Mw, which accepts on input 01 and on input 10 runs M(w). Then, D(Mw) indicates whether M accepts w, which is a problem. ⊠ Exercise 10.9. Consider the language of (M,w) such that M tries to move its head off of the left end of the input at least once when run on input w. Is it decidable? Exercise 10.10. Consider the language of (M,w), where M run on input w moves left at least once. These are undecidable and decidable, respectively, which is counterintuitive. But here’s why: Solution to Exercise 10.9. The proof will reduce from ATM to this language: on input (M,w) make a Turing machine N that shifts w over one cell, marks a special symbol on the leftmost cell, and then simulates M(w) on the tape. Then, if M moves to the marked cell but hasn’t yet accepted, then N moves back to the right, and if M accepts, then N tries to move past the end. Then, (M,w) is in ATM iff (N,w) isn’t. ⊠ Solution to Exercise 10.10. To explicitly decide it, on input (M,w), run the machine for \|QM\|+ \|w\|+1 steps; accept if it’s moved to the left, and reject otherwise. In this sense, if it hasn’t moved to the left, it won’t; this is because when it’s passed \|w\| pieces of tape, then it’s only looking at blanks, and then after a further \|QM\|+1 steps, it’s transitioned into at least one state twice by the pigeonhole principle, so it’s trapped in a loop (since it always reads blanks), so it will keep repeating, and therefore never move left. ⊠ This leads to the Rice’s theorem, which is an extremely powerful way to analyze the decidability of languages. Theorem 10.11 (Rice). Let L be a language over Turing machine descriptions M. Then, suppose L satisﬁes the following two properties. (1) L is nontrivial, i.e. there exist Turing machines MYES and MNO such that MYES ∈L and MNO ̸∈L. (2) L is semantic, i.e. for all Turing machines M1 and M2 such that L(M1) = L(M2), then M1 ∈L iff M2 ∈L. Then, L is undecidable. Basically every undecidable language we’ve seen so far is an easy consequence of applying Rice’s theorem in the right way. Another way to perceive this is to realize a property of Turing machines to be a function P{Turing machines} →{0,1}. Then, P is called nontrivial if there exists a MYES such that P(MYES) = 1 and an MNO such that P(MNO) = 0, and P is called semantic if whenever L(M1) = L(M2), then P(M1) = P(M2). Then, Rice’s theorem states that the language L p{M \| P(M) = 1} is undecidable. The notion of a semantic property, one that depends only on the language of a given Turing machine, is hard to intuitively understand, so here are some examples. • M accepts 0. • For all w, M(w) accepts iff M(wR) accepts (the reverse of the string). • L(M) = {0}. 31 • L(M) is empty. • L(M) is regular. • M accepts exactly 163 strings. In all of these cases, the languages for which these properties are true are undecidable! Here are some properties which aren’t semantic. • M halts and rejects 0. • M tries to move its head off the left end of the tape on input 0. • M never moves its head left on input 0. • M has exactly 154 states. • M halts on all input. Notice that they do not depend on L(M). Proof of Theorem 10.11. The goal is to reduce ATM to the given language L. Let M; be a Turing machine that never halts, which as seen before can be constructed. Suppose ﬁrst that M; ̸∈L, and choose an MYES ∈L (which must exist, because L is nontrivial). Then, reduce from ATM: on input (M,w), output the Turing machine Mw(x) which accepts if M accepts w and MYES accepts x, and rejects otherwise. Then, L(Mw) = L(MYES) ∈L. Conversely, if M doesn’t accept w, then L(Mw) = ; = L(M), but M̸ ∈L, so Mw ̸∈L. Thus, this is a reduction. In the other case, where M; ∈L, reduce instead ¬ATM to L, by producing an MNO ̸∈L (since L is nontrivial), and Mw accepts iff M accepts w and MNO accepts x. This is a reduction by almost exactly the same proof. In this case we know the slightly stronger statement that L isn’t recognizable. ⊠ Self-Reference and the Recursion Theorem. This part of the lecture, about Turing machines that think about themselves, is probably the trippiest part of the class. Lemma 10.12. There is a computable function q : Σ∗→Σ∗such that for any string w, q(w) is the description of a Turing machine Pw that one very input, prints out w and accepts. This is easily shown: just make the Turing machine which reads w, and writes the Turing machine that ignores its input and writes w. Getting weirder: Theorem 10.13. There is a Turing machine which prints its own code. Proof. First, deﬁne a Turing machine B which, on input M, produces the Turing machine which: on input w, calculate M(PM(w)), which ignores w and feeds M to itself. Now, consider the Turing machine B(PB). On input w, it prints M(PM), where M = B, so it prints B(PB). Whoa. ⊠ Another way of looking at this is that there is a program which prints its own description or source, or a sentence which describes itself, e.g. “Print this sentence.” Another one, which corresponds to the machine B in the proof, is: “Print two copies of the following, the second in quotes: ‘Print two copies of the following, the second in quotes.’ ” There’s yet another way to approach this, using lambda calculus. The self-printing function is (λx. x x)(λx. x x), i.e. evaluates the second argument on the ﬁrst, but it is its own second argument. 32 This leads to the really weird recursion theorem: Theorem 10.14 (Recursion). For every Turing machine T computing a function t : Σ∗→Σ∗→ Σ∗, there is a Turing machine R computing a function r : Σ∗→Σ∗given by r(w) = t(R,w). In other words, R computes on itself, so one can think of T as obtaining its own description! Thus, Turing machine pseudocode is allowed to just obtain its own description and use it. Notice that the Church-Turing thesis means that this has implications to AI and free will. . . Proof of Theorem 10.14. Suppose T computes t(a,b). Let B be the Turing machine that, on input N (a Turing machine that takes 2 inputs), runs N(PN(w),w) on input w. Then, let M be deﬁned as T(B(N),w). Finally, deﬁne R = T(B(PM(w)),w). Thus, w SM →M B →S →T(S,w). So, what’s S? S is what happens when (M,w) is fed into B, so S ends up just computing T(B(PM(w)),w)! So S = R and therefore R satisﬁes the required property. ⊠ Puzzle: can you decide (M1,w1), (M2,w2), and (M3.w3) (i.e. whether all three halt) with only two calls to an oracle to HALTTM? 11. SELF-REFERENCE AND THE FOUNDATIONS OF MATHEMATICS: 2/11/14 “I promise to never again squeeze humor out of self-reference.” - xkcd This lecture has an ambitious title, but indeed is an ambitious lecture. If one embraces the Church-Turing thesis that everything computable is represented by a Turing machine, then computability theory should say something about the foundations of mathematics. One helpful way to understand statements like the Recursion Theorem is to draw pictures: For example, the Turing machine which prints its own description looks like M / B w / PM M / M / in the notation used last lecture. It’s counterintuitive, but ﬁddle with it enough and it ought to make sense. But then, it can be used to allow Turing machines to access their own descriptions. This often leads to faster proofs, e.g. that ATM is undecidable: Proof. Suppose ATM is decidable, and let H be a decider for it. Then, let B be the machine that, on input w, obtains its own description B, runs H on (B,w), and then ﬂips the output. This leads to a paradox. ⊠ This can be seen as a formalization or resolution of the free will paradox: no single deterministic machine can determine what an arbitrary Turing machine can do, by way of doing the opposite of what it reported. Theorem 11.1 (The Fixed-Point Theorem). Let t : Σ∗→Σ∗be computable. Then, there exists a Turing machine F such that t(F) outputs the description of a Turing machine G such that L(F) = L(G). Proof. Here’s the pseudocode for F: on input w, (1) Run t(F) and obtain an output string G. Then, interpret G as a description of a Turing machine. (2) Run G on input w, and accept iff G does. ⊠ This can be used to given an alternate proof of Rice’s theorem: 33 Proof of Theorem 10.11. Suppose one could decide a nontrivial semantic L. Then, t(M), which outputs a Turing machine MNO ̸∈L if M ∈L and a Turing machine MYES ∈L if M ̸∈L, is a computable function. Thus, it must have a ﬁxed point M, but if M ∈L, then t(M) = MNO ̸∈L, but L(M) = L(MNO), so L can’t be semantic (and again with the same idea for M ̸∈L). ⊠ Now, for the ambitious part of the lecture: what can computability theory say about the foundations of mathematics itself? This will be relatively informal. Deﬁnition 11.2. A formal system describes a formal language for writing ﬁnite mathemati-cal statements in which: • there is a deﬁnition of what statements are “true,” and • there is a notion of a proof of a statement. Example 11.3. Every Turing machine M deﬁnes a formal system F, where the mathemat-ical statements are elements of Σ∗. Then, the string w represents the statement that M accepts w. Then, the true statements in F are exactly L(M), and a proof that M accepts w is deﬁned to be an accepting computation history for w. ( Deﬁnition 11.4. • A formal system F is consistent (or sound) if no false statement in F has a valid proof in F (i.e. provability implies truth). • A formal system F is complete if every true statement in F has a valid proof in F (i.e. truth implies provability). These are nice notions to have for formal systems, and there are certainly many that are both consistent and complete, but we also want formal systems that has implications on mathematics. Deﬁnition 11.5. A formal system F is interesting if (1) Any mathematical statement about computation can be described as a statement of F. That is, given (M,w), there is an SM,w ∈F such that SM,w is true in F iff M accepts w. (2) Proofs are convincing, i.e. it should be possible to check that a proof of a theorem is correct. In other words, the set {(S,P) \| P is a proof of S in F} is decidable. (3) If S ∈F and there is a proof of S describable in terms of computation, then there is a proof of S in F. This means the system is expressive enough to understand computation histories and the like. In other words, if M accepts w, then there is a proof P in F of SM,w. There are plenty of uninteresting systems, such as the ﬁrst-order theory of numbers with only addition; no notion of prime numbers, etc. exist. But it’s consistent and complete. But there are also interesting systems. The theorems described below, which place some limitations on mathematics, are powerful and have had huge implications on the mathematics of the past century. Their proofs are quite tricky, so sketches will be given. Theorem 11.6 (Gödel 1931). For every consistent and interesting system F, F is incomplete: there are mathematical statements in F that are true, but cannot be proven in F. 34 It’s not immediately clear how something can be known to be true if it isn’t provable; but the idea is that a system has a set of true statements and a set of proofs and the two don’t match. It would also be nice to be able to prove that mathematics is consistent, but. . . Theorem 11.7 (Gödel 1931). The consistency of F cannot be proven in F. Theorem 11.8 (Church-Turing 1936). The problem of checking whether a given statement in F has a proof is undecidable. Given a statement and a proof, it’s decidable, but with only a statement there’s in general nothing we can do. Proof sketch of Theorem 11.6. Let SM,w ∈F be true iff M accepts w. Then, deﬁne a Turing machine G(x) which: (1) First, uses the Recursion Theorem to obtain its own description. (2) Constructs a statement S′ = ¬SG,ε. (3) Searches for a proof of S′ in F over all ﬁnite-length strings. Accept if such a proof is found. S′ effectively says there is no proof of S′ in F, which can’t be provable in F, but is thus true. If such a proof exists, that G doesn’t accept the empty string, then G does accept ε, which is a problem. ⊠ Notice that we must be looking in from some other system. . . this is quite a diversion to get into, and is outside of the scope of this lecture. Proof of Theorem 11.7. Suppose it is possible to prove that F is consistent within F. Then, let S(=)¬SG,ε as before, which is true, but has no proof in F (since G doesn’t accept ε). Thus, since F is consistent, then there’s a proof in F that ¬SG,ε is true: if SG,ε is true, then there is a proof in F of ¬SG,ε, and therefore, since F is consistent, then ¬SG,ε is true. But SG,ε and ¬SG,ε cannot both be true, so there’s a contradiction, and therefore ¬SG,ε is true (i.e. the initial assumption was false). But this is a contradiction, since it was just seen in the previous proof to be undecidable! ⊠ Finally, we can turn undecidability to mathematics. Proof of Theorem 11.8. Let PROVABLEF be the set of S such that F has a proof for S or for ¬S. Suppose PROVABLEF is decidable with a decider P. Then, one can decide ATM as follows: • On input (M,w), run P on SM,w. • If P accepts, then a proof exists, so examine all possible proofs in F and accept when a proof of SM,w is found. • If P accepts and a proof of ¬SM,w is found, then reject. (This doesn’t conﬂict with the above, because F is consistent.) • If P rejects, then reject. This is reasonable because if M accepts w, then there is always a proof of that. ⊠ 12. A UNIVERSAL THEORY OF DATA COMPRESSION: KOLMOGOROV COMPLEXITY: 2/18/14 “For every technical topic, there is a relevant xkcd.” 35 Kolmogorov complexity is a very general theory of data compression. As the Church-Turing thesis claims that everyone’s intuitive notion of algorithms corresponds to Turing machines, one might want to formulate a universal theory of information. In particular, can one quantify how much information is contained in a string? Consider two strings A = 010101010101010101010101010101010101 and B = 11001001110111010110100 Intuitively, A doesn’t contain as much information as B. In some sense, A can be compressed into a smaller space than B, which intuits that it contains less information. Thus, the thesis is that the amount of information contained in a string is the length of the shortest way to describe it. But how should strings be described? So far, the answer has usually been Turing machines. Deﬁnition 12.1. If x ∈{0,1}∗, then the shortest description of x, denoted d(x), is the lexico-graphically shortest string 〈M,w〉such that M(w) halts with only x on its tape. Note that different representations of Turing machines might give different speciﬁc exam-ples of lexicographically shortest strings. The notation 〈M,w〉represents a pairing of M and w, i.e. some sort of concatenation. Theorem 12.2. There is a one-to-one computable function 〈·,·〉and computable functions π1,π2 : Σ∗→Σ∗such that z = 〈M,w〉iff π1(z) = M and π2(z) = w. Proof. As an example, let Z(x1 ···xk) = 0x10x20···0xk1. Then, 〈M,w〉= Z(M)w is a pairing, and one can write Turing machines which calculate π1 and π2. ⊠ Notice that in the above example, \|〈M,w〉\| = 2\|M\|+\|w\|+1. It would be perfect to have a pairing which has minimal size \|M\|+\|w\|, but this isn’t possible in real life. It is possible to create better pairings; for example, let b(n) represent the binary encoding of an n ∈N, and let Z be as above. Then, deﬁne 〈M,w〉= Z(b(\|M\|))Mw. This concatenates M and w, and tacks on the length of M, so that one can reconstruct the original information later. This asymptotically takes less space, since it’s logarithmic. For example, to encode 〈10110,101〉, b(\|10110\|) = 101, so the pairing is 〈10110,101〉= 010001110110101. Now, \|〈M,w〉\| ≤2log\|M\|+\|M\|+\|w\|+1. It’s possible to still do better, with a loglog\|M\| term, and so on. Deﬁnition 12.3. The Kolmogorov complexity of x (or the K-complexity of x) is K(x) = \|d(x)\|, where d(x) is the lexicographically shortest description of x as above. This is given in the context of some ﬁxed pairing function. This is in some sense the maximal data compression for x. It’s also quite abstract; examples will develop more naturally from properties of the Kolmogorov complexity. Theorem 12.4. There exists a ﬁxed c such that for all x ∈{0,1}∗, K(x) ≤\|x\|+ c. Intuitively, this says that the amount of information in x isn’t much more than \|x\|. Proof of Theorem 12.4. Deﬁne a Turing machine M which simply halts on any input w. Then, on any string x, M(x) halts with x on its tape, so if c = 2\|M\|+1, then K(x) ≤\|〈M,x〉\| ≤2\|M\|+\|x\|+1 ≤\|x\|+ c. ⊠ This is just a rough upper bound; it might be possible to make it smaller. The intuition for the following theorem is that the information in xx isn’t that much more than that of x. 36 Theorem 12.5. There is a ﬁxed constant c such that for all x ∈{0,1}∗, K(xx) ≤K(x)+ c. Proof. Let N be the Turing machine which on input 〈M,w〉, computes M(w) and prints it twice (i.e. s = M(w), and it prints s). Then, if 〈M,w〉is the shortest description of x, then 〈N,〈M,w〉〉is a description of xx, so K(xx) ≤\|〈N,〈M,x〉〉\| ≤2\|N\|+\|〈M,x〉\|+1 ≤2\|N\|+ K(x)+1 ≤c + K(x) where c = 2\|N\|+1. ⊠ This generalizes: there isn’t all that much more information in xn than in x. Theorem 12.6. There is a ﬁxed constant c such that for all n ≥2 and x ∈{0,1}∗, K(xn) ≤ K(x)+ clogn. Proof. Let N be the Turing machine which on input 〈n,M,w〉, assigns x = M(w) and prints x n times, and let 〈M,w〉be the shortest description for x. Then, K(xn) ≤K(〈N,〈n,M,w〉〉) ≤ 2\|N\|+ dlogn+ K(x) ≤clogn+ K(x) for some constants c and d. ⊠ In general, it’s not really possible to do better than logn here. But this allows us to understand K((01)n), with the example string given earlier: here, K((01)n) ≤c+dlogn, which is much nicer than average. One disadvantage of basing everything on Turing machines is that representations might be shorter in other languages. But this can also be formally addressed. Deﬁnition 12.7. An interpreter is a semi-computable7 function p : Σ∗→Σ∗. Intuitively, it takes programs as input and (may) print their outputs. Deﬁnition 12.8. Then, the shortest description of x ∈{0,1}∗relative to p is the shortest description of any string w such that p(w) = x, and is denoted dp(x). The Kolmogorov complexity relative to p is K p(x) = \|dp(x)\|. Theorem 12.9. For every interpreter p, there is a c ∈Z such that for all x ∈{0,1}∗, K(x) ≤ K p(x)+ c. This intuitively says that interpreters might be faster, but not much more so asymptotically. Proof of Theorem 12.9. Let M be the Turing machine which on input w, simulates p(w) and writes its output to the tape. Then, 〈M,dp(x)〉is a description of x, and K(x) ≤\|〈M,dp(x)〉\| ≤2\|M\|+ K p(x)+1 ≤c + K p(x). ⊠ This really is a universal notion, up to an additive constant. There’s no reason we had to start with Turing machines; the results are always the same. Theorem 12.10. There exist incompressible strings of every length; for all n, there is an x ∈{0,1}n such that K(x) ≥n. Proof. There are 2n binary strings of length n, and 2n−1 binary descriptions of length strictly less than n. And since strings to descriptions must be bijective, then there must be at least one n-bit string which doesn’t have a description of length less than n. ⊠ 7This means that it is given by a Turing machine which may compute the answer, and may not halt. 37 Though it seems like this is a major ﬂaw in this particular data compression algorithm, it’s actually a very general statement about data compression: every compression algorithm must have some incompressible strings. Worse off, a randomly chosen string is highly likely to be incompressible. In the intuition of the following theorem, n is large and c is small. Theorem 12.11. For all n ∈N and c ≥1, Prx∈{0,1}n[K(x) ≥n−c] ≥1−1/2c. Proof. There are 2n string of length n, and 2n−c −1 descriptions of length less than n −c. Hence, the probability that a random x satisﬁes K(x) < n−c is at most (2n−c−1)/2n < 1/2c. ⊠ The fact that there exist incompressible strings is surprisingly useful, even outside of this corner of theory. For example, consider the language L′ = {ww \| w ∈{0,1}∗}, which is not regular. There’s a proof of this using Kolmogorov complexity: suppose L′ is regular and D is a DFA such that L(D) = L′. Then, choose a string x ∈{0,1}n such that K(x) ≥n. After reading in x, D is in state qx. Deﬁne a Turing machine M, which on input (D, q,n) tries all possible paths p of length n through D from state q; if p ends in an accept state, it prints the string that caused it to accept from q. But then, 〈M,〈D, qx,n〉〉is a description of x, because D accepts xx, and if y ̸= x, then xy isn’t rejected. But its length is at most c1 + c2logn, because D and M have ﬁnite, constant descriptions that don’t depend on x, and n takes just logn bits. But this means K(x) = O(logn), which is a contradiction, because x was chosen to be incompressible. Thus, L′ is not regular. Of course, you can use Myhill-Nerode to show this, but this proof is an example of a fundamentally different style of argument. Exercise 12.12. Give short algorithms for generating the following strings: (1) 01000110110000010100111001011101110000 (2) 123581321345589144233377610987 (3) 126241207205040403203628803628800. Solution. The ﬁrst string just counts upwards from 1 in binary; the second concatenates the ﬁrst several Fibonacci numbers; and the last concatenates the ﬁrst ten factorials. These lead to efﬁcient algorithms. Well, this seems kind of hard in general. There’s a formal way to think about this.8 Does there exist an algorithm to perform optimal compression, and can algorithms tell us if a given string is compressible? Theorem 12.13. The language COMPRESS = {(x, c) \| K(x) ≤c} is undecidable. This proof is different from most of the proofs we’ve seen before for undecidability: if it were decidable, then inutitively, one could construct an algorithm that prints the shortest incompressible string of length n, and then such a string could be succinctly described by the code of the algorithm. Barry’s Paradox. The smallest integer that cannot be described in fewer than thirteen words. But this is a description on twelve words! Self-reference is mystifying. 8Mentioned in lecture today: 38 Proof of Theorem 12.13. Suppose that COMPRESS is decidable. Then, let M be the Turing machine which, on input x ∈{0,1}∗, interprets x as a number N. Then, for all y ∈{0,1}n, considered in lexicographical order, if (y,N) ∈COMPRESS, then print y and halt. This algorithm ﬁnds the shortest string that y′ such that K(y′) > n. But then, 〈M,x〉de-scribes y′, and \|〈M,x〉\| ≤c+logN, so N < K(y′) ≤c+2logN, which is a problem for sufﬁciently large N. ⊠ The proof encodes Barry’s paradox, in some sense. This can be used to give (yet) another proof that ATM is undecidable, by reducing from COMPRESS. Given a pair (x, c), construct a Turing machine Mx,c, which, on input w: • For all pairs 〈M′,w′〉with \|〈M′,w′〉\| ≤c, simulate each M′ on w′ in sequence. • If some M′ halts and prints x, then accept. Then, K(x) ≤c iff Mx,c accepts ε (or any string, really, because M ignores its input), Kolmogorov complexity is still a very active topic, e.g. in Bauwens et al., Short lists with short concepts in short time, in the IEEE Conference on Computational Complexity, 2013, it is shown that the following problem is computable: given an input x, print a list of \|x\|2 strings such that at least one string y on the list encodes a description of x, and where y has length at most K(x)+ c. The fact that one has a list of strings allows one to approximate the Kolmogorov complexity, even though the exact problem is undecidable. More on Formal Systems. Kolmogorov complexity leads to another view of the formal-systems approach to the foundation of mathematics. Theorem 12.14. For every interesting, consistent formal system F, there exists a t such that, for all x, K(x) > t is an unprovable statement within F. Proof. Deﬁne a Turing machine which reads its input as an integer: M, which on input k searches over all strings x and proofs P for a proof P in F that K(x) > k. Then, output x if it is found. If M(k) halts, then it must print some output x′. Then, K(x′) = K(〈M,k〉) ≤c+\|k\| ≤c+logk for some c. But since F is consistent, then K(x′)k is true, but k < c +logk doesn’t hold for large enough k, so choose a t for which this doesn’t hold. Then, M(t) cannot halt, so there is no proof that K(x) > t. ⊠ The theorem statement is even more absurd when one realizes that almost all strings of sufﬁciently large length are incompressible. But that’s just life. Also, this leads to a huge number of randomly generated statements with no proof, but that are true with very high probability, a far reach from Gödel’s abstruse statements. Part 3. Complexity Theory: The Modern Models 13. TIME COMPLEXITY: 2/20/14 “This is the way real math is done: ﬁrst you do the proof, then you ﬁgure out what the theorem says.” Complexity theory asks not just what can or can’t be computed in principle, but also what can and can’t be computed with limited resources, e.g. time and space. Clearly, this is a bit more practical than these abstract notions of decidability and undecidability; something may be abstractly solvable, but still take forever to complete. 39 Complexity theory is safely enriched with many, many important open problems, and nobody has any idea how to solve them. Asymptotics and Time Complexity. If f , g : N →R+, then recall that f (n) = O(g(n)) if there exist positive integers c and n0 such that for every n ≥n0, f (n) ≤cg(n), i.e. as n gets large enough, f is bounded above by a ﬁxed multiple of g. In the case where f = O(g), one says that g is an upper bound of f . For example, 2n4.1 +200284n4 +2 = O(n4.1), and 3nlog2 n+5nlog2log2 n = O(nlog2 n), and nlog10 n78 = O(nlogn), because log(n78) = 78logn, and the constant can be swallowed up. In fact, since the change of base rule for logarithms involves scaling by a constant factor, the base of a logarithm is irrelevant with respect to this: O(logn) = O(log2 n) = O(log10 n), so the base is usually unspeciﬁed. Time complexity is measured by counting the number of steps it takes for a Turing machine to halt. Since we want the Turing machine to halt on all inputs, then we will only look at decidable problems right now. Deﬁnition 13.1. Let M be a Turing machine that halts on all of its inputs. Then, the running time or time complexity of M is the function T : N →N, where T(n) is the maximum number of steps taken by M over all inputs of length n. Deﬁnition 13.2. The time complexity class of a function t(n) is TIME(t(n)), the set of languages L′ for which there is a Turing machine M that decides L′ and has time complexity O(t(n). Equivalently, one would say that M has time complexity at most c · t(n), or that M has O(t(n)) running time. For example, the standard algorithm for deciding A = {0k1k \| k ≥0}, which moves back and forth to the midpoint, is in TIME(n2). If REGULARTM denotes the set of regular languages, then REGULARTM ⊆TIME(n), because a DFA necessarily runs in the number of steps equal to the size of the string. It’s also possible to place A ∈TIME(nlogn). This is given by a Turing machine M which, on input w, • If w isn’t of the form 0∗1∗, reject. (This takes time O(n).) • Then, repeat the following until all bits of w are crossed out: – If the parity of 0s and 1s are different (i.e. the numbers mod 2), then reject. – Cross out every other 0 and every other 1. • Then, accept. Knowing the parity at every step allows one to know exactly how many 0s and 1s there are, akin to reading the bits in the binary description of the number. Then, crossing out every other 0 or 1 pushes it to the next-order bit. Can one do better than nlogn? It can be proven that a single-tape Turing machine cannot decide A in time less than O(nlogn). Exercise 13.3 (Extra credit). Suppose that f (n) = o(nlogn) (i.e. is strictly bounded above).9 Then, show that TIME(f (n)) contains only regular languages! Note that this doesn’t imply the nonexistence of nloglogn algorithms; this is just single-tape Turing machines, so more interesting algorithms still exist. But this collapse of time complexity is very interesting. 9More precisely, f (n) = o(g(n)) iff lim n→∞f (n)/g(n) = 0. 40 Theorem 13.4. A = {0k1k \| k ≥0} can be decided in time O(n) with a two-tape Turing machine. Proof idea. Scan all 0s and copy them to the second tape. Then, scan all 1s, and for each 1 scanned, cross off a 0 from the second tape. Then, if the numbers disagree, then reject; else, accept. ⊠ Notice a difference with computability theory: multiple-tape Turing machines make no difference as to whether a language is decidable, but produce different results in complexity theory. Theorem 13.5. Let t : N →N satisfy t(n) ≥n for all n. Then, every t(n)-time multi-tape Turing machine has an equivalent O(t(n)2)-time single-tape Turing machine (that is, they decide the same language). This follows by tracing out the complexity of their proof of (decidability) equivalence: every step of the multitape machine corresponds to updating all of the previous data on the single tape, causing T(n)2 steps. Universal Turing machines are also efﬁcient (i.e. take quadratic time). Theorem 13.6. There is a (single-tape) Turing machine U which takes as input (M,w), where M is a Turing machine and w is an input string, and accepts (M,w) in O(\|M\|2t2) steps iff M accepts w in t steps. Proof sketch. Once again, this follows by tracing the proof of the existence of the universal Turing machine for computability (i.e. the recognizer for ATM): we had a two-tape Turing machine representing the inputs and the states, and so in t steps U updates O(\|M\|t) steps to update both of these: \|M\| is intuitively there because we need to update the conﬁguration. ⊠ This paves the way to something called the Time Hierarchy Theorem, which intuits that with more time to compute, one can solve strictly more problems. As referenced by Exercise 13.3, there are gaps. Theorem 13.7 (Time Hierarchy). For “reasonable” f , g : N →N where f (n) = O(g(n)1/2), TIME(f (n)) ⊊TIME(g(n)). The exact notion of “reasonable” will be explained within the proof, which is easier than stating it beforehand. Proof. Deﬁne a Turing machine N as follows: if w ̸= M10k for some Turing machine M and integer k, reject. Then, let U be the universal Turing machine, and run it on input (M,w) for f (\|w\|) steps, which takes O(f (\|w\|)2) time. Then, accept if M has not accepted w; otherwise, reject. Note that the above argument depended on the following three assumptions: (1) g(n) ≥n, so that the multi-tape Turing machine can be converted into a single-tape Turing machine. (2) f (\|w\|2) ≤O(g(n)). (3) The value f (n) can be efﬁciently computed, i.e. in O(g(n)) time. The number may be large, but there’s a nice algorithm for computing it. Most functions anyone would care about satisfy this. This is what is meant by “reasonable.” Now, we have seen that L(N) ∈TIME(g(n)), and it remains to show it’s not in TIME(f (n)). What happens when we feed N to itself? Speciﬁcally, suppose N(N10k) runs in time O(f (n)). 41 Then, it outputs the opposite value to what it should if it halts, so it can’t halt in time f (n). ⊠ The conditions can be improved slightly. There is a much more difﬁcult proof (e.g. in CS 254) of the following. Theorem 13.8. For “reasonable” f and g where f (n) = O(g(n)/log2 g(n)), TIME(f (n)) ⊊ TIME(g(n)). Thus, there is an inﬁnite hierarchy of decidable, but increasingly time-consuming problems. One important question is whether there exist natural problems (as opposed to the contrived one cooked up for the proof) that sit at every level of the time hierarchy. One example of a natural problem is 3-Sum, which given an S ⊆N, asks if there exist a,b, c ∈S such that a + b = c. This has an O(n2) solution that is popular for job interviews, but it’s open as to whether this problem has a better solution (e.g. linear time). Yet this is a useful problem. The notion of an efﬁcient computation in complexity theory is a problem solvable in polynomial time: P = [ k∈N TIME(nk). This leads to a notion of intuition about efﬁcient algorithms. The Extended Church-Turing Thesis. Everyone’s intuitive notion of efﬁcient algorithms corresponds to the formalism of polynomial-time Turing machines. This is much more controversial (or as another way of thinking about it, open) thesis: randomized algorithms might be efﬁcient, but it’s open as to whether they can be efﬁciently deterministically simulated. Additionally, TIME(n1000000) ⊆P, but most people wouldn’t call these efﬁcient. Finally, do quantum algorithms that cannot be efﬁciently encoded as classical algorithms quality as efﬁcient? These algorithms could undermine cryptography as we know it. Deﬁnition 13.9. A nondeterminstic Turing machine is a Turing machine that can intuitively choose between several state transitions at each step, and accepts a string if any of these choices leads to an accept state. Just like with NFAs, the formal deﬁnition is very similar to that of a deterministic Turing machine, except that the transition function is again represented by a function δ from the states to sets of state–tape-direction triples, where each element of the set is one possible transition. Then, one can deﬁne accepting computation histories in the same manner: where each conﬁguration can yield the next one (even if it could also yield other ones), and they lead to acceptance. Then, the Turing machine accepts a string if such a computation history exists. Finally, N has time complexity T(n) if for all n, all inputs of length n, and all histories, N halts in T(n) time. If one views the possible paths through these choices as a tree, then this says that the maximal depth of the tree is T(n). Deﬁnition 13.10. NTIME(t(n)) is the set of languages decided by an O(t(n))-time nondeter-ministic Turing machine. Since determinism is a special case of nondeterminism, then TIME(t(n)) ⊆NTIME(t(n)). But the reverse inclusion is open for most t(n): it’s known that TIME(t(n)) ⊊NTIME(t(n)), but in general this question is open. 42 A Boolean formula is a function on booleans using ¬, ∧, ∨, and parentheses. A satisfying assignment for such a formula is an assignment to the variables that makes it true (when evaluated in the normal manner). For example, φ = a∧b ∧c ∧¬d is satisﬁable (take a = b = c = 1 and d = 0). But ¬(x∨y)∧x is not satisﬁable. The set of satisﬁable Boolean formulas is called SAT. A restriction of this problem: a 3cnf-formula has a form like (x1 ∨¬x2 ∨x3)∧(x4 ∨x2 ∨x5)∧(x3 ∨¬x2 ∨¬x1). The inside groupings, separated by ∨symbols, are called literals; the greater structures (triples of literals) are called clauses, and are separated by ∧. Then, it is possible to reduce some formulas into 3cnf-form. 3SAT denotes the set of satisﬁable 3cnf-formulas. Theorem 13.11. 3SAT ∈NTIME(n2). Proof. On input φ, ﬁrst check if the formula is in 3cnf-form, which takes O(n2) time. Then, for each variable v in φ, nondeterministically substitute 0 or 1 in place of v, Then, evaluate the formula for these choices of the variables, and accept if it’s true. ⊠ Thus, 3SAT ∈NP, where NP = [ k∈N NTIME(nk). Theorem 13.12. L ∈NP iff there is a constant k and polynomial-time Turing machine V such that (13.13) L = {x \| there exists a y ∈Σ∗such that \|y\| ≤\|x\|k and V(x, y) accepts}. 14. MORE ON P VERSUS NP AND THE COOK-LEVIN THEOREM: 2/25/14 At this point, the words “problem” and “language” really mean the same thing: the set of solutions to a problem forms a language, and so on. More modern terminology tends to favor the former, though. Recalling Theorem 13.12, contrast this with the fact that a language A is recognizable iff there exists a deciding Turing machine V(x, y) such that A = {x ∈Σ∗\| there exists a y ∈ Σ∗such that V(x, y) accepts}. The change was that for NP, V is required to run in polynomial time (which also restricts the length of the input). In some sense, we’re looking at P within NP just like recognizability within decidability; P and recognizability are problems for which veriﬁcation is feasible, and decidability and NP are those problems for which solutions can be feasibly found. Much of complexity theory has been guided by this analogy.10 Proof sketch of Theorem 13.12. Suppose L can be written in the form (13.13); then, a nonde-terministic algorithm to decide L is to nondeterministically guess a y and then run V(x, y). Since this computation takes polynomial time, L ∈NP. Conversely, if L ∈NP, then there exists a nondeterministic polynomial-time Turing ma-chine that decides L. Then, deﬁne V(x, y) to accept iff y encodes an accepting computation history of N on x. ⊠ 10The analogy isn’t perfect; the diagonalization proof that there are recognizable, but undecidable, languages doesn’t map across the analogy. 43 Thus, we can say that L ∈NP iff there are polynomial-length proofs for membership in L (these proofs come from computation histories). One can look at 3SAT and SAT, which are hugely important (on the order of multiple Turing awards given for progress made, even in practical areas). Another way to think of NP is that it is easy to verify “yes” answers, but not necessarily “no” answers (just like recognizability). Thus, SAT ∈NP, because if π ∈SAT, there is a short proof given the correct variables to satisfy the expression, and verifying that proof can be done quickly. Proving that something is not satisﬁable, though, is harder; it doesn’t seem possible to do better than trying all 2n variable combinations, which takes exponential time. So SAT ∈NP. The Hamiltonian path problem is another example, coming from the very large class of problems of ﬁnding something small in something larger. It’s very useful in lots of settings, e.g. printing circuit boards. The problem itself is to, given a graph, determine whether a Hamiltonian path (i.e. a path that traverses every node exactly once) exists in it. Making this more formal, assume some reasonable encoding of graphs (e.g. the adjacency matrix); then, deﬁne HAMPATH = {(G,s,t) \| G is a directed graph with a Hamiltonian path from s to t}. Theorem 14.1. HAMPATH ∈NP. Intuitively, given a graph and a purported Hamiltonian path, it is possible to verify that it actually is a Hamiltonian path in polynomial time. Another graph-theoretic problem is the k-clique problem. A k-clique on a graph is a com-plete subgraph of k nodes (e.g. ﬁve friends on a social network who all know each other). This problem can be formalized as CLIQUE = {(G,k) \| G is an undirected graph with a k-clique}. Theorem 14.2. CLIQUE ∈NP. Proof. A k-clique in G is certainly a proof that (G,k) ∈CLIQUE, and given a subset S of k nodes from G, it’s possible to efﬁciently check that all possible edges are present in G between the nodes of S. ⊠ Notice that there’s no formalism of Turing machines in these proofs; it’s perfectly ﬁne to use these higher-level methods to prove things. Once again: P is the set of problems that can be efﬁciently solved, and NP is the set of problems where solutions can be efﬁciently veriﬁed. So, the question on everyone’s tongues: is P = NP? Heuristically, this would mean that problem solving can be automated in some sense; every problem that can be efﬁciently veriﬁed can be efﬁciently solved. This is (as is unlikely to be a surprise) an open question. Clearly, at least P ⊆NP. But if you can get the other direction, it’s a Millennium Prize problem, meaning a correct solution is worth $1,000,000. It might even be possible to use it to check the existence of short proofs of the other open problems. . . Notice that for ﬁnite automata, these are equal (i.e. veriﬁcation and problem-solving are the same), and for Turing machines, recognzability and decidability are known to be different notions. So complexity theory lies somewhere between automata theory and computability theory. The general academic consensus is that it’s probably the case that P ̸= NP. This is because if P = NP, then: • Mathematicians might be out of a job; while provability is in general undecidable, short proofs (no more than quadratic or cubic in the length of the theorem) are a 44 question of complexity theory rather than computability theory. Verifying a short proof is a polynomial-time problem, so it’s in NP, so if P = NP, then one could construct an algorithm to generate short proofs (assuming the constant multiplier isn’t completely atrocious). Note that if the proof is sufﬁciently nonconstructive, it might not be possible to extract an algorithm. • Cryptography as we know it would be impossible; speciﬁcally, a one-way function is a function which is easy to compute but hard to invert (i.e. given a y, it’s hard to ﬁnd a x such that f (x) = y, but given an x, it’s easy to calculate f (x)). Clearly, this requires P ̸= NP. • There are grander conjectures of implications to efﬁciently and globally optimizing every aspect of daily life, though these seem farther aﬁeld. It’s also possible that this question is undecidable, and people have asked this, but right now the techniques complexity theorists have just don’t work for it. Maybe in 40 years someone should try again. Polynomial-Time Reducibility. Deﬁnition 14.3. f : Σ∗→Σ∗is a polynomial-time computable function if there is a polynomial-time Turing machine M such that for every input w, M halts with f (w) on its tape. This looks just like the previous deﬁnitions of reducibility, replaced with the condition on M. Deﬁnition 14.4. A language A is polynomial-time reducible to another language B, denoted A ≤P B, if there is a polynomial-time computable f : Σ∗→Σ∗such that w ∈A iff f (w) ∈B. Then, one says f is a polynomial-time reduction from A to B. In the above case, there is a k such that \|f (w)\| ≤\|w\|k. Theorem 14.5. If A ≤P B and B ≤P C, then A ≤P C. The proof is a little more involved, but boils down to the fact that the composition of two polynomials is a polynomial. Theorem 14.6. If A ≤P B and B ∈P, then A ∈P. Proof. let MB be a polynomial-time Turing machine that decides B and f be a polynomial-time reduction from A to V. Then, build a polynomial-time decider for A that computes f on its input w, and then runs B on f (w). ⊠ Theorem 14.7. If A ≤P B and B ∈NP, then A ∈NP. The proof is the same as the proof for Theorem 14.6, but with nondeterministic polynomial-time deciders. One way to understand this class of languages better is to understand its maximal elements under ≤P. Deﬁnition 14.8. • If every A ∈NP is polynomial-time reducible to B (i.e. A ≤P B), then B is called NP-hard. • If B ∈NP and B is NP-hard, then B is called NP-complete. 45 In computability theory, we saw that for every recognizable language L, L ≤m ATM; this can be said as ATM is complete for all recognizable languages. Note that if L is NP-complete and L ∈P, then P = NP, and if L ̸∈P, then P ̸= NP, of course. Thus, the whole problem boils down to these NP-complete problems, and makes this whole theory more compelling. Suppose L is NP-complete and P ̸= NP; then, L isn’t decidable in time nk for every k. The following theorem seems speciﬁc, but is very useful for other problems in complexity theory or elsewhere, e.g. by reducing a general problem (e.g. bug ﬁnding) to a SAT problem, and then using a SAT solver to deal with it. Theorem 14.9 (Cook-Levin). SAT and 3SAT are NP-complete. Proof. First, 3SAT ∈NP, because a satisfying assignment is a “proof” that a solution exists. Then, why is every language A ∈NP polynomial-time reducible to 3SAT? For any A ∈NP, let N be a nondeterministic Turing machine deciding A in nk time. The idea is to encode the entire tree of possible paths of execution of N as a 3cnf-formula with length polynomial in n, but the naïve way of doing this (encoding the entire tree) has exponential size.11 Deﬁnition 14.10. • If L(N) ∈NTIME(nk), then a tableau for N on w is an nk × nk table whose rows are conﬁgurations of some possible computation history of N on w. Since one can use at most nk space in nk time, then having only nk columns (each the next square of tape) is reasonable. • A tableau is accepting if some row of the tableau has an accept state. In this case, all rows after that row are taken to preserve this accept state. • Each of the (nk)2 entries in the tableau is called a cell; the cell in row i and column j is denoted cell[i, j], and is the jth symbol in the ith conﬁguration. Then, the idea is to, given w, construct a 3cnf formula φ describing all of the logical constraints that every accepting tableau for N on w must satisfy; this φ will be satisﬁable iff there is an accepting tableau for N on w. (There are lots of proof of this theorem, but this one seems easier to understand, and is in Sipser besides.) Let C = Q ∪Γ∪{#}. Then, for a given tableau, for 1 ≤i, j ≤nk and s ∈C, let xi,j,s be the Boolean variable that cell[i, j] = s. These xi,j,s will be the variables for the φ that we wish to construct, i.e. cell[i, j] = s iff xi,j,s = 1. φ will be given by four subformulas: φ = φcell ∧φstart ∧φaccept ∧φmove, given as: • φcell checks that for all i and j, there is exactly one s ∈C such that xi,j,s = 1. • φstart checks that the ﬁrst row of the table is the start conﬁguration of N on input w. • φaccept checks that the last row of the table has an accept state. • φmove checks that every row is a conﬁguration that legally follows from the previous conﬁguration. More explicitly, φcell = ^ 1≤i,j≤nk    Ã _ s∈C xi,j,s ! ∧    ^ s,t∈C s̸=t (¬xi,j,s ∨¬xi,j,t)      . 11Notice that this strategy worked for automata: you could encode all possible paths of an NFA into a DFA. 46 These aren’t technically 3cnf, but can be easily converted into them (clearly, they’re general satisﬁability problems). The last, hardest one is φmove, which needs to show that every 2×3 “window” of cells in the tableau is legal, i.e. consistent with the transition functions of N. This is sufﬁcient because in a single step, a Turing machine can only move a small distance (i.e. 1). For example, if δ(q1,a) = {(q1,b,R)} and δ(q1,b) = {(q2, c,L),(q2,a,R)}, then a q1 b q2 a c is a legal window. The proof will be continued next lecture. 15. NP-COMPLETE PROBLEMS: 2/27/14 “NP-complete problems are like the zombies of computer science. They’re everywhere, and they’re really hard to kill.” . . . though we’re still looking at the proof that 3SAT is NP-complete, let alone anything else. We needed to construct φmove, by looking at every 2×3 window and checking that it’s consistent with the transition function. Then, the key lemma is that if it’s consistent locally, then it’s consistent globally: if every window is legal and the top row is the start conﬁguraton, then each row is a conﬁguration that yields the next row. This is proven by induction on the number rows: if one row doesn’t yield the rest, then there must be some illegal window in the last row. Finally, we can deﬁne φmove as φmove = ^ 1≤i≤nk−1 1≤j≤nk−2 W(i, j), where W(i, j) is the formula that the window whose upper left corner is at position (i, j) is legal, i.e. W(i, j) = _ legal windows (a1,...,a6) (xi,j,a1 ∧···∧xi+1,j+2,a6), but this isn’t in 3cnf form, so you have to take its complement, which turns the outside ∧ into a ∨, which is OK. It’s also necessary to convert longer strings into shorter ones: a1∨a2∨···∨at can be written in 3cnf by introducing extra variables zi. Once this is done, there are O(nk) or O(n2k) clauses in each of these formulas, so φ has O(n2k) clauses and was generated efﬁciently from A. ⊠ An alternate proof deﬁnes a problem CIRCUIT-SAT, which is easier to reduce to SAT, but can also be shown to be NP-complete. Is it possible that 3SAT can be solved in O(n) time on a multitape Turing machine? It seems unlikely, because this would not just imply P = NP, but that there would be a “dream machine” that cranks out short proofs of theorems, quickly optimize a lot of things, and in general produce quality work from a formula for recognizing it. Nonetheless, this is an open question! There are thousands of NP-complete problems, within every topic of CS and even within the other sciences too, and if one were determined in either way, then all of them would. These are huge within theoretical computer science. 47 Corollary 15.1. SAT ∈P iff P = NP. Given some problem Π ∈NP, one now has a simpler recipe to show that languages are NP-complete: (1) Take a problem Σ known to be NP-hard (i.e. 3SAT). (2) Construct a polynomial-time reduction Σ ≤P Π. For example, we deﬁned CLIQUE, the clique problem on graphs. Theorem 15.2. CLIQUE is NP-complete. Proof. This proof will show 3SAT ≤P CLIQUE, i.e. efﬁciently transform a 3cnf formula into a graph with the speciﬁc clique properties. Let k be the number of clauses of φ, and construct a graph G with k groups of 3 nodes each; intuitively, the ith group of nodes corresponds to the ith clause Ci of φ, and join nodes v and v′ by an edge if they aren’t in the same group and they don’t correspond to opposite literals (i.e. xi and ¬xi). Given an SAT assignment A of φ, then for every clause C there is at least one literal set true by A (since the entire thing needs to come out as true). For every clause C, let vC be a vertex in G corresponding to a literal of C satisﬁed by A. Then, there is a k-clique S = {vC : C ∈φ}, because if (vC,vC′) ̸∈E, then vC and v′ C label in-consistent literals, but that means that A can’t satisfy both of them, which is a contradiction. Hence, S is a k-clique, so (G,k) ∈CLIQUE. In the reverse direction, suppose (G,k) ∈CLIQUE was given by φ, so the goal is to construct a satisfying assignment A for φ. Let S be an m-clique of G; then, S contains one variable from each group (since there are no edges between two things of the same group). Now, for each variable x of φ, make an assignment A where x = 1 iff there is a vertex v ∈S with label x. Thus, A satisﬁes at least one literal in the ith clause of φ, so it is a satisfying assignment. ⊠ Related to the clique problem is the independent-set problem IS, which, given a graph G = (V,E) and an integer k, is there an independent set of size k in the graph, i.e. a subgraph with no vertices of size at least k? Clearly, CLIQUE ≤P IS, by the algorithm that takes in a graph and produces its complement, so IS is NP-complete. Deﬁnition 15.3. A vertex cover of a graph G is a set of nodes C such that every edge borders some node in C. For example, in the following graph, the nodes marked c form a vertex cover: c c These have applications to networks of all kinds. Clearly, the entire graph is its own vertex cover, but we want a more minimal example. Then, VERTEX-COVER is the problem that, given a graph G and an integer k, determines whether there is a vertex cover of size \|G\|−k. This is a minimization problem, in the sense that the other graph problems here have been maximization problems. Then, IS ≤P VERTEX-COVER: 48 Claim 15.4. For every graph G = (V,E) and subset S ⊂V, S is an independent set iff V \ S is a vertex cover. Proof. S is an independent set iff for all u,v ∈S, (u,v) ̸∈E, so if (u,v) ∈E, then one of u or v is not in S. Thus, V \ S is a vertex cover of size \|G\|−k. ⊠ Now, one can create the polynomial-time reduction (G,k) 7→(G,\|G\|−k), so VERTEX-COVER is NP-complete. But there are also NP-complete problems out of number theory, which look completely different and illustrate how deep the notion of NP-completeness is. The subset-sum problem is given a set S = {a1,...,an} ⊂N and a t ∈N. Then, the question is whether there exists an S′ ⊆S such that t = X i∈S′ ai = t. Then, SUBSET-SUM is the problem as to whether such a S′ exists, given S and t. The following result was proven in CS 161, using dynamic programming. Theorem 15.5. There is an algorithmm polynomial in n and t, that solves SUBSET-SUM. Yet since t can be speciﬁed in logt bits, this isn’t polynomial in the length of the input! It will be possible to reduce VERTEX-COVER ≤P SUBSET-SUM by reducing a graph to a speciﬁc set of numbers. Given (G,k), let E = {e0,..., em−1} and V = {1,...,n}. Then, the subset-sum instance (S,t) will have \|S\| = m+ n, given by two clases of numbers: • The “edge numbers:” if e j ∈E, then add b j = 4j in S. • The “vertex numbers:” for each i ∈V, put ai = 4m + X j:i∈e j 4j in S. Then, let t = k ·4m + m−1 X j=0 (2·4j). Claim 15.6. If (G,k) ∈VERTEX-COVER, then (S,t) ∈SUBSET-SUM. This is because if C ⊆V is a vertex cover with k vertices, then let S′ = {ai : i ∈C}∪{b j : \|e j ∩C\| = 1} ⊆S. Then, the sum of the numbers in S′ will be equal to t. The way to intuit this is to think of these numbers being vectors in “base 4,”12 i.e. 4j = b j ∼(0,...,0,1,0...,0), and ai is also a binary vector with a 1 in the jth slot wherever vertex i is contained in edge j (and a 1 in the top slot). Then, t = (k,2,2,...,2) (so not quite in base 4, but nearly so). Then, for each component of the vector, the sum over a vertex cover C is exactly 2 for the edges, because for any edge, either e j is hit once by C, in which the value is 2, or it’s hit twice by C, in which case you still get 2. Then, throwing in the vertices implies that the subset-sum solution exists. 12Or maybe in Z/4Z? 49 In the opposite direction, if (S,t) ∈SUBSET-SUM, I want to show that (G,k) ∈VERTEX-COVER. But suppose C ⊆V and F ⊆satisfy X i∈C ai + X e j∈F b j = t; then, C is a vertex cover of size k. Once again, this follows because the numbers can be thought of as vertices in base 4, so C must have size k, because each vertex contributes a 1 to the ﬁrst component, but the ﬁrst component of t is k. There’s a whole wing of people on the 4th ﬂoor of Gates who study the subset-sum problem (cryptographers), but, oddly enough, they don’t really care about solving vertex-cover, even though the problems are the same from this viewpoint. If G is a graph and s and t are nodes of G, we can ask the shortest (longest) path problem: is there a simple path of length less than k (resp. more than k) from s to t? Using Dijkstra’s algorithm, the shortest-path problem is in P, but the longest-path problem ends up in NP. 16. NP-COMPLETE PROBLEMS, PART II: 3/4/14 Related to the subset-sum problem is something in economics called the knapsack problem, in which one is given a cost budget C, a proﬁt target P, and a set S = {(p1, c1),...,(pn, cn)} of pairs of positive integers.13 Then, the goal is to choose items out of S such that their costs (ﬁrst item of the pair) are no more than C, but their proﬁts (second entry) are higher than P. In some sense, what’s the greatest utility I can get out of stufﬁng heavy items into a knapsack? Theorem 16.1. KNAPSACK is NP-complete. Proof. This will be pretty easy after reducing from SUBSET-SUM. First, KNAPSACK ∈NP, because it has a pretty reasonable polynomial-time veriﬁer (in the same was as everything else we’ve done these few days). Then, it will be possible to reduce SUBSET-SUM ≤P KNAPSACK as follows: given an instance (S = {a1,...,an},t) of SUBSET-SUM, create an instance of KNAPSACK by setting (pi, ci) := (ai,ai) for all i, and then S′ = {(p1, c1),...,(p,cn)}. Finally, let P = C = t. ⊠ Another somewhat economic problem is the partition problem: given a set S = {a1,...,an} ⊂ N, is there an S′ ⊆S such that X i∈S′ ai = X a∈S\S′ ai? In other words, PARTITION is the problem as to whether there is a way to partition S “fairly” into two parts. Theorem 16.2. PARTITION is NP-complete. Proof. This seems intuitively easier than SUBSET-SUM, since there’s no ﬁxed target to reach. However, a reduction still exists. Clearly, there’s a polynomial-time veriﬁer for this problem, but we can reduce SUBSET-SUM ≤P PARTITION: given a set S = {a1,...,an} ⊂N and a target t, output T = {a1,...,an,2A −t, A + t}, where A = P i∈S ai. Now, we want to show that there’s a solution to (S,t) ∈SUBSET-SUM iff there’s a partition of T. The sum of all of the numbers in T is 4A, so T has a partition iff there exists a T′ ⊆T 13Sometimes, one can take fractional amounts of each item, in which case this becomes a linear programming problem. 50 summing to 2A. Thus, intersecting it with {a1,...,an} yields a solution to the subset-sum problem, because exactly one of {2A −t.A + t} ∈T′ (or else there would be too much there). Thus, what remains, i.e. T′ {2A −t} is a subset of S if that’s in T′, and if A + t ∈T′, then instead take T \ T′, which has the same sum (since T′ is a solution of T) and contains only 2A −t and elements of S. In the reverse direction, if (S,t) ∈SUBSET-SUM, then one can take a subset S′ that sums to t, and add A −t to it to get a partition. ⊠ Right now, we have the following chain of reductions, where A →B indicates that A ≤P B. SAT / 3SAT / CLIQUE IS PARTITION VERTEX-COVER / SUBSET-SUM O KNAPSACK Another problem is the bin-packing problem: given a set S = {a1,...,an} of positive integers, a bin capacity B, and a target K ∈N, can one partition S into K subsets such that each subset sums to at most B? One can imagine lots of applications of this to shippiing and to optimization. Sometimes, this is formally denoted BINPACKING. Claim 16.3. PARTITION ≤P BINPACKING, and in particular the latter is NP-complete. Proof. Clearly, bin packing is in NP, since it can be veriﬁed in polynomial time. But given an instance S = {a1,...,an} of the partition problem, it can be worded as a bin packing problem by creating two bins. Then, a solution for one exists iff there is a solution for the other. ⊠ Recall that last lecture we looked at the shortest-path and longest=path problems; the former was shown to be in P by Dijkstra’s algorithm. This can be reduced from HAMPATH (since if there is a Hamiltonian path, it’s the longest), which we already saw in Theorem 14.1 was in NP. Theorem 16.4. HAMPATH is NP-complete. Proof. The proof will reduce 3SAT ≤P HAMPATH. Speciﬁcally, let F = C1 ∧···∧Cn be a 3cnf formula, with each ci = xi ∨xj ∨xk. Then, we can construct graphs Gi that look like this: xi ( -( h ··· ( h ( h h q xi+1 51 Speciﬁcally, there are 3m+1 nodes in the second row, corresponding to allowable paths. Then, let G be the union of these graphs (joining the node xi from Gi−1 to xi in Gi). Add some more edges to G: speciﬁcally, add a node for each of c1,..., cm, and in the second row, consider each pair of arrows from left to right to correspond to C1,...,Cm, connecting (say) nodes c j,i and c j,f for each j. Then, if clause j contains xj, add arrows from c j,i →C j →c j,f , and if it contains ¬xj, add arrows c j,f →C j →c f ,i; thus, in some sense, truth represents heading to the right along the second row, and falsity to the left. Then, F is satisﬁable iff G has a Hamiltonian path; now that the graph is set up, this is some amount of tracing through the logic. ⊠ Eew. . . incredibly, this this is the nicest reduction the professor knows of, from any NP-complete problem. Maybe it’s possible to do better. It at least has the property that it’s agnostic to the variables. Don’t think in terms of “let’s suppose I have a solution.” Thus, the longest-path problem is also NP-hard. coNP and Friends. There’s another class coNP = {L \| ¬L ∈NP}. This also looks like nondeterminism, but with the following twist: a co-nondeterministic machine has multiple paths of computation, but the machine accepts if all paths reach an accept state, and rejects if at least one path rejects. For the same reason that P ⊆NP, P ⊆coNP: every deterministic machine can be seen as a co-nondeterministic one. However, NP and coNP are a little more interesting: it’s believed they aren’t equal, but this is a very open problem. Of course, P = coP, since a deterministic decider for A can be turned into one for ¬A by ﬂipping the accept and reject states. One can draw Venn diagrams relating these sets, though there are so many aspects of these diagrams that are open problems. Deﬁnition 16.5. A language is coNP-complete if: (1) B ∈coNP, and (2) For every A ∈coNP, A ≤P B (i.e. B is coNP-hard). Let’s look at ¬SAT (sometimes called UNSAT), the set of Boolean formulas φ such that no variable assignment satisﬁes φ. Theorem 16.6. ¬SAT is coNP-complete. Proof. Since SAT ∈NP, then ¬SAT ∈coNP. To show that it’s coNP-hard, use the Cook-Levin theorem. Let A ∈coNP. Then, on input w, transform w into a formula φ using the reducer for ¬A ≤P SAT (guaranteed by Cook-Levin); then, w ∈A iff φ ∈¬SAT. ⊠ We also have the set of tautologies: TAUT = {φ \| ¬φ ∈¬SAT} (i.e. things satisﬁed by every assignment of variables). TAUT is coNP-complete, by a similar line of reasoning to the above theorem. Another worthwhile question is whether P = NP∩coNP. In computability theory, the anal-ogous statement (i.e. a language is decidable iff it is both recognizable and co-recognizable) is true. The idea is that if this is true, then short proofs of positive and negative answers imply it’s possible to solve these kinds of problems quickly. Of course, this is an open problem, and one direction would imply P = NP. However, there are interesting problems in NP∩coNP, e.g. prime factorization! One can formalize this into a decision problem by asking for the set of integers (m,n) (each greater than 1) such that 52 there is a prime factor p of m greater than n. But if this factoring problem is in P, then most public-key cryptography would be breakable, e.g. RSA. However, we can say that factoring is in NP∩coNP. Deﬁning PRIMES to be the set of prime numbers, it’s rather nontrivial and interesting to show the following. Theorem 16.7 (Pratt). PRIMES ∈NP∩coNP. In fact, PRIMES ∈P, which was open for a while, and shown relatively recently. Theorem 16.8. FACTORING ∈NP∩coNP. Proof. The idea is that it’s easy to check the prime factorization of a number: given a factorization pe1 1 ··· pek k of m, one can efﬁciently check that each pi is prime, and then whether they multiply together to make m. ⊠ Note that several of these problems, including prime factorization, rely on quantum computers, which aren’t strictly deterministic, nor quite nondeterministic. This class of problems is called BQP, and there are lots of open questions about it too. 17. POLYTIME AND ORACLES, SPACE COMPLEXITY: 3/6/14 “Let’s talk about something a little more down-to-Earth, space complexity.” Last time, we saw that TAUT ∈coNP, so if a formula isn’t a tautology (a tautology is a formula that is always satisﬁable for every assignment), then there is a short proof (i.e. an assignment that doesn’t work). However, whether the complement is true (do there exist short proofs that something’s a tautology?) is open, i.e. whether coNP = NP. Thus, TAUT is coNP-complete, as we saw last time. You can play this game with lots of coNP problems. We saw yet another open question, as to whether P = NP∩coNP. If this is true, then there are unfortunate implications for the RSA algorithm for cryptography. NP-complete problems lead to coNP-problems, e.g. ¬SAT, ¬HAMPATH. But nobody knows if there are any problems that are (NP∩coNP)-complete! Intuitively, P, NP, coNP, and so on can all be deﬁned with speciﬁc machine models. For every machine, we can get an encoding, thanks to the framework of computabulity theory. But NP∩coNP is just the intersection of two complexity classes, so there’s no easy description that we know of for these sorts of problems. Oracle Turing Machines. Recall from computability theory we have the notion of oracle Turing machines, which can magically answer queries to some problem; then, one asks what else one could solve with this power. The key to using them here is that such oracle Turing machines can ask their oracles in polynomial time (in fact, in one step). Deﬁnition 17.1. For some language B, the complexity class PB is the set of languages that can be decided by a polynomial-time Turing machine with an oracle for B. For example, we might want to know what can be solved in polynomial time with a SAT solver, which would be PSAT, and PNP is the class of language decidable by some polynomial-time Turing machine, with an oracle for some B ∈NP. But of course, since SAT is NP-complete, then PSAT = PNP: using the Cook-Levin theorem, for any B ∈NP, an oracle for B can be simulated in polynomial time by an oracle for SAT. Maybe the polynomial is slightly larger, but it’s still polynomial. 53 If B ∈P, then PB = P, because an oracle for B can be straight-up simulated in polynomial time. More interesting question: is NP ⊆PNP? This is also true: for any L ∈NP, one can create an oracle for L, and then just ask it. We also have coNP ⊆PNP, for the same reason: the oracle decides problems in NP, and therefore also decides coNP: for any L ∈coNP, one can create an oracle for ¬L, and then ﬂip its answer. This corresponds to the notion that the complements of recognizable languages can be decided by an oracle for the halting problem, even though the halting problem isn’t co-recognizable. In general, this implies that PNP = PcoNP, even though we don’t know if NP = coNP! This complexity class has the very natural deﬁnition of the class of problems decidable in polynomial time if SAT solvers work, and this is something which causes it to appear in a reasonable number of applications.14 A typical problem in PNP looks like FIRST-SAT, the set of pairs (φ, i) such that φ ∈SAT and the ith bit of the lexicographically ﬁrst15 satisﬁable assignment of φ is 1. One can compute the lexicographically ﬁrst satisfying assignment using only polynomially many calls to a SAT oracle, in a manner similar to this week’s homework (i.e. reducing the problem, one variable at a time). But notice that this problem isn’t obviously in either NP or coNP; it isn’t thought to be, but it might. In the same manner, one can deﬁne NPB and coNPB, which are what you would expect (languages decidable by a nondeterministic or co-nondeterministic Turing machine with an oracle for B in polynomial time, respectively). Now we can start asking some interesting-looking questions: is NP = NPNP? How about coNPNP = NPNP? These are both open questions, even if one assumes that P ̸= NP, but the general belief is that both of these are untrue. In fact, there are heuristic arguments that something is unlikely by reducing it to one of these equalities! In the same way as before, NPNP = NPcoNP, because an oracle for an L ∈NP can be turned into an oracle for ¬L ∈coNP and vice versa in polynomial time. There are very natural problems in some of these seemingly abstract complexity classes. For example, here’s one in coNPNP. Deﬁnition 17.2. • Two Boolean formulas φ and ψ under the same number of variables are equivalent if they have the same value on every assignment to the variables, e.g. x and x∨x. • A Boolean formula is minimal if every equivalent formula has at least as many symbols as it (i.e. ∧, ∨, and ¬). This is a very useful problem in lots of ﬁelds. Let EQUIV = {(φ,ψ) \| φ and ψ are equivalent}, and MIN-FORMULA denote the set of minimal formulas. Theorem 17.3. MIN-FORMULA ∈coNPNP. Proof. Notice that EQUIV ∈coNP, because if two formulas aren’t equivalent, then there’s a short proof of that. In fact, it’s coNP-hard, because it can be reduced to TAUT (is φ equivalent to the trivial tautology?). Thus, EQUIV can be decided by an oracle for SAT. 14One might ask whether PNP = NP∪coNP. This is an open problem, but the answer is believed to be no. 15Viewing the assignment as a binary string, with one bit for each variable, then this is the dictionary order 0n < 0n−11 < ··· < 1n. 54 Here’s a coNPEQUIV Turing machine that efﬁciently decides MIN-FORMULA, given an input formula φ: for every formula ψ such that ¯ ¯ψ ¯ ¯ < ¯ ¯φ ¯ ¯, determine whether (φ,ψ) ∈EQUIV. If so, reject; then, after testing all of these, accept. ⊠ However, we believe this problem isn’t in NP or coNP, so it (probably) lies within interest-ing intersections of complexity classes. Note that the time hierarchy still exists when one adds in an oracle. This suggests that oracles can be used to learn about the relationship between P and NP, but it’s not as useful as one might hope: Theorem 17.4. (1) There is an oracle B for which PB = NPB. (2) There is an oracle A for which PA ̸= NPA. (1) is pretty easy: just take B to be an oracle for the halting problem. But the second part is much trickier; see Sipser, §9.2. Space Complexity. Space complexity seems less natural than time complexity,16 though it has an interesting theory and useful applications, even touching on some aspects of game theory. Deﬁnition 17.5. • Let M be a deterministic Turing machine that halts on all inputs. Then, the space complexity of M is the function f : N →N, where f (n) is the index of the furthest tape cell reached by M on any input of length n. • Let M be a nondeterministic Turing machine that halts on all inputs. Then, the space complexity of M is the function f : N →N, where f (n) is the index of the furthest tape cell reached by M on any input of length n and any computation path through M. Basically, what’s the largest amount of space that this algorithm can take? Deﬁnition 17.6. • SPACE(s(n)) is the set of languages L decided by a Turing machine with O(s(n)) space complexity. • NSPACE(s(n)) is the set of languages L decided by a nondeterministic Turing machine with O(s(n)) space complexity. The idea is that SPACE(s(n)) formalizes the class of problems solvable by computers with bounded memory. This sounds very much like problems we face in the real world. One important question is how time and space relate to each other in computing. Going from time to space isn’t too bad, but SPACE(n2) problems could take much longer than n2 time to run, because, in some sense, you can reuse space, but not time. Even SPACE(n) is very powerful; for example, 3SAT ∈SPACE(n), because one can try every single assignment within linear space, and thus decide 3SAT (albeit in exponential time). If M is a halting, non-deterministic Turing machine which on an input x uses S space on all paths, then the number of time steps is at most the total number of conﬁgurations, 16Though I suppose if you’re traveling at relativistic speeds they’re the same thing. 55 because if a conﬁguration repeats, then the machine loops. But since a conﬁguration can be written in O(s(n)) bits, then this takes time at most 2O(S). In other words, SPACE(s(n)) ⊆ [ c∈N TIME(2c·s(n)). In fact, it’s also the case that NSPACE(s(n)) ⊆ [ c∈N TIME(2c·s(n)). This can be shown by viewing it as a graph problem, where the nodes are the conﬁgurations of a given Turing machine M on its given input x, and an edge from u →v implies that the conﬁguration u can yield conﬁguration v in one step. Now, this is just a graph reachability problem of size 2O(S), which can be solved in O(n2) time. Deﬁnition 17.7. The notions of efﬁcient computation and veriﬁcation are PSPACE = [ k∈N SPACE(nk) and NPSPACE = [ k∈N NSPACE(nk). Notice that P ⊆PSPACE, and that NP ⊆PSPACE as well, because we saw that 3SAT ∈ PSPACE. In fact, every complexity class we’ve talked about so far is contained in PSPACE, even the exotic ones like coNPNP. Things do get weirder: the complexity class RLOGSPACE of problems that can be solved by a randomized algorithm with high probability in logarithmic space is believed to be equal to SPACE(logn), though it’s an open problem. This is more generally the case: we have randomized classes RP versus P and so on, and they seem like they should be equal, but we aren’t certain. 18. SPACE COMPLEXITY, SAVITCH’S THEOREM, AND PSPACE: 3/11/14 “Every time I hear about it, the thesis gets shorter.” Recall that we were talking about the space complexity of a Turing machine M, the function f (n) that on n returns the largest amount of space (i.e. tape) M takes when running on all inputs of length n. We also saw that computations done in S(n) space take at most 2O(S(n)) time, because of a bound on the number of conﬁgurations, and that in a slightly more complex proof, this is also true of nondeterministic computations. It’s true that NPNP ⊆PSPACE, because an oracle for NP can be simulated in polynomial space (even if it takes a lot of time); instead of nondeterministically guessing something, loop through all possible choices (more concretely, this could involve reducing via SAT or something, and testing all possible assignments, which is deterministic, uses exponential time, and uses polynomial space). Thus, this takes polynomial space, and the original (non-oracle) part can be simulated in the same way. In general, reasoning about these can be more tricky; for example, P ̸= SPACE(O(n)), but this proof is somewhat involved. Since a deterministic Turing machine M that halts on all inputs using at most s(n) space has an upper bound on running time of s(n)\|Q\|\|Γ\|s(n) = 2O(s(n)), then it makes sense to deﬁne EXPTIME = [ k∈N O ³ 2nk´ . Then, it’s not too bad to show (sort of like we did above) that PSPACE ⊆EXPTIME and NPSPACE ⊆EXPTIME. Thus, we know that P ⊆NP ⊆PSPACE ⊆EXPTIME, and by the Time Hierarchy theorem, P ̸= EXPTIME, since P ⊊TIME(2n). But which of the three 56 inclusions is proper? That’s also open. Or is more than one true? Proving any of these would win a Turing award. . . Similar to the Time Hierarchy theorem, there’s a space hierarchy, but it’s much more pow-erful; we saw that TIME(nloglogn) = TIME(O(n)) (which is the class of regular languages), but SPACE(nloglogn) ⊋SPACE(O(n)). Theorem 18.1 (Space Hierarchy). For any unbounded function a : N →N and “reasonable” functions s : N →N, SPACE(s(n)) ⊊SPACE(a(n)s(n)). Proof idea. This is basically the same proof as of Theorem 13.7, using diagonalization and the fact that the halting problem is unrecognizable. One really surprising result shows that PSPACE = NPSPACE, among other things. Theorem 18.2 (Savitch). For functions s(n) where s(n) ≥n, NSPACE(s(n)) ⊆SPACE(s(n)2). Proof. Given a nondeterministic Turing machine N with space complexity s(n), one might construct a deterministic machine M that tries every possible branch of s(n). But this is too much data to keep track of, as there are 22s(n) branches to keep track of, so one needs at least that much space. The intuition here is to look at breadth-ﬁrst and depth-ﬁrst searching algorithms. So we’ll need a different algorithm. Given conﬁgurations C1 and C2 of an s(n)-space (nondeterministic) Turing machine N, and a number t = 2k, the goal is to determine whether N can get from C1 to C2 within t steps. This procedure is called SIM(C1,C2,t). • If t = 1, then accept if C1 = C2, or C1 yields C2 within one step, which takes space O(s(n)). • More interesting is the inductive step, which guesses the midpoint of the path between C1 and C2. Speciﬁcally, for all conﬁgurations Cm using s(n) space, run SIM(C1,Cm,t/2) and SIM(Cm,C2,t/2), and accept if they both accept. • If no such Cm leads to existence, then reject. This algorithm has O(logt) levels of recursion, and each level of recursion takes O(s(n)) space. But we only need to keep track of the one path through everything, so failed guesses for the midpoint can be overwritten. Thus, the total space needed is O(s(n)logt) space. Now, let d > 0 be the number such that the maximum running time of N(w) on any computation path is at most 2ds(\|w\|), and let Cacc be the unique accepting conﬁguration on inputs of length \|w\|.17. Now, simulate N by M, which on input w runs SIM(q0w,Cacc,2ds(\|w\|)) and accepts iff it does. This can be calculated to take O(s(n)2) space, since logt = O(S(n)2). ⊠ Another surprising result is the existence of PSPACE-complete problems. These offer general strategies for solving problems in PSPACE, and intuitively are a lot broader than NP-complete problems and such. Deﬁnition 18.3. A language B is PSPACE-complete if: (1) B ∈PSPACE, and (2) For any A ∈PSPACE, A ≤P B. 17This hasn’t really been discussed yet, but can be done without loss of generality by, for example, clearing the tape before accepting. 57 Deﬁnition 18.4. A fully quantiﬁed Boolean formula is a Boolean formula where every variable is quantiﬁed (i.e. there is an ∀or an ∃for each variable) These formulas are either true or false, and are much more general than SAT or ¬SAT e.g. ϕ ∈SAT iff ∃x1 ···∃xn[ϕ], and ϕ ∈¬SAT iff ∀x1 ···∀xn[¬ϕ] is true. For example, some more such fully quantiﬁed Boolean formulas are ∃x∃y[x∨¬y], ∀x[x∨¬x], and ∀x[x]. Let TQBF denote the set of true fully quantiﬁed Boolean formulas. Theorem 18.5 (Meyer-Stockmeyer). TQBF is PSPACE-complete. Proof. That TQBF is in PSPACE isn’t yet obvious, so consider the following algorithm QBF-SOLVER: (1) If φ has no quantiﬁers, then it is an expression with only constants, so it can be straightforwardly evaluated. Accept iff φ evaluates to 1. (2) If φ = ∃xψ (it’s existentially quantiﬁed), recursively call this algorithm on ψ twice, ﬁrst with x set to 0, and second with x set to 1. Accept iff at least one of these calls accepts. (3) If φ = ∀xψ (it’s universally quantiﬁed), then recursively call this algorithm twice, ﬁrst with x set to 0, and second with x set to 1. Then, accept iff both of these calls accept. What’s the space complexity of this algorithm? In the same way as for the proof of Savitch’s theorem, it’s possible to reuse space, because the algorithm only needs to store one path through the tree of recursion at once, so if \|φ\| = K, then this takes O(K · n) ≤O(K2) (where there are n variables). Now, why is TQBF PSPACE-hard? This will use tableaux in the same way as the Cook-Levin theorem, but will need to be modiﬁed in order to work properly: the algorithm used above creates a tableau that has width nk and height 2O(nk), which s a priori too large. However, a cleverly designed φ ∈TQBF will do the trick, i.e. be true iff M accepts w. Let s(n) = nk and b ≥0 be an integer. Using two collections of s(n) Boolean variables denoted c and d representing two conﬁgurations and an integer t ≥0, construct the quantiﬁed Boolean formula φc,d,t which is true iff M starting in conﬁguration c reaches conﬁguration d in at most t steps. Then, set φ = φstart,cacc,h, where cacc is the unique accept state and h is the exponential time bound. Similarly to Savitch’s theorem, φc,d,t claims that there is a c′ such that φc,c′,t/2 and φc′,d,t/2 are both true. In the base case, φc,d,1 means that c = d or d follows from c in a single step of M; the former is expressed by saying each of the bs(n) variables representing c are equal to the corresponding ones in d. To show that d follows from c in a single step of M, use 2×3 windows in the same way as in the Cook-Levin theorem, and write a cnf formula. Now, we can construct φc,d,t recursively for t > 1: φc,d,t = ∃m[φc,m,t/2 ∧φc,m,t/2]. Here, M = ∃x1∃x2 ···∃xS where S = bnk. But this is too long; every level of recursion cuts t in half, but doubles the size of the formula! So the length will be O(t), which is exponential. So far, the proof has only used existential quantiﬁers, so maybe we should introduce a ∀in there somewhere. Instead, modify the formula to be φc,d,t = ∃m∀x, y[((x, y) = (c,m)∨(x, y) = (m,d)) = ⇒φx,y,t/2]. 58 This folds the two recursive copies of φc,d,t into one. Then, size(s,t) = O(S(n))+size(s,t/2), so the total size is bounded above by O(s(n)logt) (the number of calls is logarithmic in t). This is much shorter, only quadratic in the space bound! ⊠ 19. PSPACE-COMPLETENESS AND RANDOMIZED COMPLEXITY: 3/13/14 There’s a general intuition that NP corresponds to ﬁnding optimal strategies in hard solitare games, and PSPACE corresponds to ﬁnding optimal strategies in two-player games. Accordingly, a recent paper shows that Candy Crush is NP-hard,18 and for formalizations of many popular two-player games, it’s PSPACE-complete to decide who has a winning strategy on a game board. For example, TQBF can be interpreted as a game between two players, E and A. Given a fully quantiﬁed Boolean formula, such as ∀x∃y[(x ∨y) ∧(¬x ∨¬y)], E and A alternate choosing values for variables, where E chooses for existential quantiﬁers and A for universal quantiﬁers. Then, E wins if the result is an assignment, and A wins if it’s not valid. Theorem 19.1. If FG is the set of formulas in which E has a winning strategy, then FG is PSPACE-complete. Proof. FG = TQBF, because if there’s an assignment to all of the existentially quantiﬁed variables such that all choices of universally quantiﬁed variables work, then that formula is in TQBF, and this choice is a winning strategy for E. ⊠ Another game is the geography game, where two players take turns naming cities from anywhere in the world, but each city chosen must begin with the same letter the previous city ended with, and cities can’t be repeated. For example, one might go from Austin to Newark to Kalamazoo to Opelika and so on. The game ends when one player can’t think of another choice of words. This looks like a nice Hamiltonian path problem. The generalized geography problem considers a directed graph and a start state a; Player 1 moves from here to some other state, and they Player 2 makes another move on the graph, and so on, subject to the constraint that they cannot revisit previously visited nodes. Then, game ends when one player has no moves, and thus loses. For example, in the following graph, Player 2 has a winning strategy of not doing anything. / Here’s a graph where Player 1 wins: / / 18Walsh, Toby. “Candy Crush is NP-hard.” 11 March 2014. 59 Now things can get more interesting: d $ b : $ g # / a : $ e : $ i { c : h z f d Who has a winning strategy here? Theorem 19.2. GG, the set of generalized-geography games in which Player 1 has a winning strategy, is PSPACE-complete. Proof. First, why is it even in PSPACE? Here’s an algorithm M accepting a graph G and a start node a. • If node a has no outgoing edges, then reject. • For all nodes a1,...,ak pointed to by a, recursively call M(G {a},ai). • If all of the calls accept, then reject; otherwise, accept (since a turn ﬂips the roles of Players 1 and 2; they’re symmetric except for who goes ﬁrst). If there are n nodes, one needs O(n2) space to store each stack frame, but there are at most O(n) stack frames, and thus this takes polynomial space. This is interesting, given that writing down a complete winning strategy could take exponential space (since it would involve lots of branching, based on whatever player 2 does). To show that GGis PSPACE-hard, we will show that FG ≤P GG by transforming a formula φ into a pointed graph (i.e. graph with start node) (G,a) such that E has a winning strategy in φ iff player 1 does in (G,a). The idea is that, given some formula, without loss of generality write it as a bunch of cnf clauses separated by ∧. Then, build a graph that starts like this: G(x1) = a T F corresponding to x1 (the left side corresponds to setting it to true, and the right side to false). Then, deﬁne G(x2),...,G(xk) similarly, and connect G(xi) →G(xi+1). Then, from G(xk) lead out to a node c, which is connected to nodes c1,..., cn corresponding to each clause. Then, each clause has arrows to each of its literals (e.g. if C1 = x1 ∨¬x2 ∨x3, there are arrows to nodes (C1,x1), (C1,¬x2), and (C1,x3), so different clauses have different nodes). Then, from each node (Ci,xi), add a node to the T node of G(xi), and for (Ci,¬xi), add an arrow to the F 60 node of G(xi). Tracing through this spaghetti graph,19 it can be shown that E can satisfy the formula iff Player 1 has a winning strategy on this graph. ⊠ Here’s an interesting question: is chess PSPACE-complete? The crux is, of course, how you deﬁne chess as a class of problems. Conventional 8×8 chess has a winning strategy that can be determined in cosntant time. . . for some incredibly painful constant. But generalized versions of chess, checkers, and Go can be shown to be PSPACE-hard. Randomized Complexity. Randomized algorithms are useful to understand, since they happen a lot in practice and are useful, but so little is known about them theoretically. Deﬁnition 19.3. A probabilistic Turing machine is a nondeterministic Turing machine where each nondeterministic step is called a “coin ﬂip,” and is set up such that each nondeterministic step has only two legal next moves (heads or tails, in some sense). Then, this Turing machine accepts if the path of coin ﬂips leads it to an accept state. . . which means the same probabilistic Turing machine might accept and reject the same input on different trials, and the question is how probable these are, not whether they are black-and-white true or false. The probability of reaching a state after k coin ﬂips is 2−k. These are very useful objects to study: often, there is a simpler probabilistic randomized algorithm for a problem than a deterministic algorithm for that problem, and often it’s more efﬁcient. However, it’s completely open as to whether randomized algorithms can be used to solve problems much faster than deterministic ones in general. Deﬁnition 19.4. The probability that a randomized Turing machine machine M accepts a string w is Pr[M accepts w] = X H∈H Pr[M(w) has history H], where H is the set of accepting computation histories of M on w. Theorem 19.5. A ∈NP iff there exists a nondeterministic polynomial-time Turing machine M such that Pr[M accepts w] > 0 for all w ∈A and is zero for w ̸∈A. Deﬁnition 19.6. A probabilistic Turing machine M recognizes a language A with error ε if for all w ∈A, Pr[M accepts w] ≥1−ε and for w ̸∈A, Pr[M doesn’t accept w] ≥1−ε. Lemma 19.7 (Error Reduction). Let 0 < ε < 1/2 and f : N →N be a function (often taken to be a polynomial). If M1 is a randomized Turing machine that has error ε and runs in t(n) time, the there is an equivalent randomized Turing machine M2 such that M −2 has error 2−f (n) and runs in O(f (n)t(n)) time. Proof. Good old Monte Carlo simulation: M2 runs M1 for 11f (n) times, and then chooses the majority response. ⊠ Deﬁnition 19.8. BPP denotes the complexity class of languages L that can be recognized by a probabilistic polynomial-time Turing machine with error at most 1/3. By the lemma, though, any number between 0 and 1/2 would work, and the same class BPP would be obtained. 19Not a technical term. 61 Deﬁnition 19.9. An arithmetic formula is a formula in some ﬁnite number of variables, and symbols +, −, and ∗, so that one can make polynomials. Then, ZERO-POLY denotes the set of formulas that are identically zero, e.g. (x −y)·(x + y)−x · x −y· y−2· x · y, which can be notationally simpliﬁed, of course, into (x+ y)2 −x2 −y2 −2xy. There’s a rich history of polynomial identities in mathematics (somehow involving Gauss, to nobody’s surprise), and these are also useful for veriﬁcation of sufﬁciently restricted programs. This can be thought of as an arithmetic analogue of SAT. Let p ∈Z[x] (i.e. a polynomial over Z in one variable), but such that we can’t see the coefﬁcients, and can only query it. However, we can test whether p = 0 by querying d +1 values, because this is enough to uniquely determine such a polynomial. So how does this generalize to Z[x1,...,xn]? If a p(x1,...,xn) ∈Z[x1,...,xn] is a product of m single-variable polynomials, each of which has at most t terms, then expanding out everything would take tm time, which is icky. But choosing large random values to evaluate p at is a probabilistic algorithm that works. Theorem 19.10 (Schwartz-Zippel). Let p(x1,...,xn) ∈Z[x1,...,xn] be nonzero, where each variable has degree at most d. Then, if F ⊂Z is a ﬁnite set and a1,...,an ∈F are selected randomly (in the uniform distribution), then Pr[p(a1,...,an) = 0] ≤dn \|F\|. Proof. Proceed by induction on n; we saw already the idea for n = 1 (though it’s missing a step or two). In general, assume it’s true for n−1. Then, we can “factor out” the last variable, writing p = p0+xnp1+···+xd n pd, where p0,..., pd are polynomials in x1,...,xn−1.20 Then, if p(a1,...,an) = 0, then there are two possibilities: either pi(a1,...,an−1 = 0 for all i, or an is a root of the overall polynomial, which eventually leads to the inductive bound. ⊠ Theorem 19.11. ZERO-POLY ∈BPP. Proof idea. Let p be an arithmetic formula, and suppose \|p\| = n. Then, p has k ≤n variables, and the degree of each variable is O(n). Then, for all i = 1,...,k, choose ri randomly from {1,...,n2}, and of p(r1,...,rk) = 0, then claim that p = 0; otherwise, report that it’s nonzero. Then, by Theorem 19.10, this works! ⊠ Note that we have no idea if ZERO-POLY ∈P or not; if we could prove it, then there would be new, nicer lower bounds. Next question: is BPP ⊆NP? It seems like it would be reasonable, because probabilistic and nondeterministic Turing machines are cousins, but this is an open question! We want BPP to model efﬁcient computation, but really have no idea. It’s true that BPP ⊆NPNP, which is a story for CS 254. Another open question is whether NP ⊆BPP, or whether BPP ⊆PNP. But we do know that BPP ⊆PSPACE, because a polynomial-time probabilistic Turing machine can be simulated by trying all possible coin ﬂips in the same space. We also don’t know whether BPP ? = EXPTIME. . . though if this is ever proven, it will probably be negative; the question itself is totally ridiculous. Basically, we think P = BPP, but can’t rule out that BPP = EXPTIME! 20In algebra, this is just using the fact that Z[x1,...,xn] = Z[x1,...,xn−1][xn]. 62 Here one might be tempted to draw a Venn diagram reﬂecting inclusions between different classes of languages, P, NP, coNP, NPNP, coNPNP, PSPACE, EXPTIME, PSPACE, NEXP (nondeterministic exponential time, which might be equal yet to BPP), BPP, and so on. But there are so few known strict inclusions that this seems hopeless. 63
16311	https://math.colorado.edu/~rmg/2001/BookOfProof.pdf	Book of Proof Richard Hammack Virginia Commonwealth University Richard Hammack (publisher) Department of Mathematics & Applied Mathematics P.O. Box 842014 Virginia Commonwealth University Richmond, Virginia, 23284 Book of Proof Edition 2.2 © 2013 by Richard Hammack This work is licensed under the Creative Commons Attribution-No Derivative Works 3.0 License Typeset in 11pt T EX Gyre Schola using PDFL AT EX To my students Contents Preface vii Introduction viii I Fundamentals 1. Sets 3 1.1. Introduction to Sets 3 1.2. The Cartesian Product 8 1.3. Subsets 11 1.4. Power Sets 14 1.5. Union, Intersection, Diﬀerence 17 1.6. Complement 19 1.7. Venn Diagrams 21 1.8. Indexed Sets 24 1.9. Sets that Are Number Systems 29 1.10. Russell’s Paradox 31 2. Logic 33 2.1. Statements 34 2.2. And, Or, Not 38 2.3. Conditional Statements 41 2.4. Biconditional Statements 44 2.5. Truth Tables for Statements 46 2.6. Logical Equivalence 49 2.7. Quantiﬁers 51 2.8. More on Conditional Statements 54 2.9. Translating English to Symbolic Logic 55 2.10. Negating Statements 57 2.11. Logical Inference 61 2.12. An Important Note 62 3. Counting 63 3.1. Counting Lists 63 3.2. Factorials 70 3.3. Counting Subsets 73 3.4. Pascal’s Triangle and the Binomial Theorem 78 3.5. Inclusion-Exclusion 81 v II How to Prove Conditional Statements 4. Direct Proof 87 4.1. Theorems 87 4.2. Deﬁnitions 89 4.3. Direct Proof 92 4.4. Using Cases 98 4.5. Treating Similar Cases 99 5. Contrapositive Proof 102 5.1. Contrapositive Proof 102 5.2. Congruence of Integers 105 5.3. Mathematical Writing 107 6. Proof by Contradiction 111 6.1. Proving Statements with Contradiction 112 6.2. Proving Conditional Statements by Contradiction 115 6.3. Combining Techniques 116 6.4. Some Words of Advice 117 III More on Proof 7. Proving Non-Conditional Statements 121 7.1. If-and-Only-If Proof 121 7.2. Equivalent Statements 123 7.3. Existence Proofs; Existence and Uniqueness Proofs 124 7.4. Constructive Versus Non-Constructive Proofs 128 8. Proofs Involving Sets 131 8.1. How to Prove a ∈A 131 8.2. How to Prove A ⊆B 133 8.3. How to Prove A = B 136 8.4. Examples: Perfect Numbers 139 9. Disproof 146 9.1. Counterexamples 148 9.2. Disproving Existence Statements 150 9.3. Disproof by Contradiction 152 10. Mathematical Induction 154 10.1. Proof by Strong Induction 161 10.2. Proof by Smallest Counterexample 165 10.3. Fibonacci Numbers 167 vi IV Relations, Functions and Cardinality 11. Relations 175 11.1. Properties of Relations 179 11.2. Equivalence Relations 184 11.3. Equivalence Classes and Partitions 188 11.4. The Integers Modulo n 191 11.5. Relations Between Sets 194 12. Functions 196 12.1. Functions 196 12.2. Injective and Surjective Functions 201 12.3. The Pigeonhole Principle 205 12.4. Composition 208 12.5. Inverse Functions 211 12.6. Image and Preimage 214 13. Cardinality of Sets 217 13.1. Sets with Equal Cardinalities 217 13.2. Countable and Uncountable Sets 223 13.3. Comparing Cardinalities 228 13.4. The Cantor-Bernstein-Schröeder Theorem 232 Conclusion 239 Solutions 240 Index 301 Preface I n writing this book I have been motivated by the desire to create a high-quality textbook that costs almost nothing. The book is available on my web page for free, and the paperback version (produced through an on-demand press) costs considerably less than comparable traditional textbooks. Any revisions or new editions will be issued solely for the purpose of correcting mistakes and clarifying exposition. New exercises may be added, but the existing ones will not be unnecessarily changed or renumbered. This text is an expansion and reﬁnement of lecture notes I developed while teaching proofs courses over the past fourteen years at Virginia Commonwealth University (a large state university) and Randolph-Macon College (a small liberal arts college). I found the needs of these two audiences to be nearly identical, and I wrote this book for them. But I am mindful of a larger audience. I believe this book is suitable for almost any undergraduate mathematics program. This second edition incorporates many minor corrections and additions that were suggested by readers around the world. In addition, several new examples and exercises have been added, and a section on the Cantor-Bernstein-Schröeder theorem has been added to Chapter 13. Richard Hammack Richmond, Virginia May 25, 2013 Introduction T his is a book about how to prove theorems. Until this point in your education, mathematics has probably been presented as a primarily computational discipline. You have learned to solve equations, compute derivatives and integrals, multiply matrices and ﬁnd determinants; and you have seen how these things can answer practical questions about the real world. In this setting, your primary goal in using mathematics has been to compute answers. But there is another side of mathematics that is more theoretical than computational. Here the primary goal is to understand mathematical structures, to prove mathematical statements, and even to invent or discover new mathematical theorems and theories. The mathematical techniques and procedures that you have learned and used up until now are founded on this theoretical side of mathematics. For example, in computing the area under a curve, you use the fundamental theorem of calculus. It is because this theorem is true that your answer is correct. However, in learning calculus you were probably far more concerned with how that theorem could be applied than in understanding why it is true. But how do we know it is true? How can we convince ourselves or others of its validity? Questions of this nature belong to the theoretical realm of mathematics. This book is an introduction to that realm. This book will initiate you into an esoteric world. You will learn and apply the methods of thought that mathematicians use to verify theorems, explore mathematical truth and create new mathematical theories. This will prepare you for advanced mathematics courses, for you will be better able to understand proofs, write your own proofs and think critically and inquisitively about mathematics. ix The book is organized into four parts, as outlined below. PART I Fundamentals • Chapter 1: Sets • Chapter 2: Logic • Chapter 3: Counting Chapters 1 and 2 lay out the language and conventions used in all advanced mathematics. Sets are fundamental because every mathematical structure, object or entity can be described as a set. Logic is fundamental because it allows us to understand the meanings of statements, to deduce information about mathematical structures and to uncover further structures. All subsequent chapters will build on these ﬁrst two chapters. Chapter 3 is included partly because its topics are central to many branches of mathematics, but also because it is a source of many examples and exercises that occur throughout the book. (However, the course instructor may choose to omit Chapter 3.) PART II Proving Conditional Statements • Chapter 4: Direct Proof • Chapter 5: Contrapositive Proof • Chapter 6: Proof by Contradiction Chapters 4 through 6 are concerned with three main techniques used for proving theorems that have the “conditional” form “If P, then Q.” PART III More on Proof • Chapter 7: Proving Non-Conditional Statements • Chapter 8: Proofs Involving Sets • Chapter 9: Disproof • Chapter 10: Mathematical Induction These chapters deal with useful variations, embellishments and conse-quences of the proof techniques introduced in Chapters 4 through 6. PART IV Relations, Functions and Cardinality • Chapter 11: Relations • Chapter 12: Functions • Chapter 13: Cardinality of Sets These ﬁnal chapters are mainly concerned with the idea of functions, which are central to all of mathematics. Upon mastering this material you will be ready for advanced mathematics courses such as combinatorics, abstract algebra, theory of computation, analysis and topology. x Introduction To the instructor. The book is designed for a three credit course. Here is a possible timetable for a fourteen-week semester. Week Monday Wednesday Friday 1 Section 1.1 Section 1.2 Sections 1.3, 1.4 2 Sections 1.5, 1.6, 1.7 Section 1.8 Sections 1.9∗, 2.1 3 Section 2.2 Sections 2.3, 2.4 Sections 2.5, 2.6 4 Section 2.7 Sections 2.8∗, 2.9 Sections 2.10, 2.11∗, 2.12∗ 5 Sections 3.1, 3.2 Section 3.3 Sections 3.4, 3.5∗ 6 EXAM Sections 4.1, 4.2, 4.3 Sections 4.3, 4.4, 4.5∗ 7 Sections 5.1, 5.2, 5.3∗ Section 6.1 Sections 6.2 6.3∗ 8 Sections 7.1, 7.2∗, 7.3 Sections 8.1, 8.2 Section 8.3 9 Section 8.4 Sections 9.1, 9.2, 9.3∗ Section 10.0 10 Sections 10.0, 10.3∗ Sections 10.1, 10.2 EXAM 11 Sections 11.0, 11.1 Sections 11.2, 11.3 Sections 11.4, 11.5 12 Section 12.1 Section 12.2 Section 12.2 13 Sections 12.3, 12.4∗ Section 12.5 Sections 12.5, 12.6∗ 14 Section 13.1 Section 13.2 Sections 13.3, 13.4∗ Sections marked with ∗may require only the briefest mention in class, or may be best left for the students to digest on their own. Some instructors may prefer to omit Chapter 3. Acknowledgments. I thank my students in VCU’s MATH 300 courses for oﬀering feedback as they read the ﬁrst edition of this book. Thanks especially to Cory Colbert and Lauren Pace for rooting out typographical mistakes and inconsistencies. I am especially indebted to Cory for reading early drafts of each chapter and catching numerous mistakes before I posted the ﬁnal draft on my web page. Cory also created the index, suggested some interesting exercises, and wrote some solutions. Thanks to Andy Lewis and Sean Cox for suggesting many improvements while teaching from the book. I am indebted to Lon Mitchell, whose expertise with typesetting and on-demand publishing made the print version of this book a reality. And thanks to countless readers all over the world who contacted me concerning errors and omissions. Because of you, this is a better book. Part I Fundamentals CHAPTER 1 Sets A ll of mathematics can be described with sets. This becomes more and more apparent the deeper into mathematics you go. It will be apparent in most of your upper level courses, and certainly in this course. The theory of sets is a language that is perfectly suited to describing and explaining all types of mathematical structures. 1.1 Introduction to Sets A set is a collection of things. The things in the collection are called elements of the set. We are mainly concerned with sets whose elements are mathematical entities, such as numbers, points, functions, etc. A set is often expressed by listing its elements between commas, en-closed by braces. For example, the collection © 2,4,6,8 ª is a set which has four elements, the numbers 2,4,6 and 8. Some sets have inﬁnitely many elements. For example, consider the collection of all integers, © ...,−4,−3,−2,−1,0,1,2,3,4,... ª . Here the dots indicate a pattern of numbers that continues forever in both the positive and negative directions. A set is called an inﬁnite set if it has inﬁnitely many elements; otherwise it is called a ﬁnite set. Two sets are equal if they contain exactly the same elements. Thus © 2,4,6,8 ª = © 4,2,8,6 ª because even though they are listed in a diﬀerent order, the elements are identical; but © 2,4,6,8 ª ̸= © 2,4,6,7 ª. Also © ...−4,−3,−2,−1,0,1,2,3,4... ª = © 0,−1,1,−2,2,−3,3,−4,4,... ª . We often let uppercase letters stand for sets. In discussing the set © 2,4,6,8 ª we might declare A = © 2,4,6,8 ª and then use A to stand for © 2,4,6,8 ª. To express that 2 is an element of the set A, we write 2 ∈A, and read this as “2 is an element of A,” or “2 is in A,” or just “2 in A.” We also have 4 ∈A, 6 ∈A and 8 ∈A, but 5 ∉A. We read this last expression as “5 is not an element of A,” or “5 not in A.” Expressions like 6,2 ∈A or 2,4,8 ∈A are used to indicate that several things are in a set. 4 Sets Some sets are so signiﬁcant and prevalent that we reserve special symbols for them. The set of natural numbers (i.e., the positive whole numbers) is denoted by N, that is, N = © 1,2,3,4,5,6,7,... ª . The set of integers Z = © ...,−3,−2,−1,0,1,2,3,4,... ª is another fundamental set. The symbol R stands for the set of all real numbers, a set that is undoubtedly familiar to you from calculus. Other special sets will be listed later in this section. Sets need not have just numbers as elements. The set B = © T,F ª consists of two letters, perhaps representing the values “true” and “false.” The set C = © a, e, i,o,u ª consists of the lowercase vowels in the English alphabet. The set D = © (0,0),(1,0),(0,1),(1,1) ª has as elements the four corner points of a square on the x-y coordinate plane. Thus (0,0) ∈D, (1,0) ∈D, etc., but (1,2) ∉D (for instance). It is even possible for a set to have other sets as elements. Consider E = © 1, © 2,3 ª , © 2,4 ªª, which has three elements: the number 1, the set © 2,3 ª and the set © 2,4 ª. Thus 1 ∈E and © 2,3 ª ∈E and © 2,4 ª ∈E. But note that 2 ∉E, 3 ∉E and 4 ∉E. Consider the set M = ©£0 0 0 0 ¤ , £1 0 0 1 ¤ , £1 0 1 1 ¤ª of three two-by-two matrices. We have £0 0 0 0 ¤ ∈M, but £1 1 0 1 ¤ ∉M. Letters can serve as symbols denoting a set’s elements: If a = £0 0 0 0 ¤, b = £1 0 0 1 ¤ and c = £1 0 1 1 ¤, then M = © a,b, c ª. If X is a ﬁnite set, its cardinality or size is the number of elements it has, and this number is denoted as \|X\|. Thus for the sets above, \|A\| = 4, \|B\| = 2, \|C\| = 5, \|D\| = 4, \|E\| = 3 and \|M\| = 3. There is a special set that, although small, plays a big role. The empty set is the set ©ª that has no elements. We denote it as ;, so ; = ©ª. Whenever you see the symbol ;, it stands for ©ª. Observe that \|;\| = 0. The empty set is the only set whose cardinality is zero. Be careful in writing the empty set. Don’t write © ; ª when you mean ;. These sets can’t be equal because ; contains nothing while © ; ª contains one thing, namely the empty set. If this is confusing, think of a set as a box with things in it, so, for example, © 2,4,6,8 ª is a “box” containing four numbers. The empty set ; = ©ª is an empty box. By contrast, © ; ª is a box with an empty box inside it. Obviously, there’s a diﬀerence: An empty box is not the same as a box with an empty box inside it. Thus ; ̸= © ; ª. (You might also note \|;\| = 0 and ¯ ¯© ; ª¯ ¯ = 1 as additional evidence that ; ̸= © ; ª.) Introduction to Sets 5 This box analogy can help us think about sets. The set F = © ;, © ; ª , ©© ; ªªª may look strange but it is really very simple. Think of it as a box containing three things: an empty box, a box containing an empty box, and a box containing a box containing an empty box. Thus \|F\| = 3. The set G = © N,Z ª is a box containing two boxes, the box of natural numbers and the box of integers. Thus \|G\| = 2. A special notation called set-builder notation is used to describe sets that are too big or complex to list between braces. Consider the inﬁnite set of even integers E = © ...,−6,−4,−2,0,2,4,6,... ª. In set-builder notation this set is written as E = © 2n : n ∈Z ª . We read the ﬁrst brace as “the set of all things of form,” and the colon as “such that.” So the expression E = © 2n : n ∈Z ª is read as “E equals the set of all things of form 2n, such that n is an element of Z.” The idea is that E consists of all possible values of 2n, where n takes on all values in Z. In general, a set X written with set-builder notation has the syntax X = ©expression : rule ª , where the elements of X are understood to be all values of “expression” that are speciﬁed by “rule.” For example, the set E above is the set of all values the expression 2n that satisfy the rule n ∈Z. There can be many ways to express the same set. For example, E = © 2n : n ∈Z ª = © n : n is an even integer ª = © n : n = 2k,k ∈Z ª. Another common way of writing it is E = © n ∈Z : n is even ª , read “E is the set of all n in Z such that n is even.” Some writers use a bar instead of a colon; for example, E = © n ∈Z \| n is even ª. We use the colon. Example 1.1 Here are some further illustrations of set-builder notation. 1. © n : n is a prime number ª = © 2,3,5,7,11,13,17,... ª 2. © n ∈N : n is prime ª = © 2,3,5,7,11,13,17,... ª 3. © n2 : n ∈Z ª = © 0,1,4,9,16,25,... ª 4. © x ∈R : x2 −2 = 0 ª = ©p 2,− p 2 ª 5. © x ∈Z : x2 −2 = 0 ª = ; 6. © x ∈Z : \|x\| < 4 ª = © −3,−2,−1,0,1,2,3 ª 7. © 2x : x ∈Z,\|x\| < 4 ª = © −6,−4,−2,0,2,4,6 ª 8. © x ∈Z : \|2x\| < 4 ª = © −1,0,1 ª 6 Sets These last three examples highlight a conﬂict of notation that we must always be alert to. The expression \|X\| means absolute value if X is a number and cardinality if X is a set. The distinction should always be clear from context. Consider © x ∈Z : \|x\| < 4 ª in Example 1.1 (6) above. Here x ∈Z, so x is a number (not a set), and thus the bars in \|x\| must mean absolute value, not cardinality. On the other hand, suppose A = ©© 1,2 ª , © 3,4,5,6 ª , © 7 ªª and B = © X ∈A : \|X\| < 3 ª. The elements of A are sets (not numbers), so the \|X\| in the expression for B must mean cardinality. Therefore B = ©© 1,2 ª , © 7 ªª. We close this section with a summary of special sets. These are sets or types of sets that come up so often that they are given special names and symbols. • The empty set: ; = ©ª • The natural numbers: N = © 1,2,3,4,5,... ª • The integers: Z = © ...,−3,−2,−1,0,1,2,3,4,5,... ª • The rational numbers: Q = © x : x = m n , where m,n ∈Z and n ̸= 0 ª • The real numbers: R (the set of all real numbers on the number line) Notice that Q is the set of all numbers that can be expressed as a fraction of two integers. You are surely aware that Q ̸= R, as p 2 ∉Q but p 2 ∈R. Following are some other special sets that you will recall from your study of calculus. Given two numbers a,b ∈R with a < b, we can form various intervals on the number line. • Closed interval: [a,b] = © x ∈R : a ≤x ≤b ª • Half open interval: (a,b] = © x ∈R : a < x ≤b ª • Half open interval: [a,b) = © x ∈R : a ≤x < b ª • Open interval: (a,b) = © x ∈R : a < x < b ª • Inﬁnite interval: (a,∞) = © x ∈R : a < x ª • Inﬁnite interval: [a,∞) = © x ∈R : a ≤x ª • Inﬁnite interval: (−∞,b) = © x ∈R : x < b ª • Inﬁnite interval: (−∞,b] = © x ∈R : x ≤b ª Remember that these are intervals on the number line, so they have in-ﬁnitely many elements. The set (0.1,0.2) contains inﬁnitely many numbers, even though the end points may be close together. It is an unfortunate notational accident that (a,b) can denote both an interval on the line and a point on the plane. The diﬀerence is usually clear from context. In the next section we will see still another meaning of (a,b). Introduction to Sets 7 Exercises for Section 1.1 A. Write each of the following sets by listing their elements between braces. 1. © 5x−1 : x ∈Z ª 2. © 3x+2 : x ∈Z ª 3. © x ∈Z : −2 ≤x < 7 ª 4. © x ∈N : −2 < x ≤7 ª 5. © x ∈R : x2 = 3 ª 6. © x ∈R : x2 = 9 ª 7. © x ∈R : x2 +5x = −6 ª 8. © x ∈R : x3 +5x2 = −6x ª 9. © x ∈R : sinπx = 0 ª 10. © x ∈R : cosx = 1 ª 11. © x ∈Z : \|x\| < 5 ª 12. © x ∈Z : \|2x\| < 5 ª 13. © x ∈Z : \|6x\| < 5 ª 14. © 5x : x ∈Z,\|2x\| ≤8 ª 15. © 5a+2b : a,b ∈Z ª 16. © 6a+2b : a,b ∈Z ª B. Write each of the following sets in set-builder notation. 17. © 2,4,8,16,32,64... ª 18. © 0,4,16,36,64,100,... ª 19. © ...,−6,−3,0,3,6,9,12,15,... ª 20. © ...,−8,−3,2,7,12,17,... ª 21. © 0,1,4,9,16,25,36,... ª 22. © 3,6,11,18,27,38,... ª 23. © 3,4,5,6,7,8 ª 24. © −4,−3,−2,−1,0,1,2 ª 25. © ..., 1 8, 1 4, 1 2,1,2,4,8,... ª 26. © ..., 1 27, 1 9, 1 3,1,3,9,27,... ª 27. © ...,−π,−π 2 ,0, π 2 ,π, 3π 2 ,2π, 5π 2 ,... ª 28. © ...,−3 2,−3 4,0, 3 4, 3 2, 9 4,3, 15 4 , 9 2,... ª C. Find the following cardinalities. 29. ¯ ¯©© 1 ª , © 2, © 3,4 ªª ,; ª¯ ¯ 30. ¯ ¯©© 1,4 ª ,a,b, ©© 3,4 ªª , © ; ªª¯ ¯ 31. ¯ ¯©©© 1 ª , © 2, © 3,4 ªª ,; ªª¯ ¯ 32. ¯ ¯©©© 1,4 ª ,a,b, ©© 3,4 ªª , © ; ªªª¯ ¯ 33. ¯ ¯© x ∈Z : \|x\| < 10 ª¯ ¯ 34. ¯ ¯© x ∈N : \|x\| < 10 ª¯ ¯ 35. ¯ ¯© x ∈Z : x2 < 10 ª¯ ¯ 36. ¯ ¯© x ∈N : x2 < 10 ª¯ ¯ 37. ¯ ¯© x ∈N : x2 < 0 ª¯ ¯ 38. ¯ ¯© x ∈N : 5x ≤20 ª¯ ¯ D. Sketch the following sets of points in the x-y plane. 39. © (x, y) : x ∈[1,2], y ∈[1,2] ª 40. © (x, y) : x ∈[0,1], y ∈[1,2] ª 41. © (x, y) : x ∈[−1,1], y = 1 ª 42. © (x, y) : x = 2, y ∈[0,1] ª 43. © (x, y) : \|x\| = 2, y ∈[0,1] ª 44. © (x,x2) : x ∈R ª 45. © (x, y) : x, y ∈R,x2 + y2 = 1 ª 46. © (x, y) : x, y ∈R,x2 + y2 ≤1 ª 47. © (x, y) : x, y ∈R, y ≥x2 −1 ª 48. © (x, y) : x, y ∈R,x > 1 ª 49. © (x,x+ y) : x ∈R, y ∈Z ª 50. © (x, x2 y ) : x ∈R, y ∈N ª 51. © (x, y) ∈R2 : (y−x)(y+ x) = 0 ª 52. © (x, y) ∈R2 : (y−x2)(y+ x2) = 0 ª 8 Sets 1.2 The Cartesian Product Given two sets A and B, it is possible to “multiply” them to produce a new set denoted as A ×B. This operation is called the Cartesian product. To understand it, we must ﬁrst understand the idea of an ordered pair. Deﬁnition 1.1 An ordered pair is a list (x, y) of two things x and y, enclosed in parentheses and separated by a comma. For example, (2,4) is an ordered pair, as is (4,2). These ordered pairs are diﬀerent because even though they have the same things in them, the order is diﬀerent. We write (2,4) ̸= (4,2). Right away you can see that ordered pairs can be used to describe points on the plane, as was done in calculus, but they are not limited to just that. The things in an ordered pair don’t have to be numbers. You can have ordered pairs of letters, such as (m,ℓ), ordered pairs of sets such as ( © 2,5 ª , © 3,2 ª ), even ordered pairs of ordered pairs like ((2,4),(4,2)). The following are also ordered pairs: (2, © 1,2,3 ª ), (R,(0,0)). Any list of two things enclosed by parentheses is an ordered pair. Now we are ready to deﬁne the Cartesian product. Deﬁnition 1.2 The Cartesian product of two sets A and B is another set, denoted as A ×B and deﬁned as A ×B = © (a,b) : a ∈A,b ∈B ª. Thus A × B is a set of ordered pairs of elements from A and B. For example, if A = © k,ℓ,m ª and B = © q,r ª, then A ×B = © (k, q),(k,r),(ℓ, q),(ℓ,r),(m, q),(m,r) ª . Figure 1.1 shows how to make a schematic diagram of A ×B. Line up the elements of A horizontally and line up the elements of B vertically, as if A and B form an x- and y-axis. Then ﬁll in the ordered pairs so that each element (x, y) is in the column headed by x and the row headed by y. B A q r (k,r) (ℓ,r) (m,r) (k, q) (ℓ, q) (m, q) k ℓ m A ×B Figure 1.1. A diagram of a Cartesian product The Cartesian Product 9 For another example, © 0,1 ª × © 2,1 ª = © (0,2),(0,1),(1,2),(1,1) ª. If you are a visual thinker, you may wish to draw a diagram similar to Figure 1.1. The rectangular array of such diagrams give us the following general fact. Fact 1.1 If A and B are ﬁnite sets, then \|A ×B\| = \|A\|·\|B\|. The set R×R = © (x, y) : x, y ∈R ª should be very familiar. It can be viewed as the set of points on the Cartesian plane, and is drawn in Figure 1.2(a). The set R×N = © (x, y) : x ∈R, y ∈N ª can be regarded as all of the points on the Cartesian plane whose second coordinate is a natural number. This is illustrated in Figure 1.2(b), which shows that R×N looks like inﬁnitely many horizontal lines at integer heights above the x axis. The set N×N can be visualized as the set of all points on the Cartesian plane whose coordinates are both natural numbers. It looks like a grid of dots in the ﬁrst quadrant, as illustrated in Figure 1.2(c). x x x y y y (a) (b) (c) R×R R×N N×N Figure 1.2. Drawings of some Cartesian products It is even possible for one factor of a Cartesian product to be a Cartesian product itself, as in R×(N×Z) = © (x,(y, z)) : x ∈R, (y, z) ∈N×Z ª. We can also deﬁne Cartesian products of three or more sets by moving beyond ordered pairs. An ordered triple is a list (x, y, z). The Cartesian product of the three sets R, N and Z is R×N×Z = © (x, y, z) : x ∈R, y ∈N, z ∈Z ª. Of course there is no reason to stop with ordered triples. In general, A1 × A2 ×···× An = © (x1,x2,...,xn) : xi ∈Ai for each i = 1,2,...,n ª . Be mindful of parentheses. There is a slight diﬀerence between R×(N×Z) and R×N×Z. The ﬁrst is a Cartesian product of two sets; its elements are ordered pairs (x,(y, z)). The second is a Cartesian product of three sets; its elements look like (x, y, z). To be sure, in many situations there is no harm in blurring the distinction between expressions like (x,(y, z)) and (x, y, z), but for now we consider them as diﬀerent. 10 Sets We can also take Cartesian powers of sets. For any set A and positive integer n, the power An is the Cartesian product of A with itself n times: An = A × A ×···× A = © (x1,x2,...,xn) : x1,x2,...,xn ∈A ª . In this way, R2 is the familiar Cartesian plane and R3 is three-dimensional space. You can visualize how, if R2 is the plane, then Z2 = © (m,n) : m,n ∈Z ª is a grid of points on the plane. Likewise, as R3 is 3-dimensional space, Z3 = © (m,n, p) : m,n, p ∈Z ª is a grid of points in space. In other courses you may encounter sets that are very similar to Rn, but yet have slightly diﬀerent shades of meaning. Consider, for example, the set of all two-by-three matrices with entries from R: M = ©£ u v w x y z ¤ : u,v,w,x, y, z ∈R ª . This is not really all that diﬀerent from the set R6 = © (u,v,w,x, y, z) : u,v,w,x, y, z ∈R ª . The elements of these sets are merely certain arrangements of six real numbers. Despite their similarity, we maintain that M ̸= R6, for two-by-three matrices are not the same things as sequences of six numbers. Exercises for Section 1.2 A. Write out the indicated sets by listing their elements between braces. 1. Suppose A = © 1,2,3,4 ª and B = © a, c ª. (a) A ×B (b) B × A (c) A × A (d) B ×B (e) ;×B (f) (A ×B)×B (g) A ×(B ×B) (h) B3 2. Suppose A = © π, e,0 ª and B = © 0,1 ª. (a) A ×B (b) B × A (c) A × A (d) B ×B (e) A ×; (f) (A ×B)×B (g) A ×(B ×B) (h) A ×B ×B 3. © x ∈R : x2 = 2 ª × © a, c, e ª 4. © n ∈Z : 2 < n < 5 ª × © n ∈Z : \|n\| = 5 ª 5. © x ∈R : x2 = 2 ª × © x ∈R : \|x\| = 2 ª 6. © x ∈R : x2 = x ª × © x ∈N : x2 = x ª 7. © ; ª × © 0,; ª × © 0,1 ª 8. © 0,1 ª4 B. Sketch these Cartesian products on the x-y plane R2 (or R3 for the last two). 9. © 1,2,3 ª × © −1,0,1 ª 10. © −1,0,1 ª × © 1,2,3 ª 11. [0,1]×[0,1] 12. [−1,1]×[1,2] 13. © 1,1.5,2 ª ×[1,2] 14. [1,2]× © 1,1.5,2 ª 15. © 1 ª ×[0,1] 16. [0,1]× © 1 ª 17. N×Z 18. Z×Z 19. [0,1]×[0,1]×[0,1] 20. © (x, y) ∈R2 : x2 + y2 ≤1 ª ×[0,1] Subsets 11 1.3 Subsets It can happen that every element of some set A is also an element of another set B. For example, each element of A = © 0,2,4 ª is also an element of B = © 0,1,2,3,4 ª. When A and B are related this way we say that A is a subset of B. Deﬁnition 1.3 Suppose A and B are sets. If every element of A is also an element of B, then we say A is a subset of B, and we denote this as A ⊆B. We write A ̸⊆B if A is not a subset of B, that is, if it is not true that every element of A is also an element of B. Thus A ̸⊆B means that there is at least one element of A that is not an element of B. Example 1.2 Be sure you understand why each of the following is true. 1. © 2,3,7 ª ⊆ © 2,3,4,5,6,7 ª 2. © 2,3,7 ª ̸⊆ © 2,4,5,6,7 ª 3. © 2,3,7 ª ⊆ © 2,3,7 ª 4. © 2n : n ∈Z ª ⊆Z 5. © (x,sin(x)) : x ∈R ª ⊆R2 6. © 2,3,5,7,11,13,17,... ª ⊆N 7. N ⊆Z ⊆Q ⊆R 8. R×N ⊆R×R This brings us to a signiﬁcant fact: If B is any set whatsoever, then ; ⊆B. To see why this is true, look at the last sentence of Deﬁnition 1.3. It says that ; ̸⊆B would mean that there is at least one element of ; that is not an element of B. But this cannot be so because ; contains no elements! Thus it is not the case that ; ̸⊆B, so it must be that ; ⊆B. Fact 1.2 The empty set is a subset of every set, that is, ; ⊆B for any set B. Here is another way to look at it. Imagine a subset of B as a thing you make by starting with braces ©ª, then ﬁlling them with selections from B. For example, to make one particular subset of B = © a,b, c ª, start with ©ª, select b and c from B and insert them into ©ª to form the subset © b, c ª. Alternatively, you could have chosen just a to make © a ª, and so on. But one option is to simply select nothing from B. This leaves you with the subset ©ª. Thus ©ª ⊆B. More often we write it as ; ⊆B. 12 Sets This idea of “making” a subset can help us list out all the subsets of a given set B. As an example, let B = © a,b, c ª. Let’s list all of its subsets. One way of approaching this is to make a tree-like structure. Begin with the subset ©ª, which is shown on the left of Figure 1.3. Considering the element a of B, we have a choice: insert it or not. The lines from ©ª point to what we get depending whether or not we insert a, either ©ª or © a ª. Now move on to the element b of B. For each of the sets just formed we can either insert or not insert b, and the lines on the diagram point to the resulting sets ©ª, © b ª, © a ª, or © a,b ª. Finally, to each of these sets, we can either insert c or not insert it, and this gives us, on the far right-hand column, the sets ©ª, © c ª, © b ª, © b, c ª, © a ª, © a, c ª, © a,b ª and © a,b, c ª. These are the eight subsets of B = © a,b, c ª. Insert a? Insert b? Insert c? No Yes Yes Yes Yes Yes Yes Yes No No No No No No ©ª ©ª ©ª ©ª © c ª © b ª © b, c ª © a ª © a, c ª © a,b ª © a,b, c ª © b ª © a ª © a,b ª © a ª Figure 1.3. A “tree” for listing subsets We can see from the way this tree branches out that if it happened that B = © a ª, then B would have just two subsets, those in the second column of the diagram. If it happened that B = © a,b ª, then B would have four subsets, those listed in the third column, and so on. At each branching of the tree, the number of subsets doubles. Thus in general, if \|B\| = n, then B must have 2n subsets. Fact 1.3 If a ﬁnite set has n elements, then it has 2n subsets. Subsets 13 For a slightly more complex example, consider listing the subsets of B = © 1,2, © 1,3 ªª. This B has just three elements: 1, 2 and © 1,3 ª. At this point you probably don’t even have to draw a tree to list out B’s subsets. You just make all the possible selections from B and put them between braces to get ©ª , © 1 ª , © 2 ª , ©© 1,3 ªª , © 1,2 ª , © 1, © 1,3 ªª , © 2, © 1,3 ªª , © 1,2, © 1,3 ªª . These are the eight subsets of B. Exercises like this help you identify what is and isn’t a subset. You know immediately that a set such as © 1,3 ª is not a subset of B because it can’t be made by selecting elements from B, as the 3 is not an element of B and thus is not a valid selection. Notice that although © 1,3 ª ̸⊆B, it is true that © 1,3 ª ∈B. Also, ©© 1,3 ªª ⊆B. Example 1.3 Be sure you understand why the following statements are true. Each illustrates an aspect of set theory that you’ve learned so far. 1. 1 ∈ © 1, © 1 ªª. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1 is the ﬁrst element listed in © 1, © 1 ªª 2. 1 ̸⊆ © 1, © 1 ªª . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . because 1 is not a set 3. © 1 ª ∈ © 1, © 1 ªª . . . . . . . . . . . . . . . . . . . . . . . . . © 1 ª is the second element listed in © 1, © 1 ªª 4. © 1 ª ⊆ © 1, © 1 ªª . . . . . . . . . . . . . . . . . . . . . . . make subset © 1 ª by selecting 1 from © 1, © 1 ªª 5. ©© 1 ªª ∉ © 1, © 1 ªª. . . . . . . . . . .because © 1, © 1 ªª contains only 1 and © 1 ª, and not ©© 1 ªª 6. ©© 1 ªª ⊆ © 1, © 1 ªª. . . . . . . . . . . . . . . . . .make subset ©© 1 ªª by selecting © 1 ª from © 1, © 1 ªª 7. N ∉N. . . . . . . . .because N is a set (not a number) and N contains only numbers 8. N ⊆N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . because X ⊆X for every set X 9. ; ∉N. . . . . . . . . . . . . . . . . . . .because the set N contains only numbers and no sets 10. ; ⊆N. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .because ; is a subset of every set 11. N ∈ © N ª. . . . . . . . . . . . . . . . . . . . . . . . . . .because © N ª has just one element, the set N 12. N ̸⊆ © N ª . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .because, for instance, 1 ∈N but 1 ∉ © N ª 13. ; ∉ © N ª . . . . . . . . . . . . . . . . . . . . . note that the only element of © N ª is N, and N ̸= ; 14. ; ⊆ © N ª . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . because ; is a subset of every set 15. ; ∈ © ;,N ª. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ; is the ﬁrst element listed in © ;,N ª 16. ; ⊆ © ;,N ª. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .because ; is a subset of every set 17. © N ª ⊆ © ;,N ª . . . . . . . . . . . . . . . . . . . . . . . make subset © N ª by selecting N from © ;,N ª 18. © N ª ̸⊆ © ;, © N ªª . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . because N ∉ © ;, © N ªª 19. © N ª ∈ © ;, © N ªª. . . . . . . . . . . . . . . . . . . . . . © N ª is the second element listed in © ;, © N ªª 20. © (1,2),(2,2),(7,1) ª ⊆N×N . . . . . . . . . . . . . . . . . . . each of (1,2), (2,2), (7,1) is in N×N Though they should help you understand the concept of subset, the above examples are somewhat artiﬁcial. But in general, subsets arise very 14 Sets naturally. For instance, consider the unit circle C = © (x, y) ∈R2 : x2 + y2 = 1 ª. This is a subset C ⊆R2. Likewise the graph of a function y = f (x) is a set of points G = © (x, f (x)) : x ∈R ª, and G ⊆R2. Surely sets such as C and G are more easily understood or visualized when regarded as subsets of R2. Mathematics is ﬁlled with such instances where it is important to regard one set as a subset of another. Exercises for Section 1.3 A. List all the subsets of the following sets. 1. © 1,2,3,4 ª 2. © 1,2,; ª 3. ©© R ªª 4. ; 5. © ; ª 6. © R,Q,N ª 7. © R, © Q,N ªª 8. ©© 0,1 ª , © 0,1, © 2 ªª , © 0 ªª B. Write out the following sets by listing their elements between braces. 9. © X : X ⊆ © 3,2,a ª and \|X\| = 2 ª 10. © X ⊆N : \|X\| ≤1 ª 11. © X : X ⊆ © 3,2,a ª and \|X\| = 4 ª 12. © X : X ⊆ © 3,2,a ª and \|X\| = 1 ª C. Decide if the following statements are true or false. Explain. 13. R3 ⊆R3 14. R2 ⊆R3 15. © (x, y) : x−1 = 0 ª ⊆ © (x, y) : x2 −x = 0 ª 16. © (x, y) : x2 −x = 0 ª ⊆ © (x, y) : x−1 = 0 ª 1.4 Power Sets Given a set, you can form a new set with the power set operation, deﬁned as follows. Deﬁnition 1.4 If A is a set, the power set of A is another set, denoted as P(A) and deﬁned to be the set of all subsets of A. In symbols, P(A) = © X : X ⊆A ª. For example, suppose A = © 1,2,3 ª. The power set of A is the set of all subsets of A. We learned how to ﬁnd these subsets in the previous section, and they are ©ª, © 1 ª, © 2 ª, © 3 ª, © 1,2 ª, © 1,3 ª, © 2,3 ª and © 1,2,3 ª. Therefore the power set of A is P(A) = © ;, © 1 ª , © 2 ª , © 3 ª , © 1,2 ª , © 1,3 ª , © 2,3 ª , © 1,2,3 ª ª . As we saw in the previous section, if a ﬁnite set A has n elements, then it has 2n subsets, and thus its power set has 2n elements. Power Sets 15 Fact 1.4 If A is a ﬁnite set, then \|P(A)\| = 2\|A\|. Example 1.4 You should examine the following statements and make sure you understand how the answers were obtained. In particular, notice that in each instance the equation \|P(A)\| = 2\|A\| is true. 1. P ¡© 0,1,3 ª¢ = © ;, © 0 ª , © 1 ª , © 3 ª , © 0,1 ª , © 0,3 ª , © 1,3 ª , © 0,1,3 ª ª 2. P ¡© 1,2 ª¢ = © ;, © 1 ª , © 2 ª , © 1,2 ª ª 3. P ¡© 1 ª¢ = © ;, © 1 ª ª 4. P (;) = © ; ª 5. P ¡© a ª¢ = © ;, © a ª ª 6. P ¡© ; ª¢ = © ;, © ; ª ª 7. P ¡© a ª¢ ×P ¡© ; ª¢ = © (;,;), ¡ ;, © ; ª¢ , ¡© a ª ,; ¢ , ¡© a ª , © ; ª¢ ª 8. P ¡P ¡© ; ª¢¢ = © ;, © ; ª , ©© ; ªª , © ;, © ; ªª ª 9. P ¡© 1, © 1,2 ªª¢ = © ;, © 1 ª , ©© 1,2 ªª , © 1, © 1,2 ªª ª 10. P ¡© Z,N ª¢ = © ;, © Z ª , © N ª , © Z,N ª ª Next are some that are wrong. See if you can determine why they are wrong and make sure you understand the explanation on the right. 11. P(1) = © ;, © 1 ª ª . . . . . . . . . . . . . . . . . . . . . . . . . . . meaningless because 1 is not a set 12. P ¡© 1, © 1,2 ªª¢ = © ;, © 1 ª , © 1,2 ª , © 1, © 1,2 ªªª . . . . . . . . wrong because © 1,2 ª ̸⊆ © 1, © 1,2 ªª 13. P ¡© 1, © 1,2 ªª¢ = © ;, ©© 1 ªª , ©© 1,2 ªª , © ;, © 1,2 ªªª. . . . .wrong because ©© 1 ªª ̸⊆ © 1, © 1,2 ªª If A is ﬁnite, it is possible (though maybe not practical) to list out P(A) between braces as was done in the above example. That is not possible if A is inﬁnite. For example, consider P(N). If you start listing its elements you quickly discover that N has inﬁnitely many subsets, and it’s not clear how (or if) they could be arranged as a list with a deﬁnite pattern: P(N) = © ;, © 1 ª , © 2 ª ,..., © 1,2 ª , © 1,3 ª ,..., © 39,47 ª , ..., © 3,87,131 ª ,..., © 2,4,6,8,... ª ,... ? ... ª . The set P(R2) is mind boggling. Think of R2 = © (x, y) : x, y ∈R ª as the set of all points on the Cartesian plane. A subset of R2 (that is, an element of P(R2)) is a set of points in the plane. Let’s look at some of these sets. Since © (0,0),(1,1) ª ⊆R2, we know that © (0,0),(1,1) ª ∈P(R2). We can even draw a picture of this subset, as in Figure 1.4(a). For another example, the graph of the equation y = x2 is the set of points G = © (x,x2) : x ∈R ª and this is a subset of R2, so G ∈P(R2). Figure 1.4(b) is a picture of G. Because this can be done for any function, the graph of any imaginable function f : R →R is an element of P(R2). 16 Sets x x x y y y (a) (b) (c) Figure 1.4. Three of the many, many sets in P(R2) In fact, any black-and-white image on the plane can be thought of as a subset of R2, where the black points belong to the subset and the white points do not. So the text “INFINITE” in Figure 1.4(c) is a subset of R2 and therefore an element of P(R2). By that token, P(R2) contains a copy of the page you are reading now. Thus in addition to containing every imaginable function and every imaginable black-and-white image, P(R2) also contains the full text of every book that was ever written, those that are yet to be written and those that will never be written. Inside of P(R2) is a detailed biography of your life, from beginning to end, as well as the biographies of all of your unborn descendants. It is startling that the ﬁve symbols used to write P(R2) can express such an incomprehensibly large set. Homework: Think about P(P(R2)). Exercises for Section 1.4 A. Find the indicated sets. 1. P ¡©© a,b ª , © c ªª¢ 2. P ¡© 1,2,3,4 ª¢ 3. P ¡©© ; ª ,5 ª¢ 4. P ¡© R,Q ª¢ 5. P ¡P ¡© 2 ª¢¢ 6. P ¡© 1,2 ª¢ ×P ¡© 3 ª¢ 7. P ¡© a,b ª¢ ×P ¡© 0,1 ª¢ 8. P ¡© 1,2 ª × © 3 ª¢ 9. P ¡© a,b ª × © 0 ª¢ 10. © X ∈P ¡© 1,2,3 ª¢ : \|X\| ≤1 ª 11. © X ⊆P ¡© 1,2,3 ª¢ : \|X\| ≤1 ª 12. © X ∈P ¡© 1,2,3 ª¢ : 2 ∈X ª B. Suppose that \|A\| = m and \|B\| = n. Find the following cardinalities. 13. \|P(P(P(A)))\| 14. \|P(P(A))\| 15. \|P(A ×B)\| 16. \|P(A)×P(B)\| 17. ¯ ¯© X ∈P(A) : \|X\| ≤1 ª¯ ¯ 18. \|P(A ×P(B))\| 19. \|P(P(P(A ×;)))\| 20. ¯ ¯© X ⊆P(A) : \|X\| ≤1 ª¯ ¯ Union, Intersection, Diﬀerence 17 1.5 Union, Intersection, Diﬀerence Just as numbers are combined with operations such as addition, subtrac-tion and multiplication, there are various operations that can be applied to sets. The Cartesian product (deﬁned in Section 1.2) is one such operation; given sets A and B, we can combine them with × to get a new set A ×B. Here are three new operations called union, intersection and diﬀerence. Deﬁnition 1.5 Suppose A and B are sets. The union of A and B is the set A ∪B = © x : x ∈A or x ∈B ª. The intersection of A and B is the set A ∩B = © x : x ∈A and x ∈B ª. The diﬀerence of A and B is the set A −B = © x : x ∈A and x ∉B ª. In words, the union A ∪B is the set of all things that are in A or in B (or in both). The intersection A ∩B is the set of all things in both A and B. The diﬀerence A −B is the set of all things that are in A but not in B. Example 1.5 Suppose A = © a,b, c,d, e ª, B = © d, e, f ª and C = © 1,2,3 ª. 1. A ∪B = © a,b, c,d, e, f ª 2. A ∩B = © d, e ª 3. A −B = © a,b, c ª 4. B −A = © f ª 5. (A −B)∪(B −A) = © a,b, c, f ª 6. A ∪C = © a,b, c,d, e,1,2,3 ª 7. A ∩C = ; 8. A −C = © a,b, c,d, e ª 9. (A ∩C)∪(A −C) = © a,b, c,d, e ª 10. (A ∩B)×B = © (d,d),(d, e),(d, f ),(e,d),(e, e),(e, f ) ª 11. (A ×C)∩(B ×C) = © (d,1),(d,2),(d,3),(e,1),(e,2),(e,3) ª Observe that for any sets X and Y it is always true that X ∪Y = Y ∪X and X ∩Y = Y ∩X, but in general X −Y ̸= Y −X. Continuing the example, parts 12–15 below use the interval notation discussed in Section 1.1, so [2,5] = © x ∈R : 2 ≤x ≤5 ª, etc. Sketching these examples on the number line may help you understand them. 12. [2,5]∪[3,6] = [2,6] 13. [2,5]∩[3,6] = [3,5] 14. [2,5]−[3,6] = [2,3) 15. [0,3]−[1,2] = [0,1)∪(2,3] 18 Sets A B A ∪B A ∩B A −B (a) (b) (c) (d) Figure 1.5. The union, intersection and diﬀerence of sets A and B Example 1.6 Let A = © (x,x2) : x ∈R ª be the graph of the equation y = x2 and let B = © (x,x+2) : x ∈R ª be the graph of the equation y = x+2. These sets are subsets of R2. They are sketched together in Figure 1.5(a). Figure 1.5(b) shows A ∪B, the set of all points (x, y) that are on one (or both) of the two graphs. Observe that A ∩B = © (−1,1),(2,4) ª consists of just two elements, the two points where the graphs intersect, as illustrated in Figure 1.5(c). Figure 1.5(d) shows A−B, which is the set A with “holes” where B crossed it. In set builder notation, we could write A∪B = © (x, y) : x ∈R, y = x2 or y = x+2 ª and A −B = © (x,x2) : x ∈R− © −1,2 ªª. Exercises for Section 1.5 1. Suppose A = © 4,3,6,7,1,9 ª, B = © 5,6,8,4 ª and C = © 5,8,4 ª . Find: (a) A ∪B (b) A ∩B (c) A −B (d) A −C (e) B −A (f) A ∩C (g) B ∩C (h) B ∪C (i) C −B 2. Suppose A = © 0,2,4,6,8 ª, B = © 1,3,5,7 ª and C = © 2,8,4 ª . Find: (a) A ∪B (b) A ∩B (c) A −B (d) A −C (e) B −A (f) A ∩C (g) B ∩C (h) C −A (i) C −B 3. Suppose A = © 0,1 ª and B = © 1,2 ª. Find: (a) (A ×B)∩(B ×B) (b) (A ×B)∪(B ×B) (c) (A ×B)−(B ×B) (d) (A ∩B)× A (e) (A ×B)∩B (f) P(A)∩P(B) (g) P(A)−P(B) (h) P(A ∩B) (i) P(A ×B) 4. Suppose A = © b, c,d ª and B = © a,b ª. Find: (a) (A ×B)∩(B ×B) (b) (A ×B)∪(B ×B) (c) (A ×B)−(B ×B) (d) (A ∩B)× A (e) (A ×B)∩B (f) P(A)∩P(B) (g) P(A)−P(B) (h) P(A ∩B) (i) P(A)×P(B) Complement 19 5. Sketch the sets X = [1,3]×[1,3] and Y = [2,4]×[2,4] on the plane R2. On separate drawings, shade in the sets X ∪Y, X ∩Y, X −Y and Y −X. (Hint: X and Y are Cartesian products of intervals. You may wish to review how you drew sets like [1,3]×[1,3] in the exercises for Section 1.2.) 6. Sketch the sets X = [−1,3] × [0,2] and Y = [0,3] × [1,4] on the plane R2. On separate drawings, shade in the sets X ∪Y, X ∩Y, X −Y and Y −X. 7. Sketch the sets X = © (x, y) ∈R2 : x2 + y2 ≤1 ª and Y = © (x, y) ∈R2 : x ≥0 ª on R2. On separate drawings, shade in the sets X ∪Y, X ∩Y, X −Y and Y −X. 8. Sketch the sets X = © (x, y) ∈R2 : x2 + y2 ≤1 ª and Y = © (x, y) ∈R2 : −1 ≤y ≤0 ª on R2. On separate drawings, shade in the sets X ∪Y, X ∩Y, X −Y and Y −X. 9. Is the statement (R×Z)∩(Z×R) = Z×Z true or false? What about the statement (R×Z)∪(Z×R) = R×R? 10. Do you think the statement (R−Z)×N = (R×N)−(Z×N) is true, or false? Justify. 1.6 Complement This section introduces yet another set operation, called the set complement. The deﬁnition requires the idea of a universal set, which we now discuss. When dealing with a set, we almost always regard it as a subset of some larger set. For example, consider the set of prime numbers P = © 2,3,5,7,11,13,... ª. If asked to name some things that are not in P, we might mention some composite numbers like 4 or 6 or 423. It probably would not occur to us to say that Vladimir Putin is not in P. True, Vladimir Putin is not in P, but he lies entirely outside of the discussion of what is a prime number and what is not. We have an unstated assumption that P ⊆N because N is the most natural setting in which to discuss prime numbers. In this context, anything not in P should still be in N. This larger set N is called the universal set or universe for P. Almost every useful set in mathematics can be regarded as having some natural universal set. For instance, the unit circle is the set C = © (x, y) ∈R2 : x2 + y2 = 1 ª, and since all these points are in the plane R2 it is natural to regard R2 as the universal set for C. In the absence of speciﬁcs, if A is a set, then its universal set is often denoted as U. We are now ready to deﬁne the complement operation. Deﬁnition 1.6 Let A be a set with a universal set U. The complement of A, denoted A, is the set A = U −A. 20 Sets Example 1.7 If P is the set of prime numbers, then P = N−P = © 1,4,6,8,9,10,12,... ª . Thus P is the set of composite numbers and 1. Example 1.8 Let A = © (x,x2) : x ∈R ª be the graph of the equation y = x2. Figure 1.6(a) shows A in its universal set R2. The complement of A is A = R2 −A = © (x, y) ∈R2 : y ̸= x2ª, illustrated by the shaded area in Figure 1.6(b). A A (a) (b) Figure 1.6. A set and its complement Exercises for Section 1.6 1. Let A = © 4,3,6,7,1,9 ª and B = © 5,6,8,4 ª have universal set U = © 0,1,2,...,10 ª. Find: (a) A (b) B (c) A ∩A (d) A ∪A (e) A −A (f) A −B (g) A −B (h) A ∩B (i) A ∩B 2. Let A = © 0,2,4,6,8 ª and B = © 1,3,5,7 ª have universal set U = © 0,1,2,...,8 ª. Find: (a) A (b) B (c) A ∩A (d) A ∪A (e) A −A (f) A ∪B (g) A ∩B (h) A ∩B (i) A ×B 3. Sketch the set X = [1,3]×[1,2] on the plane R2. On separate drawings, shade in the sets X and X ∩([0,2]×[0,3]). 4. Sketch the set X = [−1,3]×[0,2] on the plane R2. On separate drawings, shade in the sets X and X ∩([−2,4]×[−1,3]). 5. Sketch the set X = © (x, y) ∈R2 : 1 ≤x2 + y2 ≤4 ª on the plane R2. On a separate drawing, shade in the set X. 6. Sketch the set X = © (x, y) ∈R2 : y < x2ª on R2. Shade in the set X. Venn Diagrams 21 1.7 Venn Diagrams In thinking about sets, it is sometimes helpful to draw informal, schematic diagrams of them. In doing this we often represent a set with a circle (or oval), which we regard as enclosing all the elements of the set. Such diagrams can illustrate how sets combine using various operations. For example, Figures 1.7(a–c) show two sets A and B that overlap in a middle region. The sets A ∪B, A ∩B and A −B are shaded. Such graphical representations of sets are called Venn diagrams, after their inventor, British logician John Venn, 1834–1923. A A A B B B A ∪B A ∩B A −B (a) (b) (c) Figure 1.7. Venn diagrams for two sets Though you are unlikely to draw Venn diagrams as a part of a proof of any theorem, you will probably ﬁnd them to be useful “scratch work” devices that help you to understand how sets combine, and to develop strategies for proving certain theorems or solving certain problems. The remainder of this section uses Venn diagrams to explore how three sets can be combined using ∪and ∩. Let’s begin with the set A ∪B ∪C. Our deﬁnitions suggest this should consist of all elements which are in one or more of the sets A, B and C. Figure 1.8(a) shows a Venn diagram for this. Similarly, we think of A∩B∩C as all elements common to each of A, B and C, so in Figure 1.8(b) the region belonging to all three sets is shaded. A A B B C C A ∪B ∪C A ∩B ∩C (a) (b) Figure 1.8. Venn diagrams for three sets 22 Sets We can also think of A ∩B ∩C as the two-step operation (A ∩B)∩C. In this expression the set A ∩B is represented by the region common to both A and B, and when we intersect this with C we get Figure 1.8(b). This is a visual representation of the fact that A ∩B ∩C = (A ∩B)∩C. Similarly, we have A ∩B ∩C = A ∩(B ∩C). Likewise, A ∪B ∪C = (A ∪B)∪C = A ∪(B ∪C). Notice that in these examples, where the expression either contains only the symbol ∪or only the symbol ∩, the placement of the parentheses is irrelevant, so we are free to drop them. It is analogous to the situations in algebra involving expressions (a+ b)+ c = a+(b + c) or (a· b)· c = a·(b · c). We tend to drop the parentheses and write simply a+ b + c or a· b · c. By contrast, in an expression like (a + b) · c the parentheses are absolutely essential because (a+ b)· c and a+(b · c) are generally not equal. Now let’s use Venn diagrams to help us understand the expressions (A ∪B)∩C and A ∪(B ∩C), which use a mix of ∪and ∩. Figure 1.9 shows how to draw a Venn diagram for (A∪B)∩C. In the drawing on the left, the set A ∪B is shaded with horizontal lines, while C is shaded with vertical lines. Thus the set (A ∪B)∩C is represented by the cross-hatched region where A ∪B and C overlap. The superﬂuous shadings are omitted in the drawing on the right showing the set (A ∪B)∩C. A A B B C C Figure 1.9. How to make a Venn diagram for (A ∪B)∩C Now think about A ∪(B ∩C). In Figure 1.10 the set A is shaded with horizontal lines, and B∩C is shaded with vertical lines. The union A∪(B∩C) is represented by the totality of all shaded regions, as shown on the right. A A B B C C Figure 1.10. How to make a Venn diagram for A ∪(B ∩C) Venn Diagrams 23 Compare the diagrams for (A ∪B)∩C and A ∪(B∩C) in Figures 1.9 and 1.10. The fact that the diagrams are diﬀerent indicates that (A ∪B)∩C ̸= A ∪(B ∩C) in general. Thus an expression such as A ∪B ∩C is absolutely meaningless because we can’t tell whether it means (A∪B)∩C or A∪(B∩C). In summary, Venn diagrams have helped us understand the following. Important Points: • If an expression involving sets uses only ∪, then parentheses are optional. • If an expression involving sets uses only ∩, then parentheses are optional. • If an expression uses both ∪and ∩, then parentheses are essential. In the next section we will study types of expressions that use only ∪ or only ∩. These expressions will not require the use of parentheses. Exercises for Section 1.7 1. Draw a Venn diagram for A. 2. Draw a Venn diagram for B −A. 3. Draw a Venn diagram for (A −B)∩C. 4. Draw a Venn diagram for (A ∪B)−C. 5. Draw Venn diagrams for A∪(B∩C) and (A∪B)∩(A∪C). Based on your drawings, do you think A ∪(B ∩C) = (A ∪B)∩(A ∪C)? 6. Draw Venn diagrams for A∩(B∪C) and (A∩B)∪(A∩C). Based on your drawings, do you think A ∩(B ∪C) = (A ∩B)∪(A ∩C)? 7. Suppose sets A and B are in a universal set U. Draw Venn diagrams for A ∩B and A ∪B. Based on your drawings, do you think it’s true that A ∩B = A ∪B? 8. Suppose sets A and B are in a universal set U. Draw Venn diagrams for A ∪B and A ∩B. Based on your drawings, do you think it’s true that A ∪B = A ∩B? 9. Draw a Venn diagram for (A ∩B)−C. 10. Draw a Venn diagram for (A −B)∪C. Following are Venn diagrams for expressions involving sets A,B and C. Write the corresponding expression. 11. A B C 12. A B C 13. A B C 14. A B C 24 Sets 1.8 Indexed Sets When a mathematical problem involves lots of sets, it is often convenient to keep track of them by using subscripts (also called indices). Thus instead of denoting three sets as A,B and C, we might instead write them as A1, A2 and A3. These are called indexed sets. Although we deﬁned union and intersection to be operations that combine two sets, you by now have no diﬃculty forming unions and intersections of three or more sets. (For instance, in the previous section we drew Venn diagrams for the intersection and union of three sets.) But let’s take a moment to write down careful deﬁnitions. Given sets A1, A2,..., An, the set A1 ∪A2 ∪A3 ∪···∪An consists of everything that is in at least one of the sets Ai. Likewise A1 ∩A2 ∩A3 ∩···∩An consists of everything that is common to all of the sets Ai. Here is a careful deﬁnition. Deﬁnition 1.7 Suppose A1, A2,..., An are sets. Then A1 ∪A2 ∪A3 ∪···∪An = © x : x ∈Ai for at least one set Ai, for 1 ≤i ≤n ª , A1 ∩A2 ∩A3 ∩···∩An = © x : x ∈Ai for every set Ai, for 1 ≤i ≤n ª . But if the number n of sets is large, these expressions can get messy. To overcome this, we now develop some notation that is akin to sigma notation. You already know that sigma notation is a convenient symbolism for expressing sums of many numbers. Given numbers a1,a2,a3,...,an, then n X i=1 ai = a1 + a2 + a3 +···+ an. Even if the list of numbers is inﬁnite, the sum ∞ X i=1 ai = a1 + a2 + a3 +···+ ai +··· is often still meaningful. The notation we are about to introduce is very similar to this. Given sets A1, A2, A3, ..., An, we deﬁne n [ i=1 Ai = A1 ∪A2 ∪A3 ∪···∪An and n \ i=1 Ai = A1 ∩A2 ∩A3 ∩···∩An. Example 1.9 Suppose A1 = © 0,2,5 ª, A2 = © 1,2,5 ª and A3 = © 2,5,7 ª. Then 3 [ i=1 Ai = A1 ∪A2 ∪A3 = © 0,1,2,5,7 ª and 3 \ i=1 Ai = A1 ∩A2 ∩A3 = © 2,5 ª . Indexed Sets 25 This notation is also used when the list of sets A1, A2, A3, ... is inﬁnite: ∞ [ i=1 Ai = A1 ∪A2 ∪A3 ∪··· = © x : x ∈Ai for at least one set Ai with 1 ≤i ª . ∞ \ i=1 Ai = A1 ∩A2 ∩A3 ∩··· = © x : x ∈Ai for every set Ai with 1 ≤i ª . Example 1.10 This example involves the following inﬁnite list of sets. A1 = © −1,0,1 ª , A2 = © −2,0,2 ª , A3 = © −3,0,3 ª , ··· , Ai = © −i,0, i ª , ··· Observe that ∞ [ i=1 Ai = Z, and ∞ \ i=1 Ai = © 0 ª. Here is a useful twist on our new notation. We can write 3 [ i=1 Ai = [ i∈{1,2,3} Ai, as this takes the union of the sets Ai for i = 1,2,3. Likewise: 3 \ i=1 Ai = \ i∈{1,2,3} Ai ∞ [ i=1 Ai = [ i∈N Ai ∞ \ i=1 Ai = \ i∈N Ai Here we are taking the union or intersection of a collection of sets Ai where i is an element of some set, be it © 1,2,3 ª or N. In general, the way this works is that we will have a collection of sets Ai for i ∈I, where I is the set of possible subscripts. The set I is called an index set. It is important to realize that the set I need not even consist of integers. (We could subscript with letters or real numbers, etc.) Since we are programmed to think of i as an integer, let’s make a slight notational change: We use α, not i, to stand for an element of I. Thus we are dealing with a collection of sets Aα for α ∈I. This leads to the following deﬁnition. Deﬁnition 1.8 If we have a set Aα for every α in some index set I, then [ α∈I Aα = © x : x ∈Aα for at least one set Aα with α ∈I ª \ α∈I Aα = © x : x ∈Aα for every set Aα with α ∈I ª . 26 Sets Example 1.11 Here the sets Aα will be subsets of R2. Let I = [0,2] = © x ∈R : 0 ≤x ≤2 ª. For each number α ∈I, let Aα = © (x,α) : x ∈R,1 ≤x ≤2 ª. For instance, given α = 1 ∈I the set A1 = © (x,1) : x ∈R,1 ≤x ≤2 ª is a horizontal line segment one unit above the x-axis and stretching between x = 1 and x = 2, as shown in Figure 1.11(a). Likewise Ap 2 = © (x, p 2) : x ∈R,1 ≤x ≤2 ª is a horizontal line segment p 2 units above the x-axis and stretching between x = 1 and x = 2. A few other of the Aα are shown in Figure 1.11(a), but they can’t all be drawn because there is one Aα for each of the inﬁnitely many numbers α ∈[0,2]. The totality of them covers the shaded region in Figure 1.11(b), so this region is the union of all the Aα. Since the shaded region is the set © (x, y) ∈R2 : 1 ≤x ≤2,0 ≤y ≤2 ª = [1,2]×[0,2], it follows that [ α∈[0,2] Aα = [1,2]×[0,2]. Likewise, since there is no point (x, y) that is in every set Aα, we have \ α∈[0,2] Aα = ;. x x y y 1 1 2 2 1 1 2 2 A0.25 A0.5 A1 A2 Ap 2 [ α∈[0,2] Aα (a) (b) Figure 1.11. The union of an indexed collection of sets One ﬁnal comment. Observe that Aα = [1,2]× © α ª, so the above expres-sions can be written as [ α∈[0,2] [1,2]× © α ª = [1,2]×[0,2] and \ α∈[0,2] [1,2]× © α ª = ;. Indexed Sets 27 Example 1.12 In this example our sets are indexed by R2. For any (a,b) ∈R2, let P(a,b) be the following subset of R3: P(a,b) = © (x, y, z) ∈R3 : ax+ by = 0 ª . In words, given a point (a,b) ∈R2, the corresponding set P(a,b) consists of all points (x, y, z) in R3 that satisfy the equation ax+ by = 0. From previous math courses you will recognize this as a plane in R3, that is, P(a,b) is a plane in R3. Moreover, since any point (0,0, z) on the z-axis automatically satisﬁes ax+ by = 0, each P(a,b) contains the z-axis. Figure 1.12 (left) shows the set P(1,2) = © (x, y, z) ∈R3 : x+2y = 0 ª. It is the vertical plane that intersects the xy-plane at the line x+2y = 0. x y z (−2,1,0) P(1,2) Figure 1.12. The sets P(a,b) are vertical planes containing the z-axis. For any point (a,b) ∈R2 with (a,b) ̸= (0,0), we can visualize P(a,b) as the vertical plane that cuts the xy-plane at the line ax + by = 0. Figure 1.12 (right) shows a few of the P(a,b). Since any two such planes intersect along the z-axis, and because the z-axis is a subset of every P(a,b), it is immediately clear that \ (a,b)∈R2 P(a,b) = © (0,0, z) : z ∈R ª = “the z-axis”. For the union, note that any given point (a,b, c) ∈R3 belongs to the set P(−b,a) because (x, y, z) = (a,b, c) satisﬁes the equation −bx+ ay = 0. (In fact, any (a,b, c) belongs to the special set P(0,0) = R3, which is the only P(a,b) that is not a plane.) Since any point in R3 belongs to some P(a,b) we have [ (a,b)∈R2 P(a,b) = R3. 28 Sets Exercises for Section 1.8 1. Suppose A1 = © a,b,d, e, g, f ª, A2 = © a,b, c,d ª, A3 = © b,d,a ª and A4 = © a,b,h ª. (a) 4 [ i=1 Ai = (b) 4 \ i=1 Ai = 2. Suppose    A1 = © 0,2,4,8,10,12,14,16,18,20,22,24 ª , A2 = © 0,3,6,9,12,15,18,21,24 ª , A3 = © 0,4,8,12,16,20,24 ª . (a) 3 [ i=1 Ai = (b) 3 \ i=1 Ai = 3. For each n ∈N, let An = © 0,1,2,3,...,n ª. (a) [ i∈N Ai = (b) \ i∈N Ai = 4. For each n ∈N, let An = © −2n,0,2n ª. (a) [ i∈N Ai = (b) \ i∈N Ai = 5. (a) [ i∈N [i, i +1] = (b) \ i∈N [i, i +1] = 6. (a) [ i∈N [0, i +1] = (b) \ i∈N [0, i +1] = 7. (a) [ i∈N R×[i, i +1] = (b) \ i∈N R×[i, i +1] = 8. (a) [ α∈R © α ª ×[0,1] = (b) \ α∈R © α ª ×[0,1] = 9. (a) [ X∈P(N) X = (b) \ X∈P(N) X = 10. (a) [ x∈[0,1] [x,1]×[0,x2] = (b) \ x∈[0,1] [x,1]×[0,x2] = 11. Is \ α∈I Aα ⊆ [ α∈I Aα always true for any collection of sets Aα with index set I? 12. If \ α∈I Aα = [ α∈I Aα, what do you think can be said about the relationships between the sets Aα? 13. If J ̸= ; and J ⊆I, does it follow that [ α∈J Aα ⊆ [ α∈I Aα? What about \ α∈J Aα ⊆ \ α∈I Aα? 14. If J ̸= ; and J ⊆I, does it follow that \ α∈I Aα ⊆ \ α∈J Aα? Explain. Sets that Are Number Systems 29 1.9 Sets that Are Number Systems In practice, the sets we tend to be most interested in often have special properties and structures. For example, the sets Z, Q and R are familiar number systems: Given such a set, any two of its elements can be added (or multiplied, etc.) together to produce another element in the set. These operations obey the familiar commutative, associative and distributive properties that we all have dealt with for years. Such properties lead to the standard algebraic techniques for solving equations. Even though we are concerned with the idea of proof, we will not ﬁnd it necessary to deﬁne, prove or verify such properties and techniques; we will accept them as the ground rules upon which our further deductions are based. We also accept as fact the natural ordering of the elements of N,Z,Q and R, so that (for example) the meaning of “5 < 7” is understood and does not need to be justiﬁed or explained. Similarly, if x ≤y and a ̸= 0, we know that ax ≤ay or ax ≥ay, depending on whether a is positive or negative. Another thing that our ingrained understanding of the ordering of numbers tells us is that any non-empty subset of N has a smallest element. In other words, if A ⊆N and A ̸= ;, then there is an element x0 ∈A that is smaller than every other element of A. (To ﬁnd it, start at 1, then move in increments to 2, 3, 4, etc., until you hit a number x0 ∈A; this is the smallest element of A.) Similarly, given an integer b, any non-empty subset A ⊆ © b, b +1, b +2, b +3,... ª has a smallest element. This fact is sometimes called the well-ordering principle. There is no need to remember this term, but do be aware that we will use this simple, intuitive idea often in proofs, usually without a second thought. The well-ordering principle seems innocent enough, but it actually says something very fundamental and special about the positive integers N. In fact, the corresponding statement about the positive real numbers is false: The subset A = © 1 n : n ∈N ª of the positive reals has no smallest element because for any x0 = 1 n ∈A that we might pick, there is always a smaller element 1 n+1 ∈A. One consequence of the well-ordering principle (as we will see below) is the familiar fact that any integer a can be divided by a non-zero integer b, resulting in a quotient q and remainder r. For example, b = 3 goes into a = 17 q = 5 times with remainder r = 2. In symbols, 17 = 5·3+2, or a = qb + r. This signiﬁcant fact is called the division algorithm. Fact 1.5 (The Division Algorithm) Given integers a and b with b > 0, there exist integers q and r for which a = qb + r and 0 ≤r < b. 30 Sets Although there is no harm in accepting the division algorithm without proof, note that it does follow from the well-ordering principle. Here’s how: Given integers a,b with b > 0, form the set A = © a−xb : x ∈Z, 0 ≤a−xb ª ⊆ © 0,1,2,3,... ª . (For example, if a = 17 and b = 3 then A = © 2,5,8,11,14,17,20,... ª is the set of positive integers obtained by adding multiples of 3 to 17. Notice that the remainder r = 2 of 17÷3 is the smallest element of this set.) In general, let r be the smallest element of the set A = © a−xb : x ∈Z, 0 ≤a−xb ª. Then r = a −qb for some x = q ∈Z, so a = qb + r. Moreover, 0 ≤r < b, as follows. The fact that r ∈A ⊆ © 0,1,2,3... ª implies 0 ≤r. In addition, it cannot happen that r ≥b: If this were the case, then the non-negative number r−b = (a−qb)−b = a−(q+1)b having form a−xb would be a smaller element of A than r, and r was explicitly chosen as the smallest element of A. Since it is not the case that r ≥b, it must be that r < b. Therefore 0 ≤r < b. We’ve now produced a q and an r for which a = qb + r and 0 ≤r < b. Moving on, it is time to clarify a small issue. This chapter asserted that all of mathematics can be described with sets. But at the same time we maintained that some mathematical entities are not sets. (For instance, our approach was to say that an individual number, such as 5, is not itself a set, though it may be an element of a set.) We have made this distinction because we need a place to stand as we explore sets: After all, it would appear suspiciously circular to declare that every mathematical entity is a set, and then go on to deﬁne a set as a collection whose members are sets! But to most mathematicians, saying “The number 5 is not a set,” is like saying “The number 5 is not a number.” The truth is that any number can itself be understood as a set. One way to do this is to begin with the identiﬁcation 0 = ;. Then 1 = {;} = {0}, and 2 = © ;,{;} ª = {0,1}, and 3 = © ;,{;},{;,{;}} ª = {0,1,2}. In general the natural number n is the set n = {0,1,2,...,n−1} of the n numbers (which are themselves sets) that come before it. We will not undertake such a study here, but the elements of the number systems Z, Q and R can all be deﬁned in terms of sets. (Even the operations of addition, multiplication, etc., can be deﬁned in set-theoretic terms.) In fact, mathematics itself can be regarded as the study of things that can be described as sets. Any mathematical entity is a set, whether or not we choose to think of it that way. Russell’s Paradox 31 1.10 Russell’s Paradox This section contains some background information that may be interesting, but is not used in the remainder of the book. The philosopher and mathematician Bertrand Russell (1872–1970) did groundbreaking work on the theory of sets and the foundations of mathematics. He was probably among the ﬁrst to understand how the misuse of sets can lead to bizarre and paradoxical situations. He is famous for an idea that has come to be known as Russell’s paradox. Russell’s paradox involves the following set of sets: A = © X : X is a set and X ∉X ª . (1.1) In words, A is the set of all sets that do not include themselves as elements. Most sets we can think of are in A. The set Z of integers is not an integer (i.e., Z ∉Z) and therefore Z ∈A. Also ; ∈A because ; is a set and ; ∉;. Is there a set that is not in A? Consider B = ©©©© ... ªªªª. Think of B as a box containing a box, containing a box, containing a box, and so on, forever. Or a set of Russian dolls, nested one inside the other, endlessly. The curious thing about B is that it has just one element, namely B itself: B = © ©©© ... ªªª \| {z } B ª . Thus B ∈B. As B does not satisfy B ∉B, Equation (1.1) says B ∉A. Russell’s paradox arises from the question “Is A an element of A?” For a set X, Equation (1.1) says X ∈A means the same thing as X ∉X. So for X = A, the previous line says A ∈A means the same thing as A ∉A. Conclusion: if A ∈A is true, then it is false; if A ∈A is false, then it is true. This is Russell’s paradox. Initially Russell’s paradox sparked a crisis among mathematicians. How could a mathematical statement be both true and false? This seemed to be in opposition to the very essence of mathematics. The paradox instigated a very careful examination of set theory and an evaluation of what can and cannot be regarded as a set. Eventually mathematicians settled upon a collection of axioms for set theory—the so-called Zermelo-Fraenkel axioms. One of these axioms is the well-ordering principle of the previous section. Another, the axiom of foundation, states that no non-empty set X is allowed to have the property X ∩x ̸= ; for all its elements x. This rules out such circularly deﬁned “sets” as X = © X ª introduced above. If we adhere to these axioms, then situations 32 Sets like Russell’s paradox disappear. Most mathematicians accept all this on faith and happily ignore the Zermelo-Fraenkel axioms. Paradoxes like Russell’s do not tend to come up in everyday mathematics—you have to go out of your way to construct them. Still, Russell’s paradox reminds us that precision of thought and lan-guage is an important part of doing mathematics. The next chapter deals with the topic of logic, a codiﬁcation of thought and language. Additional Reading on Sets. For a lively account of Bertrand Russell’s life and work (including his paradox), see the graphic novel Logicomix: An Epic Search For Truth, by Apostolos Doxiadis and Christos Papadimitriou. Also see cartoonist Jessica Hagy’s online strip Indexed—it is based largely on Venn diagrams. CHAPTER 2 Logic L ogic is a systematic way of thinking that allows us to deduce new infor-mation from old information and to parse the meanings of sentences. You use logic informally in everyday life and certainly also in doing mathe-matics. For example, suppose you are working with a certain circle, call it “Circle X,” and you have available the following two pieces of information. 1. Circle X has radius equal to 3. 2. If any circle has radius r, then its area is πr2 square units. You have no trouble putting these two facts together to get: 3. Circle X has area 9π square units. In doing this you are using logic to combine existing information to produce new information. Because deducing new information is central to mathematics, logic plays a fundamental role. This chapter is intended to give you a suﬃcient mastery of it. It is important to realize that logic is a process of deducing information correctly, not just deducing correct information. For example, suppose we were mistaken and Circle X actually had a radius of 4, not 3. Let’s look at our exact same argument again. 1. Circle X has radius equal to 3. 2. If any circle has radius r, then its area is πr2 square units. 3. Circle X has area 9π square units. The sentence “Circle X has radius equal to 3.” is now untrue, and so is our conclusion “Circle X has area 9π square units.” But the logic is perfectly correct; the information was combined correctly, even if some of it was false. This distinction between correct logic and correct information is signiﬁcant because it is often important to follow the consequences of an incorrect assumption. Ideally, we want both our logic and our information to be correct, but the point is that they are diﬀerent things. 34 Logic In proving theorems, we apply logic to information that is considered obviously true (such as “Any two points determine exactly one line.”) or is already known to be true (e.g., the Pythagorean theorem). If our logic is correct, then anything we deduce from such information will also be true (or at least as true as the “obviously true” information we began with). 2.1 Statements The study of logic begins with statements. A statement is a sentence or a mathematical expression that is either deﬁnitely true or deﬁnitely false. You can think of statements as pieces of information that are either correct or incorrect. Thus statements are pieces of information that we might apply logic to in order to produce other pieces of information (which are also statements). Example 2.1 Here are some examples of statements. They are all true. If a circle has radius r, then its area is πr2 square units. Every even number is divisible by 2. 2 ∈Z p 2 ∉Z N ⊆Z The set {0,1,2} has three elements. Some right triangles are isosceles. Example 2.2 Here are some additional statements. They are all false. All right triangles are isosceles. 5 = 2 p 2 ∉R Z ⊆N {0,1,2}∩N = ; Example 2.3 Here we pair sentences or expressions that are not state-ments with similar expressions that are statements. NOT Statements: Statements: Add 5 to both sides. Adding 5 to both sides of x−5 = 37 gives x = 42. Z 42 ∈Z 42 42 is not a number. What is the solution of 2x = 84? The solution of 2x = 84 is 42. Statements 35 Example 2.4 We will often use the letters P, Q, R and S to stand for speciﬁc statements. When more letters are needed we can use subscripts. Here are more statements, designated with letters. You decide which of them are true and which are false. P : For every integer n > 1, the number 2n −1 is prime. Q : Every polynomial of degree n has at most n roots. R : The function f (x) = x2 is continuous. S1 : Z ⊆; S2 : {0,−1,−2}∩N = ; Designating statements with letters (as was done above) is a very useful shorthand. In discussing a particular statement, such as “The function f (x) = x2 is continuous,” it is convenient to just refer to it as R to avoid having to write or say it many times. Statements can contain variables. Here is an example. P : If an integer x is a multiple of 6, then x is even. This is a sentence that is true. (All multiples of 6 are even, so no matter which multiple of 6 the integer x happens to be, it is even.) Since the sentence P is deﬁnitely true, it is a statement. When a sentence or statement P contains a variable such as x, we sometimes denote it as P(x) to indicate that it is saying something about x. Thus the above statement can be denoted as P(x) : If an integer x is a multiple of 6, then x is even. A statement or sentence involving two variables might be denoted P(x, y), and so on. It is quite possible for a sentence containing variables to not be a statement. Consider the following example. Q(x) : The integer x is even. Is this a statement? Whether it is true or false depends on just which integer x is. It is true if x = 4 and false if x = 7, etc. But without any stipulations on the value of x it is impossible to say whether Q(x) is true or false. Since it is neither deﬁnitely true nor deﬁnitely false, Q(x) cannot be a statement. A sentence such as this, whose truth depends on the value of one or more variables, is called an open sentence. The variables in an open sentence (or statement) can represent any type of entity, not just numbers. Here is an open sentence where the variables are functions: 36 Logic R(f , g) : The function f is the derivative of the function g. This open sentence is true if f (x) = 2x and g(x) = x2. It is false if f (x) = x3 and g(x) = x2, etc. We point out that a sentence such as R(f , g) (that involves variables) can be denoted either as R(f , g) or just R. We use the expression R(f , g) when we want to emphasize that the sentence involves variables. We will have more to say about open sentences later, but for now let’s return to statements. Statements are everywhere in mathematics. Any result or theorem that has been proved true is a statement. The quadratic formula and the Pythagorean theorem are both statements: P : The solutions of the equation ax2 + bx+ c = 0 are x = −b ± p b2 −4ac 2a . Q : If a right triangle has legs of lengths a and b and hypotenuse of length c, then a2 + b2 = c2. Here is a very famous statement, so famous, in fact, that it has a name. It is called Fermat’s last theorem after Pierre Fermat, a seventeenth-century French mathematician who scribbled it in the margin of a notebook. R : For all numbers a,b, c,n ∈N with n > 2, it is the case that an+bn ̸= cn. Fermat believed this statement was true. He noted that he could prove it was true, except his notebook’s margin was too narrow to contain his proof. It is doubtful that he really had a correct proof in mind, for after his death generations of brilliant mathematicians tried unsuccessfully to prove that his statement was true (or false). Finally, in 1993, Andrew Wiles of Princeton University announced that he had devised a proof. Wiles had worked on the problem for over seven years, and his proof runs through hundreds of pages. The moral of this story is that some true statements are not obviously true. Here is another statement famous enough to be named. It was ﬁrst posed in the eighteenth century by the German mathematician Christian Goldbach, and thus is called the Goldbach conjecture: S : Every even integer greater than 2 is a sum of two prime numbers. You must agree that S is either true or false. It appears to be true, because when you examine even numbers that are bigger than 2, they seem to be sums of two primes: 4 = 2+2, 6 = 3+3, 8 = 3+5, 10 = 5+5, 12 = 5+7, Statements 37 100 = 17+83 and so on. But that’s not to say there isn’t some large even number that’s not the sum of two primes. If such a number exists, then S is false. The thing is, in the over 260 years since Goldbach ﬁrst posed this problem, no one has been able to determine whether it’s true or false. But since it is clearly either true or false, S is a statement. This book is about the methods that can be used to prove that S (or any other statement) is true or false. To prove that a statement is true, we start with obvious statements (or other statements that have been proven true) and use logic to deduce more and more complex statements until ﬁnally we obtain a statement such as S. Of course some statements are more diﬃcult to prove than others, and S appears to be notoriously diﬃcult; we will concentrate on statements that are easier to prove. But the point is this: In proving that statements are true, we use logic to help us understand statements and to combine pieces of information to produce new pieces of information. In the next several sections we explore some standard ways that statements can be combined to form new statements, or broken down into simpler statements. Exercises for Section 2.1 Decide whether or not the following are statements. In the case of a statement, say if it is true or false, if possible. 1. Every real number is an even integer. 2. Every even integer is a real number. 3. If x and y are real numbers and 5x = 5y, then x = y. 4. Sets Z and N. 5. Sets Z and N are inﬁnite. 6. Some sets are ﬁnite. 7. The derivative of any polynomial of degree 5 is a polynomial of degree 6. 8. N ∉P(N). 9. cos(x) = −1 10. (R×N)∩(N×R) = N×N 11. The integer x is a multiple of 7. 12. If the integer x is a multiple of 7, then it is divisible by 7. 13. Either x is a multiple of 7, or it is not. 14. Call me Ishmael. 15. In the beginning, God created the heaven and the earth. 38 Logic 2.2 And, Or, Not The word “and” can be used to combine two statements to form a new statement. Consider for example the following sentence. R1 : The number 2 is even and the number 3 is odd. We recognize this as a true statement, based on our common-sense under-standing of the meaning of the word “and.” Notice that R1 is made up of two simpler statements: P : The number 2 is even. Q : The number 3 is odd. These are joined together by the word “and” to form the more complex statement R1. The statement R1 asserts that P and Q are both true. Since both P and Q are in fact true, the statement R1 is also true. Had one or both of P and Q been false, then R1 would be false. For instance, each of the following statements is false. R2 : The number 1 is even and the number 3 is odd. R3 : The number 2 is even and the number 4 is odd. R4 : The number 3 is even and the number 2 is odd. From these examples we see that any two statements P and Q can be combined to form a new statement “P and Q.” In the spirit of using letters to denote statements, we now introduce the special symbol ∧to stand for the word “and.” Thus if P and Q are statements, P ∧Q stands for the statement “P and Q.” The statement P ∧Q is true if both P and Q are true; otherwise it is false. This is summarized in the following table, called a truth table. P Q P ∧Q T T T T F F F T F F F F In this table, T stands for “True,” and F stands for “False.” (T and F are called truth values.) Each line lists one of the four possible combinations or truth values for P and Q, and the column headed by P ∧Q tells whether the statement P ∧Q is true or false in each case. And, Or, Not 39 Statements can also be combined using the word “or.” Consider the following four statements. S1 : The number 2 is even or the number 3 is odd. S2 : The number 1 is even or the number 3 is odd. S3 : The number 2 is even or the number 4 is odd. S4 : The number 3 is even or the number 2 is odd. In mathematics, the assertion “P or Q” is always understood to mean that one or both of P and Q is true. Thus statements S1, S2, S3 are all true, while S4 is false. The symbol ∨is used to stand for the word “or.” So if P and Q are statements, P ∨Q represents the statement “P or Q.” Here is the truth table. P Q P ∨Q T T T T F T F T T F F F It is important to be aware that the meaning of “or” expressed in the above table diﬀers from the way it is often used in everyday conversation. For example, suppose a university oﬃcial makes the following threat: You pay your tuition or you will be withdrawn from school. You understand that this means that either you pay your tuition or you will be withdrawn from school, but not both. In mathematics we never use the word “or” in such a sense. For us “or” means exactly what is stated in the table for ∨. Thus P ∨Q being true means one or both of P and Q is true. If we ever need to express the fact that exactly one of P and Q is true, we use one of the following constructions: P or Q, but not both. Either P or Q. Exactly one of P or Q. If the university oﬃcial were a mathematician, he might have qualiﬁed his statement in one of the following ways. Pay your tuition or you will be withdrawn from school, but not both. Either you pay your tuition or you will be withdrawn from school. 40 Logic To conclude this section, we mention another way of obtaining new statements from old ones. Given any statement P, we can form the new statement “It is not true that P.” For example, consider the following statement. The number 2 is even. This statement is true. Now change it by inserting the words “It is not true that” at the beginning: It is not true that the number 2 is even. This new statement is false. For another example, starting with the false statement “2 ∈;,” we get the true statement “It is not true that 2 ∈;.” We use the symbol ∼to stand for the words “It’s not true that,” so ∼P means “It’s not true that P.” We often read ∼P simply as “not P.” Unlike ∧and ∨, which combine two statements, the symbol ∼just alters a single statement. Thus its truth table has just two lines, one for each possible truth value of P. P ∼P T F F T The statement ∼P is called the negation of P. The negation of a speciﬁc statement can be expressed in numerous ways. Consider P : The number 2 is even. Here are several ways of expressing its negation. ∼P : It’s not true that the number 2 is even. ∼P : It is false that the number 2 is even. ∼P : The number 2 is not even. In this section we’ve learned how to combine or modify statements with the operations ∧, ∨and ∼. Of course we can also apply these operations to open sentences or a mixture of open sentences and statements. For example, (x is an even integer)∧(3 is an odd integer) is an open sentence that is a combination of an open sentence and a statement. Conditional Statements 41 Exercises for Section 2.2 Express each statement or open sentence in one of the forms P ∧Q, P ∨Q, or ∼P. Be sure to also state exactly what statements P and Q stand for. 1. The number 8 is both even and a power of 2. 2. The matrix A is not invertible. 3. x ̸= y 4. x < y 5. y ≥x 6. There is a quiz scheduled for Wednesday or Friday. 7. The number x equals zero, but the number y does not. 8. At least one of the numbers x and y equals 0. 9. x ∈A −B 10. x ∈A ∪B 11. A ∈ © X ∈P(N) : \|X\| < ∞ ª 12. Happy families are all alike, but each unhappy family is unhappy in its own way. (Leo Tolstoy, Anna Karenina) 13. Human beings want to be good, but not too good, and not all the time. (George Orwell) 14. A man should look for what is, and not for what he thinks should be. (Albert Einstein) 2.3 Conditional Statements There is yet another way to combine two statements. Suppose we have in mind a speciﬁc integer a. Consider the following statement about a. R : If the integer a is a multiple of 6, then a is divisible by 2. We immediately spot this as a true statement based on our knowledge of integers and the meanings of the words “if” and “then.” If integer a is a multiple of 6, then a is even, so therefore a is divisible by 2. Notice that R is built up from two simpler statements: P : The integer a is a multiple of 6. Q : The integer a is divisible by 2. R : If P, then Q. In general, given any two statements P and Q whatsoever, we can form the new statement “If P, then Q.” This is written symbolically as P ⇒Q which we read as “If P, then Q,” or “P implies Q.” Like ∧and ∨, the symbol ⇒has a very speciﬁc meaning. When we assert that the statement P ⇒Q is true, we mean that if P is true then Q must also be true. (In other words we mean that the condition P being true forces Q to be true.) A statement of form P ⇒Q is called a conditional statement because it means Q will be true under the condition that P is true. 42 Logic You can think of P ⇒Q as being a promise that whenever P is true, Q will be true also. There is only one way this promise can be broken (i.e. be false) and that is if P is true but Q is false. Thus the truth table for the promise P ⇒Q is as follows: P Q P ⇒Q T T T T F F F T T F F T Perhaps you are bothered by the fact that P ⇒Q is true in the last two lines of this table. Here’s an example to convince you that the table is correct. Suppose your professor makes the following promise: If you pass the ﬁnal exam, then you will pass the course. Your professor is making the promise (You pass the exam) ⇒(You pass the course). Under what circumstances did she lie? There are four possible scenarios, depending on whether or not you passed the exam and whether or not you passed the course. These scenarios are tallied in the following table. You pass exam You pass course (You pass exam) ⇒(You pass course) T T T T F F F T T F F T The ﬁrst line describes the scenario where you pass the exam and you pass the course. Clearly the professor kept her promise, so we put a T in the third column to indicate that she told the truth. In the second line, you passed the exam, but your professor gave you a failing grade in the course. In this case she broke her promise, and the F in the third column indicates that what she said was untrue. Now consider the third row. In this scenario you failed the exam but still passed the course. How could that happen? Maybe your professor felt sorry for you. But that doesn’t make her a liar. Her only promise was that if you passed the exam then you would pass the course. She did not say Conditional Statements 43 passing the exam was the only way to pass the course. Since she didn’t lie, then she told the truth, so there is a T in the third column. Finally look at the fourth row. In that scenario you failed the exam and you failed the course. Your professor did not lie; she did exactly what she said she would do. Hence the T in the third column. In mathematics, whenever we encounter the construction “If P, then Q” it means exactly what the truth table for ⇒expresses. But of course there are other grammatical constructions that also mean P ⇒Q. Here is a summary of the main ones. If P, then Q. Q if P. Q whenever P. Q, provided that P. Whenever P, then also Q. P is a suﬃcient condition for Q. For Q, it is suﬃcient that P. Q is a necessary condition for P. For P, it is necessary that Q. P only if Q.                                    P ⇒Q These can all be used in the place of (and mean exactly the same thing as) “If P, then Q.” You should analyze the meaning of each one and convince yourself that it captures the meaning of P ⇒Q. For example, P ⇒Q means the condition of P being true is enough (i.e., suﬃcient) to make Q true; hence “P is a suﬃcient condition for Q.” The wording can be tricky. Often an everyday situation involving a conditional statement can help clarify it. For example, consider your professor’s promise: (You pass the exam) ⇒(You pass the course) This means that your passing the exam is a suﬃcient (though perhaps not necessary) condition for your passing the course. Thus your professor might just as well have phrased her promise in one of the following ways. Passing the exam is a suﬃcient condition for passing the course. For you to pass the course, it is suﬃcient that you pass the exam. However, when we want to say “If P, then Q” in everyday conversation, we do not normally express this as “Q is a necessary condition for P” or “P only if Q.” But such constructions are not uncommon in mathematics. To understand why they make sense, notice that P ⇒Q being true means 44 Logic that it’s impossible that P is true but Q is false, so in order for P to be true it is necessary that Q is true; hence “Q is a necessary condition for P.” And this means that P can only be true if Q is true, i.e., “P only if Q.” Exercises for Section 2.3 Without changing their meanings, convert each of the following sentences into a sentence having the form “If P, then Q.” 1. A matrix is invertible provided that its determinant is not zero. 2. For a function to be continuous, it is suﬃcient that it is diﬀerentiable. 3. For a function to be integrable, it is necessary that it is continuous. 4. A function is rational if it is a polynomial. 5. An integer is divisible by 8 only if it is divisible by 4. 6. Whenever a surface has only one side, it is non-orientable. 7. A series converges whenever it converges absolutely. 8. A geometric series with ratio r converges if \|r\| < 1. 9. A function is integrable provided the function is continuous. 10. The discriminant is negative only if the quadratic equation has no real solutions. 11. You fail only if you stop writing. (Ray Bradbury) 12. People will generally accept facts as truth only if the facts agree with what they already believe. (Andy Rooney) 13. Whenever people agree with me I feel I must be wrong. (Oscar Wilde) 2.4 Biconditional Statements It is important to understand that P ⇒Q is not the same as Q ⇒P. To see why, suppose that a is some integer and consider the statements (a is a multiple of 6) ⇒ (a is divisible by 2), (a is divisible by 2) ⇒ (a is a multiple of 6). The ﬁrst statement asserts that if a is a multiple of 6 then a is divisible by 2. This is clearly true, for any multiple of 6 is even and therefore divisible by 2. The second statement asserts that if a is divisible by 2 then it is a multiple of 6. This is not necessarily true, for a = 4 (for instance) is divisible by 2, yet not a multiple of 6. Therefore the meanings of P ⇒Q and Q ⇒P are in general quite diﬀerent. The conditional statement Q ⇒P is called the converse of P ⇒Q, so a conditional statement and its converse express entirely diﬀerent things. Biconditional Statements 45 But sometimes, if P and Q are just the right statements, it can happen that P ⇒Q and Q ⇒P are both necessarily true. For example, consider the statements (a is even) ⇒ (a is divisible by 2), (a is divisible by 2) ⇒ (a is even). No matter what value a has, both of these statements are true. Since both P ⇒Q and Q ⇒P are true, it follows that (P ⇒Q)∧(Q ⇒P) is true. We now introduce a new symbol ⇔to express the meaning of the statement (P ⇒Q)∧(Q ⇒P). The expression P ⇔Q is understood to have exactly the same meaning as (P ⇒Q)∧(Q ⇒P). According to the previous section, Q ⇒P is read as “P if Q,” and P ⇒Q can be read as “P only if Q.” Therefore we pronounce P ⇔Q as “P if and only if Q.” For example, given an integer a, we have the true statement (a is even) ⇔(a is divisible by 2), which we can read as “Integer a is even if and only if a is divisible by 2.” The truth table for ⇔is shown below. Notice that in the ﬁrst and last rows, both P ⇒Q and Q ⇒P are true (according to the truth table for ⇒), so (P ⇒Q) ∧(Q ⇒P) is true, and hence P ⇔Q is true. However, in the middle two rows one of P ⇒Q or Q ⇒P is false, so (P ⇒Q)∧(Q ⇒P) is false, making P ⇔Q false. P Q P ⇔Q T T T T F F F T F F F T Compare the statement R : (a is even) ⇔(a is divisible by 2) with this truth table. If a is even then the two statements on either side of ⇔ are true, so according to the table R is true. If a is odd then the two statements on either side of ⇔are false, and again according to the table R is true. Thus R is true no matter what value a has. In general, P ⇔Q being true means P and Q are both true or both false. Not surprisingly, there are many ways of saying P ⇔Q in English. The following constructions all mean P ⇔Q: 46 Logic P if and only if Q. P is a necessary and suﬃcient condition for Q. For P it is necessary and suﬃcient that Q. If P, then Q, and conversely.          P ⇔Q The ﬁrst three of these just combine constructions from the previous section to express that P ⇒Q and Q ⇒P. In the last one, the words “...and conversely” mean that in addition to “If P, then Q” being true, the converse statement “If Q, then P” is also true. Exercises for Section 2.4 Without changing their meanings, convert each of the following sentences into a sentence having the form “P if and only if Q.” 1. For matrix A to be invertible, it is necessary and suﬃcient that det(A) ̸= 0. 2. If a function has a constant derivative then it is linear, and conversely. 3. If xy = 0 then x = 0 or y = 0, and conversely. 4. If a ∈Q then 5a ∈Q, and if 5a ∈Q then a ∈Q. 5. For an occurrence to become an adventure, it is necessary and suﬃcient for one to recount it. (Jean-Paul Sartre) 2.5 Truth Tables for Statements You should now know the truth tables for ∧, ∨, ∼, ⇒and ⇔. They should be internalized as well as memorized. You must understand the symbols thoroughly, for we now combine them to form more complex statements. For example, suppose we want to convey that one or the other of P and Q is true but they are not both true. No single symbol expresses this, but we could combine them as (P ∨Q)∧∼(P ∧Q), which literally means: P or Q is true, and it is not the case that both P and Q are true. This statement will be true or false depending on the truth values of P and Q. In fact we can make a truth table for the entire statement. Begin as usual by listing the possible true/false combinations of P and Q on four lines. The statement (P ∨Q)∧∼(P ∧Q) contains the individual statements (P ∨Q) and (P ∧Q), so we next tally their truth values in the third and fourth columns. The ﬁfth column lists values for ∼(P ∧Q), and these Truth Tables for Statements 47 are just the opposites of the corresponding entries in the fourth column. Finally, combining the third and ﬁfth columns with ∧, we get the values for (P ∨Q)∧∼(P ∧Q) in the sixth column. P Q (P ∨Q) (P ∧Q) ∼(P ∧Q) (P ∨Q)∧∼(P ∧Q) T T T T F F T F T F T T F T T F T T F F F F T F This truth table tells us that (P ∨Q)∧∼(P ∧Q) is true precisely when one but not both of P and Q are true, so it has the meaning we intended. (Notice that the middle three columns of our truth table are just “helper columns” and are not necessary parts of the table. In writing truth tables, you may choose to omit such columns if you are conﬁdent about your work.) For another example, consider the following familiar statement con-cerning two real numbers x and y: The product xy equals zero if and only if x = 0 or y = 0. This can be modeled as (xy = 0) ⇔(x = 0 ∨y = 0). If we introduce letters P,Q and R for the statements xy = 0, x = 0 and y = 0, it becomes P ⇔(Q∨R). Notice that the parentheses are necessary here, for without them we wouldn’t know whether to read the statement as P ⇔(Q ∨R) or (P ⇔Q)∨R. Making a truth table for P ⇔(Q∨R) entails a line for each T/F combina-tion for the three statements P, Q and R. The eight possible combinations are tallied in the ﬁrst three columns of the following table. P Q R Q ∨R P ⇔(Q ∨R) T T T T T T T F T T T F T T T T F F F F F T T T F F T F T F F F T T F F F F F T We ﬁll in the fourth column using our knowledge of the truth table for ∨. Finally the ﬁfth column is ﬁlled in by combining the ﬁrst and fourth columns with our understanding of the truth table for ⇔. The resulting table gives the true/false values of P ⇔(Q ∨R) for all values of P,Q and R. 48 Logic Notice that when we plug in various values for x and y, the statements P : xy = 0, Q : x = 0 and R : y = 0 have various truth values, but the statement P ⇔(Q ∨R) is always true. For example, if x = 2 and y = 3, then P,Q and R are all false. This scenario is described in the last row of the table, and there we see that P ⇔(Q ∨R) is true. Likewise if x = 0 and y = 7, then P and Q are true and R is false, a scenario described in the second line of the table, where again P ⇔(Q ∨R) is true. There is a simple reason why P ⇔(Q∨R) is true for any values of x and y: It is that P ⇔(Q∨R) represents (xy = 0) ⇔(x = 0 ∨y = 0), which is a true mathematical statement. It is absolutely impossible for it to be false. This may make you wonder about the lines in the table where P ⇔(Q∨R) is false. Why are they there? The reason is that P ⇔(Q ∨R) can also represent a false statement. To see how, imagine that at the end of the semester your professor makes the following promise. You pass the class if and only if you get an “A” on the ﬁnal or you get a “B” on the ﬁnal. This promise has the form P ⇔(Q ∨R), so its truth values are tabulated in the above table. Imagine it turned out that you got an “A” on the exam but failed the course. Then surely your professor lied to you. In fact, P is false, Q is true and R is false. This scenario is reﬂected in the sixth line of the table, and indeed P ⇔(Q ∨R) is false (i.e., it is a lie). The moral of this example is that people can lie, but true mathematical statements never lie. We close this section with a word about the use of parentheses. The symbol ∼is analogous to the minus sign in algebra. It negates the expression it precedes. Thus ∼P ∨Q means (∼P) ∨Q, not ∼(P ∨Q). In ∼(P ∨Q), the value of the entire expression P ∨Q is negated. Exercises for Section 2.5 Write a truth table for the logical statements in problems 1–9: 1. P ∨(Q ⇒R) 2. (Q ∨R) ⇔(R ∧Q) 3. ∼(P ⇒Q) 4. ∼(P ∨Q)∨(∼P) 5. (P∧∼P)∨Q 6. (P∧∼P)∧Q 7. (P∧∼P) ⇒Q 8. P ∨(Q∧∼R) 9. ∼(∼P∨∼Q) 10. Suppose the statement ((P ∧Q)∨R) ⇒(R ∨S) is false. Find the truth values of P,Q,R and S. (This can be done without a truth table.) 11. Suppose P is false and that the statement (R ⇒S) ⇔(P ∧Q) is true. Find the truth values of R and S. (This can be done without a truth table.) Logical Equivalence 49 2.6 Logical Equivalence In contemplating the truth table for P ⇔Q, you probably noticed that P ⇔Q is true exactly when P and Q are both true or both false. In other words, P ⇔Q is true precisely when at least one of the statements P ∧Q or ∼P∧∼Q is true. This may tempt us to say that P ⇔Q means the same thing as (P ∧Q)∨(∼P∧∼Q). To see if this is really so, we can write truth tables for P ⇔Q and (P ∧Q) ∨(∼P∧∼Q). In doing this, it is more eﬃcient to put these two statements into the same table, as follows. (This table has helper columns for the intermediate expressions ∼P, ∼Q, (P ∧Q) and (∼P∧∼Q).) P Q ∼P ∼Q (P ∧Q) (∼P∧∼Q) (P ∧Q)∨(∼P∧∼Q) P ⇔Q T T F F T F T T T F F T F F F F F T T F F F F F F F T T F T T T The table shows that P ⇔Q and (P ∧Q)∨(∼P∧∼Q) have the same truth value, no matter the values P and Q. It is as if P ⇔Q and (P∧Q)∨(∼P∧∼Q) are algebraic expressions that are equal no matter what is “plugged into” variables P and Q. We express this state of aﬀairs by writing P ⇔Q = (P ∧Q)∨(∼P∧∼Q) and saying that P ⇔Q and (P ∧Q)∨(∼P∧∼Q) are logically equivalent. In general, two statements are logically equivalent if their truth values match up line-for-line in a truth table. Logical equivalence is important because it can give us diﬀerent (and potentially useful) ways of looking at the same thing. As an example, the following table shows that P ⇒Q is logically equivalent to (∼Q) ⇒(∼P). P Q ∼P ∼Q (∼Q) ⇒(∼P) P ⇒Q T T F F T T T F F T F F F T T F T T F F T T T T The fact that P ⇒Q = (∼Q) ⇒(∼P) is useful because so many theorems have the form P ⇒Q. As we will see in Chapter 5, proving such a theorem may be easier if we express it in the logically equivalent form (∼Q) ⇒(∼P). 50 Logic There are two pairs of logically equivalent statements that come up again and again throughout this book and beyond. They are prevalent enough to be digniﬁed by a special name: DeMorgan’s laws. Fact 2.1 (DeMorgan’s Laws) 1. ∼(P ∧Q) = (∼P)∨(∼Q) 2. ∼(P ∨Q) = (∼P)∧(∼Q) The ﬁrst of DeMorgan’s laws is veriﬁed by the following table. You are asked to verify the second in one of the exercises. P Q ∼P ∼Q P ∧Q ∼(P ∧Q) (∼P)∨(∼Q) T T F F T F F T F F T F T T F T T F F T T F F T T F T T DeMorgan’s laws are actually very natural and intuitive. Consider the statement ∼(P ∧Q), which we can interpret as meaning that it is not the case that both P and Q are true. If it is not the case that both P and Q are true, then at least one of P or Q is false, in which case (∼P)∨(∼Q) is true. Thus ∼(P ∧Q) means the same thing as (∼P)∨(∼Q). DeMorgan’s laws can be very useful. Suppose we happen to know that some statement having form ∼(P ∨Q) is true. The second of DeMorgan’s laws tells us that (∼Q)∧(∼P) is also true, hence ∼P and ∼Q are both true as well. Being able to quickly obtain such additional pieces of information can be extremely useful. Here is a summary of some signiﬁcant logical equivalences. Those that are not immediately obvious can be veriﬁed with a truth table. P ⇒Q = (∼Q) ⇒(∼P) Contrapositive law (2.1) ∼(P ∧Q) = ∼P∨∼Q ∼(P ∨Q) = ∼P∧∼Q o DeMorgan’s laws (2.2) P ∧Q = Q ∧P P ∨Q = Q ∨P o Commutative laws (2.3) P ∧(Q ∨R) = (P ∧Q)∨(P ∧R) P ∨(Q ∧R) = (P ∨Q)∧(P ∨R) o Distributive laws (2.4) P ∧(Q ∧R) = (P ∧Q)∧R P ∨(Q ∨R) = (P ∨Q)∨R o Associative laws (2.5) Notice how the distributive law P ∧(Q ∨R) = (P ∧Q) ∨(P ∧R) has the same structure as the distributive law p ·(q + r) = p · q + p · r from algebra. Quantiﬁers 51 Concerning the associative laws, the fact that P∧(Q∧R) = (P∧Q)∧R means that the position of the parentheses is irrelevant, and we can write this as P ∧Q ∧R without ambiguity. Similarly, we may drop the parentheses in an expression such as P ∨(Q ∨R). But parentheses are essential when there is a mix of ∧and ∨, as in P ∨(Q ∧R). Indeed, P ∨(Q ∧R) and (P ∨Q)∧R are not logically equivalent. (See Exercise 13 for Section 2.6, below.) Exercises for Section 2.6 A. Use truth tables to show that the following statements are logically equivalent. 1. P ∧(Q ∨R) = (P ∧Q)∨(P ∧R) 2. P ∨(Q ∧R) = (P ∨Q)∧(P ∨R) 3. P ⇒Q = (∼P)∨Q 4. ∼(P ∨Q) = (∼P)∧(∼Q) 5. ∼(P ∨Q ∨R) = (∼P)∧(∼Q)∧(∼R) 6. ∼(P ∧Q ∧R) = (∼P)∨(∼Q)∨(∼R) 7. P ⇒Q = (P∧∼Q) ⇒(Q∧∼Q) 8. ∼P ⇔Q = (P ⇒∼Q)∧(∼Q ⇒P) B. Decide whether or not the following pairs of statements are logically equivalent. 9. P ∧Q and ∼(∼P∨∼Q) 10. (P ⇒Q)∨R and ∼((P∧∼Q)∧∼R) 11. (∼P)∧(P ⇒Q) and ∼(Q ⇒P) 12. ∼(P ⇒Q) and P∧∼Q 13. P ∨(Q ∧R) and (P ∨Q)∧R 14. P ∧(Q∨∼Q) and (∼P) ⇒(Q∧∼Q) 2.7 Quantiﬁers Using symbols ∧, ∨, ∼, ⇒and ⇔, we can deconstruct many English sentences into a symbolic form. As we have seen, this symbolic form can help us understand the logical structure of sentences and how diﬀerent sentences may actually have the same meaning (as in logical equivalence). But these symbols alone are not powerful enough to capture the full meaning of every statement. To help overcome this defect, we introduce two new symbols that correspond to common mathematical phrases. The symbol “∀” stands for the phrase “For all” or “For every.” The symbol “∃” stands for the phrase “There exists a” or “There is a.” Thus the statement For every n ∈Z, 2n is even, can be expressed in either of the following ways: ∀n ∈Z, 2n is even, ∀n ∈Z, E(2n). 52 Logic Likewise, a statement such as There exists a subset X of N for which \|X\| = 5. can be translated as ∃X,(X ⊆N)∧(\|X\| = 5) or ∃X ⊆N,\|X\| = 5 or ∃X ∈P(N),\|X\| = 5. The symbols ∀and ∃are called quantiﬁers because they refer in some sense to the quantity (i.e., all or some) of the variable that follows them. Symbol ∀is called the universal quantiﬁer and ∃is called the existen-tial quantiﬁer. Statements which contain them are called quantiﬁed statements. A statement beginning with ∀is called a universally quan-tiﬁed statement, and one beginning with ∃is called an existentially quantiﬁed statement. Example 2.5 The following English statements are paired with their translations into symbolic form. Every integer that is not odd is even. ∀n ∈Z,∼(n is odd ) ⇒(n is even), or ∀n ∈Z,∼O(n) ⇒E(n). There is an integer that is not even. ∃n ∈Z,∼E(n). For every real number x, there is a real number y for which y3 = x. ∀x ∈R,∃y ∈R, y3 = x. Given any two rational numbers a and b, it follows that ab is rational. ∀a,b ∈Q,ab ∈Q. Given a set S (such as, but not limited to, N, Z, Q etc.), a quantiﬁed statement of form ∀x ∈S,P(x) is understood to be true if P(x) is true for every x ∈S. If there is at least one x ∈S for which P(x) is false, then ∀x ∈S,P(x) is a false statement. Similarly, ∃x ∈S,P(x) is true provided that P(x) is true for at least one element x ∈S; otherwise it is false. Thus each statement in Example 2.5 is true. Here are some examples of quantiﬁed statements that are false: Example 2.6 The following false quantiﬁed statements are paired with their translations. Every integer is even. ∀n ∈Z,E(n). Quantiﬁers 53 There is an integer n for which n2 = 2. ∃n ∈Z,n2 = 2. For every real number x, there is a real number y for which y2 = x. ∀x ∈R,∃y ∈R, y2 = x. Given any two rational numbers a and b, it follows that p ab is rational. ∀a,b ∈Q, p ab ∈Q. Example 2.7 When a statement contains two quantiﬁers you must be very alert to their order, for reversing the order can change the meaning. Consider the following statement from Example 2.5. ∀x ∈R,∃y ∈R, y3 = x. This statement is true, for no matter what number x is there exists a number y = 3 px for which y3 = x. Now reverse the order of the quantiﬁers to get the new statement ∃y ∈R,∀x ∈R, y3 = x. This new statement says that there exists a particular number y with the property that y3 = x for every real number x. Since no number y can have this property, the statement is false. The two statements above have entirely diﬀerent meanings. Quantiﬁed statements are often misused in casual conversation. Maybe you’ve heard someone say “All students do not pay full tuition.” when they mean “Not all students pay full tuition.” While the mistake is perhaps marginally forgivable in casual conversation, it must never be made in a mathematical context. Do not say “All integers are not even.” because that means there are no even integers. Instead, say “Not all integers are even.” Exercises for Section 2.7 Write the following as English sentences. Say whether they are true or false. 1. ∀x ∈R,x2 > 0 2. ∀x ∈R,∃n ∈N,xn ≥0 3. ∃a ∈R,∀x ∈R,ax = x 4. ∀X ∈P(N), X ⊆R 5. ∀n ∈N,∃X ∈P(N),\|X\| < n 6. ∃n ∈N,∀X ∈P(N),\|X\| < n 7. ∀X ⊆N,∃n ∈Z,\|X\| = n 8. ∀n ∈Z, ∃X ⊆N,\|X\| = n 9. ∀n ∈Z,∃m ∈Z,m = n+5 10. ∃m ∈Z,∀n ∈Z,m = n+5 54 Logic 2.8 More on Conditional Statements It is time to address a very important point about conditional statements that contain variables. To motivate this, let’s return to the following example concerning integers x: (x is a multiple of 6) ⇒(x is even). As noted earlier, since every multiple of 6 is even, this is a true statement no matter what integer x is. We could even underscore this fact by writing this true statement as ∀x ∈Z, (x is a multiple of 6) ⇒(x is even). But now switch things around to get the diﬀerent statement (x is even) ⇒(x is a multiple of 6). This is true for some values of x such as −6, 12, 18, etc., but false for others (such as 2, 4, etc.). Thus we do not have a statement, but rather an open sentence. (Recall from Section 2.1 that an open sentence is a sentence whose truth value depends on the value of a certain variable or variables.) However, by putting a universal quantiﬁer in front we get ∀x ∈Z, (x is even) ⇒(x is a multiple of 6), which is deﬁnitely false, so this new expression is a statement, not an open sentence. In general, given any two open sentences P(x) and Q(x) about integers x, the expression ∀x ∈Z, P(x) ⇒Q(x) is either true or false, so it is a statement, not an open sentence. Now we come to the very important point. In mathematics, whenever P(x) and Q(x) are open sentences concerning elements x in some set S (depending on context), an expression of form P(x) ⇒Q(x) is understood to be the statement ∀x ∈S, P(x) ⇒Q(x). In other words, if a conditional statement is not explicitly quantiﬁed then there is an implied universal quantiﬁer in front of it. This is done because statements of the form ∀x ∈S, P(x) ⇒Q(x) are so common in mathematics that we would get tired of putting the ∀x ∈S in front of them. Thus the following sentence is a true statement (as it is true for all x). If x is a multiple of 6, then x is even. Translating English to Symbolic Logic 55 Likewise, the next sentence is a false statement (as it is not true for all x). If x is even, then x is a multiple of 6. This leads to the following signiﬁcant interpretation of a conditional statement, which is more general than (but consistent with) the interpre-tation from Section 2.3. Deﬁnition 2.1 If P and Q are statements or open sentences, then “If P, then Q,” is a statement. This statement is true if it’s impossible for P to be true while Q is false. It is false if there is at least one instance in which P is true but Q is false. Thus the following are true statements: If x ∈R, then x2 +1 > 0. If a function f is diﬀerentiable on R, then f is continuous on R. Likewise, the following are false statements: If p is a prime number, then p is odd. (2 is prime.) If f is a rational function, then f has an asymptote. (x2 is rational.) 2.9 Translating English to Symbolic Logic In writing (and reading) proofs of theorems, we must always be alert to the logical structure and meanings of the sentences. Sometimes it is necessary or helpful to parse them into expressions involving logic symbols. This may be done mentally or on scratch paper, or occasionally even explicitly within the body of a proof. The purpose of this section is to give you suﬃcient practice in translating English sentences into symbolic form so that you can better understand their logical structure. Here are some examples: Example 2.8 Consider the Mean Value Theorem from Calculus: If f is continuous on the interval [a,b] and diﬀerentiable on (a,b), then there is a number c ∈(a,b) for which f ′(c) = f (b)−f (a) b−a . Here is a translation to symbolic form: ³¡ f cont. on [a,b] ¢ ∧ ¡ f is diﬀ. on (a,b) ¢´ ⇒ ³ ∃c ∈(a,b), f ′(c) = f (b)−f (a) b−a ´ . 56 Logic Example 2.9 Consider Goldbach’s conjecture, from Section 2.1: Every even integer greater than 2 is the sum of two primes. This can be translated in the following ways, where P is the set of prime numbers and S = {4,6,8,10,...} is the set of even integers greater than 2. ¡ n ∈S ¢ ⇒ ¡ ∃p, q ∈P, n = p + q ¢ ∀n ∈S, ∃p, q ∈P, n = p + q These translations of Goldbach’s conjecture illustrate an important point. The ﬁrst has the basic structure (n ∈S) ⇒Q(n) and the second has structure ∀n ∈S, Q(n), yet they have exactly the same meaning. This is signiﬁcant. Every universally quantiﬁed statement can be expressed as a conditional statement. Fact 2.2 Suppose S is a set and Q(x) is a statement about x for each x ∈S. The following statements mean the same thing: ∀x ∈S, Q(x) (x ∈S) ⇒Q(x). This fact is signiﬁcant because so many theorems have the form of a conditional statement. (The Mean Value Theorem is an example!) In proving a theorem we have to think carefully about what it says. Sometimes a theorem will be expressed as a universally quantiﬁed statement but it will be more convenient to think of it as a conditional statement. Understanding the above fact allows us to switch between the two forms. We close this section with some ﬁnal points. In translating a state-ment, be attentive to its intended meaning. Don’t jump into, for example, automatically replacing every “and” with ∧and “or” with ∨. An example: At least one of the integers x and y is even. Don’t be led astray by the presence of the word “and.” The meaning of the statement is that one or both of the numbers is even, so it should be translated with “or,” not “and”: (x is even) ∨(y is even). Finally, the logical meaning of “but” can be captured by “and.” The sentence “The integer x is even, but the integer y is odd,” is translated as (x is even) ∧(y is odd). Negating Statements 57 Exercises for Section 2.9 Translate each of the following sentences into symbolic logic. 1. If f is a polynomial and its degree is greater than 2, then f ′ is not constant. 2. The number x is positive but the number y is not positive. 3. If x is prime then px is not a rational number. 4. For every prime number p there is another prime number q with q > p. 5. For every positive number ε, there is a positive number δ for which \|x−a\| < δ implies \|f (x)−f (a)\| < ε. 6. For every positive number ε there is a positive number M for which \|f (x)−b\| < ε, whenever x > M. 7. There exists a real number a for which a+ x = x for every real number x. 8. I don’t eat anything that has a face. 9. If x is a rational number and x ̸= 0, then tan(x) is not a rational number. 10. If sin(x) < 0, then it is not the case that 0 ≤x ≤π. 11. There is a Providence that protects idiots, drunkards, children and the United States of America. (Otto von Bismarck) 12. You can fool some of the people all of the time, and you can fool all of the people some of the time, but you can’t fool all of the people all of the time. (Abraham Lincoln) 13. Everything is funny as long as it is happening to somebody else. (Will Rogers) 2.10 Negating Statements Given a statement R, the statement ∼R is called the negation of R. If R is a complex statement, then it is often the case that its negation ∼R can be written in a simpler or more useful form. The process of ﬁnding this form is called negating R. In proving theorems it is often necessary to negate certain statements. We now investigate how to do this. We have already examined part of this topic. DeMorgan’s laws ∼(P ∧Q) = (∼P)∨(∼Q) (2.6) ∼(P ∨Q) = (∼P)∧(∼Q) (2.7) (from Section 2.6) can be viewed as rules that tell us how to negate the statements P ∧Q and P ∨Q. Here are some examples that illustrate how DeMorgan’s laws are used to negate statements involving “and” or “or.” 58 Logic Example 2.10 Consider negating the following statement. R : You can solve it by factoring or with the quadratic formula. Now, R means (You can solve it by factoring) ∨(You can solve it with Q.F.), which we will denote as P ∨Q. The negation of this is ∼(P ∨Q) = (∼P)∧(∼Q). Therefore, in words, the negation of R is ∼R : You can’t solve it by factoring and you can’t solve it with the quadratic formula. Maybe you can ﬁnd ∼R without invoking DeMorgan’s laws. That is good; you have internalized DeMorgan’s laws and are using them unconsciously. Example 2.11 We will negate the following sentence. R : The numbers x and y are both odd. This statement means (x is odd) ∧(y is odd), so its negation is ∼ ¡ (x is odd)∧(y is odd) ¢ = ∼(x is odd) ∨∼(y is odd) = (x is even)∨(y is even). Therefore the negation of R can be expressed in the following ways: ∼R : The number x is even or the number y is even. ∼R : At least one of x and y is even. Now let’s move on to a slightly diﬀerent kind of problem. It’s often necessary to ﬁnd the negations of quantiﬁed statements. For example, consider ∼(∀x ∈N, P(x)). Reading this in words, we have the following: It is not the case that P(x) is true for all natural numbers x. This means P(x) is false for at least one x. In symbols, this is ∃x ∈N, ∼P(x). Thus ∼(∀x ∈N, P(x)) = ∃x ∈N, ∼P(x). Similarly, you can reason out that ∼(∃x ∈N, P(x)) = ∀x ∈N, ∼P(x). In general: ∼(∀x ∈S, P(x)) = ∃x ∈S, ∼P(x), (2.8) ∼(∃x ∈S, P(x)) = ∀x ∈S, ∼P(x). (2.9) Negating Statements 59 Example 2.12 Consider negating the following statement. R : The square of every real number is non-negative. Symbolically, R can be expressed as ∀x ∈R, x2 ≥0, and thus its negation is ∼(∀x ∈R, x2 ≥0) = ∃x ∈R, ∼(x2 ≥0) = ∃x ∈R, x2 < 0. In words, this is ∼R : There exists a real number whose square is negative. Observe that R is true and ∼R is false. You may be able to get ∼R immediately, without using Equation (2.8) as we did above. If so, that is good; if not, you will probably be there soon. If a statement has multiple quantiﬁers, negating it will involve several iterations of Equations (2.8) and (2.9). Consider the following: S : For every real number x there is a real number y for which y3 = x. This statement asserts any real number x has a cube root y, so it’s true. Symbolically S can be expressed as ∀x ∈R,∃y ∈R, y3 = x. Let’s work out the negation of this statement. ∼(∀x ∈R,∃y ∈R, y3 = x) = ∃x ∈R,∼(∃y ∈R, y3 = x) = ∃x ∈R,∀y ∈R, ∼(y3 = x) = ∃x ∈R,∀y ∈R, y3 ̸= x. Therefore the negation is the following (false) statement. ∼S : There is a real number x for which y3 ̸= x for all real numbers y. In writing proofs you will sometimes have to negate a conditional statement P ⇒Q. The remainder of this section describes how to do this. To begin, look at the expression ∼(P ⇒Q), which literally says “P ⇒Q is false.” You know from the truth table for ⇒that the only way that P ⇒Q can be false is if P is true and Q is false. Therefore ∼(P ⇒Q) = P∧∼Q. ∼(P ⇒Q) = P∧∼Q (2.10) (In fact, in Exercise 12 of Section 2.6, you used a truth table to verify that these two statements are indeed logically equivalent.) 60 Logic Example 2.13 Negate the following statement about a particular (i.e., constant) number a. R : If a is odd then a2 is odd. Using Equation (2.10), we get the following negation. ∼R : a is odd and a2 is not odd. Example 2.14 This example is like the previous one, but the constant a is replaced by a variable x. We will negate the following statement. R : If x is odd then x2 is odd. As discussed in Section 2.8, we interpret this as the universally quantiﬁed statement R : ∀x ∈Z, (x odd) ⇒(x2 odd). By Equations (2.8) and (2.10), we get the following negation for R. ∼ ¡ ∀x ∈Z, (x odd) ⇒(x2 odd) ¢ = ∃x ∈Z,∼ ¡ (x odd) ⇒(x2 odd) ¢ = ∃x ∈Z,(x odd)∧∼(x2 odd). Translating back into words, we have ∼R : There is an odd integer x whose square is not odd. Notice that R is true and ∼R is false. The above Example 2.14 showed how to negate a conditional statement P(x) ⇒Q(x). This type of problem can sometimes be embedded in more complex negation. See Exercise 5 below (and its solution). Exercises for Section 2.10 Negate the following sentences. 1. The number x is positive, but the number y is not positive. 2. If x is prime, then px is not a rational number. 3. For every prime number p, there is another prime number q with q > p. 4. For every positive number ε, there is a positive number δ such that \|x−a\| < δ implies \|f (x)−f (a)\| < ε. 5. For every positive number ε, there is a positive number M for which \|f (x)−b\| < ε whenever x > M. 6. There exists a real number a for which a+ x = x for every real number x. Logical Inference 61 7. I don’t eat anything that has a face. 8. If x is a rational number and x ̸= 0, then tan(x) is not a rational number. 9. If sin(x) < 0, then it is not the case that 0 ≤x ≤π. 10. If f is a polynomial and its degree is greater than 2, then f ′ is not constant. 11. You can fool all of the people all of the time. 12. Whenever I have to choose between two evils, I choose the one I haven’t tried yet. (Mae West) 2.11 Logical Inference Suppose we know that a statement of form P ⇒Q is true. This tells us that whenever P is true, Q will also be true. By itself, P ⇒Q being true does not tell us that either P or Q is true (they could both be false, or P could be false and Q true). However if in addition we happen to know that P is true then it must be that Q is true. This is called a logical inference: Given two true statements we can infer that a third statement is true. In this instance true statements P ⇒Q and P are “added together” to get Q. This is described below with P ⇒Q and P stacked one atop the other with a line separating them from Q. The intended meaning is that P ⇒Q combined with P produces Q. P ⇒Q P Q P ⇒Q ∼Q ∼P P ∨Q ∼P Q Two other logical inferences are listed above. In each case you should convince yourself (based on your knowledge of the relevant truth tables) that the truth of the statements above the line forces the statement below the line to be true. Following are some additional useful logical inferences. The ﬁrst expresses the obvious fact that if P and Q are both true then the statement P ∧Q will be true. On the other hand, P ∧Q being true forces P (also Q) to be true. Finally, if P is true, then P ∨Q must be true, no matter what statement Q is. P Q P ∧Q P ∧Q P P P ∨Q These inferences are so intuitively obvious that they scarcely need to be mentioned. However, they represent certain patterns of reasoning that we will frequently apply to sentences in proofs, so we should be cognizant of the fact that we are using them. 62 Logic 2.12 An Important Note It is important to be aware of the reasons that we study logic. There are three very signiﬁcant reasons. First, the truth tables we studied tell us the exact meanings of the words such as “and,” “or,” “not” and so on. For instance, whenever we use or read the “If..., then” construction in a mathematical context, logic tells us exactly what is meant. Second, the rules of inference provide a system in which we can produce new information (statements) from known information. Finally, logical rules such as DeMorgan’s laws help us correctly change certain statements into (potentially more useful) statements with the same meaning. Thus logic helps us understand the meanings of statements and it also produces new meaningful statements. Logic is the glue that holds strings of statements together and pins down the exact meaning of certain key phrases such as the “If..., then” or “For all” constructions. Logic is the common language that all mathematicians use, so we must have a ﬁrm grip on it in order to write and understand mathematics. But despite its fundamental role, logic’s place is in the background of what we do, not the forefront. From here on, the beautiful symbols ∧, ∨, ⇒, ⇔, ∼, ∀and ∃are rarely written. But we are aware of their meanings constantly. When reading or writing a sentence involving mathematics we parse it with these symbols, either mentally or on scratch paper, so as to understand the true and unambiguous meaning. CHAPTER 3 Counting I t may seem peculiar that a college-level text has a chapter on counting. At its most basic level, counting is a process of pointing to each object in a collection and calling oﬀ“one, two, three,...” until the quantity of objects is determined. How complex could that be? Actually, counting can become quite subtle, and in this chapter we explore some of its more sophisticated aspects. Our goal is still to answer the question “How many?” but we introduce mathematical techniques that bypass the actual process of counting individual objects. Almost every branch of mathematics uses some form of this “sophisti-cated counting.” Many such counting problems can be modeled with the idea of a list, so we start there. 3.1 Counting Lists A list is an ordered sequence of objects. A list is denoted by an opening parenthesis, followed by the objects, separated by commas, followed by a closing parenthesis. For example (a,b, c,d, e) is a list consisting of the ﬁrst ﬁve letters of the English alphabet, in order. The objects a,b, c,d, e are called the entries of the list; the ﬁrst entry is a, the second is b, and so on. If the entries are rearranged we get a diﬀerent list, so, for instance, (a,b, c,d, e) ̸= (b,a, c,d, e). A list is somewhat like a set, but instead of being a mere collection of objects, the entries of a list have a deﬁnite order. Note that for sets we have © a,b, c,d, e ª = © b,a, c,d, e ª , but—as noted above—the analogous equality for lists does not hold. Unlike sets, lists are allowed to have repeated entries. For example (5,3,5,4,3,3) is a perfectly acceptable list, as is (S,O,S). The number of entries in a list is called its length. Thus (5,3,5,4,3,3) has length six, and (S,O,S) has length three. 64 Counting Occasionally we may get sloppy and write lists without parentheses and commas; for instance, we may express (S,O,S) as SOS if there is no danger of confusion. But be alert that doing this can lead to ambiguity. Is it reasonable that (9,10,11) should be the same as 91011? If so, then (9,10,11) = 91011 = (9,1,0,1,1), which makes no sense. We will thus almost always adhere to the parenthesis/comma notation for lists. Lists are important because many real-world phenomena can be de-scribed and understood in terms of them. For example, your phone number (with area code) can be identiﬁed as a list of ten digits. Order is essential, for rearranging the digits can produce a diﬀerent phone number. A byte is another important example of a list. A byte is simply a length-eight list of 0’s and 1’s. The world of information technology revolves around bytes. To continue our examples of lists, (a,15) is a list of length two. Likewise (0,(0,1,1)) is a list of length two whose second entry is a list of length three. The list (N,Z,R) has length three, and each of its entries is a set. We emphasize that for two lists to be equal, they must have exactly the same entries in exactly the same order. Consequently if two lists are equal, then they must have the same length. Said diﬀerently, if two lists have diﬀerent lengths, then they are not equal. For example, (0,0,0,0,0,0) ̸= (0,0,0,0,0). For another example note that ( g,r,o, c, e,r, y, l, i,s,t) bread milk eggs mustard coﬀee ̸= ¡ ¢ because the list on the left has length eleven but the list on the right has just one entry (a piece of paper with some words on it). There is one very special list which has no entries at all. It is called the empty list, and is denoted (). It is the only list whose length is zero. One often needs to count up the number of possible lists that satisfy some condition or property. For example, suppose we need to make a list of length three having the property that the ﬁrst entry must be an element of the set © a,b, c ª, the second entry must be in © 5,7 ª and the third entry must be in © a,x ª. Thus (a,5,a) and (b,5,a) are two such lists. How many such lists are there all together? To answer this question, imagine making the list by selecting the ﬁrst element, then the second and ﬁnally the third. This is described in Figure 3.1. The choices for the ﬁrst list entry are a,b or c, and the left of the diagram branches out in three directions, one for each choice. Once this choice is made there are two choices (5 or 7) for the second entry, and this is described graphically by two branches from each of the three choices for the ﬁrst entry. This pattern continues Counting Lists 65 for the choice for the third entry, which is either a or x. Thus, in the diagram there are 3·2·2 = 12 paths from left to right, each corresponding to a particular choice for each entry in the list. The corresponding lists are tallied at the far-right end of each path. So, to answer our original question, there are 12 possible lists with the stated properties. ﬁrst choice second choice third choice Resulting list a b c 5 7 5 7 5 7 a x a x a x x a x a x a (a,5,a) (a,5,x) (a,7,a) (a,7,x) (b,5,a) (b,5,x) (b,7,a) (b,7,x) (c,5,a) (c,5,x) (c,7,a) (c,7,x) Figure 3.1. Constructing lists of length 3 We summarize the type of reasoning used above in an important fact called the multiplication principle. Fact 3.1 (Multiplication Principle) Suppose in making a list of length n there are a1 possible choices for the ﬁrst entry, a2 possible choices for the second entry, a3 possible choices for the third entry and so on. Then the total number of diﬀerent lists that can be made this way is the product a1 · a2 · a3 ····· an. So, for instance, in the above example we had a1 = 3,a2 = 2 and a3 = 2, so the total number of lists was a1 · a2 · a3 = 3·2·2 = 12. Now let’s look at some additional examples of how the multiplication principle can be used. Example 3.1 A standard license plate consists of three letters followed by four numbers. For example, JRB-4412 and MMX-8901 are two standard license plates. (Vanity plates such as LV2COUNT are not included among the standard plates.) How many diﬀerent standard license plates are possible? 66 Counting To answer this question, note that any standard license plate such as JRB-4412 corresponds to a length-7 list (J,R,B,4,4,1,2), so the question can be answered by counting how many such lists are possible. We use the multiplication principle. There are a1 = 26 possibilities (one for each letter of the alphabet) for the ﬁrst entry of the list. Similarly, there are a2 = 26 possibilities for the second entry and a3 = 26 possibilities for the third entry. There are a4 = 10 possibilities for the fourth entry, and likewise a5 = a6 = a7 = 10. Therefore there are a total of a1 · a2 · a3 · a4 · a5 · a6 · a7 = 26·26·26·10·10·10·10 = 175,760,000 possible standard license plates. There are two types of list-counting problems. On one hand, there are situations in which the same symbol or symbols may appear multiple times in diﬀerent entries of the list. For example, license plates or telephone numbers can have repeated symbols. The sequence CCX-4144 is a perfectly valid license plate in which the symbols C and 4 appear more than once. On the other hand, for some lists repeated symbols do not make sense or are not allowed. For instance, imagine drawing 5 cards from a standard 52-card deck and laying them in a row. Since no 2 cards in the deck are identical, this list has no repeated entries. We say that repetition is allowed in the ﬁrst type of list and repetition is not allowed in the second kind of list. (Often we call a list in which repetition is not allowed a non-repetitive list.) The following example illustrates the diﬀerence. Example 3.2 Consider making lists from symbols A, B, C, D, E, F, G. (a) How many length-4 lists are possible if repetition is allowed? (b) How many length-4 lists are possible if repetition is not allowed? (c) How many length-4 lists are possible if repetition is not allowed and the list must contain an E? (d) How many length-4 lists are possible if repetition is allowed and the list must contain an E? Solutions: (a) Imagine the list as containing four boxes that we ﬁll with selections from the letters A,B,C,D,E,F and G, as illustrated below. , , , ( ) 7 choices 7 choices 7 choices 7 choices There are seven possibilities for the contents of each box, so the total number of lists that can be made this way is 7·7·7·7 = 2401. Counting Lists 67 (b) This problem is the same as the previous one except that repetition is not allowed. We have seven choices for the ﬁrst box, but once it is ﬁlled we can no longer use the symbol that was placed in it. Hence there are only six possibilities for the second box. Once the second box has been ﬁlled we have used up two of our letters, and there are only ﬁve left to choose from in ﬁlling the third box. Finally, when the third box is ﬁlled we have only four possible letters for the last box. , , , ( ) 7 choices 6 choices 5 choices 4 choices Thus the answer to our question is that there are 7·6·5·4 = 840 lists in which repetition does not occur. (c) We are asked to count the length-4 lists in which repetition is not allowed and the symbol E must appear somewhere in the list. Thus E occurs once and only once in each such list. Let us divide these lists into four categories depending on whether the E occurs as the ﬁrst, second, third or fourth entry. These four types of lists are illustrated below. , , , , , , , , , , , , E E E E Type 1 Type 2 Type 3 Type 4 ( ( ( ( ) ) ) ) 6 choices 6 choices 6 choices 6 choices 5 choices 5 choices 5 choices 5 choices 4 choices 4 choices 4 choices 4 choices Consider lists of the ﬁrst type, in which the E appears in the ﬁrst entry. We have six remaining choices (A,B,C,D,F or G) for the second entry, ﬁve choices for the third entry and four choices for the fourth entry. Hence there are 6·5·4 = 120 lists having an E in the ﬁrst entry. As indicated in the above diagram, there are also 6·5·4 = 120 lists having an E in the second, third or fourth entry. Thus there are 120+120+120+120 = 480 such lists all together. (d) Now we must ﬁnd the number of length-four lists where repetition is allowed and the list must contain an E. Our strategy is as follows. By Part (a) of this exercise there are 7 ·7 ·7 ·7 = 74 = 2401 lists where repetition is allowed. Obviously this is not the answer to our current question, for many of these lists contain no E. We will subtract from 2401 the number of lists that do not contain an E. In making a list that does not contain an E, we have six choices for each list entry (because 68 Counting we can choose any one of the six letters A,B,C,D,F or G). Thus there are 6·6·6·6 = 64 = 1296 lists that do not have an E. Therefore the ﬁnal answer to our question is that there are 2401 −1296 = 1105 lists with repetition allowed that contain at least one E. Perhaps you wondered if Part (d) of Example 3.2 could be solved with a setup similar to that of Part (c). Let’s try doing it that way. We want to count the length-4 lists (with repetition allowed) that contain at least one E. The following diagram is adapted from Part (c), the only diﬀerence being that there are now seven choices in each slot because we are allowed to repeat any of the seven letters. , , , , , , , , , , , , E E E E Type 1 Type 2 Type 3 Type 4 ( ( ( ( ) ) ) ) 7 choices 7 choices 7 choices 7 choices 7 choices 7 choices 7 choices 7 choices 7 choices 7 choices 7 choices 7 choices This gives a total of 73 + 73 + 73 + 73 = 1372 lists, an answer that is substantially larger than the (correct) value of 1105 that we got in our solution to Part (d) above. It is not hard to see what went wrong. The list (E,E, A,B) is of type 1 and type 2, so it got counted twice. Similarly (E,E,C,E) is of type 1, 3 and 4, so it got counted three times. In fact, you can ﬁnd many similar lists that were counted multiple times. In solving counting problems, we must always be careful to avoid this kind of double-counting or triple-counting, or worse. Exercises for Section 3.1 Note: A calculator may be helpful for some of the exercises in this chapter. This is the only chapter for which a calculator may be helpful. (As for the exercises in the other chapters, a calculator makes them harder.) 1. Consider lists made from the letters T,H,E,O,R,Y, with repetition allowed. (a) How many length-4 lists are there? (b) How many length-4 lists are there that begin with T ? (c) How many length-4 lists are there that do not begin with T ? 2. Airports are identiﬁed with 3-letter codes. For example, the Richmond, Virginia airport has the code RIC, and Portland, Oregon has PDX. How many diﬀerent 3-letter codes are possible? 3. How many lists of length 3 can be made from the symbols A,B,C,D,E,F if... Counting Lists 69 (a) ... repetition is allowed. (b) ... repetition is not allowed. (c) ... repetition is not allowed and the list must contain the letter A. (d) ... repetition is allowed and the list must contain the letter A. 4. Five cards are dealt oﬀof a standard 52-card deck and lined up in a row. How many such line-ups are there in which all 5 cards are of the same suit? 5. Five cards are dealt oﬀof a standard 52-card deck and lined up in a row. How many such line-ups are there in which all 5 cards are of the same color (i.e., all black or all red)? 6. Five cards are dealt oﬀof a standard 52-card deck and lined up in a row. How many such line-ups are there in which exactly one of the 5 cards is a queen? 7. This problem involves 8-digit binary strings such as 10011011 or 00001010 (i.e., 8-digit numbers composed of 0’s and 1’s). (a) How many such strings are there? (b) How many such strings end in 0? (c) How many such strings have the property that their second and fourth digits are 1’s? (d) How many such strings have the property that their second or fourth digits are 1’s? 8. This problem concerns lists made from the symbols A,B,C,D,E. (a) How many such length-5 lists have at least one letter repeated? (b) How many such length-6 lists have at least one letter repeated? 9. This problem concerns 4-letter codes made from the letters A,B,C,D,...,Z. (a) How many such codes can be made? (b) How many such codes have no two consecutive letters the same? 10. This problem concerns lists made from the letters A,B,C,D,E,F,G,H,I,J. (a) How many length-5 lists can be made from these letters if repetition is not allowed and the list must begin with a vowel? (b) How many length-5 lists can be made from these letters if repetition is not allowed and the list must begin and end with a vowel? (c) How many length-5 lists can be made from these letters if repetition is not allowed and the list must contain exactly one A? 11. This problem concerns lists of length 6 made from the letters A,B,C,D,E,F,G,H. How many such lists are possible if repetition is not allowed and the list contains two consecutive vowels? 12. Consider the lists of length six made with the symbols P, R, O, F, S, where repetition is allowed. (For example, the following is such a list: (P,R,O,O,F,S).) How many such lists can be made if the list must end in an S and the symbol O is used more than once? 70 Counting 3.2 Factorials In working the examples from Section 3.1, you may have noticed that often we need to count the number of non-repetitive lists of length n that are made from n symbols. In fact, this particular problem occurs with such frequency that a special idea, called a factorial, is introduced to handle it. The table below motivates this idea. The ﬁrst column lists successive integer values n (beginning with 0) and the second column contains a set © A,B,··· ª of n symbols. The third column contains all the possible non-repetitive lists of length n which can be made from these symbols. Finally, the last column tallies up how many lists there are of that type. Notice that when n = 0 there is only one list of length 0 that can be made from 0 symbols, namely the empty list (). Thus the value 1 is entered in the last column of that row. n Symbols Non-repetitive lists of length n made from the symbols n! 0 ©ª ( ) 1 1 © A ª (A) 1 2 © A,B ª (A,B),(B, A) 2 3 © A,B,C ª (A,B,C),(A,C,B),(B,C, A),(B, A,C),(C, A,B),(C,B, A) 6 4 © A,B,C,D ª (A,B,C,D),(A,B,D,C),(A,C,B,D),(A,C,D,B),(A,D,B,C),(A,D,C,B) (B,A,C,D),(B,A,D,C),(B,C,A,D),(B,C,D,A),(B,D, A,C),(B,D,C,A) (C,A,B,D),(C,A,D,B),(C,B,A,D),(C,B,D,A),(C,D,A,B),(C,D,B,A) (D,A,B,C),(D,A,C,B),(D,B,A,C),(D,B,C,A),(D,C,A,B),(D,C,B,A) 24 . . . . . . . . . . . . For n > 0, the number that appears in the last column can be computed using the multiplication principle. The number of non-repetitive lists of length n that can be made from n symbols is n(n−1)(n−2)···3·2·1. Thus, for instance, the number in the last column of the row for n = 4 is 4·3·2·1 = 24. The number that appears in the last column of Row n is called the factorial of n. It is denoted as n! (read “n factorial”). Here is the deﬁnition: Deﬁnition 3.1 If n is a non-negative integer, then the factorial of n, denoted n!, is the number of non-repetitive lists of length n that can be made from n symbols. Thus 0! = 1 and 1! = 1. If n > 1, then n! = n(n−1)(n−2)···3·2·1. Factorials 71 It follows that 0! = 1 1! = 1 2! = 2·1 = 2 3! = 3·2·1 = 6 4! = 4·3·2·1 = 24 5! = 5·4·3·2·1 = 120 6! = 6·5·4·3·2·1 = 720, and so on. Students are often tempted to say 0! = 0, but this is wrong. The correct value is 0! = 1, as the above deﬁnition and table tell us. Here is another way to see that 0! must equal 1: Notice that 5! = 5·4·3·2·1 = 5·(4·3·2·1) = 5·4!. Also 4! = 4·3·2·1 = 4·(3·2·1) = 4·3!. Generalizing this reasoning, we have the following formula. n! = n·(n−1)! (3.1) Plugging in n = 1 gives 1! = 1·(1−1)! = 1·0!, that is, 1! = 1·0!. If we mistakenly thought 0! were 0, this would give the incorrect result 1! = 0. We round out our discussion of factorials with an example. Example 3.3 This problem involves making lists of length seven from the symbols 0,1,2,3,4,5 and 6. (a) How many such lists are there if repetition is not allowed? (b) How many such lists are there if repetition is not allowed and the ﬁrst three entries must be odd? (c) How many such lists are there in which repetition is allowed, and the list must contain at least one repeated number? To answer the ﬁrst question, note that there are seven symbols, so the number of lists is 7! = 5040. To answer the second question, notice that the set © 0,1,2,3,4,5,6 ª contains three odd numbers and four even numbers. Thus in making the list the ﬁrst three entries must be ﬁlled by odd numbers and the ﬁnal four must be ﬁlled with even numbers. By the multiplication principle, the number of such lists is 3·2·1·4·3·2·1 = 3!4! = 144. To answer the third question, notice that there are 77 = 823,543 lists in which repetition is allowed. The set of all such lists includes lists that are non-repetitive (e.g., (0,6,1,2,4,3,5)) as well as lists that have some repetition (e.g., (6,3,6,2,0,0,0)). We want to compute the number of lists that have at least one repeated number. To ﬁnd the answer we can subtract the number of non-repetitive lists of length seven from the total number of possible lists of length seven. Therefore the answer is 77 −7! = 823,543−5040 = 818,503. 72 Counting We close this section with a formula that combines the ideas of the ﬁrst and second sections of the present chapter. One of the main problems of Section 3.1 was as follows: Given n symbols, how many non-repetitive lists of length k can be made from the n symbols? We learned how to apply the multiplication principle to obtain the answer n(n−1)(n−2)···(n−k +1). Notice that by cancellation this value can also be written as n(n−1)(n−2)···(n−k +1)(n−k)(n−k −1)···3·2·1 (n−k)(n−k −1)···3·2·1 = n! (n−k)!. We summarize this as follows: Fact 3.2 The number of non-repetitive lists of length k whose entries are chosen from a set of n possible entries is n! (n−k)!. For example, consider ﬁnding the number of non-repetitive lists of length ﬁve that can be made from the symbols 1,2,3,4,5,6,7,8. We will do this two ways. By the multiplication principle, the answer is 8·7·6·5·4 = 6720. Using the formula from Fact 3.2, the answer is 8! (8−5)! = 8! 3! = 40,320 6 = 6720. The new formula isn’t really necessary, but it is a nice repackaging of an old idea and will prove convenient in the next section. Exercises for Section 3.2 1. What is the smallest n for which n! has more than 10 digits? 2. For which values of n does n! have n or fewer digits? 3. How many 5-digit positive integers are there in which there are no repeated digits and all digits are odd? 4. Using only pencil and paper, ﬁnd the value of 100! 95! . 5. Using only pencil and paper, ﬁnd the value of 120! 118!. 6. There are two 0’s at the end of 10! = 3,628,800. Using only pencil and paper, determine how many 0’s are at the end of the number 100!. 7. Compute how many 9-digit numbers can be made from the digits 1,2,3,4,5,6,7,8,9 if repetition is not allowed and all the odd digits occur ﬁrst (on the left) followed by all the even digits (i.e. as in 137598264, but not 123456789). 8. Compute how many 7-digit numbers can be made from the digits 1,2,3,4,5,6,7 if there is no repetition and the odd digits must appear in an unbroken sequence. (Examples: 3571264 or 2413576 or 2467531, etc., but not 7234615.) Counting Subsets 73 9. There is a very interesting function Γ : [0,∞) →R called the gamma function. It is deﬁned as Γ(x) = R∞ 0 tx−1e−tdt. It has the remarkable property that if x ∈N, then Γ(x) = (x−1)!. Check that this is true for x = 1,2,3,4. Notice that this function provides a way of extending factorials to numbers other than integers. Since Γ(n) = (n−1)! for all n ∈N, we have the formula n! = Γ(n+1). But Γ can be evaluated at any number in [0,∞), not just at integers, so we have a formula for n! for any n ∈[0,∞). Extra credit: Compute π!. 10. There is another signiﬁcant function called Stirling’s formula that provides an approximation to factorials. It states that n! ≈ p 2πn ¡ n e ¢n. It is an approximation to n! in the sense that n! p 2πn ¡ n e ¢n approaches 1 as n approaches ∞. Use Stirling’s formula to ﬁnd approximations to 5!, 10!, 20! and 50!. 3.3 Counting Subsets The previous two sections were concerned with counting the number of lists that can be made by selecting k entries from a set of n possible entries. We turn now to a related question: How many subsets can be made by selecting k elements from a set with n elements? To highlight the diﬀerences between these two problems, look at the set A = © a,b, c,d, e ª. First, think of the non-repetitive lists that can be made from selecting two entries from A. By Fact 3.2 (on the previous page), there are 5! (5−2)! = 5! 3! = 120 6 = 20 such lists. They are as follows. (a,b), (a, c), (a,d), (a, e), (b, c), (b,d), (b, e), (c,d), (c, e) (d, e) (b,a), (c,a), (d,a), (e,a), (c,b), (d,b), (e,b), (d, c), (e, c) (e,d) Next consider the subsets of A that can made from selecting two ele-ments from A. There are only ten such subsets, as follows. © a,b ª , © a, c ª , © a,d ª , © a, e ª , © b, c ª , © b,d ª , © b, e ª , © c,d ª , © c, e ª , © d, e ª . The reason that there are more lists than subsets is that changing the order of the entries of a list produces a diﬀerent list, but changing the order of the elements of a set does not change the set. Using elements a,b ∈A, we can make two lists (a,b) and (b,a), but only one subset © a,b ª. In this section we are concerned not with counting lists, but with counting subsets. As was noted above, the basic question is this: How many subsets can be made by choosing k elements from an n-element set? We begin with some notation that gives a name to the answer to this question. 74 Counting Deﬁnition 3.2 If n and k are integers, then ¡n k ¢ denotes the number of subsets that can be made by choosing k elements from a set with n elements. The symbol ¡n k ¢ is read “n choose k.” (Some textbooks write C(n,k) instead of ¡n k ¢.) To illustrate this deﬁnition, the following table computes the values of ¡4 k ¢ for various values of k by actually listing all the subsets of the 4-element set A = © a,b, c,d ª that have cardinality k. The values of k appear in the far-left column. To the right of each k are all of the subsets (if any) of A of size k. For example, when k = 1, set A has four subsets of size k, namely © a ª, © b ª, © c ª and © d ª. Therefore ¡4 1 ¢ = 4. Similarly, when k = 2 there are six subsets of size k so ¡4 2 ¢ = 6. k k-element subsets of © a,b, c,d ª ¡4 k ¢ −1 ¡ 4 −1 ¢ = 0 0 ; ¡4 0 ¢ = 1 1 © a ª , © b ª , © c ª , © d ª ¡4 1 ¢ = 4 2 © a,b ª , © a, c ª , © a,d ª , © b, c ª , © b,d ª , © c,d ª ¡4 2 ¢ = 6 3 © a,b, c ª , © a,b,d ª , © a, c,d ª , © b, c,d ª ¡4 3 ¢ = 4 4 © a,b, c,d ª ¡4 4 ¢ = 1 5 ¡4 5 ¢ = 0 6 ¡4 6 ¢ = 0 When k = 0, there is only one subset of A that has cardinality k, namely the empty set, ;. Therefore ¡4 0 ¢ = 1. Notice that if k is negative or greater than \|A\|, then A has no subsets of cardinality k, so ¡4 k ¢ = 0 in these cases. In general ¡n k ¢ = 0 whenever k < 0 or k > n. In particular this means ¡n k ¢ = 0 if n is negative. Although it was not hard to work out the values of ¡4 k ¢ by writing out subsets in the above table, this method of actually listing sets would not be practical for computing ¡n k ¢ when n and k are large. We need a formula. To ﬁnd one, we will now carefully work out the value of ¡5 3 ¢ in such a way that a pattern will emerge that points the way to a formula for any ¡n k ¢. Counting Subsets 75 To begin, note that ¡5 3 ¢ is the number of 3-element subsets of © a,b, c,d, e ª. These are listed in the following table. We see that in fact ¡5 3 ¢ = 10. ©a,b,c ª©a,b,d ª©a,b,e ª©a,c,d ª ©a,c,e ª ©a,d,e ª©b,c,d ª ©b,c,e ª ©b,d,e ª ©c,d,e ª ¡5 3 ¢ 3! The formula will emerge when we expand this table as follows. Taking any one of the ten 3-element sets above, we can make 3! diﬀerent non-repetitive lists from its elements. For example, consider the ﬁrst set © a,b, c ª. The ﬁrst column of the following table tallies the 3! = 6 diﬀerent lists that can be the letters © a,b, c ª. The second column tallies the lists that can be made from © a,b,d ª, and so on. abc abd abe acd ace ade bcd bce bde cde acb adb aeb adc aec aed bdc bec bed ced bac bad bae cad cae dae cbd cbe dbe dce bca bda bea cda cea dea cdb ceb deb dec cba dba eba dca eca eda dcb ecb edb edc cab dab eab dac eac ead dbc ebc ebd ecd 3! ¡5 3 ¢ This table has ¡5 3 ¢ columns and 3! rows, so it has a total of 3! ¡5 3 ¢ lists. But notice also that the table consists of every non-repetitive length-3 list that can be made from the symbols © a,b, c,d, e ª. We know from Fact 3.2 that there are 5! (5−3)! such lists. Thus the total number of lists in the table is 3! ¡5 3 ¢ = 5! (5−3)!. Dividing both sides of this equation by 3!, we get Ã 5 3 ! = 5! 3!(5−3)!. Working this out, you will ﬁnd that it does give the correct value of 10. But there was nothing special about the values 5 and 3. We could do the above analysis for any ¡n k ¢ instead of ¡5 3 ¢. The table would have ¡n k ¢ columns and k! rows. We would get Ã n k ! = n! k!(n−k)!. We summarize this as follows: 76 Counting Fact 3.3 If n,k ∈Z and 0 ≤k ≤n, then Ã n k ! = n! k!(n−k)!. Otherwise Ã n k ! = 0. Let’s now use our new knowledge to work some exercises. Example 3.4 How many 4-element subsets does © 1,2,3,4,5,6,7,8,9 ª have? The answer is ¡9 4 ¢ = 9! 4!(9−4)! = 9! 4!5! = 9·8·7·6·5! 4!5! = 9·8·7·6 4! = 9·8·7·6 24 = 126. Example 3.5 A single 5-card hand is dealt oﬀof a standard 52-card deck. How many diﬀerent 5-card hands are possible? To answer this, think of the deck as being a set D of 52 cards. Then a 5-card hand is just a 5-element subset of D. For example, here is one of many diﬀerent 5-card hands that might be dealt from the deck. ½ 7 ♣, 2 ♣, 3 ♥, A ♠, 5 ♦ ¾ The total number of possible hands equals the number of 5-element subsets of D, that is Ã 52 5 ! = 52! 5!·47! = 52·51·50·49·48·47! 5!·47! = 52·51·50·49·48 5! = 2,598,960. Thus the answer to our question is that there are 2,598,960 diﬀerent ﬁve-card hands that can be dealt from a deck of 52 cards. Example 3.6 This problem concerns 5-card hands that can be dealt oﬀ of a 52-card deck. How many such hands are there in which two of the cards are clubs and three are hearts? Solution: Think of such a hand as being described by a list of length two of the form µ ½ ∗ ♣, ∗ ♣ ¾ , ½ ∗ ♥, ∗ ♥, ∗ ♥ ¾ ¶ , where the ﬁrst entry is a 2-element subset of the set of 13 club cards, and the second entry is a 3-element subset of the set of 13 heart cards. There are ¡13 2 ¢ choices for the ﬁrst entry and ¡13 3 ¢ choices for the second entry, so by the multiplication principle there are ¡13 2 ¢¡13 3 ¢ = 13! 2!11! 13! 3!10! = 22,308 such lists. Answer: There are 22,308 possible 5-card hands with two clubs and three hearts. Example 3.7 Imagine a lottery that works as follows. A bucket contains 36 balls numbered 1,2,3,4,...,36. Six of these balls will be drawn randomly. For $1 you buy a ticket that has six blanks: □□□□□□. You ﬁll in the blanks with six diﬀerent numbers between 1 and 36. You win $1,000,000 Counting Subsets 77 if you chose the same numbers that are drawn, regardless of order. What are your chances of winning? Solution: In ﬁlling out the ticket you are choosing six numbers from a set of 36 numbers. Thus there are ¡36 6 ¢ = 36! 6!(36−6)! = 1,947,792 diﬀerent combinations of numbers you might write. Only one of these will be a winner. Your chances of winning are one in 1,947,792. Exercises for Section 3.3 1. Suppose a set A has 37 elements. How many subsets of A have 10 elements? How many subsets have 30 elements? How many have 0 elements? 2. Suppose A is a set for which \|A\| = 100. How many subsets of A have 5 elements? How many subsets have 10 elements? How many have 99 elements? 3. A set X has exactly 56 subsets with 3 elements. What is the cardinality of X? 4. Suppose a set B has the property that ¯ ¯© X : X ∈P(B),\|X\| = 6 ª¯ ¯ = 28. Find \|B\|. 5. How many 16-digit binary strings contain exactly seven 1’s? (Examples of such strings include 0111000011110000 and 0011001100110010, etc.) 6. ¯ ¯© X ∈P( © 0,1,2,3,4,5,6,7,8,9 ª ) : \|X\| = 4 ª¯ ¯ = 7. ¯ ¯© X ∈P( © 0,1,2,3,4,5,6,7,8,9 ª ) : \|X\| < 4 ª¯ ¯ = 8. This problem concerns lists made from the symbols A,B,C,D,E,F,G,H,I. (a) How many length-5 lists can be made if repetition is not allowed and the list is in alphabetical order? (Example: BDEFI or ABCGH, but not BACGH.) (b) How many length-5 lists can be made if repetition is not allowed and the list is not in alphabetical order? 9. This problem concerns lists of length 6 made from the letters A,B,C,D,E,F, without repetition. How many such lists have the property that the D occurs before the A? 10. A department consists of 5 men and 7 women. From this department you select a committee with 3 men and 2 women. In how many ways can you do this? 11. How many positive 10-digit integers contain no 0’s and exactly three 6’s? 12. Twenty-one people are to be divided into two teams, the Red Team and the Blue Team. There will be 10 people on Red Team and 11 people on Blue Team. In how many ways can this be done? 13. Suppose n and k are integers for which 0 ≤k ≤n. Use the formula ¡n k ¢ = n! k!(n−k)! to show that ¡n k ¢ = ¡ n n−k ¢. 14. Suppose n,k ∈Z, and 0 ≤k ≤n. Use Deﬁnition 3.2 alone (without using Fact 3.3) to show that ¡n k ¢ = ¡ n n−k ¢. 78 Counting 3.4 Pascal’s Triangle and the Binomial Theorem There are some beautiful and signiﬁcant patterns among the numbers ¡n k ¢. This section investigates a pattern based on one equation in particular. It happens that Ã n+1 k ! = Ã n k −1 ! + Ã n k ! (3.2) for any integers n and k with 1 ≤k ≤n. To see why this is true, recall that ¡n+1 k ¢ equals the number of k-element subsets of a set with n+1 elements. Now, the set A = © 0,1,2,3,...,n ª has n+1 elements, so ¡n+1 k ¢ equals the number of k-element subsets of A. Such subsets can be divided into two types: those that contain 0 and those that do not contain 0. To make a k-element subset that contains 0 we can start with © 0 ª and then append to this set an additional k −1 numbers selected from © 1,2,3,...,n ª. There are ¡ n k−1 ¢ ways to make this selection, so there are ¡ n k−1 ¢ k-element subsets of A that contain 0. Concerning the k-element subsets of A that do not contain 0, there are ¡n k ¢ of these sets, for we can form them by selecting k elements from the n-element set © 1,2,3,...,n ª. In light of all this, Equation (3.2) just expresses the obvious fact that the number of k-element subsets of A equals the number of k-element subsets that contain 0 plus the number of k-element subsets that do not contain 0. ¡0 0 ¢ ¡1 0 ¢ ¡1 1 ¢ ¡2 0 ¢ ¡2 1 ¢ ¡2 2 ¢ ¡3 0 ¢ ¡3 1 ¢ ¡3 2 ¢ ¡3 3 ¢ ¡4 0 ¢ ¡4 1 ¢ ¡4 2 ¢ ¡4 3 ¢ ¡4 4 ¢ ¡5 0 ¢ ¡5 1 ¢ ¡5 2 ¢ ¡5 3 ¢ ¡5 4 ¢ ¡5 5 ¢ ¡6 0 ¢ ¡6 1 ¢ ¡6 2 ¢ ¡6 3 ¢ ¡6 4 ¢ ¡6 5 ¢ ¡6 6 ¢ ¡7 0 ¢ ¡7 1 ¢ ¡7 2 ¢ ¡7 3 ¢ ¡7 4 ¢ ¡7 5 ¢ ¡7 6 ¢ ¡7 7 ¢ ... . . . ... 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 ... . . . ... Figure 3.2. Pascal’s triangle Now that we have seen why Equation (3.2) is true, we are going to arrange the numbers ¡n k ¢ in a triangular pattern that highlights various relationships among them. The left-hand side of Figure 3.2 shows numbers ¡n k ¢ arranged in a pyramid with ¡0 0 ¢ at the apex, just above a row containing ¡1 k ¢ with k = 0 and k = 1. Below this is a row listing the values of ¡2 k ¢ for k = 0,1,2. In general, each row listing the numbers ¡n k ¢ is just above a row listing the numbers ¡n+1 k ¢. Pascal’s Triangle and the Binomial Theorem 79 Any number ¡n+1 k ¢ for 0 < k < n in this pyramid is immediately below and between the the two numbers ¡ n k−1 ¢ and ¡n k ¢ in the previous row. But Equation 3.2 says ¡n+1 k ¢ = ¡ n k−1 ¢ + ¡n k ¢, and therefore any number (other than 1) in the pyramid is the sum of the two numbers immediately above it. This pattern is especially evident on the right of Figure 3.2, where each ¡n k ¢ is worked out. Notice how 21 is the sum of the numbers 6 and 15 above it. Similarly, 5 is the sum of the 1 and 4 above it and so on. The arrangement on the right of Figure 3.2 is called Pascal’s triangle. (It is named after Blaise Pascal, 1623–1662, a French mathematician and philosopher who discovered many of its properties.) Although we have written only the ﬁrst eight rows of Pascal’s triangle (beginning with Row 0 at the apex), it obviously could be extended downward indeﬁnitely. We could add an additional row at the bottom by placing a 1 at each end and obtaining each remaining number by adding the two numbers above its position. Doing this would give the following row: 1 8 28 56 70 56 28 8 1 This row consists of the numbers ¡8 k ¢ for 0 ≤k ≤8, and we have computed them without the formula ¡8 k ¢ = 8! k!(8−k)!. Any ¡n k ¢ can be computed this way. The very top row (containing only 1) is called Row 0. Row 1 is the next down, followed by Row 2, then Row 3, etc. With this labeling, Row n consists of the numbers ¡n k ¢ for 0 ≤k ≤n. Notice that Row n appears to be a list of the coeﬃcients of (x + y)n. For example (x+ y)2 = 1x2 +2xy+1y2, and Row 2 lists the coeﬃcients 1 2 1. Similarly (x + y)3 = 1x3 +3x2y +3xy2 +1y3, and Row 3 is 1 3 3 1. Pascal’s triangle is shown on the left of Figure 3.3 and on the right are the expansions of (x+ y)n for 0 ≤n ≤5. In every case (at least as far as you care to check) the numbers in Row n match up with the coeﬃcients of (x+ y)n. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 ... . . . ... 1 1x + 1y 1x2 + 2xy + 1y2 1x3 + 3x2 y + 3xy2 + 1y3 1x4 + 4x3 y + 6x2 y2 + 4xy3 + 1y4 1x5 + 5x4 y +10x3 y2+10x2 y3+ 5xy4 + 1y5 ... . . . ... Figure 3.3. The nth row of Pascal’s triangle lists the coeﬃcients of (x+ y)n 80 Counting In fact this turns out to be true for every n. This result is known as the binomial theorem, and it is worth mentioning here. It tells how to raise a binomial x+ y to a non-negative integer power n. Theorem 3.1 (Binomial Theorem) If n is a non-negative integer, then (x+ y)n = ¡n 0 ¢ xn + ¡n 1 ¢ xn−1y+ ¡n 2 ¢ xn−2y2 + ¡n 3 ¢ xn−3y3 +···+ ¡ n n−1 ¢ xyn−1 + ¡n n ¢ yn. For now we will be content to accept the binomial theorem without proof. (You will be asked to prove it in an exercise in Chapter 10.) You may ﬁnd it useful from time to time. For instance, you can apply it if you ever need to expand an expression such as (x+ y)7. To do this, look at Row 7 of Pascal’s triangle in Figure 3.2 and apply the binomial theorem to get (x+ y)7 = x7 +7x6y+21x5y2 +35x4y3 +35x3y4 +21x2y5 +7xy6 + y7. For another example, (2a−b)4 = ((2a)+(−b))4 = (2a)4 +4(2a)3(−b)+6(2a)2(−b)2 +4(2a)(−b)3 +(−b)4 = 16a4 −32a3b +24a2b2 −8ab3 + b4. Exercises for Section 3.4 1. Write out Row 11 of Pascal’s triangle. 2. Use the binomial theorem to ﬁnd the coeﬃcient of x8 y5 in (x+ y)13. 3. Use the binomial theorem to ﬁnd the coeﬃcient of x8 in (x+2)13. 4. Use the binomial theorem to ﬁnd the coeﬃcient of x6 y3 in (3x−2y)9. 5. Use the binomial theorem to show Pn k=0 ¡n k ¢ = 2n. 6. Use Deﬁnition 3.2 (page 74) and Fact 1.3 (page 12) to show Pn k=0 ¡n k ¢ = 2n. 7. Use the binomial theorem to show Pn k=0 3k¡n k ¢ = 4n. 8. Use Fact 3.3 (page 76) to derive Equation 3.2 (page 78). 9. Use the binomial theorem to show ¡n 0 ¢ − ¡n 1 ¢ + ¡n 2 ¢ − ¡n 3 ¢ + ¡n 4 ¢ −···+(−1)n¡n n ¢ = 0. 10. Show that the formula k ¡n k ¢ = n ¡n−1 k−1 ¢ is true for all integers n,k with 0 ≤k ≤n. 11. Use the binomial theorem to show 9n = Pn k=0(−1)k¡n k ¢ 10n−k. 12. Show that ¡n k ¢¡ k m ¢ = ¡ n m ¢¡n−m k−m ¢. 13. Show that ¡n 3 ¢ = ¡2 2 ¢ + ¡3 2 ¢ + ¡4 2 ¢ + ¡5 2 ¢ +···+ ¡n−1 2 ¢. 14. The ﬁrst ﬁve rows of Pascal’s triangle appear in the digits of powers of 11: 110 = 1, 111 = 11, 112 = 121, 113 = 1331 and 114 = 14641. Why is this so? Why does the pattern not continue with 115? Inclusion-Exclusion 81 3.5 Inclusion-Exclusion Many counting problems involve computing the cardinality of a union A∪B of two ﬁnite sets. We examine this kind of problem now. First we develop a formula for \|A ∪B\|. It is tempting to say that \|A ∪B\| must equal \|A\|+\|B\|, but that is not quite right. If we count the elements of A and then count the elements of B and add the two ﬁgures together, we get \|A\|+\|B\|. But if A and B have some elements in common, then we have counted each element in A ∩B twice. A B Therefore \|A\| + \|B\| exceeds \|A ∪B\| by \|A ∩B\|, and consequently \|A ∪B\| = \|A\|+\|B\|−\|A ∩B\|. This can be a useful equation. \|A ∪B\| = \|A\|+\|B\|−\|A ∩B\| (3.3) Notice that the sets A, B and A∩B are all generally smaller than A∪B, so Equation (3.3) has the potential of reducing the problem of determining \|A ∪B\| to three simpler counting problems. It is sometimes called an inclusion-exclusion formula because elements in A ∩B are included (twice) in \|A\|+\|B\|, then excluded when \|A∩B\| is subtracted. Notice that if A∩B = ;, then we do in fact get \|A ∪B\| = \|A\|+\|B\|; conversely if \|A ∪B\| = \|A\|+\|B\|, then it must be that A ∩B = ;. Example 3.8 A 3-card hand is dealt oﬀof a standard 52-card deck. How many diﬀerent such hands are there for which all 3 cards are red or all three cards are face cards? Solution: Let A be the set of 3-card hands where all three cards are red (i.e., either ♥or ♦). Let B be the set of 3-card hands in which all three cards are face cards (i.e., J,K or Q of any suit). These sets are illustrated below. A = (( 5 ♥, K ♦, 2 ♥ ) , ( K ♥, J ♥, Q ♥ ) , ( A ♦, 6 ♦, 6 ♥ ) ,... ) (Red cards) B = (( K ♠, K ♦, J ♣ ) , ( K ♥, J ♥, Q ♥ ) , ( Q ♦, Q ♣, Q ♥ ) ,... ) (Face cards) 82 Counting We seek the number of 3-card hands that are all red or all face cards, and this number is \|A ∪B\|. By Formula (3.3), \|A ∪B\| = \|A\| + \|B\| −\|A ∩B\|. Let’s examine \|A\|,\|B\| and \|A ∩B\| separately. Any hand in A is formed by selecting three cards from the 26 red cards in the deck, so \|A\| = ¡26 3 ¢. Similarly, any hand in B is formed by selecting three cards from the 12 face cards in the deck, so \|B\| = ¡12 3 ¢. Now think about A ∩B. It contains all the 3-card hands made up of cards that are red face cards. A ∩B = (( K ♥, K ♦, J ♥ ) , ( K ♥, J ♥, Q ♥ ) , ( , Q ♦, J ♦, Q ♥ ) ,... ) (Red face cards) The deck has only 6 red face cards, so \|A ∩B\| = ¡6 3 ¢. Now we can answer our question. The number of 3-card hands that are all red or all face cards is \|A ∪B\| = \|A\|+\|B\|−\|A ∩B\| = ¡26 3 ¢ + ¡12 3 ¢ − ¡6 3 ¢ = 2600+220−20 = 2800. There is an analogue to Equation (3.3) that involves three sets. Consider three sets A, B and C, as represented in the following Venn Diagram. A B C Using the same kind of reasoning that resulted in Equation (3.3), you can convince yourself that \|A ∪B ∪C\| = \|A\|+\|B\|+\|C\|−\|A ∩B\|−\|A ∩C\|−\|B ∩C\|+\|A ∩B ∩C\|. (3.4) There’s probably not much harm in ignoring this one for now, but if you ﬁnd this kind of thing intriguing you should deﬁnitely take a course in combinatorics. (Ask your instructor!) As we’ve noted, Equation (3.3) becomes \|A ∪B\| = \|A\|+\|B\| if it happens that A ∩B = ;. Also, in Equation (3.4), note that if A ∩B = ;, A ∩C = ; and B ∩C = ;, we get the simple formula \|A ∪B ∪C\| = \|A\|+\|B\|+\|C\|. In general, we have the following formula for n sets, none of which overlap. It is sometimes called the addition principle. Fact 3.4 (Addition Principle) If A1, A2,..., An are sets with Ai ∩A j = ; whenever i ̸= j, then \|A1 ∪A2 ∪···∪An\| = \|A1\|+\|A2\|+···+\|An\|. Inclusion-Exclusion 83 Example 3.9 How many 7-digit binary strings (0010100, 1101011, etc.) have an odd number of 1’s? Solution: Let A be the set of all 7-digit binary strings with an odd number of 1’s, so the answer to the question will be \|A\|. To compute \|A\|, we break A up into smaller parts. Notice any string in A will have either one, three, ﬁve or seven 1’s. Let A1 be the set of 7-digit binary strings with only one 1. Let A3 be the set of 7-digit binary strings with three 1’s. Let A5 be the set of 7-digit binary strings with ﬁve 1’s, and let A7 be the set of 7-digit binary strings with seven 1’s. Therefore A = A1 ∪A3 ∪A5 ∪A7. Notice that any two of the sets Ai have empty intersection, so Fact 3.4 gives \|A\| = \|A1\|+\|A3\|+\|A5\|+\|A7\|. Now the problem is to ﬁnd the values of the individual terms of this sum. For instance take A3, the set of 7-digit binary strings with three 1’s. Such a string can be formed by selecting three out of seven positions for the 1’s and putting 0’s in the other spaces. Therefore \|A3\| = ¡7 3 ¢. Similarly \|A1\| = ¡7 1 ¢, \|A5\| = ¡7 5 ¢, and \|A7\| = ¡7 7 ¢. Finally the answer to our question is \|A\| = \|A1\|+\|A3\|+\|A5\|+\|A7\| = ¡7 1 ¢ + ¡7 3 ¢ + ¡7 5 ¢ + ¡7 7 ¢ = 7+35+21+1 = 64. There are 64 seven-digit binary strings with an odd number of 1’s. You may already have been using the Addition Principle intuitively, without thinking of it as a free-standing result. For instance, we used it in Example 3.2(c) when we divided lists into four types and computed the number of lists of each type. Exercises for Section 3.5 1. At a certain university 523 of the seniors are history majors or math majors (or both). There are 100 senior math majors, and 33 seniors are majoring in both history and math. How many seniors are majoring in history? 2. How many 4-digit positive integers are there for which there are no repeated digits, or for which there may be repeated digits, but all are odd? 3. How many 4-digit positive integers are there that are even or contain no 0’s? 4. This problem involves lists made from the letters T,H,E,O,R,Y, with repetition allowed. (a) How many 4-letter lists are there that don’t begin with T, or don’t end in Y? (b) How many 4-letter lists are there in which the sequence of letters T,H,E appears consecutively? (c) How many 5-letter lists are there in which the sequence of letters T,H,E appears consecutively? 84 Counting 5. How many 7-digit binary strings begin in 1 or end in 1 or have exactly four 1’s? 6. Is the following statement true or false? Explain. If A1 ∩A2 ∩A3 = ;, then \|A1 ∪A2 ∪A3\| = \|A1\|+\|A2\|+\|A3\|. 7. This problem concerns 4-card hands dealt oﬀof a standard 52-card deck. How many 4-card hands are there for which all 4 cards are of the same suit or all 4 cards are red? 8. This problem concerns 4-card hands dealt oﬀof a standard 52-card deck. How many 4-card hands are there for which all 4 cards are of diﬀerent suits or all 4 cards are red? 9. A 4-letter list is made from the letters L,I,S,T,E,D according to the following rule: Repetition is allowed, and the ﬁrst two letters on the list are vowels or the list ends in D. How many such lists are possible? 10. A 5-card poker hand is called a ﬂush if all cards are the same suit. How many diﬀerent ﬂushes are there? Part II How to Prove Conditional Statements CHAPTER 4 Direct Proof I t is time to prove some theorems. There are various strategies for doing this; we now examine the most straightforward approach, a technique called direct proof. As we begin, it is important to keep in mind the meanings of three key terms: Theorem, proof and deﬁnition. A theorem is a mathematical statement that is true and can be (and has been) veriﬁed as true. A proof of a theorem is a written veriﬁcation that shows that the theorem is deﬁnitely and unequivocally true. A proof should be understandable and convincing to anyone who has the requisite background and knowledge. This knowledge includes an understanding of the meanings of the mathematical words, phrases and symbols that occur in the theorem and its proof. It is crucial that both the writer of the proof and the readers of the proof agree on the exact meanings of all the words, for otherwise there is an intolerable level of ambiguity. A deﬁnition is an exact, unambiguous explanation of the meaning of a mathematical word or phrase. We will elaborate on the terms theorem and deﬁnition in the next two sections, and then ﬁnally we will be ready to begin writing proofs. 4.1 Theorems A theorem is a statement that is true and has been proved to be true. You have encountered many theorems in your mathematical education. Here are some theorems taken from an undergraduate calculus text. They will be familiar to you, though you may not have read all the proofs. Theorem: Let f be diﬀerentiable on an open interval I and let c ∈I. If f (c) is the maximum or minimum value of f on I, then f ′(c) = 0. Theorem: If P∞ k=1 ak converges, then limk→∞ak = 0. Theorem: Suppose f is continuous on the interval [a,b]. Then f is integrable on [a,b]. Theorem: Every absolutely convergent series converges. 88 Direct Proof Observe that each of these theorems either has the conditional form “If P, then Q,” or can be put into that form. The ﬁrst theorem has an initial sentence “Let f be diﬀerentiable on an open interval I, and let c ∈I,” which sets up some notation, but a conditional statement follows it. The third theorem has form “Suppose P. Then Q,” but this means the same thing as “If P, then Q.” The last theorem can be re-expressed as “If a series is absolutely convergent, then it is convergent.” A theorem of form “If P, then Q,” can be regarded as a device that produces new information from P. Whenever we are dealing with a situation in which P is true, then the theorem guarantees that, in addition, Q is true. Since this kind of expansion of information is useful, theorems of form “If P, then Q,” are very common. But not every theorem is a conditional statement. Some have the form of the biconditional P ⇔Q, but, as we know, that can be expressed as two conditional statements. Other theorems simply state facts about speciﬁc things. For example, here is another theorem from your study of calculus. Theorem: The series 1+ 1 2 + 1 3 + 1 4 + 1 5 +··· diverges. It would be diﬃcult (or at least awkward) to restate this as a conditional statement. Still, it is true that most theorems are conditional statements, so much of this book will concentrate on that type of theorem. It is important to be aware that there are a number of words that mean essentially the same thing as the word “theorem,” but are used in slightly diﬀerent ways. In general the word “theorem” is reserved for a statement that is considered important or signiﬁcant (the Pythagorean theorem, for example). A statement that is true but not as signiﬁcant is sometimes called a proposition. A lemma is a theorem whose main purpose is to help prove another theorem. A corollary is a result that is an immediate consequence of a theorem or proposition. It is not important that you remember all these words now, for their meanings will become clear with usage. Our main task is to learn how to prove theorems. As the above examples suggest, proving theorems requires a clear understanding of the meaning of the conditional statement, and that is the primary reason we studied it so extensively in Chapter 2. In addition, it is also crucial to understand the role of deﬁnitions. Deﬁnitions 89 4.2 Deﬁnitions A proof of a theorem should be absolutely convincing. Ambiguity must be avoided. Everyone must agree on the exact meaning of each mathematical term. In Chapter 1 we deﬁned the meanings of the sets N, Z, R, Q and ;, as well as the meanings of the symbols ∈and ⊆, and we shall make frequent use of these things. Here is another deﬁnition that we use often. Deﬁnition 4.1 An integer n is even if n = 2a for some integer a ∈Z. Thus, for example, 10 is even because 10 = 2·5. Also, according to the deﬁnition, 7 is not even because there is no integer a for which 7 = 2a. While there would be nothing wrong with deﬁning an integer to be odd if it’s not even, the following deﬁnition is more concrete. Deﬁnition 4.2 An integer n is odd if n = 2a+1 for some integer a ∈Z. Thus 7 is odd because 7 = 2·3+1. We will use these deﬁnitions whenever the concept of even or odd numbers arises. If in a proof a certain number turns out to be even, the deﬁnition allows us to write it as 2a for an appropriate integer a. If some quantity has form 2b + 1 where b is an integer, then the deﬁnition tells us the quantity is odd. Deﬁnition 4.3 Two integers have the same parity if they are both even or they are both odd. Otherwise they have opposite parity. Thus 5 and −17 have the same parity, as do 8 and 0; but 3 and 4 have opposite parity. Two points about deﬁnitions are in order. First, in this book the word or term being deﬁned appears in boldface type. Second, it is common to express deﬁnitions as conditional statements even though the biconditional would more appropriately convey the meaning. Consider the deﬁnition of an even integer. You understand full well that if n is even then n = 2a (for a ∈Z), and if n = 2a, then n is even. Thus, technically the deﬁnition should read “An integer n is even if and only if n = 2a for some a ∈Z.” However, it is an almost-universal convention that deﬁnitions are phrased in the conditional form, even though they are interpreted as being in the biconditional form. There is really no good reason for this, other than economy of words. It is the standard way of writing deﬁnitions, and we have to get used to it. Here is another deﬁnition that we will use often. 90 Direct Proof Deﬁnition 4.4 Suppose a and b are integers. We say that a divides b, written a \| b, if b = ac for some c ∈Z. In this case we also say that a is a divisor of b, and that b is a multiple of a. For example, 5 divides 15 because 15 = 5 · 3. We write this as 5 \| 15. Similarly 8 \| 32 because 32 = 8·4, and −6 \| 6 because 6 = −6·−1. However, 6 does not divide 9 because there is no integer c for which 9 = 6· c. We express this as 6 ∤9, which we read as “6 does not divide 9.” Be careful of your interpretation of the symbols. There is a big diﬀerence between the expressions a \| b and a/b. The expression a \| b is a statement, while a/b is a fraction. For example, 8 \| 16 is true and 8 \| 20 is false. By contrast, 8/16 = 0.5 and 8/20 = 0.4 are numbers, not statements. Be careful not to write one when you mean the other. Every integer has a set of integers that divide it. For example, the set of divisors of 6 is © a ∈Z : a \| 6 ª = © −6,−3,−2,−1,1,2,3,6 ª. The set of divisors of 5 is © −5,−1,1,5 ª. The set of divisors of 0 is Z. This brings us to the following deﬁnition, with which you are already familiar. Deﬁnition 4.5 A natural number n is prime if it has exactly two positive divisors, 1 and n. For example, 2 is prime, as are 5 and 17. The deﬁnition implies that 1 is not prime, as it only has one (not two) positive divisor, namely 1. An integer n is composite if it factors as n = ab where a,b > 1. Deﬁnition 4.6 The greatest common divisor of integers a and b, denoted gcd(a,b), is the largest integer that divides both a and b. The least common multiple of non-zero integers a and b, denoted lcm(a,b), is smallest positive integer that is a multiple of both a and b. So gcd(18,24) = 6, gcd(5,5) = 5 and gcd(32,−8) = 8. Also gcd(50,18) = 2, but gcd(50,9) = 1. Note that gcd(0,6) = 6, because, although every integer divides 0, the largest divisor of 6 is 6. The expression gcd(0,0) is problematic. Every integer divides 0, so the only conclusion is that gcd(0,0) = ∞. We circumvent this irregularity by simply agreeing to consider gcd(a,b) only when a and b are not both zero. Continuing our examples, lcm(4,6) = 12, and lcm(7,7) = 7. Of course not all terms can be deﬁned. If every word in a deﬁnition were deﬁned, there would be separate deﬁnitions for the words that appeared in those deﬁnitions, and so on, until the chain of deﬁned terms became circular. Thus we accept some ideas as being so intuitively clear that they require no deﬁnitions or veriﬁcations. For example, we will not ﬁnd Deﬁnitions 91 it necessary to deﬁne what an integer (or a real number) is. Nor will we deﬁne addition, multiplication, subtraction and division, though we will use these operations freely. We accept and use such things as the distributive and commutative properties of addition and multiplication, as well as other standard properties of arithmetic and algebra. As mentioned in Section 1.9, we accept as fact the natural ordering of the elements of N,Z,Q and R, so that (for example) statements such as “5 < 7,” and “x < y implies −x > −y,” do not need to be justiﬁed. In addition, we accept the following fact without justiﬁcation or proof. Fact 4.1 Suppose a and b are integers. Then: • a+ b ∈Z • a−b ∈Z • ab ∈Z These three statements can be combined. For example, we see that if a,b and c are integers, then a2b −ca+ b is also an integer. We will also accept as obvious the fact that any integer a can be divided by a non-zero integer b, resulting in a unique quotient q and remainder r. For example, b = 3 goes into a = 17 q = 5 times with remainder r = 2. In symbols, 17 = 5·3+2, or a = qb + r. This fact, called the division algorithm, was mentioned on page 29. (The Division Algorithm) Given integers a and b with b > 0, there exist unique integers q and r for which a = qb + r and 0 ≤r < b. Another fact that we will accept without proof (at least for now) is that every natural number greater than 1 has a unique factorization into primes. For example, the number 1176 can be factored into primes as 1176 = 2·2·2·3·7·7 = 23 ·3·72. By unique we mean that any factorization of 1176 into primes will have exactly the same factors (i.e., three 2’s, one 3 and two 7’s). Thus, for example, there is no valid factorization of 1176 that has a factor of 5. You may be so used to factoring numbers into primes that it seems obvious that there cannot be diﬀerent prime factorizations of the same number, but in fact this is a fundamental result whose proof is not transparent. Nonetheless, we will be content to assume that every natural number greater than 1 has a unique factorization into primes. (We will revisit the issue of a proof in Section 10.2.) We will introduce other accepted facts, as well as deﬁnitions, as needed. 92 Direct Proof 4.3 Direct Proof This section explains a simple way to prove theorems or propositions that have the form of conditional statements. The technique is called direct proof. To simplify the discussion, our ﬁrst examples will involve proving statements that are almost obviously true. Thus we will call the statements propositions rather than theorems. (Remember, a proposition is a statement that, although true, is not as signiﬁcant as a theorem.) To understand how the technique of direct proof works, suppose we have some proposition of the following form. Proposition If P, then Q. This proposition is a conditional statement of form P ⇒Q. Our goal is to show that this conditional statement is true. To see how to proceed, look at the truth table. P Q P ⇒Q T T T T F F F T T F F T The table shows that if P is false, the statement P ⇒Q is automatically true. This means that if we are concerned with showing P ⇒Q is true, we don’t have to worry about the situations where P is false (as in the last two lines of the table) because the statement P ⇒Q will be automatically true in those cases. But we must be very careful about the situations where P is true (as in the ﬁrst two lines of the table). We must show that the condition of P being true forces Q to be true also, for that means the second line of the table cannot happen. This gives a fundamental outline for proving statements of the form P ⇒Q. Begin by assuming that P is true (remember, we don’t need to worry about P being false) and show this forces Q to be true. We summarize this as follows. Outline for Direct Proof Proposition If P, then Q. Proof. Suppose P. . . . Therefore Q. ■ Direct Proof 93 So the setup for direct proof is remarkably simple. The ﬁrst line of the proof is the sentence “Suppose P.” The last line is the sentence “Therefore Q.” Between the ﬁrst and last line we use logic, deﬁnitions and standard math facts to transform the statement P to the statement Q. It is common to use the word “Proof” to indicate the beginning of a proof, and the symbol to indicate the end. As our ﬁrst example, let’s prove that if x is odd then x2 is also odd. (Granted, this is not a terribly impressive result, but we will move on to more signiﬁcant things in due time.) The ﬁrst step in the proof is to ﬁll in the outline for direct proof. This is a lot like painting a picture, where the basic structure is sketched in ﬁrst. We leave some space between the ﬁrst and last line of the proof. The following series of frames indicates the steps you might take to ﬁll in this space with a logical chain of reasoning. Proposition If x is odd, then x2 is odd. Proof. Suppose x is odd. Therefore x2 is odd. ■ Now that we have written the ﬁrst and last lines, we need to ﬁll in the space with a chain of reasoning that shows that x being odd forces x2 to be odd. In doing this it’s always advisable to use any deﬁnitions that apply. The ﬁrst line says x is odd, and by Deﬁnition 4.2 it must be that x = 2a+1 for some a ∈Z, so we write this in as our second line. Proposition If x is odd, then x2 is odd. Proof. Suppose x is odd. Then x = 2a+1 for some a ∈Z, by deﬁnition of an odd number. Therefore x2 is odd. ■ Now jump down to the last line, which says x2 is odd. Think about what the line immediately above it would have to be in order for us to conclude that x2 is odd. By the deﬁnition of an odd number, we would have to have x2 = 2a+1 for some a ∈Z. However, the symbol a now appears earlier in the proof in a diﬀerent context, so we should use a diﬀerent symbol, say b. 94 Direct Proof Proposition If x is odd, then x2 is odd. Proof. Suppose x is odd. Then x = 2a+1 for some a ∈Z, by deﬁnition of an odd number. Thus x2 = 2b +1 for an integer b. Therefore x2 is odd, by deﬁnition of an odd number. ■ We are almost there. We can bridge the gap as follows. Proposition If x is odd, then x2 is odd. Proof. Suppose x is odd. Then x = 2a+1 for some a ∈Z, by deﬁnition of an odd number. Thus x2 = (2a+1)2 = 4a2 +4a+1 = 2(2a2 +2a)+1. So x2 = 2b +1 where b is the integer b = 2a2 +2a. Thus x2 = 2b +1 for an integer b. Therefore x2 is odd, by deﬁnition of an odd number. ■ Finally, we may wish to clean up our work and write the proof in paragraph form. Here is our ﬁnal version. Proposition If x is odd, then x2 is odd. Proof. Suppose x is odd. Then x = 2a+1 for some a ∈Z, by deﬁnition of an odd number. Thus x2 = (2a+1)2 = 4a2+4a+1 = 2(2a2+2a)+1, so x2 = 2b +1 where b = 2a2 +2a ∈Z. Therefore x2 is odd, by deﬁnition of an odd number. ■ At least initially, it’s generally a good idea to write the ﬁrst and last line of your proof ﬁrst, and then ﬁll in the gap, sometimes jumping alternately between top and bottom until you meet in the middle, as we did above. This way you are constantly reminded that you are aiming for the statement at the bottom. Sometimes you will leave too much space, sometimes not enough. Sometimes you will get stuck before ﬁguring out what to do. This is normal. Mathematicians do scratch work just as artists do sketches for their paintings. Direct Proof 95 Here is another example. Consider proving following proposition. Proposition Let a,b and c be integers. If a \| b and b \| c, then a \| c. Let’s apply the basic outline for direct proof. To clarify the procedure we will write the proof in stages again. Proposition Let a,b and c be integers. If a \| b and b \| c, then a \| c. Proof. Suppose a \| b and b \| c. Therefore a \| c. ■ Our ﬁrst step is to apply Deﬁnition 4.4 to the ﬁrst line. The deﬁnition says a \| b means b = ac for some integer c, but since c already appears in a diﬀerent context on the ﬁrst line, we must use a diﬀerent letter, say d. Similarly let’s use a new letter e in the deﬁnition of b \| c. Proposition Let a,b and c be integers. If a \| b and b \| c, then a \| c. Proof. Suppose a \| b and b \| c. By Deﬁnition 4.4, we know a \| b means there is an integer d with b = ad. Likewise, b \| c means there is an integer e for which c = be. Therefore a \| c. ■ We have almost bridged the gap. The line immediately above the last line should show that a \| c. According to Deﬁnition 4.4, this line should say that c = ax for some integer x. We can get this equation from the lines at the top, as follows. Proposition Let a,b and c be integers. If a \| b and b \| c, then a \| c. Proof. Suppose a \| b and b \| c. By Deﬁnition 4.4, we know a \| b means there is an integer d with b = ad. Likewise, b \| c means there is an integer e for which c = be. Thus c = be = (ad)e = a(de), so c = ax for the integer x = de. Therefore a \| c. ■ The next example is presented all at once rather than in stages. 96 Direct Proof Proposition If x is an even integer, then x2 −6x+5 is odd. Proof. Suppose x is an even integer. Then x = 2a for some a ∈Z, by deﬁnition of an even integer. So x2−6x+5 = (2a)2−6(2a)+5 = 4a2−12a+5 = 4a2−12a+4+1 = 2(2a2−6a+2)+1. Therefore we have x2 −6x+5 = 2b +1, where b = 2a2 −6a+2 ∈Z. Consequently x2 −6x+5 is odd, by deﬁnition of an odd number. ■ One doesn’t normally use a separate line for each sentence in a proof, but for clarity we will often do this in the ﬁrst few chapters of this book. Our next example illustrates a standard technique for showing two quantities are equal. If we can show m ≤n and n ≤m then it follows that m = n. In general, the reasoning involved in showing m ≤n can be quite diﬀerent from that of showing n ≤m. Recall Deﬁnition 4.6 of a least common multiple on page 90. Proposition If a,b, c ∈N, then lcm(ca, cb) = c ·lcm(a,b). Proof. Assume a,b, c ∈N. Let m = lcm(ca, cb) and n = c ·lcm(a,b). We will show m = n. By deﬁnition, lcm(a,b) is a multiple of both a and b, so lcm(a,b) = ax = by for some x, y ∈Z. From this we see that n = c ·lcm(a,b) = cax = cby is a multiple of both ca and cb. But m = lcm(ca, cb) is the smallest multiple of both ca and cb. Thus m ≤n. On the other hand, as m = lcm(ca, cb) is a multiple of both ca and cb, we have m = cax = cby for some x, y ∈Z. Then 1 c m = ax = by is a multiple of both a and b. Therefore lcm(a,b) ≤1 c m, so c ·lcm(a,b) ≤m, that is, n ≤m. We’ve shown m ≤n and n ≤m, so m = n. The proof is complete. ■ The examples we’ve looked at so far have all been proofs of statements about integers. In our next example, we are going to prove that if x and y are positive real numbers for which x ≤y, then px ≤py. You may feel that the proof is not as “automatic” as the proofs we have done so far. Finding the right steps in a proof can be challenging, and that is part of the fun. Proposition Let x and y be positive numbers. If x ≤y, then px ≤py. Proof. Suppose x ≤y. Subtracting y from both sides gives x−y ≤0. This can be written as px2 −py2 ≤0. Factor this to get (px−py)(px+py) ≤0. Dividing both sides by the positive number px+py produces px−py ≤0. Adding py to both sides gives px ≤py. ■ Direct Proof 97 This proposition tells us that whenever x ≤y, we can take the square root of both sides and be assured that px ≤py. This can be useful, as we will see in our next proposition. That proposition will concern the expression 2pxy ≤x+ y. Notice when you substitute random positive values for the variables, the expression is true. For example, for x = 6 and y = 4, the left side is 2 p 6·4 = 4 p 6 ≈9.79, which is less than the right side 6+4 = 10. Is it true that 2pxy ≤x+ y for any positive x and y? How could we prove it? To see how, let’s ﬁrst cast this into the form of a conditional statement: If x and y are positive real numbers, then 2pxy ≤x+ y. The proof begins with the assumption that x and y are positive, and ends with 2pxy ≤x+ y. In mapping out a strategy, it can be helpful to work backwards, working from 2pxy ≤x+ y to something that is obviously true. Then the steps can be reversed in the proof. In this case, squaring both sides of 2pxy ≤x+ y gives us 4xy ≤x2 +2xy+ y2. Now subtract 4xy from both sides and factor. 0 ≤ x2 −2xy+ y2 0 ≤ (x−y)2 But this last line is clearly true, since the square of x−y cannot be negative! This gives us a strategy for the proof, which follows. Proposition If x and y are positive real numbers, then 2pxy ≤x+ y. Proof. Suppose x and y are positive real numbers. Then 0 ≤(x−y)2, that is, 0 ≤x2 −2xy+ y2. Adding 4xy to both sides gives 4xy ≤x2 +2xy+ y2. Factoring yields 4xy ≤(x+ y)2. Previously we proved that such an inequality still holds after taking the square root of both sides; doing so produces 2pxy ≤x+ y. ■ Notice that in the last step of the proof we took the square root of both sides of 4xy ≤(x+ y)2 and got p 4xy ≤ p (x+ y)2, and the fact that this did not reverse the symbol ≤followed from our previous proposition. This is an important point. Often the proof of a proposition or theorem uses another proposition or theorem (that has already been proved). 98 Direct Proof 4.4 Using Cases In proving a statement is true, we sometimes have to examine multiple cases before showing the statement is true in all possible scenarios. This section illustrates a few examples. Our examples will concern the expression 1+(−1)n(2n −1). Here is a table showing its value for various integers for n. Notice that 1+(−1)n(2n−1) is a multiple of 4 in every line. n 1+(−1)n(2n−1) 1 0 2 4 3 −4 4 8 5 −8 6 12 Is 1+(−1)n(2n−1) always a multiple of 4? We prove the answer is “yes” in our next example. Notice, however, that the expression 1+(−1)n(2n−1) behaves diﬀerently depending on whether n is even or odd, for in the ﬁrst case (−1)n = 1, and in the second (−1)n = −1. Thus the proof must examine these two possibilities separately. Proposition If n ∈N, then 1+(−1)n(2n−1) is a multiple of 4. Proof. Suppose n ∈N. Then n is either even or odd. Let’s consider these two cases separately. Case 1. Suppose n is even. Then n = 2k for some k ∈Z, and (−1)n = 1. Thus 1+(−1)n(2n−1) = 1+(1)(2·2k −1) = 4k, which is a multiple of 4. Case 2. Suppose n is odd. Then n = 2k +1 for some k ∈Z, and (−1)n = −1. Thus 1+(−1)n(2n−1) = 1−(2(2k +1)−1) = −4k, which is a multiple of 4. These cases show that 1+(−1)n(2n−1) is always a multiple of 4. ■ Now let’s examine the ﬂip side of the question. We just proved that 1+(−1)n(2n−1) is always a multiple of 4, but can we get every multiple of 4 this way? The following proposition and proof give an aﬃrmative answer. Treating Similar Cases 99 Proposition Every multiple of 4 equals 1+(−1)n(2n−1) for some n ∈N. Proof. In conditional form, the proposition is as follows: If k is a multiple of 4, then there is an n ∈N for which 1+(−1)n(2n−1) = k. What follows is a proof of this conditional statement. Suppose k is a multiple of 4. This means k = 4a for some integer a. We must produce an n ∈N for which 1+(−1)n(2n−1) = k. This is done by cases, depending on whether a is zero, positive or negative. Case 1. Suppose a = 0. Let n = 1. Then 1+(−1)n(2n−1) = 1+(−1)1(2−1) = 0 = 4·0 = 4a = k. Case 2. Suppose a > 0. Let n = 2a, which is in N because a is positive. Also n is even, so (−1)n = 1. Thus 1+(−1)n(2n−1) = 1+(2n−1) = 2n = 2(2a) = 4a = k. Case 3. Suppose a < 0. Let n = 1−2a, which is an element of N because a is negative, making 1−2a positive. Also n is odd, so (−1)n = −1. Thus 1+(−1)n(2n−1) = 1−(2n−1) = 1−(2(1−2a)−1) = 4a = k. The above cases show that no matter whether a multiple k = 4a of 4 is zero, positive or negative, k = 1+(−1)n(2n−1) for some n ∈N. ■ 4.5 Treating Similar Cases Occasionally two or more cases in a proof will be so similar that writing them separately seems tedious or unnecessary. Here is an example. Proposition If two integers have opposite parity, then their sum is odd. Proof. Suppose m and n are two integers with opposite parity. We need to show that m+ n is odd. This is done in two cases, as follows. Case 1. Suppose m is even and n is odd. Thus m = 2a and n = 2b +1 for some integers a and b. Therefore m+ n = 2a+2b +1 = 2(a+ b)+1, which is odd (by Deﬁnition 4.2). Case 2. Suppose m is odd and n is even. Thus m = 2a+1 and n = 2b for some integers a and b. Therefore m+ n = 2a+1+2b = 2(a+ b)+1, which is odd (by Deﬁnition 4.2). In either case, m+ n is odd. ■ The two cases in this proof are entirely alike except for the order in which the even and odd terms occur. It is entirely appropriate to just do one case and indicate that the other case is nearly identical. The phrase “Without loss of generality...” is a common way of signaling that the proof is treating just one of several nearly identical cases. Here is a second version of the above example. 100 Direct Proof Proposition If two integers have opposite parity, then their sum is odd. Proof. Suppose m and n are two integers with opposite parity. We need to show that m+ n is odd. Without loss of generality, suppose m is even and n is odd. Thus m = 2a and n = 2b +1 for some integers a and b. Therefore m+n = 2a+2b+1 = 2(a+b)+1, which is odd (by Deﬁnition 4.2). ■ In reading proofs in other texts, you may sometimes see the phrase “Without loss of generality” abbreviated as “WLOG.” However, in the interest of transparency we will avoid writing it this way. In a similar spirit, it is advisable—at least until you become more experienced in proof writing—that you write out all cases, no matter how similar they appear to be. Please check your understanding by doing the following exercises. The odd numbered problems have complete proofs in the Solutions section in the back of the text. Exercises for Chapter 4 Use the method of direct proof to prove the following statements. 1. If x is an even integer, then x2 is even. 2. If x is an odd integer, then x3 is odd. 3. If a is an odd integer, then a2 +3a+5 is odd. 4. Suppose x, y ∈Z. If x and y are odd, then xy is odd. 5. Suppose x, y ∈Z. If x is even, then xy is even. 6. Suppose a,b, c ∈Z. If a \| b and a \| c, then a \| (b + c). 7. Suppose a,b ∈Z. If a \| b, then a2 \| b2. 8. Suppose a is an integer. If 5 \| 2a, then 5 \| a. 9. Suppose a is an integer. If 7 \| 4a, then 7 \| a. 10. Suppose a and b are integers. If a \| b, then a \| (3b3 −b2 +5b). 11. Suppose a,b, c,d ∈Z. If a \| b and c \| d, then ac \| bd. 12. If x ∈R and 0 < x < 4, then 4 x(4−x) ≥1. 13. Suppose x, y ∈R. If x2 +5y = y2 +5x, then x = y or x+ y = 5. 14. If n ∈Z, then 5n2 +3n+7 is odd. (Try cases.) 15. If n ∈Z, then n2 +3n+4 is even. (Try cases.) 16. If two integers have the same parity, then their sum is even. (Try cases.) 17. If two integers have opposite parity, then their product is even. 18. Suppose x and y are positive real numbers. If x < y, then x2 < y2. Treating Similar Cases 101 19. Suppose a,b and c are integers. If a2 \| b and b3 \| c, then a6 \| c. 20. If a is an integer and a2 \| a, then a ∈ © −1,0,1 ª. 21. If p is prime and k is an integer for which 0 < k < p, then p divides ¡p k ¢. 22. If n ∈N, then n2 = 2 ¡n 2 ¢ + ¡n 1 ¢. (You may need a separate case for n = 1.) 23. If n ∈N, then ¡2n n ¢ is even. 24. If n ∈N and n ≥2, then the numbers n!+2, n!+3, n!+4, n!+5, ..., n!+ n are all composite. (Thus for any n ≥2, one can ﬁnd n consecutive composite numbers. This means there are arbitrarily large “gaps” between prime numbers.) 25. If a,b, c ∈N and c ≤b ≤a, then ¡a b ¢¡b c ¢ = ¡ a b−c ¢¡a−b+c c ¢. 26. Every odd integer is a diﬀerence of two squares. (Example 7 = 42 −32, etc.) 27. Suppose a,b ∈N. If gcd(a,b) > 1, then b \| a or b is not prime. 28. If a,b, c ∈Z, then c ·gcd(a,b) ≤gcd(ca, cb). CHAPTER 5 Contrapositive Proof W e now examine an alternative to direct proof called contrapositive proof. Like direct proof, the technique of contrapositive proof is used to prove conditional statements of the form “If P, then Q.” Although it is possible to use direct proof exclusively, there are occasions where contrapositive proof is much easier. 5.1 Contrapositive Proof To understand how contrapositive proof works, imagine that you need to prove a proposition of the following form. Proposition If P, then Q. This is a conditional statement of form P ⇒Q. Our goal is to show that this conditional statement is true. Recall that in Section 2.6 we observed that P ⇒Q is logically equivalent to ∼Q ⇒∼P. For convenience, we duplicate the truth table that veriﬁes this fact. P Q ∼Q ∼P P ⇒Q ∼Q ⇒∼P T T F F T T T F T F F F F T F T T T F F T T T T According to the table, statements P ⇒Q and ∼Q ⇒∼P are diﬀerent ways of expressing exactly the same thing. The expression ∼Q ⇒∼P is called the contrapositive form of P ⇒Q.1 1Do not confuse the words contrapositive and converse. Recall from Section 2.4 that the converse of P ⇒Q is the statement Q ⇒P, which is not logically equivalent to P ⇒Q. Contrapositive Proof 103 Since P ⇒Q is logically equivalent to ∼Q ⇒∼P, it follows that to prove P ⇒Q is true, it suﬃces to instead prove that ∼Q ⇒∼P is true. If we were to use direct proof to show ∼Q ⇒∼P is true, we would assume ∼Q is true use this to deduce that ∼P is true. This in fact is the basic approach of contrapositive proof, summarized as follows. Outline for Contrapositive Proof Proposition If P, then Q. Proof. Suppose ∼Q. . . . Therefore ∼P. ■ So the setup for contrapositive proof is very simple. The ﬁrst line of the proof is the sentence “Suppose Q is not true.” (Or something to that eﬀect.) The last line is the sentence “Therefore P is not true.” Between the ﬁrst and last line we use logic and deﬁnitions to transform the statement ∼Q to the statement ∼P. To illustrate this new technique, and to contrast it with direct proof, we now prove a proposition in two ways: ﬁrst with direct proof and then with contrapositive proof. Proposition Suppose x ∈Z. If 7x+9 is even, then x is odd. Proof. (Direct) Suppose 7x+9 is even. Thus 7x+9 = 2a for some integer a. Subtracting 6x+9 from both sides, we get x = 2a−6x−9. Thus x = 2a−6x−9 = 2a−6x−10+1 = 2(a−3x−5)+1. Consequently x = 2b +1, where b = a−3x−5 ∈Z. Therefore x is odd. ■ Here is a contrapositive proof of the same statement: Proposition Suppose x ∈Z. If 7x+9 is even, then x is odd. Proof. (Contrapositive) Suppose x is not odd. Thus x is even, so x = 2a for some integer a. Then 7x+9 = 7(2a)+9 = 14a+8+1 = 2(7a+4)+1. Therefore 7x+9 = 2b +1, where b is the integer 7a+4. Consequently 7x+9 is odd. Therefore 7x+9 is not even. ■ 104 Contrapositive Proof Though the proofs are of equal length, you may feel that the con-trapositive proof ﬂowed more smoothly. This is because it is easier to transform information about x into information about 7x+9 than the other way around. For our next example, consider the following proposition concerning an integer x: Proposition If x2 −6x+5 is even, then x is odd. A direct proof would be problematic. We would begin by assuming that x2 −6x+5 is even, so x2 −6x+5 = 2a. Then we would need to transform this into x = 2b +1 for b ∈Z. But it is not quite clear how that could be done, for it would involve isolating an x from the quadratic expression. However the proof becomes very simple if we use contrapositive proof. Proposition Suppose x ∈Z. If x2 −6x+5 is even, then x is odd. Proof. (Contrapositive) Suppose x is not odd. Thus x is even, so x = 2a for some integer a. So x2−6x+5 = (2a)2−6(2a)+5 = 4a2−12a+5 = 4a2−12a+4+1 = 2(2a2−6a+2)+1. Therefore x2 −6x+5 = 2b +1, where b is the integer 2a2 −6a+2. Consequently x2 −6x+5 is odd. Therefore x2 −6x+5 is not even. ■ In summary, since x being not odd (∼Q) resulted in x2−6x+5 being not even (∼P), then x2 −6x+5 being even (P) means that x is odd (Q). Thus we have proved P ⇒Q by proving ∼Q ⇒∼P. Here is another example: Proposition Suppose x, y ∈R. If y3 + yx2 ≤x3 + xy2, then y ≤x. Proof. (Contrapositive) Suppose it is not true that y ≤x, so y > x. Then y−x > 0. Multiply both sides of y−x > 0 by the positive value x2 + y2. (y−x)(x2 + y2) > 0(x2 + y2) yx2 + y3 −x3 −xy2 > 0 y3 + yx2 > x3 + xy2 Therefore y3 + yx2 > x3 + xy2, so it is not true that y3 + yx2 ≤x3 + xy2. ■ Proving “If P, then Q,” with the contrapositive approach necessarily involves the negated statements ∼P and ∼Q. In working with these we may have to use the techniques for negating statements (e.g., DeMorgan’s laws) discussed in Section 2.10. We consider such an example next. Congruence of Integers 105 Proposition Suppose x, y ∈Z. If 5 ∤xy, then 5 ∤x and 5 ∤y. Proof. (Contrapositive) Suppose it is not true that 5 ∤x and 5 ∤y. By DeMorgan’s law, it is not true that 5 ∤x or it is not true that 5 ∤y. Therefore 5 \| x or 5 \| y. We consider these possibilities separately. Case 1. Suppose 5 \| x. Then x = 5a for some a ∈Z. From this we get xy = 5(ay), and that means 5 \| xy. Case 2. Suppose 5 \| y. Then y = 5a for some a ∈Z. From this we get xy = 5(ax), and that means 5 \| xy. The above cases show that 5 \| xy, so it is not true that 5 ∤xy. ■ 5.2 Congruence of Integers This is a good time to introduce a new deﬁnition. It is not necessarily related to contrapositive proof, but introducing it now ensures that we have a suﬃcient variety of exercises to practice all our proof techniques on. This new deﬁnition occurs in many branches of mathematics, and it will surely play a role in some of your later courses. But our primary reason for introducing it is that it will give us more practice in writing proofs. Deﬁnition 5.1 Given integers a and b and an n ∈N, we say that a and b are congruent modulo n if n \| (a−b). We express this as a ≡b (mod n). If a and b are not congruent modulo n, we write this as a ̸≡b (mod n). Example 5.1 Here are some examples: 1. 9 ≡1 (mod 4) because 4 \| (9−1). 2. 6 ≡10 (mod 4) because 4 \| (6−10). 3. 14 ̸≡8 (mod 4) because 4 ∤(14−8). 4. 20 ≡4 (mod 8) because 8 \| (20−4). 5. 17 ≡−4 (mod 3) because 3 \| (17−(−4)). In practical terms, a ≡b (mod n) means that a and b have the same remainder when divided by n. For example, we saw above that 6 ≡10 (mod 4) and indeed 6 and 10 both have remainder 2 when divided by 4. Also we saw 14 ̸≡8 (mod 4), and sure enough 14 has remainder 2 when divided by 4, while 8 has remainder 0. To see that this is true in general, note that if a and b both have the same remainder r when divided by n, then it follows that a = kn+ r and b = ℓn + r for some k,ℓ∈Z. Then a −b = (kn + r) −(ℓn + r) = n(k −ℓ). But a −b = n(k −ℓ) means n \| (a −b), so a ≡b (mod n). Conversely, one of the exercises for this chapter asks you to show that if a ≡b (mod n), then a and b have the same remainder when divided by n. 106 Contrapositive Proof We conclude this section with several proofs involving congruence of integers, but you will also test your skills with other proofs in the exercises. Proposition Let a,b ∈Z and n ∈N. If a ≡b (mod n), then a2 ≡b2 (mod n). Proof. We will use direct proof. Suppose a ≡b (mod n). By deﬁnition of congruence of integers, this means n \| (a−b). Then by deﬁnition of divisibility, there is an integer c for which a−b = nc. Now multiply both sides of this equation by a+ b. a−b = nc (a−b)(a+ b) = nc(a+ b) a2 −b2 = nc(a+ b) Since c(a+ b) ∈Z, the above equation tells us n \| (a2 −b2). According to Deﬁnition 5.1, this gives a2 ≡b2 (mod n). ■ Let’s pause to consider this proposition’s meaning. It says a ≡b (mod n) implies a2 ≡b2 (mod n). In other words, it says that if integers a and b have the same remainder when divided by n, then a2 and b2 also have the same remainder when divided by n. As an example of this, 6 and 10 have the same remainder (2) when divided by n = 4, and their squares 36 and 100 also have the same remainder (0) when divided by n = 4. The proposition promises this will happen for all a, b and n. In our examples we tend to concentrate more on how to prove propositions than on what the propositions mean. This is reasonable since our main goal is to learn how to prove statements. But it is helpful to sometimes also think about the meaning of what we prove. Proposition Let a,b, c ∈Z and n ∈N. If a ≡b (mod n), then ac ≡bc (mod n). Proof. We employ direct proof. Suppose a ≡b (mod n). By Deﬁnition 5.1, it follows that n \| (a−b). Therefore, by deﬁnition of divisibility, there exists an integer k for which a −b = nk. Multiply both sides of this equation by c to get ac −bc = nkc. Thus ac −bc = n(kc) where kc ∈Z, which means n \| (ac −bc). By Deﬁnition 5.1, we have ac ≡bc (mod n). ■ Contrapositive proof seems to be the best approach in the next example, since it will eliminate the symbols ∤and ̸≡. Mathematical Writing 107 Proposition Suppose a,b ∈Z and n ∈N. If 12a ̸≡12b (mod n), then n ∤12. Proof. (Contrapositive) Suppose n \| 12, so there is an integer c for which 12 = nc. Now reason as follows. 12 = nc 12(a−b) = nc(a−b) 12a−12b = n(ca−cb) Since ca−cb ∈Z, the equation 12a−12b = n(ca−cb) implies n \| (12a−12b). This in turn means 12a ≡12b (mod n). ■ 5.3 Mathematical Writing Now that we have begun writing proofs, it is a good time to contemplate the craft of writing. Unlike logic and mathematics, where there is a clear-cut distinction between what is right or wrong, the diﬀerence between good and bad writing is sometimes a matter of opinion. But there are some standard guidelines that will make your writing clearer. Some of these are listed below. 1. Begin each sentence with a word, not a mathematical symbol. The reason is that sentences begin with capital letters, but mathematical symbols are case sensitive. Because x and X can have entirely diﬀerent meanings, putting such symbols at the beginning of a sentence can lead to ambiguity. Here are some examples of bad usage (marked with ×) and good usage (marked with ✓). A is a subset of B. × The set A is a subset of B. ✓ x is an integer, so 2x+5 is an integer. × Because x is an integer, 2x+5 is an integer. ✓ x2 −x+2 = 0 has two solutions. × X 2 −x+2 = 0 has two solutions. × (and silly too) The equation x2 −x+2 = 0 has two solutions. ✓ 2. End each sentence with a period, even when the sentence ends with a mathematical symbol or expression. Euler proved that ∞ X k=1 1 ks = Y p∈P 1 1−1 ps × Euler proved that ∞ X k=1 1 ks = Y p∈P 1 1−1 ps . ✓ 108 Contrapositive Proof Mathematical statements (equations, etc.) are like English phrases that happen to contain special symbols, so use normal punctuation. 3. Separate mathematical symbols and expressions with words. Not doing this can cause confusion by making distinct expressions appear to merge into one. Compare the clarity of the following examples. Because x2 −1 = 0, x = 1 or x = −1. × Because x2 −1 = 0, it follows that x = 1 or x = −1. ✓ Unlike A ∪B, A ∩B equals ;. × Unlike A ∪B, the set A ∩B equals ;. ✓ 4. Avoid misuse of symbols. Symbols such as =, ≤, ⊆, ∈, etc., are not words. While it is appropriate to use them in mathematical expressions, they are out of place in other contexts. Since the two sets are =, one is a subset of the other. × Since the two sets are equal, one is a subset of the other. ✓ The empty set is a ⊆of every set. × The empty set is a subset of every set. ✓ Since a is odd and x odd ⇒x2 odd, a2 is odd. × Since a is odd and any odd number squared is odd, then a2 is odd.✓ 5. Avoid using unnecessary symbols. Mathematics is confusing enough without them. Don’t muddy the water even more. No set X has negative cardinality. × No set has negative cardinality. ✓ 6. Use the ﬁrst person plural. In mathematical writing, it is common to use the words “we” and “us” rather than “I,” “you” or “me.” It is as if the reader and writer are having a conversation, with the writer guiding the reader through the details of the proof. 7. Use the active voice. This is just a suggestion, but the active voice makes your writing more lively. The value x = 3 is obtained through the division of both sides by 5.× Dividing both sides by 5, we get the value x = 3. ✓ 8. Explain each new symbol. In writing a proof, you must explain the meaning of every new symbol you introduce. Failure to do this can lead to ambiguity, misunderstanding and mistakes. For example, consider the following two possibilities for a sentence in a proof, where a and b have been introduced on a previous line. Mathematical Writing 109 Since a \| b, it follows that b = ac. × Since a \| b, it follows that b = ac for some integer c. ✓ If you use the ﬁrst form, then a reader who has been carefully following your proof may momentarily scan backwards looking for where the c entered into the picture, not realizing at ﬁrst that it came from the deﬁnition of divides. 9. Watch out for “it.” The pronoun “it” can cause confusion when it is unclear what it refers to. If there is any possibility of confusion, you should avoid the word “it.” Here is an example: Since X ⊆Y, and 0 < \|X\|, we see that it is not empty. × Is “it” X or Y? Either one would make sense, but which do we mean? Since X ⊆Y, and 0 < \|X\|, we see that Y is not empty. ✓ 10. Since, because, as, for, so. In proofs, it is common to use these words as conjunctions joining two statements, and meaning that one statement is true and as a consequence the other true. The following statements all mean that P is true (or assumed to be true) and as a consequence Q is true also. Q since P Q because P Q, as P Q, for P P, so Q Since P, Q Because P, Q as P, Q Notice that the meaning of these constructions is diﬀerent from that of “If P, then Q,” for they are asserting not only that P implies Q, but also that P is true. Exercise care in using them. It must be the case that P and Q are both statements and that Q really does follow from P. x ∈N, so Z × x ∈N, so x ∈Z ✓ 11. Thus, hence, therefore consequently. These adverbs precede a statement that follows logically from previous sentences or clauses. Be sure that a statement follows them. Therefore 2k +1. × Therefore a = 2k +1. ✓ 12. Clarity is the gold standard of mathematical writing. If you believe breaking a rule makes your writing clearer, then break the rule. Your mathematical writing will evolve with practice useage. One of the best ways to develop a good mathematical writing style is to read other people’s proofs. Adopt what works and avoid what doesn’t. 110 Contrapositive Proof Exercises for Chapter 5 A. Use the method of contrapositive proof to prove the following statements. (In each case you should also think about how a direct proof would work. You will ﬁnd in most cases that contrapositive is easier.) 1. Suppose n ∈Z. If n2 is even, then n is even. 2. Suppose n ∈Z. If n2 is odd, then n is odd. 3. Suppose a,b ∈Z. If a2(b2 −2b) is odd, then a and b are odd. 4. Suppose a,b, c ∈Z. If a does not divide bc, then a does not divide b. 5. Suppose x ∈R. If x2 +5x < 0 then x < 0. 6. Suppose x ∈R. If x3 −x > 0 then x > −1. 7. Suppose a,b ∈Z. If both ab and a+ b are even, then both a and b are even. 8. Suppose x ∈R. If x5 −4x4 +3x3 −x2 +3x−4 ≥0, then x ≥0. 9. Suppose n ∈Z. If 3 ∤n2, then 3 ∤n. 10. Suppose x, y, z ∈Z and x ̸= 0. If x ∤yz, then x ∤y and x ∤z. 11. Suppose x, y ∈Z. If x2(y+3) is even, then x is even or y is odd. 12. Suppose a ∈Z. If a2 is not divisible by 4, then a is odd. 13. Suppose x ∈R. If x5 +7x3 +5x ≥x4 + x2 +8, then x ≥0. B. Prove the following statements using either direct or contrapositive proof. Sometimes one approach will be much easier than the other. 14. If a,b ∈Z and a and b have the same parity, then 3a+7 and 7b −4 do not. 15. Suppose x ∈Z. If x3 −1 is even, then x is odd. 16. Suppose x ∈Z. If x+ y is even, then x and y have the same parity. 17. If n is odd, then 8 \| (n2 −1). 18. For any a,b ∈Z, it follows that (a+ b)3 ≡a3 + b3 (mod 3). 19. Let a,b ∈Z and n ∈N. If a ≡b (mod n) and a ≡c (mod n), then c ≡b (mod n). 20. If a ∈Z and a ≡1 (mod 5), then a2 ≡1 (mod 5). 21. Let a,b ∈Z and n ∈N. If a ≡b (mod n), then a3 ≡b3 (mod n) 22. Let a ∈Z, n ∈N. If a has remainder r when divided by n, then a ≡r (mod n). 23. Let a,b, c ∈Z and n ∈N. If a ≡b (mod n), then ca ≡cb (mod n). 24. If a ≡b (mod n) and c ≡d (mod n), then ac ≡bd (mod n). 25. If n ∈N and 2n −1 is prime, then n is prime. 26. If n = 2k −1 for k ∈N, then every entry in Row n of Pascal’s Triangle is odd. 27. If a ≡0 (mod 4) or a ≡1 (mod 4), then ¡ a 2 ¢ is even. 28. If n ∈Z, then 4 ∤(n2 −3). 29. If integers a and b are not both zero, then gcd(a,b) = gcd(a−b,b). 30. If a ≡b (mod n), then gcd(a,n) = gcd(b,n). 31. Suppose the division algorithm applied to a and b yields a = qb + r. Then gcd(a,b) = gcd(r,b). CHAPTER 6 Proof by Contradiction W e now explore a third method of proof: proof by contradiction. This method is not limited to proving just conditional statements— it can be used to prove any kind of statement whatsoever. The basic idea is to assume that the statement we want to prove is false, and then show that this assumption leads to nonsense. We are then led to conclude that we were wrong to assume the statement was false, so the statement must be true. As an example, consider the following proposition and its proof. Proposition If a,b ∈Z, then a2 −4b ̸= 2. Proof. Suppose this proposition is false. This conditional statement being false means there exist numbers a and b for which a,b ∈Z is true, but a2 −4b ̸= 2 is false. In other words, there exist integers a,b ∈Z for which a2 −4b = 2. From this equation we get a2 = 4b +2 = 2(2b +1), so a2 is even. Because a2 is even, it follows that a is even, so a = 2c for some integer c. Now plug a = 2c back into the boxed equation to get (2c)2 −4b = 2, so 4c2 −4b = 2. Dividing by 2, we get 2c2 −2b = 1. Therefore 1 = 2(c2 −b), and because c2 −b ∈Z, it follows that 1 is even. We know 1 is not even, so something went wrong. But all the logic after the ﬁrst line of the proof is correct, so it must be that the ﬁrst line was incorrect. In other words, we were wrong to assume the proposition was false. Thus the proposition is true. ■ You may be a bit suspicious of this line of reasoning, but in the next section we will see that it is logically sound. For now, notice that at the end of the proof we deduced that 1 is even, which conﬂicts with our knowledge that 1 is odd. In essence, we have obtained the statement (1 is odd)∧∼(1 is odd), which has the form C∧∼C. Notice that no matter what statement C is, and whether or not it is true, the statement C∧∼C is false. A statement—like this one—that cannot be true is called a contradiction. Contradictions play a key role in our new technique. 112 Proof by Contradiction 6.1 Proving Statements with Contradiction Let’s now see why the proof on the previous page is logically valid. In that proof we needed to show that a statement P : (a,b ∈Z) ⇒(a2 −4b ̸= 2) was true. The proof began with the assumption that P was false, that is that ∼P was true, and from this we deduced C∧∼C. In other words we proved that ∼P being true forces C∧∼C to be true, and this means that we proved that the conditional statement (∼P) ⇒(C ∧∼C) is true. To see that this is the same as proving P is true, look at the following truth table for (∼P) ⇒(C ∧∼C). Notice that the columns for P and (∼P) ⇒(C ∧∼C) are exactly the same, so P is logically equivalent to (∼P) ⇒(C ∧∼C). P C ∼P C ∧∼C (∼P) ⇒(C ∧∼C) T T F F T T F F F T F T T F F F F T F F Therefore to prove a statement P, it suﬃces to instead prove the conditional statement (∼P) ⇒(C ∧∼C). This can be done with direct proof: Assume ∼P and deduce C ∧∼C. Here is the outline: Outline for Proof by Contradiction Proposition P. Proof. Suppose ∼P. . . . Therefore C ∧∼C. ■ One slightly unsettling feature of this method is that we may not know at the beginning of the proof what the statement C is going to be. In doing the scratch work for the proof, you assume that ∼P is true, then deduce new statements until you have deduced some statement C and its negation ∼C. If this method seems confusing, look at it this way. In the ﬁrst line of the proof we suppose ∼P is true, that is we assume P is false. But if P is really true then this contradicts our assumption that P is false. But we haven’t yet proved P to be true, so the contradiction is not obvious. We use logic and reasoning to transform the non-obvious contradiction ∼P to an obvious contradiction C∧∼C. Proving Statements with Contradiction 113 The idea of proof by contradiction is quite ancient, and goes back at least as far as the Pythagoreans, who used it to prove that certain numbers are irrational. Our next example follows their logic to prove that p 2 is irrational. Recall that a number is rational if it equals a fraction of two integers, and it is irrational if it cannot be expressed as a fraction of two integers. Here is the exact deﬁnition. Deﬁnition 6.1 A real number x is rational if x = a b for some a,b ∈Z. Also, x is irrational if it is not rational, that is if x ̸= a b for every a,b ∈Z. We are now ready to use contradiction to prove that p 2 is irrational. According to the outline, the ﬁrst line of the proof should be “Suppose that it is not true that p 2 is irrational.” But it is helpful (though not mandatory) to tip our reader oﬀto the fact that we are using proof by contradiction. One standard way of doing this is to make the ﬁrst line “Suppose for the sake of contradiction that it is not true that p 2 is irrational." Proposition The number p 2 is irrational. Proof. Suppose for the sake of contradiction that it is not true that p 2 is irrational. Then p 2 is rational, so there are integers a and b for which p 2 = a b. (6.1) Let this fraction be fully reduced; in particular, this means that a and b are not both even. (If they were both even, the fraction could be further reduced by factoring 2’s from the numerator and denominator and canceling.) Squaring both sides of Equation 6.1 gives 2 = a2 b2 , and therefore a2 = 2b2. (6.2) From this it follows that a2 is even. But we proved earlier (Exercise 1 on page 110) that a2 being even implies a is even. Thus, as we know that a and b are not both even, it follows that b is odd. Now, since a is even there is an integer c for which a = 2c. Plugging this value for a into Equation (6.2), we get (2c)2 = 2b2, so 4c2 = 2b2, and hence b2 = 2c2. This means b2 is even, so b is even also. But previously we deduced that b is odd. Thus we have the contradiction b is even and b is odd. ■ To appreciate the power of proof by contradiction, imagine trying to prove that p 2 is irrational without it. Where would we begin? What would be our initial assumption? There are no clear answers to these questions. 114 Proof by Contradiction Proof by contradiction gives us a starting point: Assume p 2 is rational, and work from there. In the above proof we got the contradiction (b is even) ∧∼(b is even) which has the form C∧∼C. In general, your contradiction need not necessarily be of this form. Any statement that is clearly false is suﬃcient. For example 2 ̸= 2 would be a ﬁne contradiction, as would be 4 \| 2, provided that you could deduce them. Here is another ancient example, dating back at least as far as Euclid: Proposition There are inﬁnitely many prime numbers. Proof. For the sake of contradiction, suppose there are only ﬁnitely many prime numbers. Then we can list all the prime numbers as p1, p2, p3,... pn, where p1 = 2, p2 = 3, p3 = 5, p4 = 7 and so on. Thus pn is the nth and largest prime number. Now consider the number a = (p1p2p3 ··· pn)+1, that is, a is the product of all prime numbers, plus 1. Now a, like any natural number greater than 1, has at least one prime divisor, and that means pk \| a for at least one of our n prime numbers pk. Thus there is an integer c for which a = cpk, which is to say (p1p2p3 ··· pk−1pk pk+1 ··· pn)+1 = cpk. Dividing both sides of this by pk gives us (p1p2p3 ··· pk−1pk+1 ··· pn)+ 1 pk = c, so 1 pk = c −(p1p2p3 ··· pk−1pk+1 ··· pn). The expression on the right is an integer, while the expression on the left is not an integer. This is a contradiction. ■ Proof by contradiction often works well in proving statements of the form ∀x,P(x). The reason is that the proof set-up involves assuming ∼∀x,P(x), which as we know from Section 2.10 is equivalent to ∃x,∼P(x). This gives us a speciﬁc x for which ∼P(x) is true, and often that is enough to produce a contradiction. Here is an example: Proposition For every real number x ∈[0,π/2], we have sinx+cosx ≥1. Proof. Suppose for the sake of contradiction that this is not true. Then there exists an x ∈[0,π/2] for which sinx+cosx < 1. Proving Conditional Statements by Contradiction 115 Since x ∈[0,π/2], neither sinx nor cosx is negative, so 0 ≤sinx+cosx < 1. Thus 02 ≤(sinx+cosx)2 < 12, which gives 02 ≤sin2 x+2sinxcosx+cos2 x < 12. As sin2 x+cos2 x = 1, this becomes 0 ≤1+2sinxcosx < 1, so 1+2sinxcosx < 1. Subtracting 1 from both sides gives 2sinxcosx < 0. But this contradicts the fact that neither sinx nor cosx is negative. ■ 6.2 Proving Conditional Statements by Contradiction Since the previous two chapters dealt exclusively with proving conditional statements, we now formalize the procedure in which contradiction is used to prove a conditional statement. Suppose we want to prove a proposition of the following form. Proposition If P, then Q. Thus we need to prove that P ⇒Q is a true statement. Proof by contradiction begins with the assumption that ∼(P ⇒Q) is true, that is, that P ⇒Q is false. But we know that P ⇒Q being false means that it is possible that P can be true while Q is false. Thus the ﬁrst step in the proof is to assume P and ∼Q. Here is an outline: Outline for Proving a Conditional Statement with Contradiction Proposition If P, then Q. Proof. Suppose P and ∼Q. . . . Therefore C ∧∼C. ■ To illustrate this new technique, we revisit a familiar result: If a2 is even, then a is even. According to the outline, the ﬁrst line of the proof should be “For the sake of contradiction, suppose a2 is even and a is not even.” Proposition Suppose a ∈Z. If a2 is even, then a is even. Proof. For the sake of contradiction, suppose a2 is even and a is not even. Then a2 is even, and a is odd. Since a is odd, there is an integer c for which a = 2c +1. Then a2 = (2c +1)2 = 4c2 +4c +1 = 2(2c2 +2c)+1, so a2 is odd. Thus a2 is even and a2 is not even, a contradiction. ■ 116 Proof by Contradiction Here is another example. Proposition If a,b ∈Z and a ≥2, then a ∤b or a ∤(b +1). Proof. Suppose for the sake of contradiction there exist a,b ∈Z with a ≥2, and for which it is not true that a ∤b or a ∤(b +1). By DeMorgan’s law, we have a \| b and a \| (b +1). The deﬁnition of divisibility says there are c,d ∈Z with b = ac and b+1 = ad. Subtracting one equation from the other gives ad −ac = 1, so a(d −c) = 1. Since a is positive, d−c is also positive (otherwise a(d−c) would be negative). Then d −c is a positive integer and a(d −c) = 1, so a = 1/(d −c) < 2. Thus we have a ≥2 and a < 2, a contradiction. ■ 6.3 Combining Techniques Often, especially in more complex proofs, several proof techniques are combined within a single proof. For example, in proving a conditional statement P ⇒Q, we might begin with direct proof and thus assume P to be true with the aim of ultimately showing Q is true. But the truth of Q might hinge on the truth of some other statement R which—together with P—would imply Q. We would then need to prove R, and we would use whichever proof technique seems most appropriate. This can lead to “proofs inside of proofs.” Consider the following example. The overall approach is direct, but inside the direct proof is a separate proof by contradiction. Proposition Every non-zero rational number can be expressed as a product of two irrational numbers. Proof. This proposition can be reworded as follows: If r is a non-zero rational number, then r is a product of two irrational numbers. In what follows, we prove this with direct proof. Suppose r is a non-zero rational number. Then r = a b for integers a and b. Also, r can be written as a product of two numbers as follows: r = p 2· r p 2 . We know p 2 is irrational, so to complete the proof we must show r/ p 2 is also irrational. To show this, assume for the sake of contradiction that r/ p 2 is rational. This means r p 2 = c d Some Words of Advice 117 for integers c and d, so p 2 = r d c . But we know r = a/b, which combines with the above equation to give p 2 = r d c = a b d c = ad bc . This means p 2 is rational, which is a contradiction because we know it is irrational. Therefore r/ p 2 is irrational. Consequently r = p 2· r/ p 2 is a product of two irrational numbers. ■ For another example of a proof-within-a-proof, try Exercise 5 at the end of this chapter (or see its solution). Exercise 5 asks you to prove that p 3 is irrational. This turns out to be slightly trickier than proving that p 2 is irrational. 6.4 Some Words of Advice Despite the power of proof by contradiction, it’s best to use it only when the direct and contrapositive approaches do not seem to work. The reason for this is that a proof by contradiction can often have hidden in it a simpler contrapositive proof, and if this is the case it’s better to go with the simpler approach. Consider the following example. Proposition Suppose a ∈Z. If a2 −2a+7 is even, then a is odd. Proof. To the contrary, suppose a2 −2a+7 is even and a is not odd. That is, suppose a2 −2a+7 is even and a is even. Since a is even, there is an integer c for which a = 2c. Then a2 −2a+7 = (2c)2 −2(2c)+7 = 2(2c2 −2c +3)+1, so a2 −2a+7 is odd. Thus a2 −2a+7 is both even and odd, a contradiction. ■ Though there is nothing really wrong with this proof, notice that part of it assumes a is not odd and deduces that a2 −2a+7 is not even. That is the contrapositive approach! Thus it would be more eﬃcient to proceed as follows, using contrapositive proof. Proposition Suppose a ∈Z. If a2 −2a+7 is even, then a is odd. Proof. (Contrapositive) Suppose a is not odd. Then a is even, so there is an integer c for which a = 2c. Then a2 −2a+7 = (2c)2 −2(2c)+7 = 2(2c2 −2c +3)+1, so a2 −2a+7 is odd. Thus a2 −2a+7 is not even. ■ 118 Proof by Contradiction Exercises for Chapter 6 A. Use the method of proof by contradiction to prove the following statements. (In each case, you should also think about how a direct or contrapositive proof would work. You will ﬁnd in most cases that proof by contradiction is easier.) 1. Suppose n ∈Z. If n is odd, then n2 is odd. 2. Suppose n ∈Z. If n2 is odd, then n is odd. 3. Prove that 3 p 2 is irrational. 4. Prove that p 6 is irrational. 5. Prove that p 3 is irrational. 6. If a,b ∈Z, then a2 −4b −2 ̸= 0. 7. If a,b ∈Z, then a2 −4b −3 ̸= 0. 8. Suppose a,b, c ∈Z. If a2 + b2 = c2, then a or b is even. 9. Suppose a,b ∈R. If a is rational and ab is irrational, then b is irrational. 10. There exist no integers a and b for which 21a+30b = 1. 11. There exist no integers a and b for which 18a+6b = 1. 12. For every positive x ∈Q, there is a positive y ∈Q for which y < x. 13. For every x ∈[π/2,π], sinx−cosx ≥1. 14. If A and B are sets, then A ∩(B −A) = ;. 15. If b ∈Z and b ∤k for every k ∈N, then b = 0. 16. If a and b are positive real numbers, then a+ b ≥2 p ab. 17. For every n ∈Z, 4 ∤(n2 +2). 18. Suppose a,b ∈Z. If 4 \| (a2 + b2), then a and b are not both odd. B. Prove the following statements using any method from Chapters 4, 5 or 6. 19. The product of any ﬁve consecutive integers is divisible by 120. (For example, the product of 3,4,5,6 and 7 is 2520, and 2520 = 120·21.) 20. We say that a point P = (x, y) in R2 is rational if both x and y are rational. More precisely, P is rational if P = (x, y) ∈Q2. An equation F(x, y) = 0 is said to have a rational point if there exists x0, y0 ∈Q such that F(x0, y0) = 0. For example, the curve x2 + y2 −1 = 0 has rational point (x0, y0) = (1,0). Show that the curve x2 + y2 −3 = 0 has no rational points. 21. Exercise 20 (above) involved showing that there are no rational points on the curve x2 + y2 −3 = 0. Use this fact to show that p 3 is irrational. 22. Explain why x2 + y2 −3 = 0 not having any rational solutions (Exercise 20) implies x2 + y2 −3k = 0 has no rational solutions for k an odd, positive integer. 23. Use the above result to prove that p 3k is irrational for all odd, positive k. 24. The number log2 3 is irrational. Part III More on Proof CHAPTER 7 Proving Non-Conditional Statements T he last three chapters introduced three major proof techniques: direct, contrapositive and contradiction. These three techniques are used to prove statements of the form “If P, then Q.” As we know, most theorems and propositions have this conditional form, or they can be reworded to have this form. Thus the three main techniques are quite important. But some theorems and propositions cannot be put into conditional form. For example, some theorems have form “P if and only if Q.” Such theorems are biconditional statements, not conditional statements. In this chapter we examine ways to prove them. In addition to learning how to prove if-and-only-if theorems, we will also look at two other types of theorems. 7.1 If-and-Only-If Proof Some propositions have the form P if and only if Q. We know from Section 2.4 that this statement asserts that both of the following conditional statements are true: If P, then Q. If Q, then P. So to prove “P if and only if Q,” we must prove two conditional statements. Recall from Section 2.4 that Q ⇒P is called the converse of P ⇒Q. Thus we need to prove both P ⇒Q and its converse. Since these are both condi-tional statements we may prove them with either direct, contrapositive or contradiction proof. Here is an outline: Outline for If-and-Only-If Proof Proposition P if and only if Q. Proof. [Prove P ⇒Q using direct, contrapositive or contradiction proof.] [Prove Q ⇒P using direct, contrapositive or contradiction proof.] 122 Proving Non-Conditional Statements Let’s start with a very simple example. You already know that an integer n is odd if and only if n2 is odd, but let’s prove it anyway, just to illustrate the outline. In this example we prove (n is odd)⇒(n2 is odd) using direct proof and (n2 is odd)⇒(n is odd) using contrapositive proof. Proposition The integer n is odd if and only if n2 is odd. Proof. First we show that n being odd implies that n2 is odd. Suppose n is odd. Then, by deﬁnition of an odd number, n = 2a+1 for some integer a. Thus n2 = (2a+1)2 = 4a2 +4a+1 = 2(2a2 +2a)+1. This expresses n2 as twice an integer, plus 1, so n2 is odd. Conversely, we need to prove that n2 being odd implies that n is odd. We use contrapositive proof. Suppose n is not odd. Then n is even, so n = 2a for some integer a (by deﬁnition of an even number). Thus n2 = (2a)2 = 2(2a2), so n2 is even because it’s twice an integer. Thus n2 is not odd. We’ve now proved that if n is not odd, then n2 is not odd, and this is a contrapositive proof that if n2 is odd then n is odd. ■ In proving “P if and only if Q,” you should begin a new paragraph when starting the proof of Q ⇒P. Since this is the converse of P ⇒Q, it’s a good idea to begin the paragraph with the word “Conversely” (as we did above) to remind the reader that you’ve ﬁnished the ﬁrst part of the proof and are moving on to the second. Likewise, it’s a good idea to remind the reader of exactly what statement that paragraph is proving. The next example uses direct proof in both parts of the proof. Proposition Suppose a and b are integers. Then a ≡b (mod 6) if and only if a ≡b (mod 2) and a ≡b (mod 3). Proof. First we prove that if a ≡b (mod 6), then a ≡b (mod 2) and a ≡b (mod 3). Suppose a ≡b (mod 6). This means 6 \| (a−b), so there is an integer n for which a−b = 6n. From this we get a−b = 2(3n), which implies 2 \| (a−b), so a ≡b (mod 2). But we also get a−b = 3(2n), which implies 3 \| (a−b), so a ≡b (mod 3). Therefore a ≡b (mod 2) and a ≡b (mod 3). Conversely, suppose a ≡b (mod 2) and a ≡b (mod 3). Since a ≡b (mod 2) we get 2 \| (a−b), so there is an integer k for which a−b = 2k. Therefore a−b is even. Also, from a ≡b (mod 3) we get 3 \| (a−b), so there is an integer ℓ for which a−b = 3ℓ. Equivalent Statements 123 But since we know a −b is even, it follows that ℓmust be even also, for if it were odd then a −b = 3ℓwould be odd (because a −b would be the product of two odd integers). Hence ℓ= 2m for some integer m. Thus a−b = 3ℓ= 3·2m = 6m. This means 6 \| (a−b), so a ≡b (mod 6). ■ Since if-and-only-if proofs simply combine methods with which we are already familiar, we will not do any further examples in this section. However, it is of utmost importance that you practice your skill on some of this chapter’s exercises. 7.2 Equivalent Statements In other courses you will sometimes encounter a certain kind of theorem that is neither a conditional nor a biconditional statement. Instead, it asserts that a list of statements is “equivalent.” You saw this (or will see it) in your linear algebra textbook, which featured the following theorem: Theorem Suppose A is an n × n matrix. The following statements are equivalent: (a) The matrix A is invertible. (b) The equation Ax = b has a unique solution for every b ∈Rn. (c) The equation Ax = 0 has only the trivial solution. (d) The reduced row echelon form of A is In. (e) det(A) ̸= 0. (f) The matrix A does not have 0 as an eigenvalue. When a theorem asserts that a list of statements is “equivalent,” it is asserting that either the statements are all true, or they are all false. Thus the above theorem tells us that whenever we are dealing with a particular n×n matrix A, then either the statements (a) through (f) are all true for A, or statements (a) through (f) are all false for A. For example, if we happen to know that det(A) ̸= 0, the theorem assures us that in addition to statement (e) being true, all the statements (a) through (f) are true. On the other hand, if it happens that det(A) = 0, the theorem tells us that all statements (a) through (f) are false. In this way, the theorem multiplies our knowledge of A by a factor of six. Obviously that can be very useful. What method would we use to prove such a theorem? In a certain sense, the above theorem is like an if-and-only-if theorem. An if-and-only-if theorem of form P ⇔Q asserts that P and Q are either both true or both false, that is, that P and Q are equivalent. To prove P ⇔Q we prove P ⇒Q followed by Q ⇒P, essentially making a “cycle” of implications from P to Q 124 Proving Non-Conditional Statements and back to P. Similarly, one approach to proving the theorem about the n × n matrix would be to prove the conditional statement (a) ⇒(b), then (b) ⇒(c), then (c) ⇒(d), then (d) ⇒(e), then (e) ⇒(f ) and ﬁnally (f ) ⇒(a). This pattern is illustrated below. (a) = ⇒ (b) = ⇒ (c) ⇑ ⇓ (f ) ⇐ = (e) ⇐ = (d) Notice that if these six implications have been proved, then it really does follow that the statements (a) through (f) are either all true or all false. If one of them is true, then the circular chain of implications forces them all to be true. On the other hand, if one of them (say (c)) is false, the fact that (b) ⇒(c) is true forces (b) to be false. This combined with the truth of (a) ⇒(b) makes (a) false, and so on counterclockwise around the circle. Thus to prove that n statements are equivalent, it suﬃces to prove n conditional statements showing each statement implies another, in circular pattern. But it is not necessary that the pattern be circular. The following schemes would also do the job: (a) = ⇒ (b) ⇐ ⇒ (c) ⇑ ⇓ (f ) ⇐ = (e) ⇐ ⇒ (d) (a) ⇐ ⇒ (b) ⇐ ⇒ (c) ⇕ (f ) ⇐ ⇒ (e) ⇐ ⇒ (d) But a circular pattern yields the fewest conditional statements that must be proved. Whatever the pattern, each conditional statement can be proved with either direct, contrapositive or contradiction proof. Though we shall not do any of these proofs in this text, you are sure to encounter them in subsequent courses. 7.3 Existence Proofs; Existence and Uniqueness Proofs Up until this point, we have dealt with proving conditional statements or with statements that can be expressed with two or more conditional statements. Generally, these conditional statements have form P(x) ⇒Q(x). (Possibly with more than one variable.) We saw in Section 2.8 that this can be interpreted as a universally quantiﬁed statement ∀x,P(x) ⇒Q(x). Existence Proofs; Existence and Uniqueness Proofs 125 Thus, conditional statements are universally quantiﬁed statements, so in proving a conditional statement—whether we use direct, contrapositive or contradiction proof—we are really proving a universally quantiﬁed statement. But how would we prove an existentially quantiﬁed statement? What technique would we employ to prove a theorem of the following form? ∃x,R(x) This statement asserts that there exists some speciﬁc object x for which R(x) is true. To prove ∃x,R(x) is true, all we would have to do is ﬁnd and display an example of a speciﬁc x that makes R(x) true. Though most theorems and propositions are conditional (or if-and-only-if) statements, a few have the form ∃x,R(x). Such statements are called existence statements, and theorems that have this form are called existence theorems. To prove an existence theorem, all you have to do is provide a particular example that shows it is true. This is often quite simple. (But not always!) Here are some examples: Proposition There exists an even prime number. Proof. Observe that 2 is an even prime number. ■ Admittedly, this last proposition was a bit of an oversimpliﬁcation. The next one is slightly more challenging. Proposition There exists an integer that can be expressed as the sum of two perfect cubes in two diﬀerent ways. Proof. Consider the number 1729. Note that 13 +123 = 1729 and 93 +103 = 1729. Thus the number 1729 can be expressed as the sum of two perfect cubes in two diﬀerent ways. ■ Sometimes in the proof of an existence statement, a little veriﬁcation is needed to show that the example really does work. For example, the above proof would be incomplete if we just asserted that 1729 can be written as a sum of two cubes in two ways without showing how this is possible. WARNING: Although an example suﬃces to prove an existence statement, a single example does not prove a conditional statement. 126 Proving Non-Conditional Statements Often an existence statement will be embedded inside of a conditional statement. Consider the following. (Recall the deﬁnition of gcd on page 90.) If a,b ∈N, then there exist integers k and ℓfor which gcd(a,b) = ak+bℓ. This is a conditional statement that has the form a,b ∈N = ⇒ ∃k,ℓ∈Z, gcd(a,b) = ak + bℓ. To prove it with direct proof, we would ﬁrst assume that a,b ∈N, then prove the existence statement ∃k,ℓ∈Z, gcd(a,b) = ak + bℓ. That is, we would produce two integers k and ℓ(which depend on a and b) for which gcd(a,b) = ak+bℓ. Let’s carry out this plan. (We will use this fundamental proposition several times later, so it is given a number.) Proposition 7.1 If a,b ∈N, then there exist integers k and ℓfor which gcd(a,b) = ak + bℓ. Proof. (Direct) Suppose a,b ∈N. Consider the set A = © ax + by : x, y ∈Z ª. This set contains both positive and negative integers, as well as 0. (Reason: Let y = 0 and let x range over all integers. Then ax+ by = ax ranges over all multiples of a, both positive, negative and zero.) Let d be the smallest positive element of A. Then, because d is in A, it must have the form d = ak + bℓfor some speciﬁc k,ℓ∈Z. To ﬁnish, we will show d = gcd(a,b). We will ﬁrst argue that d is a common divisor of a and b, and then that it is the greatest common divisor. To see that d \| a, use the division algorithm (page 29) to write a = qd+r for integers q and r with 0 ≤r < d. The equation a = qd + r yields r = a−qd = a−q(ak + bℓ) = a(1−qk)+ b(−qℓ). Therefore r has form r = ax + by, so it belongs to A. But 0 ≤r < d and d is the smallest positive number in A, so r can’t be positive; hence r = 0. Updating our equation a = qd + r, we get a = qd, so d \| a. Repeating this argument with b = qd + r shows d \| b. Thus d is indeed a common divisor of a and b. It remains to show that it is the greatest common divisor. As gcd(a,b) divides a and b, we have a = gcd(a,b)·m and b = gcd(a,b)·n for some m,n ∈Z. So d = ak+bℓ= gcd(a,b)·mk+gcd(a,b)·nℓ= gcd(a,b) ¡ mk+nℓ ¢, and thus d is a multiple of gcd(a,b). Therefore d ≥gcd(a,b). But d can’t be a larger common divisor of a and b than gcd(a,b), so d = gcd(a,b). ■ Existence Proofs; Existence and Uniqueness Proofs 127 We conclude this section with a discussion of so-called uniqueness proofs. Some existence statements have form “There is a unique x for which P(x).” Such a statement asserts that there is exactly one example x for which P(x) is true. To prove it, you must produce an example x = d for which P(d) is true, and you must show that d is the only such example. The next proposition illustrates this. In essence, it asserts that the set © ax+ by : x, y ∈Z ª consists precisely of all the multiples of gcd(a,b). Proposition Suppose a,b ∈N. Then there exists a unique d ∈N for which: An integer m is a multiple of d if and only if m = ax+ by for some x, y ∈Z. Proof. Suppose a,b ∈N. Let d = gcd(a,b). We now show that an integer m is a multiple of d if and only if m = ax+by for some x, y ∈Z. Let m = dn be a multiple of d. By Proposition 7.1 (on the previous page), there are integers k and ℓfor which d = ak + bℓ. Then m = dn = (ak + bℓ)n = a(kn)+ b(ℓn), so m = ax+ by for integers x = kn and y = ℓn. Conversely, suppose m = ax+ by for some x, y ∈Z. Since d = gcd(a,b) is a divisor of both a and b, we have a = dc and b = de for some c, e ∈Z. Then m = ax+ by = dcx+ dey = d(cx+ ey), and this is a multiple of d. We have now shown that there is a natural number d with the property that m is a multiple of d if and only if m = ax + by for some x, y ∈Z. It remains to show that d is the unique such natural number. To do this, suppose d′ is any natural number with the property that d has: m is a multiple of d′ ⇐ ⇒m = ax+ by for some x, y ∈Z. (7.1) We next argue that d′ = d; that is, d is the unique natural number with the stated property. Because of (7.1), m = a·1+ b ·0 = a is a multiple of d′. Likewise m = a · 0 + b · 1 = b is a multiple of d′. Hence a and b are both multiples of d′, so d′ is a common divisor of a and b, and therefore d′ ≤gcd(a,b) = d. But also, by (7.1), the multiple m = d′ · 1 = d′ of d′ can be expressed as d′ = ax+by for some x, y ∈Z. As noted in the second paragraph of the proof, a = dc and b = de for some c, e ∈Z. Thus d′ = ax+ by = dcx+ dey = d(cx+ ey), so d′ is a multiple d. As d′ and d are both positive, it follows that d ≤d′. We’ve now shown that d′ ≤d and d ≤d′, so d = d′. The proof is complete. ■ 128 Proving Non-Conditional Statements 7.4 Constructive Versus Non-Constructive Proofs Existence proofs fall into two categories: constructive and non-constructive. Constructive proofs display an explicit example that proves the theorem; non-constructive proofs prove an example exists without actually giving it. We illustrate the diﬀerence with two proofs of the same fact: There exist irrational numbers x and y (possibly equal) for which xy is rational. Proposition There exist irrational numbers x, y for which xy is rational. Proof. Let x = p 2 p 2 and y = p 2. We know y is irrational, but it is not clear whether x is rational or irrational. On one hand, if x is irrational, then we have an irrational number to an irrational power that is rational: xy = µp 2 p 2¶p 2 = p 2 p 2 p 2 = p 2 2 = 2. On the other hand, if x is rational, then yy = p 2 p 2 = x is rational. Either way, we have a irrational number to an irrational power that is rational. ■ The above is a classic example of a non-constructive proof. It shows that there exist irrational numbers x and y for which xy is rational without actually producing (or constructing) an example. It convinces us that one of ¡p 2 p 2¢p 2 or p 2 p 2 is an irrational number to an irrational power that is rational, but it does not say which one is the correct example. It thus proves that an example exists without explicitly stating one. Next comes a constructive proof of this statement, one that produces (or constructs) two explicit irrational numbers x, y for which xy is rational. Proposition There exist irrational numbers x, y for which xy is rational. Proof. Let x = p 2 and y = log2 9. Then xy = p 2 log2 9 = p 2 log2 32 = p 2 2log2 3 = ³p 2 2´log2 3 = 2log2 3 = 3. As 3 is rational, we have shown that xy = 3 is rational. We know that x = p 2 is irrational. The proof will be complete if we can show that y = log2 9 is irrational. Suppose for the sake of contradiction that log2 9 is rational, so there are integers a and b for which a b = log2 9. This means 2a/b = 9, so ¡ 2a/b¢b = 9b, which reduces to 2a = 9b. But 2a is even, while 9b is odd (because it is the product of the odd number 9 with itself b times). This is a contradiction; the proof is complete. ■ Constructive Versus Non-Constructive Proofs 129 This existence proof has inside of it a separate proof (by contradiction) that log2 9 is irrational. Such combinations of proof techniques are, of course, typical. Be alert to constructive and non-constructive proofs as you read proofs in other books and articles, as well as to the possibility of crafting such proofs of your own. Exercises for Chapter 7 Prove the following statements. These exercises are cumulative, covering all techniques addressed in Chapters 4–7. 1. Suppose x ∈Z. Then x is even if and only if 3x+5 is odd. 2. Suppose x ∈Z. Then x is odd if and only if 3x+6 is odd. 3. Given an integer a, then a3 + a2 + a is even if and only if a is even. 4. Given an integer a, then a2 +4a+5 is odd if and only if a is even. 5. An integer a is odd if and only if a3 is odd. 6. Suppose x, y ∈R. Then x3 + x2 y = y2 + xy if and only if y = x2 or y = −x. 7. Suppose x, y ∈R. Then (x+ y)2 = x2 + y2 if and only if x = 0 or y = 0. 8. Suppose a,b ∈Z. Prove that a ≡b (mod 10) if and only if a ≡b (mod 2) and a ≡b (mod 5). 9. Suppose a ∈Z. Prove that 14 \| a if and only if 7 \| a and 2 \| a. 10. If a ∈Z, then a3 ≡a (mod 3). 11. Suppose a,b ∈Z. Prove that (a−3)b2 is even if and only if a is odd or b is even. 12. There exist a positive real number x for which x2 < px. 13. Suppose a,b ∈Z. If a+ b is odd, then a2 + b2 is odd. 14. Suppose a ∈Z. Then a2 \| a if and only if a ∈ © −1,0,1 ª. 15. Suppose a,b ∈Z. Prove that a+ b is even if and only if a and b have the same parity. 16. Suppose a,b ∈Z. If ab is odd, then a2 + b2 is even. 17. There is a prime number between 90 and 100. 18. There is a set X for which N ∈X and N ⊆X. 19. If n ∈N, then 20 +21 +22 +23 +24 +···+2n = 2n+1 −1. 20. There exists an n ∈N for which 11 \| (2n −1). 21. Every real solution of x3 + x+3 = 0 is irrational. 22. If n ∈Z, then 4 \| n2 or 4 \| (n2 −1). 23. Suppose a,b and c are integers. If a \| b and a \| (b2 −c), then a \| c. 24. If a ∈Z, then 4 ∤(a2 −3). 130 Proving Non-Conditional Statements 25. If p > 1 is an integer and n ∤p for each integer n for which 2 ≤n ≤pp, then p is prime. 26. The product of any n consecutive positive integers is divisible by n!. 27. Suppose a,b ∈Z. If a2 + b2 is a perfect square, then a and b are not both odd. 28. Prove the division algorithm: If a,b ∈N, there exist unique integers q,r for which a = bq + r, and 0 ≤r < b. (A proof of existence is given in Section 1.9, but uniqueness needs to be established too.) 29. If a \| bc and gcd(a,b) = 1, then a \| c. (Suggestion: Use the proposition on page 126.) 30. Suppose a,b, p ∈Z and p is prime. Prove that if p \| ab then p \| a or p \| b. (Suggestion: Use the proposition on page 126.) 31. If n ∈Z, then gcd(n,n+1) = 1. 32. If n ∈Z, then gcd(n,n+2) ∈ © 1,2 ª. 33. If n ∈Z, then gcd(2n+1,4n2 +1) = 1. 34. If gcd(a, c) = gcd(b, c) = 1, then gcd(ab, c) = 1. (Suggestion: Use the proposition on page 126.) 35. Suppose a,b ∈N. Then a = gcd(a,b) if and only if a \| b. 36. Suppose a,b ∈N. Then a = lcm(a,b) if and only if b \| a. CHAPTER 8 Proofs Involving Sets S tudents in their ﬁrst advanced mathematics classes are often surprised by the extensive role that sets play and by the fact that most of the proofs they encounter are proofs about sets. Perhaps you’ve already seen such proofs in your linear algebra course, where a vector space was deﬁned to be a set of objects (called vectors) that obey certain properties. Your text proved many things about vector spaces, such as the fact that the intersection of two vector spaces is also a vector space, and the proofs used ideas from set theory. As you go deeper into mathematics, you will encounter more and more ideas, theorems and proofs that involve sets. The purpose of this chapter is to give you a foundation that will prepare you for this new outlook. We will discuss how to show that an object is an element of a set, how to prove one set is a subset of another and how to prove two sets are equal. As you read this chapter, you may need to occasionally refer back to Chapter 1 to refresh your memory. For your convenience, the main deﬁnitions from Chapter 1 are summarized below. If A and B are sets, then: A ×B = © (x, y) : x ∈A, y ∈B ª , A ∪B = © x : (x ∈A)∨(x ∈B) ª , A ∩B = © x : (x ∈A)∧(x ∈B) ª , A −B = © x : (x ∈A)∧(x ∉B) ª , A = U −A. Recall that A ⊆B means that every element of A is also an element of B. 8.1 How to Prove a ∈A We will begin with a review of set-builder notation, and then review how to show that a given object a is an element of some set A. 132 Proofs Involving Sets Generally, a set A will be expressed in set-builder notation A = © x : P(x) ª, where P(x) is some statement (or open sentence) about x. The set A is understood to have as elements all those things x for which P(x) is true. For example, © x : x is an odd integer ª = © ...,−5,−3,−1,1,3,5,... ª . A common variation of this notation is to express a set as A = © x ∈S : P(x) ª. Here it is understood that A consists of all elements x of the (predetermined) set S for which P(x) is true. Keep in mind that, depending on context, x could be any kind of object (integer, ordered pair, set, function, etc.). There is also nothing special about the particular variable x; any reasonable symbol x, y, k, etc., would do. Some examples follow. © n ∈Z : n is odd ª = © ...,−5,−3,−1,1,3,5,... ª © x ∈N : 6\|x ª = © 6,12,18,24,30,... ª © (a,b) ∈Z×Z : b = a+5 ª = © ...,(−2,3),(−1,4),(0,5),(1,6),... ª © X ∈P(Z) : \|X\| = 1 ª = © ..., © −1 ª , © 0 ª , © 1 ª , © 2 ª , © 3 ª , © 4 ª ,... ª Now it should be clear how to prove that an object a belongs to a set © x : P(x) ª. Since © x : P(x) ª consists of all things x for which P(x) is true, to show that a ∈ © x : P(x) ª we just need to show that P(a) is true. Likewise, to show a ∈ © x ∈S : P(x) ª, we need to conﬁrm that a ∈S and that P(a) is true. These ideas are summarized below. However, you should not memorize these methods, you should understand them. With contemplation and practice, using them becomes natural and intuitive. How to show a ∈ © x : P(x) ª How to show a ∈ © x ∈S : P(x) ª Show that P(a) is true. 1. Verify that a ∈S. 2. Show that P(a) is true. Example 8.1 Let’s investigate elements of A = © x : x ∈N and 7\|x ª. This set has form A = © x : P(x) ª where P(x) is the open sentence (x ∈N)∧(7\|x). Thus 21 ∈A because P(21) is true. Similarly, 7,14,28,35, etc., are all elements of A. But 8 ∉A (for example) because P(8) is false. Likewise −14 ∉A because P(−14) is false. Example 8.2 Consider the set A = © X ∈P(N) : \|X\| = 3 ª. We know that © 4,13,45 ª ∈A because © 4,13,45 ª ∈P(N) and ¯ ¯© 4,13,45 ª¯ ¯ = 3. Also © 1,2,3 ª ∈A, © 10,854,3 ª ∈A, etc. However © 1,2,3,4 ª ∉A because ¯ ¯© 1,2,3,4 ª¯ ¯ ̸= 3. Further, © −1,2,3 ª ∉A because © −1,2,3 ª ∉P(N). How to Prove A ⊆B 133 Example 8.3 Consider the set B = © (x, y) ∈Z × Z : x ≡y (mod 5) ª. Notice (8,23) ∈B because (8,23) ∈Z×Z and 8 ≡23 (mod 5). Likewise, (100,75) ∈B, (102,77) ∈B, etc., but (6,10) ∉B. Now suppose n ∈Z and consider the ordered pair (4n+3,9n−2). Does this ordered pair belong to B? To answer this, we ﬁrst observe that (4n+3,9n−2) ∈Z×Z. Next, we observe that (4n+3)−(9n−2) = −5n+5 = 5(1−n), so 5\| ¡ (4n+3)−(9n−2) ¢, which means (4n+3) ≡(9n−2) (mod 5). Therefore we have established that (4n+3,9n−2) meets the requirements for belonging to B, so (4n+3,9n−2) ∈B for every n ∈Z. Example 8.4 This illustrates another common way of deﬁning a set. Consider the set C = © 3x3 +2 : x ∈Z ª. Elements of this set consist of all the values 3x3 +2 where x is an integer. Thus −22 ∈C because −22 = 3(−2)3 +2. You can conﬁrm −1 ∈C and 5 ∈C, etc. Also 0 ∉C and 1 2 ∉C, etc. 8.2 How to Prove A ⊆B In this course (and more importantly, beyond it) you will encounter many circumstances where it is necessary to prove that one set is a subset of an-other. This section explains how to do this. The methods we discuss should improve your skills in both writing your own proofs and in comprehending the proofs that you read. Recall (Deﬁnition 1.3) that if A and B are sets, then A ⊆B means that every element of A is also an element of B. In other words, it means if a ∈A, then a ∈B. Therefore to prove that A ⊆B, we just need to prove that the conditional statement “If a ∈A, then a ∈B” is true. This can be proved directly, by assuming a ∈A and deducing a ∈B. The contrapositive approach is another option: Assume a ∉B and deduce a ∉A. Each of these two approaches is outlined below. How to Prove A ⊆B How to Prove A ⊆B (Direct approach) (Contrapositive approach) Proof. Suppose a ∈A. . . . Therefore a ∈B. Thus a ∈A implies a ∈B, so it follows that A ⊆B. ■ Proof. Suppose a ∉B. . . . Therefore a ∉A. Thus a ∉B implies a ∉A, so it follows that A ⊆B. ■ 134 Proofs Involving Sets In practice, the direct approach usually results in the most straight-forward and easy proof, though occasionally the contrapositive is the most expedient. (You can even prove A ⊆B by contradiction: Assume (a ∈A)∧(a ∉B), and deduce a contradiction.) The remainder of this section consists of examples with occasional commentary. Unless stated otherwise, we will use the direct approach in all proofs; pay special attention to how the above outline for the direct approach is used. Example 8.5 Prove that © x ∈Z : 18\|x ª ⊆ © x ∈Z : 6\|x ª. Proof. Suppose a ∈ © x ∈Z : 18\|x ª. This means that a ∈Z and 18\|a. By deﬁnition of divisibility, there is an integer c for which a = 18c. Consequently a = 6(3c), and from this we deduce that 6\|a. Therefore a is one of the integers that 6 divides, so a ∈ © x ∈Z : 6\|x ª. We’ve shown a ∈ © x ∈Z : 18\|x ª implies a ∈ © n ∈Z : 6\|x ª, so it follows that © x ∈Z : 18\|x ª ⊆ © x ∈Z : 6\|x ª. ■ Example 8.6 Prove that © x ∈Z : 2\|x ª ∩ © x ∈Z : 9\|x ª ⊆ © x ∈Z : 6\|x ª. Proof. Suppose a ∈ © x ∈Z : 2\|x ª ∩ © x ∈Z : 9\|x ª. By deﬁnition of intersection, this means a ∈ © x ∈Z : 2\|x ª and a ∈ © x ∈Z : 9\|x ª. Since a ∈ © x ∈Z : 2\|x ª we know 2\|a, so a = 2c for some c ∈Z. Thus a is even. Since a ∈ © x ∈Z : 9\|x ª we know 9\|a, so a = 9d for some d ∈Z. As a is even, a = 9d implies d is even. (Otherwise a = 9d would be odd.) Then d = 2e for some integer e, and we have a = 9d = 9(2e) = 6(3e). From a = 6(3e), we conclude 6\|a, and this means a ∈ © x ∈Z : 6\|x ª. We have shown that a ∈ © x ∈Z : 2\|x ª ∩ © x ∈Z : 9\|x ª implies a ∈ © x ∈Z : 6\|x ª, so it follows that © x ∈Z : 2\|x ª ∩ © x ∈Z : 9\|x ª ⊆ © x ∈Z : 6\|x ª. ■ Example 8.7 Show © (x, y) ∈Z×Z : x ≡y (mod 6) ª ⊆ © (x, y) ∈Z×Z : x ≡y (mod 3) ª. Proof. Suppose (a,b) ∈ © (x, y) ∈Z×Z : x ≡y (mod 6) ª. This means (a,b) ∈Z×Z and a ≡b (mod 6). Consequently 6\|(a−b), so a−b = 6c for some integer c. It follows that a−b = 3(2c), and this means 3\|(a−b), so a ≡b (mod 3). Thus (a,b) ∈ © (x, y) ∈Z×Z : x ≡y (mod 3) ª. We’ve now seen that (a,b) ∈ © (x, y) ∈Z×Z : x ≡y (mod 6) ª implies (a,b) ∈ © (x, y) ∈Z×Z : x ≡y (mod 3) ª, so it follows that © (x, y) ∈Z×Z : x ≡y (mod 6) ª ⊆ © (x, y) ∈Z×Z : x ≡y (mod 3) ª. ■ How to Prove A ⊆B 135 Some statements involving subsets are transparent enough that we often accept (and use) them without proof. For example, if A and B are any sets, then it’s very easy to conﬁrm A ∩B ⊆A. (Reason: Suppose x ∈A ∩B. Then x ∈A and x ∈B by deﬁnition of intersection, so in particular x ∈A. Thus x ∈A ∩B implies x ∈A, so A ∩B ⊆A.) Other statements of this nature include A ⊆A ∪B and A −B ⊆A, as well as conditional statements such as ¡ (A ⊆B)∧(B ⊆C) ¢ ⇒(A ⊆C) and (X ⊆A) ⇒(X ⊆A ∪B). Our point of view in this text is that we do not need to prove such obvious statements unless we are explicitly asked to do so in an exercise. (Still, you should do some quick mental proofs to convince yourself that the above statements are true. If you don’t see that A ∩B ⊆A is true but that A ⊆A ∩B is not necessarily true, then you need to spend more time on this topic.) The next example will show that if A and B are sets, then P(A)∪P(B) ⊆ P(A ∪B). Before beginning our proof, let’s look at an example to see if this statement really makes sense. Suppose A = © 1,2 ª and B = © 2,3 ª. Then P(A)∪P(B) = © ;, © 1 ª , © 2 ª , © 1,2 ªª ∪ © ;, © 2 ª , © 3 ª , © 2,3 ªª = © ;, © 1 ª , © 2 ª , © 3 ª , © 1,2 ª , © 2,3 ªª . Also P(A∪B) = P( © 1,2,3 ª ) = © ;, © 1 ª , © 2 ª , © 3 ª , © 1,2 ª , © 2,3 ª , © 1,3 ª , © 1,2,3 ªª. Thus, even though P(A)∪P(B) ̸= P(A∪B), it is true that P(A)∪P(B) ⊆P(A∪B) for this particular A and B. Now let’s prove P(A)∪P(B) ⊆P(A ∪B) no matter what sets A and B are. Example 8.8 Prove that if A and B are sets, then P(A)∪P(B) ⊆P(A∪B). Proof. Suppose X ∈P(A)∪P(B). By deﬁnition of union, this means X ∈P(A) or X ∈P(B). Therefore X ⊆A or X ⊆B (by deﬁnition of power sets). We consider cases. Case 1. Suppose X ⊆A. Then X ⊆A ∪B, and this means X ∈P(A ∪B). Case 2. Suppose X ⊆B. Then X ⊆A ∪B, and this means X ∈P(A ∪B). (We do not need to consider the case where X ⊆A and X ⊆B because that is taken care of by either of cases 1 or 2.) The above cases show that X ∈P(A ∪B). Thus we’ve shown that X ∈P(A)∪P(B) implies X ∈P(A ∪B), and this completes the proof that P(A)∪P(B) ⊆P(A ∪B). ■ In our next example, we prove a conditional statement. Direct proof is used, and in the process we use our new technique for showing A ⊆B. 136 Proofs Involving Sets Example 8.9 Suppose A and B are sets. If P(A) ⊆P(B), then A ⊆B. Proof. We use direct proof. Assume P(A) ⊆P(B). Based on this assumption, we must now show that A ⊆B. To show A ⊆B, suppose that a ∈A. Then the one-element set © a ª is a subset of A, so © a ª ∈P(A). But then, since P(A) ⊆P(B), it follows that © a ª ∈P(B). This means that © a ª ⊆B, hence a ∈B. We’ve shown that a ∈A implies a ∈B, so therefore A ⊆B. ■ 8.3 How to Prove A = B In proofs it is often necessary to show that two sets are equal. There is a standard way of doing this. Suppose we want to show A = B. If we show A ⊆B, then every element of A is also in B, but there is still a possibility that B could have some elements that are not in A, so we can’t conclude A = B. But if in addition we also show B ⊆A, then B can’t contain anything that is not in A, so A = B. This is the standard procedure for proving A = B: Prove both A ⊆B and B ⊆A. How to Prove A = B Proof. [Prove that A ⊆B.] [Prove that B ⊆A.] Therefore, since A ⊆B and B ⊆A, it follows that A = B. ■ Example 8.10 Prove that © n ∈Z : 35\|n ª = © n ∈Z : 5\|n ª ∩ © n ∈Z : 7\|n ª. Proof. First we show © n ∈Z : 35\|n ª ⊆ © n ∈Z : 5\|n ª ∩ © n ∈Z : 7\|n ª. Suppose a ∈ © n ∈Z : 35\|n ª. This means 35\|a, so a = 35c for some c ∈Z. Thus a = 5(7c) and a = 7(5c). From a = 5(7c) it follows that 5\|a, so a ∈ © n ∈Z : 5\|n ª. From a = 7(5c) it follows that 7\|a, which means a ∈ © n ∈Z : 7\|n ª. As a belongs to both © n ∈Z : 5\|n ª and © n ∈Z : 7\|n ª, we get a ∈ © n ∈Z : 5\|n ª ∩ © n ∈Z : 7\|n ª. Thus we’ve shown that © n ∈Z : 35\|n ª ⊆ © n ∈Z : 5\|n ª ∩ © n ∈Z : 7\|n ª. Next we show © n ∈Z : 5\|n ª ∩ © n ∈Z : 7\|n ª ⊆ © n ∈Z : 35\|n ª. Suppose that a ∈ © n ∈Z : 5\|n ª ∩ © n ∈Z : 7\|n ª. By deﬁnition of intersection, this means that a ∈ © n ∈Z : 5\|n ª and a ∈ © n ∈Z : 7\|n ª. Therefore it follows that 5\|a and 7\|a. By deﬁnition of divisibility, there are integers c and d with a = 5c and a = 7d. Then a has both 5 and 7 as prime factors, so the prime factorization of a How to Prove A = B 137 must include factors of 5 and 7. Hence 5·7 = 35 divides a, so a ∈ © n ∈Z : 35\|n ª. We’ve now shown that © n ∈Z : 5\|n ª ∩ © n ∈Z : 7\|n ª ⊆ © n ∈Z : 35\|n ª. At this point we’ve shown that © n ∈Z : 35\|n ª ⊆ © n ∈Z : 5\|n ª ∩ © n ∈Z : 7\|n ª and © n ∈Z : 5\|n ª ∩ © n ∈Z : 7\|n ª ⊆ © n ∈Z : 35\|n ª, so we’ve proved © n ∈Z : 35\|n ª = © n ∈Z : 5\|n ª ∩ © n ∈Z : 7\|n ª. ■ You know from algebra that if c ̸= 0 and ac = bc, then a = b. The next example shows that an analogous statement holds for sets A,B and C. The example asks us to prove a conditional statement. We will prove it with direct proof. In carrying out the process of direct proof, we will have to use the new techniques from this section. Example 8.11 Suppose A, B, and C are sets, and C ̸= ;. Prove that if A ×C = B ×C, then A = B. Proof. Suppose A ×C = B ×C. We must now show A = B. First we will show A ⊆B. Suppose a ∈A. Since C ̸= ;, there exists an element c ∈C. Thus, since a ∈A and c ∈C, we have (a, c) ∈A × C, by deﬁnition of the Cartesian product. But then, since A×C = B×C, it follows that (a, c) ∈B ×C. Again by deﬁnition of the Cartesian product, it follows that a ∈B. We have shown a ∈A implies a ∈B, so A ⊆B. Next we show B ⊆A. We use the same argument as above, with the roles of A and B reversed. Suppose a ∈B. Since C ̸= ;, there exists an element c ∈C. Thus, since a ∈B and c ∈C, we have (a, c) ∈B ×C. But then, since B × C = A × C, we have (a, c) ∈A × C. It follows that a ∈A. We have shown a ∈B implies a ∈A, so B ⊆A. The previous two paragraphs have shown A ⊆B and B ⊆A, so A = B. In summary, we have shown that if A ×C = B ×C, then A = B. This completes the proof. ■ Now we’ll look at another way that set operations are similar to oper-ations on numbers. From algebra you are familiar with the distributive property a·(b + c) = a· b + a· c. Replace the numbers a,b, c with sets A,B,C, and replace · with × and + with ∩. We get A ×(B ∩C) = (A × B)∩(A × C). This statement turns out to be true, as we now prove. Example 8.12 Given sets A, B, and C, prove A×(B∩C) = (A×B)∩(A×C). Proof. First we will show that A ×(B ∩C) ⊆(A ×B)∩(A ×C). Suppose (a,b) ∈A ×(B ∩C). By deﬁnition of the Cartesian product, this means a ∈A and b ∈B ∩C. By deﬁnition of intersection, it follows that b ∈B and b ∈C. 138 Proofs Involving Sets Thus, since a ∈A and b ∈B, it follows that (a,b) ∈A ×B (by deﬁnition of ×). Also, since a ∈A and b ∈C, it follows that (a,b) ∈A ×C (by deﬁnition of ×). Now we have (a,b) ∈A ×B and (a,b) ∈A ×C, so (a,b) ∈(A ×B)∩(A ×C). We’ve shown that (a,b) ∈A × (B ∩C) implies (a,b) ∈(A × B) ∩(A × C) so we have A ×(B ∩C) ⊆(A ×B)∩(A ×C). Next we will show that (A ×B)∩(A ×C) ⊆A ×(B ∩C). Suppose (a,b) ∈(A ×B)∩(A ×C). By deﬁnition of intersection, this means (a,b) ∈A ×B and (a,b) ∈A ×C. By deﬁnition of the Cartesian product, (a,b) ∈A ×B means a ∈A and b ∈B. By deﬁnition of the Cartesian product, (a,b) ∈A ×C means a ∈A and b ∈C. We now have b ∈B and b ∈C, so b ∈B ∩C, by deﬁnition of intersection. Thus we’ve deduced that a ∈A and b ∈B ∩C, so (a,b) ∈A ×(B ∩C). In summary, we’ve shown that (a,b) ∈(A×B)∩(A×C) implies (a,b) ∈A×(B∩C) so we have (A ×B)∩(A ×C) ⊆A ×(B ∩C). The previous two paragraphs show that A×(B∩C) ⊆(A×B)∩(A×C) and (A×B)∩(A×C) ⊆A×(B∩C), so it follows that (A×B)∩(A×C) = A×(B∩C). ■ Occasionally you can prove two sets are equal by working out a series of equalities leading from one set to the other. This is analogous to showing two algebraic expressions are equal by manipulating one until you obtain the other. We illustrate this in the following example, which gives an alternate solution to the previous example. You are cautioned that this approach is sometimes diﬃcult to apply, but when it works it can shorten a proof dramatically. Before beginning the example, a note is in order. Notice that any statement P is logically equivalent to P ∧P. (Write out a truth table if you are in doubt.) At one point in the following example we will replace the expression x ∈A with the logically equivalent statement (x ∈A)∧(x ∈A). Example 8.13 Given sets A, B, and C, prove A×(B∩C) = (A×B)∩(A×C). Proof. Just observe the following sequence of equalities. A ×(B ∩C) = © (x, y) : (x ∈A)∧(y ∈B ∩C) ª (def. of ×) = © (x, y) : (x ∈A)∧(y ∈B)∧(y ∈C) ª (def. of ∩) = © (x, y) : (x ∈A)∧(x ∈A)∧(y ∈B)∧(y ∈C) ª (P = P ∧P) = © (x, y) : ((x ∈A)∧(y ∈B))∧((x ∈A)∧(y ∈C)) ª (rearrange) = © (x, y) : (x ∈A)∧(y ∈B) ª ∩ © (x, y) : (x ∈A)∧(y ∈C) ª (def. of ∩) = (A ×B)∩(A ×C) (def. of ×) The proof is complete. ■ Examples: Perfect Numbers 139 The equation A×(B∩C) = (A×B)∩(A×C) just obtained is a fundamental law that you may actually use fairly often as you continue with mathematics. Some similar equations are listed below. Each of these can be proved with this section’s techniques, and the exercises will ask that you do so. A ∩B = A ∪B A ∪B = A ∩B ) DeMorgan’s laws for sets A ∩(B ∪C) = (A ∩B)∪(A ∩C) A ∪(B ∩C) = (A ∪B)∩(A ∪C) ¾ Distributive laws for sets A ×(B ∪C) = (A ×B)∪(A ×C) A ×(B ∩C) = (A ×B)∩(A ×C) ¾ Distributive laws for sets It is very good practice to prove these equations. Depending on your learning style, it is probably not necessary to commit them to memory. But don’t forget them entirely. They may well be useful later in your mathematical education. If so, you can look them up or re-derive them on the spot. If you go on to study mathematics deeply, you will at some point realize that you’ve internalized them without even being cognizant of it. 8.4 Examples: Perfect Numbers Sometimes it takes a good bit of work and creativity to show that one set is a subset of another or that they are equal. We illustrate this now with examples from number theory involving what are called perfect numbers. Even though this topic is quite old, dating back more than 2000 years, it leads to some questions that are unanswered even today. The problem involves adding up the positive divisors of a natural number. To begin the discussion, consider the number 12. If we add up the positive divisors of 12 that are less than 12, we obtain 1+2+3+4+6 = 16, which is greater than 12. Doing the same thing for 15, we get 1+3+5 = 9 which is less than 15. For the most part, given a natural number p, the sum of its positive divisors less than itself will either be greater than p or less than p. But occasionally the divisors add up to exactly p. If this happens, then p is said to be a perfect number. Deﬁnition 8.1 A number p ∈N is perfect if it equals the sum of its positive divisors less than itself. Some examples follow. • The number 6 is perfect since 6 = 1+2+3. • The number 28 is perfect since 28 = 1+2+4+7+14. • The number 496 is perfect since 496 = 1+2+4+8+16+31+62+124+248. 140 Proofs Involving Sets Though it would take a while to ﬁnd it by trial-and-error, the next perfect number after 496 is 8128. You can check that 8128 is perfect. Its divisors are 1, 2, 4, 8, 16, 32, 64, 127, 254, 508, 1016, 2032, 4064 and indeed 8128 = 1+2+4+8+16+32+64+127+254+508+1016+2032+4064. Are there other perfect numbers? How can they be found? Do they obey any patterns? These questions fascinated the ancient Greek mathematicians. In what follows we will develop an idea—recorded by Euclid—that partially answers these questions. Although Euclid did not use sets,1 we will nonetheless phrase his idea using the language of sets. Since our goal is to understand what numbers are perfect, let’s deﬁne the following set: P = © p ∈N : p is perfect ª . Therefore P = © 6,28,496,8128,... ª, but it is unclear what numbers are in P other than the ones listed. Our goal is to gain a better understanding of just which numbers the set P includes. To do this, we will examine the following set A. It looks more complicated than P, but it will be very helpful for understanding P, as we will soon see. A = © 2n−1(2n −1) : n ∈N, and 2n −1 is prime ª In words, A consists of every natural number of form 2n−1(2n −1), where 2n −1 is prime. To get a feel for what numbers belong to A, look at the following table. For each natural number n, it tallies the corresponding numbers 2n−1 and 2n −1. If 2n −1 happens to be prime, then the product 2n−1(2n −1) is given; otherwise that entry is labeled with an ∗. n 2n−1 2n −1 2n−1(2n −1) 1 1 1 ∗ 2 2 3 6 3 4 7 28 4 8 15 ∗ 5 16 31 496 6 32 63 ∗ 7 64 127 8128 8 128 255 ∗ 9 256 511 ∗ 10 512 1023 ∗ 11 1024 2047 ∗ 12 2048 4095 ∗ 13 4096 8191 33,550,336 1Set theory was invented over 2000 years after Euclid died. Examples: Perfect Numbers 141 Notice that the ﬁrst four entries of A are the perfect numbers 6, 28, 496 and 8128. At this point you may want to jump to the conclusion that A = P. But it is a shocking fact that in over 2000 years no one has ever been able to determine whether or not A = P. But it is known that A ⊆P, and we will now prove it. In other words, we are going to show that every element of A is perfect. (But by itself, that leaves open the possibility that there may be some perfect numbers in P that are not in A.) The main ingredient for the proof will be the formula for the sum of a geometric series with common ratio r. You probably saw this most recently in Calculus II. The formula is n X k=0 rk = rn+1 −1 r −1 . We will need this for the case r = 2, which is n X k=0 2k = 2n+1 −1. (8.1) (See the solution for Exercise 19 in Section 7.4 for a proof of this for-mula.) Now we are ready to prove our result. Let’s draw attention to its signiﬁcance by calling it a theorem rather than a proposition. Theorem 8.1 If A = © 2n−1(2n −1) : n ∈N, and 2n −1 is prime ª and P = © p ∈N : p is perfect ª, then A ⊆P. Proof. Assume A and P are as stated. To show A ⊆P, we must show that p ∈A implies p ∈P. Thus suppose p ∈A. By deﬁnition of A, this means p = 2n−1(2n −1) (8.2) for some n ∈N for which 2n −1 is prime. We want to show that p ∈P, that is, we want to show p is perfect. Thus, we need to show that the sum of the positive divisors of p that are less than p add up to p. Notice that since 2n −1 is prime, any divisor of p = 2n−1(2n −1) must have the form 2k or 2k(2n −1) for 0 ≤k ≤n−1. Thus the positive divisors of p are as follows: 20, 21, 22, ... 2n−2, 2n−1, 20(2n −1), 21(2n −1), 22(2n −1), ... 2n−2(2n −1), 2n−1(2n −1). Notice that this list starts with 20 = 1 and ends with 2n−1(2n −1) = p. 142 Proofs Involving Sets If we add up all these divisors except for the last one (which equals p) we get the following: n−1 X k=0 2k + n−2 X k=0 2k(2n −1) = n−1 X k=0 2k +(2n −1) n−2 X k=0 2k = (2n −1)+(2n −1)(2n−1 −1) (by Equation (8.1)) = 1+(2n−1 −1) = 2n−1(2n −1) = p (by Equation (8.2)). This shows that the positive divisors of p that are less than p add up to p. Therefore p is perfect, by deﬁnition of a perfect number. Thus p ∈P, by deﬁnition of P. We have shown that p ∈A implies p ∈P, which means A ⊆P. ■ Combined with the chart on the previous page, this theorem gives us a new perfect number! The element p = 213−1(213 −1) = 33,550,336 in A is perfect. Observe also that every element of A is a multiple of a power of 2, and therefore even. But this does not necessarily mean every perfect number is even, because we’ve only shown A ⊆P, not A = P. For all we know there may be odd perfect numbers in P −A that are not in A. Are there any odd perfect numbers? No one knows. In over 2000 years, no one has ever found an odd perfect number, nor has anyone been able to prove that there are none. But it is known that the set A does contain every even perfect number. This fact was ﬁrst proved by Euler, and we duplicate his reasoning in the next theorem, which proves that A = E, where E is the set of all even perfect numbers. It is a good example of how to prove two sets are equal. For convenience, we are going to use a slightly diﬀerent deﬁnition of a perfect number. A number p ∈N is perfect if its positive divisors add up to 2p. For example, the number 6 is perfect since the sum of its divisors is 1+2+3+6 = 2·6. This deﬁnition is simpler than the ﬁrst one because we do not have to stipulate that we are adding up the divisors that are less than p. Instead we add in the last divisor p, and that has the eﬀect of adding an additional p, thereby doubling the answer. Examples: Perfect Numbers 143 Theorem 8.2 If A = © 2n−1(2n −1) : n ∈N, and 2n −1 is prime ª and E = © p ∈N : p is perfect and even ª, then A = E. Proof. To show that A = E, we need to show A ⊆E and E ⊆A. First we will show that A ⊆E. Suppose p ∈A. This means p is even, because the deﬁnition of A shows that every element of A is a multiple of a power of 2. Also, p is a perfect number because Theorem 8.1 states that every element of A is also an element of P, hence perfect. Thus p is an even perfect number, so p ∈E. Therefore A ⊆E. Next we show that E ⊆A. Suppose p ∈E. This means p is an even perfect number. Write the prime factorization of p as p = 2k3n15n27n2 ..., where some of the powers n1, n2, n3 ... may be zero. But, as p is even, the power k must be greater than zero. It follows p = 2kq for some positive integer k and an odd integer q. Now, our aim is to show that p ∈A, which means we must show p has form p = 2n−1(2n−1). To get our current p = 2kq closer to this form, let n = k +1, so we now have p = 2n−1q. (8.3) List the positive divisors of q as d1,d2,d3,...,dm. (Where d1 = 1 and dm = q.) Then the divisors of p are: 20d1 20d2 20d3 ... 20dm 21d1 21d2 21d3 ... 21dm 22d1 22d2 22d3 ... 22dm 23d1 23d2 23d3 ... 23dm . . . . . . . . . . . . 2n−1d1 2n−1d2 2n−1d3 ... 2n−1dm Since p is perfect, these divisors add up to 2p. By Equation (8.3), their sum is 2p = 2(2n−1q) = 2nq. Adding the divisors column-by-column, we get n−1 X k=0 2kd1 + n−1 X k=0 2kd2 + n−1 X k=0 2kd3 +···+ n−1 X k=0 2kdm = 2nq. Applying Equation (8.1), this becomes (2n −1)d1 +(2n −1)d2 +(2n −1)d3 +···+(2n −1)dm = 2nq (2n −1)(d1 + d2 + d3 +···+ dm) = 2nq d1 + d2 + d3 +···+ dm = 2nq 2n −1, 144 Proofs Involving Sets so that d1 + d2 + d3 +···+ dm = (2n −1+1)q 2n −1 = (2n −1)q + q 2n −1 = q + q 2n −1. From this we see that q 2n−1 is an integer. It follows that both q and q 2n−1 are positive divisors of q. Since their sum equals the sum of all positive divisors of q, it follows that q has only two positive divisors, q and q 2n−1. Since one of its divisors must be 1, it must be that q 2n−1 = 1, which means q = 2n −1. Now a number with just two positive divisors is prime, so q = 2n −1 is prime. Plugging this into Equation (8.3) gives p = 2n−1(2n −1), where 2n −1 is prime. This means p ∈A, by deﬁnition of A. We have now shown that p ∈E implies p ∈A, so E ⊆A. Since A ⊆E and E ⊆A, it follows that A = E. ■ Do not be alarmed if you feel that you wouldn’t have thought of this proof. It took the genius of Euler to discover this approach. We’ll conclude this chapter with some facts about perfect numbers. • The sixth perfect number is p = 217−1(217 −1) = 8589869056. • The seventh perfect number is p = 219−1(219 −1) = 137438691328. • The eighth perfect number is p = 231−1(231 −1) = 2305843008139952128. • The twentieth perfect number is p = 24423−1(24423 −1). It has 2663 digits. • The twenty-third perfect number is p = 211213−1(211213 −1). It has 6957 digits. As mentioned earlier, no one knows whether or not there are any odd perfect numbers. It is not even known whether there are ﬁnitely many or inﬁnitely many perfect numbers. It is known that the last digit of every even perfect number is either a 6 or an 8. Perhaps this is something you’d enjoy proving. We’ve seen that perfect numbers are closely related to prime numbers that have the form 2n −1. Such prime numbers are called Mersenne primes, after the French scholar Marin Mersenne (1588–1648), who popularized them. The ﬁrst several Mersenne primes are 22 −1 = 3, 23 −1 = 7, 25 −1 = 31, 27 −1 = 127 and 213 −1 = 8191. To date, only 48 Mersenne primes are known, the largest of which is 257,885,161 −1. There is a substantial cash prize for anyone who ﬁnds a 49th. (See You will probably have better luck with the exercises. Examples: Perfect Numbers 145 Exercises for Chapter 8 Use the methods introduced in this chapter to prove the following statements. 1. Prove that © 12n : n ∈Z ª ⊆ © 2n : n ∈Z ª ∩ © 3n : n ∈Z ª. 2. Prove that © 6n : n ∈Z ª = © 2n : n ∈Z ª ∩ © 3n : n ∈Z ª. 3. If k ∈Z, then © n ∈Z : n\|k ª ⊆ © n ∈Z : n\|k2ª. 4. If m,n ∈Z, then © x ∈Z : mn\|x ª ⊆ © x ∈Z : m\|x ª ∩ © x ∈Z : n\|x ª. 5. If p and q are positive integers, then © pn : n ∈N ª ∩ © qn : n ∈N ª ̸= ;. 6. Suppose A,B and C are sets. Prove that if A ⊆B, then A −C ⊆B −C. 7. Suppose A,B and C are sets. If B ⊆C, then A ×B ⊆A ×C. 8. If A,B and C are sets, then A ∪(B ∩C) = (A ∪B)∩(A ∪C). 9. If A,B and C are sets, then A ∩(B ∪C) = (A ∩B)∪(A ∩C). 10. If A and B are sets in a universal set U, then A ∩B = A ∪B. 11. If A and B are sets in a universal set U, then A ∪B = A ∩B. 12. If A,B and C are sets, then A −(B ∩C) = (A −B)∪(A −C). 13. If A,B and C are sets, then A −(B ∪C) = (A −B)∩(A −C). 14. If A,B and C are sets, then (A ∪B)−C = (A −C)∪(B −C). 15. If A,B and C are sets, then (A ∩B)−C = (A −C)∩(B −C). 16. If A,B and C are sets, then A ×(B ∪C) = (A ×B)∪(A ×C). 17. If A,B and C are sets, then A ×(B ∩C) = (A ×B)∩(A ×C). 18. If A,B and C are sets, then A ×(B −C) = (A ×B)−(A ×C). 19. Prove that © 9n : n ∈Z ª ⊆ © 3n : n ∈Z ª, but © 9n : n ∈Z ª ̸= © 3n : n ∈Z ª 20. Prove that © 9n : n ∈Q ª = © 3n : n ∈Q ª. 21. Suppose A and B are sets. Prove A ⊆B if and only if A −B = ;. 22. Let A and B be sets. Prove that A ⊆B if and only if A ∩B = A. 23. For each a ∈R, let Aa = © (x,a(x2−1)) ∈R2 : x ∈R ª. Prove that \ a∈R Aa = © (−1,0),(1,0) ª. 24. Prove that \ x∈R [3−x2,5+ x2] = [3,5]. 25. Suppose A,B,C and D are sets. Prove that (A ×B)∪(C × D) ⊆(A ∪C)×(B ∪D). 26. Prove © 4k +5 : k ∈Z ª = © 4k +1 : k ∈Z ª. 27. Prove © 12a+4b : a,b ∈Z ª = © 4c : c ∈Z ª. 28. Prove © 12a+25b : a,b ∈Z ª = Z. 29. Suppose A ̸= ;. Prove that A ×B ⊆A ×C, if and only if B ⊆C. 30. Prove that (Z×N)∩(N×Z) = N×N. 31. Suppose B ̸= ; and A ×B ⊆B ×C. Prove A ⊆C. CHAPTER 9 Disproof E ver since Chapter 4 we have dealt with one major theme: Given a statement, prove that is it true. In every example and exercise we were handed a true statement and charged with the task of proving it. Have you ever wondered what would happen if you were given a false statement to prove? The answer is that no (correct) proof would be possible, for if it were, the statement would be true, not false. But how would you convince someone that a statement is false? The mere fact that you could not produce a proof does not automatically mean the statement is false, for you know (perhaps all too well) that proofs can be diﬃcult to construct. It turns out that there is a very simple and utterly convincing procedure that proves a statement is false. The process of carrying out this procedure is called disproof. Thus, this chapter is concerned with disproving statements. Before describing the new method, we will set the stage with some relevant background information. First, we point out that mathematical statements can be divided into three categories, described below. One category consists of all those statements that have been proved to be true. For the most part we regard these statements as signiﬁcant enough to be designated with special names such as “theorem,” “proposition,” “lemma” and “corollary.” Some examples of statements in this category are listed in the left-hand box in the diagram on the following page. There are also some wholly uninteresting statements (such as 2 = 2) in this category, and although we acknowledge their existence we certainly do not dignify them with terms such as “theorem” or “proposition.” At the other extreme is a category consisting of statements that are known to be false. Examples are listed in the box on the right. Since mathematicians are not very interested in them, these types of statements do not get any special names, other than the blanket term “false statement.” But there is a third (and quite interesting) category between these two extremes. It consists of statements whose truth or falsity has not been determined. Examples include things like “Every perfect number 147 is even,” or “Every even integer greater than 2 is the sum of two primes.” (The latter statement is called the Goldbach conjecture. See Section 2.1.) Mathematicians have a special name for the statements in this category that they suspect (but haven’t yet proved) are true. Such statements are called conjectures. Three Types of Statements: Known to be true Truth unknown Known to be false (Theorems & propositions) (Conjectures) Examples: • Pythagorean theorem • Fermat’s last theorem (Section 2.1) • The square of an odd number is odd. • The series ∞ X k=1 1 k diverges. Examples: • All perfect numbers are even. • Any even number greater than 2 is the sum of two primes. (Goldbach’s conjecture, Section 2.1) • There are inﬁnitely many prime numbers of form 2n −1, with n ∈N. Examples: • All prime numbers are odd. • Some quadratic equations have three solutions. • 0 = 1 • There exist natural numbers a,b and c for which a3 + b3 = c3. Mathematicians spend much of their time and energy attempting to prove or disprove conjectures. (They also expend considerable mental energy in creating new conjectures based on collected evidence or intuition.) When a conjecture is proved (or disproved) the proof or disproof will typically appear in a published paper, provided the conjecture is of suﬃcient interest. If it is proved, the conjecture attains the status of a theorem or proposition. If it is disproved, then no one is really very interested in it anymore—mathematicians do not care much for false statements. Most conjectures that mathematicians are interested in are quite diﬃcult to prove or disprove. We are not at that level yet. In this text, the “conjectures” that you will encounter are the kinds of statements that an experienced mathematician would immediately spot as true or false, but you may have to do some work before ﬁguring out a proof or disproof. But in keeping with the cloud of uncertainty that surrounds conjectures at the advanced levels of mathematics, most exercises in this chapter (and many beyond it) will ask you to prove or disprove statements without giving any hint as to whether they are true or false. Your job will be to decide whether or not they are true and to either prove or disprove them. The examples in this chapter will illustrate the processes one typically goes through in 148 Disproof deciding whether a statement is true or false, and then verifying that it’s true or false. You know the three major methods of proving a statement: direct proof, contrapositive proof and proof by contradiction. Now we are ready to understand the method of disproving a statement. Suppose you want to disprove a statement P. In other words you want to prove that P is false. The way to do this is to prove that ∼P is true, for if ∼P is true, it follows immediately that P has to be false. How to disprove P: Prove ∼P. Our approach is incredibly simple. To disprove P, prove ∼P. In theory, this proof can be carried out by direct, contrapositive or contradiction approaches. However, in practice things can be even easier than that if we are disproving a universally quantiﬁed statement or a conditional statement. That is our next topic. 9.1 Disproving Universal Statements: Counterexamples A conjecture may be described as a statement that we hope is a theorem. As we know, many theorems (hence many conjectures) are universally quantiﬁed statements. Thus it seems reasonable to begin our discussion by investigating how to disprove a universally quantiﬁed statement such as ∀x ∈S,P(x). To disprove this statement, we must prove its negation. Its negation is ∼(∀x ∈S,P(x)) = ∃x ∈S,∼P(x). The negation is an existence statement. To prove the negation is true, we just need to produce an example of an x ∈S that makes ∼P(x) true, that is, an x that makes P(x) false. This leads to the following outline for disproving a universally quantiﬁed statement. How to disprove ∀x ∈S,P(x). Produce an example of an x ∈S that makes P(x) false. Counterexamples 149 Things are even simpler if we want to disprove a conditional statement P(x) ⇒Q(x). This statement asserts that for every x that makes P(x) true, Q(x) will also be true. The statement can only be false if there is an x that makes P(x) true and Q(x) false. This leads to our next outline for disproof. How to disprove P(x) ⇒Q(x). Produce an example of an x that makes P(x) true and Q(x) false. In both of the above outlines, the statement is disproved simply by exhibiting an example that shows the statement is not always true. (Think of it as an example that proves the statement is a promise that can be broken.) There is a special name for an example that disproves a statement: It is called a counterexample. Example 9.1 As our ﬁrst example, we will work through the process of deciding whether or not the following conjecture is true. Conjecture: For every n ∈Z, the integer f (n) = n2 −n+11 is prime. In resolving the truth or falsity of a conjecture, it’s a good idea to gather as much information about the conjecture as possible. In this case let’s start by making a table that tallies the values of f (n) for some integers n. n −3 −2 −1 0 1 2 3 4 5 6 7 8 9 10 f (n) 23 17 13 11 11 13 17 23 31 41 53 67 83 101 In every case, f (n) is prime, so you may begin to suspect that the conjecture is true. Before attempting a proof, let’s try one more n. Unfortunately, f (11) = 112−11+11 = 112 is not prime. The conjecture is false because n = 11 is a counterexample. We summarize our disproof as follows: Disproof. The statement “For every n ∈Z, the integer f (n) = n2 −n +11 is prime,” is false. For a counterexample, note that for n = 11, the integer f (11) = 121 = 11·11 is not prime. ■ In disproving a statement with a counterexample, it is important to explain exactly how the counterexample makes the statement false. Our work would not have been complete if we had just said “for a counterexample, consider n = 11,” and left it at that. We need to show that the answer f (11) is not prime. Showing the factorization f (11) = 11·11 suﬃces for this. 150 Disproof Example 9.2 Either prove or disprove the following conjecture. Conjecture If A, B and C are sets, then A −(B ∩C) = (A −B)∩(A −C). Disproof. This conjecture is false because of the following counterexample. Let A = © 1,2,3 ª, B = © 1,2 ª and C = © 2,3 ª. Notice that A −(B ∩C) = © 1,3 ª and (A −B)∩(A −C) = ;, so A −(B ∩C) ̸= (A −B)∩(A −C). ■ (To see where this counterexample came from, draw Venn diagrams for A−(B∩C) and (A−B)∩(A−C). You will see that the diagrams are diﬀerent. The numbers 1, 2 and 3 can then be inserted into the regions of the diagrams in such a way as to create the above counterexample.) 9.2 Disproving Existence Statements We have seen that we can disprove a universally quantiﬁed statement or a conditional statement simply by ﬁnding a counterexample. Now let’s turn to the problem of disproving an existence statement such as ∃x ∈S,P(x). Proving this would involve simply ﬁnding an example of an x that makes P(x) true. To disprove it, we have to prove its negation ∼(∃x ∈S,P(x)) = ∀x ∈S,∼P(x). But this negation is universally quantiﬁed. Proving it involves showing that ∼P(x) is true for all x ∈S, and for this an example does not suﬃce. Instead we must use direct, contrapositive or contradiction proof to prove the conditional statement “If x ∈S, then ∼P(x).” As an example, here is a conjecture to either prove or disprove. Example 9.3 Either prove or disprove the following conjecture. Conjecture: There is a real number x for which x4 < x < x2. This may not seem like an unreasonable statement at ﬁrst glance. After all, if the statement were asserting the existence of a real number for which x3 < x < x2, then it would be true: just take x = −2. But it asserts there is an x for which x4 < x < x2. When we apply some intelligent guessing to locate such an x we run into trouble. If x = 1 2, then x4 < x, but we don’t have x < x2; similarly if x = 2, we have x < x2 but not x4 < x. Since ﬁnding an x with x4 < x < x2 seems problematic, we may begin to suspect that the given statement is false. Let’s see if we can disprove it. According to our strategy for disproof, to disprove it we must prove its negation. Symbolically, the statement is Disproving Existence Statements 151 ∃x ∈R,x4 < x < x2, so its negation is ∼(∃x ∈R,x4 < x < x2) = ∀x ∈R,∼(x4 < x < x2). Thus, in words the negation is: For every real number x, it is not the case that x4 < x < x2. This can be proved with contradiction, as follows. Suppose for the sake of contradiction that there is an x for which x4 < x < x2. Then x must be positive since it’s greater than the non-negative number x4. Dividing all parts of x4 < x < x2 by the positive number x produces x3 < 1 < x. Now subtract 1 from all parts of x3 < 1 < x to obtain x3 −1 < 0 < x−1 and reason as follows: x3 −1 < 0 < x−1 (x−1)(x2 + x+1) < 0 < (x−1) x2 + x+1 < 0 < 1 (Division by x−1 did not reverse the inequality < because the second line above shows 0 < x−1, that is, x−1 is positive.) Now we have x2 + x+1 < 0, which is a contradiction because x being positive forces x2 + x+1 > 0 We summarize our work as follows. The statement “There is a real number x for which x4 < x < x2” is false because we have proved its negation “For every real number x, it is not the case that x4 < x < x2.” As you work the exercises, keep in mind that not every conjecture will be false. If one is true, then a disproof is impossible and you must produce a proof. Here is an example: Example 9.4 Either prove or disprove the following conjecture. Conjecture There exist three integers x, y, z, all greater than 1 and no two equal, for which xy = yz. This conjecture is true. It is an existence statement, so to prove it we just need to give an example of three integers x, y, z, all greater than 1 and no two equal, so that xy = yz. A proof follows. Proof. Note that if x = 2, y = 16 and z = 4, then xy = 216 = (24)4 = 164 = yz. ■ 152 Disproof 9.3 Disproof by Contradiction Contradiction can be a very useful way to disprove a statement. To see how this works, suppose we wish to disprove a statement P. We know that to disprove P, we must prove ∼P. To prove ∼P with contradiction, we assume ∼∼P is true and deduce a contradiction. But since ∼∼P = P, this boils down to assuming P is true and deducing a contradiction. Here is an outline: How to disprove P with contradiction: Assume P is true, and deduce a contradiction. To illustrate this, let’s revisit Example 9.3 but do the disproof with contradiction. You will notice that the work duplicates much of what we did in Example 9.3, but is it much more streamlined because here we do not have to negate the conjecture. Example 9.5 Disprove the following conjecture. Conjecture: There is a real number x for which x4 < x < x2. Disproof. Suppose for the sake of contradiction that this conjecture is true. Let x be a real number for which x4 < x < x2. Then x is positive, since it is greater than the non-negative number x4. Dividing all parts of x4 < x < x2 by the positive number x produces x3 < 1 < x. Now subtract 1 from all parts of x3 < 1 < x to obtain x3 −1 < 0 < x−1 and reason as follows: x3 −1 < 0 < x−1 (x−1)(x2 + x+1) < 0 < (x−1) x2 + x+1 < 0 < 1 Now we have x2 + x+1 < 0, which is a contradiction because x is positive. Thus the conjecture must be false. ■ Exercises for Chapter 9 Each of the following statements is either true or false. If a statement is true, prove it. If a statement is false, disprove it. These exercises are cumulative, covering all topics addressed in Chapters 1–9. 1. If x, y ∈R, then \|x+ y\| = \|x\|+\|y\|. 2. For every natural number n, the integer 2n2 −4n+31 is prime. Disproof by Contradiction 153 3. If n ∈Z and n5 −n is even, then n is even. 4. For every natural number n, the integer n2 +17n+17 is prime. 5. If A, B,C and D are sets, then (A ×B)∪(C × D) = (A ∪C)×(B ∪D). 6. If A, B,C and D are sets, then (A ×B)∩(C × D) = (A ∩C)×(B ∩D). 7. If A, B and C are sets, and A ×C = B ×C, then A = B. 8. If A, B and C are sets, then A −(B ∪C) = (A −B)∪(A −C). 9. If A and B are sets, then P(A)−P(B) ⊆P(A −B). 10. If A and B are sets and A ∩B = ;, then P(A)−P(B) ⊆P(A −B). 11. If a,b ∈N, then a+ b < ab. 12. If a,b, c ∈N and ab, bc and ac all have the same parity, then a,b and c all have the same parity. 13. There exists a set X for which R ⊆X and ; ∈X. 14. If A and B are sets, then P(A)∩P(B) = P(A ∩B). 15. Every odd integer is the sum of three odd integers. 16. If A and B are ﬁnite sets, then \|A ∪B\| = \|A\|+\|B\|. 17. For all sets A and B, if A −B = ;, then B ̸= ;. 18. If a,b, c ∈N, then at least one of a−b, a+ c and b −c is even. 19. For every r,s ∈Q with r < s, there is an irrational number u for which r < u < s. 20. There exist prime numbers p and q for which p −q = 1000. 21. There exist prime numbers p and q for which p −q = 97. 22. If p and q are prime numbers for which p < q, then 2p + q2 is odd. 23. If x, y ∈R and x3 < y3, then x < y. 24. The inequality 2x ≥x+1 is true for all positive real numbers x. 25. For all a,b, c ∈Z, if a\|bc, then a\|b or a\| c. 26. Suppose A, B and C are sets. If A = B −C, then B = A ∪C. 27. The equation x2 = 2x has three real solutions. 28. Suppose a,b ∈Z. If a\|b and b\|a, then a = b. 29. If x, y ∈R and \|x+ y\| = \|x−y\|, then y = 0. 30. There exist integers a and b for which 42a+7b = 1. 31. No number (other than 1) appears in Pascal’s triangle more than four times. 32. If n,k ∈N and ¡ n k ¢ is a prime number, then k = 1 or k = n−1. 33. Suppose f (x) = a0 + a1x+ a2x2 +···+ anxn is a polynomial of degree 1 or greater, and for which each coeﬃcient ai is in N. Then there is an n ∈N for which the integer f (n) is not prime. 34. If X ⊆A ∪B, then X ⊆A or X ⊆B. CHAPTER 10 Mathematical Induction T his chapter explains a powerful proof technique called mathematical induction (or just induction for short). To motivate the discussion, let’s ﬁrst examine the kinds of statements that induction is used to prove. Consider the following statement. Conjecture. The sum of the ﬁrst n odd natural numbers equals n2. The following table illustrates what this conjecture says. Each row is headed by a natural number n, followed by the sum of the ﬁrst n odd natural numbers, followed by n2. n sum of the ﬁrst n odd natural numbers n2 1 1 = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2 1+3 = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 3 1+3+5 = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 4 1+3+5+7 = . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 5 1+3+5+7+9 = . . . . . . . . . . . . . . . . . . . . . . . . 25 . . . . . . . . . n 1+3+5+7+9+11+···+(2n−1) = . . . . . . . n2 . . . . . . . . . Note that in the ﬁrst ﬁve lines of the table, the sum of the ﬁrst n odd numbers really does add up to n2. Notice also that these ﬁrst ﬁve lines indicate that the nth odd natural number (the last number in each sum) is 2n −1. (For instance, when n = 2, the second odd natural number is 2·2−1 = 3; when n = 3, the third odd natural number is 2·3−1 = 5, etc.) The table raises a question. Does the sum 1+3+5+7+···+(2n−1) really always equal n2? In other words, is the conjecture true? Let’s rephrase this as follows. For each natural number n (i.e., for each line of the table), we have a statement Sn, as follows: 155 S1 : 1 = 12 S2 : 1+3 = 22 S3 : 1+3+5 = 32 . . . Sn : 1+3+5+7+···+(2n−1) = n2 . . . Our question is: Are all of these statements true? Mathematical induction is designed to answer just this kind of question. It is used when we have a set of statements S1,S2,S3,...,Sn,..., and we need to prove that they are all true. The method is really quite simple. To visualize it, think of the statements as dominoes, lined up in a row. Imagine you can prove the ﬁrst statement S1, and symbolize this as domino S1 being knocked down. Additionally, imagine that you can prove that any statement Sk being true (falling) forces the next statement Sk+1 to be true (to fall). Then S1 falls, and knocks down S2. Next S2 falls and knocks down S3, then S3 knocks down S4, and so on. The inescapable conclusion is that all the statements are knocked down (proved true). The Simple Idea Behind Mathematical Induction Statements are lined up like dominoes. (1) Suppose the ﬁrst statement falls (i.e. is proved true); (2) Suppose the kth falling always causes the (k +1)th to fall; Then all must fall (i.e. all statements are proved true). S1 S2 S3 S4 S5 S6 S1 S2 S3 S4 S5 S6 Sk Sk+1 Sk+2 Sk+3 Sk Sk Sk+1 Sk+1 Sk+2 Sk+2 Sk+2 Sk+3 Sk+3 Sk+3 Sk+4 Sk+4 Sk+4 ··· ··· ··· ··· ··· ··· ··· ··· S1 Sk Sk+1 S1 S2 S3 S4 S5 S6 S2 S3 S4 S5 S6 156 Mathematical Induction This picture gives our outline for proof by mathematical induction. Outline for Proof by Induction Proposition The statements S1,S2,S3,S4,... are all true. Proof. (Induction) (1) Prove that the ﬁrst statement S1 is true. (2) Given any integer k ≥1, prove that the statement Sk ⇒Sk+1 is true. It follows by mathematical induction that every Sn is true. ■ In this setup, the ﬁrst step (1) is called the basis step. Because S1 is usually a very simple statement, the basis step is often quite easy to do. The second step (2) is called the inductive step. In the inductive step direct proof is most often used to prove Sk ⇒Sk+1, so this step is usually carried out by assuming Sk is true and showing this forces Sk+1 to be true. The assumption that Sk is true is called the inductive hypothesis. Now let’s apply this technique to our original conjecture that the sum of the ﬁrst n odd natural numbers equals n2. Our goal is to show that for each n ∈N, the statement Sn : 1+3+5+7+···+(2n−1) = n2 is true. Before getting started, observe that Sk is obtained from Sn by plugging k in for n. Thus Sk is the statement Sk : 1+3+5+7+···+(2k−1) = k2. Also, we get Sk+1 by plugging in k+1 for n, so that Sk+1 : 1+3+5+7+···+(2(k+1)−1) = (k+1)2. Proposition If n ∈N, then 1+3+5+7+···+(2n−1) = n2. Proof. We will prove this with mathematical induction. (1) Observe that if n = 1, this statement is 1 = 12, which is obviously true. (2) We must now prove Sk ⇒Sk+1 for any k ≥1. That is, we must show that if 1+3+5+7+···+(2k−1) = k2, then 1+3+5+7+···+(2(k+1)−1) = (k+1)2. We use direct proof. Suppose 1+3+5+7+···+(2k −1) = k2. Then 1+3+5+7+···············+(2(k +1)−1) = 1+3+5+7+···+ (2k −1) +(2(k +1)−1) = ¡ 1+3+5+7+···+(2k −1) ¢ +(2(k +1)−1) = k2 +(2(k +1)−1) = k2 +2k +1 = (k +1)2. Thus 1+3+5+7+···+(2(k +1)−1) = (k +1)2. This proves that Sk ⇒Sk+1. It follows by induction that 1+3+5+7+···+(2n−1) = n2 for every n ∈N. ■ 157 In induction proofs it is usually the case that the ﬁrst statement S1 is indexed by the natural number 1, but this need not always be so. Depending on the problem, the ﬁrst statement could be S0, or Sm for any other integer m. In the next example the statements are S0,S1,S2,S3,... The same outline is used, except that the basis step veriﬁes S0, not S1. Proposition If n is a non-negative integer, then 5 \| (n5 −n). Proof. We will prove this with mathematical induction. Observe that the ﬁrst non-negative integer is 0, so the basis step involves n = 0. (1) If n = 0, this statement is 5 \| (05 −0) or 5 \| 0, which is obviously true. (2) Let k ≥0. We need to prove that if 5 \| (k5 −k), then 5 \| ((k +1)5 −(k +1)). We use direct proof. Suppose 5 \| (k5 −k). Thus k5 −k = 5a for some a ∈Z. Observe that (k +1)5 −(k +1) = k5 +5k4 +10k3 +10k2 +5k +1−k −1 = (k5 −k)+5k4 +10k3 +10k2 +5k = 5a+5k4 +10k3 +10k2 +5k = 5(a+ k4 +2k3 +2k2 + k). This shows (k+1)5−(k+1) is an integer multiple of 5, so 5 \| ((k+1)5−(k+1)). We have now shown that 5 \| (k5 −k) implies 5 \| ((k +1)5 −(k +1)). It follows by induction that 5 \| (n5 −n) for all non-negative integers n. ■ As noted, induction is used to prove statements of the form ∀n ∈N,Sn. But notice the outline does not work for statements of form ∀n ∈Z,Sn (where n is in Z, not N). The reason is that if you are trying to prove ∀n ∈Z,Sn by induction, and you’ve shown S1 is true and Sk ⇒Sk+1, then it only follows from this that Sn is true for n ≥1. You haven’t proved that any of the statements S0,S−1,S−2,... are true. If you ever want to prove ∀n ∈Z,Sn by induction, you have to show that some Sa is true and Sk ⇒Sk+1 and Sk ⇒Sk−1. Unfortunately, the term mathematical induction is sometimes confused with inductive reasoning, that is, the process of reaching the conclusion that something is likely to be true based on prior observations of similar circumstances. Please note that that mathematical induction, as intro-duced here, is a rigorous method that proves statements with absolute certainty. 158 Mathematical Induction To round out this section, we present four additional induction proofs. Proposition If n ∈Z and n ≥0, then n X i=0 i · i! = (n+1)!−1. Proof. We will prove this with mathematical induction. (1) If n = 0, this statement is 0 X i=0 i · i! = (0+1)!−1. Since the left-hand side is 0·0! = 0, and the right-hand side is 1!−1 = 0, the equation P0 i=0 i · i! = (0+1)!−1 holds, as both sides are zero. (2) Consider any integer k ≥0. We must show that Sk implies Sk+1. That is, we must show that k X i=0 i · i! = (k +1)!−1 implies k+1 X i=0 i · i! = ((k +1)+1)!−1. We use direct proof. Suppose k X i=0 i · i! = (k +1)!−1. Observe that k+1 X i=0 i · i! = Ã k X i=0 i · i! ! +(k +1)(k +1)! = ³ (k +1)!−1 ´ +(k +1)(k +1)! = (k +1)!+(k +1)(k +1)!−1 = ¡ 1+(k +1) ¢ (k +1)!−1 = (k +2)(k +1)!−1 = (k +2)!−1 = ((k +1)+1)!−1. Therefore k+1 X i=0 i · i! = ((k +1)+1)!−1. It follows by induction that n X i=0 i · i! = (n+1)!−1 for every integer n ≥0. ■ 159 The next example illustrates a trick that is occasionally useful. You know that you can add equal quantities to both sides of an equation without violating equality. But don’t forget that you can add unequal quantities to both sides of an inequality, as long as the quantity added to the bigger side is bigger than the quantity added to the smaller side. For example, if x ≤y and a ≤b, then x+a ≤y+ b. Similarly, if x ≤y and b is positive, then x ≤y+ b. This oft-forgotten fact is used in the next proof. Proposition For each n ∈N, it follows that 2n ≤2n+1 −2n−1 −1. Proof. We will prove this with mathematical induction. (1) If n = 1, this statement is 21 ≤21+1 −21−1 −1, which simpliﬁes to 2 ≤4−1−1, which is obviously true. (2) Suppose k ≥1. We need to show that 2k ≤2k+1 −2k−1 −1 implies 2k+1 ≤2(k+1)+1−2(k+1)−1−1. We use direct proof. Suppose 2k ≤2k+1−2k−1−1, and reason as follows: 2k ≤ 2k+1 −2k−1 −1 2(2k) ≤ 2(2k+1 −2k−1 −1) (multiply both sides by 2) 2k+1 ≤ 2k+2 −2k −2 2k+1 ≤ 2k+2 −2k −2+1 (add 1 to the bigger side) 2k+1 ≤ 2k+2 −2k −1 2k+1 ≤ 2(k+1)+1 −2(k+1)−1 −1. It follows by induction that 2n ≤2n+1 −2n−1 −1 for each n ∈N. ■ We next prove that if n ∈N, then the inequality (1+ x)n ≥1+ nx holds for all x ∈R with x > −1. Thus we will need to prove that the statement Sn : (1+ x)n ≥1+ nx for every x ∈R with x > −1 is true for every natural number n. This is (only) slightly diﬀerent from our other examples, which proved statements of the form ∀n ∈N, P(n), where P(n) is a statement about the number n. This time we are proving something of form ∀n ∈N, P(n,x), where the statement P(n,x) involves not only n, but also a second variable x. (For the record, the inequality (1 + x)n ≥1 + nx is known as Bernoulli’s inequality.) 160 Mathematical Induction Proposition If n ∈N, then (1+ x)n ≥1+ nx for all x ∈R with x > −1. Proof. We will prove this with mathematical induction. (1) For the basis step, notice that when n = 1 the statement is (1+ x)1 ≥ 1+1· x , and this is true because both sides equal 1+ x. (2) Assume that for some k ≥1, the statement (1+ x)k ≥1+ kx is true for all x ∈R with x > −1. From this we need to prove (1+ x)k+1 ≥1+(k +1)x. Now, 1+ x is positive because x > −1, so we can multiply both sides of (1+ x)k ≥1+ kx by (1+ x) without changing the direction of the ≥. (1+ x)k(1+ x) ≥ (1+ kx)(1+ x) (1+ x)k+1 ≥ 1+ x+ kx+ kx2 (1+ x)k+1 ≥ 1+(k +1)x+ kx2 The above term kx2 is positive, so removing it from the right-hand side will only make that side smaller. Thus we get (1+ x)k+1 ≥1+(k +1)x. ■ Next, an example where the basis step involves more than routine checking. (It will be used later, so it is numbered for reference.) Proposition 10.1 Suppose a1,a2,...,an are n integers, where n ≥2. If p is prime and p \| (a1 · a2 · a3 ···an), then p \| ai for at least one of the ai. Proof. The proof is induction on n. (1) The basis step involves n = 2. Let p be prime and suppose p \| (a1a2). We need to show that p \| a1 or p \| a2, or equivalently, if p ∤a1, then p \| a2. Thus suppose p ∤a1. Since p is prime, it follows that gcd(p,a1) = 1. By Proposition 7.1 (on page 126), there are integers k and ℓfor which 1 = pk + a1ℓ. Multiplying this by a2 gives a2 = pka2 + a1a2ℓ. As we are assuming that p divides a1a2, it is clear that p divides the expression pka2+a1a2ℓon the right; hence p \| a2. We’ve now proved that if p \| (a1a2), then p \| a1 or p \| a2. This completes the basis step. (2) Suppose that k ≥2, and p \| (a1 · a2 ···ak) implies then p \| ai for some ai. Now let p \| (a1 · a2 ···ak · ak+1). Then p \| ¡ (a1 · a2 ···ak)· ak+1 ¢. By what we proved in the basis step, it follows that p \| (a1 · a2 ···ak) or p \| ak+1. This and the inductive hypothesis imply that p divides one of the ai. ■ Please test your understanding now by working a few exercises. Proof by Strong Induction 161 10.1 Proof by Strong Induction This section describes a useful variation on induction. Occasionally it happens in induction proofs that it is diﬃcult to show that Sk forces Sk+1 to be true. Instead you may ﬁnd that you need to use the fact that some “lower” statements Sm (with m < k) force Sk+1 to be true. For these situations you can use a slight variant of induction called strong induction. Strong induction works just like regular induction, except that in Step (2) instead of assuming Sk is true and showing this forces Sk+1 to be true, we assume that all the statements S1,S2,...,Sk are true and show this forces Sk+1 to be true. The idea is that if it always happens that the ﬁrst k dominoes falling makes the (k +1)th domino fall, then all the dominoes must fall. Here is the outline. Outline for Proof by Strong Induction Proposition The statements S1,S2,S3,S4,... are all true. Proof. (Strong induction) (1) Prove the ﬁrst statement S1. (Or the ﬁrst several Sn.) (2) Given any integer k ≥1, prove (S1 ∧S2 ∧S3 ∧···∧Sk) ⇒Sk+1. ■ Strong induction can be useful in situations where assuming Sk is true does not neatly lend itself to forcing Sk+1 to be true. You might be better served by showing some other statement (Sk−1 or Sk−2 for instance) forces Sk to be true. Strong induction says you are allowed to use any (or all) of the statements S1,S2,...,Sk to prove Sk+1. As our ﬁrst example of strong induction, we are going to prove that 12 \| (n4 −n2) for any n ∈N. But ﬁrst, let’s look at how regular induction would be problematic. In regular induction we would start by showing 12 \| (n4 −n2) is true for n = 1. This part is easy because it reduces to 12 \| 0, which is clearly true. Next we would assume that 12 \| (k4 −k2) and try to show this implies 12 \| ((k+1)4−(k+1)2). Now, 12 \| (k4−k2) means k4−k2 = 12a for some a ∈Z. Next we use this to try to show (k +1)4 −(k +1)2 = 12b for some integer b. Working out (k +1)4 −(k +1)2, we get (k +1)4 −(k +1)2 = (k4 +4k3 +6k2 +4k +1)−(k2 +2k +1) = (k4 −k2)+4k3 +6k2 +6k = 12a+4k3 +6k2 +6k. At this point we’re stuck because we can’t factor out a 12. Now let’s see how strong induction can get us out of this bind. 162 Mathematical Induction Strong induction involves assuming each of statements S1,S2,...,Sk is true, and showing that this forces Sk+1 to be true. In particular, if S1 through Sk are true, then certainly Sk−5 is true, provided that 1 ≤k−5 < k. The idea is then to show Sk−5 ⇒Sk+1 instead of Sk ⇒Sk+1. For this to make sense, our basis step must involve checking that S1,S2,S3,S4,S5,S6 are all true. Once this is established, Sk−5 ⇒Sk+1 will imply that the other Sk are all true. For example, if k = 6, then Sk−5 ⇒Sk+1 is S1 ⇒S7, so S7 is true; for k = 7, then Sk−5 ⇒Sk+1 is S2 ⇒S8, so S8 is true, etc. Proposition If n ∈N, then 12 \| (n4 −n2). Proof. We will prove this with strong induction. (1) First note that the statement is true for the ﬁrst six positive integers: If n = 1, 12 divides n4 −n2 = 14 −12 = 0. If n = 2, 12 divides n4 −n2 = 24 −22 = 12. If n = 3, 12 divides n4 −n2 = 34 −32 = 72. If n = 4, 12 divides n4 −n2 = 44 −42 = 240. If n = 5, 12 divides n4 −n2 = 54 −52 = 600. If n = 6, 12 divides n4 −n2 = 64 −62 = 1260. (2) Let k ≥6 and assume 12 \| (m4 −m2) for 1 ≤m ≤k. (That is, assume statements S1,S2,...,Sk are all true.) We must show 12 \| ¡ (k+1)4−(k+1)2¢. (That is, we must show that Sk+1 is true.) Since Sk−5 is true, we have 12 \| ((k −5)4 −(k −5)2). For simplicity, let’s set m = k −5, so we know 12 \| (m4−m2), meaning m4 −m2 = 12a for some integer a. Observe that: (k +1)4 −(k +1)2 = (m+6)4 −(m+6)2 = m4 +24m3 +216m2 +864m+1296−(m2 +12m+36) = (m4 −m2)+24m3 +216m2 +852m+1260 = 12a+24m3 +216m2 +852m+1260 = 12 ¡ a+2m3 +18m2 +71m+105 ¢ . As (a+2m3 +18m2 +71m+105) is an integer, we get 12 \| ((k+1)4 −(k+1)2). This shows by strong induction that 12 \| (n4 −n2) for every n ∈N. ■ Proof by Strong Induction 163 Our next example involves mathematical objects called graphs. In mathematics, the word graph is used in two contexts. One context involves the graphs of equations and functions from algebra and calculus. In the other context, a graph is a conﬁguration consisting of points (called vertices) and edges which are lines connecting the vertices. Following are some pictures of graphs. Let’s agree that all of our graphs will be in “one piece,” that is, you can travel from any vertex of a graph to any other vertex by traversing a route of edges from one vertex to the other. v0 v1 v2 v3 v4 Figure 10.1. Examples of Graphs A cycle in a graph is a sequence of distinct edges in the graph that form a route that ends where it began. For example, the graph on the far left of Figure 10.1 has a cycle that starts at vertex v1, then goes to v2, then to v3, then v4 and ﬁnally back to its starting point v1. You can ﬁnd cycles in both of the graphs on the left, but the two graphs on the right do not have cycles. There is a special name for a graph that has no cycles; it is called a tree. Thus the two graphs on the right of Figure 10.1 are trees, but the two graphs on the left are not trees. Figure 10.2. A tree Note that the trees in Figure 10.1 both have one fewer edge than vertex. The tree on the far right has 5 vertices and 4 edges. The one next to it has 6 vertices and 5 edges. Draw any tree; you will ﬁnd that if it has n vertices, then it has n−1 edges. We now prove that this is always true. 164 Mathematical Induction Proposition If a tree has n vertices, then it has n−1 edges. Proof. Notice that this theorem asserts that for any n ∈N, the following statement is true: Sn : A tree with n vertices has n−1 edges. We use strong induction to prove this. (1) Observe that if a tree has n = 1 vertex then it has no edges. Thus it has n−1 = 0 edges, so the theorem is true when n = 1. (2) Now take an integer k ≥1. We must show (S1 ∧S2 ∧···∧Sk) ⇒Sk+1. In words, we must show that if it is true that any tree with m vertices has m−1 edges, where 1 ≤m ≤k, then any tree with k +1 vertices has (k +1)−1 = k edges. We will use direct proof. Suppose that for each integer m with 1 ≤m ≤k, any tree with m vertices has m−1 edges. Now let T be a tree with k +1 vertices. Single out an edge of T and label it e, as illustrated below. ··· ··· ··· ··· T1 T2 T e ··· ··· ··· ··· Now remove the edge e from T, but leave the two endpoints of e. This leaves two smaller trees that we call T1 and T2. Let’s say T1 has x vertices and T2 has y vertices. As each of these two smaller trees has fewer than k +1 vertices, our inductive hypothesis guarantees that T1 has x−1 edges, and T2 has y−1 edges. Think about our original tree T. It has x+ y vertices. It has x−1 edges that belong to T1 and y−1 edges that belong to T2, plus it has the additional edge e that belongs to neither T1 nor T2. Thus, all together, the number of edges that T has is (x−1)+(y−1)+1 = (x+ y)−1. In other words, T has one fewer edges than it has vertices. Thus it has (k +1)−1 = k edges. It follows by strong induction that a tree with n vertices has n−1 edges. ■ Notice that it was absolutely essential that we used strong induction in the above proof because the two trees T1 and T2 will not both have k vertices. At least one will have fewer than k vertices. Thus the statement Sk is not enough to imply Sk+1. We need to use the assumption that Sm will be true whenever m ≤k, and strong induction allows us to do this. Proof by Smallest Counterexample 165 10.2 Proof by Smallest Counterexample This section introduces yet another proof technique, called proof by small-est counterexample. It is a hybrid of induction and proof by contradiction. It has the nice feature that it leads you straight to a contradiction. It is therefore more “automatic” than the proof by contradiction that was introduced in Chapter 6. Here is the outline: Outline for Proof by Smallest Counterexample Proposition The statements S1,S2,S3,S4,... are all true. Proof. (Smallest counterexample) (1) Check that the ﬁrst statement S1 is true. (2) For the sake of contradiction, suppose not every Sn is true. (3) Let k > 1 be the smallest integer for which Sk is false. (4) Then Sk−1 is true and Sk is false. Use this to get a contradiction. ■ Notice that this setup leads you to a point where Sk−1 is true and Sk is false. It is here, where true and false collide, that you will ﬁnd a contradiction. Let’s do an example. Proposition If n ∈N, then 4 \| (5n −1). Proof. We use proof by smallest counterexample. (We will number the steps to match the outline, but that is not usually done in practice.) (1) If n = 1, then the statement is 4 \| (51 −1), or 4 \| 4, which is true. (2) For sake of contradiction, suppose it’s not true that 4 \| (5n −1) for all n. (3) Let k > 1 be the smallest integer for which 4 ∤(5k −1). (4) Then 4 \| (5k−1−1), so there is an integer a for which 5k−1−1 = 4a. Then: 5k−1 −1 = 4a 5(5k−1 −1) = 5·4a 5k −5 = 20a 5k −1 = 20a+4 5k −1 = 4(5a+1) This means 4 \| (5k−1), a contradiction, because 4 ∤(5k−1) in Step 3. Thus, we were wrong in Step 2 to assume that it is untrue that 4 \| (5n −1) for every n. Therefore 4 \| (5n −1) is true for every n. ■ 166 Mathematical Induction We next prove the fundamental theorem of arithmetic, which says any integer greater than 1 has a unique prime factorization. For example, 12 factors into primes as 12 = 2·2·3, and moreover any factorization of 12 into primes uses exactly the primes 2, 2 and 3. Our proof combines the techniques of induction, cases, minimum counterexample and the idea of uniqueness of existence outlined at the end of Section 7.3. We dignify this fundamental result with the label of “Theorem.” Theorem 10.1 (Fundamental Theorem of Arithmetic) Any integer n > 1 has a unique prime factorization. That is, if n = p1 · p2 · p3 ··· pk and n = a1 ·a2 ·a3 ···aℓare two prime factorizations of n, then k = ℓ, and the primes pi and ai are the same, except that they may be in a diﬀerent order. Proof. Suppose n > 1. We ﬁrst use strong induction to show that n has a prime factorization. For the basis step, if n = 2, it is prime, so it is already its own prime factorization. Let n ≥2 and assume every integer between 2 and n (inclusive) has a prime factorization. Consider n+1. If it is prime, then it is its own prime factorization. If it is not prime, then it factors as n+1 = ab with a,b > 1. Because a and b are both less than n+1 they have prime factorizations a = p1 · p2 · p3 ··· pk and b = p′ 1 · p′ 2 · p′ 3 ··· p′ ℓ. Then n+1 = ab = (p1 · p2 · p3 ··· pk)(p′ 1 · p′ 2 · p′ 3 ··· p′ ℓ) is a prime factorization of n+1. This competes the proof by strong induction that every integer greater than 1 has a prime factorization. Next we use proof by smallest counterexample to prove that the prime factorization of any n ≥2 is unique. If n = 2, then n clearly has only one prime factorization, namely itself. Assume for the sake of contradiction that there is an n > 2 that has diﬀerent prime factorizations n = p1 · p2 · p3 ··· pk and n = a1·a2·a3 ···aℓ. Assume n is the smallest number with this property. From n = p1 · p2 · p3 ··· pk, we see that p1 \| n, so p1 \| (a1 · a2 · a3 ···aℓ). By Proposition 10.1 (page 160), p1 divides one of the primes ai. As ai is prime, we have p1 = ai. Dividing n = p1 · p2 · p3 ··· pk = a1 · a2 · a3 ···aℓby p1 = ai yields p2 · p3 ··· pk = a1 · a2 · a3 ···ai−1 · ai+1 ···aℓ. These two factorizations are diﬀerent, because the two prime factorizations of n were diﬀerent. (Remember: the primes p1 and ai are equal, so the diﬀerence appears in the remaining factors, displayed above.) But also the above number p2 · p3 ··· pk is smaller than n, and this contradicts the fact that n was the smallest number with two diﬀerent prime factorizations. ■ Fibonacci Numbers 167 One word of warning about proof by smallest counterexample. In proofs in other textbooks or in mathematical papers, it often happens that the writer doesn’t tell you up front that proof by smallest counterexample is being used. Instead, you will have to read through the proof to glean from context that this technique is being used. In fact, the same warning applies to all of our proof techniques. If you continue with mathematics, you will gradually gain through experience the ability to analyze a proof and understand exactly what approach is being used when it is not stated explicitly. Frustrations await you, but do not be discouraged by them. Frustration is a natural part of anything that’s worth doing. 10.3 Fibonacci Numbers Leonardo Pisano, now known as Fibonacci, was a mathematician born around 1175 in what is now Italy. His most signiﬁcant work was a book Liber Abaci, which is recognized as a catalyst in medieval Europe’s slow transition from Roman numbers to the Hindu-Arabic number system. But he is best known today for a number sequence that he described in his book and that bears his name. The Fibonacci sequence is 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377,... The numbers that appear in this sequence are called Fibonacci numbers. The ﬁrst two numbers are 1 and 1, and thereafter any entry is the sum of the previous two entries. For example 3+5 = 8, and 5+8 = 13, etc. We denote the nth term of this sequence as Fn. Thus F1 = 1, F2 = 1, F3 = 2, F4 = 3, F7 = 13 and so on. Notice that the Fibonacci Sequence is entirely determined by the rules F1 = 1, F2 = 1, and Fn = Fn−1 + Fn−2. We introduce Fibonacci’s sequence here partly because it is something everyone should know about, but also because it is a great source of induc-tion problems. This sequence, which appears with surprising frequency in nature, is ﬁlled with mysterious patterns and hidden structures. Some of these structures will be revealed to you in the examples and exercises. We emphasize that the condition Fn = Fn−1+Fn−2 (or equivalently Fn+1 = Fn + Fn−1) is the perfect setup for induction. It suggests that we can determine something about Fn by looking at earlier terms of the sequence. In using induction to prove something about the Fibonacci sequence, you should expect to use the equation Fn = Fn−1 + Fn−2 somewhere. For our ﬁrst example we will prove that F2 n+1 −Fn+1Fn −F2 n = (−1)n for any natural number n. For example, if n = 5 we have F2 6 −F6F5 −F2 5 = 82 −8·5−52 = 64−40−25 = −1 = (−1)5. 168 Mathematical Induction Proposition The Fibonacci sequence obeys F2 n+1 −Fn+1Fn −F2 n = (−1)n. Proof. We will prove this with mathematical induction. (1) If n = 1 we have F2 n+1−Fn+1Fn−F2 n = F2 2 −F2F1−F2 1 = 12−1·1−12 = −1 = (−1)1 = (−1)n, so indeed F2 n+1 −Fn+1Fn −F2 n = (−1)n is true when n = 1. (2) Take any integer k ≥1. We must show that if F2 k+1−Fk+1Fk−F2 k = (−1)k, then F2 k+2 −Fk+2Fk+1 −F2 k+1 = (−1)k+1. We use direct proof. Suppose F2 k+1 −Fk+1Fk −F2 k = (−1)k. Now we are going to carefully work out the expression F2 k+2 −Fk+2Fk+1 −F2 k+1 and show that it really does equal (−1)k+1. In so doing, we will use the fact that Fk+2 = Fk+1 + Fk. F2 k+2 −Fk+2Fk+1 −F2 k+1 = (Fk+1 + Fk)2 −(Fk+1 + Fk)Fk+1 −F2 k+1 = F2 k+1 +2Fk+1Fk + F2 k −F2 k+1 −FkFk+1 −F2 k+1 = −F2 k+1 + Fk+1Fk + F2 k = −(F2 k+1 −Fk+1Fk −F2 k) = −(−1)k (inductive hypothesis) = (−1)1(−1)k = (−1)k+1 Therefore F2 k+2 −Fk+2Fk+1 −F2 k+1 = (−1)k+1. It follows by induction that F2 n+1 −Fn+1Fn −F2 n = (−1)n for every n ∈N. ■ Let’s pause for a moment and think about what the result we just proved means. Dividing both sides of F2 n+1−Fn+1Fn −F2 n = (−1)n by F2 n gives µFn+1 Fn ¶2 −Fn+1 Fn −1 = (−1)n F2 n . For large values of n, the right-hand side is very close to zero, and the left-hand side is Fn+1/Fn plugged into the polynomial x2 −x−1. Thus, as n increases, the ratio Fn+1/Fn approaches a root of x2 −x −1 = 0. By the quadratic formula, the roots of x2−x−1 are 1± p 5 2 . As Fn+1/Fn > 1, this ratio must be approaching the positive root 1+ p 5 2 . Therefore lim n→∞ Fn+1 Fn = 1+ p 5 2 . (10.1) For a quick spot check, note that F13/F12 ≈1.618025, while 1+ p 5 2 ≈1.618033. Even for the small value n = 12, the numbers match to four decimal places. Fibonacci Numbers 169 The number Φ = 1+ p 5 2 is sometimes called the golden ratio, and there has been much speculation about its occurrence in nature as well as in classical art and architecture. One theory holds that the Parthenon and the Great Pyramids of Egypt were designed in accordance with this number. But we are here concerned with things that can be proved. We close by observing how the Fibonacci sequence in many ways resembles a geometric sequence. Recall that a geometric sequence with ﬁrst term a and common ratio r has the form a, ar, ar2, ar3, ar4, ar5, ar6, ar7, ar8,... where any term is obtained by multiplying the previous term by r. In general its nth term is Gn = arn, and Gn+1/Gn = r. Equation (10.1) tells us that Fn+1/Fn ≈Φ. Thus even though it is not a geometric sequence, the Fibonacci sequence tends to behave like a geometric sequence with common ratio Φ, and the further “out” you go, the higher the resemblance. Exercises for Chapter 10 Prove the following statements with either induction, strong induction or proof by smallest counterexample. 1. For every integer n ∈N, it follows that 1+2+3+4+···+ n = n2 + n 2 . 2. For every integer n ∈N, it follows that 12 +22 +32 +42 +···+ n2 = n(n+1)(2n+1) 6 . 3. For every integer n ∈N, it follows that 13 +23 +33 +43 +···+ n3 = n2(n+1)2 4 . 4. If n ∈N, then 1·2+2·3+3·4+4·5+···+ n(n+1) = n(n+1)(n+2) 3 . 5. If n ∈N, then 21 +22 +23 +···+2n = 2n+1 −2. 6. For every natural number n, it follows that n X i=1 (8i −5) = 4n2 −n. 7. If n ∈N, then 1·3+2·4+3·5+4·6+···+ n(n+2) = n(n+1)(2n+7) 6 . 8. If n ∈N, then 1 2! + 2 3! + 3 4! +···+ n (n+1)! = 1− 1 (n+1)! 9. For any integer n ≥0, it follows that 24 \| (52n −1). 10. For any integer n ≥0, it follows that 3 \| (52n −1). 11. For any integer n ≥0, it follows that 3 \| (n3 +5n+6). 12. For any integer n ≥0, it follows that 9 \| (43n +8). 170 Mathematical Induction 13. For any integer n ≥0, it follows that 6 \| (n3 −n). 14. Suppose a ∈Z. Prove that 5 \| 2na implies 5 \| a for any n ∈N. 15. If n ∈N, then 1 1·2 + 1 2·3 + 1 3·4 + 1 4·5 +···+ 1 n(n+1) = 1− 1 n+1. 16. For every natural number n, it follows that 2n +1 ≤3n. 17. Suppose A1, A2,... An are sets in some universal set U, and n ≥2. Prove that A1 ∩A2 ∩···∩An = A1 ∪A2 ∪···∪An. 18. Suppose A1, A2,... An are sets in some universal set U, and n ≥2. Prove that A1 ∪A2 ∪···∪An = A1 ∩A2 ∩···∩An. 19. Prove that 1 1 + 1 4 + 1 9 +···+ 1 n2 ≤2−1 n. 20. Prove that (1+2+3+···+ n)2 = 13 +23 +33 +···+ n3 for every n ∈N. 21. If n ∈N, then 1 1 + 1 2 + 1 3 + 1 4 + 1 5 +···+ 1 2n −1 + 1 2n ≥1+ n 2 . (Note: This problem asserts that the sum of the ﬁrst 2n terms of the harmonic series is at least 1+ n/2. It thus implies that the harmonic series diverges.) 22. If n ∈N, then µ 1−1 2 ¶µ 1−1 4 ¶µ 1−1 8 ¶µ 1−1 16 ¶ ··· µ 1−1 2n ¶ ≥1 4 + 1 2n+1 . 23. Use mathematical induction to prove the binomial theorem (Theorem 3.1 on page 80). You may ﬁnd that you need Equation (3.2) on page 78. 24. Prove that n X k=1 k ¡ n k ¢ = n2n−1 for each natural number n. 25. Concerning the Fibonacci sequence, prove that F1+F2+F3+F4+...+Fn = Fn+2−1. 26. Concerning the Fibonacci sequence, prove that n X k=1 F2 k = FnFn+1. 27. Concerning the Fibonacci sequence, prove that F1+F3+F5+F7+...+F2n−1 = F2n. 28. Concerning the Fibonacci sequence, prove that F2 + F4 + F6 + F8 + ... + F2n = F2n+1 −1. 29. In this problem n ∈N and Fn is the nth Fibonacci number. Prove that ¡ n 0 ¢ + ¡ n−1 1 ¢ + ¡ n−2 2 ¢ + ¡ n−3 3 ¢ +···+ ¡ 0 n ¢ = Fn+1. (For example, ¡6 0 ¢ + ¡5 1 ¢ + ¡4 2 ¢ + ¡3 3 ¢ + ¡2 4 ¢ + ¡1 5 ¢ + ¡0 6 ¢ = 1+5+6+1+0+0+0 = 13 = F6+1.) 30. Here Fn is the nth Fibonacci number. Prove that Fn = ³ 1+ p 5 2 ´n − ³ 1− p 5 2 ´n p 5 . 31. Prove that n X k=0 ¡ k r ¢ = ¡ n+1 r+1 ¢, where 1 ≤r ≤n. 32. Prove that the number of n-digit binary numbers that have no consecutive 1’s is the Fibonacci number Fn+2. For example, for n = 2 there are three such Fibonacci Numbers 171 numbers (00, 01, and 10), and 3 = F2+2 = F4. Also, for n = 3 there are ﬁve such numbers (000, 001, 010, 100, 101), and 5 = F3+2 = F5. 33. Suppose n (inﬁnitely long) straight lines lie on a plane in such a way that no two of the lines are parallel, and no three of the lines intersect at a single point. Show that this arrangement divides the plane into n2+n+2 2 regions. 34. Prove that 31 +32 +33 +34 +···+3n = 3n+1 −3 2 for every n ∈N. 35. Prove that if n,k ∈N, and n is even and k is odd, then ¡n k ¢ is even. 36. Prove that if n = 2k−1 for some k ∈N, then every entry in the nth row of Pascal’s triangle is odd. The remaining odd-numbered exercises below are not solved in the back of the book. 37. Prove that if m,n ∈N, then n P k=0 k ¡m+k m ¢ = n ¡m+n+1 m+1 ¢ − ¡m+n+1 m+2 ¢. 38. Prove that if n is a positive integer, then ¡n 0 ¢2 + ¡n 1 ¢2 + ¡n 2 ¢2 +···+ ¡n n ¢2 = ¡2n n ¢. 39. Prove that if n is a positive integer, then ¡n+0 0 ¢ + ¡n+1 1 ¢ + ¡n+2 2 ¢ +···+ ¡n+k k ¢ = ¡n+k+1 k ¢. 40. Prove that p P k=0 ¡m k ¢¡ n p−k ¢ = ¡m+n p ¢ for positive integers m,n and p. 41. Prove that m P k=0 ¡m k ¢¡ n p+k ¢ = ¡m+n m+p ¢ for positive integers m,n and p. Part IV Relations, Functions and Cardinality CHAPTER 11 Relations I n mathematics there are endless ways that two entities can be related to each other. Consider the following mathematical statements. 5 < 10 5 ≤5 6 = 30 5 5 \| 80 7 > 4 x ̸= y 8 ∤3 a ≡b ( mod n) 6 ∈Z X ⊆Y π ≈3.14 0 ≥−1 p 2 ∉Z Z ̸⊆N In each case two entities appear on either side of a symbol, and we interpret the symbol as expressing some relationship between the two entities. Symbols such as <,≤,=,\|,∤,≥,>, ∈and ⊂, etc., are called relations because they convey relationships among things. Relations are signiﬁcant. In fact, you would have to admit that there would be precious little left of mathematics if we took away all the relations. Therefore it is important to have a ﬁrm understanding of them, and this chapter is intended to develop that understanding. Rather than focusing on each relation individually (an impossible task anyway since there are inﬁnitely many diﬀerent relations), we will develop a general theory that encompasses all relations. Understanding this general theory will give us the conceptual framework and language needed to understand and discuss any speciﬁc relation. Before stating the theoretical deﬁnition of a relation, let’s look at a motivational example. This example will lead naturally to our deﬁnition. Consider the set A = © 1,2,3,4,5 ª. (There’s nothing special about this particular set; any set of numbers would do for this example.) Elements of A can be compared to each other by the symbol “<.” For example, 1 < 4, 2 < 3, 2 < 4, and so on. You have no trouble understanding this because the notion of numeric order is so ingrained. But imagine you had to explain it to an idiot savant, one with an obsession for detail but absolutely no understanding of the meaning of (or relationships between) integers. You might consider writing down for your student the following set: R = © (1,2),(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,4),(3,5),(4,5) ª . 176 Relations The set R encodes the meaning of the < relation for elements in A. An ordered pair (a,b) appears in the set if and only if a < b. If asked whether or not it is true that 3 < 4, your student could look through R until he found the ordered pair (3,4); then he would know 3 < 4 is true. If asked about 5 < 2, he would see that (5,2) does not appear in R, so 5 ̸< 2. The set R, which is a subset of A × A, completely describes the relation < for A. Though it may seem simple-minded at ﬁrst, this is exactly the idea we will use for our main deﬁnition. This deﬁnition is general enough to describe not just the relation < for the set A = © 1,2,3,4,5 ª, but any relation for any set A. Deﬁnition 11.1 A relation on a set A is a subset R ⊆A × A. We often abbreviate the statement (x, y) ∈R as xR y. The statement (x, y) ∉R is abbreviated as x̸R y. Notice that a relation is a set, so we can use what we know about sets to understand and explore relations. But before getting deeper into the theory of relations, let’s look at some examples of Deﬁnition 11.1. Example 11.1 Let A = © 1,2,3,4 ª, and consider the following set: R = © (1,1),(2,1),(2,2),(3,3),(3,2),(3,1),(4,4),(4,3),(4,2),(4,1) ª ⊆A × A. The set R is a relation on A, by Deﬁnition 11.1. Since (1,1) ∈R, we have 1R1. Similarly 2R1 and 2R2, and so on. However, notice that (for example) (3,4) ∉R, so 3̸R 4. Observe that R is the familiar relation ≥for the set A. Chapter 1 proclaimed that all of mathematics can be described with sets. Just look at how successful this program has been! The greater-than-or-equal-to relation is now a set R. (We might even express this in the rather cryptic form ≥= R.) Example 11.2 Let A = © 1,2,3,4 ª, and consider the following set: S = © (1,1),(1,3),(3,1),(3,3),(2,2),(2,4),(4,2),(4,4) ª ⊆A × A. Here we have 1S1, 1S3, 4S2, etc., but 3 ̸S4 and 2 ̸S1. What does S mean? Think of it as meaning “has the same parity as.” Thus 1S1 reads “1 has the same parity as 1,” and 4S2 reads “4 has the same parity as 2.” Example 11.3 Consider relations R and S of the previous two examples. Note that R∩S = © (1,1),(2,2),(3,3),(3,1),(4,4),(4,2) ª ⊆A×A is a relation on A. The expression x(R ∩S)y means “x ≥y, and x, y have the same parity.” 177 Example 11.4 Let B = © 0,1,2,3,4,5 ª, and consider the following set: U = © (1,3),(3,3),(5,2),(2,5),(4,2) ª ⊆B ×B. Then U is a relation on B because U ⊆B × B. You may be hard-pressed to invent any “meaning” for this particular relation. A relation does not have to have any meaning. Any random subset of B ×B is a relation on B, whether or not it describes anything familiar. Some relations can be described with pictures. For example, we can depict the above relation U on B by drawing points labeled by elements of B. The statement (x, y) ∈U is then represented by an arrow pointing from x to y, a graphic symbol meaning “x relates to y.” Here’s a picture of U: 0 1 2 3 4 5 The next picture illustrates the relation R on the set A = © a,b, c,d ª, where xR y means x comes before y in the alphabet. According to Deﬁnition 11.1, as a set this relation is R = © (a,b),(a, c),(a,d),(b, c),(b,d),(c,d) ª. You may feel that the picture conveys the relation better than the set does. They are two diﬀerent ways of expressing the same thing. In some instances pictures are more convenient than sets for discussing relations. d c b a Although such diagrams can help us visualize relations, they do have their limitations. If A and R were inﬁnite, then the diagram would be impossible to draw, but the set R might be easily expressed in set-builder notation. Here are some examples. Example 11.5 Consider the set R = © (x, y) ∈Z×Z : x−y ∈N ª ⊆Z×Z. This is the > relation on the set A = Z. It is inﬁnite because there are inﬁnitely many ways to have x > y where x and y are integers. Example 11.6 The set R = © (x,x) : x ∈R ª ⊆R×R is the relation = on the set R, because xR y means the same thing as x = y. Thus R is a set that expresses the notion of equality of real numbers. 178 Relations Exercises for Section 11.0 1. Let A = © 0,1,2,3,4,5 ª. Write out the relation R that expresses > on A. Then illustrate it with a diagram. 2. Let A = © 1,2,3,4,5,6 ª. Write out the relation R that expresses \| (divides) on A. Then illustrate it with a diagram. 3. Let A = © 0,1,2,3,4,5 ª. Write out the relation R that expresses ≥on A. Then illustrate it with a diagram. 4. Here is a diagram for a relation R on a set A. Write the sets A and R. 0 1 2 3 4 5 5. Here is a diagram for a relation R on a set A. Write the sets A and R. 0 1 2 3 4 5 6. Congruence modulo 5 is a relation on the set A = Z. In this relation xR y means x ≡y (mod 5). Write out the set R in set-builder notation. 7. Write the relation < on the set A = Z as a subset R of Z×Z. This is an inﬁnite set, so you will have to use set-builder notation. 8. Let A = © 1,2,3,4,5,6 ª. Observe that ; ⊆A × A, so R = ; is a relation on A. Draw a diagram for this relation. 9. Let A = © 1,2,3,4,5,6 ª. How many diﬀerent relations are there on the set A? 10. Consider the subset R = (R×R)− © (x,x) : x ∈R ª ⊆R×R. What familiar relation on R is this? Explain. 11. Given a ﬁnite set A, how many diﬀerent relations are there on A? In the following exercises, subsets R of R2 = R×R or Z2 = Z×Z are indicated by gray shading. In each case, R is a familiar relation on R or Z. State it. 12. R 13. R 14. 15. Properties of Relations 179 11.1 Properties of Relations A relational expression xR y is a statement (or an open sentence); it is either true or false. For example, 5 < 10 is true, and 10 < 5 is false. (Thus an operation like + is not a relation, because, for instance, 5+10 has a numeric value, not a T/F value.) Since relational expressions have T/F values, we can combine them with logical operators; for example, xR y ⇒yRx is a statement or open sentence whose truth or falsity may depend on x and y. With this in mind, note that some relations have properties that others don’t have. For example, the relation ≤on Z satisﬁes x ≤x for every x ∈Z. But this is not so for < because x < x is never true. The next deﬁnition lays out three particularly signiﬁcant properties that relations may have. Deﬁnition 11.2 Suppose R is a relation on a set A. 1. Relation R is reﬂexive if xRx for every x ∈A. That is, R is reﬂexive if ∀x ∈A,xRx. 2. Relation R is symmetric if xR y implies yRx for all x, y ∈A That is, R is symmetric if ∀x, y ∈A, xR y ⇒yRx. 3. Relation R is transitive if whenever xR y and yRz, then also xRz. That is, R is transitive if ∀x, y, z ∈A, ¡ (xR y)∧(yRz) ¢ ⇒xRz. To illustrate this, let’s consider the set A = Z. Examples of reﬂexive relations on Z include ≤, =, and \|, because x ≤x, x = x and x\|x are all true for any x ∈Z. On the other hand, >, <, ̸= and ∤are not reﬂexive for none of the statements x < x, x > x, x ̸= x and x ∤x is ever true. The relation ̸= is symmetric, for if x ̸= y, then surely y ̸= x also. Also, the relation = is symmetric because x = y always implies y = x. The relation ≤is not symmetric, as x ≤y does not necessarily imply y ≤x. For instance 5 ≤6 is true, but 6 ≤5 is false. Notice (x ≤y) ⇒(y ≤x) is true for some x and y (for example, it is true when x = 2 and y = 2), but still ≤is not symmetric because it is not the case that (x ≤y) ⇒(y ≤x) is true for all integers x and y. The relation ≤is transitive because whenever x ≤y and y ≤z, it also is true that x ≤z. Likewise <,≥,> and = are all transitive. Examine the following table and be sure you understand why it is labeled as it is. Relations on Z: < ≤ = \| ∤ ̸= Reﬂexive no yes yes yes no no Symmetric no no yes no no yes Transitive yes yes yes yes no no 180 Relations Example 11.7 Here A = © b, c,d, e ª, and R is the following relation on A: R = © (b,b),(b, c),(c,b),(c, c),(d,d),(b,d),(d,b),(c,d),(d, c) ª. This relation is not reﬂexive, for although bRb, cRc and dRd, it is not true that eRe. For a relation to be reﬂexive, xRx must be true for all x ∈A. The relation R is symmetric, because whenever we have xR y, it follows that yRx too. Observe that bRc and cRb; bRd and dRb; dRc and cRd. Take away the ordered pair (c,b) from R, and R is no longer symmetric. The relation R is transitive, but it takes some work to check it. We must check that the statement (xR y∧yRz) ⇒xRz is true for all x, y, z ∈A. For example, taking x = b, y = c and z = d, we have (bRc ∧cRd) ⇒bRd, which is the true statement (T ∧T) ⇒T. Likewise, (bRd ∧dRc) ⇒bRc is the true statement (T ∧T) ⇒T. Take note that if x = b, y = e and z = c, then (bRe ∧eRc) ⇒bRc becomes (F ∧F) ⇒T, which is still true. It’s not much fun, but going through all the combinations, you can verify that (xR y∧yRz) ⇒xRz is true for all choices x, y, z ∈A. (Try at least a few of them.) The relation R from Example 11.7 has a meaning. You can think of xR y as meaning that x and y are both consonants. Thus bRc because b and c are both consonants; but b̸Re because it’s not true that b and e are both consonants. Once we look at it this way, it’s immediately clear that R has to be transitive. If x and y are both consonants and y and z are both consonants, then surely x and z are both consonants. This illustrates a point that we will see again later in this section: Knowing the meaning of a relation can help us understand it and prove things about it. Here is a picture of R. Notice that we can immediately spot several properties of R that may not have been so clear from its set description. For instance, we see that R is not reﬂexive because it lacks a loop at e, hence e̸Re. b c d e Figure 11.1. The relation R from Example 11.7 Properties of Relations 181 In what follows, we summarize how to spot the various properties of a relation from its diagram. Compare these with Figure 11.1. 1. A relation is reﬂexive if for each point x ... x ...there is a loop at x: x 2. A relation is symmetric if whenever there is an arrow from x to y ... x y ...there is also an arrow from y back to x: x y 3. A relation is transitive if whenever there are arrows from x to y and y to z ... x y z ...there is also an arrow from x to z: x y z (If x = z, this means that if there are arrows from x to y and from y to x ... x y ...there is also a loop from x back to x.) x y Consider the bottom diagram in Box 3, above. The transitive property demands (xR y∧yRx) ⇒xRx. Thus, if xR y and yRx in a transitive relation, then also xRx, so there is a loop at x. In this case (yRx∧xR y) ⇒yRy, so there will be a loop at y too. Although these visual aids can be illuminating, their use is limited be-cause many relations are too large and complex to be adequately described as diagrams. For example, it would be impossible to draw a diagram for the relation ≡(mod n), where n ∈N. Such a relation would best be explained in a more theoretical (and less visual) way. We next prove that ≡(mod n) is reﬂexive, symmetric and transitive. Obviously we will not glean this from a drawing. Instead we will prove it from the properties of ≡(mod n) and Deﬁnition 11.2. Pay attention to this example. It illustrates how to prove things about relations. 182 Relations Example 11.8 Prove the following proposition. Proposition Let n ∈N. The relation ≡(mod n) on the set Z is reﬂexive, symmetric and transitive. Proof. First we will show that ≡(mod n) is reﬂexive. Take any integer x ∈Z, and observe that n\|0, so n \| (x−x). By deﬁnition of congruence modulo n, we have x ≡x (mod n). This shows x ≡x (mod n) for every x ∈Z, so ≡(mod n) is reﬂexive. Next, we will show that ≡(mod n) is symmetric. For this, we must show that for all x, y ∈Z, the condition x ≡y (mod n) implies that y ≡x (mod n). We use direct proof. Suppose x ≡y (mod n). Thus n \| (x−y) by deﬁnition of congruence modulo n. Then x −y = na for some a ∈Z by deﬁnition of divisibility. Multiplying both sides by −1 gives y −x = n(−a). Therefore n \| (y−x), and this means y ≡x (mod n). We’ve shown that x ≡y (mod n) implies that y ≡x (mod n), and this means ≡(mod n) is symmetric. Finally we will show that ≡(mod n) is transitive. For this we must show that if x ≡y (mod n) and y ≡z (mod n), then x ≡z (mod n). Again we use direct proof. Suppose x ≡y (mod n) and y ≡z (mod n). This means n \| (x −y) and n \| (y−z). Therefore there are integers a and b for which x −y = na and y −z = nb. Adding these two equations, we obtain x−z = na+nb. Consequently, x−z = n(a+b), so n \| (x−z), hence x ≡z (mod n). This completes the proof that ≡(mod n) is transitive. The past three paragraphs have shown that ≡(mod n) is reﬂexive, symmetric and transitive, so the proof is complete. ■ As you continue with mathematics you will ﬁnd that the reﬂexive, symmetric and transitive properties take on special signiﬁcance in a variety of settings. In preparation for this, the next section explores further consequences of these properties. But ﬁrst work some of the following exercises. Exercises for Section 11.1 1. Consider the relation R = © (a,a),(b,b),(c, c),(d,d),(a,b),(b,a) ª on set A = © a,b, c,d ª. Is R reﬂexive? Symmetric? Transitive? If a property does not hold, say why. 2. Consider the relation R = © (a,b),(a, c),(c, c),(b,b),(c,b),(b, c) ª on the set A = © a,b, c ª. Is R reﬂexive? Symmetric? Transitive? If a property does not hold, say why. 3. Consider the relation R = © (a,b),(a, c),(c,b),(b, c) ª on the set A = © a,b, c ª. Is R reﬂexive? Symmetric? Transitive? If a property does not hold, say why. Properties of Relations 183 4. Let A = © a,b, c,d ª. Suppose R is the relation R = © (a,a),(b,b),(c, c),(d,d),(a,b),(b,a),(a, c),(c,a), (a,d),(d,a),(b, c),(c,b),(b,d),(d,b),(c,d),(d, c) ª . Is R reﬂexive? Symmetric? Transitive? If a property does not hold, say why. 5. Consider the relation R = © (0,0),( p 2,0),(0, p 2),( p 2, p 2) ª on R. Is R reﬂexive? Symmetric? Transitive? If a property does not hold, say why. 6. Consider the relation R = © (x,x) : x ∈Z ª on Z. Is R reﬂexive? Symmetric? Transitive? If a property does not hold, say why. What familiar relation is this? 7. There are 16 possible diﬀerent relations R on the set A = © a,b ª. Describe all of them. (A picture for each one will suﬃce, but don’t forget to label the nodes.) Which ones are reﬂexive? Symmetric? Transitive? 8. Deﬁne a relation on Z as xR y if \|x−y\| < 1. Is R reﬂexive? Symmetric? Transitive? If a property does not hold, say why. What familiar relation is this? 9. Deﬁne a relation on Z by declaring xR y if and only if x and y have the same parity. Is R reﬂexive? Symmetric? Transitive? If a property does not hold, say why. What familiar relation is this? 10. Suppose A ̸= ;. Since ; ⊆A × A, the set R = ; is a relation on A. Is R reﬂexive? Symmetric? Transitive? If a property does not hold, say why. 11. Suppose A = © a,b, c,d ª and R = © (a,a),(b,b),(c, c),(d,d) ª. Is R reﬂexive? Symmet-ric? Transitive? If a property does not hold, say why. 12. Prove that the relation \| (divides) on the set Z is reﬂexive and transitive. (Use Example 11.8 as a guide if you are unsure of how to proceed.) 13. Consider the relation R = © (x, y) ∈R×R : x−y ∈Z ª on R. Prove that this relation is reﬂexive, symmetric and transitive. 14. Suppose R is a symmetric and transitive relation on a set A, and there is an element a ∈A for which aRx for every x ∈A. Prove that R is reﬂexive. 15. Prove or disprove: If a relation is symmetric and transitive, then it is also reﬂexive. 16. Deﬁne a relation R on Z by declaring that xR y if and only if x2 ≡y2 (mod 4). Prove that R is reﬂexive, symmetric and transitive. 17. Modifying the above Exercise 8 (above) slightly, deﬁne a relation ∼on Z as x ∼y if and only if \|x−y\| ≤1. Say whether ∼is reﬂexive. Is it symmetric? Transitive? 18. The table on page 179 shows that relations on Z may obey various combinations of the reﬂexive, symmetric and transitive properties. In all, there are 23 = 8 possible combinations, and the table shows 5 of them. (There is some redundancy, as ≤and \| have the same type.) Complete the table by ﬁnding examples of relations on Z for the three missing combinations. 184 Relations 11.2 Equivalence Relations The relation = on the set Z (or on any set A) is reﬂexive, symmetric and transitive. There are many other relations that are also reﬂexive, symmetric and transitive. Relations that have all three of these properties occur very frequently in mathematics and often play quite signiﬁcant roles. (For instance, this is certainly true of the relation =.) Such relations are given a special name. They are called equivalence relations. Deﬁnition 11.3 A relation R on a set A is an equivalence relation if it is reﬂexive, symmetric and transitive. As an example, Figure 11.2 shows four diﬀerent equivalence relations R1,R2,R3 and R4 on the set A = © −1,1,2,3,4 ª. Each one has its own meaning, as labeled. For example, in the second row the relation R2 literally means “has the same parity as.” So 1R2 3 means “1 has the same parity as 3,” etc. Relation R Diagram Equivalence classes (see next page) “is equal to” (=) R1 = © (−1,−1),(1,1),(2,2),(3,3),(4,4) ª −1 3 1 4 2 © −1 ª, © 1 ª, © 2 ª, © 3 ª, © 4 ª “has same parity as” R2 = © (−1,−1),(1,1),(2,2),(3,3),(4,4), (−1,1),(1,−1),(−1,3),(3,−1), (1,3),(3,1),(2,4),(4,2) ª −1 3 1 4 2 © −1,1,3 ª, © 2,4 ª “has same sign as” R3 = © (−1,−1),(1,1),(2,2),(3,3),(4,4), (1,2),(2,1),(1,3),(3,1),(1,4),(4,1),(3,4), (4,3),(2,3),(3,2),(2,4),(4,2),(1,3),(3,1) ª −1 3 1 4 2 © −1 ª, © 1,2,3,4 ª “has same parity and sign as” R4 = © (−1,−1),(1,1),(2,2),(3,3),(4,4), (1,3),(3,1),(2,4),(4,2) ª −1 3 1 4 2 © −1 ª, © 1,3 ª, © 2,4 ª Figure 11.2. Examples of equivalence relations on the set A = © −1,1,2,3,4 ª Equivalence Relations 185 The above diagrams make it easy to check that each relation is reﬂexive, symmetric and transitive, i.e., that each is an equivalence relation. For example, R1 is symmetric because xR1y ⇒yR1x is always true: When x = y it becomes T ⇒T (true), and when x ̸= y it becomes F ⇒F (also true). In a similar fashion, R1 is transitive because (xR1y∧yR1z) ⇒xR1z is always true: It always works out to one of T ⇒T, F ⇒T or F ⇒F. (Check this.) As you can see from the examples in Figure 11.2, equivalence relations on a set tend to express some measure of “sameness” among the elements of the set, whether it is true equality or something weaker (like having the same parity). It’s time to introduce an important deﬁnition. Whenever you have an equivalence relation R on a set A, it divides A into subsets called equivalence classes. Here is the deﬁnition: Deﬁnition 11.4 Suppose R is an equivalence relation on a set A. Given any element a ∈A, the equivalence class containing a is the subset © x ∈A : xRa ª of A consisting of all the elements of A that relate to a. This set is denoted as [a]. Thus the equivalence class containing a is the set [a] = © x ∈A : xRa ª. Example 11.9 Consider the relation R1 in Figure 11.2. The equivalence class containing 2 is the set = © x ∈A : xR12 ª. Because in this relation the only element that relates to 2 is 2 itself, we have = © 2 ª. Other equivalence classes for R1 are [−1] = © −1 ª, = © 1 ª, = © 3 ª and = © 4 ª. Thus this relation has ﬁve separate equivalence classes. Example 11.10 Consider the relation R2 in Figure 11.2. The equivalence class containing 2 is the set = © x ∈A : xR22 ª. Because only 2 and 4 relate to 2, we have = © 2,4 ª. Observe that we also have = © x ∈A : xR24 ª = © 2,4 ª, so = . Another equivalence class for R2 is = © x ∈A : xR21 ª = © −1,1,3 ª. In addition, note that = [−1] = = © −1,1,3 ª. Thus this relation has just two equivalence classes, namely © 2,4 ª and © −1,1,3 ª. Example 11.11 The relation R4 in Figure 11.2 has three equivalence classes. They are [−1] = © −1 ª and = = © 1,3 ª and = = © 2,4 ª. Don’t be misled by Figure 11.2. It’s important to realize that not every equivalence relation can be drawn as a diagram involving nodes and arrows. Even the simple relation R = © (x,x) : x ∈R ª, which expresses equality in the set R, is too big to be drawn. Its picture would involve a point for every real number and a loop at each point. Clearly that’s too many points and loops to draw. 186 Relations We close this section with several other examples of equivalence rela-tions on inﬁnite sets. Example 11.12 Let P be the set of all polynomials with real coeﬃcients. Deﬁne a relation R on P as follows. Given f (x), g(x) ∈P, let f (x)R g(x) mean that f (x) and g(x) have the same degree. Thus (x2 +3x−4)R (3x2 −2) and (x3 +3x2 −4) ̸R (3x2 −2), for example. It takes just a quick mental check to see that R is an equivalence relation. (Do it.) It’s easy to describe the equivalence classes of R. For example, [3x2+2] is the set of all polynomials that have the same degree as 3x2 +2, that is, the set of all polynomials of degree 2. We can write this as [3x2 +2] = © ax2 + bx+ c : a,b, c ∈R,a ̸= 0 ª. Example 11.8 proved that for a given n ∈N the relation ≡(mod n) is reﬂexive, symmetric and transitive. Thus, in our new parlance, ≡(mod n) is an equivalence relation on Z. Consider the case n = 3. Let’s ﬁnd the equivalence classes of the equivalence relation ≡(mod 3). The equivalence class containing 0 seems like a reasonable place to start. Observe that = © x ∈Z : x ≡0(mod 3) ª = © x ∈Z : 3 \| (x−0) ª = © x ∈Z : 3 \| x ª = © ...,−3,0,3,6,9,... ª . Thus the class consists of all the multiples of 3. (Or, said diﬀerently, consists of all integers that have a remainder of 0 when divided by 3). Note that = = = , etc. The number 1 does not show up in the set so let’s next look at the equivalence class : = © x ∈Z : x ≡1(mod 3) ª = © x ∈Z : 3 \| (x−1) ª = © ...,−5,−2,1,4,7,10,... ª . The equivalence class consists of all integers that give a remainder of 1 when divided by 3. The number 2 is in neither of the sets or , so we next look at the equivalence class : = © x ∈Z : x ≡2(mod 3) ª = © x ∈Z : 3 \| (x−2) ª = © ...,−4,−1,2,5,8,11,... ª . The equivalence class consists of all integers that give a remainder of 2 when divided by 3. Observe that any integer is in one of the sets , or , so we have listed all of the equivalence classes. Thus ≡(mod 3) has exactly three equivalence classes, as described above. Similarly, you can show that the equivalence relation ≡(mod n) has n equivalence classes ,,,..., [n−1]. Equivalence Relations 187 Exercises for Section 11.2 1. Let A = © 1,2,3,4,5,6 ª, and consider the following equivalence relation on A: R = © (1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(2,3),(3,2),(4,5),(5,4),(4,6),(6,4),(5,6),(6,5) ª List the equivalence classes of R. 2. Let A = © a,b, c,d, e ª. Suppose R is an equivalence relation on A. Suppose R has two equivalence classes. Also aRd, bRc and eRd. Write out R as a set. 3. Let A = © a,b, c,d, e ª. Suppose R is an equivalence relation on A. Suppose R has three equivalence classes. Also aRd and bRc. Write out R as a set. 4. Let A = © a,b, c,d, e ª. Suppose R is an equivalence relation on A. Suppose also that aRd and bRc, eRa and cRe. How many equivalence classes does R have? 5. There are two diﬀerent equivalence relations on the set A = © a,b ª. Describe them. Diagrams will suﬃce. 6. There are ﬁve diﬀerent equivalence relations on the set A = © a,b, c ª. Describe them all. Diagrams will suﬃce. 7. Deﬁne a relation R on Z as xR y if and only if 3x −5y is even. Prove R is an equivalence relation. Describe its equivalence classes. 8. Deﬁne a relation R on Z as xR y if and only if x2 + y2 is even. Prove R is an equivalence relation. Describe its equivalence classes. 9. Deﬁne a relation R on Z as xR y if and only if 4\|(x+3y). Prove R is an equivalence relation. Describe its equivalence classes. 10. Suppose R and S are two equivalence relations on a set A. Prove that R ∩S is also an equivalence relation. (For an example of this, look at Figure 11.2. Observe that for the equivalence relations R2,R3 and R4, we have R2 ∩R3 = R4.) 11. Prove or disprove: If R is an equivalence relation on an inﬁnite set A, then R has inﬁnitely many equivalence classes. 12. Prove or disprove: If R and S are two equivalence relations on a set A, then R ∪S is also an equivalence relation on A. 13. Suppose R is an equivalence relation on a ﬁnite set A, and every equivalence class has the same cardinality m. Express \|R\| in terms of m and \|A\|. 14. Suppose R is a reﬂexive and symmetric relation on a ﬁnite set A. Deﬁne a relation S on A by declaring xSy if and only if for some n ∈N there are elements x1,x2,...,xn ∈A satisfying xRx1, x1Rx2, x2Rx3, x3Rx4,...,xn−1Rxn, and xnRy. Show that S is an equivalence relation and R ⊆S. Prove that S is the unique smallest equivalence relation on A containing R. 15. Suppose R is an equivalence relation on a set A, with four equivalence classes. How many diﬀerent equivalence relations S on A are there for which R ⊆S? 188 Relations 11.3 Equivalence Classes and Partitions This section collects several properties of equivalence classes. Our ﬁrst result proves that [a] = [b] if and only if aRb. This is useful because it assures us that whenever we are in a situation where [a] = [b], we also have aRb, and vice versa. Being able to switch back and forth between these two pieces of information can be helpful in a variety of situations, and you may ﬁnd yourself using this result a lot. Be sure to notice that the proof uses all three properties (reﬂexive, symmetric and transitive) of equivalence relations. Notice also that we have to use some Chapter 8 techniques in dealing with the sets [a] and [b]. Theorem 11.1 Suppose R is an equivalence relation on a set A. Suppose also that a,b ∈A. Then [a] = [b] if and only if aRb. Proof. Suppose [a] = [b]. Note that aRa by the reﬂexive property of R, so a ∈ © x ∈A : xRa ª = [a] = [b] = © x ∈A : xRb ª. But a belonging to © x ∈A : xRb ª means aRb. This completes the ﬁrst part of the if-and-only-if proof. Conversely, suppose aRb. We need to show [a] = [b]. We will do this by showing [a] ⊆[b] and [b] ⊆[a]. First we show [a] ⊆[b]. Suppose c ∈[a]. As c ∈[a] = © x ∈A : xRa ª, we get cRa. Now we have cRa and aRb, so cRb because R is transitive. But cRb implies c ∈ © x ∈A : xRb ª = [b]. This demonstrates that c ∈[a] implies c ∈[b], so [a] ⊆[b]. Next we show [b] ⊆[a]. Suppose c ∈[b]. As c ∈[b] = © x ∈A : xRb ª, we get cRb. Remember that we are assuming aRb, so bRa because R is symmetric. Now we have cRb and bRa, so cRa because R is transitive. But cRa implies c ∈ © x ∈A : xRa ª = [a]. This demonstrates that c ∈[b] implies c ∈[a]; hence [b] ⊆[a]. The previous two paragraphs imply that [a] = [b]. ■ To illustrate Theorem 11.1, recall how we worked out the equivalence classes of ≡(mod 3) at the end of Section 11.2. We observed that [−3] = = © ...,−3,0,3,6,9,... ª . Note that [−3] = and −3 ≡9 (mod 3), just as Theorem 11.1 predicts. The theorem assures us that this will work for any equivalence relation. In the future you may ﬁnd yourself using the result of Theorem 11.1 often. Over time it may become natural and familiar; you will use it automatically, without even thinking of it as a theorem. Equivalence Classes and Partitions 189 Our next topic addresses the fact that an equivalence relation on a set A divides A into various equivalence classes. There is a special word for this kind of situation. We address it now, as you are likely to encounter it in subsequent mathematics classes. Deﬁnition 11.5 A partition of a set A is a set of non-empty subsets of A, such that the union of all the subsets equals A, and the intersection of any two diﬀerent subsets is ;. Example 11.13 Let A = © a,b, c,d ª. One partition of A is ©© a,b ª , © c ª , © d ªª. This is a set of three subsets © a,b ª, © c ª and © d ª of A. The union of the three subsets equals A; the intersection of any two subsets is ;. Other partitions of A are ©© a,b ª , © c,d ªª , ©© a, c ª , © b ª , © d ªª , ©© a ª , © b ª , © c ª© d ªª , ©© a,b, c,d ªª , to name a few. Intuitively, a partition is just a dividing up of A into pieces. Example 11.14 Consider the equivalence relations in Figure 11.2. Each of these is a relation on the set A = © −1,1,2,3,4 ª. The equivalence classes of each relation are listed on the right side of the ﬁgure. Observe that, in each case, the set of equivalence classes forms a partition of A. For example, the relation R1 yields the partition ©© −1 ª , © 1 ª , © 2 ª , © 3 ª , © 4 ªª of A. Likewise the equivalence classes of R2 form the partition ©© −1,1,3 ª , © 2,4 ªª. Example 11.15 Recall that we worked out the equivalence classes of the equivalence relation ≡(mod 3) on the set Z. These equivalence classes give the following partition of Z: ©© ...,−3,0,3,6,9,... ª , © ...,−2,1,4,7,10,... ª , © ...,−1,2,5,8,11,... ªª . We can write it more compactly as © ,, ª. Our examples and experience suggest that the equivalence classes of an equivalence relation on a set form a partition of that set. This is indeed the case, and we now prove it. Theorem 11.2 Suppose R is an equivalence relation on a set A. Then the set © [a] : a ∈A ª of equivalence classes of R forms a partition of A. Proof. To show that © [a] : a ∈A ª is a partition of A we need to show two things: We need to show that the union of all the sets [a] equals A, and we need to show that if [a] ̸= [b], then [a]∩[b] = ;. 190 Relations Notationally, the union of all the sets [a] is S a∈A[a], so we need to prove S a∈A[a] = A. Suppose x ∈S a∈A[a]. This means x ∈[a] for some a ∈A. Since [a] ⊆A, it then follows that x ∈A. Thus S a∈A[a] ⊆A. On the other hand, suppose x ∈A. As x ∈[x], we know x ∈[a] for some a ∈A (namely a = x). Therefore x ∈S a∈A[a], and this shows A ⊆S a∈A[a]. Since S a∈A[a] ⊆A and A ⊆S a∈A[a], it follows that S a∈A[a] = A. Next we need to show that if [a] ̸= [b] then [a] ∩[b] = ;. Let’s use contrapositive proof. Suppose it’s not the case that [a]∩[b] = ;, so there is some element c with c ∈[a]∩[b]. Thus c ∈[a] and c ∈[b]. Now, c ∈[a] means cRa, and then aRc since R is symmetric. Also c ∈[b] means cRb. Now we have aRc and cRb, so aRb (because R is transitive). By Theorem 11.1, aRb implies [a] = [b]. Thus [a] ̸= [b] is not true. We’ve now shown that the union of all the equivalence classes is A, and the intersection of two diﬀerent equivalence classes is ;. Therefore the set of equivalence classes is a partition of A. ■ Theorem 11.2 says the equivalence classes of any equivalence relation on a set A form a partition of A. Conversely, any partition of A describes an equivalence relation R where xR y if and only if x and y belong to the same set in the partition. (See Exercise 4 for this section, below.) Thus equivalence relations and partitions are really just two diﬀerent ways of looking at the same thing. In your future mathematical studies, you may ﬁnd yourself easily switching between these two points of view. Exercises for Section 11.3 1. List all the partitions of the set A = © a,b ª. Compare your answer to the answer to Exercise 5 of Section 11.2. 2. List all the partitions of the set A = © a,b, c ª. Compare your answer to the answer to Exercise 6 of Section 11.2. 3. Describe the partition of Z resulting from the equivalence relation ≡(mod 4). 4. Suppose P is a partition of a set A. Deﬁne a relation R on A by declaring xR y if and only if x, y ∈X for some X ∈P. Prove R is an equivalence relation on A. Then prove that P is the set of equivalence classes of R. 5. Consider the partition P = ©© ...,−4,−2,0,2,4,... ª , © ...,−5,−3,−1,1,3,5,... ªª of Z. Let R be the equivalence relation whose equivalence classes are the two ele-ments of P. What familiar equivalence relation is R? The Integers Modulo n 191 11.4 The Integers Modulo n Example 11.8 proved that for a given n ∈N, the relation ≡(mod n) is reﬂexive, symmetric and transitive, so it is an equivalence relation. This is a particularly signiﬁcant equivalence relation in mathematics, and in the present section we deduce some of its properties. To make matters simpler, let’s pick a concrete n, say n = 5. Let’s begin by looking at the equivalence classes of the relation ≡(mod 5). There are ﬁve equivalence classes, as follows: = © x ∈Z : x ≡0 (mod 5)) ª = © x ∈Z : 5 \| (x−0) ª = © ...,−10,−5,0,5,10,15,... ª , = © x ∈Z : x ≡1 (mod 5)) ª = © x ∈Z : 5 \| (x−1) ª = © ..., −9,−4,1,6,11,16,... ª , = © x ∈Z : x ≡2 (mod 5)) ª = © x ∈Z : 5 \| (x−2) ª = © ..., −8,−3,2,7,12,17,... ª , = © x ∈Z : x ≡3 (mod 5)) ª = © x ∈Z : 5 \| (x−3) ª = © ..., −7,−2,3,8,13,18,... ª , = © x ∈Z : x ≡4 (mod 5)) ª = © x ∈Z : 5 \| (x−4) ª = © ..., −6,−1,4,9,14,19,... ª . Notice how these equivalence classes form a partition of the set Z. We label the ﬁve equivalence classes as ,,, and , but you know of course that there are other ways to label them. For example, = = = , and so on; and = = [−4], etc. Still, for this discussion we denote the ﬁve classes as ,,, and . These ﬁve classes form a set, which we shall denote as Z5. Thus Z5 = © ,,,, ª is a set of ﬁve sets. The interesting thing about Z5 is that even though its elements are sets (and not numbers), it is possible to add and multiply them. In fact, we can deﬁne the following rules that tell how elements of Z5 can be added and multiplied. [a]+[b] = [a+ b] [a]·[b] = [a· b] For example, + = [2 + 1] = , and · = [2 · 2] = . We stress that in doing this we are adding and multiplying sets (more precisely equivalence classes), not numbers. We added (or multiplied) two elements of Z5 and obtained another element of Z5. Here is a trickier example. Observe that + = . This time we added elements , ∈Z5, and got the element ∈Z5. That was easy, except where is our answer in the set Z5 = © ,,,, ª? Since = , it is more appropriate to write + = . 192 Relations In a similar vein, · = would be written as · = because = . Test your skill with this by verifying the following addition and multiplication tables for Z5. + · We call the set Z5 = © ,,,, ª the integers modulo 5. As our tables suggest, Z5 is more than just a set: It is a little number system with its own addition and multiplication. In this way it is like the familiar set Z which also comes equipped with an addition and a multiplication. Of course, there is nothing special about the number 5. We can also deﬁne Zn for any natural number n. Here is the deﬁnition: Deﬁnition 11.6 Let n ∈N. The equivalence classes of the equivalence relation ≡(mod n) are ,,,...,[n−1]. The integers modulo n is the set Zn = © ,,,...,[n −1] ª. Elements of Zn can be added by the rule [a]+[b] = [a+ b] and multiplied by the rule [a]·[b] = [ab]. Given a natural number n, the set Zn is a number system containing n elements. It has many of the algebraic properties that Z,R and Q possess. For example, it is probably obvious to you already that elements of Zn obey the commutative laws [a]+[b] = [b]+[a] and [a]·[b] = [b]·[a]. You can also verify the distributive law [a]·([b]+[c]) = [a]·[b]+[a]·[c], as follows: [a]·([b]+[c]) = [a]·[b + c] = [a(b + c)] = [ab + ac] = [ab]+[ac] = [a]·[b]+[a]·[c]. The integers modulo n are signiﬁcant because they more closely ﬁt certain applications than do other number systems such as Z or R. If you go on to The Integers Modulo n 193 take a course in abstract algebra, then you will work extensively with Zn as well as other, more exotic, number systems. (In such a course you will also use all of the proof techniques that we have discussed, as well as the ideas of equivalence relations.) To close this section we take up an issue that may have bothered you earlier. It has to do with our deﬁnitions of addition [a]+[b] = [a + b] and multiplication [a] · [b] = [ab]. These deﬁnitions deﬁne addition and multiplication of equivalence classes in terms of representatives a and b in the equivalence classes. Since there are many diﬀerent ways to choose such representatives, we may well wonder if addition and multiplication are consistently deﬁned. For example, suppose two people, Alice and Bob, want to multiply the elements and in Z5. Alice does the calculation as · = = , so her ﬁnal answer is . Bob does it diﬀerently. Since = and = , he works out · as · = . Since 56 ≡1 (mod 5), Bob’s answer is = , and that agrees with Alice’s answer. Will their answers always agree or did they just get lucky (with the arithmetic)? The fact is that no matter how they do the multiplication in Zn, their answers will agree. To see why, suppose Alice and Bob want to multiply the elements [a],[b] ∈Zn, and suppose [a] = [a′] and [b] = [b′]. Alice and Bob do the multiplication as follows: Alice: [a]·[b] = [ab], Bob: [a′]·[b′] = [a′b′]. We need to show that their answers agree, that is, we need to show [ab] = [a′b′]. Since [a] = [a′], we know by Theorem 11.1 that a ≡a′ (mod n). Thus n \| (a−a′), so a−a′ = nk for some integer k. Likewise, as [b] = [b′], we know b ≡b′ (mod n), or n \| (b −b′), so b −b′ = nℓfor some integer ℓ. Thus we get a = a′ + nk and b = b′ + nℓ. Therefore: ab = (a′ + nk)(b′ + nℓ) = a′b′ + a′nℓ+ nkb′ + n2kℓ, hence ab −a′b′ = n(a′ℓ+ kb′ + nkℓ). This shows n \| (ab −a′b′), so ab ≡a′b′ (mod n), and from that we conclude [ab] = [a′b′]. Consequently Alice and Bob really do get the same answer, so we can be assured that the deﬁnition of multiplication in Zn is consistent. Exercise 8 below asks you to show that addition in Zn is similarly consistent. 194 Relations Exercises for Section 11.4 1. Write the addition and multiplication tables for Z2. 2. Write the addition and multiplication tables for Z3. 3. Write the addition and multiplication tables for Z4. 4. Write the addition and multiplication tables for Z6. 5. Suppose [a],[b] ∈Z5 and [a]·[b] = . Is it necessarily true that either [a] = or [b] = ? 6. Suppose [a],[b] ∈Z6 and [a]·[b] = . Is it necessarily true that either [a] = or [b] = ? 7. Do the following calculations in Z9, in each case expressing your answer as [a] with 0 ≤a ≤8. (a) + (b) + (c) · (d) · 8. Suppose [a],[b] ∈Zn, and [a] = [a′] and [b] = [b′]. Alice adds [a] and [b] as [a]+[b] = [a+ b]. Bob adds them as [a′]+[b′] = [a′ + b′]. Show that their answers [a+ b] and [a′ + b′] are the same. 11.5 Relations Between Sets In the beginning of this chapter, we deﬁned a relation on a set A to be a subset R ⊆A × A. This created a framework that could model any situation in which elements of A are compared to themselves. In this setting, the statement xR y has elements x and y from A on either side of the R because R compares elements from A. But there are other relational symbols that don’t work this way. Consider ∈. The statement 5 ∈Z expresses a relationship between 5 and Z (namely that the element 5 is in the set Z) but 5 and Z are not in any way naturally regarded as both elements of some set A. To overcome this diﬃculty, we generalize the idea of a relation on A to a relation from A to B. Deﬁnition 11.7 A relation from a set A to a set B is a subset R ⊆A×B. We often abbreviate the statement (x, y) ∈R as xR y. The statement (x, y) ∉R is abbreviated as x̸R y. Example 11.16 Suppose A = © 1,2 ª and B = P(A) = © ;, © 1 ª , © 2 ª , © 1,2 ªª. Then R = © (1, © 1 ª ),(2, © 2 ª ),(1, © 1,2 ª ),(2, © 1,2 ª ) ª ⊆A×B is a relation from A to B. Note that we have 1R © 1 ª, 2R © 2 ª, 1R © 1,2 ª and 2R © 1,2 ª. The relation R is the familiar relation ∈for the set A, that is, xR X means exactly the same thing as x ∈X. Relations Between Sets 195 Diagrams for relations from A to B diﬀer from diagrams for relations on A. Since there are two sets A and B in a relation from A to B, we have to draw labeled nodes for each of the two sets. Then we draw arrows from x to y whenever xR y. The following ﬁgure illustrates this for Example 11.16. ; © 1 ª © 2 ª © 1,2 ª 1 2 A B Figure 11.3. A relation from A to B The ideas from this chapter show that any relation (whether it is a familiar one like ≥, ≤, =, \|, ∈or ⊆, or a more exotic one) is really just a set. Therefore the theory of relations is a part of the theory of sets. In the next chapter, we will see that this idea touches on another important mathematical construction, namely functions. We will deﬁne a function to be a special kind of relation from one set to another, and in this context we will see that any function is really just a set. CHAPTER 12 Functions Y ou know from calculus that functions play a fundamental role in math-ematics. You likely view a function as a kind of formula that describes a relationship between two (or more) quantities. You certainly understand and appreciate the fact that relationships between quantities are impor-tant in all scientiﬁc disciplines, so you do not need to be convinced that functions are important. Still, you may not be aware of the full signiﬁcance of functions. Functions are more than merely descriptions of numeric relationships. In a more general sense, functions can compare and relate diﬀerent kinds of mathematical structures. You will see this as your understanding of mathematics deepens. In preparation of this deepening, we will now explore a more general and versatile view of functions. The concept of a relation between sets (Deﬁnition 11.7) plays a big role here, so you may want to quickly review it. 12.1 Functions Let’s start on familiar ground. Consider the function f (x) = x2 from R to R. Its graph is the set of points R = © (x,x2) : x ∈R ª ⊆R×R. R R (x,x2) x Figure 12.1. A familiar function Having read Chapter 11, you may see f in a new light. Its graph R ⊆R × R is a relation on the set R. In fact, as we shall see, functions are just special kinds of relations. Before stating the exact deﬁnition, we Functions 197 look at another example. Consider the function f (n) = \|n\|+2 that converts integers n into natural numbers \|n\|+2. Its graph is R = © (n,\|n\|+2) : n ∈Z ª ⊆Z×N. N Z −4 −3 −2 −1 0 1 2 3 4 1 2 3 4 5 6 Figure 12.2. The function f : Z →N, where f (n) = \|n\|+2 Figure 12.2 shows the graph R as darkened dots in the grid of points Z×N. Notice that in this example R is not a relation on a single set. The set of input values Z is diﬀerent from the set N of output values, so the graph R ⊆Z×N is a relation from Z to N. This example illustrates three things. First, a function can be viewed as sending elements from one set A to another set B. (In the case of f , A = Z and B = N.) Second, such a function can be regarded as a relation from A to B. Third, for every input value n, there is exactly one output value f (n). In your high school algebra course, this was expressed by the vertical line test: Any vertical line intersects a function’s graph at most once. It means that for any input value x, the graph contains exactly one point of form (x, f (x)). Our main deﬁnition, given below, incorporates all of these ideas. Deﬁnition 12.1 Suppose A and B are sets. A function f from A to B (denoted as f : A →B) is a relation f ⊆A × B from A to B, satisfying the property that for each a ∈A the relation f contains exactly one ordered pair of form (a,b). The statement (a,b) ∈f is abbreviated f (a) = b. Example 12.1 Consider the function f graphed in Figure 12.2. According to Deﬁnition 12.1, we regard f as the set of points in its graph, that is, f = © (n,\|n\|+2) : n ∈Z ª ⊆Z×N. This is a relation from Z to N, and indeed given any a ∈Z the set f contains exactly one ordered pair (a,\|a\|+2) whose ﬁrst coordinate is a. Since (1,3) ∈f , we write f (1) = 3; and since (−3,5) ∈f we write f (−3) = 5, etc. In general, (a,b) ∈f means that f sends the input 198 Functions value a to the output value b, and we express this as f (a) = b. This function can be expressed by a formula: For each input value n, the output value is \|n\|+2, so we may write f (n) = \|n\|+2. All this agrees with the way we thought of functions in algebra and calculus; the only diﬀerence is that now we also think of a function as a relation. Deﬁnition 12.2 For a function f : A →B, the set A is called the domain of f . (Think of the domain as the set of possible “input values” for f .) The set B is called the codomain of f . The range of f is the set © f (a) : a ∈A ª = © b : (a,b) ∈f ª. (Think of the range as the set of all possible “output values” for f . Think of the codomain as a sort of “target” for the outputs.) Continuing Example 12.1, the domain of f is Z and its codomain is N. Its range is © f (a) : a ∈Z ª = © \|a\|+2 : a ∈Z ª = © 2,3,4,5,... ª. Notice that the range is a subset of the codomain, but it does not (in this case) equal the codomain. In our examples so far, the domains and codomains are sets of numbers, but this needn’t be the case in general, as the next example indicates. Example 12.2 Let A = © p, q,r,s ª and B = © 0,1,2 ª, and f = © (p,0),(q,1),(r,2),(s,2) ª ⊆A ×B. This is a function f : A →B because each element of A occurs exactly once as a ﬁrst coordinate of an ordered pair in f . We have f (p) = 0, f (q) = 1, f (r) = 2 and f (s) = 2. The domain of f is © p, q,r,s ª, and the codomain and range are both © 0,1,2 ª. (p,0) (p,1) (p,2) (q,0) (q,1) (q,2) (r,0) (r,1) (r,2) (s,0) (s,1) (s,2) 0 1 2 p q r s A B (a) 0 1 2 p q r s (b) A B Figure 12.3. Two ways of drawing the function f = © (p,0),(q,1),(r,2),(s,2) ª Functions 199 If A and B are not both sets of numbers it can be diﬃcult to draw a graph of f : A →B in the traditional sense. Figure 12.3(a) shows an attempt at a graph of f from Example 12.2. The sets A and B are aligned roughly as x- and y-axes, and the Cartesian product A × B is ﬁlled in accordingly. The subset f ⊆A ×B is indicated with dashed lines, and this can be regarded as a “graph” of f . A more natural visual description of f is shown in 12.3(b). The sets A and B are drawn side-by-side, and arrows point from a to b whenever f (a) = b. In general, if f : A →B is the kind of function you may have encountered in algebra or calculus, then conventional graphing techniques oﬀer the best visual description of it. On the other hand, if A and B are ﬁnite or if we are thinking of them as generic sets, then describing f with arrows is often a more appropriate way of visualizing it. We emphasize that, according to Deﬁnition 12.1, a function is really just a special kind of set. Any function f : A →B is a subset of A ×B. By contrast, your calculus text probably deﬁned a function as a certain kind of “rule.” While that intuitive outlook is adequate for the ﬁrst few semesters of calculus, it does not hold up well to the rigorous mathematical standards necessary for further progress. The problem is that words like “rule” are too vague. Deﬁning a function as a set removes the ambiguity. It makes a function into a concrete mathematical object. Still, in practice we tend to think of functions as rules. Given f : Z →N where f (x) = \|x\|+2, we think of this as a rule that associates any number n ∈Z to the number \|n\| + 2 in N, rather than a set containing ordered pairs (n,\|n\|+2). It is only when we have to understand or interpret the theoretical nature of functions (as we do in this text) that Deﬁnition 12.1 comes to bear. The deﬁnition is a foundation that gives us license to think about functions in a more informal way. The next example brings up a point about notation. Consider a function such as f : Z2 →Z, whose domain is a Cartesian product. This function takes as input an ordered pair (m,n) ∈Z2 and sends it to a number f ((m,n)) ∈ Z. To simplify the notation, it is common to write f (m,n) instead of f ((m,n)), even though this is like writing f x instead of f (x). We also remark that although we’ve been using the letters f , g and h to denote functions, any other reasonable symbol could be used. Greek letters such as ϕ and θ are common. Example 12.3 Say a function ϕ : Z2 →Z is deﬁned as ϕ(m,n) = 6m −9n. Note that as a set, this function is ϕ = ©¡ (m,n),6m−9n ¢ : (m,n) ∈Z2ª ⊆Z2×Z. What is the range of ϕ? 200 Functions To answer this, ﬁrst observe that for any (m,n) ∈Z2, the value f (m,n) = 6m−9n = 3(2m−3n) is a multiple of 3. Thus every number in the range is a multiple of 3, so the range is a subset of the set of all multiples of 3. On the other hand if b = 3k is a multiple of 3 we have ϕ(−k,−k) = 6(−k)−9(−k) = 3k = b, which means any multiple of 3 is in the range of ϕ. Therefore the range of ϕ is the set © 3k : k ∈Z ª of all multiples of 3. To conclude this section, let’s use Deﬁnition 12.1 to help us understand what it means for two functions f : A →B and g : C →D to be equal. According to our deﬁnition, functions f and g are subsets f ⊆A ×B and g ⊆C × D. It makes sense to say that f and g are equal if f = g, that is, if they are equal as sets. Thus the two functions f = © (1,a),(2,a),(3,b) ª and g = © (3,b),(2,a),(1,a) ª are equal because the sets f and g are equal. Notice that the domain of both functions is A = © 1,2,3 ª, the set of ﬁrst elements x in the ordered pairs (x, y) ∈f = g. In general, equal functions must have equal domains. Observe also that the equality f = g means f (x) = g(x) for every x ∈A. We repackage these ideas in the following deﬁnition. Deﬁnition 12.3 Two functions f : A →B and g : A →D are equal if f (x) = g(x) for every x ∈A. Observe that f and g can have diﬀerent codomains and still be equal. Consider the functions f : Z →N and g : Z →Z deﬁned as f (x) = \|x\|+2 and g(x) = \|x\|+2. Even though their codomains are diﬀerent, the functions are equal because f (x) = g(x) for every x in the domain. Exercises for Section 12.1 1. Suppose A = © 0,1,2,3,4 ª, B = © 2,3,4,5 ª and f = © (0,3),(1,3),(2,4),(3,2),(4,2) ª. State the domain and range of f . Find f (2) and f (1). 2. Suppose A = © a,b, c,d ª, B = © 2,3,4,5,6 ª and f = © (a,2),(b,3),(c,4),(d,5) ª. State the domain and range of f . Find f (b) and f (d). 3. There are four diﬀerent functions f : © a,b ª → © 0,1 ª. List them all. Diagrams will suﬃce. 4. There are eight diﬀerent functions f : © a,b, c ª → © 0,1 ª. List them all. Diagrams will suﬃce. 5. Give an example of a relation from © a,b, c,d ª to © d, e ª that is not a function. 6. Suppose f : Z →Z is deﬁned as f = © (x,4x+5) : x ∈Z ª. State the domain, codomain and range of f . Find f (10). Injective and Surjective Functions 201 7. Consider the set f = © (x, y) ∈Z×Z : 3x + y = 4 ª. Is this a function from Z to Z? Explain. 8. Consider the set f = © (x, y) ∈Z×Z : x +3y = 4 ª. Is this a function from Z to Z? Explain. 9. Consider the set f = © (x2,x) : x ∈R ª. Is this a function from R to R? Explain. 10. Consider the set f = © (x3,x) : x ∈R ª. Is this a function from R to R? Explain. 11. Is the set θ = © (X,\|X\|) : X ⊆Z5 ª a function? If so, what is its domain and range? 12. Is the set θ = ©¡ (x, y),(3y,2x,x+ y) ¢ : x, y ∈R ª a function? If so, what is its domain, codomain and range? 12.2 Injective and Surjective Functions You may recall from algebra and calculus that a function may be one-to-one and onto, and these properties are related to whether or not the function is invertible. We now review these important ideas. In advanced mathematics, the word injective is often used instead of one-to-one, and surjective is used instead of onto. Here are the exact deﬁnitions: Deﬁnition 12.4 A function f : A →B is: 1. injective (or one-to-one) if for every x, y ∈A, x ̸= y implies f (x) ̸= f (y); 2. surjective (or onto) if for every b ∈B there is an a ∈A with f (a) = b; 3. bijective if f is both injective and surjective. Below is a visual description of Deﬁnition 12.4. In essence, injective means that unequal elements in A always get sent to unequal elements in B. Surjective means that every element of B has an arrow pointing to it, that is, it equals f (a) for some a in the domain of f . A A A A B B B B b b a x x y y Injective means that for any two x, y ∈A, this happens... ...and not this: Surjective means that for any b ∈B... ...this happens: 202 Functions For more concrete examples, consider the following functions from R to R. The function f (x) = x2 is not injective because −2 ̸= 2, but f (−2) = f (2). Nor is it surjective, for if b = −1 (or if b is any negative number), then there is no a ∈R with f (a) = b. On the other hand, g(x) = x3 is both injective and surjective, so it is also bijective. There are four possible injective/surjective combinations that a function may possess. This is illustrated in the following ﬁgure showing four functions from A to B. Functions in the ﬁrst column are injective, those in the second column are not injective. Functions in the ﬁrst row are surjective, those in the second row are not. A A A A B B B B a a a a b b b b c c c 1 1 1 1 2 2 2 2 3 3 3 (bijective) Injective Not injective Surjective Not surjective We note in passing that, according to the deﬁnitions, a function is surjective if and only if its codomain equals its range. Often it is necessary to prove that a particular function f : A →B is injective. For this we must prove that for any two elements x, y ∈A, the conditional statement (x ̸= y) ⇒ ¡ f (x) ̸= f (y) ¢ is true. The two main approaches for this are summarized below. How to show a function f : A →B is injective: Direct approach: Suppose x, y ∈A and x ̸= y. . . . Therefore f (x) ̸= f (y). Contrapositive approach: Suppose x, y ∈A and f (x) = f (y). . . . Therefore x = y. Of these two approaches, the contrapositive is often the easiest to use, especially if f is deﬁned by an algebraic formula. This is because the contrapositive approach starts with the equation f (x) = f (y) and proceeds Injective and Surjective Functions 203 to the equation x = y. In algebra, as you know, it is usually easier to work with equations than inequalities. To prove that a function is not injective, you must disprove the statement (x ̸= y) ⇒ ¡ f (x) ̸= f (y) ¢. For this it suﬃces to ﬁnd example of two elements x, y ∈A for which x ̸= y and f (x) = f (y). Next we examine how to prove that f : A →B is surjective. According to Deﬁnition 12.4, we must prove the statement ∀b ∈B,∃a ∈A, f (a) = b. In words, we must show that for any b ∈B, there is at least one a ∈A (which may depend on b) having the property that f (a) = b. Here is an outline. How to show a function f : A →B is surjective: Suppose b ∈B. [Prove there exists a ∈A for which f (a) = b.] In the second step, we have to prove the existence of an a for which f (a) = b. For this, just ﬁnding an example of such an a would suﬃce. (How to ﬁnd such an example depends on how f is deﬁned. If f is given as a formula, we may be able to ﬁnd a by solving the equation f (a) = b for a. Sometimes you can ﬁnd a by just plain common sense.) To show f is not surjective, we must prove the negation of ∀b ∈B,∃a ∈A, f (a) = b, that is, we must prove ∃b ∈B,∀a ∈A, f (a) ̸= b. The following examples illustrate these ideas. (For the ﬁrst example, note that the set R− © 0 ª is R with the number 0 removed.) Example 12.4 Show that the function f : R− © 0 ª →R deﬁned as f (x) = 1 x +1 is injective but not surjective. We will use the contrapositive approach to show that f is injective. Suppose x, y ∈R− © 0 ª and f (x) = f (y). This means 1 x +1 = 1 y +1. Subtracting 1 from both sides and inverting produces x = y. Therefore f is injective. Function f is not surjective because there exists an element b = 1 ∈R for which f (x) = 1 x +1 ̸= 1 for every x ∈R− © 0 ª. Example 12.5 Show that the function g : Z × Z →Z × Z deﬁned by the formula g(m,n) = (m+ n,m+2n), is both injective and surjective. We will use the contrapositive approach to show that g is injective. Thus we need to show that g(m,n) = g(k,ℓ) implies (m,n) = (k,ℓ). Suppose (m,n),(k,ℓ) ∈Z×Z and g(m,n) = g(k,ℓ). Then (m+n,m+2n) = (k+ℓ,k+2ℓ). It follows that m+ n = k+ℓand m+2n = k+2ℓ. Subtracting the ﬁrst equation from the second gives n = ℓ. Next, subtract n = ℓfrom m+ n = k +ℓto get m = k. Since m = k and n = ℓ, it follows that (m,n) = (k,ℓ). Therefore g is injective. 204 Functions To see that g is surjective, consider an arbitrary element (b, c) ∈Z×Z. We need to show that there is some (x, y) ∈Z×Z for which g(x, y) = (b, c). To ﬁnd (x, y), note that g(x, y) = (b, c) means (x+ y,x+2y) = (b, c). This leads to the following system of equations: x + y = b x + 2y = c. Solving gives x = 2b −c and y = c −b. Then (x, y) = (2b −c, c −b). We now have g(2b −c, c −b) = (b, c), and it follows that g is surjective. Example 12.6 Consider function h : Z×Z →Q deﬁned as h(m,n) = m \|n\|+1. Determine whether this is injective and whether it is surjective. This function is not injective because of the unequal elements (1,2) and (1,−2) in Z×Z for which h(1,2) = h(1,−2) = 1 3. However, h is surjective: Take any element b ∈Q. Then b = c d for some c,d ∈Z. Notice we may assume d is positive by making c negative, if necessary. Then h(c,d−1) = c \|d−1\|+1 = c d = b. Exercises for Section 12.2 1. Let A = © 1,2,3,4 ª and B = © a,b, c ª. Give an example of a function f : A →B that is neither injective nor surjective. 2. Consider the logarithm function ln : (0,∞) →R. Decide whether this function is injective and whether it is surjective. 3. Consider the cosine function cos : R →R. Decide whether this function is injective and whether it is surjective. What if it had been deﬁned as cos : R →[−1,1]? 4. A function f : Z →Z × Z is deﬁned as f (n) = (2n,n + 3). Verify whether this function is injective and whether it is surjective. 5. A function f : Z →Z is deﬁned as f (n) = 2n+1. Verify whether this function is injective and whether it is surjective. 6. A function f : Z × Z →Z is deﬁned as f (m,n) = 3n −4m. Verify whether this function is injective and whether it is surjective. 7. A function f : Z × Z →Z is deﬁned as f (m,n) = 2n −4m. Verify whether this function is injective and whether it is surjective. 8. A function f : Z×Z →Z×Z is deﬁned as f (m,n) = (m+ n,2m+ n). Verify whether this function is injective and whether it is surjective. 9. Prove that the function f : R− © 2 ª →R− © 5 ª deﬁned by f (x) = 5x+1 x−2 is bijective. 10. Prove the function f : R− © 1 ª →R− © 1 ª deﬁned by f (x) = µ x+1 x−1 ¶3 is bijective. 11. Consider the function θ : © 0,1 ª ×N →Z deﬁned as θ(a,b) = (−1)ab. Is θ injective? Is it surjective? Bijective? Explain. The Pigeonhole Principle 205 12. Consider the function θ : © 0,1 ª ×N →Z deﬁned as θ(a,b) = a−2ab+b. Is θ injective? Is it surjective? Bijective? Explain. 13. Consider the function f : R2 →R2 deﬁned by the formula f (x, y) = (xy,x3). Is f injective? Is it surjective? Bijective? Explain. 14. Consider the function θ : P(Z) →P(Z) deﬁned as θ(X) = X. Is θ injective? Is it surjective? Bijective? Explain. 15. This question concerns functions f : © A,B,C,D,E,F,G ª → © 1,2,3,4,5,6,7 ª. How many such functions are there? How many of these functions are injective? How many are surjective? How many are bijective? 16. This question concerns functions f : © A,B,C,D,E ª → © 1,2,3,4,5,6,7 ª. How many such functions are there? How many of these functions are injective? How many are surjective? How many are bijective? 17. This question concerns functions f : © A,B,C,D,E,F,G ª → © 1,2 ª. How many such functions are there? How many of these functions are injective? How many are surjective? How many are bijective? 18. Prove that the function f : N →Z deﬁned as f (n) = (−1)n(2n−1)+1 4 is bijective. 12.3 The Pigeonhole Principle Here is a simple but useful idea. Imagine there is a set A of pigeons and a set B of pigeon-holes, and all the pigeons ﬂy into the pigeon-holes. You can think of this as describing a function f : A →B, where pigeon X ﬂies into pigeon-hole f (X). Figure 12.4 illustrates this. Pigeons Pigeon-holes (a) f Pigeons Pigeon-holes (b) f Figure 12.4. The pigeonhole principle In Figure 12.4(a) there are more pigeons than pigeon-holes, and it is obvious that in such a case at least two pigeons have to ﬂy into the same pigeon-hole, meaning that f is not injective. In Figure 12.4(b) there are fewer pigeons than pigeon-holes, so clearly at least one pigeon-hole remains empty, meaning that f is not surjective. 206 Functions Although the underlying idea expressed by these ﬁgures has little to do with pigeons, it is nonetheless called the pigeonhole principle: The Pigeonhole Principle Suppose A and B are ﬁnite sets and f : A →B is any function. Then: 1. If \|A\| > \|B\|, then f is not injective. 2. If \|A\| < \|B\|, then f is not surjective. Though the pigeonhole principle is obvious, it can be used to prove some things that are not so obvious. Example 12.7 Prove the following proposition. Proposition If A is any set of 10 integers between 1 and 100, then there exist two diﬀerent subsets X ⊆A and Y ⊆A for which the sum of elements in X equals the sum of elements in Y. To illustrate what this proposition is saying, consider the random set A = © 5,7,12,11,17,50,51,80,90,100 ª of 10 integers between 1 and 100. Notice that A has subsets X = © 5,80 ª and Y = © 7,11,17,50 ª for which the sum of the elements in X equals the sum of those in Y. If we tried to “mess up” A by changing the 5 to a 6, we get A = © 6,7,12,11,17,50,51,80,90,100 ª which has subsets X = © 7,12,17,50 ª and Y = © 6,80 ª both of whose elements add up to the same number (86). The proposition asserts that this is always possible, no matter what A is. Here is a proof: Proof. Suppose A ⊆ © 1,2,3,4,...,99,100 ª and \|A\| = 10, as stated. Notice that if X ⊆A, then X has no more than 10 elements, each between 1 and 100, and therefore the sum of all the elements of X is less than 100·10 = 1000. Consider the function f : P(A) → © 0,1,2,3,4,...,1000 ª where f (X) is the sum of the elements in X. (Examples: f ¡© 3,7,50 ª¢ = 60; f ¡© 1,70,80,95 ª¢ = 246.) As \|P(A)\| = 210 = 1024 > 1001 = ¯ ¯© 0,1,2,3,...,1000 ª¯ ¯, it follows from the pigeonhole principle that f is not injective. Therefore there are two unequal sets X,Y ∈P(A) for which f (X) = f (Y ). In other words, there are subsets X ⊆A and Y ⊆A for which the sum of elements in X equals the sum of elements in Y. ■ The Pigeonhole Principle 207 Example 12.8 Prove the following proposition. Proposition There are at least two Texans with the same number of hairs on their heads. Proof. We will use two facts. First, the population of Texas is more than twenty million. Second, it is a biological fact that every human head has fewer than one million hairs. Let A be the set of all Texans, and let B = © 0,1,2,3,4,...,1000000 ª. Let f : A →B be the function for which f (x) equals the number of hairs on the head of x. Since \|A\| > \|B\|, the pigeonhole principle asserts that f is not injective. Thus there are two Texans x and y for whom f (x) = f (y), meaning that they have the same number of hairs on their heads. ■ Proofs that use the pigeonhole principle tend to be inherently non-constructive, in the sense discussed in Section 7.4. For example, the above proof does not explicitly give us of two Texans with the same number of hairs on their heads; it only shows that two such people exist. If we were to make a constructive proof, we could ﬁnd examples of two bald Texans. Then they have the same number of head hairs, namely zero. Exercises for Section 12.3 1. Prove that if six numbers are chosen at random, then at least two of them will have the same remainder when divided by 5. 2. Prove that if a is a natural number, then there exist two unequal natural numbers k and ℓfor which ak −aℓis divisible by 10. 3. Prove that if six natural numbers are chosen at random, then the sum or diﬀerence of two of them is divisible by 9. 4. Consider a square whose side-length is one unit. Select any ﬁve points from inside this square. Prove that at least two of these points are within p 2 2 units of each other. 5. Prove that any set of seven distinct natural numbers contains a pair of numbers whose sum or diﬀerence is divisible by 10. 6. Given a sphere S, a great circle of S is the intersection of S with a plane through its center. Every great circle divides S into two parts. A hemisphere is the union of the great circle and one of these two parts. Prove that if ﬁve points are placed arbitrarily on S, then there is a hemisphere that contains four of them. 7. Prove or disprove: Any subset X ⊆ © 1,2,3,...,2n ª with \|X\| > n contains two (unequal) elements a,b ∈X for which a \| b or b \| a. 208 Functions 12.4 Composition You should be familiar with the notion of function composition from algebra and calculus. Still, it is worthwhile to revisit it now with our more sophisticated ideas about functions. Deﬁnition 12.5 Suppose f : A →B and g : B →C are functions with the property that the codomain of f equals the domain of g. The composition of f with g is another function, denoted as g◦f and deﬁned as follows: If x ∈A, then g◦f (x) = g(f (x)). Therefore g◦f sends elements of A to elements of C, so g◦f : A →C. The following ﬁgure illustrates the deﬁnition. Here f : A →B, g : B →C, and g◦f : A →C. We have, for example, g◦f (0) = g(f (0)) = g(2) = 4. Be very careful with the order of the symbols. Even though g comes ﬁrst in the symbol g◦f , we work out g◦f (x) as g(f (x)), with f acting on x ﬁrst, followed by g acting on f (x). A A C C B 3 2 1 0 3 2 1 0 7 6 5 4 7 6 5 4 3 2 1 f g g ◦f Figure 12.5. Composition of two functions Notice that the composition g ◦f also makes sense if the range of f is a subset of the domain of g. You should take note of this fact, but to keep matters simple we will continue to emphasize situations where the codomain of f equals the domain of g. Example 12.9 Suppose A = © a,b, c ª, B = © 0,1 ª, C = © 1,2,3 ª. Let f : A →B be the function f = © (a,0),(b,1),(c,0) ª, and let g : B →C be the function g = © (0,3),(1,1) ª. Then g ◦f = © (a,3),(b,1),(c,3) ª. Example 12.10 Suppose A = © a,b, c ª, B = © 0,1 ª, C = © 1,2,3 ª. Let f : A →B be the function f = © (a,0),(b,1),(c,0) ª, and let g : C →B be the function g = © (1,0),(2,1),(3,1) ª. In this situation the composition g ◦f is not deﬁned because the codomain B of f is not the same set as the domain C of g. Composition 209 Remember: In order for g◦f to make sense, the codomain of f must equal the domain of g. (Or at least be a subset of it.) Example 12.11 Let f : R →R be deﬁned as f (x) = x2 + x, and g : R →R be deﬁned as g(x) = x + 1. Then g ◦f : R →R is the function deﬁned by the formula g ◦f (x) = g(f (x)) = g(x2 + x) = x2 + x+1. Since the domains and codomains of g and f are the same, we can in this case do a composition in the other order. Note that f ◦g : R →R is the function deﬁned as f ◦g(x) = f (g(x)) = f (x+1) = (x+1)2 +(x+1) = x2 +3x+2. This example illustrates that even when g◦f and f ◦g are both deﬁned, they are not necessarily equal. We can express this fact by saying function composition is not commutative. We close this section by proving several facts about function composition that you are likely to encounter in your future study of mathematics. First, we note that, although it is not commutative, function composition is associative. Theorem 12.1 Composition of functions is associative. That is if f : A →B, g : B →C and h : C →D, then (h◦g)◦f = h◦(g ◦f ). Proof. Suppose f , g,h are as stated. It follows from Deﬁnition 12.5 that both (h◦g)◦f and h◦(g ◦f ) are functions from A to D. To show that they are equal, we just need to show ³ (h◦g)◦f ´ (x) = ³ h◦(g ◦f ) ´ (x) for every x ∈A. Note that Deﬁnition 12.5 yields ³ (h◦g)◦f ´ (x) = (h◦g)(f (x)) = h(g(f (x)). Also ³ h◦(g ◦f ) ´ (x) = h(g ◦f (x)) = h(g(f (x))). Thus ³ (h◦g)◦f ´ (x) = ³ h◦(g ◦f ) ´ (x), as both sides equal h(g(f (x))). ■ Theorem 12.2 Suppose f : A →B and g : B →C. If both f and g are injective, then g◦f is injective. If both f and g are surjective, then g◦f is surjective. 210 Functions Proof. First suppose both f and g are injective. To see that g◦f is injective, we must show that g◦f (x) = g◦f (y) implies x = y. Suppose g◦f (x) = g◦f (y). This means g(f (x)) = g(f (y)). It follows that f (x) = f (y). (For otherwise g wouldn’t be injective.) But since f (x) = f (y) and f is injective, it must be that x = y. Therefore g ◦f is injective. Next suppose both f and g are surjective. To see that g◦f is surjective, we must show that for any element c ∈C, there is a corresponding element a ∈A for which g ◦f (a) = c. Thus consider an arbitrary c ∈C. Because g is surjective, there is an element b ∈B for which g(b) = c. And because f is surjective, there is an element a ∈A for which f (a) = b. Therefore g(f (a)) = g(b) = c, which means g ◦f (a) = c. Thus g ◦f is surjective. ■ Exercises for Section 12.4 1. Suppose A = © 5,6,8 ª, B = © 0,1 ª, C = © 1,2,3 ª. Let f : A →B be the function f = © (5,1),(6,0),(8,1) ª, and g : B →C be g = © (0,1),(1,1) ª. Find g ◦f . 2. Suppose A = © 1,2,3,4 ª, B = © 0,1,2 ª, C = © 1,2,3 ª. Let f : A →B be f = © (1,0),(2,1),(3,2),(4,0) ª , and g : B →C be g = © (0,1),(1,1),(2,3) ª. Find g ◦f . 3. Suppose A = © 1,2,3 ª. Let f : A →A be the function f = © (1,2),(2,2),(3,1) ª, and let g : A →A be the function g = © (1,3),(2,1),(3,2) ª. Find g ◦f and f ◦g. 4. Suppose A = © a,b, c ª. Let f : A →A be the function f = © (a, c),(b, c),(c, c) ª, and let g : A →A be the function g = © (a,a),(b,b),(c,a) ª. Find g ◦f and f ◦g. 5. Consider the functions f , g : R →R deﬁned as f (x) = 3 p x+1 and g(x) = x3. Find the formulas for g ◦f and f ◦g. 6. Consider the functions f , g : R →R deﬁned as f (x) = 1 x2+1 and g(x) = 3x+2. Find the formulas for g ◦f and f ◦g. 7. Consider the functions f , g : Z × Z →Z × Z deﬁned as f (m,n) = (mn,m2) and g(m,n) = (m+1,m+ n). Find the formulas for g ◦f and f ◦g. 8. Consider the functions f , g : Z×Z →Z×Z deﬁned as f (m,n) = (3m −4n,2m + n) and g(m,n) = (5m+ n,m). Find the formulas for g ◦f and f ◦g. 9. Consider the functions f : Z×Z →Z deﬁned as f (m,n) = m+ n and g : Z →Z×Z deﬁned as g(m) = (m,m). Find the formulas for g ◦f and f ◦g. 10. Consider the function f : R2 →R2 deﬁned by the formula f (x, y) = (xy,x3). Find a formula for f ◦f . Inverse Functions 211 12.5 Inverse Functions You may recall from calculus that if a function f is injective and surjective, then it has an inverse function f −1 that “undoes” the eﬀect of f in the sense that f −1(f (x)) = x for every x in the domain. (For example, if f (x) = x3, then f −1(x) = 3 px.) We now review these ideas. Our approach uses two ingredients, outlined in the following deﬁnitions. Deﬁnition 12.6 Given a set A, the identity function on A is the func-tion i A : A →A deﬁned as i A(x) = x for every x ∈A. Example: If A = © 1,2,3 ª, then i A = © (1,1),(2,2),(3,3) ª. Also iZ = © (n,n) : n ∈Z ª. The identity function on a set is the function that sends any element of the set to itself. Notice that for any set A, the identity function i A is bijective: It is injective because i A(x) = i A(y) immediately reduces to x = y. It is surjective because if we take any element b in the codomain A, then b is also in the domain A, and i A(b) = b. Deﬁnition 12.7 Given a relation R from A to B, the inverse relation of R is the relation from B to A deﬁned as R−1 = © (y,x) : (x, y) ∈R ª. In other words, the inverse of R is the relation R−1 obtained by interchanging the elements in every ordered pair in R. For example, let A = © a,b, c ª and B = © 1,2,3 ª, and suppose f is the relation f = © (a,2),(b,3),(c,1) ª from A to B. Then f −1 = © (2,a),(3,b),(1, c) ª and this is a relation from B to A. Notice that f is actually a function from A to B, and f −1 is a function from B to A. These two relations are drawn below. Notice the drawing for relation f −1 is just the drawing for f with arrows reversed. A A B B c b a c b a 3 2 1 3 2 1 f = © (a,2),(b,3),(c,1) ª f −1 = © (2,a),(3,b),(1, c) ª For another example, let A and B be the same sets as above, but consider the relation g = © (a,2),(b,3),(c,3) ª from A to B. Then g−1 = © (2,a),(3,b),(3, c) ª is a relation from B to A. These two relations are sketched below. 212 Functions A A B B c b a c b a 3 2 1 3 2 1 g = © (a,2),(b,3),(c,3) ª g−1 = © (2,a),(3,b),(3, c) ª This time, even though the relation g is a function, its inverse g−1 is not a function because the element 3 occurs twice as a ﬁrst coordinate of an ordered pair in g−1. In the above examples, relations f and g are both functions, and f −1 is a function and g−1 is not. This raises a question: What properties does f have and g lack that makes f −1 a function and g−1 not a function? The answer is not hard to see. Function g is not injective because g(b) = g(c) = 3, and thus (b,3) and (c,3) are both in g. This causes a problem with g−1 because it means (3,b) and (3, c) are both in g−1, so g−1 can’t be a function. Thus, in order for g−1 to be a function, it would be necessary that g be injective. But that is not enough. Function g also fails to be surjective because no element of A is sent to the element 1 ∈B. This means g−1 contains no ordered pair whose ﬁrst coordinate is 1, so it can’t be a function from B to A. If g−1 were to be a function it would be necessary that g be surjective. The previous two paragraphs suggest that if g is a function, then it must be bijective in order for its inverse relation g−1 to be a function. Indeed, this is easy to verify. Conversely, if a function is bijective, then its inverse relation is easily seen to be a function. We summarize this in the following theorem. Theorem 12.3 Let f : A →B be a function. Then f is bijective if and only if the inverse relation f −1 is a function from B to A. Suppose f : A →B is bijective, so according to the theorem f −1 is a function. Observe that the relation f contains all the pairs (x, f (x)) for x ∈A, so f −1 contains all the pairs (f (x),x). But (f (x),x) ∈f −1 means f −1(f (x)) = x. Therefore f −1◦f (x) = x for every x ∈A. From this we get f −1◦f = i A. Similar reasoning produces f ◦f −1 = iB. This leads to the following deﬁnitions. Deﬁnition 12.8 If f : A →B is bijective then its inverse is the function f −1 : B →A. Functions f and f −1 obey the equations f −1 ◦f = i A and f ◦f −1 = iB. Inverse Functions 213 You probably recall from algebra and calculus at least one technique for computing the inverse of a bijective function f : to ﬁnd f −1, start with the equation y = f (x). Then interchange variables to get x = f (y). Solving this equation for y (if possible) produces y = f −1(x). The next two examples illustrate this. Example 12.12 The function f : R →R deﬁned as f (x) = x3 +1 is bijective. Find its inverse. We begin by writing y = x3 +1. Now interchange variables to obtain x = y3 +1. Solving for y produces y = 3 p x−1. Thus f −1(x) = 3 p x−1. (You can check your answer by computing f −1(f (x)) = 3 p f (x)−1 = 3 p x3 +1−1 = x. Therefore f −1(f (x)) = x. Any answer other than x indicates a mistake.) We close with one ﬁnal example. Example 12.5 showed that the function g : Z×Z →Z×Z deﬁned by the formula g(m,n) = (m+ n,m+2n) is bijective. Let’s ﬁnd its inverse. The approach outlined above should work, but we need to be careful to keep track of coordinates in Z × Z. We begin by writing (x, y) = g(m,n), then interchanging the variables (x, y) and (m,n) to get (m,n) = g(x, y). This gives (m,n) = (x+ y,x+2y), from which we get the following system of equations: x + y = m x + 2y = n. Solving this system using techniques from algebra with which you are familiar, we get x = 2m−n y = n−m. Then (x, y) = (2m−n,n−m), so g−1(m,n) = (2m−n,n−m). 214 Functions We can check our work by conﬁrming that g−1(g(m,n)) = (m,n). Doing the math, g−1(g(m,n)) = g−1(m+ n,m+2n) = ¡ 2(m+ n)−(m+2n),(m+2n)−(m+ n) ¢ = (m,n). Exercises for Section 12.5 1. Check that the function f : Z →Z deﬁned by f (n) = 6 −n is bijective. Then compute f −1. 2. In Exercise 9 of Section 12.2 you proved that f : R − © 2 ª →R − © 5 ª deﬁned by f (x) = 5x+1 x−2 is bijective. Now ﬁnd its inverse. 3. Let B = © 2n : n ∈Z ª = © ..., 1 4, 1 2,1,2,4,8,... ª. Show that the function f : Z →B deﬁned as f (n) = 2n is bijective. Then ﬁnd f −1. 4. The function f : R →(0,∞) deﬁned as f (x) = ex3+1 is bijective. Find its inverse. 5. The function f : R →R deﬁned as f (x) = πx−e is bijective. Find its inverse. 6. The function f : Z×Z →Z×Z deﬁned by the formula f (m,n) = (5m+4n,4m+3n) is bijective. Find its inverse. 7. Show that the function f : R2 →R2 deﬁned by the formula f (x, y) = ((x2 +1)y,x3) is bijective. Then ﬁnd its inverse. 8. Is the function θ : P(Z) →P(Z) deﬁned as θ(X) = X bijective? If so, what is its inverse? 9. Consider the function f : R×N →N×R deﬁned as f (x, y) = (y,3xy). Check that this is bijective; ﬁnd its inverse. 10. Consider f : N →Z deﬁned as f (n) = (−1)n(2n−1)+1 4 . This function is bijective by Exercise 18 in Section 12.2. Find its inverse. 12.6 Image and Preimage It is time to take up a matter of notation that you will encounter in future mathematics classes. Suppose we have a function f : A →B. If X ⊆A, the expression f (X) has a special meaning. It stands for the set © f (x) : x ∈X ª. Similarly, if Y ⊆B then f −1(Y ) has a meaning even if f is not invertible. The expression f −1(Y ) stands for the set © x ∈A : f (x) ∈Y ª. Here are the precise deﬁnitions. Image and Preimage 215 Deﬁnition 12.9 Suppose f : A →B is a function. 1. If X ⊆A, the image of X is the set f (X) = © f (x) : x ∈X ª ⊆B. 2. If Y ⊆B, the preimage of Y is the set f −1(Y ) = © x ∈A : f (x) ∈Y ª ⊆A. In words, the image f (X) of X is the set of all things in B that f sends elements of X to. (Roughly speaking, you might think of f (X) as a kind of distorted “copy” or “image” of X in B.) The preimage f −1(Y ) of Y is the set of all things in A that f sends into Y. Maybe you have already encountered these ideas in linear algebra, in a setting involving a linear transformation T : V →W between two vector spaces. If X ⊆V is a subspace of V, then its image T(X) is a subspace of W. If Y ⊆W is a subspace of W, then its preimage T−1(Y ) is a subspace of V. (If this does not sound familiar, then ignore it.) Example 12.13 Let f : © s,t,u,v,w,x, y, z ª → © 0,1,2,3,4,5,6,7,8,9 ª, where f = © (s,4),(t,8),(u,8),(v,1),(w,2),(x,4),(y,6),(z,4) ª . Notice that f is neither injective nor surjective, so it certainly is not invertible. Be sure you understand the following statements. 1. f ¡© s,t,u, z ª¢ = © 8,4 ª 2. f ¡© s,x, z ª¢ = © 4 ª 3. f ¡© s,v,w, y ª¢ = © 1,2,4,6 ª 4. f −1¡© 4 ª¢ = © s,x, z ª 5. f −1¡© 4,9 ª¢ = © s,x, z ª 6. f −1¡© 9 ª¢ = ; 7. f −1¡© 1,4,8 ª¢ = © s,t,u,v,x, z ª It is important to realize that the X and Y in Deﬁnition 12.9 are subsets (not elements!) of A and B. Note that in the above example we had f −1¡© 4 ª¢ = © s,x, z ª, while f −1(4) has absolutely no meaning because the inverse function f −1 does not exist. Likewise, there is a subtle diﬀerence between f ¡© s ª¢ = © 4 ª and f (s) = 4. Be careful. Example 12.14 Consider the function f : R →R deﬁned as f (x) = x2. Note that f ¡© 0,1,2 ª¢ = © 0,1,4 ª and f −1¡© 0,1,4 ª¢ = © −2,−1,0,1,2 ª. This shows f −1(f (X)) ̸= X in general. Using the same f , now check your understanding of the following statements involving images and preimages of intervals: f ([−2,3]) = [0,9], and f −1([0,9]) = [−3,3]. Also f (R) = [0,∞) and f −1([−2,−1]) = ;. 216 Functions If you continue with mathematics you are likely to encounter the following results. For now, you are asked to prove them in the exercises. Theorem 12.4 Suppose f : A →B is a function. Let W, X ⊆A, and Y ,Z ⊆B. Then: 1. f (W ∩X) ⊆f (W)∩f (X) 2. f (W ∪X) = f (W)∪f (X) 3. f −1(Y ∩Z) = f −1(Y )∩f −1(Z) 4. f −1(Y ∪Z) = f −1(Y )∪f −1(Z) 5. X ⊆f −1(f (X)) 6. f (f −1(Y )) ⊆Y. Exercises for Section 12.6 1. Consider the function f : R →R deﬁned as f (x) = x2 + 3. Find f ([−3,5]) and f −1([12,19]). 2. Consider the function f : © 1,2,3,4,5,6,7 ª → © 0,1,2,3,4,5,6,7,8,9 ª given as f = © (1,3),(2,8),(3,3),(4,1),(5,2),(6,4),(7,6) ª . Find: f ¡© 1,2,3 ª¢, f ¡© 4,5,6,7 ª¢, f (;), f −1¡© 0,5,9 ª¢ and f −1¡© 0,3,5,9 ª¢. 3. This problem concerns functions f : © 1,2,3,4,5,6,7 ª → © 0,1,2,3,4 ª. How many such functions have the property that ¯ ¯f −1¡© 3 ª¢¯ ¯ = 3? 4. This problem concerns functions f : © 1,2,3,4,5,6,7,8 ª → © 0,1,2,3,4,5,6 ª. How many such functions have the property that ¯ ¯f −1¡© 2 ª¢¯ ¯ = 4? 5. Consider a function f : A →B and a subset X ⊆A. We observed in Section 12.6 that f −1(f (X)) ̸= X in general. However X ⊆f −1(f (X)) is always true. Prove this. 6. Given a function f : A →B and a subset Y ⊆B, is f (f −1(Y )) = Y always true? Prove or give a counterexample. 7. Given a function f : A →B and subsets W, X ⊆A, prove f (W ∩X) ⊆f (W)∩f (X). 8. Given a function f : A →B and subsets W, X ⊆A, then f (W ∩X) = f (W)∩f (X) is false in general. Produce a counterexample. 9. Given a function f : A →B and subsets W, X ⊆A, prove f (W ∪X) = f (W)∪f (X). 10. Given f : A →B and subsets Y ,Z ⊆B, prove f −1(Y ∩Z) = f −1(Y )∩f −1(Z). 11. Given f : A →B and subsets Y ,Z ⊆B, prove f −1(Y ∪Z) = f −1(Y )∪f −1(Z). 12. Consider f : A →B. Prove that f is injective if and only if X = f −1(f (X)) for all X ⊆A. Prove that f is surjective if and only if f (f −1(Y )) = Y for all Y ⊆B. 13. Let f : A →B be a function, and X ⊆A. Prove or disprove: f ¡ f −1(f (X)) ¢ = f (X). 14. Letf : A →B be a function, and Y ⊆B. Prove or disprove: f −1¡ f (f −1(Y )) ¢ = f −1(Y ). CHAPTER 13 Cardinality of Sets T his chapter is all about cardinality of sets. At ﬁrst this looks like a very simple concept. To ﬁnd the cardinality of a set, just count its elements. If A = © a,b, c,d ª, then \|A\| = 4; if B = © n ∈Z : −5 ≤n ≤5 ª, then \|B\| = 11. In this case \|A\| < \|B\|. What could be simpler than that? Actually, the idea of cardinality becomes quite subtle when the sets are inﬁnite. The main point of this chapter is to explain how there are numerous diﬀerent kinds of inﬁnity, and some inﬁnities are bigger than others. Two sets A and B can both have inﬁnite cardinality, yet \|A\| < \|B\|. 13.1 Sets with Equal Cardinalities We begin with a discussion of what it means for two sets to have the same cardinality. Up until this point we’ve said \|A\| = \|B\| if A and B have the same number of elements: Count the elements of A, then count the elements of B. If you get the same number, then \|A\| = \|B\|. Although this is a ﬁne strategy if the sets are ﬁnite (and not too big!), it doesn’t apply to inﬁnite sets because we’d never be done counting their elements. We need a new approach that applies to both ﬁnite and inﬁnite sets. Here it is: Deﬁnition 13.1 Two sets A and B have the same cardinality, written \|A\| = \|B\|, if there exists a bijective function f : A →B. If no such bijective function exists, then the sets have unequal cardinalities, that is, \|A\| ̸= \|B\|. e d c b a 4 3 2 1 0 A B f The above picture illustrates our deﬁnition. There is a bijective function f : A →B, so \|A\| = \|B\|. The function f matches up A with B. Think of f as describing how to overlay A onto B so that they ﬁt together perfectly. 218 Cardinality of Sets On the other hand, if A and B are as indicated in either of the following ﬁgures, then there can be no bijection f : A →B. (The best we can do is a function that is either injective or surjective, but not both). Therefore the deﬁnition says \|A\| ̸= \|B\| in these cases. d c b a 4 3 2 1 0 A B f d c b a e 3 2 1 0 A B f Example 13.1 The sets A = © n ∈Z : 0 ≤n ≤5 ª and B = © n ∈Z : −5 ≤n ≤0 ª have the same cardinality because there is a bijective function f : A →B given by the rule f (n) = −n. Several comments are in order. First, if \|A\| = \|B\|, there can be lots of bijective functions from A to B. We only need to ﬁnd one of them in order to conclude \|A\| = \|B\|. Second, as bijective functions play such a big role here, we use the word bijection to mean bijective function. Thus the function f (n) = −n from Example 13.1 is a bijection. Also, an injective function is called an injection and a surjective function is called a surjection. We emphasize and reiterate that Deﬁnition 13.1 applies to ﬁnite as well as inﬁnite sets. If A and B are inﬁnite, then \|A\| = \|B\| provided there exists a bijection f : A →B. If no such bijection exists, then \|A\| ̸= \|B\|. Example 13.2 This example shows that \|N\| = \|Z\|. To see why this is true, notice that the following table describes a bijection f : N →Z. n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 . . . f (n) 0 1 −1 2 −2 3 −3 4 −4 5 −5 6 −6 7 −7 . . . Notice that f is described in such a way that it is both injective and surjective. Every integer appears exactly once on the inﬁnitely long second row. Thus, according to the table, given any b ∈Z there is some natural number n with f (n) = b, so f is surjective. It is injective because the way the table is constructed forces f (m) ̸= f (n) whenever m ̸= n. Because of this bijection f : N →Z, we must conclude from Deﬁnition 13.1 that \|N\| = \|Z\|. Example 13.2 may seem slightly unsettling. On one hand it makes sense that \|N\| = \|Z\| because N and Z are both inﬁnite, so their cardinalities are both “inﬁnity.” On the other hand, Z may seem twice as large as Sets with Equal Cardinalities 219 N because Z has all the negative integers as well as the positive ones. Deﬁnition 13.1 settles the issue. Because the bijection f : N →Z matches up N with Z, it follows that \|N\| = \|Z\|. We summarize this with a theorem. Theorem 13.1 There exists a bijection f : N →Z. Therefore \|N\| = \|Z\|. The fact that N and Z have the same cardinality might prompt us compare the cardinalities of other inﬁnite sets. How, for example, do N and R compare? Let’s turn our attention to this. In fact, \|N\| ̸= \|R\|. This was ﬁrst recognized by Georg Cantor (1845–1918), who devised an ingenious argument to show that there are no surjective functions f : N →R. (This in turn implies that there can be no bijections f : N →R, so \|N\| ̸= \|R\| by Deﬁnition 13.1.) We now describe Cantor’s argument for why there are no surjections f : N →R. We will reason informally, rather than writing out an exact proof. Take any arbitrary function f : N →R. Here’s why f can’t be surjective: Imagine making a table for f , where values of n in N are in the left-hand column and the corresponding values f (n) are on the right. The ﬁrst few entries might look something as follows. In this table, the real numbers f (n) are written with all their decimal places trailing oﬀto the right. Thus, even though f (1) happens to be the real number 0.4, we write it as 0.40000000...., etc. n f (n) 1 0 . 4 0 0 0 0 0 0 0 0 0 0 0 0 0. . . 2 8 . 5 0 0 6 0 7 0 8 6 6 6 9 0 0. . . 3 7 . 5 0 5 0 0 9 4 0 0 4 4 1 0 1. . . 4 5 . 5 0 7 0 4 0 0 8 0 4 8 0 5 0. . . 5 6 . 9 0 0 2 6 0 0 0 0 0 0 5 0 6. . . 6 6 . 8 2 8 0 9 5 8 2 0 5 0 0 2 0. . . 7 6 . 5 0 5 0 5 5 5 0 6 5 5 8 0 8. . . 8 8 . 7 2 0 8 0 6 4 0 0 0 0 4 4 8. . . 9 0 . 5 5 0 0 0 0 8 8 8 8 0 0 7 7. . . 10 0 . 5 0 0 2 0 7 2 2 0 7 8 0 5 1. . . 11 2 . 9 0 0 0 0 8 8 0 0 0 0 9 0 0. . . 12 6 . 5 0 2 8 0 0 0 8 0 0 9 6 7 1. . . 13 8 . 8 9 0 0 8 0 2 4 0 0 8 0 5 0. . . 14 8 . 5 0 0 0 8 7 4 2 0 8 0 2 2 6. . . . . . . . . 220 Cardinality of Sets There is a diagonal shaded band in the table. For each n ∈N, this band covers the nth decimal place of f (n): The 1st decimal place of f (1) is the 1st entry on the diagonal. The 2nd decimal place of f (2) is the 2nd entry on the diagonal. The 3rd decimal place of f (3) is the 3rd entry on the diagonal. The 4th decimal place of f (4) is the 4th entry on the diagonal, etc. The diagonal helps us construct a number b ∈R that is unequal to any f (n). Just let the nth decimal place of b diﬀer from the nth entry of the diagonal. Then the nth decimal place of b diﬀers from the nth decimal place of f (n). In order to be deﬁnite, deﬁne b to be the positive number less than 1 whose nth decimal place is 0 if the nth decimal place of f (n) is not 0, and whose nth decimal place is 1 if the nth decimal place of f (n) equals 0. Thus, for the function f illustrated in the above table, we have b = 0.01010001001000... and b has been deﬁned so that, for any n ∈N, its nth decimal place is unequal to the nth decimal place of f (n). Therefore f (n) ̸= b for every natural number n, meaning f is not surjective. Since this argument applies to any function f : N →R (not just the one in the above example) we conclude that there exist no bijections f : N →R, so \|N\| ̸= \|R\| by Deﬁnition 13.1. We summarize this as a theorem. Theorem 13.2 There exists no bijection f : N →R. Therefore \|N\| ̸= \|R\|. This is our ﬁrst indication of how there are diﬀerent kinds of inﬁnities. Both N and R are inﬁnite sets, yet \|N\| ̸= \|R\|. We will continue to develop this theme throughout this chapter. The next example shows that the intervals (0,∞) and (0,1) on R have the same cardinality. ∞ 1          1 x P 0 −1 f (x) Figure 13.1. A bijection f : (0,∞) →(0,1) Sets with Equal Cardinalities 221 Example 13.3 Show that \|(0,∞)\| = \|(0,1)\|. To accomplish this, we need to show that there is a bijection f : (0,∞) →(0,1). We describe this function geometrically. Consider the interval (0,∞) as the positive x-axis of R2. Let the interval (0,1) be on the y-axis as illustrated in Figure 13.1, so that (0,∞) and (0,1) are perpendicular to each other. The ﬁgure also shows a point P = (−1,1). Deﬁne f (x) to be the point on (0,1) where the line from P to x ∈(0,∞) intersects the y-axis. By similar triangles, we have 1 x+1 = f (x) x , and therefore f (x) = x x+1. If it is not clear from the ﬁgure that f : (0,∞) →(0,1) is bijective, then you can verify it using the techniques from Section 12.2. (Exercise 16, below.) It is important to note that equality of cardinalities is an equivalence relation on sets: it is reﬂexive, symmetric and transitive. Let us conﬁrm this. Given a set A, the identity function A →A is a bijection, so \|A\| = \|A\|. (This is the reﬂexive property.) For the symmetric property, if \|A\| = \|B\|, then there is a bijection f : A →B, and its inverse is a bijection f −1 : B →A, so \|B\| = \|A\|. For transitivity, suppose \|A\| = \|B\| and \|B\| = \|C\|. Then there are bijections f : A →B and g : B →C. The composition g ◦f : A →C is a bijection (Theorem 12.2), so \|A\| = \|C\|. The transitive property can be useful. If, in trying to show two sets A and C have the same cardinality, we can produce a third set B for which \|A\| = \|B\| and \|B\| = \|C\|, then transitivity assures us that indeed \|A\| = \|C\|. The next example uses this idea. Example 13.4 Show that \|R\| = \|(0,1)\|. Because of the bijection g : R →(0,∞) where g(x) = 2x, we have \|R\| = \|(0,∞)\|. Also, Example 13.3 shows that \|(0,∞)\| = \|(0,1)\|. Therefore \|R\| = \|(0,1)\|. So far in this chapter we have declared that two sets have “the same cardinality” if there is a bijection between them. They have “diﬀerent cardinalities” if there exists no bijection between them. Using this idea, we showed that \|Z\| = \|N\| ̸= \|R\| = \|(0,∞)\| = \|(0,1)\|. So, we have a means of determining when two sets have the same or diﬀerent cardinalities. But we have neatly avoided saying exactly what cardinality is. For example, we can say that \|Z\| = \|N\|, but what exactly is \|Z\|, or \|N\|? What exactly are these things that are equal? Certainly not numbers, for they are too big. 222 Cardinality of Sets And saying they are “inﬁnity” is not accurate, because we now know that there are diﬀerent types of inﬁnity. So just what kind of mathematical entity is \|Z\|? In general, given a set X, exactly what is its cardinality \|X\|? This is a lot like asking what a number is. A number, say 5, is an abstraction, not a physical thing. Early in life we instinctively grouped together certain sets of things (ﬁve apples, ﬁve oranges, etc.) and conceived of 5 as the thing common to all such sets. In a very real sense, the number 5 is an abstraction of the fact that any two of these sets can be matched up via a bijection. That is, it can be identiﬁed with a certain equivalence class of sets under the "has the same cardinality as" relation. (Recall that this is an equivalence relation.) This is easy to grasp because our sense of numeric quantity is so innate. But in exactly the same way we can say that the cardinality of a set X is what is common to all sets that can be matched to X via a bijection. This may be harder to grasp, but it is really no diﬀerent from the idea of the magnitude of a (ﬁnite) number. In fact, we could be concrete and deﬁne \|X\| to be the equivalence class of all sets whose cardinality is the same as that of X. This has the advantage of giving an explicit meaning to \|X\|. But there is no harm in taking the intuitive approach and just interpreting the cardinality \|X\| of a set X to be a measure the “size” of X. The point of this section is that we have a means of deciding whether two sets have the same size or diﬀerent sizes. Exercises for Section 13.1 A. Show that the two given sets have equal cardinality by describing a bijection from one to the other. Describe your bijection with a formula (not as a table). 1. R and (0,∞) 2. R and ( p 2,∞) 3. R and (0,1) 4. The set of even integers and the set of odd integers 5. A = © 3k : k ∈Z ª and B = © 7k : k ∈Z ª 6. N and S = © p 2 n : n ∈N ª 7. Z and S = © ..., 1 8, 1 4, 1 2,1,2,4,8,16,... ª 8. Z and S = © x ∈R : sinx = 1 ª 9. © 0,1 ª ×N and N 10. © 0,1 ª ×N and Z 11. [0,1] and (0,1) 12. N and Z (Suggestion: use Exercise 18 of Section 12.2.) 13. P(N) and P(Z) (Suggestion: use Exercise 12, above.) 14. N×N and © (n,m) ∈N×N : n ≤m ª B. Answer the following questions concerning bijections from this section. 15. Find a formula for the bijection f in Example 13.2 (page 218). 16. Verify that the function f in Example 13.3 is a bijection. Countable and Uncountable Sets 223 13.2 Countable and Uncountable Sets Let’s summarize the main points from the previous section. 1. \|A\| = \|B\| if and only if there exists a bijection A →B. 2. \|N\| = \|Z\| because there exists a bijection N →Z. 3. \|N\| ̸= \|R\| because there exists no bijection N →R. Thus, even though N, Z and R are all inﬁnite sets, their cardinalities are not all the same. The sets N and Z have the same cardinality, but R’s cardinality is diﬀerent from that of both the other sets. This means inﬁnite sets can have diﬀerent sizes. We now make some deﬁnitions to put words and symbols to this phenomenon. In a certain sense you can count the elements of N; you can count its elements oﬀas 1,2,3,4,..., but you’d have to continue this process forever to count the whole set. Thus we will call N a countably inﬁnite set, and the same term is used for any set whose cardinality equals that of N. Deﬁnition 13.2 Suppose A is a set. Then A is countably inﬁnite if \|N\| = \|A\|, that is, if there exists a bijection N →A. The set A is uncountable if A is inﬁnite and \|N\| ̸= \|A\|, that is, if A is inﬁnite and there exists no bijection N →A. Thus Z is countably inﬁnite but R is uncountable. This section deals mainly with countably inﬁnite sets. Uncountable sets are treated later. If A is countably inﬁnite, then \|N\| = \|A\|, so there is a bijection f : N →A. You can think of f as “counting” the elements of A. The ﬁrst element of A is f (1), followed by f (2), then f (3) and so on. It makes sense to think of a countably inﬁnite set as the smallest type of inﬁnite set, because if the counting process stopped, the set would be ﬁnite, not inﬁnite; a countably inﬁnite set has the fewest elements that a set can have and still be inﬁnite. It is common to reserve the special symbol ℵ0 to stand for the cardinality of countably inﬁnite sets. Deﬁnition 13.3 The cardinality of the natural numbers is denoted as ℵ0. That is, \|N\| = ℵ0. Thus any countably inﬁnite set has cardinality ℵ0. (The symbol ℵis the ﬁrst letter in the Hebrew alphabet, and is pronounced “aleph.” The symbol ℵ0 is pronounced “aleph naught.”) The summary of facts at the beginning of this section shows \|Z\| = ℵ0 and \|R\| ̸= ℵ0. Example 13.5 Let E = © 2k : k ∈Z ª be the set of even integers. The function f : Z →E deﬁned as f (n) = 2n is easily seen to be a bijection, so we have \|Z\| = \|E\|. Thus, as \|N\| = \|Z\| = \|E\|, the set E is countably inﬁnite and \|E\| = ℵ0. 224 Cardinality of Sets Here is a signiﬁcant fact: The elements of any countably inﬁnite set A can be written in an inﬁnitely long list a1,a2,a3,a4,... that begins with some element a1 ∈A and includes every element of A. For example, the set E in the above example can be written in list form as 0,2,−2,4,−4,6,−6,8,−8,... The reason that this can be done is as follows. Since A is countably inﬁnite, Deﬁnition 13.2 says there is a bijection f : N →A. This allows us to list out the set A as an inﬁnite list f (1), f (2), f (3), f (4),... Conversely, if the elements of A can be written in list form as a1,a2,a3,..., then the function f : N →A deﬁned as f (n) = an is a bijection, so A is countably inﬁnite. We summarize this as follows. Theorem 13.3 A set A is countably inﬁnite if and only if its elements can be arranged in an inﬁnite list a1,a2,a3,a4,... As an example of how this theorem might be used, let P denote the set of all prime numbers. Since we can list its elements as 2,3,5,7,11,13,..., it follows that the set P is countably inﬁnite. As another consequence of Theorem 13.3, note that we can interpret the fact that the set R is not countably inﬁnite as meaning that it is impossible to write out all the elements of R in an inﬁnite list. (After all, we tried to do that in the table on page 219, and failed!) This raises a question. Is it also impossible to write out all the elements of Q in an inﬁnite list? In other words, is the set Q of rational numbers countably inﬁnite or uncountable? If you start plotting the rational num-bers on the number line, they seem to mostly ﬁll up R. Sure, some numbers such as p 2, π and e will not be plotted, but the dots representing rational numbers seem to predominate. We might thus expect Q to be uncountable. However, it is a surprising fact that Q is countable. The proof presented below arranges all the rational numbers in an inﬁnitely long list. Theorem 13.4 The set Q of rational numbers is countably inﬁnite. Proof. To prove this, we just need to show how to write the set Q in list form. Begin by arranging all rational numbers in an inﬁnite array. This is done by making the following chart. The top row has a list of all integers, beginning with 0, then alternating signs as they increase. Each column headed by an integer k contains all the fractions (in reduced form) with numerator k. For example, the column headed by 2 contains the fractions 2 1, 2 3, 2 5, 2 7,..., and so on. It does not contain 2 2, 2 4, 2 6, etc., because those are not reduced, and in fact their reduced forms appear in the column headed by 1. You should examine this table and convince yourself that it contains all rational numbers in Q. Countable and Uncountable Sets 225 0 1 −1 2 −2 3 −3 4 −4 5 −5 ··· 0 1 1 1 −1 1 2 1 −2 1 3 1 −3 1 4 1 −4 1 5 1 −5 1 ··· 1 2 −1 2 2 3 −2 3 3 2 −3 2 4 3 −4 3 5 2 −5 2 ··· 1 3 −1 3 2 5 −2 5 3 4 −3 4 4 5 −4 5 5 3 −5 3 ··· 1 4 −1 4 2 7 −2 7 3 5 −3 5 4 7 −4 7 5 4 −5 4 ··· 1 5 −1 5 2 9 −2 9 3 7 −3 7 4 9 −4 9 5 6 −5 6 ··· 1 6 −1 6 2 11 −2 11 3 8 −3 8 4 11 −4 11 5 7 −5 7 ··· 1 7 −1 7 2 13 −2 13 3 10 −3 10 4 13 −4 13 5 8 −5 8 ··· . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... Next, draw an inﬁnite path in this array, beginning at 0 1 and snaking back and forth as indicated below. Every rational number is on this path. 0 1 −1 2 −2 3 −3 4 −4 5 −5 ··· 0 1 1 1 −1 1 2 1 −2 1 3 1 −3 1 4 1 −4 1 5 1 −5 1 ··· 1 2 −1 2 2 3 −2 3 3 2 −3 2 4 3 −4 3 5 2 −5 2 ··· 1 3 −1 3 2 5 −2 5 3 4 −3 4 4 5 −4 5 5 3 −5 3 ··· 1 4 −1 4 2 7 −2 7 3 5 −3 5 4 7 −4 7 5 4 −5 4 ··· 1 5 −1 5 2 9 −2 9 3 7 −3 7 4 9 −4 9 5 6 −5 6 ··· 1 6 −1 6 2 11 −2 11 3 8 −3 8 4 11 −4 11 5 7 −5 7 ··· 1 7 −1 7 2 13 −2 13 3 10 −3 10 4 13 −4 13 5 8 −5 8 ··· 1 8 −1 8 2 15 −2 15 3 11 −3 11 4 15 −4 15 5 9 −5 9 ··· 1 9 −1 9 2 17 −2 17 3 13 −3 13 4 17 −4 17 5 11 −5 11 ··· . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... 226 Cardinality of Sets Beginning at 0 1 and following the path, we get an inﬁnite list of all rational numbers: 0, 1, 1 2, −1 2, −1, 2, 2 3, 2 5, −1 3, 1 3, 1 4, −1 4, 2 7, −2 7, −2 5, −2 3, −2 3, −2, 3, 3 2, ... By Theorem 13.3, it follows that Q is countably inﬁnite, that is, \|Q\| = \|N\|. ■ It is also true that the Cartesian product of two countably inﬁnite sets is itself countably inﬁnite, as our next theorem states. Theorem 13.5 If A and B are both countably inﬁnite, then so is A ×B. Proof. Suppose A and B are both countably inﬁnite. By Theorem 13.3, we know we can write A and B in list form as A = © a1,a2,a3,a4,... ª , B = © b1,b2,b3,b4,... ª . Figure 13.2 shows how to form an inﬁnite path winding through all of A×B. Therefore A ×B can be written in list form, so it is countably inﬁnite. ■ (a1,b1) (a1,b2) (a1,b3) (a1,b4) (a1,b5) (a1,b6) (a1,b7) (a2,b1) (a2,b2) (a2,b3) (a2,b4) (a2,b5) (a2,b6) (a2,b7) (a3,b1) (a3,b2) (a3,b3) (a3,b4) (a3,b5) (a3,b6) (a3,b7) (a4,b1) (a4,b2) (a4,b3) (a4,b4) (a4,b5) (a4,b6) (a4,b7) (a5,b1) (a5,b2) (a5,b3) (a5,b4) (a5,b5) (a5,b6) (a5,b7) (a6,b1) (a6,b2) (a6,b3) (a6,b4) (a6,b5) (a6,b6) (a6,b7) (a7,b1) (a7,b2) (a7,b3) (a7,b4) (a7,b5) (a7,b6) (a7,b7) a1 a2 a3 a4 a5 a6 a7 b1 b2 b3 b4 b5 b6 b7 . . . . . . . . . . . . . . . . . . . . . . . . ··· ··· ··· ··· ··· ··· ··· ··· A B Figure 13.2. A product of two countably inﬁnite sets is countably inﬁnite Countable and Uncountable Sets 227 As an example of a consequence of this theorem, notice that since Q is countably inﬁnite, the set Q×Q is also countably inﬁnite. Recall that the word “corollary” means a result that follows easily from some other result. We have the following corollary of Theorem 13.5. Corollary 13.1 Given n countably inﬁnite sets A1, A2, A3,..., An, with n ≥2, the Cartesian product A1 × A2 × A3 ×···× An is also countably inﬁnite. Proof. The proof is by induction on n. For the basis step, notice that when n = 2 the statement asserts that for countably inﬁnite sets A1 and A2, the product A1 × A2 is countably inﬁnite, and this is true by Theorem 13.5. Assume that for k ≥2, any product A1 × A2 × A3 ×··· ×Ak of countably inﬁnite sets is countably inﬁnite. Consider a product A1×A2×A3×···×Ak+1 of k +1 countably inﬁnite sets. It is easily conﬁrmed that the function f : A1 × A2 × A3 ×···× Ak × Ak+1 − → (A1 × A2 × A3 ×···× Ak)× Ak+1 f (x1,x2,...,xk,xk+1) = ¡ (x1,x2,...,xk), xk+1 ¢ is bijective, so \|A1 × A2 × A3 ×···× Ak × Ak+1\| = \|(A1 × A2 × A3 ×···× Ak)× Ak+1\|. By the induction hypothesis, (A1 × A2 × A3 ×···× Ak)× Ak+1 is a product of two countably inﬁnite sets, so it is countably inﬁnite by Theorem 13.5. As noted above, A1 × A2 × A3 ×···× Ak × Ak+1 has the same cardinality, so it too is countably inﬁnite. ■ Theorem 13.6 If A and B are both countably inﬁnite, then A ∪B is countably inﬁnite. Proof. Suppose A and B are both countably inﬁnite. By Theorem 13.3, we know we can write A and B in list form as A = © a1,a2,a3,a4,... ª , B = © b1,b2,b3,b4,... ª . We can “shuﬄe” A and B into one inﬁnite list for A ∪B as follows. A ∪B = © a1,b1,a2,b2,a3,b3,a4,b4,... ª . (We agree not to list an element twice if it belongs to both A and B.) Therefore, by Theorem 13.3, it follows that A ∪B is countably inﬁnite. ■ 228 Cardinality of Sets Exercises for Section 13.2 1. Prove that the set A = © ln(n) : n ∈N ª ⊆R is countably inﬁnite. 2. Prove that the set A = © (m,n) ∈N×N : m ≤n ª is countably inﬁnite. 3. Prove that the set A = © (5n,−3n) : n ∈Z ª is countably inﬁnite. 4. Prove that the set of all irrational numbers is uncountable. (Suggestion: Consider proof by contradiction using Theorems 13.4 and 13.6.) 5. Prove or disprove: There exists a countably inﬁnite subset of the set of irrational numbers. 6. Prove or disprove: There exists a bijective function f : Q →R. 7. Prove or disprove: The set Q100 is countably inﬁnite. 8. Prove or disprove: The set Z×Q is countably inﬁnite. 9. Prove or disprove: The set © 0,1 ª ×N is countably inﬁnite. 10. Prove or disprove: The set A = © p 2 n : n ∈N ª countably inﬁnite. 11. Describe a partition of N that divides N into eight countably inﬁnite subsets. 12. Describe a partition of N that divides N into ℵ0 countably inﬁnite subsets. 13. Prove or disprove: If A = {X ⊆N : X is ﬁnite}, then \|A\| = ℵ0. 14. Suppose A = © (m,n) ∈N×R : n = πm ª. Is it true that \|N\| = \|A\|? 15. Theorem 13.5 implies that N×N is countably inﬁnite. Construct an alternate proof of this fact by showing that the function ϕ : N×N →N deﬁned as ϕ(m,n) = 2n−1(2m−1) is bijective. 13.3 Comparing Cardinalities At this point we know that there are at least two diﬀerent kinds of inﬁnity. On one hand, there are countably inﬁnite sets such as N, of cardinality ℵ0. Then there is the uncountable set R. Are there other kinds of inﬁnity beyond these two kinds? The answer is “yes,” but to see why we ﬁrst need to introduce some new deﬁnitions and theorems. Our ﬁrst task will be to formulate a deﬁnition for what we mean by \|A\| < \|B\|. Of course if A and B are ﬁnite we know exactly what this means: \|A\| < \|B\| means that when the elements of A and B are counted, A is found to have fewer elements than B. But this process breaks down if A or B is inﬁnite, for then the elements can’t be counted. The language of functions helps us overcome this diﬃculty. Notice that for ﬁnite sets A and B it is intuitively clear that \|A\| < \|B\| if and only if there exists an injective function f : A →B but there are no surjective functions f : A →B. The following diagram illustrates this: Comparing Cardinalities 229 d c b a 4 3 2 1 0 A B f We will use this idea to deﬁne what is meant by \|A\| < \|B\| and \|A\| ≤\|B\|. For emphasis, the following deﬁnition also restates what is meant by \|A\| = \|B\|. Deﬁnition 13.4 Suppose A and B are sets. (1) \|A\| = \|B\| means there is a bijection A →B. (2) \|A\| < \|B\| means there is an injection A →B, but no surjection A →B. (3) \|A\| ≤\|B\| means \|A\| < \|B\| or \|A\| = \|B\|. For example, consider N and R. The function f : N →R deﬁned as f (n) = n is clearly injective, but it is not surjective because given the element 1 2 ∈R, we have f (n) ̸= 1 2 for every n ∈N. In fact, Theorem 13.2 of Section 13.1 asserts that there is no surjection N →R. Deﬁnition 13.4 yields \|N\| < \|R\|. (13.1) Said diﬀerently, ℵ0 < \|R\|. Is there a set X for which \|R\| < \|X\|? The answer is “yes,” and the next theorem explains why. It implies \|R\| < \|P(R)\|. (Recall that P(A) denotes the power set of A.) Theorem 13.7 If A is any set, then \|A\| < \|P(A)\|. Proof. Before beginning the proof, we remark that this statement is obvious if A is ﬁnite, for then \|A\| < 2\|A\| = \|P(A)\|. But our proof must apply to all sets A, both ﬁnite and inﬁnite, so it must use Deﬁnition 13.4. We prove the theorem with direct proof. Let A be an arbitrary set. According to Deﬁnition 13.4, to prove \|A\| < \|P(A)\| we must show that there is an injection f : A →P(A), but no surjection f : A →P(A). To see that there is an injection f : A →P(A), deﬁne f by the rule f (x) = © x ª. In words, f sends any element x of A to the one-element set © x ª ∈P(A). Then f : A →P(A) is injective, as follows. Suppose f (x) = f (y). Then © x ª = © y ª. Now, the only way that © x ª and © y ª can be equal is if x = y, so it follows that x = y. Thus f is injective. Next we need to show that there exists no surjection f : A →P(A). Suppose for the sake of contradiction that there does exist a surjection 230 Cardinality of Sets f : A →P(A). Notice that for any element x ∈A, we have f (x) ∈P(A), so f (x) is a subset of A. Thus f is a function that sends elements of A to subsets of A. It follows that for any x ∈A, either x is an element of the subset f (x) or it is not. Using this idea, deﬁne the following subset B of A: B = © x ∈A : x ∉f (x) ª ⊆A. Now since B ⊆A we have B ∈P(A), and since f is surjective there is an a ∈A for which f (a) = B. Now, either a ∈B or a ∉B. We will consider these two cases separately, and show that each leads to a contradiction. Case 1. If a ∈B, then the deﬁnition of B implies a ∉f (a), and since f (a) = B we have a ∉B, which is a contradiction. Case 2. If a ∉B, then the deﬁnition of B implies a ∈f (a), and since f (a) = B we have a ∈B, again a contradiction. Since the assumption that there is a surjection f : A →P(A) leads to a contradiction, we conclude that there are no such surjective functions. In conclusion, we have seen that there exists an injection A →P(A) but no surjection A →P(A), so Deﬁnition 13.4 implies that \|A\| < \|P(A)\|. ■ Beginning with the set A = N and applying Theorem 13.7 over and over again, we get the following chain of inﬁnite cardinalities. ℵ0 = \|N\| < \|P(N)\| < \|P(P(N))\| < \|P(P(P(N)))\| < ··· (13.2) Thus there is an inﬁnite sequence of diﬀerent types of inﬁnity, starting with ℵ0 and becoming ever larger. The set N is countable, and all the sets P(N), P(P(N)), etc., are uncountable. In the next section we will prove that \|P(N)\| = \|R\|. Thus \|N\| and \|R\| are the ﬁrst two entries in the chain (13.2) above. They are are just two relatively tame inﬁnities in a long list of other wild and exotic inﬁnities. Unless you plan on studying advanced set theory or the foundations of mathematics, you are unlikely to ever encounter any types of inﬁnity beyond ℵ0 and \|R\|. Still you will in future mathematics courses need to distinguish between countably inﬁnite and uncountable sets, so we close with two ﬁnal theorems that can help you do this. Theorem 13.8 An inﬁnite subset of a countably inﬁnite set is countably inﬁnite. Proof. Suppose A is an inﬁnite subset of the countably inﬁnite set B. Because B is countably inﬁnite, its elements can be written in a list Comparing Cardinalities 231 b1,b2,b3,b4,... Then we can also write A’s elements in list form by proceed-ing through the elements of B, in order, and selecting those that belong to A. Thus A can be written in list form, and since A is inﬁnite, its list will be inﬁnite. Consequently A is countably inﬁnite. ■ Theorem 13.9 If U ⊆A, and U is uncountable, then A is uncountable. Proof. Suppose for the sake of contradiction that U ⊆A, and U is uncount-able but A is not uncountable. Then since U ⊆A and U is inﬁnite, then A must be inﬁnite too. Since A is inﬁnite, and not uncountable, it must be countably inﬁnite. Then U is an inﬁnite subset of a countably inﬁnite set A, so U is countably inﬁnite by Theorem 13.8. Thus U is both uncountable and countably inﬁnite, a contradiction. ■ Theorems 13.8 and 13.9 can be useful when we need to decide whether a set is countably inﬁnite or uncountable. They sometimes allow us to decide its cardinality by comparing it to a set whose cardinality is known. For example, suppose we want to decide whether or not the set A = R2 is uncountable. Since the x-axis U = © (x,0) : x ∈R ª ⊆R2 has the same cardinality as R, it is uncountable. Theorem 13.9 implies that R2 is uncountable. Other examples can be found in the exercises. Exercises for Section 13.3 1. Suppose B is an uncountable set and A is a set. Given that there is a surjective function f : A →B, what can be said about the cardinality of A? 2. Prove that the set C of complex numbers is uncountable. 3. Prove or disprove: If A is uncountable, then \|A\| = \|R\|. 4. Prove or disprove: If A ⊆B ⊆C and A and C are countably inﬁnite, then B is countably inﬁnite. 5. Prove or disprove: The set © 0,1 ª ×R is uncountable. 6. Prove or disprove: Every inﬁnite set is a subset of a countably inﬁnite set. 7. Prove or disprove: If A ⊆B and A is countably inﬁnite and B is uncountable, then B −A is uncountable. 8. Prove or disprove: The set © (a1,a2,a3,...) : ai ∈Z} of inﬁnite sequences of integers is countably inﬁnite. 9. Prove that if A and B are ﬁnite sets with \|A\| = \|B\|, then any injection f : A →B is also a surjection. Show this is not necessarily true if A and B are not ﬁnite. 10. Prove that if A and B are ﬁnite sets with \|A\| = \|B\|, then any surjection f : A →B is also an injection. Show this is not necessarily true if A and B are not ﬁnite. 232 Cardinality of Sets 13.4 The Cantor-Bernstein-Schröeder Theorem An often used property of numbers is that if a ≤b and b ≤a, then a = b. It is reasonable to ask if the same property applies to cardinality. If \|A\| ≤\|B\| and \|B\| ≤\|A\|, is it true that \|A\| = \|B\|? This is in fact true, and this section’s goal is to prove it. This will yield an alternate (and highly eﬀective) method of proving that two sets have the same cardianlity. Recall (Deﬁnition 13.4) that \|A\| ≤\|B\| means that \|A\| < \|B\| or \|A\| = \|B\|. If \|A\| < \|B\| then (by Deﬁnition 13.4) there is an injection A →B. On the other hand, if \|A\| = \|B\|, then there is a bijection (hence also an injection) A →B. Thus \|A\| ≤\|B\| implies that there is an injection f : A →B. Likewise, \|B\| ≤\|A\| implies that there is an injection g : B →A. Our aim is to show that if \|A\| ≤\|B\| and \|B\| ≤\|A\|, then \|A\| = \|B\|. In other words, we aim to show that if there are injections f : A →B and g : B →A, then there is a bijection h : A →B. The proof of this fact, though not particularly diﬃcult, is not entirely trivial, either. The fact that f and g guarantee that such an h exists is called the the Cantor-Bernstein-Schröeder theorem. This theorem is very useful for proving two sets A and B have the same cardinality: it says that instead of ﬁnding a bijection A →B, it suﬃces to ﬁnd injections A →B and B →A. This is useful because injections are often easier to ﬁnd than bijections. We will prove the Cantor-Bernstein-Schröeder theorem, but before doing so let’s work through an informal visual argument that will guide us through (and illustrate) the proof. Suppose there are injections f : A →B and g : B →A. We want to use them to produce a bijection h : A →B. Sets A and B are sketched below. For clarity, each has the shape of the letter that denotes it, and to help distinguish them the set A is shaded. A B Figure 13.3. The sets A and B The injections f : A →B and g : B →A are illustrated in Figure 13.4. Think of f as putting a “copy” f (A) = © f (x) : x ∈A ª of A into B, as illustrated. This copy, the range of f , does not ﬁll up all of B (unless f happens to be surjective). Likewise, g puts a “copy” g(B) of B into A. Because they are The Cantor-Bernstein-Schröeder Theorem 233 not necessarily bijective, neither f nor g is guaranteed to have an inverse. But the map g : B →g(B) from B to g(B) = {g(x) : x ∈B} is bijective, so there is an inverse g−1 : g(B) →B. (We will need this inverse soon.) f g g−1 Figure 13.4. The injections f : A →B and g : B →A Consider the chain of injections illustrated in Figure 13.5. On the left, g puts a copy of B into A. Then f puts a copy of A (containing the copy of B) into B. Next, g puts a copy of this B-containing-A-containing-B into A, and so on, always alternating g and f . g g g f f f ··· Figure 13.5. An inﬁnite chain of injections The ﬁrst time A occurs in this sequence, it has a shaded region A−g(B). In the second occurrence of A, the shaded region is (A−g(B))∪(g◦f )(A−g(B)). In the third occurrence of A, the shaded region is (A −g(B)) ∪(g ◦f )(A −g(B)) ∪(g ◦f ◦g ◦f )(A −g(B)). To tame the notation, let’s say (g ◦f )2 = (g ◦f ) ◦(g ◦f ), and (g ◦f )3 = (g◦f )◦(g◦f )◦(g◦f ), and so on. Let’s also agree that (g◦f )0 = ιA, that is, it is the identity function on A. Then the shaded region of the nth occurrence of A in the sequence is n−1 [ k=0 (g ◦f )k(A −g(B)). This process divides A into gray and white regions: the gray region is G = ∞ [ k=0 (g ◦f )k(A −g(B)), 234 Cardinality of Sets and the white region is A −G. (See Figure 13.6.) Figure 13.6 suggests our desired bijection h : A →B. The injection f sends the gray areas on the left bijectively to the gray areas on the right. The injection g−1 : g(B) →B sends the white areas on the left bijectively to the white areas on the right. We can thus deﬁne h : A →B so that h(x) = f (x) if x is a gray point, and h(x) = g−1(x) if x is a white point. f g−1 f g−1 . . . A B Figure 13.6. The bijection h : A →B This informal argument suggests that given injections f : A →B and g : B →A, there is a bijection h : A →B. But it is not a proof. We now present this as a theorem and tighten up our reasoning in a careful proof, with the above diagrams and ideas as a guide. Theorem 13.10 (The Cantor-Bernstein-Schröeder Theorem) If \|A\| ≤\|B\| and \|B\| ≤\|A\|, then \|A\| = \|B\|. In other words, if there are injections f : A →B and g : B →A, then there is a bijection h : A →B. Proof. (Direct) Suppose there are injections f : A →B and g : B →A. Then, in particular, g : B →g(B) is a bijection from B onto the range of g, so it has an inverse g−1 : g(B) →B. (Note that g : B →A itself has no inverse g−1 : A →B unless g is surjective.) Consider the subset G = ∞ [ k=0 (g ◦f )k(A −g(B)) ⊆A. The Cantor-Bernstein-Schröeder Theorem 235 Let W = A −G, so A = G ∪W is partitioned into two sets G (think gray) and W (think white). Deﬁne a function h : A →B as h(x) = ( f (x) if x ∈G g−1(x) if x ∈W. Notice that this makes sense: if x ∈W, then x ∉G, so x ∉A−g(B) ⊆G, hence x ∈g(B), so g−1(x) is deﬁned. To ﬁnish the proof, we must show that h is both injective and surjective. For injective, we assume h(x) = h(y), and deduce x = y. There are three cases to consider. First, if x and y are both in G, then h(x) = h(y) means f (x) = f (y), so x = y because f is injective. Second, if x and y are both in W, then h(x) = h(y) means g−1(x) = g−1(y), and applying g to both sides gives x = y. In the third case, one of x and y is in G and the other is in W. Say x ∈G and y ∈W. The deﬁnition of G gives x = (g ◦f )k(z) for some k ≥0 and z ∈A −g(B). Note h(x) = h(y) now implies f (x) = g−1(y), that is, f ((g ◦f )k(z)) = g−1(y). Applying g to both sides gives (g ◦f )k+1(z) = y, which means y ∈G. But this is impossible, as y ∈W. Thus this third case cannot happen. But in the ﬁrst two cases h(x) = h(y) implies x = y, so h is injective. To see that h is surjective, take any b ∈B. We will ﬁnd an x ∈A with h(x) = b. Note that g(b) ∈A, so either g(b) ∈W or g(b) ∈G. In the ﬁrst case, h(g(b)) = g−1(g(b)) = b, so we have an x = g(b) ∈A for which h(x) = b. In the second case, g(b) ∈G. The deﬁnition of G shows g(b) = (g ◦f )k(z) for some k > 0, and z ∈A −g(B). Thus g(b) = (g ◦f )◦(g ◦f )k−1(z). Rewriting this, g(b) = g ³ f ¡ (g ◦f )k−1(z) ¢´ . Because g is injective, this implies b = f ¡ (g ◦f )k−1(z) ¢ . Let x = (g ◦f )k−1(z), so x ∈G by deﬁnition of G. Observe that h(x) = f (x) = f ¡ (g ◦f )k−1(z) ¢ = b. We have now seen that for any b ∈B, there is an x ∈A for which h(x) = b. Thus h is surjective. Since h : A →B is both injective and surjective, it is also bijective. ■ 236 Cardinality of Sets Here are some examples illustrating how the Cantor-Bernstein-Schröeder theorem can be used. This includes a proof that \|R\| = \|P(N)\|. Example 13.6 The intervals [0,1) and (0,1) in R have equal cardinalities. Surely this fact is plausible, for the two intervals are identical except for the endpoint 0. Yet concocting a bijection [0,1) →(0,1) is tricky. (Though not particularly diﬃcult: see the solution of Exercise 11 of Section 13.1.) For a simpler approach, note that f (x) = 1 4+ 1 2 x is an injection [0,1) →(0,1). Also, g(x) = x is an injection (0,1) →[0,1). The Cantor-Bernstein-Schröeder theorem guarantees a bijection h : [0,1) →(0,1), so \|[0,1)\| = \|(0,1)\|. Theorem 13.11 The sets R and P(N) have the same cardinality. Proof. Example 13.4 shows that \|R\| = \|(0,1)\|, and Example 13.6 shows \|(0,1)\| = \|[0,1)\|. Thus \|R\| = \|[0,1)\|, so to prove the theorem we just need to show that \|[0,1)\| = \|P(N)\|. By the Cantor-Bernstein-Schröeder theorem, it suﬃces to ﬁnd injections f : [0,1) →P(N) and g : P(N) →[0,1). To deﬁne f : [0,1) →P(N), we use the fact that any number in [0,1) has a unique decimal representation 0.b1b2b3b4 ..., where each bi one of the digits 0,1,2,...,9, and there is not a repeating sequence of 9’s at the end. (Recall that, e.g., 0.359999 = 0.360, etc.) Deﬁne f : [0,1) →P(N) as f ¡ 0.b1b2b3b4 ... ¢ = © 10b1, 102b2, 103b3, ... ª . For example, f (0.121212) = © 10,200,1000,20000,100000,... ª, and f (0.05) = © 0,500 ª. Also f (0.5) = f (0.50) = © 0,50 ª. To see that f is injective, take two unequal numbers 0.b1b2b3b4 ... and 0.d1d2d3d4 ... in [0,1). Then bi ̸= di for some index i. Hence bi10i ∈f (0.b1b2b3b4 ...) but bi10i ∉f (0.d1d2d3d4 ...), so f (0.b1b2b3b4 ...) ̸= f (0.d1d2d3d4 ...). Consequently f is injective. Next, deﬁne g : P(N) →[0,1), where g(X) = 0.b1b2b3b4 ... is the number for which bi = 1 if i ∈X and bi = 0 if i ∉X. For example, g ¡© 1,3 ª¢ = 0.101000, and g ¡© 2,4,6,8,... ª¢ = 0.01010101. Also g(;) = 0 and g(N) = 0.1111. To see that g is injective, suppose X ̸= Y. Then there is at least one integer i that belongs to one of X or Y, but not the other. Consequently g(X) ̸= g(Y ) because they diﬀer in the ith decimal place. This shows g is injective. From the injections f : [0,1) →P(N) and g : P(N) →[0,1), the Cantor-Bernstein-Schröeder theorem guarantees a bijection h : [0,1) →P(N). Hence \|[0,1)\| = \|P(N)\|. As \|R\| = \|[0,1)\|, we conclude \|R\| = \|P(N)\|. ■ The Cantor-Bernstein-Schröeder Theorem 237 We know that \|R\| ̸= \|N\|. But we just proved \|R\| = \|P(N)\|. This suggests that the cardinality of R is not “too far” from \|N\| = ℵ0. We close with a few informal remarks on this mysterious relationship between ℵ0 and \|R\|. We established earlier in this chapter that ℵ0 < \|R\|. For nearly a century after Cantor formulated his theories on inﬁnite sets, mathematicians struggled with the question of whether or not there exists a set A for which ℵ0 < \|A\| < \|R\|. It was commonly suspected that no such set exists, but no one was able to prove or disprove this. The assertion that no such A exists came to be called the continuum hypothesis. Theorem 13.11 states that \|R\| = \|P(N)\|. Placing this in the context of the chain (13.2) on page 230, we have the following relationships. ℵ0 \|R\| = = \|N\| < \|P(N)\| < \|P(P(N))\| < \|P(P(P(N)))\| < ··· From this, we can see that the continuum hypothesis asserts that no set has a cardinality between that of N and its power set. Although this may seem intuitively plausible, it eluded proof since Cantor ﬁrst posed it in the 1880s. In fact, the real state of aﬀairs is almost paradoxical. In 1931, the logician Kurt Gödel proved that for any suﬃciently strong and consistent axiomatic system, there exist statements which can neither be proved nor disproved within the system. Later he proved that the negation of the continuum hypothesis cannot be proved within the standard axioms of set theory (i.e., the Zermelo-Fraenkel axioms, mentioned in Section 1.10). This meant that either the continuum hypothesis is false and cannot be proven false, or it is true. In 1964, Paul Cohen discovered another startling truth: Given the laws of logic and the axioms of set theory, no proof can deduce the continuum hypothesis. In essence he proved that the continuum hypothesis cannot be proved. Taken together, Gödel and Cohens’ results mean that the standard axioms of mathematics cannot “decide” whether the continuum hypothesis is true or false; that no logical conﬂict can arise from either asserting or denying the continuum hypothesis. We are free to either accept it as true or accept it as false, and the two choices lead to diﬀerent—but equally consistent—versions of set theory. 238 Cardinality of Sets On the face of it, this seems to undermine the foundation of logic, and everything we have done in this book. The continuum hypothesis should be a statement – it should be either true or false. How could it be both? Here is an analogy that may help make sense of this. Consider the number systems Zn. What if we asked whether = is true or false? Of course the answer depends on n. The expression = is true in Z2 and false in Z3. Moreover, if we assert that = is true, we are logically forced to the conclusion that this is taking place in the system Z2. If we assert that = is false, then we are dealing with some other Zn. The fact that = can be either true or false does not necessarily mean that there is some inherent inconsistency within the individual number systems Zn. The equation = is a true statement in the “universe” of Z2 and a false statement in the universe of (say) Z3. It is the same with the continuum hypothesis. Saying it’s true leads to one system of set theory. Saying it’s false leads to some other system of set theory. Gödel and Cohens’ discoveries mean that these two types of set theory, although diﬀerent, are equally consistent and valid mathematical universes. So what should you believe? Fortunately, it does not make much diﬀerence, because most important mathematical results do not hinge on the continuum hypothesis. (They are true in both universes.) Unless you undertake a deep study of the foundations of mathematics, you will be ﬁne accepting the continuum hypothesis as true. Most mathematicians are agnostics on this issue, but they tend to prefer the version of set theory in which the continuum hypothesis holds. The situation with the continuum hypothesis is a testament to the immense complexity of mathematics. It is a reminder of the importance of rigor and careful, systematic methods of reasoning that begin with the ideas introduced in this book. Exercises for Section 13.4 1. Show that if A ⊆B and there is an injection g : B →A, then \|A\| = \|B\|. 2. Show that \|R2\| = \|R\|. Suggestion: Begin by showing \|(0,1)×(0,1)\| = \|(0,1)\|. 3. Let F be the set of all functions N → © 0,1 ª. Show that \|R\| = \|F\|. 4. Let F be the set of all functions R → © 0,1 ª. Show that \|R\| < \|F\|. 5. Consider the subset B = © (x, y) : x2 + y2 ≤1 ª ⊆R2. Show that \|B\| = \|R2\|. 6. Show that \|P(N×N)\| = \|P(N)\|. 7. Prove or disprove: If there is a injection f : A →B and a surjection g : A →B, then there is a bijection h : A →B. Conclusion I f you have internalized the ideas in this book, then you have a set of rhetorical tools for deciphering and communicating mathematics. These tools are indispensable at any level. But of course it takes more than mere tools to build something. Planning, creativity, inspiration, skill, talent, intuition, passion and persistence are also vitally important. It is safe to say that if you have come this far, then you probably possess a suﬃcient measure of these traits. The quest to understand mathematics has no end, but you are well equipped for the journey. It is my hope that the things you have learned from this book will lead you to a higher plane of understanding, creativity and expression. Good luck and best wishes. R.H. Solutions Chapter 1 Exercises Section 1.1 1. {5x−1 : x ∈Z} = {...−11,−6,−1,4,9,14,19,24,29,...} 3. {x ∈Z : −2 ≤x < 7} = {−2,−1,0,1,2,3,4,5,6} 5. © x ∈R : x2 = 3 ª = © − p 3, p 3 ª 7. © x ∈R : x2 +5x = −6 ª = {−2,−3} 9. {x ∈R : sinπx = 0} = {...,−2,−1,0,1,2,3,4,...} = Z 11. {x ∈Z : \|x\| < 5} = {−4,−3,−2,−1,0,1,2,3,4} 13. {x ∈Z : \|6x\| < 5} = {0} 15. {5a+2b : a,b ∈Z} = {...,−2,−1,0,1,2,3,...} = Z 17. {2,4,8,16,32,64...} = {2x : x ∈N} 19. {...,−6,−3,0,3,6,9,12,15,...} = {3x : x ∈Z} 21. {0,1,4,9,16,25,36,...} = © x2 : x ∈Z ª 23. {3,4,5,6,7,8} = {x ∈Z : 3 ≤x ≤8} = {x ∈N : 3 ≤x ≤8} 25. © ..., 1 8, 1 4, 1 2,1,2,4,8,... ª = {2n : n ∈Z} 27. © ...,−π,−π 2 ,0, π 2 ,π, 3π 2 ,2π, 5π 2 ,... ª = n kπ 2 : k ∈Z o 29. \|{{1},{2,{3,4}},;}\| = 3 31. \|{{{1},{2,{3,4}},;}}\| = 1 33. \|{x ∈Z : \|x\| < 10}\| = 19 35. \| © x ∈Z : x2 < 10 ª \| = 7 37. \| © x ∈N : x2 < 0 ª \| = 0 39. {(x, y) : x ∈[1,2], y ∈[1,2]} −3 −2 −1 1 2 3 −2 −1 1 2 41. {(x, y) : x ∈[−1,1], y = 1} −3 −2 −1 1 2 3 −2 −1 1 2 43. {(x, y) : \|x\| = 2, y ∈[0,1]} −3 −2 −1 1 2 3 −2 −1 1 2 45. © (x, y) : x, y ∈R,x2 + y2 = 1 ª −3 −2 −1 1 2 3 −2 −1 1 2 241 47. © (x, y) : x, y ∈R, y ≥x2 −1 ª −3 −2 −1 1 2 3 −3 −2 −1 1 2 3 49. {(x,x+ y) : x ∈R, y ∈Z} −3 −2 −1 1 2 3 −3 −2 −1 1 2 3 51. {(x, y) ∈R2 : (y−x)(y+ x) = 0} −3 −2 −1 1 2 3 −3 −2 −1 1 2 3 Section 1.2 1. Suppose A = {1,2,3,4} and B = {a, c}. (a) A ×B = {(1,a),(1, c),(2,a),(2, c),(3,a),(3, c),(4,a),(4, c)} (b) B × A = {(a,1),(a,2),(a,3),(a,4),(c,1),(c,2),(c,3),(c,4)} (c) A × A = {(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4), (3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4)} (d) B ×B = {(a,a),(a, c),(c,a),(c, c)} (e) ;×B = {(a,b) : a ∈;,b ∈B} = ; (There are no ordered pairs (a,b) with a ∈;.) (f) (A ×B)×B = {((1,a),a),((1, c),a),((2,a),a),((2, c),a),((3,a),a),((3, c),a),((4,a),a),((4, c),a), ((1,a), c),((1, c), c),((2,a), c),((2, c), c),((3,a), c),((3, c), c),((4,a), c),((4, c), c)} (g) A ×(B ×B) = © (1,(a,a)),(1,(a, c)),(1,(c,a)),(1,(c, c)), (2,(a,a)),(2,(a, c)),(2,(c,a)),(2,(c, c)), (3,(a,a)),(3,(a, c)),(3,(c,a)),(3,(c, c)), (4,(a,a)),(4,(a, c)),(4,(c,a)),(4,(c, c)) ª (h) B3 = {(a,a,a),(a,a, c),(a, c,a),(a, c, c),(c,a,a),(c,a, c),(c, c,a),(c, c, c)} 3. © x ∈R : x2 = 2 ª ×{a, c, e} = © (− p 2,a),( p 2,a),(− p 2, c),( p 2, c),(− p 2, e),( p 2, e) ª 5. © x ∈R : x2 = 2 ª ×{x ∈R : \|x\| = 2} = © (− p 2,−2),( p 2,2),(− p 2,2),( p 2,−2) ª 7. {;}×{0,;}×{0,1} = {(;,0,0),(;,0,1),(;,;,0),(;,;,1)} 242 Solutions Sketch the following Cartesian products on the x-y plane. 9. {1,2,3}×{−1,0,1} −3 −2 −1 1 2 3 −2 −1 1 2 11. [0,1]×[0,1] −3 −2 −1 1 2 3 −2 −1 1 2 13. {1,1.5,2}×[1,2] −3 −2 −1 1 2 3 −2 −1 1 2 15. {1}×[0,1] −3 −2 −1 1 2 3 −2 −1 1 2 17. N×Z −3 −2 −1 1 2 3 −2 −1 1 2 19. [0,1]×[0,1]×[0,1] −3 −2 −1 1 2 3 −2 −1 1 2 Section 1.3 A. List all the subsets of the following sets. 1. The subsets of {1,2,3,4} are: {}, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}. 3. The subsets of {{R}} are: {} and {{R}}. 5. The subsets of {;} are {} and {;}. 7. The subsets of {R,{Q,N}} are {}, {R},{{Q,N}}, {R,{Q,N}}. B. Write out the following sets by listing their elements between braces. 9. © X : X ⊆{3,2,a} and \|X\| = 2 ª = {{3,2},{3,a},{2,a}} 11. © X : X ⊆{3,2,a} and \|X\| = 4 ª = {} = ; C. Decide if the following statements are true or false. 13. R3 ⊆R3 is true because any set is a subset of itself. 15. © (x, y) : x−1 = 0 ª ⊆ © (x, y) : x2 −x = 0 ª. This is true. (The even-numbered ones are both false. You have to explain why.) 243 Section 1.4 A. Find the indicated sets. 1. P({{a,b},{c}}) = {;,{{a,b}},{{c}},{{a,b},{c}}} 3. P({{;},5}) = {;,{{;}},{5},{{;},5}} 5. P(P({2})) = {;,{;},{{2}},{;,{2}}} 7. P({a,b})×P({0,1}) = © (;,;), (;,{0}), (;,{1}), (;,{0,1}), ({a},;), ({a},{0}), ({a},{1}), ({a},{0,1}), ({b},;), ({b},{0}), ({b},{1}), ({b},{0,1}), ({a,b},;), ({a,b},{0}), ({a,b},{1}), ({a,b},{0,1}) ª 9. P({a,b}×{0}) = {;,{(a,0)},{(b,0)},{(a,0),(b,0)}} 11. {X ⊆P({1,2,3}) : \|X\| ≤1} = {;,{;},{{1}},{{2}},{{3}},{{1,2}},{{1,3}},{{2,3}},{{1,2,3}}} B. Suppose that \|A\| = m and \|B\| = n. Find the following cardinalities. 13. \|P(P(P(A)))\| = 2 ³ 2(2m)´ 15. \|P(A ×B)\| = 2mn 17. \|{X ∈P(A) : \|X\| ≤1}\| = m+1 19. \|P(P(P(A ×;)))\| = \|P(P(P(;)))\| = 4 Section 1.5 1. Suppose A = {4,3,6,7,1,9}, B = {5,6,8,4} and C = {5,8,4}. Find: (a) A ∪B = {1,3,4,5,6,7,8,9} (b) A ∩B = {4,6} (c) A −B = {3,7,1,9} (d) A −C = {3,6,7,1,9} (e) B −A = {5,8} (f) A ∩C = {4} (g) B ∩C = {5,8,4} (h) B ∪C = {5,6,8,4} (i) C −B = ; 3. Suppose A = {0,1} and B = {1,2}. Find: (a) (A ×B)∩(B ×B) = {(1,1),(1,2)} (b) (A ×B)∪(B ×B) = {(0,1),(0,2),(1,1),(1,2),(2,1),(2,2)} (c) (A ×B)−(B ×B) = {(0,1),(0,2)} (d) (A ∩B)× A = {(1,0),(1,1)} (e) (A ×B)∩B = ; (f) P(A)∩P(B) = {;,{1}} (g) P(A)−P(B) = {{0},{0,1}} (h) P(A ∩B) = {{},{1}} (i) © ;,{(0,1)},{(0,2)},{(1,1)},{(1,2)},{(0,1),(0,2)},{(0,1),(1,1)},{(0,1),(1,2)},{(0,2),(1,1)}, {(0,2),(1,2)},{(1,1),(1,2)},{(0,2),(1,1),(1,2)},{(0,1),(1,1),(1,2)},{(0,1),(0,2),(1,2)}, {(0,1),(0,2),(1,1)},{(0,1),(0,2),(1,1),(1,2)} ª 244 Solutions 5. Sketch the sets X = [1,3]×[1,3] and Y = [2,4]×[2,4] on the plane R2. On separate drawings, shade in the sets X ∪Y, X ∩Y, X −Y and Y −X. (Hint: X and Y are Cartesian products of intervals. You may wish to review how you drew sets like [1,3]×[1,3] in the Section 1.2.) Y X X ∪Y X ∩Y X −Y Y −X 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 7. Sketch the sets X = © (x, y) ∈R2 : x2 + y2 ≤1 ª and Y = © (x, y) ∈R2 : x ≥0 ª on R2. On separate drawings, shade in the sets X ∪Y, X ∩Y, X −Y and Y −X. X Y X ∪Y X ∩Y X −Y Y −X −2 −1 1 2 −2 −1 1 2 −2 −1 1 2 −2 −1 1 2 −2 −1 1 2 −2 −1 1 2 −2 −1 1 2 −2 −1 1 2 −2 −1 1 2 −2 −1 1 2 9. The ﬁrst statement is true. (A picture should convince you; draw one if necessary.) The second statement is false: Notice for instance that (0.5,0.5) is in the right-hand set, but not the left-hand set. Section 1.6 1. Suppose A = {4,3,6,7,1,9} and B = {5,6,8,4} have universal set U = {n ∈Z : 0 ≤n ≤10}. (a) A = {0,2,5,8,10} (b) B = {0,1,2,3,7,9,10} (c) A ∩A = ; (d) A ∪A = {0,1,2,3,4,5,6,7,8,9,10} = U (e) A −A = A (f) A −B = {4,6} (g) A −B = {5,8} (h) A ∩B = {5,8} (i) A ∩B = {0,1,2,3,4,6,7,9,10} 3. Sketch the set X = [1,3]×[1,2] on the plane R2. On separate drawings, shade in the sets X, and X ∩([0,2]×[0,3]). X X X ∩([0,2]×[0,3]) −1 1 2 3 −1 1 2 3 −1 1 2 3 −1 1 2 3 −1 1 2 3 −1 1 2 3 5. Sketch the set X = © (x, y) ∈R2 : 1 ≤x2 + y2 ≤4 ª on the plane R2. On a separate drawing, shade in the set X. 245 X X 1 2 3 1 2 3 1 2 3 1 2 3 Solution of 1.6, #5. A A (shaded) U Solution of 1.7, #1. Section 1.7 1. Draw a Venn diagram for A. (Solution above right) 3. Draw a Venn diagram for (A −B)∩C. Scratch work is shown on the right. The set A −B is indicated with vertical shading. The set C is indicated with horizontal shad-ing. The intersection of A −B and C is thus the overlapping region that is shaded with both vertical and horizontal lines. The ﬁnal answer is drawn on the far right, where the set (A −B)∩C is shaded in gray. A A B B C C 5. Draw Venn diagrams for A∪(B∩C) and (A∪B)∩(A∪C). Based on your drawings, do you think A ∪(B ∩C) = (A ∪B)∩(A ∪C)? If you do the drawings carefully, you will ﬁnd that your Venn diagrams are the same for both A ∪(B ∩C) and (A ∪B)∩(A ∪C). Each looks as illustrated on the right. Based on this, we are inclined to say that the equation A ∪(B ∩C) = (A ∪B)∩(A ∪C) holds for all sets A, B and C. A B C 7. Suppose sets A and B are in a universal set U. Draw Venn diagrams for A ∩B and A ∪B. Based on your drawings, do you think it’s true that A ∩B = A ∪B? The diagrams for A ∩B and A ∪B look exactly alike. In either case the diagram is the shaded region illustrated on the right. Thus we would expect that the equation A ∩B = A ∪B is true for any sets A and B. A B U 9. Draw a Venn diagram for (A ∩B)−C. A B C 11. The simplest answer is (B ∩C)−A. 13. One answer is (A ∪B ∪C)−(A ∩B ∩C). 246 Solutions Section 1.8 1. Suppose A1 = {a,b,d, e, g, f }, A2 = {a,b, c,d}, A3 = {b,d,a} and A4 = {a,b,h}. (a) 4 [ i=1 Ai = {a,b, c,d, e, f , g,h} (b) 4 \ i=1 Ai = {a,b} 3. For each n ∈N, let An = {0,1,2,3,...,n}. (a) [ i∈N Ai = {0}∪N (b) \ i∈N Ai = {0,1} 5. (a) [ i∈N [i, i +1] =[1,∞) (b) \ i∈N [i, i +1] =; 7. (a) [ i∈N R×[i, i +1] = {(x, y) : x, y ∈R, y ≥1} (b) \ i∈N R×[i, i +1] = ; 9. (a) [ X∈P(N) X = N (b) \ X∈P(N) X = ; 11. Yes, this is always true. 13. The ﬁrst is true, the second is false. Chapter 2 Exercises Section 2.1 Decide whether or not the following are statements. In the case of a statement, say if it is true or false. 1. Every real number is an even integer. (Statement, False) 3. If x and y are real numbers and 5x = 5y, then x = y. (Statement, True) 5. Sets Z and N are inﬁnite. (Statement, True) 7. The derivative of any polynomial of degree 5 is a polynomial of degree 6. (Statement, False) 9. cos(x) = −1 This is not a statement. It is an open sentence because whether it’s true or false depends on the value of x. 11. The integer x is a multiple of 7. This is an open sentence, and not a statement. 13. Either x is a multiple of 7, or it is not. This is a statement, for the sentence is true no matter what x is. 15. In the beginning God created the heaven and the earth. This is a statement, for it is either deﬁnitely true or deﬁnitely false. There is some controversy over whether it’s true or false, but no one claims that it is neither true nor false. 247 Section 2.2 Express each statement as one of the forms P ∧Q, P ∨Q, or ∼P. Be sure to also state exactly what statements P and Q stand for. 1. The number 8 is both even and a power of 2. P ∧Q P: 8 is even Q: 8 is a power of 2 Note: Do not say “Q: a power of 2,” because that is not a statement. 3. x ̸= y ∼(x = y) (Also ∼P where P : x = y.) 5. y ≥x ∼(y < x) (Also ∼P where P : y < x.) 7. The number x equals zero, but the number y does not. P∧∼Q P : x = 0 Q : y = 0 9. x ∈A −B (x ∈A)∧∼(x ∈B) 11. A ∈ © X ∈P(N) : \|X\| < ∞ ª (A ⊆N)∧(\|A\| < ∞). 13. Human beings want to be good, but not too good, and not all the time. P∧∼Q∧∼R P : Human beings want to be good. Q : Human beings want to be too good. R : Human beings want to be good all the time. Section 2.3 Without changing their meanings, convert each of the following sentences into a sentence having the form “If P, then Q.” 1. A matrix is invertible provided that its determinant is not zero. Answer: If a matrix has a determinant not equal to zero, then it is invertible. 3. For a function to be integrable, it is necessary that it is continuous. Answer: If function is integrable, then it is continuous. 5. An integer is divisible by 8 only if it is divisible by 4. Answer: If an integer is divisible by 8, then it is divisible by 4. 7. A series converges whenever it converges absolutely. Answer: If a series converges absolutely, then it converges. 9. A function is integrable provided the function is continuous. Answer: If a function is continuous, then that function is integrable. 11. You fail only if you stop writing. Answer: If you fail, then you have stopped writing. 13. Whenever people agree with me I feel I must be wrong. Answer: If people agree with me, then I feel I must be wrong. 248 Solutions Section 2.4 Without changing their meanings, convert each of the following sentences into a sentence having the form “P if and only if Q.” 1. For a matrix to be invertible, it is necessary and suﬃcient that its determinant is not zero. Answer: A matrix is invertible if and only if its determinant is not zero. 3. If xy = 0 then x = 0 or y = 0, and conversely. Answer: xy = 0 if and only if x = 0 or y = 0 5. For an occurrence to become an adventure, it is necessary and suﬃcient for one to recount it. Answer: An occurrence becomes an adventure if and only if one recounts it. Section 2.5 1. Write a truth table for P ∨(Q ⇒R) P Q R Q ⇒R P ∨(Q ⇒R) T T T T T T T F F T T F T T T T F F T T F T T T T F T F F F F F T T T F F F T T 3. Write a truth table for ∼(P ⇒Q) P Q P ⇒Q ∼(P ⇒Q) T T T F T F F T F T T F F F T F 5. Write a truth table for (P∧∼P)∨Q P Q (P∧∼P) (P∧∼P)∨Q T T F T T F F F F T F T F F F F 7. Write a truth table for (P∧∼P) ⇒Q P Q (P∧∼P) (P∧∼P) ⇒Q T T F T T F F T F T F T F F F T 9. Write a truth table for ∼(∼P∨∼Q). P Q ∼P ∼Q ∼P∨∼Q ∼(∼P∨∼Q) T T F F F T T F F T T F F T T F T F F F T T T F 249 11. Suppose P is false and that the statement (R ⇒S) ⇔(P ∧Q) is true. Find the truth values of R and S. (This can be done without a truth table.) Answer: Since P is false, it follows that (P ∧Q) is false also. But then in order for (R ⇒S) ⇔(P ∧Q) to be true, it must be that (R ⇒S) is false. The only way for (R ⇒S) to be false is if R is true and S is false. Section 2.6 A. Use truth tables to show that the following statements are logically equivalent. 1. P ∧(Q ∨R) = (P ∧Q)∨(P ∧R) P Q R Q ∨R P ∧Q P ∧R P ∧(Q ∨R) (P ∧Q)∨(P ∧R) T T T T T T T T T T F T T F T T T F T T F T T T T F F F F F F F F T T T F F F F F T F T F F F F F F T T F F F F F F F F F F F F Thus since their columns agree, the two statements are logically equivalent. 3. P ⇒Q = (∼P)∨Q P Q ∼P (∼P)∨Q P ⇒Q T T F T T T F F F F F T T T T F F T T T Thus since their columns agree, the two statements are logically equivalent. 5. ∼(P ∨Q ∨R) = (∼P)∧(∼Q)∧(∼R) P Q R P ∨Q ∨R ∼P ∼Q ∼R ∼(P ∨Q ∨R) (∼P)∧(∼Q)∧(∼R) T T T T F F F F F T T F T F F T F F T F T T F T F F F T F F T F T T F F F T T T T F F F F F T F T T F T F F F F T T T T F F F F F F F T T T T T Thus since their columns agree, the two statements are logically equivalent. 250 Solutions 7. P ⇒Q = (P∧∼Q) ⇒(Q∧∼Q) P Q ∼Q P∧∼Q Q∧∼Q (P∧∼Q) ⇒(Q∧∼Q) P ⇒Q T T F F F T T T F T T F F F F T F F F T T F F T F F T T Thus since their columns agree, the two statements are logically equivalent. B. Decide whether or not the following pairs of statements are logically equivalent. 9. By DeMorgan’s law, we have ∼(∼P∨∼Q) =∼∼P∧∼∼Q = P ∧Q. Thus the two statements are logically equivalent. 11. (∼P)∧(P ⇒Q) and ∼(Q ⇒P) P Q ∼P P ⇒Q Q ⇒P (∼P)∧(P ⇒Q) ∼(Q ⇒P) T T F T T F F T F F F T F F F T T T F T T F F T T T T F The columns for the two statements do not quite agree, thus the two state-ments are not logically equivalent. Section 2.7 Write the following as English sentences. Say whether the statements are true or false. 1. ∀x ∈R,x2 > 0 Answer: For every real number x, x2 > 0. Also: For every real number x, it follows that x2 > 0. Also: The square of any real number is positive. (etc.) This statement is FALSE. Reason: 0 is a real number, but it’s not true that 02 > 0. 3. ∃a ∈R,∀x ∈R,ax = x. Answer: There exists a real number a for which ax = x for every real number x. This statement is TRUE. Reason: Consider a = 1. 5. ∀n ∈N,∃X ∈P(N),\|X\| < n Answer: For every natural number n, there is a subset X of N with \|X\| < n. This statement is TRUE. Reason: Suppose n ∈N. Let X = ;. Then \|X\| = 0 < n. 251 7. ∀X ⊆N,∃n ∈Z,\|X\| = n Answer: For any subset X of N, there exists an integer n for which \|X\| = n. This statement is FALSE. For example, the set X = {2,4,6,8,...} of all even natural numbers is inﬁnite, so there does not exist any integer n for which \|X\| = n. 9. ∀n ∈Z,∃m ∈Z,m = n+5 Answer: For every integer n there is another integer m such that m = n+5. This statement is TRUE. Section 2.9 Translate each of the following sentences into symbolic logic. 1. If f is a polynomial and its degree is greater than 2, then f ′ is not constant. Translation: (P ∧Q) ⇒R, where P : f is a polynomial, Q : f has degree greater than 2, R : f ′ is not constant. 3. If x is prime then px is not a rational number. Translation: P ⇒∼Q, where P : x is prime, Q : px is a rational number. 5. For every positive number ε, there is a positive number δ for which \|x−a\| < δ implies \|f (x)−f (a)\| < ε. Translation: ∀ε ∈R,ε > 0,∃δ ∈R,δ > 0,(\|x−a\| < δ) ⇒(\|f (x)−f (a)\| < ε) 7. There exists a real number a for which a+ x = x for every real number x. Translation: ∃a ∈R,∀x ∈R,a+ x = x 9. If x is a rational number and x ̸= 0, then tan(x) is not a rational number. Translation: ((x ∈Q)∧(x ̸= 0)) ⇒(tan(x) ∉Q) 11. There is a Providence that protects idiots, drunkards, children and the United States of America. One translation is as follows. Let R be union of the set of idiots, the set of drunkards, the set of children, and the set consisting of the USA. Let P be the open sentence P(x): x is a Providence. Let S be the open sentence S(x, y): x protects y. Then the translation is ∃x,∀y ∈R,P(x)∧S(x, y). (Notice that, although this is mathematically correct, some humor has been lost in the translation.) 13. Everything is funny as long as it is happening to somebody else. Translation: ∀x,(∼M(x)∧S(x)) ⇒F(x), where M(x): x is happening to me, S(x): x is happening to someone, and F(x) : x is funny. 252 Solutions Section 2.10 Negate the following sentences. 1. The number x is positive, but the number y is not positive. The “but” can be interpreted as “and.” Using DeMorgan’s law, the negation is: The number x is not positive or the number y is positive. 3. For every prime number p there, is another prime number q with q > p. Negation: There is a prime number p such that for every prime number q, q ≤p. Also: There exists a prime number p for which q ≤p for every prime number q. (etc.) 5. For every positive number ε there is a positive number M for which \|f (x)−b\| < ε whenever x > M. To negate this, it may be helpful to ﬁrst write it in symbolic form. The statement is ∀ε ∈(0,∞),∃M ∈(0,∞),(x > M) ⇒(\|f (x)−b\| < ε). Working out the negation, we have ∼ ¡ ∀ε ∈(0,∞),∃M ∈(0,∞),(x > M) ⇒(\|f (x)−b\| < ε) ¢ = ∃ε ∈(0,∞),∼ ¡ ∃M ∈(0,∞),(x > M) ⇒(\|f (x)−b\| < ε) ¢ = ∃ε ∈(0,∞),∀M ∈(0,∞),∼ ¡ (x > M) ⇒(\|f (x)−b\| < ε) ¢ . Finally, using the idea from Example 2.14, we can negate the conditional statement that appears here to get ∃ε ∈(0,∞),∀M ∈(0,∞),∃x, (x > M)∧∼(\|f (x)−b\| < ε) ¢ . Negation: There exists a positive number ε with the property that for every positive number M, there is a number x for which x > M and \|f (x)−b\| ≥ε. 7. I don’t eat anything that has a face. Negation: I will eat some things that have a face. (Note. If your answer was “I will eat anything that has a face.” then that is wrong, both morally and mathematically.) 9. If sin(x) < 0, then it is not the case that 0 ≤x ≤π. Negation: There exists a number x for which sin(x) < 0 and 0 ≤x ≤π. 11. You can fool all of the people all of the time. There are several ways to negate this, including: There is a person that you can’t fool all the time. or There is a person x and a time y for which x is not fooled at time y. (But Abraham Lincoln said it better.) 253 Chapter 3 Exercises Section 3.1 1. Consider lists made from the letters T, H, E, O, R, Y, with repetition allowed. (a) How many length-4 lists are there? Answer: 6·6·6·6 = 1296. (b) How many length-4 lists are there that begin with T? Answer: 1·6·6·6 = 216. (c) How many length-4 lists are there that do not begin with T? Answer: 5·6·6·6 = 1080. 3. How many ways can you make a list of length 3 from symbols a,b,c,d,e,f if... (a) ... repetition is allowed. Answer: 6·6·6 = 216. (b) ... repetition is not allowed. Answer: 6·5·4 = 120. (c) ... repetition is not allowed and the list must contain the letter a. Answer: 5·4+5·4+5·4 = 60. (d) ... repetition is allowed and the list must contain the letter a. Answer: 6·6·6−5·5·5 = 91. (Note: See Example 3.2 if a more detailed explanation is required.) 5. Five cards are dealt oﬀof a standard 52-card deck and lined up in a row. How many such line-ups are there in which all ﬁve cards are of the same color? (i.e., all black or all red.) There are 26·25·24·23·22 = 7,893,600 possible black-card line-ups and 26·25·24·23· 22 = 7,893,600 possible red-card line-ups, so the answer is 7,893,600+7,893,600 = 15,787,200. 7. This problems involves 8-digit binary strings such as 10011011 or 00001010. (i.e., 8-digit numbers composed of 0’s and 1’s.) (a) How many such strings are there? Answer: 2·2·2·2·2·2·2·2 = 256. (b) How many such strings end in 0? Answer: 2·2·2·2·2·2·2·1 = 128. (c) How many such strings have the property that their second and fourth digits are 1’s? Answer: 2·1·2·1·2·2·2·2 = 64. (d) How many such strings are such that their second or fourth digits are 1’s? Answer: These strings can be divided into three types. Type 1 consists of those strings of form ∗1∗0∗∗∗∗, Type 2 consist of strings of form ∗0∗1∗∗∗∗, and Type 3 consists of those of form ∗1 ∗1 ∗∗∗∗. By the multiplication principle there are 26 = 64 strings of each type, so there are 3·64 = 192 8-digit binary strings whose second or fourth digits are 1’s. 9. This problem concerns 4-letter codes that can be made from the letters of the English Alphabet. (a) How many such codes can be made? Answer: 26·26·26·26 = 456976 254 Solutions (b) How many such codes have no two consecutive letters the same? We use the multiplication principle. There are 26 choices for the ﬁrst letter. The second letter can’t be the same as the ﬁrst letter, so there are only 25 choices for it. The third letter can’t be the same as the second letter, so there are only 25 choices for it. The fourth letter can’t be the same as the third letter, so there are only 25 choices for it. Thus there are 26·25·25·25 = 406,250 codes with no two consecutive letters the same. 11. This problem concerns lists of length 6 made from the letters A,B,C,D,E,F,G,H. How many such lists are possible if repetition is not allowed and the list contains two consecutive vowels? Answer: There are just two vowels A and E to choose from. The lists we want to make can be divided into ﬁve types. They have one of the forms VV ∗∗∗∗, or ∗VV ∗∗∗, or ∗∗VV ∗∗, or ∗∗∗VV∗, or ∗∗∗∗VV, where V indicates a vowel and ∗indicates a consonant. By the multiplication principle, there are 2·1·6·5·4·3 = 720 lists of form VV ∗∗∗∗. In fact, that for the same reason there are 720 lists of each form. Thus the answer to the question is 5·720 = 3600 Section 3.2 1. Answer n = 14. 3. Answer: 5! = 120. 5. 120! 118! = 120·119·118! 118! = 120·119 = 14,280. 7. Answer: 5!4! = 2880. 9. The case x = 1 is straightforward. For x = 2,3 and 4, use integration by parts. For x = π, you are on your own. Section 3.3 1. Suppose a set A has 37 elements. How many subsets of A have 10 elements? How many subsets have 30 elements? How many have 0 elements? Answers: ¡37 10 ¢ = 348,330,136; ¡37 30 ¢ = 10,295,472; ¡37 0 ¢ = 1. 3. A set X has exactly 56 subsets with 3 elements. What is the cardinality of X? The answer will be n, where ¡n 3 ¢ = 56. After some trial and error, you will discover ¡8 3 ¢ = 56, so \|X\| = 8. 5. How many 16-digit binary strings contain exactly seven 1’s? Answer: Make such a string as follows. Start with a list of 16 blank spots. Choose 7 of the blank spots for the 1’s and put 0’s in the other spots. There are ¡16 7 ¢ = 114,40 ways to do this. 7. \|{X ∈P({0,1,2,3,4,5,6,7,8,9}) : \|X\| < 4}\| = ¡10 0 ¢ + ¡10 1 ¢ + ¡10 2 ¢ + ¡10 3 ¢ = 1+10+45+120 = 176. 9. This problem concerns lists of length six made from the letters A,B,C,D,E,F, without repetition. How many such lists have the property that the D occurs before the A? Answer: Make such a list as follows. Begin with six blank spaces and select two of these spaces. Put the D in the ﬁrst selected space and the A in the second. There are ¡6 2 ¢ = 15 ways of doing this. For each of these 15 choices there are 4! = 24 ways of ﬁlling in the remaining spaces. Thus the answer to the question is 15×24 = 360 such lists. 255 11. How many 10-digit integers contain no 0’s and exactly three 6’s? Answer: Make such a number as follows: Start with 10 blank spaces and choose three of these spaces for the 6’s. There are ¡10 3 ¢ = 120 ways of doing this. For each of these 120 choices we can ﬁll in the remaining seven blanks with choices from the digits 1,2,3,4,5,7,8,9, and there are 87 to do this. Thus the answer to the question is ¡10 3 ¢ ·87 = 251,658,240. 13. Assume n,k ∈Z with 0 ≤k ≤n. Then ¡n k ¢ = n! (n−k)!k! = n! k!(n−k)! = n! (n−(n−k))!(n−k)! = ¡ n n−k ¢ . Section 3.4 1. Write out Row 11 of Pascal’s triangle. Answer: 1 11 55 165 330 462 462 330 165 55 11 1 3. Use the binomial theorem to ﬁnd the coeﬃcient of x8 in (x+2)13. Answer: According to the binomial theorem, the coeﬃcient of x8 y5 in (x+ y)13 is ¡13 8 ¢ x8 y5 = 1287x8 y5. Now plug in y = 2 to get the ﬁnal answer of 41184x8. 5. Use the binomial theorem to show Pn k=0 ¡ n k ¢ = 2n. Hint: Observe that 2n = (1+1)n. Now use the binomial theorem to work out (x+ y)n and plug in x = 1 and y = 1. 7. Use the binomial theorem to show Pn k=0 3k ¡ n k ¢ = 4n. Hint: Observe that 4n = (1+3)n. Now look at the hint for the previous problem. 9. Use the binomial theorem to show ¡ n 0 ¢ − ¡ n 1 ¢ + ¡ n 2 ¢ − ¡ n 3 ¢ + ¡ n 4 ¢ − ¡ n 5 ¢ + ... ± ¡ n n ¢ = 0. Hint: Observe that 0 = 0n = (1+(−1))n. Now use the binomial theorem. 11. Use the binomial theorem to show 9n = Pn k=0(−1)k ¡ n k ¢ 10n−k. Hint: Observe that 9n = (10+(−1))n. Now use the binomial theorem. 13. Assume n ≥3. Then ¡n 3 ¢ = ¡n−1 3 ¢ + ¡n−1 2 ¢ = ¡n−2 3 ¢ + ¡n−2 2 ¢ + ¡n−1 2 ¢ = ··· = ¡2 2 ¢ + ¡3 2 ¢ +···+ ¡n−1 2 ¢. Section 3.5 1. At a certain university 523 of the seniors are history majors or math majors (or both). There are 100 senior math majors, and 33 seniors are majoring in both history and math. How many seniors are majoring in history? Answer: Let A be the set of senior math majors and B be the set of senior history majors. From \|A ∪B\| = \|A\| + \|B\| −\|A ∩B\| we get 523 = 100 + \|B\| −33, so \|B\| = 523+33−100 = 456. There are 456 history majors. 3. How many 4-digit positive integers are there that are even or contain no 0’s? Answer: Let A be the set of 4-digit even positive integers, and let B be the set of 4-digit positive integers that contain no 0’s. We seek \|A ∪B\|. By the multiplication principle \|A\| = 9·10·10·5 = 4500. (Note the ﬁrst digit cannot be 0 and the last digit must be even.) Also \|B\| = 9·9·9·9 = 6561. Further, A∩B consists of all even 4-digit integers that have no 0’s. It follows that \|A∩B\| = 9·9·9·4 = 2916. Then the answer to our question is \|A∪B\| = \|A\|+\|B\|−\|A∩B\| = 4500+6561−2916 = 8145. 256 Solutions 5. How many 7-digit binary strings begin in 1 or end in 1 or have exactly four 1’s? Answer: Let A be the set of such strings that begin in 1. Let B be the set of such strings that end in 1. Let C be the set of such strings that have exactly four 1’s. Then the answer to our question is \|A ∪B∪C\|. Using Equation (3.4) to compute this number, we have \|A∪B∪C\| = \|A\|+\|B\|+\|C\|−\|A∩B\|−\|A∩C\|−\|B∩C\|+\|A∩B∩C\| = 26 +26 + ¡7 4 ¢ −25 − ¡6 3 ¢ − ¡6 3 ¢ + ¡5 2 ¢ = 64+64+35−32−20−20+10 = 101. 7. This problem concerns 4-card hands dealt oﬀof a standard 52-card deck. How many 4-card hands are there for which all four cards are of the same suit or all four cards are red? Answer: Let A be the set of 4-card hands for which all four cards are of the same suit. Let B be the set of 4-card hands for which all four cards are red. Then A ∩B is the set of 4-card hands for which the four cards are either all hearts or all diamonds. The answer to our question is \|A∪B\| = \|A\|+\|B\|−\|A∩B\| = 4 ¡13 4 ¢ + ¡26 4 ¢ −2 ¡13 4 ¢ = 2 ¡13 4 ¢ + ¡26 4 ¢ = 1430+14950 = 16380. 9. A 4-letter list is made from the letters L,I,S,T,E,D according to the following rule: Repetition is allowed, and the ﬁrst two letters on the list are vowels or the list ends in D. Answer: Let A be the set of such lists for which the ﬁrst two letters are vowels, so \|A\| = 2·2·6·6 = 144. Let B be the set of such lists that end in D, so \|B\| = 6·6·6·1 = 216. Then A ∩B is the set of such lists for which the ﬁrst two entries are vowels and the list ends in D. Thus \|A ∩B\| = 2 ·2 ·6 ·1 = 24. The answer to our question is \|A ∪B\| = \|A\|+\|B\|−\|A ∩B\| = 144+216−24 = 336. Chapter 4 Exercises 1. If x is an even integer, then x2 is even. Proof. Suppose x is even. Thus x = 2a for some a ∈Z. Consequently x2 = (2a)2 = 4a2 = 2(2a2). Therefore x2 = 2b, where b is the integer 2a2. Thus x2 is even by deﬁnition of an even number. ■ 3. If a is an odd integer, then a2 +3a+5 is odd. Proof. Suppose a is odd. Thus a = 2c +1 for some integer c, by deﬁnition of an odd number. Then a2 +3a+5 = (2c +1)2 +3(2c +1)+5 = 4c2 +4c +1+6c +3+5 = 4c2 +10c +9 = 4c2 +10c +8+1 = 2(2c2 +5c +4)+1. This shows a2 +3a+5 = 2b +1, where b = 2c2 +5c +4 ∈Z. Therefore a2 +3a+5 is odd. ■ 5. Suppose x, y ∈Z. If x is even, then xy is even. Proof. Suppose x, y ∈Z and x is even. Then x = 2a for some integer a, by deﬁnition of an even number. Thus xy = (2a)(y) = 2(ay). Therefore xy = 2b where b is the integer ay, so xy is even. ■ 257 7. Suppose a,b ∈Z. If a \| b, then a2 \| b2. Proof. Suppose a \| b. By deﬁnition of divisibility, this means b = ac for some integer c. Squaring both sides of this equation produces b2 = a2c2. Then b2 = a2d, where d = c2 ∈Z. By deﬁnition of divisibility, this means a2 \| b2. ■ 9. Suppose a is an integer. If 7 \| 4a, then 7 \| a. Proof. Suppose 7 \| 4a. By deﬁnition of divisibility, this means 4a = 7c for some integer c. Since 4a = 2(2a) it follows that 4a is even, and since 4a = 7c, we know 7c is even. But then c can’t be odd, because that would make 7c odd, not even. Thus c is even, so c = 2d for some integer d. Now go back to the equation 4a = 7c and plug in c = 2d. We get 4a = 14d. Dividing both sides by 2 gives 2a = 7d. Now, since 2a = 7d, it follows that 7d is even, and thus d cannot be odd. Then d is even, so d = 2e for some integer e. Plugging d = 2e back into 2a = 7d gives 2a = 14e. Dividing both sides of 2a = 14e by 2 produces a = 7e. Finally, the equation a = 7e means that 7 \| a, by deﬁnition of divisibility. ■ 11. Suppose a,b, c,d ∈Z. If a \| b and c \| d, then ac \| bd. Proof. Suppose a \| b and c \| d. As a \| b, the deﬁnition of divisibility means there is an integer x for which b = ax. As c \| d, the deﬁnition of divisibility means there is an integer y for which d = cy. Since b = ax, we can multiply one side of d = cy by b and the other by ax. This gives bd = axcy, or bd = (ac)(xy). Since xy ∈Z, the deﬁnition of divisibility applied to bd = (ac)(xy) gives ac \| bd. ■ 13. Suppose x, y ∈R. If x2 +5y = y2 +5x, then x = y or x+ y = 5. Proof. Suppose x2 +5y = y2 +5x. Then x2 −y2 = 5x−5y, and factoring gives (x−y)(x+ y) = 5(x−y). Now consider two cases. Case 1. If x −y ̸= 0 we can divide both sides of (x −y)(x + y) = 5(x −y) by the non-zero quantity x−y to get x+ y = 5. Case 2. If x−y = 0, then x = y. (By adding y to both sides.) Thus x = y or x+ y = 5. ■ 258 Solutions 15. If n ∈Z, then n2 +3n+4 is even. Proof. Suppose n ∈Z. We consider two cases. Case 1. Suppose n is even. Then n = 2a for some a ∈Z. Therefore n2 +3n+4 = (2a)2 +3(2a)+4 = 4a2 +6a+4 = 2(2a2 +3a+2). So n2 +3n+4 = 2b where b = 2a2 +3a+2 ∈Z, so n2 +3n+4 is even. Case 2. Suppose n is odd. Then n = 2a+1 for some a ∈Z. Therefore n2 +3n+4 = (2a+1)2 +3(2a+1)+4 = 4a2 +4a+1+6a+3+4 = 4a2 +10a+8 = 2(2a2+5a+4). So n2+3n+4 = 2b where b = 2a2+5a+4 ∈Z, so n2+3n+4 is even. In either case n2 +3n+4 is even. ■ 17. If two integers have opposite parity, then their product is even. Proof. Suppose a and b are two integers with opposite parity. Thus one is even and the other is odd. Without loss of generality, suppose a is even and b is odd. Therefore there are integers c and d for which a = 2c and b = 2d +1. Then the product of a and b is ab = 2c(2d +1) = 2(2cd + c). Therefore ab = 2k where k = 2cd + c ∈Z. Therefore the product ab is even. ■ 19. Suppose a,b, c ∈Z. If a2 \| b and b3 \| c then a6 \| c. Proof. Since a2 \| b we have b = ka2 for some k ∈Z. Since b3 \| c we have c = hb3 for some h ∈Z. Thus c = h(ka2)3 = hk3a6. Hence a6 \| c. ■ 21. If p is prime and 0 < k < p then p \| ¡p k ¢ . Proof. From the formula ¡p k ¢ = p! (p−k)!k!, we get p! = ¡p k ¢ (p −k)!k!. Now, since the prime number p is a factor of p! on the left, it must also be a factor of ¡p k ¢ (p−k)!k! on the right. Thus the prime number p appears in the prime factorization of ¡p k ¢ (p −k)!k!. Now, k! is a product of numbers smaller than p, so its prime factorization contains no p’s. Similarly the prime factorization of (p −k)! contains no p’s. But we noted that the prime factorization of ¡p k ¢ (p−k)!k! must contain a p, so it follows that the prime factorization of ¡p k ¢ contains a p. Thus ¡p k ¢ is a multiple of p, so p divides ¡p k ¢. ■ 23. If n ∈N then ¡2n n ¢ is even. Proof. By deﬁnition, ¡2n n ¢ is the number of n-element subsets of a set A with 2n elements. For each subset X ⊆A with \|X\| = n, the complement X is a diﬀerent set, but it also has 2n−n = n elements. Imagine listing out all the n-elements subset of a set A. It could be done in such a way that the list has form X1, X1, X2, X2, X3, X3, X4, X4, X5, X5 ... This list has an even number of items, for they are grouped in pairs. Thus ¡2n n ¢ is even. ■ 259 25. If a,b, c ∈N and c ≤b ≤a then ¡a b ¢¡b c ¢ = ¡ a b−c ¢¡a−b+c c ¢ . Proof. Assume a,b, c ∈N with c ≤b ≤a. Then we have ¡a b ¢¡b c ¢ = a! (a−b)!b! b! (b−c)!c! = a! (a−b+c)!(a−b)! (a−b+c)! (b−c)!c! = a! (b−c)!(a−b+c)! (a−b+c)! (a−b)!c! = ¡ a b−c ¢¡a−b+c c ¢ . ■ 27. Suppose a,b ∈N. If gcd(a,b) > 1, then b \| a or b is not prime. Proof. Suppose gcd(a,b) > 1. Let c = gcd(a,b) > 1. Then since c is a divisor of both a and b, we have a = cx and b = cy for integers x and y. We divide into two cases according to whether or not b is prime. Case I. Suppose b is prime. Then the above equation b = cy with c > 1 forces c = b and y = 1. Then a = cx becomes a = bx, which means b \| a. We conclude that the statement “b \| a or b is not prime,” is true. Case II. Suppose b is not prime. Then the statement “b \| a or b is not prime,” is automatically true. ■ Chapter 5 Exercises 1. Proposition Suppose n ∈Z. If n2 is even, then n is even. Proof. (Contrapositive) Suppose n is not even. Then n is odd, so n = 2a+1 for some integer a, by deﬁnition of an odd number. Thus n2 = (2a+1)2 = 4a2+4a+1 = 2(2a2 +2a)+1. Consequently n2 = 2b+1, where b is the integer 2a2 +2a, so n2 is odd. Therefore n2 is not even. ■ 3. Proposition Suppose a,b ∈Z. If a2(b2 −2b) is odd, then a and b are odd. Proof. (Contrapositive) Suppose it is not the case that a and b are odd. Then, by DeMorgan’s law, at least one of a and b is even. Let us look at these cases separately. Case 1. Suppose a is even. Then a = 2c for some integer c. Thus a2(b2 −2b) = (2c)2(b2 −2b) = 2(2c2(b2 −2b)), which is even. Case 2. Suppose b is even. Then b = 2c for some integer c. Thus a2(b2 −2b) = a2((2c)2 −2(2c)) = 2(a2(2c2 −2c)), which is even. (A third case involving a and b both even is unnecessary, for either of the two cases above cover this case.) Thus in either case a2(b2 −2b) is even, so it is not odd. ■ 5. Proposition Suppose x ∈R. If x2 +5x < 0 then x < 0. Proof. (Contrapositive) Suppose it is not the case that x < 0, so x ≥0. Then neither x2 nor 5x is negative, so x2+5x ≥0. Thus it is not true that x2+5x < 0. ■ 260 Solutions 7. Proposition Suppose a,b ∈Z. If both ab and a+ b are even, then both a and b are even. Proof. (Contrapositive) Suppose it is not the case that both a and b are even. Then at least one of them is odd. There are three cases to consider. Case 1. Suppose a is even and b is odd. Then there are integers c and d for which a = 2c and b = 2d + 1. Then ab = 2c(2d + 1), which is even; and a+ b = 2c +2d +1 = 2(c + d)+1, which is odd. Thus it is not the case that both ab and a+ b are even. Case 2. Suppose a is odd and b is even. Then there are integers c and d for which a = 2c +1 and b = 2d. Then ab = (2c +1)(2d) = 2(d(2c +1)), which is even; and a + b = 2c +1+2d = 2(c + d)+1, which is odd. Thus it is not the case that both ab and a+ b are even. Case 3. Suppose a is odd and b is odd. Then there are integers c and d for which a = 2c + 1 and b = 2d + 1. Then ab = (2c + 1)(2d + 1) = 4cd + 2c + 2d + 1 = 2(2cd + c + d)+1, which is odd; and a + b = 2c +1+2d +1 = 2(c + d +1), which is even. Thus it is not the case that both ab and a+ b are even. These cases show that it is not the case that ab and a+ b are both even. (Note that unlike Exercise 3 above, we really did need all three cases here, for each case involved speciﬁc parities for both a and b.) ■ 9. Proposition Suppose n ∈Z. If 3 ∤n2, then 3 ∤n. Proof. (Contrapositive) Suppose it is not the case that 3 ∤n, so 3 \| n. This means that n = 3a for some integer a. Consequently n2 = 9a2, from which we get n2 = 3(3a2). This shows that there in an integer b = 3a2 for which n2 = 3b, which means 3 \| n2. Therefore it is not the case that 3 ∤n2. ■ 11. Proposition Suppose x, y ∈Z. If x2(y+3) is even, then x is even or y is odd. Proof. (Contrapositive) Suppose it is not the case that x is even or y is odd. Using DeMorgan’s law, this means x is not even and y is not odd, which is to say x is odd and y is even. Thus there are integers a and b for which x = 2a+1 and y = 2b. Consequently x2(y + 3) = (2a + 1)2(2b + 3) = (4a2 + 4a + 1)(2b + 3) = 8a2b +8ab +2b +12a2 +12a+3 = 8a2b +8ab +2b +12a2 +12a+2+1 = 2(4a2b+4ab+b+6a2 +6a+1)+1. This shows x2(y+3) = 2c+1 for c = 4a2b+4ab+ b +6a2 +6a+1 ∈Z. Consequently, x2(y+3) is not even. ■ 13. Proposition Suppose x ∈R. If x5 +7x3 +5x ≥x4 + x2 +8, then x ≥0. Proof. (Contrapositive) Suppose it is not true that x ≥0. Then x < 0, that is x is negative. Consequently, the expressions x5, 7x3 and 5x are all negative (note the odd powers) so x5 + 7x3 + 5x < 0. Similarly the terms x4, x2, and 8 are all positive (note the even powers), so 0 < x4 + x2 + 8. From this we get x5 +7x3 +5x < x4 + x2 +8, so it is not true that x5 +7x3 +5x ≥x4 + x2 +8. ■ 261 15. Proposition Suppose x ∈Z. If x3 −1 is even, then x is odd. Proof. (Contrapositive) Suppose x is not odd. Thus x is even, so x = 2a for some integer a. Then x3 −1 = (2a)3 −1 = 8a3 −1 = 8a3 −2+1 = 2(4a3 −1)+1. Therefore x3 −1 = 2b +1 where b = 4a3 −1 ∈Z, so x3 −1 is odd. Thus x3 −1 is not even. ■ 17. Proposition If n is odd, then 8 \| (n2 −1). Proof. (Direct) Suppose n is odd, so n = 2a+1 for some integer a. Then n2 −1 = (2a+1)2 −1 = 4a2 +4a = 4(a2 +a) = 4a(a+1). So far we have n2 −1 = 4a(a+1), but we want a factor of 8, not 4. But notice that one of a or a+1 must be even, so a(a+1) is even and hence a(a+1) = 2c for some integer c. Now we have n2 −1 = 4a(a+1) = 4(2c) = 8c. But n2 −1 = 8c means 8 \| (n2 −1). ■ 19. Proposition Let a,b ∈Z and n ∈N. If a ≡b (mod n) and a ≡c (mod n), then c ≡b (mod n). Proof. (Direct) Suppose a ≡b (mod n) and a ≡c (mod n). This means n \| (a−b) and n \| (a−c). Thus there are integers d and e for which a−b = nd and a−c = ne. Subtracting the second equation from the ﬁrst gives c −b = nd −ne. Thus c −b = n(d −e), so n \| (c −b) by deﬁnition of divisibility. Therefore c ≡b (mod n) by deﬁnition of congruence modulo n. ■ 21. Proposition Let a,b ∈Z and n ∈N. If a ≡b (mod n), then a3 ≡b3 (mod n). Proof. (Direct) Suppose a ≡b (mod n). This means n \| (a −b), so there is an integer c for which a−b = nc. Then: a−b = nc (a−b)(a2 + ab + b2) = nc(a2 + ab + b2) a3 + a2b + ab2 −ba2 −ab2 −b3 = nc(a2 + ab + b2) a3 −b3 = nc(a2 + ab + b2). Since a2 + ab + b2 ∈Z, the equation a3 −b3 = nc(a2 + ab + b2) implies n \| (a3 −b3), and therefore a3 ≡b3 (mod n). ■ 23. Proposition Let a,b, c ∈Z and n ∈N. If a ≡b (mod n), then ca ≡cb (mod n). Proof. (Direct) Suppose a ≡b (mod n). This means n \| (a −b), so there is an integer d for which a−b = nd. Multiply both sides of this by c to get ac−bc = ndc. Consequently, there is an integer e = dc for which ac −bc = ne, so n \| (ac −bc) and consequently ac ≡bc (mod n). ■ 25. If n ∈N and 2n −1 is prime, then n is prime. Proof. Assume n is not prime. Write n = ab for some a,b > 1. Then 2n −1 = 2ab−1 = ¡ 2b−1 ¢¡ 2ab−b+2ab−2b+2ab−3b+···+2ab−ab¢. Hence 2n−1 is composite. ■ 262 Solutions 27. If a ≡0 (mod 4) or a ≡1 (mod 4) then ¡a 2 ¢ is even. Proof. We prove this directly. Assume a ≡0 (mod 4). Then ¡a 2 ¢ = a(a−1) 2 . Since a = 4k for some k ∈N, we have ¡a 2 ¢ = 4k(4k−1) 2 = 2k(4k −1). Hence ¡a 2 ¢ is even. Now assume a ≡1 (mod 4). Then a = 4k+1 for some k ∈N. Hence ¡a 2 ¢ = (4k+1)(4k) 2 = 2k(4k +1). Hence, ¡a 2 ¢ is even. This proves the result. ■ 29. If integers a and b are not both zero, then gcd(a,b) = gcd(a−b,b). Proof. (Direct) Suppose integers a and b are not both zero. Let d = gcd(a,b). Because d is a divisor of both a and b, we have a = dx and b = dy for some integers x and y. Then a −b = dx −dy = d(x −y), so it follows that d is also a common divisor of a−b and b. Therefore it can’t be greater than the greatest common divisor of a−b and b, which is to say gcd(a,b) = d ≤gcd(a−b,b). Now let e = gcd(a−b,b). Then e divides both a−b and b, that is, a−b = ex and b = ey for integers x and y. Then a = (a−b)+ b = ex+ ey = e(x+ y), so now we see that e is a divisor of both a and b. Thus it is not more than their greatest common divisor, that is, gcd(a−b,b) = e ≤gcd(a,b). The above two paragraphs have given gcd(a,b) ≤gcd(a−b,b) and gcd(a−b,b) ≤ gcd(a,b). Thus gcd(a,b) = gcd(a−b,b). ■ 31. Suppose the division algorithm applied to a and b yields a = qb + r. Then gcd(a,b) = gcd(r,b). Proof. Suppose a = qb+r. Let d = gcd(a,b), so d is a common divisor of a and b; thus a = dx and b = dy for some integers x and y. Then dx = a = qb + r = qdy+ r, hence dx = qdy+ r, and so r = dx−qdy = d(x−qy). Thus d is a divisor of r (and also of b), so gcd(a,b) = d ≤gcd(r,b). On the other hand, let e = gcd(r,b), so r = ex and b = ey for some integers x and y. Then a = qb + r = qey+ ex = e(qy+ x). Hence e is a divisor of a (and of course also of b) so gcd(r,b) = e ≤gcd(a,b). We’ve now shown gcd(a,b) ≤gcd(r,b) and gcd(r,b) ≤gcd(a,b), so gcd(r,b) = gcd(a,b). ■ Chapter 6 Exercises 1. Suppose n is an integer. If n is odd, then n2 is odd. Proof. Suppose for the sake of contradiction that n is odd and n2 is not odd. Then n2 is even. Now, since n is odd, we have n = 2a +1 for some integer a. Thus n2 = (2a+1)2 = 4a2 +4a+1 = 2(2a2 +2a)+1. This shows n2 = 2b +1, where b is the integer b = 2a2 + 2a. Therefore we have n2 is odd and n2 is even, a contradiction. ■ 263 3. Prove that 3 p 2 is irrational. Proof. Suppose for the sake of contradiction that 3 p 2 is not irrational. Therefore it is rational, so there exist integers a and b for which 3 p 2 = a b. Let us assume that this fraction is reduced, so a and b are not both even. Now we have 3 p 2 3 = ¡ a b ¢3, which gives 2 = a3 b3 , or 2b3 = a3. From this we see that a3 is even, from which we deduce that a is even. (For if a were odd, then a3 = (2c +1)3 = 8c3 +12c2 +6c +1 = 2(4c3 +6c2 +3c)+1 would be odd, not even.) Since a is even, it follows that a = 2d for some integer d. The equation 2b3 = a3 from above then becomes 2b3 = (2d)3, or 2b3 = 8d3. Dividing by 2, we get b3 = 4d3, and it follows that b3 is even. Thus b is even also. (Using the same argument we used when a3 was even.) At this point we have discovered that both a and b are even, contradicting the fact (observed above) that the a and b are not both even. ■ Here is an alternative proof. Proof. Suppose for the sake of contradiction that 3 p 2 is not irrational. Therefore there exist integers a and b for which 3 p 2 = a b. Cubing both sides, we get 2 = a3 b3 . From this, a3 = b3 + b3, which contradicts Fermat’s last theorem. ■ 5. Prove that p 3 is irrational. Proof. Suppose for the sake of contradiction that p 3 is not irrational. Therefore it is rational, so there exist integers a and b for which p 3 = a b. Let us assume that this fraction is reduced, so a and b have no common factor. Notice that p 3 2 = ¡ a b ¢2, so 3 = a2 b2 , or 3b2 = a2. This means 3 \| a2. Now we are going to show that if a ∈Z and 3 \| a2, then 3 \| a. (This is a proof-within-a-proof.) We will use contrapositive proof to prove this conditional statement. Suppose 3 ∤a. Then there is a remainder of either 1 or 2 when 3 is divided into a. Case 1. There is a remainder of 1 when 3 is divided into a. Then a = 3m+1 for some integer m. Consequently, a2 = 9m2 +6m+1 = 3(3m2 +2m)+1, and this means 3 divides into a2 with a remainder of 1. Thus 3 ∤a2. Case 2. There is a remainder of 2 when 3 is divided into a. Then a = 3m+2 for some integer m. Consequently, a2 = 9m2 + 12m + 4 = 9m2 + 12m + 3 + 1 = 3(3m2+4m+1)+1, and this means 3 divides into a2 with a remainder of 1. Thus 3 ∤a2. In either case we have 3 ∤a2, so we’ve shown 3 ∤a implies 3 ∤a2. Therefore, if 3 \| a2, then 3 \| a. Now go back to 3 \| a2 in the ﬁrst paragraph. This combined with the result of the second paragraph implies 3 \| a, so a = 3d for some integer d. Now also in the ﬁrst paragraph we had 3b2 = a2, which now becomes 3b2 = (3d)2 or 3b2 = 9d2, so b2 = 3d2. But this means 3 \| b2, and the second paragraph implies 3 \| b. Thus we have concluded that 3 \| a and 3 \| b, but this contradicts the fact that the fraction a b is reduced. ■ 264 Solutions 7. If a,b ∈Z, then a2 −4b −3 ̸= 0. Proof. Suppose for the sake of contradiction that a,b ∈Z but a2−4b−3 = 0. Then we have a2 = 4b+3 = 2(2b+1)+1, which means a2 is odd. Therefore a is odd also, so a = 2c +1 for some integer c. Plugging this back into a2 −4b −3 = 0 gives us (2c +1)2 −4b −3 = 0 4c2 +4c +1−4b −3 = 0 4c2 +4c −4b = 2 2c2 +2c −2b = 1 2(c2 + c −b) = 1. From this last equation, we see that 1 is an even number, a contradiction. ■ 9. Suppose a,b ∈R and a ̸= 0. If a is rational and ab is irrational, then b is irrational. Proof. Suppose for the sake of contradiction that a is rational and ab is irra-tional and b is not irrational. Thus we have a and b rational, and ab irrational. Since a and b are rational, we know there are integers c,d, e, f for which a = c d and b = e f . Then ab = ce d f , and since both ce and d f are integers, it follows that ab is rational. But this is a contradiction because we started out with ab irrational. ■ 11. There exist no integers a and b for which 18a+6b = 1. Proof. Suppose for the sake of contradiction that there do exist integers a and b for which 18a + 6b = 1. Then 1 = 2(9a + 3b), which means 1 is even, a contradiction. ■ 13. For every x ∈[π/2,π], sinx−cosx ≥1. Proof. Suppose for the sake of contradiction that x ∈[π/2,π], but sinx−cosx < 1. Since x ∈[π/2,π], we know sinx ≥0 and cosx ≤0, so sinx −cosx ≥0. Therefore we have 0 ≤sinx −cosx < 1. Now the square of any number between 0 and 1 is still a number between 0 and 1, so we have 0 ≤(sinx −cosx)2 < 1, or 0 ≤ sin2 x−2sinxcosx+cos2 x < 1. Using the fact that sin2 x+cos2 x = 1, this becomes 0 ≤−2sinxcosx+1 < 1. Subtracting 1, we obtain −2sinxcosx < 0. But above we remarked that sinx ≥0 and cosx ≤0, and hence −2sinxcosx ≥0. We now have the contradiction −2sinxcosx < 0 and −2sinxcosx ≥0. ■ 15. If b ∈Z and b ∤k for every k ∈N, then b = 0. Proof. Suppose for the sake of contradiction that b ∈Z and b ∤k for every k ∈N, but b ̸= 0. Case 1. Suppose b > 0. Then b ∈N, so b\|b, contradicting b ∤k for every k ∈N. Case 2. Suppose b < 0. Then −b ∈N, so b\|(−b), again a contradiction ■ 265 17. For every n ∈Z, 4 ∤(n2 +2). Proof. Assume there exists n ∈Z with 4 \| (n2+2). Then for some k ∈Z, 4k = n2+2 or 2k = n2 + 2(1 −k). If n is odd, this means 2k is odd, and we’ve reached a contradiction. If n is even then n = 2j and we get k = 2j2 +1−k for some j ∈Z. Hence 2(k −j2) = 1, so 1 is even, a contradiction. ■ Remark. It is fairly easy to see that two more than a perfect square is always either 2 (mod 4) or 3 (mod 4). This would end the proof immediately. 19. The product of 5 consecutive integers is a multiple of 120. Proof. Given any collection of 5 consecutive integers, at least one must be a multiple of two, at least one must be a multiple of three, at least one must be a multiple of four and at least one must be a multiple of 5. Hence the product is a multiple of 5·4·3·2 = 120. In particular, the product is a multiple of 60. ■ 21. Hints for Exercises 20–23. For Exercises 20, ﬁrst show that the equation a2 + b2 = 3c2 has no solutions (other than the trivial solution (a,b, c) = (0,0,0)) in the integers. To do this, investigate the remainders of a sum of squares (mod 4). After you’ve done this, prove that the only solution is indeed the trivial solution. Now, assume that the equation x2 + y2 −3 = 0 has a rational solution. Use the deﬁnition of rational numbers to yield a contradiction. Chapter 7 Exercises 1. Suppose x ∈Z. Then x is even if and only if 3x+5 is odd. Proof. We ﬁrst use direct proof to show that if x is even, then 3x +5 is odd. Suppose x is even. Then x = 2n for some integer n. Thus 3x +5 = 3(2n)+5 = 6n +5 = 6n +4+1 = 2(3n +2)+1. Thus 3x +5 is odd because it has form 2k +1, where k = 3n+2 ∈Z. Conversely, we need to show that if 3x + 5 is odd, then x is even. We will prove this using contrapositive proof. Suppose x is not even. Then x is odd, so x = 2n+1 for some integer n. Thus 3x+5 = 3(2n+1)+5 = 6n+8 = 2(3n+4). This means says 3x+5 is twice the integer 3n+4, so 3x+5 is even, not odd. ■ 3. Given an integer a, then a3 + a2 + a is even if and only if a is even. Proof. First we will prove that if a3 +a2 +a is even then a is even. This is done with contrapositive proof. Suppose a is not even. Then a is odd, so there is an integer n for which a = 2n+1. Then a3 + a2 + a = (2n+1)3 +(2n+1)2 +(2n+1) = 8n3 +12n2 +6n+1+4n2 +4n+1+2n+1 = 8n3 +16n2 +12n+2+1 = 2(4n3 +8n2 +6n+1)+1. 266 Solutions This expresses a3 + a2 + a as twice an integer plus 1, so a3 + a2 + a is odd, not even. We have now shown that if a3 + a2 + a is even then a is even. Conversely, we need to show that if a is even, then a3+a2+a is even. We will use direct proof. Suppose a is even, so a = 2n for some integer n. Then a3 + a2 + a = (2n)3 +(2n)2 +2n = 8n3 +4n2 +2n = 2(4n3 +2n2 + n). Therefore, a3 + a2 + a is even because it’s twice an integer. ■ 5. An integer a is odd if and only if a3 is odd. Proof. Suppose that a is odd. Then a = 2n + 1 for some integer n, and a3 = (2n+1)3 = 8n3 +12n2 +6n+1 = 2(4n3 +6n2 +3n)+1. This shows that a3 is twice an integer, plus 1, so a3 is odd. Thus we’ve proved that if a is odd then a3 is odd. Conversely we need to show that if a3 is odd, then a is odd. For this we employ contrapositive proof. Suppose a is not odd. Thus a is even, so a = 2n for some integer n. Then a3 = (2n)3 = 8n3 = 2(4n3) is even (not odd). ■ 7. Suppose x, y ∈R. Then (x+ y)2 = x2 + y2 if and only if x = 0 or y = 0. Proof. First we prove with direct proof that if (x + y)2 = x2 + y2, then x = 0 or y = 0. Suppose (x+ y)2 = x2+ y2. From this we get x2+2xy+ y2 = x2+ y2, so 2xy = 0, and hence xy = 0. Thus x = 0 or y = 0. Conversely, we need to show that if x = 0 or y = 0, then (x + y)2 = x2 + y2. This will be done with cases. Case 1. If x = 0 then (x+ y)2 = (0+ y)2 = y2 = 02 + y2 = x2 + y2. Case 2. If y = 0 then (x+ y)2 = (x+0)2 = x2 = x2 +02 = x2 + y2. Either way, we have (x+ y)2 = x2 + y2. ■ 9. Suppose a ∈Z. Prove that 14 \| a if and only if 7 \| a and 2 \| a. Proof. First we prove that if 14 \| a, then 7 \| a and 2 \| a. Direct proof is used. Suppose 14 \| a. This means a = 14m for some integer m. Therefore a = 7(2m), which means 7 \| a, and also a = 2(7m), which means 2 \| a. Thus 7 \| a and 2 \| a. Conversely, we need to prove that if 7 \| a and 2 \| a, then 14 \| a. Once again direct proof if used. Suppose 7 \| a and 2 \| a. Since 2 \| a it follows that a = 2m for some integer m, and that in turn implies that a is even. Since 7 \| a it follows that a = 7n for some integer n. Now, since a is known to be even, and a = 7n, it follows that n is even (if it were odd, then a = 7n would be odd). Thus n = 2p for an appropriate integer p, and plugging n = 2p back into a = 7n gives a = 7(2p), so a = 14p. Therefore 14 \| a. ■ 267 11. Suppose a,b ∈Z. Prove that (a−3)b2 is even if and only if a is odd or b is even. Proof. First we will prove that if (a −3)b2 is even, then a is odd or b is even. For this we use contrapositive proof. Suppose it is not the case that a is odd or b is even. Then by DeMorgan’s law, a is even and b is odd. Thus there are integers m and n for which a = 2m and b = 2n + 1. Now observe (a−3)b2 = (2m−3)(2n+1)2 = (2m−3)(4n2+4n+1) = 8mn2+8mn+2m−12n2−12n−3 = 8mn2 +8mn +2m −12n2 −12n −4+1 = 2(4mn2 +4mn + m −6n2 −6n −2)+1. This shows (a−3)b2 is odd, so it’s not even. Conversely, we need to show that if a is odd or b is even, then (a−3)b2 is even. For this we use direct proof, with cases. Case 1. Suppose a is odd. Then a = 2m+1 for some integer m. Thus (a−3)b2 = (2m+1−3)b2 = (2m−2)b2 = 2(m−1)b2. Thus in this case (a−3)b2 is even. Case 2. Suppose b is even. Then b = 2n for some integer n. Thus (a −3)b2 = (a−3)(2n)2 = (a−3)4n2 = 2(a−3)2n2 =. Thus in this case (a−3)b2 is even. Therefore, in any event, (a−3)b2 is even. ■ 13. Suppose a,b ∈Z. If a+ b is odd, then a2 + b2 is odd. Hint: Use direct proof. Suppose a+ b is odd. Argue that this means a and b have opposite parity. Then use cases. 15. Suppose a,b ∈Z. Prove that a+ b is even if and only if a and b have the same parity. Proof. First we will show that if a+b is even, then a and b have the same parity. For this we use contrapositive proof. Suppose it is not the case that a and b have the same parity. Then one of a and b is even and the other is odd. Without loss of generality, let’s say that a is even and b is odd. Thus there are integers m and n for which a = 2m and b = 2n+1. Then a+ b = 2m+2n+1 = 2(m+ n)+1, so a+ b is odd, not even. Conversely, we need to show that if a and b have the same parity, then a+ b is even. For this, we use direct proof with cases. Suppose a and b have the same parity. Case 1. Both a and b are even. Then there are integers m and n for which a = 2m and b = 2n, so a+ b = 2m+2n = 2(m+ n) is clearly even. Case 2. Both a and b are odd. Then there are integers m and n for which a = 2m+1 and b = 2n+1, so a+ b = 2m+1+2n+1 = 2(m+ n+1) is clearly even. Either way, a+ b is even. This completes the proof. ■ 17. There is a prime number between 90 and 100. Proof. Simply observe that 97 is prime. ■ 268 Solutions 19. If n ∈N, then 20 +21 +22 +23 +24 +···+2n = 2n+1 −1. Proof. We use direct proof. Suppose n ∈N. Let S be the number S = 20 +21 +22 +23 +24 +···+2n−1 +2n. (1) In what follows, we will solve for S and show S = 2n+1 −1. Multiplying both sides of (1) by 2 gives 2S = 21 +22 +23 +24 +25 +···+2n +2n+1. (2) Now subtract Equation (1) from Equation (2) to obtain 2S −S = −20 + 2n+1, which simpliﬁes to S = 2n+1 −1. Combining this with Equation (1) produces 20 +21 +22 +23 +24 +···+2n = 2n+1 −1, so the proof is complete. ■ 21. Every real solution of x3 + x+3 = 0 is irrational. Proof. Suppose for the sake of contradiction that this polynomial has a rational solution a b. We may assume that this fraction is fully reduced, so a and b are not both even. We have ¡ a b ¢3 + a b +3 = 0. Clearing the denominator gives a3 + ab2 +3b3 = 0. Consider two cases: First, if both a and b are odd, the left-hand side is a sum of three odds, which is odd, meaning 0 is odd, a contradiction. Second, if one of a and b is odd and the other is even, then the middle term of a3 + ab2 +3b3 is even, while a3 and 3b2 have opposite parity. Then a3 + ab2 +3b3 is the sum of two evens and an odd, which is odd, again contradicting the fact that 0 is even. ■ 23. Suppose a,b and c are integers. If a \| b and a \| (b2 −c), then a \| c. Proof. (Direct) Suppose a \| b and a \| (b2 −c). This means that b = ad and b2 −c = ae for some integers d and e. Squaring the ﬁrst equation produces b2 = a2d2. Subtracting b2 −c = ae from b2 = a2d2 gives c = a2d2 −ae = a(ad2 −e). As ad2 −e ∈Z, it follows that a \| c. ■ 25. If p > 1 is an integer and n ∤p for each integer n for which 2 ≤n ≤pp, then p is prime. Proof. (Contrapositive) Suppose that p is not prime, so it factors as p = mn for 1 < m,n < p. Observe that it is not the case that both m > pp and n > pp, because if this were true the inequalities would multiply to give mn > pppp = p, which contradicts p = mn. Therefore m ≤pp or n ≤pp. Without loss of generality, say n ≤pp. Then the equation p = mn gives n \| p, with 1 < n ≤pp. Therefore it is not true that n ∤p for each integer n for which 2 ≤n ≤pp. ■ 269 27. Suppose a,b ∈Z. If a2 + b2 is a perfect square, then a and b are not both odd. Proof. (Contradiction) Suppose a2+b2 is a perfect square, and a and b are both odd. As a2 +b2 is a perfect square, say c is the integer for which c2 = a2 +b2. As a and b are odd, we have a = 2m+1 and b = 2n+1 for integers m and n. Then c2 = a2 + b2 = (2m+1)2 +(2n+1)2 = 4(m2 + n2 + mn)+2. This is even, so c is even also; let c = 2k. Now the above equation results in (2k)2 = 4(m2 +n2 +mn)+2, which simpliﬁes to 2k2 = 2(m2 +n2 +mn)+1. Thus 2k2 is both even and odd, a contradiction. ■ 29. If a \| bc and gcd(a,b) = 1, then a \| c. Proof. (Direct) Suppose a \| bc and gcd(a,b) = 1. The fact that a \| bc means bc = az for some integer z. The fact that gcd(a,b) = 1 means that ax+ by = 1 for some integers x and y (by Proposition 7.1 on page 126). From this we get acx+bcy = c; substituting bc = az yields acx+azy = c, that is, a(cx+zy) = c. Therefore a \| c. ■ 31. If n ∈Z, then gcd(n,n+1) = 1. Proof. Suppose d is a positive integer that is a common divisor of n and n+1. Then n = dx and n+1 = dy for integers x and y. Then 1 = (n+1)−n = dy−dx = d(y −x). Now, 1 = d(y −x) is only possible if d = ±1 and y −x = ±1. Thus the greatest common divisor of n and n+1 can be no greater than 1. But 1 does divide both n and n+1, so gcd(n,n+1) = 1. ■ 33. If n ∈Z, then gcd(2n+1,4n2 +1) = 1. Proof. Note that 4n2 +1 = (2n+1)(2n−1)+2. Therefore, it suﬃces to show that gcd(2n + 1,(2n + 1)(2n −1) + 2) = 1. Let d be a common positive divisor of both 2n+1 and (2n+1)(2n−1)+2, so 2n+1 = dx and (2n+1)(2n−1)+2 = dy for integers x and y. Substituting the ﬁrst equation into the second gives dx(2n−1)+2 = dy, so 2 = dy−dx(2n−1) = d(y−2nx−x). This means d divides 2, so d equals 1 or 2. But the equation 2n+1 = dx means d must be odd. Therefore d = 1, that is, gcd(2n+1,(2n+1)(2n−1)+2) = 1. ■ 35. Suppose a,b ∈N. Then a = gcd(a,b) if and only if a \| b. Proof. Suppose a = gcd(a,b). This means a is a divisor of both a and b. In particular a \| b. Conversely, suppose a \| b. Then a divides both a and b, so a ≤gcd(a,b). On the other hand, since gcd(a,b) divides a, we have a = gcd(a,b)· x for some integer x. As all integers involved are positive, it follows that a ≥gcd(a,b). It has been established that a ≤gcd(a,b) and a ≥gcd(a,b). Thus a = gcd(a,b). ■ 270 Solutions Chapter 8 Exercises 1. Prove that {12n : n ∈Z} ⊆{2n : n ∈Z}∩{3n : n ∈Z}. Proof. Suppose a ∈{12n : n ∈Z}. This means a = 12n for some n ∈Z. Therefore a = 2(6n) and a = 3(4n). From a = 2(6n), it follows that a is multiple of 2, so a ∈{2n : n ∈Z}. From a = 3(4n), it follows that a is multiple of 3, so a ∈{3n : n ∈Z}. Thus by deﬁnition of the intersection of two sets, we have a ∈{2n : n ∈Z} ∩ {3n : n ∈Z}. Thus {12n : n ∈Z} ⊆{2n : n ∈Z}∩{3n : n ∈Z}. ■ 3. If k ∈Z, then {n ∈Z : n \| k} ⊆ © n ∈Z : n \| k2ª. Proof. Suppose k ∈Z. We now need to show {n ∈Z : n \| k} ⊆ © n ∈Z : n \| k2ª. Suppose a ∈{n ∈Z : n \| k}. Then it follows that a \| k, so there is an integer c for which k = ac. Then k2 = a2c2. Therefore k2 = a(ac2), and from this the deﬁnition of divisibility gives a \| k2. But a \| k2 means that a ∈ © n ∈Z : n \| k2ª. We have now shown {n ∈Z : n \| k} ⊆ © n ∈Z : n \| k2ª. ■ 5. If p and q are integers, then {pn : n ∈N}∩{qn : n ∈N} ̸= ;. Proof. Suppose p and q are integers. Consider the integer pq. Observe that pq ∈{pn : n ∈N} and pq ∈{qn : n ∈N}, so pq ∈{pn : n ∈N} ∩{qn : n ∈N}. Therefore {pn : n ∈N}∩{qn : n ∈N} ̸= ;. ■ 7. Suppose A,B and C are sets. If B ⊆C, then A ×B ⊆A ×C. Proof. This is a conditional statement, and we’ll prove it with direct proof. Suppose B ⊆C. (Now we need to prove A ×B ⊆A ×C.) Suppose (a,b) ∈A×B. Then by deﬁnition of the Cartesian product we have a ∈A and b ∈B. But since b ∈B and B ⊆C, we have b ∈C. Since a ∈A and b ∈C, it follows that (a,b) ∈A ×C. Now we’ve shown (a,b) ∈A ×B implies (a,b) ∈A ×C, so A ×B ⊆A ×C. In summary, we’ve shown that if B ⊆C, then A ×B ⊆A ×C. This completes the proof. ■ 9. If A,B and C are sets then A ∩(B ∪C) = (A ∩B)∪(A ∩C). Proof. We use the distributive law P ∧(Q ∨R) = (P ∧Q)∨(P ∧R) from page 50. A ∩(B ∪C) = {x : x ∈A ∧x ∈B ∪C} (def. of intersection) = {x : x ∈A ∧(x ∈B ∨x ∈C)} (def. of union) = © x : ¡ x ∈A ∧x ∈B ¢ ∨ ¡ x ∈A ∧x ∈C ¢ª (distributive law) = {x : (x ∈A ∩B) ∨(x ∈A ∩C)} (def. of intersection) = (A ∩B)∪(A ∩C) (def. of union) The proof is complete. ■ 271 11. If A and B are sets in a universal set U, then A ∪B = A ∩B. Proof. Just observe the following sequence of equalities. A ∪B = U −(A ∪B) (def. of complement) = {x : (x ∈U)∧(x ∉A ∪B)} (def. of −) = {x : (x ∈U)∧∼(x ∈A ∪B)} = {x : (x ∈U)∧∼((x ∈A)∨(x ∈B))} (def. of ∪) = {x : (x ∈U)∧(∼(x ∈A)∧∼(x ∈B))} (DeMorgan) = {x : (x ∈U)∧(x ∉A)∧(x ∉B)} = {x : (x ∈U)∧(x ∈U)∧(x ∉A)∧(x ∉B)} (x ∈U) = (x ∈U)∧(x ∈U) = {x : ((x ∈U)∧(x ∉A))∧((x ∈U)∧(x ∉B))} (regroup) = {x : (x ∈U)∧(x ∉A)}∩{x : (x ∈U)∧(x ∉B)} (def. of ∩) = (U −A)∩(U −B) (def. of −) = A ∩B (def. of complement) The proof is complete. ■ 13. If A,B and C are sets, then A −(B ∪C) = (A −B)∩(A −C). Proof. Just observe the following sequence of equalities. A −(B ∪C) = {x : (x ∈A)∧(x ∉B ∪C)} (def. of −) = {x : (x ∈A)∧∼(x ∈B ∪C)} = {x : (x ∈A)∧∼((x ∈B)∨(x ∈C))} (def. of ∪) = {x : (x ∈A)∧(∼(x ∈B)∧∼(x ∈C))} (DeMorgan) = {x : (x ∈A)∧(x ∉B)∧(x ∉C)} = {x : (x ∈A)∧(x ∈A)∧(x ∉B)∧(x ∉C)} (x ∈A) = (x ∈A)∧(x ∈A) = {x : ((x ∈A)∧(x ∉B))∧((x ∈A)∧(x ∉C))} (regroup) = {x : (x ∈A)∧(x ∉B)}∩{x : (x ∈A)∧(x ∉C)} (def. of ∩) = (A −B)∩(A −C) (def. of −) The proof is complete. ■ 15. If A,B and C are sets, then (A ∩B)−C = (A −C)∩(B −C). Proof. Just observe the following sequence of equalities. (A ∩B)−C = {x : (x ∈A ∩B)∧(x ∉C)} (def. of −) = {x : (x ∈A)∧(x ∈B)∧(x ∉C)} (def. of ∩) = {x : (x ∈A)∧(x ∉C)∧(x ∈B)∧(x ∉C)} (regroup) = {x : ((x ∈A)∧(x ∉C))∧((x ∈B)∧(x ∉C))} (regroup) = {x : (x ∈A)∧(x ∉C)}∩{x : (x ∈B)∧(x ∉C)} (def. of ∩) = (A −C)∩(B −C) (def. of ∩) The proof is complete. ■ 17. If A,B and C are sets, then A ×(B ∩C) = (A ×B)∩(A ×C). Proof. See Example 8.12. ■ 272 Solutions 19. Prove that {9n : n ∈Z} ⊆{3n : n ∈Z}, but {9n : n ∈Z} ̸= {3n : n ∈Z}. Proof. Suppose a ∈{9n : n ∈Z}. This means a = 9n for some integer n ∈Z. Thus a = 9n = (32)n = 32n. This shows a is an integer power of 3, so a ∈{3n : n ∈Z}. Therefore a ∈{9n : n ∈Z} implies a ∈{3n : n ∈Z}, so {9n : n ∈Z} ⊆{3n : n ∈Z}. But notice {9n : n ∈Z} ̸= {3n : n ∈Z} as 3 ∈{3n : n ∈Z}, but 3 ∉{9n : n ∈Z}. ■ 21. Suppose A and B are sets. Prove A ⊆B if and only if A −B = ;. Proof. First we will prove that if A ⊆B, then A −B = ;. Contrapositive proof is used. Suppose that A −B ̸= ;. Thus there is an element a ∈A −B, which means a ∈A but a ∉B. Since not every element of A is in B, we have A ̸⊆B. Conversely, we will prove that if A −B = ;, then A ⊆B. Again, contrapositive proof is used. Suppose A ̸⊆B. This means that it is not the case that every element of A is an element of B, so there is an element a ∈A with a ∉B. Therefore we have a ∈A −B, so A −B ̸= ;. ■ 23. For each a ∈R, let Aa = © (x,a(x2 −1)) ∈R2 : x ∈R ª. Prove that \ a∈R Aa = {(−1,0),(1,0))}. Proof. First we will show that {(−1,0),(1,0))} ⊆ \ a∈R Aa. Notice that for any a ∈R, we have (−1,0) ∈Aa because Aa contains the ordered pair (−1,a((−1)2−1) = (−1,0). Similarly (1,0) ∈Aa. Thus each element of {(−1,0),(1,0))} belongs to every set Aa, so every element of \ a∈R Aa, so {(−1,0),(1,0))} ⊆ \ a∈R Aa. Now we will show \ a∈R Aa ⊆{(−1,0),(1,0))}. Suppose (c,d) ∈ \ a∈R Aa. This means (c,d) is in every set Aa. In particular (c,d) ∈A0 = © (x,0(x2 −1)) : x ∈R ª = {(x,0) : x ∈R}. It follows that d = 0. Then also we have (c,d) = (c,0) ∈A1 = © (x,1(x2 −1)) : x ∈R ª = © (x,x2 −1) : x ∈R ª. Therefore (c,0) has the form (c, c2 −1), that is (c,0) = (c, c2 −1). From this we get c2 −1 = 0, so c = ±1. Therefore (c,d) = (1,0) or (c,d) = (−1,0), so (c,d) ∈{(−1,0),(1,0))}. This completes the demonstration that (c,d) ∈ \ a∈R Aa implies (c,d) ∈{(−1,0),(1,0))}, so it follows that \ a∈R Aa ⊆{(−1,0),(1,0))}. Now it’s been shown that {(−1,0),(1,0))} ⊆ \ a∈R Aa and \ a∈R Aa ⊆{(−1,0),(1,0))}, so it follows that \ a∈R Aa = {(−1,0),(1,0))}. ■ 25. Suppose A,B,C and D are sets. Prove that (A ×B)∪(C × D) ⊆(A ∪C)×(B ∪D). Proof. Suppose (a,b) ∈(A ×B)∪(C × D). By deﬁnition of union, this means (a,b) ∈(A ×B) or (a,b) ∈(C × D). We examine these two cases individually. Case 1. Suppose (a,b) ∈(A × B). By deﬁnition of ×, it follows that a ∈A and b ∈B. From this, it follows from the deﬁnition of ∪that a ∈A ∪C and b ∈B ∪D. Again from the deﬁnition of ×, we get (a,b) ∈(A ∪C)×(B ∪D). 273 Case 2. Suppose (a,b) ∈(C × D). By deﬁnition of ×, it follows that a ∈C and b ∈D. From this, it follows from the deﬁnition of ∪that a ∈A ∪C and b ∈B∪D. Again from the deﬁnition of ×, we get (a,b) ∈(A ∪C)×(B ∪D). In either case, we obtained (a,b) ∈(A ∪C)×(B ∪D), so we’ve proved that (a,b) ∈(A ×B)∪(C × D) implies (a,b) ∈(A ∪C)×(B ∪D). Therefore (A ×B)∪(C × D) ⊆(A ∪C)×(B ∪D). ■ 27. Prove {12a+4b : a,b ∈Z} = {4c : c ∈Z}. Proof. First we show {12a+4b : a,b ∈Z} ⊆{4c : c ∈Z}. Suppose x ∈{12a+4b : a,b ∈Z}. Then x = 12a+4b for some integers a and b. From this we get x = 4(3a+ b), so x = 4c where c is the integer 3a+b. Consequently x ∈{4c : c ∈Z}. This establishes that {12a+4b : a,b ∈Z} ⊆{4c : c ∈Z}. Next we show {4c : c ∈Z} ⊆{12a+4b : a,b ∈Z}. Suppose x ∈{4c : c ∈Z}. Then x = 4c for some c ∈Z. Thus x = (12 + 4(−2))c = 12c + 4(−2c), and since c and −2c are integers we have x ∈{12a+4b : a,b ∈Z}. This proves that {12a+4b : a,b ∈Z} = {4c : c ∈Z}. ■ 29. Suppose A ̸= ;. Prove that A ×B ⊆A ×C, if and only if B ⊆C. Proof. First we will prove that if A×B ⊆A×C, then B ⊆C. Using contrapositive, suppose that B ̸⊆C. This means there is an element b ∈B with b ∉C. Since A ̸= ;, there exists an element a ∈A. Now consider the ordered pair (a,b). Note that (a,b) ∈A ×B, but (a,b) ̸∈A ×C. This means A ×B ̸⊆A ×C. Conversely, we will now show that if B ⊆C, then A ×B ⊆A ×C. We use direct proof. Suppose B ⊆C. Assume that (a,b) ∈A ×B. This means a ∈A and b ∈B. But, as B ⊆C, we also have b ∈C. From a ∈A and b ∈C, we get (a,b) ∈A × C. We’ve now shown (a,b) ∈A ×B implies (a,b) ∈A ×C, so A ×B ⊆A ×C. ■ 31. Suppose B ̸= ; and A ×B ⊆B ×C. Prove A ⊆C. Proof. Suppose B ̸= ; and A ×B ⊆B ×C. In what follows, we show that A ⊆C. Let x ∈A. Because B is not empty, it contains some element b. Observe that (x,b) ∈A ×B. But as A ×B ⊆B ×C, we also have (x,b) ∈B ×C, so, in particular, x ∈B. As x ∈A and x ∈B, we have (x,x) ∈A ×B. But as A ×B ⊆B ×C, it follows that (x,x) ∈B ×C. This implies x ∈C. Now we’ve shown x ∈A implies x ∈C, so A ⊆C. ■ Chapter 9 Exercises 1. If x, y ∈R, then \|x+ y\| = \|x\|+\|y\|. This is false. Disproof: Here is a counterexample: Let x = 1 and y = −1. Then \|x+ y\| = 0 and \|x\|+\|y\| = 2, so it’s not true that \|x+ y\| = \|x\|+\|y\|. 274 Solutions 3. If n ∈Z and n5 −n is even, then n is even. This is false. Disproof: Here is a counterexample: Let n = 3. Then n5 −n = 35 −3 = 240, but n is not even. 5. If A, B,C and D are sets, then (A ×B)∪(C × D) = (A ∪C)×(B ∪D). This is false. Disproof: Here is a counterexample: Let A = {1,2}, B = {1,2}, C = {2,3} and D = {2,3}. Then (A×B)∪(C×D) = {(1,1),(1,2),(2,1),(2,2)}∪{(2,2),(2,3),(3,2),(3,3)} = {(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(3,3)}. Also (A ∪C) × (B ∪D) = {1,2,3} × {1,2,3}= {(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)}, so you can see that (A × B) ∪ (C × D) ̸= (A ∪C)×(B ∪D). 7. If A, B and C are sets, and A ×C = B ×C, then A = B. This is false. Disproof: Here is a counterexample: Let A = {1}, B = {2} and C = ;. Then A ×C = B ×C = ;, but A ̸= B. 9. If A and B are sets, then P(A)−P(B) ⊆P(A −B). This is false. Disproof: Here is a counterexample: Let A = {1,2} and B = {1}. Then P(A)− P(B) = {;,{1},{2},{1,2}}−{;,{1}} = {{2},{1,2}}. Also P(A −B) = P({2}) = {;,{2}}. In this example we have P(A)−P(B) ̸⊆P(A −B). 11. If a,b ∈N, then a+ b < ab. This is false. Disproof: Here is a counterexample: Let a = 1 and b = 1. Then a+ b = 2 and ab = 1, so it’s not true that a+ b < ab. 13. There exists a set X for which R ⊆X and ; ∈X. This is true. Proof. Simply let X = R∪{;}. If x ∈R, then x ∈R∪{;} = X, so R ⊆X. Likewise, ; ∈R∪{;} = X because ; ∈{;}. ■ 15. Every odd integer is the sum of three odd integers. This is true. Proof. Suppose n is odd. Then n = n +1+(−1), and therefore n is the sum of three odd integers. ■ 17. For all sets A and B, if A −B = ;, then B ̸= ;. This is false. Disproof: Here is a counterexample: Just let A = ; and B = ;. Then A −B = ;, but it’s not true that B ̸= ;. 19. For every r,s ∈Q with r < s, there is an irrational number u for which r < u < s. This is true. Proof. (Direct) Suppose r,s ∈Q with r < s. Consider the number u = r + p 2 s−r 2 . In what follows we will show that u is irrational and r < u < s. Certainly since 275 s−r is positive, it follows that r < r + p 2 s−r 2 = u. Also, since p 2 < 2 we have u = r + p 2 s−r 2 < r +2 s−r 2 = s, and therefore u < s. Thus we can conclude r < u < s. Now we just need to show that u is irrational. Suppose for the sake of contra-diction that u is rational. Then u = a b for some integers a and b. Since r and s are rational, we have r = c d and s = e f for some c,d, e, f ∈Z. Now we have u = r + p 2 s−r 2 a b = c d + p 2 e f −c d 2 ad −bc bd = p 2 ed −cf 2d f (ad −bc)2d f bd(ed −cf ) = p 2 This expresses p 2 as a quotient of two integers, so p 2 is rational, a contradiction. Thus u is irrational. In summary, we have produced an irrational number u with r < u < s, so the proof is complete. ■ 21. There exist two prime numbers p and q for which p −q = 97. This statement is false. Disproof: Suppose for the sake of contradiction that this is true. Let p and q be prime numbers for which p −q = 97. Now, since their diﬀerence is odd, p and q must have opposite parity, so one of p and q is even and the other is odd. But there exists only one even prime number (namely 2), so either p = 2 or q = 2. If p = 2, then p −q = 97 implies q = 2−97 = −95, which is not prime. On the other hand if q = 2, then p −q = 97 implies p = 99, but that’s not prime either. Thus one of p or q is not prime, a contradiction. 23. If x, y ∈R and x3 < y3, then x < y. This is true. Proof. (Contrapositive) Suppose x ≥y. We need to show x3 ≥y3. Case 1. Suppose x and y have opposite signs, that is one of x and y is positive and the other is negative. Then since x ≥y, x is positive and y is negative. Then, since the powers are odd, x3 is positive and y3 is negative, so x3 ≥y3. Case 2. Suppose x and y do not have opposite signs. Then x2 + xy+ y2 ≥0 and also x−y ≥0 because x ≥y. Thus we have x3 −y3 = (x−y)(x2 + xy+ y2) ≥0. From this we get x3 −y3 ≥0, so x3 ≥y3. In either case we have x3 ≥y3. ■ 276 Solutions 25. For all a,b, c ∈Z, if a \| bc, then a \| b or a \| c. This is false. Disproof: Let a = 6, b = 3 and c = 4. Note that a \| bc, but a ∤b and a ∤c. 27. The equation x2 = 2x has three real solutions. Proof. By inspection, the numbers x = 2 and x = 4 are two solutions of this equation. But there is a third solution. Let m be the real number for which m2m = 1 2. Then negative number x = −2m is a solution, as follows. x2 = (−2m)2 = 4m2 = 4 µ m2m 2m ¶2 = 4 Ã 1 2 2m !2 = 1 22m = 2−2m = 2x. Therefore we have three solutions 2, 4 and m. ■ 29. If x, y ∈R and \|x+ y\| = \|x−y\|, then y = 0. This is false. Disproof: Let x = 0 and y = 1. Then \|x+ y\| = \|x−y\|, but y = 1. 31. No number appears in Pascal’s triangle more than four times. Disproof: The number 120 appears six times. Check that ¡10 3 ¢ = ¡10 7 ¢ = ¡16 2 ¢ = ¡16 14 ¢ = ¡120 1 ¢ = ¡120 119 ¢ = 120. 33. Suppose f (x) = a0 + a1x+ a2x2 +···+ anxn is a polynomial of degree 1 or greater, and for which each coeﬃcient ai is in N. Then there is an n ∈N for which the integer f (n) is not prime. Proof. (Outline) Note that, because the coeﬃcients are all positive and the degree is greater than 1, we have f (1) > 1. Let b = f (1) > 1. Now, the polynomial f (x) −b has a root 1, so f (x) −b = (x −1)g(x) for some polynomial g. Then f (x) = (x −1)g(x) + b. Now note that f (b + 1) = bg(b) + b = b(g(b) + 1). If we can now show that g(b) + 1 is an integer, then we have a nontrivial factoring f (b +1) = b(g(b)+1), and f (b +1) is not prime. To complete the proof, use the fact that f (x)−b = (x−1)g(x) has integer coeﬃcients, and deduce that g(x) must also have integer coeﬃcients. ■ Chapter 10 Exercises 1. For every integer n ∈N, it follows that 1+2+3+4+···+ n = n2 + n 2 . Proof. We will prove this with mathematical induction. (1) Observe that if n = 1, this statement is 1 = 12 +1 2 , which is obviously true. 277 (2) Consider any integer k ≥1. We must show that Sk implies Sk+1. In other words, we must show that if 1+2+3+4+···+ k = k2+k 2 is true, then 1+2+3+4+···+ k +(k +1) = (k +1)2 +(k +1) 2 is also true. We use direct proof. Suppose k ≥1 and 1+2+3+4+···+ k = k2+k 2 . Observe that 1+2+3+4+···+ k +(k +1) = (1+2+3+4+···+ k)+(k +1) = k2 + k 2 +(k +1) = k2 + k +2(k +1) 2 = k2 +2k +1 + k +1 2 = (k +1)2 +(k +1) 2 . Therefore we have shown that 1+2+3+4+···+ k +(k +1) = (k+1)2+(k+1) 2 . ■ 3. For every integer n ∈N, it follows that 13 +23 +33 +43 +···+ n3 = n2(n+1)2 4 . Proof. We will prove this with mathematical induction. (1) When n = 1 the statement is 13 = 12(1+1)2 4 = 4 4 = 1, which is true. (2) Now assume the statement is true for some integer n = k ≥1, that is assume 13 +23 +33 +43 +···+ k3 = k2(k+1)2 4 . Observe that this implies the statement is true for n = k +1. 13 +23 +33 +43 +···+ k3 +(k +1)3 = (13 +23 +33 +43 +···+ k3)+(k +1)3 = k2(k +1)2 4 +(k +1)3 = k2(k +1)2 4 + 4(k +1)3 4 = k2(k +1)2 +4(k +1)3 4 = (k +1)2(k2 +4(k +1)1) 4 = (k +1)2(k2 +4k +4) 4 = (k +1)2(k +2)2 4 = (k +1)2((k +1)+1)2 4 Therefore 13 +23 +33 +43 +···+ k3 +(k +1)3 = (k+1)2((k+1)+1)2 4 , which means the statement is true for n = k +1. ■ 278 Solutions 5. If n ∈N, then 21 +22 +23 +···+2n = 2n+1 −2. Proof. The proof is by mathematical induction. (1) When n = 1, this statement is 21 = 21+1 −2, or 2 = 4−2, which is true. (2) Now assume the statement is true for some integer n = k ≥1, that is assume 21 +22 +23 +···+2k = 2k+1 −2. Observe this implies that the statement is true for n = k +1, as follows: 21 +22 +23 +···+2k +2k+1 = (21 +22 +23 +···+2k)+2k+1 = 2k+1 −2+2k+1 = 2·2k+1 −2 = 2k+2 −2 = 2(k+1)+1 −2 Thus we have 21 +22 +23 +···+2k +2k+1 = 2(k+1)+1 −2, so the statement is true for n = k +1. Thus the result follows by mathematical induction. ■ 7. If n ∈N, then 1·3+2·4+3·5+4·6+···+ n(n+2) = n(n+1)(2n+7) 6 . Proof. The proof is by mathematical induction. (1) When n = 1, we have 1·3 = 1(1+1)(2+7) 6 , which is the true statement 3 = 18 6 . (2) Now assume the statement is true for some integer n = k ≥1, that is assume 1·3+2·4+3·5+4·6+···+ k(k +2) = k(k+1)(2k+7) 6 . Now observe that 1·3+2·4+3·5+4·6+···+ k(k +2)+(k +1)((k +1)+2) = (1·3+2·4+3·5+4·6+···+ k(k +2))+(k +1)((k +1)+2) = k(k +1)(2k +7) 6 +(k +1)((k +1)+2) = k(k +1)(2k +7) 6 + 6(k +1)(k +3) 6 = k(k +1)(2k +7)+6(k +1)(k +3) 6 = (k +1)(k(2k +7)+6(k +3)) 6 = (k +1)(2k2 +13k +18) 6 = (k +1)(k +2)(2k +9) 6 = (k +1)((k +1)+1)(2(k +1)+7) 6 Thus we have 1·3+2·4+3·5+4·6+···+k(k+2)+(k+1)((k+1)+2) = (k+1)((k+1)+1)(2(k+1)+7) 6 , and this means the statement is true for n = k +1. Thus the result follows by mathematical induction. ■ 279 9. For any integer n ≥0, it follows that 24 \| (52n −1). Proof. The proof is by mathematical induction. (1) For n = 0, the statement is 24 \| (52·0 −1). This is 24 \| 0, which is true. (2) Now assume the statement is true for some integer n = k ≥1, that is assume 24 \| (52k −1). This means 52k −1 = 24a for some integer a, and from this we get 52k = 24a+1. Now observe that 52(k+1) −1 = 52k+2 −1 = 5252k −1 = 52(24a+1)−1 = 25(24a+1)−1 = 25·24a+25−1 = 24(25a+1). This shows 52(k+1) −1 = 24(25a+1), which means 24 \| 52(k+1) −1. This completes the proof by mathematical induction. ■ 11. For any integer n ≥0, it follows that 3 \| (n3 +5n+6). Proof. The proof is by mathematical induction. (1) When n = 0, the statement is 3 \| (03 +5·0+6), or 3 \| 6, which is true. (2) Now assume the statement is true for some integer n = k ≥0, that is assume 3 \| (k3 +5k +6). This means k3 +5k +6 = 3a for some integer a. We need to show that 3 \| ((k +1)3 +5(k +1)+6). Observe that (k +1)3 +5(k +1)+6 = k3 +3k2 +3k +1+5k +5+6 = (k3 +5k +6)+3k2 +3k +6 = 3a+3k2 +3k +6 = 3(a+ k2 + k +2). Thus we have deduced (k +1)3 −(k +1) = 3(a+ k2 + k +2). Since a+ k2 + k +2 is an integer, it follows that 3 \| ((k +1)3 +5(k +1)+6). It follows by mathematical induction that 3 \| (n3 +5n+6) for every n ≥0. ■ 13. For any integer n ≥0, it follows that 6 \| (n3 −n). Proof. The proof is by mathematical induction. (1) When n = 0, the statement is 6 \| (03 −0), or 6 \| 0, which is true. 280 Solutions (2) Now assume the statement is true for some integer n = k ≥0, that is, assume 6 \| (k3 −k). This means k3 −k = 6a for some integer a. We need to show that 6 \| ((k +1)3 −(k +1)). Observe that (k +1)3 −(k +1) = k3 +3k2 +3k +1−k −1 = (k3 −k)+3k2 +3k = 6a+3k2 +3k = 6a+3k(k +1). Thus we have deduced (k+1)3 −(k+1) = 6a+3k(k+1). Since one of k or (k+1) must be even, it follows that k(k +1) is even, so k(k +1) = 2b for some integer b. Consequently (k +1)3 −(k +1) = 6a +3k(k +1) = 6a +3(2b) = 6(a + b). Since (k +1)3 −(k +1) = 6(a+ b) it follows that 6 \| ((k +1)3 −(k +1)). Thus the result follows by mathematical induction. ■ 15. If n ∈N, then 1 1·2 + 1 2·3 + 1 3·4 + 1 4·5 +···+ 1 n(n+1) = 1− 1 n+1. Proof. The proof is by mathematical induction. (1) When n = 1, the statement is 1 1(1+1) = 1− 1 1+1, which simpliﬁes to 1 2 = 1 2. (2) Now assume the statement is true for some integer n = k ≥1, that is assume 1 1·2 + 1 2·3 + 1 3·4 + 1 4·5 +···+ 1 k(k+1) = 1− 1 k+1. Next we show that the statement for n = k +1 is true. Observe that 1 1·2 + 1 2·3 + 1 3·4 + 1 4·5 +···+ 1 k(k +1) + 1 (k +1)((k +1)+1) = µ 1 1·2 + 1 2·3 + 1 3·4 + 1 4·5 +···+ 1 k(k +1) ¶ + 1 (k +1)(k +2) = µ 1− 1 k +1 ¶ + 1 (k +1)(k +2) = 1− 1 k +1 + 1 (k +1)(k +2) = 1− k +2 (k +1)(k +2) + 1 (k +1)(k +2) = 1− k +1 (k +1)(k +2) = 1− 1 k +2 = 1− 1 (k +1)+1. This establishes 1 1·2 + 1 2·3 + 1 3·4 + 1 4·5 +···+ 1 (k+1)((k+1)+1 = 1− 1 (k+1)+1, which is to say that the statement is true for n = k +1. This completes the proof by mathematical induction. ■ 281 17. Suppose A1, A2,... An are sets in some universal set U, and n ≥2. Prove that A1 ∩A2 ∩···∩An = A1 ∪A2 ∪···∪An. Proof. The proof is by strong induction. (1) When n = 2 the statement is A1 ∩A2 = A1 ∪A2. This is not an entirely obvious statement, so we have to prove it. Observe that A1 ∩A2 = {x : (x ∈U)∧(x ∉A1 ∩A2)} (deﬁnition of complement) = {x : (x ∈U)∧∼(x ∈A1 ∩A2)} = {x : (x ∈U)∧∼((x ∈A1)∧(x ∈A2))} (deﬁnition of ∩) = {x : (x ∈U)∧(∼(x ∈A1)∨∼(x ∈A2))} (DeMorgan) = {x : (x ∈U)∧((x ∉A1)∨(x ∉A2))} = {x : (x ∈U)∧(x ∉A1)∨(x ∈U)∧(x ∉A2)} (distributive prop.) = {x : ((x ∈U)∧(x ∉A1))}∪{x : ((x ∈U)∧(x ∉A2))} (def. of ∪) = A1 ∪A2 (deﬁnition of complement) (2) Let k ≥2. Assume the statement is true if it involves k or fewer sets. Then A1 ∩A2 ∩···∩Ak−1 ∩Ak ∩Ak+1 = A1 ∩A2 ∩···∩Ak−1 ∩(Ak ∩Ak+1) = A1 ∪A2 ∪···∪Ak−1 ∪Ak ∩Ak+1 = A1 ∪A2 ∪···∪Ak−1 ∪Ak ∪Ak+1 Thus the statement is true when it involves k +1 sets. This completes the proof by strong induction. ■ 19. Prove Pn k=1 1/k2 ≤2−1/n for every n. Proof. This clearly holds for n = 1. Assume it holds for some n ≥1. Then Pn+1 k=1 1/k2 ≤2−1/n+1/(n+1)2 = 2−(n+1)2−n n(n+1)2 ≤2−1/(n+1). The proof is complete. ■ 21. If n ∈N, then 1 1 + 1 2 + 1 3 +···+ 1 2n ≥1+ n 2 . Proof. If n = 1, the result is obvious. Assume the proposition holds for some n > 1. Then 1 1 + 1 2 + 1 3 +···+ 1 2n+1 = µ1 1 + 1 2 + 1 3 +···+ 1 2n ¶ + µ 1 2n +1 + 1 2n +2 + 1 2n +3 +···+ 1 2n+1 ¶ ≥ ³ 1+ n 2 ´ + µ 1 2n +1 + 1 2n +2 + 1 2n +3 +···+ 1 2n+1 ¶ . Now, the sum ³ 1 2n+1 + 1 2n+2 + 1 2n+3 +···+ 1 2n+1 ´ on the right has 2n+1−2n = 2n terms, all greater than or equal to 1 2n+1 , so the sum is greater than 2n 1 2n+1 = 1 2. Therefore we get 1 1 + 1 2 + 1 3 +···+ 1 2n+1 ≥ ¡ 1+ n 2 ¢ + ³ 1 2n+1 + 1 2n+2 + 1 2n+3 +···+ 1 2n+1 ´ ≥ ¡ 1+ n 2 ¢ + 1 2 = 1+ n+1 2 . This means the result is true for n+1, so the theorem is proved. ■ 282 Solutions 23. Use induction to prove the binomial theorem (x+ y)n = Pn i=0 ¡n i ¢ xn−i yi. Proof. Notice that when n = 1, the formula is (x + y)1 = ¡1 0 ¢ x1 y0 + ¡1 1 ¢ x0 y1 = x + y, which is true. Now assume the theorem is true for some n > 1. We will show that this implies that it is true for the power n+1. Just observe that (x+ y)n+1 = (x+ y)(x+ y)n = (x+ y) n X i=0 Ã n i ! xn−i yi = n X i=0 Ã n i ! x(n+1)−i yi + n X i=0 Ã n i ! xn−i yi+1 = n X i=0 "Ã n i ! + Ã n i −1 !# x(n+1)−i yi + yn+1 = n X i=0 Ã n+1 i ! x(n+1)−i yi + Ã n+1 n+1 ! yn+1 = n+1 X i=0 Ã n+1 i ! x(n+1)−i yi. This shows that the formula is true for (x+ y)n+1, so the theorem is proved. ■ 25. Concerning the Fibonacci sequence, prove that F1+F2+F3+F4+...+Fn = Fn+2−1. Proof. The proof is by induction. (1) When n = 1 the statement is F1 = F1+2 −1 = F3 −1 = 2−1 = 1, which is true. Also when n = 2 the statement is F1 +F2 = F2+2 −1 = F4 −1 = 3−1 = 2, which is true, as F1 + F2 = 1+1 = 2. (2) Now assume k ≥1 and F1 + F2 + F3 + F4 +...+ Fk = Fk+2 −1. We need to show F1 + F2 + F3 + F4 +...+ Fk + Fk+1 = Fk+3 −1. Observe that F1 + F2 + F3 + F4 +...+ Fk + Fk+1 = (F1 + F2 + F3 + F4 +...+ Fk)+ Fk+1 = Fk+2 −1++Fk+1 = (Fk+1 + Fk+2)−1 = Fk+3 −1. This completes the proof by induction. ■ 27. Concerning the Fibonacci sequence, prove that F1 + F3 +···+ F2n−1 = F2n. Proof. If n = 1, the result is immediate. Assume for some n > 1 we have Pn i=1 F2i−1 = F2n. Then Pn+1 i=1 F2i−1 = F2n+1+Pn i=1 F2i−1 = F2n+1+F2n = F2n+2 = F2(n+1) as desired. ■ 283 29. Prove that ¡n 0 ¢ + ¡n−1 1 ¢ + ¡n−2 2 ¢ + ¡n−3 3 ¢ +···+ ¡ 1 n−1 ¢ + ¡0 n ¢ = Fn+1. Proof. (Strong Induction) For n = 1 this is ¡1 0 ¢ + ¡0 1 ¢ = 1+0 = 1 = F2 = F1+1. Thus the assertion is true when n = 1. Now ﬁx n and assume that ¡k 0 ¢ + ¡k−1 1 ¢ + ¡k−2 2 ¢ + ¡k−3 3 ¢ +···+ ¡ 1 k−1 ¢ + ¡0 k ¢ = Fk+1 whenever k < n. In what follows we use the identity ¡n k ¢ = ¡n−1 k−1 ¢ + ¡n−1 k ¢. We also often use ¡a b ¢ = 0 whenever it is untrue that 0 ≤b ≤a. Ã n 0 ! + Ã n−1 1 ! + Ã n−2 2 ! +···+ Ã 1 n−1 ! + Ã 0 n ! = Ã n 0 ! + Ã n−1 1 ! + Ã n−2 2 ! +···+ Ã 1 n−1 ! = Ã n−1 −1 ! + Ã n−1 0 ! + Ã n−2 0 ! + Ã n−2 1 ! + Ã n−3 1 ! + Ã n−3 2 ! +···+ Ã 0 n−1 ! + Ã 0 n ! = Ã n−1 0 ! + Ã n−2 0 ! + Ã n−2 1 ! + Ã n−3 1 ! + Ã n−3 2 ! +···+ Ã 0 n−1 ! + Ã 0 n ! = "Ã n−1 0 ! + Ã n−2 1 ! +···+ Ã 0 n−1 !# + "Ã n−2 0 ! + Ã n−3 1 ! +···+ Ã 0 n−2 !# = Fn + Fn−1 = Fn This completes the proof. ■ 31. Prove that Pn k=0 ¡k r ¢ = ¡n+1 r+1 ¢ , where r ∈N. Hint: Use induction on the integer n. After doing the basis step, break up the expression ¡k r ¢ as ¡k r ¢ = ¡k−1 r−1 ¢ + ¡k−1 r ¢. Then regroup, use the induction hypothesis, and recombine using the above identity. 33. Suppose that n inﬁnitely long straight lines lie on the plane in such a way that no two are parallel, and no three intersect at a single point. Show that this arrangement divides the plane into n2+n+2 2 regions. Proof. The proof is by induction. For the basis step, suppose n = 1. Then there is one line, and it clearly divides the plane into 2 regions, one on either side of the line. As 2 = 12+1+2 2 = n2+n+2 2 , the formula is correct when n = 1. Now suppose there are n+1 lines on the plane, and that the formula is correct for when there are n lines on the plane. Single out one of the n+1 lines on the plane, and call it ℓ. Remove line ℓ, so that there are now n lines on the plane. By the induction hypothesis, these n lines divide the plane into n2+n+2 2 regions. Now add line ℓback. Doing this adds an additional n + 1 regions. (The diagram illustrates the case where n + 1 = 5. Without ℓ, there are n = 4 lines. Adding ℓback produces n+1 = 5 new regions.) ℓ 1 2 3 4 5 284 Solutions Thus, with n+1 lines there are all together (n+1)+ n2+n+2 2 regions. Observe (n+1)+ n2 + n+2 2 = 2n+2+ n2 + n+2 2 = (n+1)2 +(n+1)+2 2 . Thus, with n + 1 lines, we have (n+1)2+(n+1)+2 2 regions, which means that the formula is true for when there are n + 1 lines. We have shown that if the formula is true for n lines, it is also true for n +1 lines. This completes the proof by induction. ■ 35. If n,k ∈N, and n is even and k is odd, then ¡n k ¢ is even. Proof. Notice that if k is not a value between 0 and n, then ¡n k ¢ = 0 is even; thus from here on we can assume that 0 < k < n. We will use strong induction. For the basis case, notice that the assertion is true for the even values n = 2 and n = 4: ¡2 1 ¢ = 2; ¡4 1 ¢ = 4; ¡4 3 ¢ = 4 (even in each case). Now ﬁx and even n assume that ¡m k ¢ is even whenever m is even, k is odd, and m < n. Using the identity ¡n k ¢ = ¡n−1 k−1 ¢ + ¡n−1 k ¢ three times, we get Ã n k ! = Ã n−1 k −1 ! + Ã n−1 k ! = Ã n−2 k −2 ! + Ã n−2 k −1 ! + Ã n−2 k −1 ! + Ã n−2 k ! = Ã n−2 k −2 ! +2 Ã n−2 k −1 ! + Ã n−2 k ! . Now, n−2 is even, and k and k −2 are odd. By the inductive hypothesis, the outer terms of the above expression are even, and the middle is clearly even; thus we have expressed ¡n k ¢ as the sum of three even integers, so it is even. ■ Chapter 11 Exercises Section 11.0 Exercises 1. Let A = {0,1,2,3,4,5}. Write out the relation R that expresses > on A. Then illustrate it with a diagram. R = © (5,4),(5,3),(5,3),(5,3),(5,1),(5,0),(4,3),(4,2),(4,1), (4,0),(3,2),(3,1),(3,0),(2,1),(2,0),(1,0) ª 0 1 2 3 4 5 3. Let A = {0,1,2,3,4,5}. Write out the relation R that expresses ≥on A. Then illustrate it with a diagram. 285 R = © (5,5),(5,4),(5,3),(5,2),(5,1),(5,0), (4,4),(4,3),(4,2),(4,1),(4,0), (3,3),(3,2),(3,1),(3,0), (2,2),(2,1),(2,0),(1,1),(1,0),(0,0) ª 0 1 2 3 4 5 5. The following diagram represents a relation R on a set A. Write the sets A and R. Answer: A = {0,1,2,3,4,5}; R = {(3,3),(4,3),(4,2),(1,2),(2,5),(5,0)} 7. Write the relation < on the set A = Z as a subset R of Z×Z. This is an inﬁnite set, so you will have to use set-builder notation. Answer: R = {(x, y) ∈Z×Z : y−x ∈N} 9. How many diﬀerent relations are there on the set A = © 1,2,3,4,5,6 ª? Consider forming a relation R ⊆A × A on A. For each ordered pair (x, y) ∈A × A, we have two choices: we can either include (x, y) in R or not include it. There are 6·6 = 36 ordered pairs in A × A. By the multiplication principle, there are thus 236 diﬀerent subsets R and hence also this many relations on A. 11. Answer: 2(\|A\|2) 13. Answer: ̸= 15. Answer: ≡(mod 3) Section 11.1 Exercises 1. Consider the relation R = {(a,a),(b,b),(c, c),(d,d),(a,b),(b,a)} on the set A = {a,b, c,d}. Which of the properties reﬂexive, symmetric and transitive does R possess and why? If a property does not hold, say why. This is reﬂexive because (x,x) ∈R (i.e., xRx )for every x ∈A. It is symmetric because it is impossible to ﬁnd an (x, y) ∈R for which (y,x) ∉R. It is transitive because (xR y∧yRz) ⇒xRz always holds. 3. Consider the relation R = {(a,b),(a, c),(c,b),(b, c)} on the set A = {a,b, c}. Which of the properties reﬂexive, symmetric and transitive does R possess and why? If a property does not hold, say why. This is not reﬂexive because (a,a) ∉R (for example). It is not symmetric because (a,b) ∈R but (b,a) ∉R. It is not transitive because cRb and bRc are true, but cRc is false. 5. Consider the relation R = © (0,0),( p 2,0),(0, p 2),( p 2, p 2) ª on R. Say whether this relation is reﬂexive, symmetric and transitive. If a property does not hold, say why. This is not reﬂexive because (1,1) ∉R (for example). It is symmetric because it is impossible to ﬁnd an (x, y) ∈R for which (y,x) ∉R. It is transitive because (xR y∧yRz) ⇒xRz always holds. 7. There are 16 possible diﬀerent relations R on the set A = {a,b}. Describe all of them. (A picture for each one will suﬃce, but don’t forget to label the nodes.) Which ones are reﬂexive? Symmetric? Transitive? 286 Solutions a a a a a a a a a a a a a a a a b b b b b b b b b b b b b b b b Only the four in the right column are reﬂexive. Only the eight in the ﬁrst and fourth rows are symmetric. All of them are transitive except the ﬁrst three on the fourth row. 9. Deﬁne a relation on Z by declaring xR y if and only if x and y have the same parity. Say whether this relation is reﬂexive, symmetric and transitive. If a property does not hold, say why. What familiar relation is this? This is reﬂexive because xRx since x always has the same parity as x. It is symmetric because if x and y have the same parity, then y and x must have the same parity (that is, xR y ⇒yRx). It is transitive because if x and y have the same parity and y and z have the same parity, then x and z must have the same parity. (That is (xR y∧yRz) ⇒xRz always holds.) The relation is congruence modulo 2. 11. Suppose A = {a,b, c,d} and R = {(a,a),(b,b),(c, c),(d,d)}. Say whether this relation is reﬂexive, symmetric and transitive. If a property does not hold, say why. This is reﬂexive because (x,x) ∈R for every x ∈A. It is symmetric because it is impossible to ﬁnd an (x, y) ∈R for which (y,x) ∉R. It is transitive because (xR y∧yRz) ⇒xRz always holds. (For example (aRa∧aRa) ⇒aRa is true, etc.) 13. Consider the relation R = {(x, y) ∈R×R : x−y ∈Z} on R. Prove that this relation is reﬂexive and symmetric, and transitive. Proof. In this relation, xR y means x−y ∈Z. To see that R is reﬂexive, take any x ∈R and observe that x−x = 0 ∈Z, so xRx. Therefore R is reﬂexive. To see that R is symmetric, we need to prove xR y ⇒yRx for all x, y ∈R. We use direct proof. Suppose xR y. This means x −y ∈Z. Then it follows that −(x−y) = y−x is also in Z. But y−x ∈Z means yRx. We’ve shown xR y implies yRx, so R is symmetric. To see that R is transitive, we need to prove (xR y ∧yRz) ⇒xRz is always true. We prove this conditional statement with direct proof. Suppose xR y and yRz. Since xR y, we know x −y ∈Z. Since yRz, we know y−z ∈Z. Thus x −y and y−z are both integers; by adding these integers we get another integer (x−y)+(y−z) = x−z. Thus x−z ∈Z, and this means xRz. We’ve now shown that if xR y and yRz, then xRz. Therefore R is transitive. ■ 287 15. Prove or disprove: If a relation is symmetric and transitive, then it is also reﬂexive. This is false. For a counterexample, consider the relation R = {(a,a),(a,b),(b,a),(b,b)} on the set A = {a,b, c}. This is symmetric and transitive but it is not reﬂexive. 17. Deﬁne a relation ∼on Z as x ∼y if and only if \|x −y\| ≤1. Say whether ∼is reﬂexive, symmetric and transitive. This is reﬂexive because \|x−x\| = 0 ≤1 for all integers x. It is symmetric because x ∼y if and only if \|x−y\| ≤1, if and only if \|y−x\| ≤1, if and only if y ∼x. It is not transitive because, for example, 0 ∼1 and 1 ∼2, but is not the case that 0 ∼2. Section 11.2 Exercises 1. Let A = {1,2,3,4,5,6}, and consider the following equivalence relation on A: R = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(2,3),(3,2),(4,5),(5,4),(4,6),(6,4),(5,6),(6,5)}. List the equivalence classes of R. The equivalence classes are: = {1}; = = {2,3}; = = = {4,5,6}. 3. Let A = {a,b, c,d, e}. Suppose R is an equivalence relation on A. Suppose R has three equivalence classes. Also aRd and bRc. Write out R as a set. Answer: R = {(a,a),(b,b),(c, c),(d,d),(e, e),(a,d),(d,a),(b, c),(c,b)}. 5. There are two diﬀerent equivalence relations on the set A = {a,b}. Describe them all. Diagrams will suﬃce. Answer: R = {(a,a),(b,b)} and R = {(a,a),(b,b),(a,b),(b,a)} 7. Deﬁne a relation R on Z as xR y if and only if 3x −5y is even. Prove R is an equivalence relation. Describe its equivalence classes. To prove that R is an equivalence relation, we must show it’s reﬂexive, sym-metric and transitive. The relation R is reﬂexive for the following reason. If x ∈Z, then 3x−5x = −2x is even. But then since 3x−5x is even, we have xRx. Thus R is reﬂexive. To see that R is symmetric, suppose xR y. We must show yRx. Since xR y, we know 3x−5y is even, so 3x−5y = 2a for some integer a. Now reason as follows: 3x−5y = 2a 3x−5y+8y−8x = 2a+8y−8x 3y−5x = 2(a+4y−4x). From this it follows that 3y−5x is even, so yRx. We’ve now shown xR y implies yRx, so R is symmetric. To prove that R is transitive, assume that xR y and yRz. (We will show that this implies xRz.) Since xR y and yRz, it follows that 3x−5y and 3y−5z are both even, so 3x−5y = 2a and 3y−5z = 2b for some integers a and b. Adding these equations, we get (3x −5y) + (3y −5z) = 2a + 2b, and this simpliﬁes to 3x −5z = 2(a + b + y). 288 Solutions Therefore 3x−5z is even, so xRz. We’ve now shown that if xR y and yRz, then xRz, so R is transitive. We’ve now shown that R is reﬂexive, symmetric and transitive, so it is an equivalence relation. The completes the ﬁrst part of the problem. Now we move on the second part. To ﬁnd the equivalence classes, ﬁrst note that = {x ∈Z : xR0} = {x ∈Z : 3x−5·0 is even} = {x ∈Z : 3x is even} = {x ∈Z : x is even}. Thus the equivalence class consists of all even integers. Next, note that = {x ∈Z : xR1} = {x ∈Z : 3x−5·1 is even} = {x ∈Z : 3x−5 is even} = © x ∈Z : x is odd ª . Thus the equivalence class consists of all odd integers. Consequently there are just two equivalence classes {...,−4,−2,0,2,4,...} and {...,−3,−1,1,3,5,...}. 9. Deﬁne a relation R on Z as xR y if and only if 4 \| (x + 3y). Prove R is an equivalence relation. Describe its equivalence classes. This is reﬂexive, because for any x ∈Z we have 4 \| (x+3x), so xRx. To prove that R is symmetric, suppose xR y. Then 4 \| (x + 3y), so x + 3y = 4a for some integer a. Multiplying by 3, we get 3x + 9y = 12a, which becomes y+3x = 12a−8y. Then y+3x = 4(3a−2y), so 4 \| (y+3x), hence yRx. Thus we’ve shown xR y implies yRx, so R is symmetric. To prove transitivity, suppose xR y and yRz. Then 4\|(x +3y) and 4\|(y+3z), so x+3y = 4a and y+3z = 4b for some integers a and b. Adding these two equations produces x+4y+3z = 4a+4b, or x+3z = 4a+4b −4y = 4(a+ b −y). Consequently 4\|(x+3z), so xRz, and R is transitive. As R is reﬂexive, symmetric and transitive, it is an equivalence relation. Now let’s compute its equivalence classes. = {x ∈Z : xR0} = {x ∈Z : 4 \| (x+3·0)} = {x ∈Z : 4 \| x} = {...−4,0,4,8,12,16...} = {x ∈Z : xR1} = {x ∈Z : 4 \| (x+3·1)} = {x ∈Z : 4 \| (x+3)} = {...−3,1,5,9,13,17...} = {x ∈Z : xR2} = {x ∈Z : 4 \| (x+3·2)} = {x ∈Z : 4 \| (x+6)} = {...−2,2,6,10,14,18...} = {x ∈Z : xR3} = {x ∈Z : 4 \| (x+3·3)} = {x ∈Z : 4 \| (x+9)} = {...−1,3,7,11,15,19...} 11. Prove or disprove: If R is an equivalence relation on an inﬁnite set A, then R has inﬁnitely many equivalence classes. This is False. Counterexample: consider the relation of congruence modulo 2. It is a relation on the inﬁnite set Z, but it has only two equivalence classes. 13. Answer: m\|A\| 15. Answer: 15 289 Section 11.3 Exercises 1. List all the partitions of the set A = {a,b}. Compare your answer to the answer to Exercise 5 of Section 11.2. There are just two partitions {{a},{b}} and {{a,b}}. These correspond to the two equivalence relations R1 = {(a,a),(b,b)} and R2 = {(a,a),(a,b),(b,a),(b,b)}, respec-tively, on A. 3. Describe the partition of Z resulting from the equivalence relation ≡(mod 4). Answer: The partition is {,,,} = © {...,−4,0,4,8,12,...},{...,−3,1,5,9,13,...}, {...,−2,2,4,6,10,14,...}, {...,−1,3,7,11,15,...} ª 5. Answer: Congruence modulo 2, or “same parity.” Section 11.4 Exercises 1. Write the addition and multiplication tables for Z2. + · 3. Write the addition and multiplication tables for Z4. + · 5. Suppose [a],[b] ∈Z5 and [a]·[b] = . Is it necessarily true that either [a] = or [b] = ? The multiplication table for Z5 is shown in Section 11.4. In the body of that table, the only place that occurs is in the ﬁrst row or the ﬁrst column. That row and column are both headed by . It follows that if [a] · [b] = , then either [a] or [b] must be . 7. Do the following calculations in Z9, in each case expressing your answer as [a] with 0 ≤a ≤8. (a) + = (b) + = (c) · = (d) · = 290 Solutions Chapter 12 Exercises Section 12.1 Exercises 1. Suppose A = {0,1,2,3,4}, B = {2,3,4,5} and f = {(0,3),(1,3),(2,4),(3,2),(4,2)}. State the domain and range of f . Find f (2) and f (1). Domain is A; Range is {2,3,4}; f (2) = 4; f (1) = 3. 3. There are four diﬀerent functions f : {a,b} →{0,1}. List them all. Diagrams will suﬃce. f1 = {(a,0),(b,0)} f2 = {(a,1),(b,0)}, f3 = {(a,0),(b,1)} f4 = {(a,1),(b,1)} 5. Give an example of a relation from {a,b, c,d} to {d, e} that is not a function. One example is {(a,d),(a, e),(b,d),(c,d),(d,d)}. 7. Consider the set f = {(x, y) ∈Z×Z : 3x+ y = 4}. Is this a function from Z to Z? Explain. Yes, since 3x+ y = 4 if and only if y = 4−3x, this is the function f : Z →Z deﬁned as f (x) = 4−3x. 9. Consider the set f = © (x2,x) : x ∈R ª. Is this a function from R to R? Explain. No. This is not a function. Observe that f contains the ordered pairs (4,2) and (4,−2). Thus the real number 4 occurs as the ﬁrst coordinate of more than one element of f . 11. Is the set θ = {(X,\|X\|) : X ⊆Z5} a function? If so, what is its domain and range? Yes, this is a function. The domain is P(Z5). The range is {0,1,2,3,4,5}. Section 12.2 Exercises 1. Let A = {1,2,3,4} and B = {a,b, c}. Give an example of a function f : A →B that is neither injective nor surjective. Consider f = {(1,a),(2,a),(3,a),(4,a)}. Then f is not injective because f (1) = f (2). Also f is not surjective because it sends no element of A to the element c ∈B. 3. Consider the cosine function cos : R →R. Decide whether this function is injective and whether it is surjective. What if it had been deﬁned as cos : R →[−1,1]? The function cos : R →R is not injective because, for example, cos(0) = cos(2π). It is not surjective because if b = 5 ∈R (for example), there is no real number for which cos(x) = b. The function cos : R →[−1,1] is surjective. but not injective. 5. A function f : Z →Z is deﬁned as f (n) = 2n+1. Verify whether this function is injective and whether it is surjective. This function is injective. To see this, suppose m,n ∈Z and f (m) = f (n). This means 2m+1 = 2n+1, from which we get 2m = 2n, and then m = n. Thus f is injective. This function is not surjective. To see this notice that f (n) is odd for all n ∈Z. So given the (even) number 2 in the codomain Z, there is no n with f (n) = 2. 291 7. A function f : Z × Z →Z is deﬁned as f ((m,n)) = 2n −4m. Verify whether this function is injective and whether it is surjective. This is not injective because (0,2) ̸= (−1,0), yet f ((0,2)) = f ((−1,0)) = 4. This is not surjective because f ((m,n)) = 2n −4m = 2(n −2m) is always even. If b ∈Z is odd, then f ((m,n)) ̸= b, for all (m,n) ∈Z×Z. 9. Prove that the function f : R−{2} →R−{5} deﬁned by f (x) = 5x+1 x−2 is bijective. Proof. First, let’s check that f is injective. Suppose f (x) = f (y). Then 5x+1 x−2 = 5y+1 y−2 (5x+1)(y−2) = (5y+1)(x−2) 5xy−10x+ y−2 = 5yx−10y+ x−2 −10x+ y = −10y+ x 11y = 11x y = x. Since f (x) = f (y) implies x = y, it follows that f is injective. Next, let’s check that f is surjective. For this, take an arbitrary element b ∈R−{5}. We want to see if there is an x ∈R−{2} for which f (x) = b, or 5x+1 x−2 = b. Solving this for x, we get: 5x+1 = b(x−2) 5x+1 = bx−2b 5x−xb = −2b −1 x(5−b) = −2b −1. Since we have assumed b ∈R −{5}, the term (5 −b) is not zero, and we can divide with impunity to get x = −2b −1 5−b . This is an x for which f (x) = b, so f is surjective. Since f is both injective and surjective, it is bijective. ■ 11. Consider the function θ : {0,1}×N →Z deﬁned as θ(a,b) = (−1)ab. Is θ injective? Is it surjective? Explain. First we show that θ is injective. Suppose θ(a,b) = θ(c,d). Then (−1)ab = (−1)cd. As b and d are both in N, they are both positive. Then because (−1)ab = (−1)cd, it follows that (−1)a and (−1)c have the same sign. Since each of (−1)a and (−1)c equals ±1, we have (−1)a = (−1)c, so then (−1)ab = (−1)cd implies b = d. But also (−1)a = (−1)c means a and c have the same parity, and because a, c ∈{0,1}, it follows a = c. Thus (a,b) = (c,d), so θ is injective. Next note that θ is not surjective because θ(a,b) = (−1)ab is either positive or negative, but never zero. Therefore there exist no element (a,b) ∈{0,1}×N for which θ(a,b) = 0 ∈Z. 292 Solutions 13. Consider the function f : R2 →R2 deﬁned by the formula f (x, y) = (xy,x3). Is f injective? Is it surjective? Notice that f (0,1) = (0,0) and f (0,0) = (0,0), so f is not injective. To show that f is also not surjective, we will show that it’s impossible to ﬁnd an ordered pair (x, y) with f (x, y) = (1,0). If there were such a pair, then f (x, y) = (xy,x3) = (1,0), which yields xy = 1 and x3 = 0. From x3 = 0 we get x = 0, so xy = 0, a contradiction. 15. This question concerns functions f : {A,B,C,D,E,F,G} →{1,2,3,4,5,6,7}. How many such functions are there? How many of these functions are injective? How many are surjective? How many are bijective? Function f can described as a list (f (A), f (B), f (C), f (D), f (E), f (F), f (G)), where there are seven choices for each entry. By the multiplication principle, the total number of functions f is 77 = 823543. If f is injective, then this list can’t have any repetition, so there are 7! = 5040 injective functions. Since any injective function sends the seven elements of the domain to seven distinct elements of the codomain, all of the injective functions are surjective, and vice versa. Thus there are 5040 surjective functions and 5040 bijective functions. 17. This question concerns functions f : {A,B,C,D,E,F,G} →{1,2}. How many such functions are there? How many of these functions are injective? How many are surjective? How many are bijective? Function f can described as a list (f (A), f (B), f (C), f (D), f (E), f (F), f (G)), where there are two choices for each entry. Therefore the total number of functions is 27 = 128. It is impossible for any function to send all seven elements of {A,B,C,D,E,F,G} to seven distinct elements of {1,2}, so none of these 128 functions is injective, hence none are bijective. How many are surjective? Only two of the 128 functions are not surjective, and they are the “constant” functions {(A,1),(B,1),(C,1),(D,1),(E,1),(F,1),(G,1)} and {(A,2),(B,2),(C,2),(D,2),(E,2),(F,2),(G,2)}. So there are 126 surjective functions. Section 12.3 Exercises 1. If 6 integers are chosen at random, at least two will have the same remainder when divided by 5. Proof. Write Z as follows: Z = S4 j=0{5k + j : k ∈Z}. This is a partition of Z into 5 sets. If six integers are picked at random, by the pigeonhole principle, at least two will be in the same set. However, each set corresponds to the remainder of a number after being divided by 5 (for example, {5k +1 : k ∈Z} are all those integers that leave a remainder of 1 after being divided by 5). ■ 3. Given any six positive integers, there are two for which their sum or diﬀerence is divisible by 9. Proof. If for two of the integers n,m we had n ≡m (mod 9), then n−m ≡0 (mod 9), and we would be done. Thus assume this is not the case. Observe that the 293 only two element subsets of positive integers that sum to 9 are {1,8},{2,7},{3,6}, and {4,5}. However, since at least ﬁve of the six integers must have distinct remainders from 1, 2, ..., 8 it follows from the pigeonhole principle that two integers n,m are in the same set. Hence n+ m ≡0 (mod 9) as desired. ■ 5. Prove that any set of 7 distinct natural numbers contains a pair of numbers whose sum or diﬀerence is divisible by 10. Proof. Let S = {a1,a2,a3,a4,a5,a6,a7} be any set of 7 natural numbers. Let’s say that a1 < a2 < a3 < ··· < a7. Consider the set A = {a1 −a2, a1 −a3, a1 −a4, a1 −a5, a1 −a6, a1 −a7, a1 + a2, a1 + a3, a1 + a4, a1 + a5, a1 + a6, a1 + a7} Thus \|A\| = 12. Now let B = {0,1,2,3,4,5,6,7,8,9}, so \|B\| = 10. Let f : A →B be the function for which f (n) equals the last digit of n. (That is f (97) = 7, f (12) = 2, f (230) = 0, etc.) Then, since \|A\| > \|B\|, the pigeonhole principle guarantees that f is not injective. Thus A contains elements a1 ± ai and a1 ± a j for which f (a1 ±ai) = f (a1 ±a j). This means the last digit of a1 ±ai is the same as the last digit of a1 ±a j. Thus the last digit of the diﬀerence (a1 ±ai)−(a1 ±a j) = ±ai ±a j is 0. Hence ±ai ± a j is a sum or diﬀerence of elements of S that is divisible by 10. ■ Section 12.4 Exercises 1. Suppose A = {5,6,8}, B = {0,1}, C = {1,2,3}. Let f : A →B be the function f = {(5,1),(6,0),(8,1)}, and g : B →C be g = {(0,1),(1,1)}. Find g ◦f . g ◦f = {(5,1),(6,1),(8,1)} 3. Suppose A = {1,2,3}. Let f : A →A be the function f = {(1,2),(2,2),(3,1)}, and let g : A →A be the function g = {(1,3),(2,1),(3,2)}. Find g ◦f and f ◦g. g ◦f = {(1,1),(2,1),(3,3)}; f ◦g = {(1,1),(2,2),(3,2)}. 5. Consider the functions f , g : R →R deﬁned as f (x) = 3 p x+1 and g(x) = x3. Find the formulas for g ◦f and f ◦g. g ◦f (x) = x+1; f ◦g(x) = 3 p x3 +1 7. Consider the functions f , g : Z × Z →Z × Z deﬁned as f (m,n) = (mn,m2) and g(m,n) = (m+1,m+ n). Find the formulas for g ◦f and f ◦g. Note g ◦f (m,n) = g(f (m,n)) = g(mn,m2) = (mn+1,mn+ m2). Thus g ◦f (m,n) = (mn+1,mn+ m2). Note f ◦g(m,n) = f (g(m,n)) = f (m+1,m+ n) = ((m+1)(m+ n),(m+1)2). Thus f ◦g(m,n) = (m2 + mn+ m+ n,m2 +2m+1). 9. Consider the functions f : Z×Z →Z deﬁned as f (m,n) = m+ n and g : Z →Z×Z deﬁned as g(m) = (m,m). Find the formulas for g ◦f and f ◦g. g ◦f (m,n) = (m+ n,m+ n) f ◦g(m) = 2m 294 Solutions Section 12.5 Exercises 1. Check that the function f : Z →Z deﬁned by f (n) = 6 −n is bijective. Then compute f −1. It is injective as follows. Suppose f (m) = f (n). Then 6−m = 6−n, which reduces to m = n. It is surjective as follows. If b ∈Z, then f (6−b) = 6−(6−b) = b. Inverse: f −1(n) = 6−n. 3. Let B = {2n : n ∈Z} = © ..., 1 4, 1 2,1,2,4,8,... ª. Show that the function f : Z →B deﬁned as f (n) = 2n is bijective. Then ﬁnd f −1. It is injective as follows. Suppose f (m) = f (n), which means 2m = 2n. Taking log2 of both sides gives log2(2m) = log2(2n), which simpliﬁes to m = n. The function f is surjective as follows. Suppose b ∈B. By deﬁnition of B this means b = 2n for some n ∈Z. Then f (n) = 2n = b. Inverse: f −1(n) = log2(n). 5. The function f : R →R deﬁned as f (x) = πx−e is bijective. Find its inverse. Inverse: f −1(x) = x+ e π . 7. Show that the function f : R2 →R2 deﬁned by the formula f ((x, y) = ((x2 +1)y,x3) is bijective. Then ﬁnd its inverse. First we prove the function is injective. Assume f (x1, y1) = f (x2, y2). Then (x2 1 +1)y1 = (x2 2 +1)y2 and x3 1 = x3 2. Since the real-valued function f (x) = x3 is one-to-one, it follows that x1 = x2. Since x1 = x2, and x2 1 +1 > 0 we may divide both sides of (x2 1 +1)y1 = (x2 1 +1)y2 by (x2 1 +1) to get y1 = y2. Hence (x1, y1) = (x2, y2). Now we prove the function is surjective. Let (a,b) ∈R2. Set x = b1/3 and y = a/(b2/3 +1). Then f (x, y) = ((b2/3 +1) a b2/3+1,(b1/3)3) = (a,b). It now follows that f is bijective. Finally, we compute the inverse. Write f (x, y) = (u,v). Interchange variables to get (x, y) = f (u,v) = ((u2 +1)v,u3). Thus x = (u2 +1)v and y = u3. Hence u = y1/3 and v = x y2/3+1. Therefore f −1(x, y) = (u,v) = ³ y1/3, x y2/3+1 ´ . 9. Consider the function f : R×N →N×R deﬁned as f (x, y) = (y,3xy). Check that this is bijective; ﬁnd its inverse. To see that this is injective, suppose f (a,b) = f (c,d). This means (b,3ab) = (d,3cd). Since the ﬁrst coordinates must be equal, we get b = d. As the second coordinates are equal, we get 3ab = 3dc, which becomes 3ab = 3bc. Note that, from the deﬁnition of f , b ∈N, so b ̸= 0. Thus we can divide both sides of 3ab = 3bc by the non-zero quantity 3b to get a = c. Now we have a = c and b = d, so (a,b) = (c,d). It follows that f is injective. Next we check that f is surjective. Given any (b, c) in the codomain N×R, notice that ( c 3b,b) belongs to the domain R×N, and f ( c 3b,b) = (b, c). Thus f is surjective. As it is both injective and surjective, it is bijective; thus the inverse exists. To ﬁnd the inverse, recall that we obtained f ( c 3b,b) = (b, c). Then f −1 f ( c 3b,b) = f −1(b, c), which reduces to ( c 3b,b) = f −1(b, c). Replacing b and c with x and y, respectively, we get f −1(x, y) = ( y 3x,x). 295 Section 12.6 Exercises 1. Consider the function f : R →R deﬁned as f (x) = x2 + 3. Find f ([−3,5]) and f −1([12,19]). Answers: f ([−3,5]) = [3,28]; f −1([12,19]) = [−4,−3]∪[3,4]. 3. This problem concerns functions f : {1,2,3,4,5,6,7} →{0,1,2,3,4}. How many such functions have the property that \|f −1({3})\| = 3? Answer: 44 ¡7 3 ¢. 5. Consider a function f : A →B and a subset X ⊆A. We observed in Section 12.6 that f −1(f (X)) ̸= X in general. However X ⊆f −1(f (X)) is always true. Prove this. Proof. Suppose a ∈X. Thus f (a) ∈{f (x) : x ∈X} = f (X), that is f (a) ∈f (X). Now, by deﬁnition of preimage, we have f −1(f (X)) = {x ∈A : f (x) ∈f (X)}. Since a ∈A and f (a) ∈f (X), it follows that a ∈f −1(f (X)). This proves X ⊆f −1(f (X)). ■ 7. Given a function f : A →B and subsets W, X ⊆A, prove f (W ∩X) ⊆f (W)∩f (X). Proof. Suppose b ∈f (W ∩X). This means b ∈{f (x) : x ∈W ∩X}, that is b = f (a) for some a ∈W ∩X. Since a ∈W we have b = f (a) ∈{f (x) : x ∈W} = f (W). Since a ∈X we have b = f (a) ∈{f (x) : x ∈X} = f (X). Thus b is in both f (W) and f (X), so b ∈f (W)∩f (X). This completes the proof that f (W ∩X) ⊆f (W)∩f (X). ■ 9. Given a function f : A →B and subsets W, X ⊆A, prove f (W ∪X) = f (W)∪f (X). Proof. First we will show f (W ∪X) ⊆f (W)∪f (X). Suppose b ∈f (W ∪X). This means b ∈{f (x) : x ∈W ∪X}, that is, b = f (a) for some a ∈W ∪X. Thus a ∈W or a ∈X. If a ∈W, then b = f (a) ∈{f (x) : x ∈W} = f (W). If a ∈X, then b = f (a) ∈ {f (x) : x ∈X} = f (X). Thus b is in f (W) or f (X), so b ∈f (W)∪f (X). This completes the proof that f (W ∪X) ⊆f (W)∪f (X). Next we will show f (W)∪f (X) ⊆f (W ∪X). Suppose b ∈f (W)∪f (X). This means b ∈f (W) or b ∈f (X). If b ∈f (W), then b = f (a) for some a ∈W. If b ∈f (X), then b = f (a) for some a ∈X. Either way, b = f (a) for some a that is in W or X. That is, b = f (a) for some a ∈W ∪X. But this means b ∈f (W ∪X). This completes the proof that f (W)∪f (X) ⊆f (W ∪X). The previous two paragraphs show f (W ∪X) = f (W)∪f (X). ■ 11. Given f : A →B and subsets Y ,Z ⊆B, prove f −1(Y ∪Z) = f −1(Y )∪f −1(Z). Proof. First we will show f −1(Y ∪Z) ⊆f −1(Y )∪f −1(Z). Suppose a ∈f −1(Y ∪Z). By Deﬁnition 12.9, this means f (a) ∈Y ∪Z. Thus, f (a) ∈Y or f (a) ∈Z. If f (a) ∈Y, then a ∈f −1(Y ), by Deﬁnition 12.9. Similarly, if f (a) ∈Z, then a ∈ f −1(Z). Hence a ∈f −1(Y ) or a ∈f −1(Z), so a ∈f −1(Y ) ∪f −1(Z). Consequently f −1(Y ∪Z) ⊆f −1(Y )∪f −1(Z). Next we show f −1(Y ) ∪f −1(Z) ⊆f −1(Y ∪Z). Suppose a ∈f −1(Y ) ∪f −1(Z). This means a ∈f −1(Y ) or a ∈f −1(Z). Hence, by Deﬁnition 12.9, f (a) ∈Y or f (a) ∈Z, which means f (a) ∈Y ∪Z. But by Deﬁnition 12.9, f (a) ∈Y ∪Z means a ∈f −1(Y ∪Z). Consequently f −1(Y )∪f −1(Z) ⊆f −1(Y ∪Z). The previous two paragraphs show f −1(Y ∪Z) = f −1(Y )∪f −1(Z). ■ 296 Solutions 13. Let f : A →B be a function, and X ⊆A. Prove or disprove: f ¡ f −1(f (X)) ¢ = f (X). Proof. First we will show f ¡ f −1(f (X)) ¢ ⊆f (X). Suppose y ∈f ¡ f −1(f (X)) ¢. By deﬁnition of image, this means y = f (x) for some x ∈f −1(f (X)). But by deﬁnition of preimage, x ∈f −1(f (X)) means f (x) ∈f (X). Thus we have y = f (x) ∈f (X), as desired. Next we show f (X) ⊆f ¡ f −1(f (X)) ¢. Suppose y ∈f (X). This means y = f (x) for some x ∈X. Then f (x) = y ∈f (X), which means x ∈f −1(f (X)). Then by deﬁnition of image, f (x) ∈f (f −1(f (X))). Now we have y = f (x) ∈f (f −1(f (X))), as desired. The previous two paragraphs show f ¡ f −1(f (X)) ¢ = f (X). ■ Chapter 13 Exercises Section 13.1 Exercises 1. R and (0,∞) Observe that the function f (x) = ex sends R to (0,∞). It is injective because f (x) = f (y) implies ex = ey, and taking ln of both sides gives x = y. It is surjective because if b ∈(0,∞), then f (ln(b)) = b. Therefore, because of the bijection f : R →(0,∞), it follows that \|R\| = \|(0,∞)\|. 3. R and (0,1) Observe that the function 1 π f (x) = cot−1(x) sends R to (0,1). It is injective and surjective by elementary trigonometry. Therefore, because of the bijection f : R →(0,1), it follows that \|R\| = \|(0,1)\|. 5. A = {3k : k ∈Z} and B = {7k : k ∈Z} Observe that the function f (x) = 7 3 x sends A to B. It is injective because f (x) = f (y) implies 7 3 x = 7 3 y, and multiplying both sides by 3 7 gives x = y. It is surjective because if b ∈B, then b = 7k for some integer k. Then 3k ∈A, and f (3k) = 7k = b. Therefore, because of the bijection f : A →B, it follows that \|A\| = \|B\|. 7. Z and S = © ..., 1 8, 1 4, 1 2,1,2,4,8,16,... ª Observe that the function f : Z →S deﬁned as f (n) = 2n is bijective: It is injective because f (m) = f (n) implies 2m = 2n, and taking log2 of both sides produces m = n. It is surjective because any element b of S has form b = 2n for some integer n, and therefore f (n) = 2n = b. Because of the bijection f : Z →S, it follows that \|Z\| = \|S\|. 9. {0,1}×N and N Consider the function f : {0,1}×N →N deﬁned as f (a,n) = 2n−a. This is injective because if f (a,n) = f (b,m), then 2n−a = 2m−b. Now if a were unequal to b, one of a or b would be 0 and the other would be 1, and one side of 2n−a = 2m−b would be odd and the other even, a contradiction. Therefore a = b. Then 2n−a = 2m−b becomes 2n−a = 2m−a; add a to both sides and divide by 2 to get m = n. Thus we have a = b and m = n, so (a,n) = (b,m), so f is injective. 297 To see that f is surjective, take any b ∈N. If b is even, then b = 2n for some integer n, and f (0,n) = 2n−0 = b. If b is odd, then b = 2n+1 for some integer n. Then f (1,n+1) = 2(n+1)−1 = 2n+1 = b. Therefore f is surjective. Then f is a bijection, so \|{0,1}×N\| = \|N\|. 11. [0,1] and (0,1) Proof. Consider the subset X = © 1 n : n ∈N ª ⊆[0,1]. Let f : [0,1] →[0,1) be deﬁned as f (x) = x if x ∈[0,1]−X and f ( 1 n) = 1 n+1 for any 1 n ∈X. It is easy to check that f is a bijection. Next let Y = © 1−1 n : n ∈N ª ⊆[0,1), and deﬁne g : [0,1) →(0,1) as g(x) = x if x ∈[0,1)−Y and g(1−1 n) = 1− 1 n+1 for any 1−1 n ∈Y. As in the case of f , it is easy to check that g is a bijection. Therefore the composition g◦f : [0,1] →(0,1) is a bijection. (See Theorem 12.2.) We conclude that \|[0,1]\| = \|(0,1)\|. ■ 13. P(N) and P(Z) Outline: By Exercise 18 of Section 12.2, we have a bijection f : N →Z deﬁned as f (n) = (−1)n(2n−1)+1 4 . Now deﬁne a function Φ : P(N) →P(Z) as Φ(X) = {f (x) : x ∈X}. Check that Φ is a bijection. 15. Find a formula for the bijection f in Example 13.2. Hint: Consider the function f from Exercise 18 of Section 12.2. Section 13.2 Exercises 1. Prove that the set A = {ln(n) : n ∈N} ⊆R is countably inﬁnite. Just note that its elements can be written in inﬁnite list form as ln(1),ln(2),ln(3),···. Thus A is countably inﬁnite. 3. Prove that the set A = {(5n,−3n) : n ∈Z} is countably inﬁnite. Consider the function f : Z →A deﬁned as f (n) = (5n,−3n). This is clearly surjective, and it is injective because f (n) = f (m) gives (5n,−3n) = (5m,−3m), so 5n = 5m, hence m = n. Thus, because f is surjective, \|Z\| = \|A\|, and \|A\| = \|Z\| = ℵ0. Therefore A is countably inﬁnite. 5. Prove or disprove: There exists a countably inﬁnite subset of the set of irrational numbers. This is true. Just consider the set consisting of the irrational numbers π 1 , π 2 , π 3 , π 4 ,···. 7. Prove or disprove: The set Q100 is countably inﬁnite. This is true. Note Q100 = Q×Q×···×Q (100 times), and since Q is countably inﬁnite, it follows from the corollary of Theorem 13.5 that this product is countably inﬁnite. 9. Prove or disprove: The set {0,1}×N is countably inﬁnite. This is true. Note that {0,1}×N can be written in inﬁnite list form as (0,1),(1,1),(0,2),(1,2),(0,3),(1,3),(0,4),(1,4),···. Thus the set is countably inﬁnite. 298 Solutions 11. Partition N into 8 countably inﬁnite sets. For each i ∈{1,2,3,4,5,6,7,8}, let Xi be those natural numbers that are congruent to i modulo 8, that is, X1 = {1,9,17,25,33,...} X2 = {2,10,18,26,34,...} X3 = {3,11,19,27,35,...} X4 = {4,12,20,28,36,...} X5 = {5,13,21,29,37,...} X6 = {6,14,22,30,38,...} X7 = {7,15,13,31,39,...} X8 = {8,16,24,32,40,...} 13. If A = {X ⊂N : X is ﬁnite}, then \|A\| = ℵ0. Proof. This is true. To show this we will describe how to arrange the items of A in an inﬁnite list X1, X2, X3, X4,.... For each natural number n, let pn be the nth prime number. Thus p1 = 2, p2 = 3, p3 = 5, p4 = 7, p5 = 11, and so on. Now consider any element X ∈A. If X ̸= ;, then X = {n1,n2,n3,...,nk}, where k = \|X\| and ni ∈N for each 1 ≤i ≤k. Deﬁne a function f : A →N ∪{0} as follows: f ({n1,n2,n3,...,nk}) = pn1 pn2 ··· pnk. For example, f ({1,2,3}) = p1p2p3 = 2·3·5 = 30, and f ({3,5}) = p3p5 = 5·11 = 55, etc. Also, we should not forget that ; ∈A, and we deﬁne f (;) = 0. Note that f : A →N ∪{0} is an injection: Let X = {n1,n2,n3,...,nk} and Y = {m1,m2,m3,...,mℓ}, and X ̸= Y. Then there is an integer a that belongs to one of X or Y but not the other. Then the prime factorization of one of the numbers f (X) and f (Y ) uses the prime number pa but the prime factorization of the other does not use pa. It follows that f (X) ̸= f (Y ) by the fundamental theorem of arithmetic. Thus f is injective. So each set X ∈A is associated with an integer f (X) ≥0, and no two diﬀerent sets are associated with the same number. Thus we can list the elements in X ∈A in increasing order of the numbers f (X). The list begins as ;, {1}, {2}, {3}, {1,2}, {4}, {1,3}, {5}, {6}, {1,4}, {2,3}, {7},... It follows that A is countably inﬁnite. ■ 15. Hint: Use the fundamental theorem of arithmetic. Section 13.3 Exercises 1. Suppose B is an uncountable set and A is a set. Given that there is a surjective function f : A →B, what can be said about the cardinality of A? 299 The set A must be uncountable, as follows. For each b ∈B, let ab be an element of A for which f (ab) = b. (Such an element must exist because f is surjective.) Now form the set U = {ab : b ∈B}. Then the function f : U →B is bijective, by construction. Then since B is uncountable, so is U. Therefore U is an uncountable subset of A, so A is uncountable by Theorem 13.9. 3. Prove or disprove: If A is uncountable, then \|A\| = \|R\|. This is false. Let A = P(R). Then A is uncountable, and by Theorem 13.7, \|R\| < \|P(R)\| = \|A\|. 5. Prove or disprove: The set {0,1}×R is uncountable. This is true. To see why, ﬁrst note that the function f : R →{0}×R deﬁned as f (x) = (0,x) is a bijection. Thus \|R\| = \|{0}×R\|, and since R is uncountable, so is {0}×R. Then {0}×R is an uncountable subset of the set {0,1}×R, so {0,1}×R is uncountable by Theorem 13.9. 7. Prove or disprove: If A ⊆B and A is countably inﬁnite and B is uncountable, then B −A is uncountable. This is true. To see why, suppose to the contrary that B−A is countably inﬁnite. Then B = A ∪(B −A) is a union of countably inﬁnite sets, and thus countable, by Theorem 13.6. This contradicts the fact that B is uncountable. Exercises for Section 13.4 1. Show that if A ⊆B and there is an injection g : B →A, then \|A\| = \|B\|. Just note that the map f : A →B deﬁned as f (x) = x is an injection. Now apply the Cantor-Bernstein-Schröeder theorem. 3. Let F be the set of all functions N → © 0,1}. Show that \|R\| = \|F\|. Because \|R\| = \|P(N)\|, it suﬃces to show that \|F\| = \|P(N)\|. To do this, we will exhibit a bijection f : F →P(N). Deﬁne f as follows. Given a function ϕ ∈F, let f (ϕ) = {n ∈N : ϕ(n) = 1}. To see that f is injective, suppose f (ϕ) = f (θ). Then {n ∈N : ϕ(n) = 1} = {n ∈N : θ(n) = 1}. Put X = {n ∈N : ϕ(n) = 1}. Now we see that if n ∈X, then ϕ(n) = 1 = θ(n). And if n ∈N−X, then ϕ(n) = 0 = θ(n). Consequently ϕ(n) = θ(n) for any n ∈N, so ϕ = θ. Thus f is injective. To see that f is surjective, take any X ∈P(N). Consider the function ϕ ∈F for which ϕ(n) = 1 if n ∈X and ϕ(n) = 0 if n ∉X. Then f (ϕ) = X, so f is surjective. 5. Consider the subset B = © (x, y) : x2 + y2 ≤1 ª ⊆R2. Show that \|B\| = \|R2\|. This will follow from the Cantor-Bernstein-Schröeder theorem provided that we can ﬁnd injections f : B →R2 and g : R2 →B. The function f : B →R2 deﬁned as f (x, y) = (x, y) is clearly injective. For g : R2 →B, consider the function g(x, y) = µ x2 + y2 x2 + y2 +1 x, x2 + y2 x2 + y2 +1 y ¶ . Verify that this is an injective function g : R2 →B. 300 Solutions 7. Prove or disprove: If there is a injection f : A →B and a surjection g : A →B, then there is a bijection h : A →B. This is true. Here is an outline of a proof. Deﬁne a function g′ : B →A as follows. For each b ∈B, choose an element xb ∈g−1({x}). (That is, choose an element xb ∈A for which g(xb) = b.) Now let g′ : B →A be the function deﬁned as g′(b) = xb. Check that g′ is injective and apply the the Cantor-Bernstein-Schröeder theorem. Index C(n,k), 74 absolute value, 6 addition principle, 82 and, 38 axiom of foundation, 31 basis step, 156 biconditional statement, 44 bijection, 218 bijective function, 201 byte, 64 Cantor, Georg, 219 Cantor-Bernstein-Schröeder theorem, 234 cardinality, 4, 217 Cartesian plane, 9 Cartesian power, 10 Cartesian product, 8 closed interval, 6 codomain of a function, 198 Cohen, Paul, 237 complement of a set, 19 composite number, 90 composition of functions, 208 conditional statement, 41 conjecture, 147 constructive proof, 128 continuum hypothesis, 237 contrapositive, 102 converse of a statement, 44, 102 corollary, 88 countable set, 223 counterexample, 149 counting, 63 deﬁnition, 87 DeMorgan’s laws, 50, 57 diﬀerence of sets, 17 disproof, 146 divides, 90 division algorithm, 29, 91 divisor, 90 domain of a function, 198 Doxiadis, Apostolos, 32 element of a set, 3 empty set, 4 entries of a list, 63 equality of functions, 200 equality of lists, 64 equality of sets, 3 equivalence class, 185 equivalence relation, 184 equivalent statements, 123 Euclid, 114, 140 Euler, Leonhard, 107, 142 existence theorem, 125 existential quantiﬁer, 52 existential statement, 125 factorial, 70 false, 34 Fermat’s last theorem, 36 Fermat, Pierre de, 36 Fibonacci sequence, 167 function, 197 range of, 198 bijective, 201, 217 codomain of, 198 composition of, 208 domain of, 198 equality, 200 injective, 201 inverse, 211 notation, 199 302 Index one-to-one, 201 onto, 201 surjective, 201 function notation, 199 fundamental theorem of arithmetic, 166 fundamental theorem of calculus, viii gamma function, 73 gcd, 90 geometric sequence, 169 Goldbach’s conjecture, 36, 56 Goldbach, Christian, 36 golden ratio, 169 graph, 163 cycle, 163 edges, 163 vertices, 163 greatest common divisor, 90 Hagy, Jessica, 32 half-open interval, 6 if-and-only-if theorem, 121 image, 215 inclusion-exclusion formula, 81 index set, 25 indexed set, 24 induction, 154 strong, 161 inductive hypothesis, 156 inductive step, 156 inﬁnite interval, 6 injection, 218 injective function, 201 integers, 3, 4 congruence, 105, 181 modulo n, 192 intersection of sets, 17 interval, 6 inverse of a function, 211 inverse relation, 211 irrational number, 113 lcm, 90 least common multiple, 90 lemma, 88 length, 63 list, 63 empty, 64 entries, 63 equal, 64 non-repetitive, 66 order, 63 repetition, 66 logic, 33 contradiction, 111 equivalence, 49 inference, 61 quantiﬁer, 52 existential, 52 universal, 52 symbols, 46, 51 logical equivalence, 49 logical inference, 61 mean value theorem, 55 Mersenne prime, 144 multiple, 90 multiplication principle, 65 natural numbers, 4 necessary condition, 43 negation of a statement, 40 non-constructive proof, 128 one-to-one function, 201 onto function, 201 open interval, 6 open sentence, 35, 54 or, 39 ordered pair, 8 ordered triple, 9 Papadimitriou, Christos, 32 parity, 89 partition, 189 Pascal’s triangle, 79 Pascal, Blaise, 79 perfect number, 139, 142 pigeonhole principle, 206 Pisano, Leonardo, 167 power set, 14 power, Cartesian, 10 preimage, 215 prime number, 36, 90 proof by cases, 98 by contradiction, 111 303 by induction, 154 strong, 161 by smallest counterexample, 165 constructive, 128 contrapositive, 102 direct, 87, 92 existence, 124 involving sets, 131 non-constructive, 128 uniqueness, 124, 127 within-a-proof, 117 proposition, 88, 92 Pythagorean theorem, 36 quadratic formula, 36 quantiﬁer, 52 quotient, 29, 91 range of a function, 198 rational numbers, 6, 113 real numbers, 4 reﬂexive property of a relation, 179 relations, 175 between sets, 194 equivalence, 184 class, 185 inverse, 211 reﬂexive, 179 symmetric, 179 transitive, 179 remainder, 29, 91, 105 Russell’s paradox, 31 Russell, Bertrand, 31 set(s) builder-notation, 5, 131 cardinalities of comparison of, 228 equal, 217 unequal, 217 cardinality of, 3, 217 complement, 19 countable, 223 element of, 3 empty, 4 equal, 3 partition of, 189 subset of, 11 uncountable, 223 sigma notation, 24 size, see cardinality statement, 34 biconditional, 44 conditional, 41 necessary, 43 suﬃcient, 43 converse, 44 equivalent, 123 existential, 125 negation, 57 Stirling’s formula, 73 strong induction, 161 subset, 11 suﬃcient condition, 43 surjection, 218 surjective function, 201 symmetric property of a relation, 179 theorem, 87 existence, 125 if-and-only-if, 121 three-dimensional space, 10 transitive property of a relation, 179 tree, 163 triple, ordered, 9 true, 34 truth table, 38 value, 38 uncountable set, 223 union of sets, 17 uniqueness proof, 127 unit circle, 14, 19 universal quantiﬁer, 52 universal set, 19 variable, 35 vector space, 131 Venn diagram, 21 well-ordering principle, 29 Wiles, Andrew, 36 WLOG, 99 Zermelo-Fraenkel axioms, 31
16312	https://www.bmj.com/content/345/bmj.e4955/rapid-responses	Skip to main content Education Diagnosis and... Diagnosis and management of cellulitis Clinical Review Diagnosis and management of cellulitis BMJ 2012; 345 doi: (Published 07 August 2012) Cite this as: BMJ 2012;345:e4955 Article Related content Metrics Responses Peer review All rapid responses ### Re: Diagnosis and management of cellulitis Bilateral cellulitis of the legs: does it exist? The clinical review by Phoenix and co-authors of the diagnosis and management of cellulitis (1) was comprehensive in all respects excepting that, perhaps quite reasonably, it omitted mention of the phenomenon of ‘bilateral cellulitis’, a diagnosis of doubtful validity. It is by no means unusual to be asked to review a patient with two uncomfortable, red legs receiving intravenous antibiotic therapy following a presumed diagnosis of bilateral cellulitis. As this excellent review mentioned, cellulitis of the legs can be easily confused with florid varicose eczema and contact dermatitis, although in both instances there will usually be obvious epidermal changes that may serve to distinguish them from cellulitis. More problematic, however, is the occurrence of lipodermatosclerosis, a sclerosing panniculitis consequent on venous insufficiency. The chronic form of this common condition presents as pigmented, indurated, bound-down skin over the gaiter regions, classically resulting in an ‘inverted champagne bottle’ shape to the leg. The acute phase of lipodermatosclerosis, however, is characterized by inflamed, painful skin over and proximal to the malleolar regions, which can readily be mistaken for cellulitis (2). These potential mimics of cellulitis of the leg are frequently, although not necessarily, bilateral, whereas cellulitis itself, usually caused by Staphylococcus aureus or Group A Streptococcus, organisms that are not known for their sense of symmetry, is invariably unilateral. Cellulitis involving both legs synchronously probably can occur as a chance happening, but it must be almost as common as hens’ teeth, and a sound clinical diktat is that if a patient appears to have bilateral cellulitis of the legs they probably do not! Phoenix G, Das S, Joshi M. Diagnosis and management of cellulitis. BMJ 2012;345:38-42. Miteva M, Romanelli P, Kirsner RS. Lipodermatosclerosis. Dermatol Ther. 2010;23:375-88 Competing interests: No competing interests ### Re: Diagnosis and management of cellulitis As a microbiologist, I would like to comment that the article did not mention orbital cellulitis. This potentially serious infection requires antibiotic cover for a mixture of possible causative organisms, including anaerobes. Also, H.influenzae Pittman type b was not included as a possible cause of periorbital cellulitis, particularly in children who have not had Hib vaccine. It will not respond to penicillin &/or flucloxacillin Competing interests: No competing interests 13 September 2012 Elizabeth H Price retired microbiologist Retired Middleway, London ### Re: Diagnosis and management of cellulitis I read this article with interest. I have just been dealing with a relevant case indicating difficulty in differential diagnosis. This was a 52 year old woman who had been in hospital for over a week under the care of general physicians with a diagnosis of a possible infected knee joint and prepatellar bursa and cellulitis of the lower leg. There was no fever at any time. The CRP was 145 mg/L and the WBC was 13.0 with a neutrophilia. Urate was normal at 0.09 mmol/L. Blood cultures were negative. Fluid was not withdrawn from the knee or the bursa. She had been seen by orthopaedic surgeons but not by a rheumatologist. She was treated with intravenous clindamycin with no significant effect, and after her discharge, still on oral antibiotics, I was called in to see her. The history was of a painful knee developing over a day to extreme pain with a lot of swelling, with no injury or obvious cause, followed within a day by painful swelling of the whole of the lower leg with skin erythema. On examination the knee and prepatellar bursa were swollen and painful, there was a tender popliteal cyst, and the lower leg was very swollen and erythematous. As she told me the story I realised she was describing an acute attack of gout with a burst popliteal cyst, and I then found that her father had gout. Doppler studies after her discharge had ruled out a deep vein thrombosis. All the blood test results are compatible with acute gout, including the normal urate. The pain and swelling of the knee rapidly resolved on treatment with diclofenac 50 mg four times daily (with a stomach protective), and the swelling and erythema of the lower leg gradually went away with the help of elevation and an elastic stocking. Strikingly, the skin over the front and sides of the knee peeled off massively during recovery, which of course is typical of gout and would not be expected after an infection. Acute arthritis is quoted as a possible differential diagnosis in the article, and this case demonstrates that acute gout must certainly be considered. Patient consent obtained. Competing interests: No competing interests 01 September 2012 Michael F. Grayson Consultant rheumatologist Private practice, Royal National Orthopaeduc Hospital Royal National Orthopaedic Hospital, 45 Bolsover Street, London W1W 5AQ ### Re: Diagnosis and management of cellulitis We thank Dr Rae, Dr Logan, Dr Martin, Dr Swann and Mr Kelley for pointing out the errors in our review, which have now been corrected. We agree with Dr Rae that infectious disease specialists play an integral role in the management of patients with cellulitis, especially the more difficult cases. Microbiologists, and many other specialties including emergency department physicians and other healthcare professionals, such as specialist nurses, whom pay an integral part in the ambulatory management of many patients with cellulitis, should also be acknowledged. The list of specialists is endless, reflecting the ‘orphan’ nature of the condition; Kilburn et al (2003), in their recent cochrane review, mention the need for a ‘multiciplinary’ approach to treatment. It is impractical to distinguish between streptococcal and staphylococcal cellulitis, as stated by Dr Rae and Dr Logan. As mentioned in our review, Flucloxacillin (and not Amoxicillin, as incorrectly mentioned in our review) should be used to cover both or when the organism is unknown. However, when organisms are isolated, especially as the case for CA-MRSA, antibiotics may need to be tailored accordingly. We agree with Dr Rae that in patients with a clinical suspicion of Necrotizing Fasciitis, surgical debridement should not be delayed, especially in those that are systemically unwell or haemodynamically unstable. However, a clinical diagnosis for Necrotizing Fasciitis is not always straightforward and in type B and type C infections, in which the onset is more insidious, MRI can confirm or refute diagnosis, as well as demarcating the extent of disease. As Dr Logan also mentions, investigations are often negative and identification of an organism is problematic; the review supports this statement and recommendations based on national guidelines have been mentioned. Dr Halpern opinion that diabetes is not a risk factor is supported by higher evidence than the anecdotal and case series that support an association in our review. We have mentioned, however, that there is no evidence to suggest active management of diabetes, improves cellulitis recurrence; a more significant point in our opinion. Furthermore, we did not find evidence supporting the active management of T.Pedis in reducing recurrence of cellulitis as well, including in the references stated by Dr Halpern. As mentioned, further research is required into the effect of treating predisposing factors and the impact on cellulitis, such as recurrence rates or duration of cellulitis symptoms or signs. We agree wholeheartedly with Dr Duncan regarding OPAT services; a service that has been shown by numerous studies to be both cost-effective and clinical-effective. The UK study referenced in our review acknowledges this but it also states that not all areas in the UK offer comprehensive OPAT services as of yet. We are sure this will soon change. Dr Basak mentions the efficacy of Co-trimoxazole for CA-MRSA; Trimethoprin-Sulfamethoxazole has also been demonstrated to have good efficacy. No national guidance thus far supports a single antibiotic for CA-MRSA. We would like to point out to Dr Gada that many of the mentioned points have been addressed in the review already, which encompass both primary and secondary care. A step-by-step approach in regards to antibiotic treatment options is not currently possible on a national scale, (although we also have reccomended the use of Flucloxacillin when the organism is unknown); this may reflect the lack of strong evidence for an optimal antibiotic regime(s) , the multiple organisms that can cause cellulitis including new strains, and or differing local resources. Kilburn S, Featherstone P, Higgins B, Brindle R, Severs M (2003) Interventions for cellulitis and erysipelas (Protocol). Cochrane Database System Rev 1: CD004299. DOI:10.1002/14651858. CD004299. Carter PS, Banwell PE. Necrotising fasciitis: a new management algorithm based on clinical classification. Int Wound J 2004;1:189—198. Phoenix G, Das S, Joshi M. Diagnosis and management of cellulitis. BMJ 2012;345:e4955. (7 August.) Pappas G, Athanasoulia AP, Matthaiou DK, Falagas ME. Trimethoprim-sulfamethoxazole for methicillin-resistant Staphylococcus aureus: a forgotten alternative? J Chemother. 2009 Apr;21(2):115-26. Competing interests: No competing interests 31 August 2012 Gokulan K Phoenix Core Surgical Trainee Meera Joshi, Saroj Das Chelsea and Westminster Hospital 369 Fulham Road, London, SW10 9NH ### Re: Diagnosis and management of cellulitis I read with interest the article written by Phoenix et al, it is a good article, but it lacks some detail. As I work as a GP, I can write my views from Primary care side. I do feel that majority of the cases of cellulitis and skin & subcutaneous infections are managed in primary care by GPs. Only minority of patients with cellulitis are referred to secondary care for further management (either because of cases not responding to the drugs used, patients becoming septic, cellulitic patch spreading, patients request, etc). The commonest drug used in primary care for cases of cellulitis is flucloxacillin (in those who are not allergic to penicillin). It will be good to come up with list of common microbes causing cellulitis, common risk factors (modifiable and non-modifiable) and guidelines with step by step choice of antibiotics to be used. Number of cases of skin infections / cellulitis, seen in primary care should be easily calculated as almost all surgeries now are computerized. We look forward to see an article in future encompassing cases of cellulitis both covering primary and secondary care. Competing interests: No competing interests 27 August 2012 Siddappa Gada GP Holbrook & Shotley surgery Ipswich ### Re: Diagnosis and management of cellulitis Phoenix et al refer to the potential use of parenteral out patient antibiotic therapy (OPAT) “when available”. In fact clinical practice in the UK has progressed to the point that OPAT services are now specifically developed for this often ambulant and clinically stable patient population. Recent Department of Health guidance on antimicrobial stewardship in hospitals (“Start smart then focus”) has highlighted OPAT as one of the five antimicrobial prescribing options. Two main models of OPAT exist in the UK: as an integral part of an infection specialist led service, or as part of broader emergency department-focussed ambulatory care (also known as hospital at home). In the UK cellulitis is the most common indication for OPAT, and the majority of patients are managed without prior hospital admission. For the last 12 years we have provided OPAT services to people with cellulitis referred directly from the community or emergency departments, and have treated over 1300 patients with excellent clinical outcomes. In our practice, patients with cellulitis referred by a medical practitioner are assessed for suitability by a specialist nurses follow a clinical protocol and patient group direction. Specialist nurses make daily assessments for complications, progression, adverse events and suitability for IV to oral switch. Using this care pathway we have shown, for uncomplicated cellulitis needing IV therapy, that nurse-led management of cellulitis via OPAT is both safe and importantly is associated with reduction in overall IV antibiotic exposure, a key antimicrobial stewardship outcome. Careful patient assessment with appropriately trained staff and care pathways are essential for managing cellulitis through OPAT. This practice is supported by recently published OPAT good practice recommendations, which we commend to clinicians considering developing such a service. Specifically in cellulitis, risk assessment is critical for determining OPAT suitability, and there are patient factors (e.g. diabetes) and disease factors (e.g. bursitis or MRSA infection) that may be associated with longer courses of IV therapy or poorer OPAT outcomes. As well as the UK practice recommendations, US guidelines for appropriate patient selection are a useful resource for clinicians developing OPAT services. References: Antimicrobial stewardship: Start smart - then focus. Department of Health, 16853, 17 Nov 2011. Chapman AL, Seaton RA, Cooper MA, Hedderwick S, Goodall V, Reed C, et al. Good practice recommendations for outpatient parenteral antimicrobial therapy (OPAT) in adults in the UK: a consensus statement. J Antimicrob Chemother 2012;67(5):1053-62. Barr DA, Semple L, Seaton RA. Outpatient parenteral antimicrobial therapy (OPAT) in a teaching hospital-based practice: a retrospective cohort study describing experience and evolution over 10 years. Int J Antimicrob Agents 2012;39(5):407-13. Seaton RA, Bell E, Gourlay Y, Semple L. Nurse-led management of uncomplicated cellulitis in the community: evaluation of a protocol incorporating intravenous ceftriaxone. J Antimicrob Chemother 2005;55(5):764-7. Seaton RA, Sharp E, Bezlyak V, Weir CJ. Factors associated with outcome and duration of therapy in outpatient parenteral antibiotic therapy (OPAT) patients with skin and soft-tissue infections. Int J Antimicrob Agents 2011;38(3):243-8. Tice AD, Rehm SJ, Dalovisio JR, Bradley JS, Martinelli LP, Graham DR, et al. Practice guidelines for outpatient parenteral antimicrobial therapy. IDSA guidelines. Clin Infect Dis 2004;38(12):1651-72. Competing interests: No competing interests 24 August 2012 Christopher J.A. Duncan Specialty Registrar Andrew Seaton Gartnavel General Hospital, Glasgow 1053 Great Western Road, Glasgow G12 0YN ### Re: Diagnosis and management of cellulitis We welcome the review of cellulitis diagnosis and management by Phoenix et al.(1) In addition to the comments made by our infection colleagues we would like to the highlight some of the key issues faced by clinicians who manage this infection regularly and several recent changes in practice that the authors have not mentioned. The number of hospital admissions in England in 2008/9 was 14,152,692, of which 82,113 were for cellulitis. In addition hospital visits for abscesses and cellulitis in the paper quoted increased from 17.3 to 32.5 visits per 1000 population, a ten fold difference to that quoted by Phoenix et al.(1,2) Multiple studies have demonstrated that streptococci rather than staphylococci are the commonest cause, in contrast to the one review quoted in the paper. (3,4,5,6) The table in the review mentions some of the rarer pathogens and associated risk factors. It is important to say that even in the context of these risk factors streptococci and staphylococci are usually the major culprits. Identification of an organism is often problematic as blood cultures are often negative and skin biopsies are confounded by skin colonisation. Empiric treatment should still treat the commonest organisms. Community-acquired (CA)-MRSA has not emerged in the UK and Europe, unlike the US, (7) as a significant pathogen and empiric treatment should remain Flucloxacillin or other similar beta-lactam antibiotics at present in these countries. Of particular concern in the review was the management of necrotising fascitis. This is commonly caused by beta–haemolytic streptococci,usually Group A streptococci; anaerobes and Gram-negative organisms are also important, particularly if the site is below the umbilicus. Diagnosis is clinical and surgical debridement should not be delayed by imaging as the authors suggest. (8,9) Upper limb, torso and facial cellulitis have very different aetiologies to that of the lower limb and are often more serious. The management of these varies considerably and involvement of a specialist should be sought early. The provision of outpatient parenteral antimicrobial therapy (OPAT) for the follow-up of inpatient care has been well established.(10) However, from April 2011 provision of an ambulatory care pathway for cellulitis became one of the clinical quality indicators for all Emergency Departments in acute trusts (11). This expects acute hospital trusts to develop and implement pathways for patients who have failed on oral antibiotics or where the infection is deemed more severe to access intravenous antibiotics and specialist input without being admitted to hospital. This is a key change in approach to managing cellulitis. In our trust the pathway has been running for 16 months. There are clear exclusion criteria relating to co-morbidity, signs of systemic sepsis, and site of cellulitis, those patients are admitted. Patients on the pathway receive one to three days of intravenous therapy with ceftriaxone and are reviewed daily by an Emergency Physician. Once stabilised the patients then complete the treatment with appropriate oral agents. The pathway was developed by the Emergency Department and Infectious Diseases teams. One hundred and thirty one patients were treated on the pathway over one year. The age range was 18-91 years. The median treatment course with once daily ceftriaxone was 2 days (range 1-6 days). Sixteen (12.1%) patients subsequently required hospital admission for a median of 3 days (range 1-7 days). Fifteen (93.4%) of these patients were admitted as they had not responded appropriately to ambulatory care pathway - with progressive infection (12), development of bursitis (2), and the development of abscess (1). Only one other patient was admitted in the subsequent 28 days with a fractured ankle. Based on this data the introduction of a multidisciplinary pathway with access to Infectious Diseases physicians, and other experts can safely maintain many patients on an ambulatory care pathway. We estimate that we have saved the hospital more than £150,000 in in-patient bed stays based on days treated on the ambulatory care pathway. Ambulatory care pathways, with increased input from infection doctors and other specialist services are all set to improve the experience of patients with cellulitis. It is important that clinicians are aware of all the options available for management of this challenging, common condition. 1) Phoenix G, Das S, Joshi M. Diagnosis and management of cellulitis. BMJ 2012;345:e4955. 2.) Hersh AL, Chambers HF, Mseli JH, Gonzales R. National trends in ambulatory visits and antibiotic prescribing for skin and soft tissue infections. Arch Intern Med 2008; 168:1585-91. 3.)Peralta G, Padrón E, Roiz MP, De Benito I, Garrido JC, Talledo F, Rodríguez-Lera MJ, Ansorena L, Sánchez MB.Risk factors for bacteremia in patients with limb cellulitis.Eur J Clin Microbiol Infect Dis. 2006;25(10):619. 4.)Bernard P, Bedane C, Mounier M, Denis F, Catanzano G, Bonnetblanc JM. Streptococcal cause of erysipelas and cellulitis in adults. A microbiologic study using a direct immunofluorescence technique. Arch Dermatol. 1989;125(6):779. 5.)Semel JD, Goldin H. Association of athlete's foot with cellulitis of the lower extremities: diagnostic value of bacterial cultures of ipsilateral interdigital space samples. Clin Infect Dis. 1996;23(5):1162. 6.)Chira S, Miller LG. Staphylococcus aureus is the most common identified cause of cellulitis: A systematic review. Epidemiol Infect 2010; 138:313-7. 7.)Kock R, et al. Methicillin-resistant Staphylococcus aureus (MRSA): burden of disease and control challenges in Europe. Euro Surveill. 2010 Oct 14;15(41): 19688 8.)Brook I, Frazier EH.Clinical and microbiological features of necrotizing fasciitis.J Clin Microbiol. 1995;33(9):2382. 9.)Wong CH, Chang HC, Pasupathy S, Khin LW, Tan JL, Low CO Necrotizing fasciitis: clinical presentation, microbiology, and determinants of mortality.J Bone Joint Surg Am. 2003;85-A(8):1454. 10.)Barr DA, Semple L, Seaton A. Outpatient parenteral antimicrobial prescribing in a teachin hospital- based practice: A retrospective cohort study describing expererience and evolution over 10 years. Int J of Antimicrob Agents 39 (2012) 407-413 11.) Competing interests: No competing interests 19 August 2012 Sarah Logan Infectious Disease specialist registrar Marieke Bokhoven, Tara Sood, Susan Hopkins Royal Free Hospital Pond Stree. London NW3 2QG ### Re: Diagnosis and management of cellulitis We read with interest the article by Pheonix et al on the diagnosis and management of cellulitis.1 As Infectious Diseases physicians, interestingly not mentioned on the list of specialists who treat cellulitis given in the article, we wish to point out some important areas where we disagree with the authors, some of which are errors. 1) Clostridium perfringens classically causes so-called “gas gangrene”, a rapidly progressive infection often associated with vascular compromise due to severe penetrating trauma or crush injuries.2 Necrotising fasciitis is often polymicrobial, with common causes including group A streptococci, Staphylococcus aureus and anaerobes. Rare causes, often in the immunocompromised, include Aeromonas hydrophilia and Vibrio vulnificus.2 2) Clinically distinguishing between streptococcal and staphylococcal cellulitis is often impossible. Recommending different treatment based on this distinction is not helpful in clinical practice. All cases of cellulitis should be treated with an antibiotic with anti-staphylococcal activity, thus amoxicillin would be inappropriate therapy. This is supported by both CREST and IDSA guidelines,2,3 in contrast to the authors’ incorrect statement that amoxicillin is recommended empirical treatment in the CREST guidance. 3) Ciprofloxacin appears to have been misspelled "ciprofloxacillin". 4) Confirmed erysipeloid (caused by Erysipelothrix rhusiopathiae) should be treated with a penicillin, such as amoxicillin, as opposed to ciprofloxacin.2 5) Cross-sectional imaging is not essential in the diagnosis of necrotising fasciitis, which is mainly clinical. Delays in obtaining imaging may lead to significant deterioration and delay in definitive surgical treatment.2,3 References 1] Phoenix G, Das S, Joshi M. Diagnosis and management of cellulitis. BMJ 2012;345:e4955. 2] Stevens DL, Bisno AL, Chambers HF, Everett ED, Dellinger P, Goldstein EJC, et al. Practice Guidelines for the Diagnosis and Management of Skin and Soft-Tissue Infections. Clin Infect Dis. 2005; 41:1373–406 3] DHSS Northern Ireland. CREST (Clinical Resource Efficiency Support Team) Guidelines on the Management of Cellulitis In Adults. 2005; 1–31. Competing interests: No competing interests 16 August 2012 Nikolas Rae ST3 infectious diseases Charis Marwick consultant infectious diseases Ninewells Hospital Dundee DD1 9SY ### Re: Diagnosis and management of cellulitis Unfortunately there is a further microbiological error in this piece. The authors quote the CREST guidelines with regard to antibiotic treatment. The quotation is wrong as the CREST guidelines do not mention amoxicillin at all. Amoxicillin is ususally inactive against Staphyloccus aureus. The CREST guidance in fact suggests flucloxacillin - an agent with good activity against both beta-haemolytic streptococci and S. aureus, the two commonest pathogens causing cellulitis. Competing interests: No competing interests 16 August 2012 Andrew Swann Consultant Microbiologist University Hospitals of Leicester Infirmary Square, Leicester, LE1 5WW ### Re: Diagnosis and management of cellulitis Dear Sir/Madame, We would like to congratulate Phoenix et al., on their paper. However with regards to the micro-organisms causing necrotising fasciitis the only causative agent mentioned is C. Perfringes. However the list of potentially causative bacteria is far greater and C. Perfringes is not the sole or even leading cause. It is therefore misleading and we would suggest that that the more common cause, synergistic infection involving streptroccus to be highlighted. Below is a more comprehensive classification of necrotising fasciitis: Class I Polymicrobial Class II Monomicrobial of which group A streptocci are the most common Class III Gram Negative Bacteria including C. Perfringes Class IV Fungal This broad array of potentially causative micro-organisms has implications upon the choice of first line antibiotics for the clinician. Barker FG, Leppard BJ, Seal DV Streptococcal necrotising fasciitis: comparison between histological and clinical features. J Clin Pathol 1987;40:335-341 doi:10.1136/jcp.40.3.335 Kaul R, McGeer A, Low DE, Green K, Schwartz B, Simor AE. Population-Based Surveillance for Group A Streptococcal Necrotizing Fasciitis: Clinical Features, Prognostic Indicators, and Microbiologic Analysis of Seventy-Seven Cases. American Journal of Medicine. Volume 103, Issue 1 , Pages 18-24, July 1997 Competing interests: No competing interests 15 August 2012 James Kelly Plastic Surgery Mr Baljit Dheansa, Consultant Plastic and Burns Surgeon Queen Victoria Hospital Holtye Road East Grinstead, West Sussex RH19 3DZ Pages 1 2 next › last » Back to top Cookies and privacy We and our 216 partners store and access personal data, like browsing data or unique identifiers, on your device. Selecting I Accept enables tracking technologies to support the purposes shown under we and our partners process data to provide. Selecting Reject All or withdrawing your consent will disable them. If trackers are disabled, some content and ads you see may not be as relevant to you. You can resurface this menu to change your choices or withdraw consent at any time by clicking the Cookie settings link on the bottom of the webpage . Your choices will have effect within our Website. 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16313	https://en.khanacademy.org/science/up-class-12-chemistry/xf544292e349f3984:electrochemistry/xf544292e349f3984:nernst-equation/v/worked-example-calculating-ecell-for-the-given-cell	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Consent Leg.Interest label label label
16314	https://flexbooks.ck12.org/cbook/ck-12-interactive-middle-school-math-7-for-ccss/section/6.2/primary/lesson/area-and-circumference-of-circles-from-radius-and-diameter-msm7-ccss/	Skip to content Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? Science Grade K to 5 Earth Science Life Science Physical Science Biology Chemistry Physics Advanced Biology FlexLets Math FlexLets Science FlexLets English Writing Spelling Social Studies Economics Geography Government History World History Philosophy Sociology More Astronomy Engineering Health Photography Technology College College Algebra College Precalculus Linear Algebra College Human Biology The Universe Adult Education Basic Education High School Diploma High School Equivalency Career Technical Ed English as 2nd Language Country Bhutan Brasil Chile Georgia India Translations Spanish Korean Deutsch Chinese Greek Polski EXPLORE Flexi A FREE Digital Tutor for Every Student FlexBooks 2.0 Customizable, digital textbooks in a new, interactive platform FlexBooks Customizable, digital textbooks Schools FlexBooks from schools and districts near you Study Guides Quick review with key information for each concept Adaptive Practice Building knowledge at each student’s skill level Simulations Interactive Physics & Chemistry Simulations PLIX Play. Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign Up 6.2 Area and Circumference of Circles from Radius and Diameter Written by:Lori Jordan Fact-checked by:The CK-12 Editorial Team Last Modified: Aug 01, 2025 Lesson In This Lesson You will learn how to find the area and circumference of circles using radius and diameter. Use the interactive to help you manipulate the circles to find the equations for the area and circumference of circles. Parts of a Circle A circle is a 2-dimensional figure such that all of the points are the same distance (the radius) from a fixed point (the center). The radius is a line segment where one endpoint is the center of the circle and the other is on the circle. The diameter of a circle is a line segment with both endpoints on the circle and also passes through the center. The length of the diameter is the same as two radii (plural for radius). In mathematical terms, this would be written The distance around a circle is called the circumference. Place the vocabulary terms in the correct spot in the picture below. Discussion Question What tool do you think would be best for measuring the circumference of a circle? What challenges would you have measuring the circumference with a circle? Let's Go Bowling! Ernesto and Tyler are heading to the bowling alley. They both want to figure out how many revolutions the bowling ball takes until it gets to the end of the lane and hits the pins. The faster they can do this, the harder the ball will hit the pins and, hopefully, the better their game will get. The typical bowling ball has a diameter of 8.5 inches and a lane is 60 feet long. How many revolutions does it take the ball to travel down the lane? In the interactive, the bowling ball is at the beginning of the lane. Drag the red dot on the ball to show its path down the lane. Every time the ball has one revolution, a line will appear to mark it, to help you determine the number of revolutions. In this interactive, you defined and found the formula for circumference. It will be helpful to memorize this formula in both forms. In addition, you discovered that one revolution of a circle (or bowling ball) is the same its circumference. Let's switch to finding the area of a circle. How could you use the circumference of a circle to find its area? Orange-You Interested? Your little brother loves playing with his food. Your mom wants him to learn while playing and she decides to give you both a slice from an orange. She asks, "Can you figure out how much orange you are eating?" You, being older and wiser than your little brother, decide to measure the circumference of the slice. He rips the orange rind and unfolds all the segments so that they all look like little triangles in a row. How could you use what your brother did and the circumference to find the area of the circular slice of orange? In the interactive, you will start with the slice. Click on the blue button at the top of the orange to "unfold" or rip the slice open. Then, take half of the orange segments and rotate them so that they fit into the other segments to make a sort of parallelogram. Use this to determine the formula for the area of a circle. Hint: The area of a parallelogram is the same as that of a rectangle; base× height. Discussion Question You just derived the formula for the area of a circle using orange segments! The area of any circle is Discuss what all the parts of the formula mean. Summary A circle is a set of points that are the same distance (the radius) from a fixed point (the center). The diameter is a line segment that passes through the center and endpoints are on the circle. is defined as and is commonly approximated as 3.14. The formula for circumference is or The formula for the area is Asked by Students Here are the top questions that students are asking Flexi for this concept: Overview A circle is a set of points that are the same distance (the radius) from a fixed point (the center). The diameter is a line segment that passes through the center and endpoints are on the circle. is defined as and is commonly approximated as 3.14. The formula for circumference is @$C=2\pi r@$ or @$C=\pi d@$ The formula for the area is @$\text{A}=\pi r^2@$ Vocabulary Area Circumference radius diameter circle distance Point center line segment endpoint line measure parallelogram rectangle Height Test Your Knowledge Question 1 The diameter in the following figure is _____. a HF b OF c HE d OH The diameter is a measure of the distance across the center of a circle. The diameter is equal to twice the length of the radius. In the given image, @$\begin{align}O\end{align}@$ is the center of the circle and @$\begin{align}HF\end{align}@$ is the diameter of the circle. Question 2 Which term describes a line segment that passes through the center of, and has endpoints on, a circle? a radius b chord c diameter d circumference A diameter of a circle is any straight line segment that passes through the center of the circle and whose endpoints lie on the circle. Asked by Students Here are the top questions that students are asking Flexi for this concept: Related Content Area of a Circle Area of a Circle - Overview Area of a Circle - Example 1 Ringside Seats Examing Circles: Area of a Circle Area of a Circle: Hula Hoop Radius Area of a Circle: Campsite \| Image \| Reference \| Attributions \| --- \| \| \| Credit: Steve Snodgrass Source: \| \| \| \| Credit: Laura Guerin Source: CK-12 Foundation License: CC BY-NC \| \| \| \| Credit: Brady Holt Source: License: CC BY-NC \| \| \| \| Credit: woodleywonderworks Source: \| \| \| \| Credit: Melissa Sanders \| \| \| \| License: CC BY-NC-SA \| \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| Student Sign Up Are you a teacher? Having issues? Click here By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? 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16315	https://www.reddit.com/r/learnmath/comments/td3nrj/logistic_differential_equation/	logistic differential equation : r/learnmath Skip to main contentlogistic differential equation : r/learnmath Open menu Open navigationGo to Reddit Home r/learnmath A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to learnmath r/learnmath r/learnmath Post all of your math-learning resources here. Questions, no matter how basic, will be answered (to the best ability of the online subscribers). 403K Members Online •4 yr. ago ilovehotmoms55 logistic differential equation Hi, I'm trying to model population growth based on the logistic growth model. I have a set of data and I can't seem to find a method of calculating the carrying capacity that would result me in a positive value (since that's what I assume carrying capacity must be, as my population is growing). Is there any way to calculate it? it doesn't have to be super reliable, I'm just a high school student and i don't know what i'm doing :( Read more Share Related Answers Section Related Answers Effective strategies for mastering algebra Tips for improving mental math skills Exploring real-world uses of number theory Comparing different methods of integration Best practices for preparing for math exams New to Reddit? Create your account and connect with a world of communities. Continue with Google Continue with Google. Opens in new tab Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community 0 0 Top Posts Reddit reReddit: Top posts of March 13, 2022 Reddit reReddit: Top posts of March 2022 Reddit reReddit: Top posts of 2022 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
16316	https://www.nature.com/articles/s41598-025-89008-x	IVC treatment between primary and second TURBT may improve the prognosis of high-risk NMIBC patients receiving BCG treatment \| Scientific Reports Your privacy, your choice We use essential cookies to make sure the site can function. We also use optional cookies for advertising, personalisation of content, usage analysis, and social media. By accepting optional cookies, you consent to the processing of your personal data - including transfers to third parties. Some third parties are outside of the European Economic Area, with varying standards of data protection. See our privacy policy for more information on the use of your personal data. Manage preferences for further information and to change your choices. Accept all cookies Skip to main content Thank you for visiting nature.com. You are using a browser version with limited support for CSS. 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Advertisement View all journals Search Search Search articles by subject, keyword or author Show results from Search Advanced search Quick links Explore articles by subject Find a job Guide to authors Editorial policies Log in Explore content Explore content Research articles News & Comment Collections Subjects Follow us on Facebook Follow us on Twitter Sign up for alerts RSS feed About the journal About the journal About Scientific Reports Contact Journal policies Guide to referees Calls for Papers Editor's Choice Journal highlights Open Access Fees and Funding Publish with us Publish with us For authors Language editing services Open access funding Submit manuscript Sign up for alerts RSS feed nature scientific reports articles article IVC treatment between primary and second TURBT may improve the prognosis of high-risk NMIBC patients receiving BCG treatment Download PDF Download PDF Article Open access Published: 10 February 2025 IVC treatment between primary and second TURBT may improve the prognosis of high-risk NMIBC patients receiving BCG treatment Zhen Li1,2na1, Zewei Wang3na1, Yang Liu4na1, Lei Yang1, Ling Gu3, Hailong Li3& … Wei Wang1 Show authors Scientific Reportsvolume 15, Article number:4874 (2025) Cite this article 2541 Accesses Metrics details A Publisher Correction to this article was published on 23 April 2025 This article has been updated Abstract To explore whether intravesical chemotherapy (IVC) between primary and second transurethral resection of bladder tumor (TURBT) affects the prognosis of non-muscle invasive bladder cancer (NMIBC) patients receiving Bacillus Calmette-Guérin (BCG) treatment. NMIBC patients who underwent a second TURBT and subsequent BCG treatment between 2012 and 2023 at the Affiliated Hospital of Xuzhou Medical University were retrospectively analyzed. These patients were divided into Group A, which received IVC between TURBT, and Group B, which did not. Recurrence-free survival (RFS) was compared among the different risk subgroups. A total of 292 NMIBC patients were included in this study. In the entire cohort, IVC treatment between the primary and second TURBT was associated with longer RFS (P = 0.009). When stratified by risk groups, in intermediate-risk patients, the difference in RFS between the groups was not statistically significant (P = 0.434). By contrast, for high-risk patients, the treated group exhibited a better prognosis compared to the non-treated group (85.6% vs. 77.6%, P = 0.007). In both univariate and multivariate COX regression analyses, after adjusting for clinical factors such as tumor stage and tumor grade, the IVC between the primary and second TURBT remained an independent prognostic factor for NMIBC patients (HR 0.571, 95% CI [0.380, 0.859], p = 0.007). IVC treatment administered between the primary and second TURBT has been demonstrated to enhance RFS of high-risk NMIBC patients undergoing BCG treatment, whereas it is not applicable to intermediate-risk patients. Similar content being viewed by others Impact of upper tract urothelial carcinoma history on patients with non-muscle invasive bladder cancer undergoing intravesical chemotherapy Article Open access 18 February 2025 Assessment of prognostic implication of a panel of oncogenes in bladder cancer and identification of a 3-gene signature associated with recurrence and progression risk in non-muscle-invasive bladder cancer Article Open access 06 October 2020 Prediction of non-muscle invasive bladder cancer recurrence using machine learning of quantitative nuclear features Article Open access 29 October 2021 Introduction Bladder cancer is the tenth most malignant tumor globally and can be classified into muscle-invasive and non-muscle-invasive bladder cancer (NMIBC) based on TNM staging1,2. NMIBC includes the pTa, pT1, and pTis stages, with tumors confined to the mucosa or submucosa accounting for approximately 75% of all bladder cancer. Compared with muscle-invasive bladder cancer, NMIBC has a better prognosis, and the treatment regimen is different. Transurethral resection of a bladder tumor (TURBT) is the standard treatment for NMIBC, frequently performed in conjunction with postoperative intravesical chemotherapy (IVC) or intravesical Bacillus Calmette-Guérin (BCG) immunotherapy3. The benefits of maximal TURBT for NMIBC are well documented, and surgical removal of all visible tumors is the cornerstone of NMIBC treatment3,4,5. However, it has been reported that approximately 17–67% of patients with stage Ta and 20–71% of patients with stage T1 still have residual tumors after the initial TURBT, which are mostly located at the initial resection site6. These residual tumors worsen patient prognosis, further emphasizing the importance of TURBT7. Current guidelines for the treatment of NMIBC recommend a second TURBT at 2–6 weeks after the initial TURBT in patients with stage T1 and high-grade Ta cancer3,8,9. The objectives of re-TURBT are to guarantee complete removal of the initial tumor, reduce the patient’s tumor burden, and enhance the quality of the pathological specimen10. Numerous studies have shown that a second TURBT improves the prognosis of NMIBC patients7,11,12,13. Patients who meet the criteria for a second TURBT usually also fulfil the criteria for BCG medication3. Therefore, theoretically, patients with NMIBC who undergo a second TURBT should also receive postoperative BCG immunotherapy. IVC directly kills intraoperatively disseminated tumor cells and residual tumor cells in trauma, and an immediate single instillation within 24 h after TURBT has been shown to reduce the rate of disease recurrence in patients with NMIBC14,15,16. However, there is no consensus as to whether continuing IVC therapy between the first and second TURBT improves the prognosis of patients with NMIBC treated with BCG. Therefore, this study aimed to investigate whether IVC applied between the first and second TURBT improves the prognosis of patients with NMIBC treated with BCG. Materials and methods Study population Patients with NMIBC who underwent a second TURBT and subsequent BCG treatment between January 2012 and December 2023 at the Affiliated Hospital of Xuzhou Medical University were retrospectively analyzed. All methods in this study were carried out in strict accordance with the relevant guidelines and regulations3. The inclusion criteria were as follows: (1) pathological diagnosis of uroepithelial carcinoma of the bladder after TURBT, with pathological stage pTa, pTis, or pT1; (2) a second TURBT at our hospital 2–6 weeks after the first TURBT; (3) IVC treatment within 24 h after the first and second TURBT; and (4) intravesical BCG immunotherapy after the second TURBT. The exclusion criteria were as follows: (1) patients with incomplete or missing information, (2) patients whose first or second TURBT was not performed at our hospital, (3) patients with low-risk and very high-risk tumors, and (4) patients with other potential comorbidities. The patients were divided into groups A and B according to whether they received IVC treatment between the first and second TURBT, with group A being the group that received IVC treatment and group B being the group that did not receive IVC treatment. The endpoint was recurrence-free survival (RFS), defined as the time interval between the second TURBT and disease recurrence. Treatment schedule During the second TURBT, all visible tumors and scars were removed. An immediate single instillation was performed within 24 h for both the initial and second TURBT. The main chemotherapeutic agents used for intravesical chemotherapy after TURBT in our hospital are pirarubicin and epirubicin (pirarubicin or epirubicin, 40 mg dissolved in 40 mL of sterile injectable water, was injected into the empty bladder through a sterile catheter and retained for 0.5 2 h). Between the initial and second TURBT, the treatment group received weekly IVC treatment after an immediate single instillation, and the non-treatment group received no additional IVC treatment after an immediate single instillation. The BCG regimen was started at least 2 weeks after the second TURBT, once a week for the first eight sessions and once a month for the next ten sessions. After the second resection, follow-ups were performed every three months for two years; every six months for the following three years; and annually thereafter. Follow-up included urine cytology, radiological imaging, and cystoscopy. Clinical and pathological information The collected clinicopathological data included age, gender, body mass index (BMI), smoking history, hematuria history, tumor stage, tumor grade, tumor number, maximum tumor diameter, surgical intervals, residual tumor, and IVC treatment regimen between the first and second TURBT. Pathological tumor staging was performed according to the International Union against Cancer TNM staging system (8th edition, 2017), whereas tumor grading was performed according to the 2004 World Health Organization bladder cancer grading system. The highest pathological stages and grades after the initial and second TURBT were included in this study. All pathological specimens, especially those obtained before 2017, were reassessed by a specialized bladder cancer pathologist. Statistical analysis Categorical variables of the demographic and clinicopathological characteristics of the two groups are expressed as numbers and percentages and compared using the χ 2 test. For continuous variables expressed as means and standard deviations, the t-test was used for comparisons. Kaplan–Meier survival curves for patients were plotted and compared using log-rank tests. Univariate and multivariate Cox regression analyses were conducted to ascertain independent risk factors associated with patient prognosis. The results were expressed as 95% confidence intervals (CI) and hazard ratios (HR). All statistical analyses were performed using SPSS (version 26.0; IBM Corporation, Armonk, NY, USA) and R 4.2.3 ( All p-values were two-sided, and p< 0.05 was considered statistically significant. Results Demographic and clinicopathological characteristics Altogether, 292 patients were included with a median follow-up of 91 months (Fig.1), all the patients were treated with adequate BCG therapy. In all, 164 patients who received IVC treatment between the initial and second TURBT were included in Group (A) A total of 128 patients who did not receive IVC treatment between the initial and second TURBT were included in Group (B) There were no statistically significant differences in the variables between the groups (Table1). According to the EAU NMIBC prognostic factor risk groups, 64 patients (39%) were at intermediate risk, and 100 patients (61%) were at high risk in group A. In group B, 52 patients (40.6%) were at intermediate risk and 76 patients (59.4%) were at high risk. Fig. 1 Patient flow chart. Full size image Table 1 Demographic and clinicopathological characteristics. Full size table Kaplan–Meier survival analysis At the end of the study, 96 (32.8%) experienced tumor recurrence: 49 (29.9%) patients in group A and 47 (36.7%) patients in group B. Among all patients, there was a statistically significant difference in RFS after the second TURBT between Group A and Group B (p = 0.009) (Fig.2). The 5-year RFS rates were 87.2 and 81.8% in Groups A, and B, respectively. In intermediate-risk NMIBC patients, the difference in postoperative RFS between the two groups was not significant (p = 0.434) (Fig.3). The 5-year RFS rates in Groups A and B were 89.6 and 88.0%, respectively. In high-risk NMIBC patients, the 5-year postoperative RFS was higher in the treatment group than in the non-treatment group (85.6 vs. 77.6%, p = 0.007) (Fig.4). Fig. 2 Recurrence-free survival (RFS) curve for both groups in all NMIBC patients. Full size image Fig. 3 Recurrence-free survival curve (RFS) for both groups in intermediate-risk NMIBC patients. Full size image Fig. 4 Recurrence-free survival (RFS) curve for both groups in high-risk NMIBC patients. Full size image Univariate and multivariate Cox regression analyses for RFS In the univariate analysis, pathology T stage (p = 0.002), pathology grade (p = 0.041), IVC between the primary and second TURBT (p = 0.010), and residual tumor (p = 0.017) were significantly associated with patients’ postoperative RFS. After adjusting for the above confounders, multivariate analysis showed that pathology T stage (HR 2.445, 95% CI [1.117,5.353], p = 0.025) and receiving IVC between the primary and second TURBT (HR 0.571, 95% CI [0.380,0.859], p = 0.007) remained as independent risk factors for postoperative RFS in NMIBC patients (Table2). Table 2 Univariate and multivariate COX regression analyses for RFS. Full size table Discussion In this study, we analyzed the prognostic impact of IVC transplantation between primary and second TURBT in patients with NMIBC treated with BCG. Furthermore, we conducted an in-depth analysis of its prognostic value in the different risk subgroups. Our findings indicate that IVC treatment between the primary and second TURBT is an effective intervention for improving the prognosis of high-risk patients with NMIBC treated with BCG. However, for patients with intermediate-risk NMIBC, IVC treatment before the second TURBT did not reduce the tumor recurrence rate, and unnecessary instillation treatments could be avoided in these patients. TURBT is regarded as a low-risk procedure, and complete resection of all visible tumors is the cornerstone of NMIBC treatment and has been endorsed by various clinical treatment guidelines3,8,9,10. The benefits of maximal TURBT for NMIBC are well documented17. The rate of residual tumors in the first TURBT ranges from 4 to 78%, depending on the tumor stage, tumor number, and operator technique18,19. A second TURBT is effective in improving the prognosis of NMIBC and is usually performed 2–6 weeks after the initial TURBT10. A systematic review by Cumberbatch et al. included one randomized controlled trial and 30 non-randomized controlled trials, which provided an in-depth analysis of the efficacy of a second TURBT for NMIBC6. The study data showed that the disease recurrence rate for patients with stage Ta tumors was 16% for those who underwent a second TURBT and 58% for those who did not. For patients with stage T1 tumors, the recurrence rate was 45% for those who underwent a second TURBT and 49% for those who did not. These results further confirm the effectiveness of a second TURBT in reducing disease recurrence, especially in patients with stage Ta tumors. IVC can directly kill tumor cells, and immediate single instillation after TURBT can significantly reduce the recurrence rate of NMIBC14,15,16,20. Except for patients with contraindications to immediate single instillation after surgery, all patients with NMIBC should be treated with immediate intravesical chemotherapy within 24 h after TURBT3. However, there is no consensus on the need for further IVC treatment after primary TURBT and secondary TURBT. To the best of our knowledge, this is the first medical center to conduct this study. The efficacy of BCG immunotherapy in patients with NMIBC is remarkable, particularly in intermediate- and high-risk patients21,22,23. A recent systematic review of real-world evidence from studies on patients with high-risk NMIBC revealed significant variations in survival, with five-year RFS ranging from 17 to 89% and progression-free survival (PFS) ranging from 58–89%24. Further findings suggest that patients who receive intravesical BCG therapy have longer survival times25,26. Specifically, for patients with intermediate-risk NMIBC, the five-year survival results were particularly promising, with RFS and PFS rates of 86 and 100%, respectively. For high-risk patients, these figures were 72 and 91%, respectively27. Similar to the above findings, in our study, the five-year RFS rates of patients with intermediate- and high-risk NMIBC were 88.8 and 82.2%, respectively. Of particular note, we found that for high-risk patients, IVC treatment between the primary and second TURBT significantly improved five-year RFS, from 77.6 to 85.6% (p = 0.007). The results of this study can be analyzed at two levels. First, we speculate that the presence of post-TURBT trauma may enhance the killing effect of the IVC on residual tumors. However, over time, epithelial tissue formation and scarring can weaken the tumor-killing effect of the drug. The efficacy of an immediate single instillation after TURBT for NMIBC is widely accepted and used in all patients with NMIBC3. A systematic review conducted by Tabayoyong et al., which comprehensively analyzed 16 randomized controlled trials, found no significant benefit in terms of recurrence, progression, or survival with maintenance intravesical chemotherapy compared with induction chemotherapy alone, further emphasizing the importance of early IVC treatment28. In addition, owing to the different tumor burdens of intermediate- and high-risk patients, the intensity of drug therapy they require postoperatively also differs. These findings suggest that tumor recurrence in patients with NMIBC tends to occur at the site of primary TURBT6, which may be because of incomplete surgical resection, resulting in tumor cells remaining in the bladder and regrowth. It is also possible that surgical trauma creates conditions for tumor implantation, thereby increasing the risk of recurrence. In high-risk patients with NMIBC, multiple resections of the tumor in a single operation are often required to achieve complete tumor removal, owing to deeper tumor growth or larger and more numerous tumors. This may have led to the spread of tumor cells within the bladder. Furthermore, larger trauma increases the risk of tumor cell replantation within the bladder, making it difficult to completely eradicate all tumor cells with a single IVC treatment. Therefore, increasing the number of IVC treatments before wound healing may help eliminate residual cancer cells more effectively. By contrast, intermediate-risk patients have a lower tumor burden, and surgery usually results in more complete and less invasive removal of the tumor. An immediate single postoperative instillation and subsequent BCG therapy are effective. Additional IVC therapy may not provide significant benefits but may result in undesirable side effects. Currently, some of the discussions on this aspect are partly based on the observed trends in clinical practice and the limited clues from relevant previous studies. In the future, large-scale targeted clinical trials will still be needed to conduct a comprehensive verification. This study is the first to investigate the potential benefits of IVC applied in primary and secondary TURBT for patients with NMIBC. During the patient selection phase, we excluded patients with NMIBC who did not receive BCG therapy after the second TURBT. This was because we believe that, in addition to incomplete resection or pathology specimens that do not show muscle layers, patients with NMIBC who undergo a second TURBT should theoretically also receive postoperative BCG. Furthermore, considering the impact of factors such as tumor stage, grade, and size on prognosis, we conducted a subgroup analysis of NMIBC patients according to the EAU NMIBC prognostic factor risk groups to make the experimental results more reasonable. Two treatment options were provided to patients with NMIBC who were expected to undergo a second TURBT and subsequent BCG treatment, thereby facilitating the management of the patients’ disease for surgeons. However, this study has some limitations. First, it should be noted that this was a retrospective study, and that the non-randomized design may allow for some potential bias on the part of surgeons in the way they treat their patients. Second, as this was a single-center study, the sample size was relatively small, which may have some bias in our final results. In the future, it would be beneficial for prospective studies with larger samples to validate our findings. Finally, because the second TURBT was performed within a certain timeframe, the number of IVC treatments varied. Furthermore, chemotherapeutic agents used for IVC have not been standardized. These factors may have affected the experimental results. Future research initiatives will focus on IVC treatment for NMIBC. We will execute a multicenter, large-sample prospective study, collaborating with leading professional medical centers to pool resources and enlarge the sample size, thereby enhancing the robustness of the results. Simultaneously, we will establish a standardized diagnostic and treatment protocol. This involves precisely defining the parameters of IVC chemotherapy drugs and detailing the TURBT operational procedures. We aim to mitigate the impact of variability, standardize treatment application, enhance patient prognoses, and furnish a solid foundation for this field of study. Conclusions IVC treatment between the primary and second TURBT reduces the rate of tumor recurrence in high-risk NMIBC patients treated with BCG. However, it is not applicable for intermediate-risk NMIBC, and unnecessary IVC treatment can be avoided in these patients. Data availability The datasets used and/or analysed during the current study are available from the corresponding author on reasonable request. Change history 23 April 2025 A Correction to this paper has been published: References Jubber, I. et al. Epidemiology of bladder Cancer in 2023: A systematic review of risk factors. Eur. Urol.84(2), 176–190 (2023). ArticlePubMedGoogle Scholar Bray, F. et al. Global cancer statistics 2018: GLOBOCAN estimates of incidence and mortality worldwide for 36 cancers in 185 countries. CA-Cancer J. Clin.68(6), 394–424 (2018). Babjuk, M. et al. European Association of Urology Guidelines on non-muscle-invasive bladder Cancer (Ta, T1, and carcinoma in situ). Eur. Urol.81(1), 75–94 (2022). Teoh, J. Y. et al. An international collaborative consensus statement on en bloc resection of bladder tumorincorporating two systematic reviews, a two-round delphi survey, and a consensus meeting. Eur. Urol.78(4), 546–569 (2020). ArticleMathSciNetPubMedGoogle Scholar Richterstetter, M. et al. The value of extended transurethral resection of bladder tumor(TURBT) in the treatment of bladder cancer. BJU Int.110(2 Pt 2), E76–E79 (2012). PubMedGoogle Scholar Cumberbatch, M. et al. Repeat transurethral resection in non-muscle-invasive bladder cancer: A systematic review. Eur. Urol.73(6), 925–933 (2018). ArticlePubMedGoogle Scholar Hashine, K. et al. Results of second transurethral resection for high-grade T1 bladder cancer. Urol. Ann.8(1), 10–15 (2016). ArticleCASPubMedPubMed CentralGoogle Scholar Flaig, T. W. et al. NCCN guidelines(R) insights: Bladder cancer, version 2.2022. J. Natl. Compr. Cancer Netw.20(8), 866–878 (2022). ArticleGoogle Scholar Chang, S. S. et al. Diagnosis and treatment of non-muscle invasive bladder cancer: AUA/SUO Guideline. J. Urol.196(4), 1021–1029 (2016). ArticlePubMedGoogle Scholar Jin, Y. H. et al. Treatment and surveillance for non-muscle-invasive bladder cancer: A clinical practice guideline (2021 edition). Mil.Med. Res.9(1), 44 (2022). PubMedPubMed CentralGoogle Scholar Eroglu, A., Ekin, R. G., Koc, G. & Divrik, R. T. The prognostic value of routine second transurethral resection in patients with newly diagnosed stage pT1 non-muscle-invasive bladder cancer: Results from randomized 10-year extension trial. Int. J. Clin. Oncol.25(4), 698–704 (2020). ArticlePubMedGoogle Scholar Gordon, P. C., Thomas, F., Noon, A. P., Rosario, D. J. & Catto, J. Long-term outcomes from re-resection for high-risk non-muscle-invasive bladder cancer: A potential to rationalize use. Eur. Urol. Focus5(4), 650–657 (2019). Lee, K., Jeong, S. H., Yoo, S. H. & Ku, J. H. Evaluating the efficacy of secondary transurethral resection of the bladder for high-grade Ta tumors. Investig. Clin. Urol.63(1), 14–20 (2022). ArticlePubMedGoogle Scholar Bosschieter, J. et al. Value of an immediate intravesical instillation of Mitomycin C in patients with non-muscle-invasive bladder cancer: A prospective multicentre randomised study in 2243 patients. Eur. Urol.73(2), 226–232 (2018). ArticleCASPubMedGoogle Scholar Messing, E. M. et al. Effect of Intravesical Instillation of Gemcitabine vs saline immediately following resection of suspected low-Grade non-muscle-invasive bladder cancer on tumor recurrence: SWOG S0337 randomized clinical trial. JAMA-J. Am. Med. Assoc.319(18), 1880–1888 (2018). Sylvester, R. J. et al. Systematic review and individual patient data meta-analysis of randomized trials comparing a single immediate instillation of chemotherapy after transurethral resection with transurethral resection alone in patients with stage pTa-pT1 urothelial carcinoma of the bladder: Which patients benefit from the instillation?. Eur. Urol.69(2), 231–244 (2016). ArticlePubMedGoogle Scholar Bree, K. K. et al. Repeat transurethral resection of muscle-invasive bladder cancer prior to radical cystectomy is prognostic but not therapeutic. J. Urol.209(1), 140–149 (2023). ArticlePubMedGoogle Scholar Brauers, A., Buettner, R. & Jakse, G. Second resection and prognosis of primary high risk superficial bladder cancer: Is cystectomy often too early?. J. Urol.165(3), 808–810 (2001). ArticleCASPubMedGoogle Scholar Brausi, M. et al. Variability in the recurrence rate at first follow-up cystoscopy after TUR in stage Ta T1 transitional cell carcinoma of the bladder: A combined analysis of seven EORTC studies. Eur. Urol.41(5), 523–531 (2002). ArticlePubMedGoogle Scholar Sylvester, R. J., Oosterlinck, W. & van der Meijden, A. P. A single immediate postoperative instillation of chemotherapy decreases the risk of recurrence in patients with stage Ta T1 bladder cancer: A meta-analysis of published results of randomized clinical trials. J. Urol.171(6 Pt 1), 2186–2190 (2004). ArticlePubMedGoogle Scholar Han, J., Gu, X., Li, Y. & Wu, Q. Mechanisms of BCG in the treatment of bladder cancer-current understanding and the prospect. Biomed. Pharmacother.129, 110393 (2020). ArticleCASPubMedGoogle Scholar Marttila, T. et al. Intravesical Bacillus calmette-guerin versus combination of epirubicin and interferon-alpha2a in reducing recurrence of non-muscle-invasive bladder carcinoma: FinnBladder-6 study. Eur. Urol.70(2), 341–347 (2016). Shang, P. F. et al. Intravesical bacillus calmette-guerin versus epirubicin for Ta and T1 bladder cancer. Cochrane Datab. Syst. Rev.11(5), CD6885 (2011). Google Scholar Musat, M. G. et al. Treatment outcomes of high-risk non-muscle invasive bladder cancer (HR-NMIBC) in real-world evidence (RWE) studies: Systematic literature review (SLR). ClinicoEcon. Outcome Res.14, 35–48 (2022). Matulay, J. T. et al. Contemporary outcomes of patients with nonmuscle-invasive bladder cancer treated with bacillus calmette-guerin: Implications for clinical trial design. J. Urol.205(6), 1612–1621 (2021). ArticlePubMedGoogle Scholar Kamat, A. M. et al. Definitions, end points, and clinical trial designs for non-muscle-invasive bladder cancer: Recommendations from the international bladder cancer group. J. Clin. Oncol.34(16), 1935–1944 (2016). Grabe-Heyne, K. et al. Intermediate and high-risk non-muscle-invasive bladder cancer: An overview of epidemiology, burden, and unmet needs. Front. Oncol.13,1170124 (2023). Tabayoyong, W. B. et al. Systematic review on the utilization of maintenance intravesical chemotherapy in the management of non-muscle-invasive bladder cancer. Eur. Urol. Focus4(4), 512–521 (2018). ArticlePubMedPubMed CentralGoogle Scholar Download references Acknowledgements Everybody who made a contribution to the research has been listed as an author. Funding No. Author information Author notes 1. Zhen Li, Zewei Wang and Yang Liu contributed equally to this work. Authors and Affiliations Department of Urology, Beijing Chaoyang Hospital Affiliated Capital Medical University, 8 Gong Ti Nan Road, Chaoyang District, Beijing, 100020, People’s Republic of China Zhen Li,Lei Yang&Wei Wang Institute of Urology, Capital Medical University, 8 Gong Ti Nan Road, Chaoyang District, Beijing, 100020, People’s Republic of China Zhen Li Department of Urology, the Affiliated Hospital of Xuzhou Medical University, Xuzhou, 221000, People’s Republic of China Zewei Wang,Ling Gu&Hailong Li Department of Urology, The First Hospital of Qiqihar City, Qiqihar, 221000, People’s Republic of China Yang Liu Authors 1. Zhen LiView author publications Search author on:PubMedGoogle Scholar 2. Zewei WangView author publications Search author on:PubMedGoogle Scholar 3. Yang LiuView author publications Search author on:PubMedGoogle Scholar 4. Lei YangView author publications Search author on:PubMedGoogle Scholar 5. Ling GuView author publications Search author on:PubMedGoogle Scholar 6. Hailong LiView author publications Search author on:PubMedGoogle Scholar 7. Wei WangView author publications Search author on:PubMedGoogle Scholar Contributions Z, Wand H conceived the study, participated in its design, performed the statistical analysis, and drafted the manuscript. Zewei W, Y and L helped to collect the data and performed the statistical analysis. All authors contributed to the article and approved the submitted version. Corresponding authors Correspondence to Hailong Li or Wei Wang. Ethics declarations Competing interests The authors declare no competing interests. Ethics approval and consent to participate Ethics approval and consent to participate: This retrospective study was approved by the Ethics Committee of the Affiliated Hospital of Xuzhou Medical University (XYFY2022-KL340). All patients provided written informed consent. Consent for publication Not applicable. Additional information Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. The original online version of this Article was revised: The original version of the Article contained an error in the Author Information. “Zhen Li, Zewei Wang, Yang Liu, Lei Yang and Ling Gu contributed equally to this work.” now reads: “Zhen Li, Zewei Wang and Yang Liu contributed equally to this work.” Rights and permissions Open Access This article is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, which permits any non-commercial use, sharing, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if you modified the licensed material. 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Download citation Received: 10 November 2024 Accepted: 03 February 2025 Published: 10 February 2025 DOI: Share this article Anyone you share the following link with will be able to read this content: Get shareable link Sorry, a shareable link is not currently available for this article. Copy shareable link to clipboard Provided by the Springer Nature SharedIt content-sharing initiative Keywords Non–muscle-invasive bladder cancer Re-TURBT BCG Chemotherapy Progression Subjects Bladder cancer Outcomes research Download PDF Sections Figures References Abstract Introduction Materials and methods Results Discussion Conclusions Data availability Change history References Acknowledgements Funding Author information Ethics declarations Additional information Rights and permissions About this article Advertisement Fig. 1 View in articleFull size image Fig. 2 View in articleFull size image Fig. 3 View in articleFull size image Fig. 4 View in articleFull size image Jubber, I. et al. Epidemiology of bladder Cancer in 2023: A systematic review of risk factors. Eur. Urol.84(2), 176–190 (2023). ArticlePubMedGoogle Scholar Bray, F. et al. Global cancer statistics 2018: GLOBOCAN estimates of incidence and mortality worldwide for 36 cancers in 185 countries. CA-Cancer J. Clin.68(6), 394–424 (2018). Babjuk, M. et al. European Association of Urology Guidelines on non-muscle-invasive bladder Cancer (Ta, T1, and carcinoma in situ). Eur. Urol.81(1), 75–94 (2022). Teoh, J. Y. et al. An international collaborative consensus statement on en bloc resection of bladder tumorincorporating two systematic reviews, a two-round delphi survey, and a consensus meeting. Eur. Urol.78(4), 546–569 (2020). ArticleMathSciNetPubMedGoogle Scholar Richterstetter, M. et al. The value of extended transurethral resection of bladder tumor(TURBT) in the treatment of bladder cancer. BJU Int.110(2 Pt 2), E76–E79 (2012). PubMedGoogle Scholar Cumberbatch, M. et al. Repeat transurethral resection in non-muscle-invasive bladder cancer: A systematic review. Eur. Urol.73(6), 925–933 (2018). ArticlePubMedGoogle Scholar Hashine, K. et al. Results of second transurethral resection for high-grade T1 bladder cancer. Urol. Ann.8(1), 10–15 (2016). ArticleCASPubMedPubMed CentralGoogle Scholar Flaig, T. W. et al. NCCN guidelines(R) insights: Bladder cancer, version 2.2022. J. Natl. Compr. Cancer Netw.20(8), 866–878 (2022). ArticleGoogle Scholar Chang, S. S. et al. Diagnosis and treatment of non-muscle invasive bladder cancer: AUA/SUO Guideline. J. Urol.196(4), 1021–1029 (2016). ArticlePubMedGoogle Scholar Jin, Y. H. et al. Treatment and surveillance for non-muscle-invasive bladder cancer: A clinical practice guideline (2021 edition). Mil.Med. Res.9(1), 44 (2022). PubMedPubMed CentralGoogle Scholar Eroglu, A., Ekin, R. G., Koc, G. & Divrik, R. T. The prognostic value of routine second transurethral resection in patients with newly diagnosed stage pT1 non-muscle-invasive bladder cancer: Results from randomized 10-year extension trial. Int. J. Clin. Oncol.25(4), 698–704 (2020). ArticlePubMedGoogle Scholar Gordon, P. C., Thomas, F., Noon, A. P., Rosario, D. J. & Catto, J. Long-term outcomes from re-resection for high-risk non-muscle-invasive bladder cancer: A potential to rationalize use. Eur. Urol. Focus5(4), 650–657 (2019). Lee, K., Jeong, S. H., Yoo, S. H. & Ku, J. H. Evaluating the efficacy of secondary transurethral resection of the bladder for high-grade Ta tumors. Investig. Clin. Urol.63(1), 14–20 (2022). ArticlePubMedGoogle Scholar Bosschieter, J. et al. Value of an immediate intravesical instillation of Mitomycin C in patients with non-muscle-invasive bladder cancer: A prospective multicentre randomised study in 2243 patients. Eur. Urol.73(2), 226–232 (2018). ArticleCASPubMedGoogle Scholar Messing, E. M. et al. Effect of Intravesical Instillation of Gemcitabine vs saline immediately following resection of suspected low-Grade non-muscle-invasive bladder cancer on tumor recurrence: SWOG S0337 randomized clinical trial. JAMA-J. Am. Med. Assoc.319(18), 1880–1888 (2018). Sylvester, R. J. et al. Systematic review and individual patient data meta-analysis of randomized trials comparing a single immediate instillation of chemotherapy after transurethral resection with transurethral resection alone in patients with stage pTa-pT1 urothelial carcinoma of the bladder: Which patients benefit from the instillation?. Eur. Urol.69(2), 231–244 (2016). ArticlePubMedGoogle Scholar Bree, K. K. et al. Repeat transurethral resection of muscle-invasive bladder cancer prior to radical cystectomy is prognostic but not therapeutic. J. Urol.209(1), 140–149 (2023). ArticlePubMedGoogle Scholar Brauers, A., Buettner, R. & Jakse, G. Second resection and prognosis of primary high risk superficial bladder cancer: Is cystectomy often too early?. J. Urol.165(3), 808–810 (2001). ArticleCASPubMedGoogle Scholar Brausi, M. et al. Variability in the recurrence rate at first follow-up cystoscopy after TUR in stage Ta T1 transitional cell carcinoma of the bladder: A combined analysis of seven EORTC studies. Eur. Urol.41(5), 523–531 (2002). ArticlePubMedGoogle Scholar Sylvester, R. J., Oosterlinck, W. & van der Meijden, A. P. A single immediate postoperative instillation of chemotherapy decreases the risk of recurrence in patients with stage Ta T1 bladder cancer: A meta-analysis of published results of randomized clinical trials. J. Urol.171(6 Pt 1), 2186–2190 (2004). ArticlePubMedGoogle Scholar Han, J., Gu, X., Li, Y. & Wu, Q. Mechanisms of BCG in the treatment of bladder cancer-current understanding and the prospect. Biomed. Pharmacother.129, 110393 (2020). ArticleCASPubMedGoogle Scholar Marttila, T. et al. Intravesical Bacillus calmette-guerin versus combination of epirubicin and interferon-alpha2a in reducing recurrence of non-muscle-invasive bladder carcinoma: FinnBladder-6 study. Eur. Urol.70(2), 341–347 (2016). Shang, P. F. et al. Intravesical bacillus calmette-guerin versus epirubicin for Ta and T1 bladder cancer. Cochrane Datab. Syst. Rev.11(5), CD6885 (2011). Google Scholar Musat, M. G. et al. Treatment outcomes of high-risk non-muscle invasive bladder cancer (HR-NMIBC) in real-world evidence (RWE) studies: Systematic literature review (SLR). ClinicoEcon. Outcome Res.14, 35–48 (2022). Matulay, J. T. et al. Contemporary outcomes of patients with nonmuscle-invasive bladder cancer treated with bacillus calmette-guerin: Implications for clinical trial design. J. Urol.205(6), 1612–1621 (2021). ArticlePubMedGoogle Scholar Kamat, A. M. et al. Definitions, end points, and clinical trial designs for non-muscle-invasive bladder cancer: Recommendations from the international bladder cancer group. J. Clin. Oncol.34(16), 1935–1944 (2016). Grabe-Heyne, K. et al. Intermediate and high-risk non-muscle-invasive bladder cancer: An overview of epidemiology, burden, and unmet needs. Front. Oncol.13,1170124 (2023). Tabayoyong, W. B. et al. Systematic review on the utilization of maintenance intravesical chemotherapy in the management of non-muscle-invasive bladder cancer. Eur. Urol. Focus4(4), 512–521 (2018). 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16317	https://www.mayoclinic.org/diseases-conditions/anxiety/symptoms-causes/syc-20350961	Anxiety disorders - Symptoms and causes - Mayo Clinic This content does not have an English version. This content does not have an Arabic version. 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However, people with anxiety disorders frequently have intense, excessive and persistent worry and fear about everyday situations. Often, anxiety disorders involve repeated episodes of sudden feelings of intense anxiety and fear or terror that reach a peak within minutes (panic attacks). These feelings of anxiety and panic interfere with daily activities, are difficult to control, are out of proportion to the actual danger and can last a long time. You may avoid places or situations to prevent these feelings. Symptoms may start during childhood or the teen years and continue into adulthood. Examples of anxiety disorders include generalized anxiety disorder, social anxiety disorder (social phobia), specific phobias and separation anxiety disorder. You can have more than one anxiety disorder. Sometimes anxiety results from a medical condition that needs treatment. Whatever form of anxiety you have, treatment can help. Products & Services A Book: Ageless Aging A Book: Mayo Clinic Family Health Book Assortment of Health Products from Mayo Clinic Store Show more products from Mayo Clinic Symptoms Common anxiety signs and symptoms include: Feeling nervous, restless or tense. Having a sense of impending danger, panic or doom. Having an increased heart rate. Breathing rapidly (hyperventilation). Sweating. Trembling. Feeling weak or tired. Trouble concentrating or thinking about anything other than the present worry. Having trouble sleeping. Having an upset stomach or other problems with digestion. Having difficulty controlling worry. Having the urge to avoid things that trigger anxiety. Several types of anxiety disorders exist: Agoraphobia (ag-uh-ruh-FOE-be-uh) is a type of anxiety disorder in which you fear and often avoid places or situations that might cause you to panic and make you feel trapped, helpless or embarrassed. Anxiety disorder due to a medical condition includes symptoms of intense anxiety or panic that are directly caused by a physical health problem. Generalized anxiety disorder includes persistent and excessive anxiety and worry about activities or events — even ordinary, routine issues. The worry is out of proportion to the actual circumstance, is difficult to control and affects how you feel physically. It often occurs along with other anxiety disorders or depression. Panic disorder involves repeated episodes of sudden feelings of intense anxiety and fear or terror that reach a peak within minutes (panic attacks). You may have feelings of impending doom, shortness of breath, chest pain, or a rapid, fluttering or pounding heart (heart palpitations). These panic attacks may lead to worrying about them happening again or avoiding situations in which they've occurred. Selective mutism is a consistent failure of children to speak in certain situations, such as school, even when they can speak in other situations, such as at home with close family members. This can interfere with school, work and social functioning. Separation anxiety disorder is a childhood disorder characterized by anxiety that's excessive for the child's developmental level and related to separation from parents or others who have parental roles. Social anxiety disorder (social phobia) involves high levels of anxiety, fear and avoidance of social situations due to feelings of embarrassment, self-consciousness and concern about being judged or viewed negatively by others. Specific phobias are characterized by major anxiety when you're exposed to a specific object or situation and a desire to avoid it. Phobias provoke panic attacks in some people. Substance-induced anxiety disorder is characterized by symptoms of intense anxiety or panic that are a direct result of misusing drugs, taking medications, being exposed to a toxic substance or withdrawal from drugs. Other types of anxiety disorders. Certain types of anxiety disorders or phobias don't fit neatly into a category. But they cause major distress and disrupt daily life. When to see a doctor See your doctor if: You feel like you're worrying too much and it's interfering with your work, relationships or other parts of your life. Your fear, worry or anxiety is upsetting to you and difficult to control. You feel depressed, have trouble with alcohol or drug use, or have other mental health concerns along with anxiety. You think your anxiety could be linked to a physical health problem. You have suicidal thoughts or behaviors. If this is the case, contact your provider right away. Or contact a suicide hotline. In the U.S., call or text 988 to reach the 988 Suicide & Crisis Lifeline, which is available 24 hours a day, every day. Or use the Lifeline Chat. Services are free and private. Your worries may not go away on their own, and they may get worse over time if you don't seek help. See your doctor or a mental health provider before your anxiety gets worse. It's easier to treat if you get help early. The U.S. Preventive Services Task Force recommends screening children and teens ages 8 to 18 for anxiety. This screening is for those who don't have a diagnosis of anxiety disorder. It's important because many children and teens have high levels of anxiety but they may not show it. The task force has not yet suggested how often this screening should happen. Request an appointment There is a problem with information submitted for this request. Review/update the information highlighted below and resubmit the form. 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Sorry something went wrong with your subscription Please, try again in a couple of minutes Retry Causes The causes of anxiety disorders aren't fully understood. Life experiences such as traumatic events appear to trigger anxiety disorders in people who are already prone to anxiety. Inherited traits also can be a factor. Medical causes For some people, anxiety may be linked to an underlying health issue. In some cases, anxiety signs and symptoms are the first indicators of a medical illness. If your doctor suspects your anxiety may have a medical cause, he or she may order tests to look for signs of a problem. Examples of medical problems that can be linked to anxiety include: Heart disease. Diabetes. Thyroid problems, such as hyperthyroidism. Respiratory disorders, such as chronic obstructive pulmonary disease (COPD) and asthma. Drug misuse or withdrawal. Withdrawal from alcohol, anti-anxiety medications (benzodiazepines) or other medications. Chronic pain or irritable bowel syndrome. Rare tumors that produce certain fight-or-flight hormones. Sometimes anxiety can be a side effect of certain medications. It's possible that your anxiety may be due to an underlying medical condition if: You don't have any blood relatives (such as a parent or sibling) with an anxiety disorder. You didn't have an anxiety disorder as a child. You don't avoid certain things or situations because of anxiety. You have a sudden occurrence of anxiety that seems unrelated to life events and you didn't have a previous history of anxiety. Risk factors These factors may increase your risk of developing an anxiety disorder: Trauma. Children who endured abuse or trauma or witnessed traumatic events are at higher risk of developing an anxiety disorder at some point in life. Adults who experience a traumatic event also can develop anxiety disorders. Stress due to an illness. Having a health condition or serious illness can cause significant worry about issues such as your treatment and your future. Stress buildup. A big event or a buildup of smaller stressful life situations may trigger excessive anxiety — for example, a death in the family, work stress or ongoing worry about finances. Personality. People with certain personality types are more prone to anxiety disorders than others are. Other mental health disorders. People with other mental health disorders, such as depression, often also have an anxiety disorder. Having blood relatives with an anxiety disorder. Anxiety disorders can run in families. Drugs or alcohol. Drug or alcohol use or misuse or withdrawal can cause or worsen anxiety. Complications Having an anxiety disorder does more than make you worry. It can also lead to, or worsen, other mental and physical conditions, such as: Depression (which often occurs with an anxiety disorder) or other mental health disorders. Substance misuse. Trouble sleeping (insomnia). Digestive or bowel problems. Headaches and chronic pain. Social isolation. Problems functioning at school or work. Poor quality of life. Suicide. Prevention There's no way to predict for certain what will cause someone to develop an anxiety disorder, but you can take steps to reduce the impact of symptoms if you're anxious: Get help early. Anxiety, like many other mental health conditions, can be harder to treat if you wait. Stay active. Participate in activities that you enjoy and that make you feel good about yourself. Enjoy social interaction and caring relationships, which can lessen your worries. Avoid alcohol or drug use. Alcohol and drug use can cause or worsen anxiety. If you're addicted to any of these substances, quitting can make you anxious. If you can't quit on your own, see your doctor or find a support group to help you. By Mayo Clinic Staff Anxiety disorders care at Mayo Clinic Request an appointment Diagnosis & treatment July 29, 2025 Print Show references Anxiety disorders. In: Diagnostic and Statistical Manual of Mental Disorders DSM-5-TR. 5th ed. American Psychiatric Association; 2022. Accessed Jan. 27, 2023. Anxiety disorders. National Institute of Mental Health. Accessed Jan. 27, 2023. Brown A. Allscripts EPSi. Mayo Clinic, Rochester, Minn. March 5, 2018. Anxiety disorders. National Alliance on Mental Illness. Accessed Jan. 27, 2023. What are anxiety disorders? American Psychiatric Association. Accessed Jan. 27, 2023. Reinhold JA, et al. Pharmacological treatment for generalized anxiety disorder in adults: An update. Expert Opinion in Pharmacotherapy. 2015;16:1669. Bandelow B, et al. Efficacy of treatments for anxiety disorders: A meta-analysis. International Clinical Psychopharmacology. 2015;30:183. Find support. National Alliance on Mental Illness. Accessed Jan. 27, 2023. Bazzan AJ, et al. Current evidence regarding the management of mood and anxiety disorders using complementary and alternative medicine. Expert Review of Neurotherapeutics. 2014;14:411. Natural medicines in the clinical management of anxiety. Natural Medicines. Accessed Feb. 26, 2018. Sarris J, et al. Plant-based medicines for anxiety disorders, Part 2: A review of clinical studies with supporting preclinical evidence. CNS Drugs. 2013;27:301. Bystritsky A. Complementary and alternative treatments for anxiety symptoms and disorders: Herbs and medications. Accessed Jan. 27, 2023. Bystritsky A. Pharmacotherapy for generalized anxiety disorder in adults. Accessed Feb. 26, 2018. Medical reviewer (expert opinion). Mayo Clinic. Feb. 27, 2023. 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16318	https://fishbase.se/summary/Lutjanus_campechanus.html	This website uses cookies to enhance your browsing experience and ensure the functionality of our site. For more detailed information about the types of cookies we use and how we protect your privacy, please visit our Privacy Information page.. Accept All Accept Necessary Cookies Only Cookie Settings × Cookie Settings This website uses different types of cookies to enhance your experience. Please select your preferences below: Strictly Necessary- [x] These cookies are essential for the website to function properly. They include session cookies, which help maintain your session while you navigate the site, as well as cookies that remember your language preferences and other essential functionalities. Without these cookies, certain features of the website cannot be provided. Performance- [x] These cookies help us understand how visitors interact with our website by collecting and reporting information anonymously. 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Save and Close Lutjanus campechanus, Northern red snapper : fisheries, gamefish You can sponsor this page Common name (e.g. trout) Genus + Species (e.g. Gadus morhua) About this page More Info Plus d'info Mais info Languages Arabic Bahasa/Malay Bangla Chinese(Si) Chinese(Tr) Deutsch English Español Farsi Français Greek Hindi Italiano Japanese Lao Nederlands Português(Br) Português(Pt) Russian Swedish Thai Vietnamese User feedbacks Comments & Corrections Fish Forum Guest Book Facebook Citation Uploads Attach website Upload photo Upload video Upload references Fish Watcher Related species Species in Lutjanus Species in Lutjanidae Classification - Lutjaninae Lutjanidae Eupercaria/misc Teleostei Chordata Animalia Lutjanuscampechanus (Poey, 1860) Northern red snapper Add your observation in Fish Watcher Native range \| All suitable habitat \| Point map \| Year 2050 This map was computer-generated and has not yet been reviewed. Lutjanus campechanusAquaMaps Data sources: GBIFOBIS Upload your photosandvideos Pictures \| Videos \|Google image Lutjanus campechanus Picture by Cox, C.D. Classification / Names Common names \| Synonyms \| Catalog of Fishes(genus, species) \| ITIS \| CoL \| WoRMS \| Cloffa Teleostei (teleosts) >Eupercaria/misc (Various families in series Eupercaria) >Lutjanidae (Snappers) > Lutjaninae Etymology: Lutjanus:Malay, ikan lutjan, name of a fish. More on author: Poey. Environment: milieu / climate zone / depth range / distribution range Ecology Marine; reef-associated; depth range 10 - 190 m (Ref. 55), usually 30 - 130 m (Ref. 55). Subtropical; 43°N - 4°S, 100°W - 40°W (Ref. 55228) Distribution Countries \| FAO areas \| Ecosystems \| Occurrences \| Point map \| Introductions \| Faunafri Western Atlantic: Gulf of Mexico and eastern coast of the USA extending northward to Massachusetts, coasts of Florida (Ref. 26938), but rare north of the Carolinas. Throughout Gulf of Mexico (Ref. 26938). This species has been referred to as Lutjanus aya by previous authors, but Rivas (Ref. 6409) proved that Bodianus aya Bloch, 1790 is not a lutjanid, but probably a sciaenid. Length at first maturity / Size / Weight / Age Maturity: L m39.3, range 25 - 47.5 cm Max length : 100.0 cm TL male/unsexed; (Ref. 26938); common length : 60.0 cm TL male/unsexed; (Ref. 55); max. published weight: 22.8 kg (Ref. 40637); max. reported age: 57 years (Ref. 48779) Short description Identification keys \| Morphology \| Morphometrics Dorsalspines (total): 10; Dorsalsoft rays (total): 14; Analspines: 3; Analsoft rays: 8 - 9. Scale rows on back rising obliquely above lateral line. Specimens under 30 to 35 cm with large dark spot on the upper sides, located below the anterior soft dorsal rays. Body shape(shape guide): fusiform / normal;Cross section: oval. Biology Glossary (e.g. epibenthic) Adults are found over rocky bottoms. Juveniles inhabit shallow waters, common over sand or muddy bottoms. Feed mainly on fishes, shrimps, crabs, worms, cephalopods, and some planktonic items including urochordates and gastropods. Marketed fresh and eaten steamed, broiled and baked (Ref. 9988). Heavily exploited in American waters where it is now closely protected; shrimp fishing, accused of destroying young snappers, is currently restricted. Life cycle and mating behavior Maturity \| Reproduction \| Spawning \| Eggs \| Fecundity \| Larvae Main reference Upload your references \| References \| Coordinator \| Collaborators Allen, G.R., 1985. FAO Species Catalogue. Vol. 6. Snappers of the world. An annotated and illustrated catalogue of lutjanid species known to date. FAO Fish. Synop. 125(6):208 p. Rome: FAO. (Ref. 55) IUCN Red List Status (Ref. 130435: Version 2025-1) Vulnerable (VU)(A2bd); Date assessed: 10 August 2015 CITES Not Evaluated CMS (Ref. 116361) Not Evaluated Threat to humans Reports of ciguatera poisoning (Ref. 30303) Human uses Fisheries: commercial; gamefish: yes FAO - Fisheries: landings; Publication: search \| FishSource \| Sea Around Us More information Trophic ecology Food items (preys) Diet composition Food consumption Food rations Predators Ecology Ecology Home ranges Population dynamics Growth parameters Max. ages / sizes Length-weight rel. Length-length rel. Length-frequencies Mass conversion Recruitment Abundance Life cycle Reproduction Maturity Maturity/Gills rel. Fecundity Spawning Spawning aggregations Eggs Egg development Larvae Larval dynamics Distribution Countries FAO areas Ecosystems Occurrences Introductions BRUVS - Videos Anatomy Gill area Brain Otolith Physiology Body composition Nutrients Oxygen consumption Swimming type Swimming speed Visual pigments Fish sound Diseases & Parasites Toxicity (LC50s) Genetics Genome Genetics Heterozygosity Heritability Human related Aquaculture systems Aquaculture profiles Strains Ciguatera cases Stamps, coins, misc. Outreach Collaborators Taxonomy Common names Synonyms Morphology Morphometrics Pictures References References Tools E-book \| Field guide \| Identification keys \| Length-frequency wizard \| Life-history tool \| Point map \| Catch-MSY \| Special reports Check for Aquarium maintenance \| Check for Species Fact Sheets \| Check for Aquaculture Fact Sheets Download XML Summary page \| Point data \| Common names \| Photos Internet sources AFORO (otoliths) \| Aquatic Commons \| BHL \| Cloffa \| Websites from users \| Check FishWatcher \| CISTI \| Catalog of Fishes: genus, species \| DiscoverLife \| ECOTOX \| FAO - Fisheries: landings; Publication: search \| Faunafri \| Fishipedia \| Fishtrace \| GloBI \| GoMexSI(interaction data) \| Google Books \| Google Scholar \| Google \| IGFA World Record \| OceanAdapt \| OneZoom \| Open Tree of Life \| Otolith Atlas of Taiwan Fishes \| Public aquariums \| PubMed \| Reef Life Survey \| RFE Identification \| Socotra Atlas \| TreeBase \| Tree of Life \| Wikipedia: Go, Search \| Zoological Record Estimates based on models Preferred temperature (Ref. 123201): 19.7 - 27.5, mean 24.4 °C (based on 218 cells). Phylogenetic diversity index (Ref. 82804):PD 50 = 0.5000 [Uniqueness, from 0.5 = low to 2.0 = high]. Bayesian length-weight: a=0.01479 (0.01306 - 0.01675), b=2.97 (2.95 - 2.99), in cm total length, based on LWR estimates for this species (Ref. 93245). Trophic level (Ref. 69278):3.9 ±0.72 se; based on food items. Generation time: 6.9 (6.1 - 7.8) years. Estimated as median ln(3)/K based on 19growth studies. Resilience (Ref. 120179):Medium, minimum population doubling time 1.4 - 4.4 years (K=0.12-0.2; tmax=16; Fec > 1 million). Prior r = 0.48, 95% CL = 0.31 - 0.71, Based on 2 data-limited stock assessments. Fishing Vulnerability (Ref. 59153):Moderate to high vulnerability (54 of 100). 🛈 Climate Vulnerability (Ref. 125649):Moderate to high vulnerability (51 of 100). 🛈 Price category (Ref. 80766):Very high. Nutrients (Ref. 124155):Calcium = 13.4 [7.2, 20.3] mg/100g; Iron = 0.274 [0.163, 0.455] mg/100g; Protein = 19.2 [17.8, 20.6] %; Omega3 = 0.195 [0.122, 0.314] g/100g; Selenium = 42 [23, 74] μg/100g; VitaminA = 59 [11, 265] μg/100g; Zinc = 0.245 [0.180, 0.365] mg/100g (wet weight); Random Species Back to Search Comments & Corrections Back to Top Accessed through: Not available FishBase mirror site : localhost Page last modified by : mrius-barile - 20 July 2016 Total processing time for the page : 0.0262 seconds
16319	https://www.journalmc.org/index.php/JMC/article/view/42/57	\| \| \| --- \| \| \| Journals \| Policy \| Permission \| \| \| \| \| --- \| \| This journal has moved to jmc.elmerpub.com Please go to the new site to make submissions \| \| \| \| \| --- \| \| Journal of Medical Cases \| \| \| \| \| --- \| \| \| Journals \| Policy \| Permission \| Journals \| Policy \| Permission \| \| \| \| --- \| \| This journal has moved to jmc.elmerpub.com Please go to the new site to make submissions \| \| \| \| \| --- \| \| Journal of Medical Cases \| \| Journal of Medical Cases \| \| \| --- \| \| Username \| \| \| Password \| \| \| Remember me \| \| \| \| \| \| \| \| \| \| All Authors Title Abstract Index terms Full Text \| \| \| \| \| \| MOST READ \| MOST READ (For the last 6-month published articles) \| \| \| INDEXED IN \| \| \| \| \| INDEXED IN Scopus Web of Science (ESCI) \| \| \| ARTICLE STATISTICS \| \| Submission to First Decision \| \| 32 Days \| \| Acceptance Rate \| \| 38% \| \| Acceptance to Publication \| \| 26 Days \| \| Average article statistics from the last 12 months data \| ARTICLE STATISTICS Submission to First Decision 32 Days Acceptance Rate 38% Acceptance to Publication 26 Days Average article statistics from the last 12 months data \| \| \| Journal of Medical Cases, ISSN 1923-4155 print, 1923-4163 online, Open Access \| \| Article copyright, the authors; Journal compilation copyright, J Med Cases and Elmer Press Inc \| \| Journal website \| Case Report Volume 1, Number 3, December 2010, pages 81-83 Fatal Outcome Due to Sepsis by Mycobacterium Bovis Six Years After BCG Intravesical Instillations Miguel Alvarez-Mugicaa, e, Jesus Ma Fernandez Gomezb, Veronica Bulnes Vazquezc, Antonio Jalon Monzonb, Jorge Garcia Rodriguezb, Jose Ma Fernandez Rodriguezd, Victor Asensid, Jose Antonio Cartond aDepartment of Urology, Hospital Valle del Nalon, Langreo, Asturias, Spain bDepartment of Radiology, Hospital Alvarez Buylla, Mieres, Asturias, Spain cDepartment of Urology, Hospital Universitario Central de Asturias, Oviedo, Asturias, Spain dDepartment of Internal Medicine, Hospital Universitario Central de Asturias, Oviedo, Asturias, Spain eCorresponding author: Urology department, Valle del Nalon Hospital, Poligono de Riano s/n, Langreo, Asturias, Spain Manuscript accepted for publication September 9, 2010 Short title: Septicaemia Due to BCG doi: \| \| \| --- \| \| Abstract \| ▴Top \| Bacillus Calmette-Guerin (BCG) in intravesical instillations is the reference treatment for urothelial carcinoma with a high risk of progression. Morbidity secondary to intravesical BCG may present both locally and systemically. Most patients suffer a self-limited irritative voiding syndrome. Prevention of these complications requires implementing rules of good practice for the instillations. The undesirable side effects should be recognized early. Their treatment should be rapid and adapted to the patient. Besides the commonly seen side effects of intravesical BCG instillations, very rare complications have been reported. In some patients, infection appears early (within 3 months after instillation) and is characterized by generalized symptoms, with pneumonitis and hepatitis. Late-presentation disease occurs more than 1 year after the first BCG treatment and usually involves focal infection of the genitourinary tract (the site at which bacteria were introduced) and/or other sites that are typical for reactivation of mycobacterial disease, such as the vertebral spine or the retroperitoneal tissues. Non caseating granulomas are found in the majority of cases, whether early or late. Most patients respond to treatment with antituberculous drugs; in early-presentation disease, when features of hypersensitivity predominate, glucocorticosteroids are sometimes added. Late localized infection often requires surgical resection. Keywords: Bacillus Calmette-Guerin; Instillations; Transitional cell carcinoma; Septicaemia \| \| \| --- \| \| Introduction \| ▴Top \| Bacillus Calmette-Guerin (BCG) is a live attenuated strain of Mycobacterium bovis that has been used to treat transitional-cell carcinoma since 1976 and has been reported to eradicate disease in more than 70% of patients with in situ and stage I disease . Intravesical therapy with BCG is generally considered safe, however, serious complications including hematuria, granulomatous pneumonitis, suppurative lymphadenitis, distant intramuscular and bone abscesses, hepatitis, and life-threatening BCG sepsis are well known , although late bacteriuria and sepsis due to BCG instillations, have not been described yet. The reported incidence of other than minor complications is under 5% . Recommended treatment for disseminated BCG disease includes a combination of antituberculous medications (with the exception of pyrazinamide, to which BCG is typically resistant) and a tapering course of steroids . These complications are an absolute contraindication for further BCG instillations. Despite its toxicity, the risk-benefit ratio favors the use of BCG in patients who have moderate and high risk tumors . There are not differences of toxicity between Connaught and Pasteur strain in intravesical BCG-therapy of superficial bladder tumors . Compliance with this treatment is altered by its potentially serious locoregional or general side effects. Prevention of these complications requires implementing rules of good practice for the instillations. The undesirable side effects should be recognized early. Their treatment should be rapid and adapted to the patient. In order to prevent complications from BCG immunotherapy, a French study showed that side effects were significantly reduced by administration of ofloxacin after each instillation of BCG. The number of side effects requiring antitubercular treatment was also reduced in the patients in this study who had received ofloxacin . Puigvert Foundation in Spain developed a practical guideline for the management of the complications followed BCG instillations . Although fatal sepsis has been described before [9-11], in a search of PubMed we did not find any fatal sepsis as late complication of BCG instillations. What we report, to our knowledge, is the first case of disseminated BCG infection causing septicemia and death 6 years after intravesical treatment with BCG therapy for bladder cancer. \| \| \| --- \| \| Case Report \| ▴Top \| An 83-year-old man was hospitalized with a 2-week history of fever, malaise, anorexia and bleeding from a back drainage. The patient had previously undergone two trans-urethral resections receiving 12 intravesical bacillus Calmette-Guerin instillations without any complications for a high grade transitional cell carcinoma of the bladder. Due to tumor progression he was performed a radical cystoprostatectomy 5 years earlier of this admission. He also had an endovascular stent-graft repair of an infrarenal abdominal aortic aneurysm. Six months before this new admission, he was diagnosed from multiple retroperitoneal abscesses in which mycobacterium bovis grew. Since then, the patient received a combination of antitubercular medication. Physical examination revealed an ill-appearing man. Pulmonary and cardiac examinations were unremarkable. He complained of abdominal pain in the right flank. Laboratory tests revealed abnormal liver function tests, anemia, renal insufficiency, leukocytosis and signs of coagulation disorder. A computed tomography (CT) scan of the abdomen showed a diffused and enlarged left psoas muscle. High doses of corticosteroids were associated to the antitubercular treatment. The patient developed a severe sepsis. Blood cultures were positive for acid-fast bacilli stains. Despite partial improvement, the course was complicated with a multiple organ failure, and the patient died two weeks after admission. \| \| \| --- \| \| Discussion \| ▴Top \| Permanent increase in neoplasm incidence including bladder neoplasm makes physicians to search for new forms and methods of treatment that prevents either appearance and/or progression of the disease. Application of new preparations entails in many cases appearance of side effects which sometimes are difficult to manage and thus must be monitored constantly. To avoid complications which in case of BCG application are very burdensome and sometimes dangerous for patient, it is necessary to intervene in due time. The BCG therapy has been effectively used in the management of superficial bladder cancers. It is especially useful as an adjuvant therapy following bladder surgery for cancer. For instance, Demkow et al , in a recent study, reported that nearly 66% of the patients who received intravesical BCG therapy following complete transurethral resection of a bladder tumor were cancer free after a median follow-up of 45 months. Adverse events following intravesical BCG therapy are related to strain virulence, allergic reactions or to nosocomial urinary tract infections. BCG is a potent immunostimulator that exerts its urological effects by inducing a strong immune response and by causing cell cycle arrest at the G1/S transition phase [12, 13]. A multicenter reviewed and studied complications on 1,278 patients after bacillus Calmette-Guerin therapy for bladder cancer . Cystitis occurred in 91% of the patients. Complications identified included fever in 50 patients (3.9%), granulomatous prostatitis in 17 (1.3%), bacillus Calmette-Guerin pneumonitis or hepatitis in 12 (0.9%), arthritis or arthralgia in 6 (0.5%), hematuria requiring catheterization or transfusion in 6 (0.5%), skin rash in 5 (0.4%), skin abscess in 5 (0.4%), ureteral obstruction in 4 (0.3%), epididymo-orchitis in 2 (0.2%), bladder contracture in 2 (0.2%), hypotension in 1 (0.1%), and cytopenia in 1 (0.1%). Deaths due to BCG sepsis and the high frequency of BCG-induced cystitis have compromised the use of BCG. However, with increased experience in applying BCG, the side-effects now appear to be less prominent. Serious side-effects are encountered in fewer than 5% . Major complications can appear after systemic absorption of the drug, therefore BCG should not be administered during the first 2 weeks after transurethral resection, in patients with hematuria and after traumatic catheterization. Morbidity secondary to intravesical BCG may present both locally and systemically. Most patients suffer a self-limited irritative voiding syndrome. Prevention of these complications requires implementing rules of good practice for the instillations. The undesirable side effects should be recognized early. Their treatment should be rapid and adapted to the patient. Besides the commonly seen side effects of intravesical BCG instillations, very rare complications have been reported. \| \| \| --- \| \| References \| ▴Top \| This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 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This is an open-access journal distributed under the terms of the Creative Commons Attribution-NonCommercial 4.0 International License, which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited. Creative Commons Attribution license (Attribution-NonCommercial 4.0 International CC-BY-NC 4.0) This journal follows the International Committee of Medical Journal Editors (ICMJE) recommendations for manuscripts submitted to biomedical journals, the Committee on Publication Ethics (COPE) guidelines, and the Principles of Transparency and Best Practice in Scholarly Publishing. website: www.currentsurgery.org editorial contact: editor@currentsurgery.org elmer.editorial2@hotmail.com Address: 9225 Leslie Street, Suite 201, Richmond Hill, Ontario, L4B 3H6, Canada © Elmer Press Inc. All Rights Reserved. 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The content of this site is intended for health care professionals. This is an open-access journal distributed under the terms of the Creative Commons Attribution-NonCommercial 4.0 International License, which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited. Creative Commons Attribution license (Attribution-NonCommercial 4.0 International CC-BY-NC 4.0) This journal follows the International Committee of Medical Journal Editors (ICMJE) recommendations for manuscripts submitted to biomedical journals, the Committee on Publication Ethics (COPE) guidelines, and the Principles of Transparency and Best Practice in Scholarly Publishing. website: www.currentsurgery.org editorial contact: editor@currentsurgery.org elmer.editorial2@hotmail.com Address: 9225 Leslie Street, Suite 201, Richmond Hill, Ontario, L4B 3H6, Canada © Elmer Press Inc. All Rights Reserved. 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16320	https://www.sigmaaldrich.com/deepweb/assets/sigmaaldrich/product/documents/172/798/a0808pis.pdf?srsltid=AfmBOooLM5tg6BhA99JRwiYUz6RcvD1ToiUhvja1MOJ6fF_nO0klINQL	03/03- CKV Page 1 of 2 ACETIC ACID, GLACIAL Sigma Prod. Nos. A0808 24,285-3 is an exact replacement for A0808 CAS NUMBER: 64-19-7 SYNONYMS: acetic acid, methanecarboxylic acid, ethanoic acid, vinegar acid, ethylic acid 1 PHYSICAL DESCRIPTION: 1,2 Appearance: Clear colorless liquid Molecular formula: CH 3COOH Molecular weight: 60.05 Density: 1.049 g/mL Melting point: 16.7 ECBoiling point: 118 ECpK a = 4.75 at 25 EC 3; K a has a maximum value of 1.754 x 10 -5 at 25 EC, but decreases in value at temperatures above and below than 25 EC. (1.700 x 10 -5 at 5 EC; 1.703 x 10 -5 at 40 EC 3)Glacial acetic acid has a very pungent odor, and is a corrosive liquid. It is also flammable, with a flash point of 40 EC. 1 STORAGE / STABILITY AS SUPPLIED: Acetic acid is extremely stable stored at room temperature. It should be stored tightly sealed to prevent vapors form escaping. SOLUBILITY / STABILITY OF SOLUTIONS: Acetic acid is infinitely miscible in water, also miscible with alcohol, glycerol, ether, carbon tetrachloride. It is practically insoluble in carbon disulfide. Aqueous solutions are stable at room temperature, if well sealed. GENERAL REMARKS: By specification A0808 ("ACS Reagent") is $99.7%. In addition, A0808 is subjected to additional testing as required by the American Chemical Society. 4The name "glacial" is often used to emphasize the use of 100% acetic acid; frozen acetic acid looks like water ice. 03/03- CKV Page 2 of 2 ACETIC ACID, GLACIAL Sigma Prod. Nos. A0808 GENERAL REMARKS: (continued) Acetic acid is widely used in manufacturing either as starting material or solvent; in the food industry, dilute solutions are used for preparation and preservation. Because acetic acid is a "weak acid", it is effective as buffer at pH 4.75 " 1. This buffer can be prepared by titrating acetic acid with hydroxide or by mixing equal volumes of equimolar acetic acid and sodium acetate solutions (standard mixing tables have been published). 5,6 METHODS OF PREPARATION: Although acetic acid has historically been produced form bacterial oxidation of ethanol, on the industrial scale, it is usually obtained by destructive distillation of wood or oxidation of acetaldehyde. 1 REFERENCES: Sigma Material Safety Data Sheet. 2. Merck Index , 12th ed., #52 (1996). 3. Handbook of Chemistry and Physics, 74th ed., D.R. Lide, ed. (CRC Press, 1993), p. 8-45 and 8-47. 4. Reagent Chemicals , 8th ed. (American Chemical society, 1993), pp. 98-101. 5. Methods in Enzymology , 182, 32 (1990). 6. Data for Biochemical Research , 3rd ed., Dawson et al., ed. (Oxford Press, 1987), p. 429. Sigma warrants that its products conform to the information contained in this and other Sigma-Aldrich publications. Purchaser must determine the suitability of the product(s) for their particular use. Additional terms and conditions may apply. Please see reverse side of the invoice or packing slip.
16321	https://www.vedantu.com/question-answer/the-number-of-subsets-of-a-set-of-order-three-is-class-9-maths-cbse-5fc943af592a52370607feee	Talk to our experts 1800-120-456-456 The number of subsets of a set of order three is a) $ 3 $ b) $ 6 $ c) $ 8 $ d) $ 9 $ © 2025.Vedantu.com. All rights reserved
16322	https://www.superteacherworksheets.com/multiplication-2-digit-by-2-digit.html	Log In Become a Member Membership Info Addition (Basic) Addition (Multi-Digit) Algebra & Pre-Algebra Comparing Numbers Daily Math Review (Math Buzz) Division (Basic) Division (Long Division) Hundreds Charts Multiplication (Basic) Multiplication (Multi-Digit) Order of Operations Place Value Skip Counting Telling Time Word Problems (Daily) More Math Worksheets Reading Comprehension Reading Comprehension Pre-K/K Reading Comprehension Grade 1 Reading Comprehension Grade 2 Reading Comprehension Grade 3 Reading Comprehension Grade 4 Reading Comprehension Grade 5 Reading Comprehension Grade 6 Reading Comprehension Reading & Writing Reading Worksheets Cause & Effect Daily ELA Review (ELA Buzz) Fact & Opinion Fix the Sentences Graphic Organizers Synonyms & Antonyms Writing Prompts Writing Story Pictures Writing Worksheets More ELA Worksheets Phonics Consonant Sounds Vowel Sounds Consonant Blends Consonant Digraphs Word Families More Phonics Worksheets Early Literacy Alphabet Build Sentences Sight Word Units Sight Words (Individual) More Early Literacy Grammar Nouns Verbs Adjectives Adverbs Pronouns Punctuation Syllables Subjects and Predicates More Grammar Worksheets Spelling Lists Spelling Grade 1 Spelling Grade 2 Spelling Grade 3 Spelling Grade 4 Spelling Grade 5 Spelling Grade 6 More Spelling Worksheets Chapter Books Bunnicula Charlotte's Web Magic Tree House #1 Boxcar Children More Literacy Units Science Animal (Vertebrate) Groups Butterfly Life Cycle Electricity Human Body Matter (Solid, Liquid, Gas) Simple Machines Space - Solar System Weather More Science Worksheets Social Studies 50 States Explorers Landforms Maps (Geography) Maps (Map Skills) More Social Studies Art & Music Colors Worksheets Coloring Pages Learn to Draw Music Worksheets More Art & Music Holidays Spring Summer Back to School Labor Day More Holiday Worksheets Puzzles & Brain Teasers Brain Teasers Logic: Addition Squares Mystery Graph Pictures Number Detective Lost in the USA More Thinking Puzzles Teacher Helpers Teaching Tools Award Certificates More Teacher Helpers Pre-K and Kindergarten Alphabet (ABCs) Numbers and Counting Shapes (Basic) More Kindergarten Worksheet Generator Word Search Generator Multiple Choice Generator Fill-in-the-Blanks Generator More Generator Tools S.T.W. Full Website Index Help Files Contact Multiplication (2-Digits Times 2-Digits) The worksheets below require students to multiply 2-digit numbers by 2-digit numbers. Includes vertical and horizontal problems, as well as math riddles, task cards, a picture puzzle, a Scoot game, and word problems. 2-Digit Times 2-Digit Worksheets Multiplication 2-digit by 2-digit FREE Practice 2-digit by 2-digit vertical multiplication problems. (example: 85 x 62) 4th and 5th Grades View PDF Variety Worksheet This variety worksheet has an array of 2-digit by 2-digit math problems, including two word problems, an input/output table, and vertical and horizontal equations. 4th and 5th Grades View PDF Shape Multiplication (2-Digit by 2-Digit Numbers) At the top of this worksheet, students are shown a dozen shapes with two-digit numbers in them. They multiply similar shapes. For example: Find the product of the numbers in the hexagons. 4th through 6th Grades View PDF Find the Errors: 2-Digit by 2-Digit Numbers When you examine these two completed multiplication problems, you'll observe that each one has an error. Find the error and explain it in words. The re-solve the incorrect problem. 4th and 5th Grades View PDF Secret Code Math: 2 Digits Times 2 Digits Begin by using the cipher key to decode the secret symbols. Then multiply the factors together to calculate the products. 4th and 5th Grades View PDF Dog Multiplication: 2-digit by 2-digit FREE A practice page with a cute dog theme; 2-digit numbers by 2-digit numbers (example: 46 x 34) 4th through 6th Grades View PDF Multiplication: 2-digit by 2-digit Graph Paper Math Drills; 2-digits times 2-digits (example: 18 x 22) 3rd through 5th Grades View PDF Multiplication Word Problems (2 Digits by 2 Digits) Word problems for double-digit multiplication practice (example: 56 x 91) 4th through 6th Grades View PDF Horizontal: 2-Digit x 2-Digit Find the products of the 2-digit factors. All problems are written horizontally. Students rewrite each problem vertically and solve. (example: 76 x 23) 4th and 5th Grades View PDF Hockey Picture Puzzle (2 Digits Times 2 Digits) Solve the 2 digit by 2 digit multiplication problems. Then attach the puzzle pieces in the correct positions to reveal an ice hockey picture. 4th and 5th Grades View PDF Multi-Digit Multiplication Worksheet Generator Multiplication Worksheet Generator Here's a tool that you can use to make your own customized PDF math worksheets. You can select the number of digits in both factors. You can also toggle between horizontal and vertical problems. Games and Task Cards Game: Multiplication Scoot (2-Dig by 2-Dig) In this classroom game, students solve a series of math problems on task cards. The cards have 2-digit by two-digit multiplication problems on them. 4th through 6th Grades View PDF Task Cards (2 Digits Times 2 Digits) These printable task cards have thirty 2-digit by 2-digit multiplication problems. 4th through 6th Grades View PDF Lattice Multiplication Lattice Multiplication (2 Digits Times 2 Digits) At the top of the worksheet is a box with an explanation of how to do lattice multiplication. Then students can try to solve the eight sample problems using a lattice grid. 4th through 6th Grades View PDF Blank Lattice Grids Here are six two-by-two lattice grids. The teacher or students can input any numbers they'd like. 4th through 6th Grades View PDF See also: Multiplication Worksheets This index page can direct you to any type of multi-digit multiplication worksheets on our website. Includes multiplying by 1, 2, and 3-digit numbers. Also have money, decimal, and fraction multiplication. Sample Worksheet Images My Account Become a Member Membership Information Site License Information Login Site Information Help / FAQ Full Site Index Privacy Policy Terms of Service Contact Us Useful Links Homepage What's New? Social Media Follow Us Not a Member? For complete access to thousands of printable lessons click the button or the link below. Become a Member © 2025 Super Teacher Worksheets
16323	https://fiveable.me/key-terms/hs-honors-geometry/vector-projection	Vector projection - (Honors Geometry) - Vocab, Definition, Explanations \| Fiveable \| Fiveable new!Printable guides for educators Printable guides for educators. Bring Fiveable to your classroom ap study content toolsprintablespricing my subjectsupgrade All Key Terms Honors Geometry Vector projection 🔷honors geometry review key term - Vector projection Citation: MLA Definition Vector projection is the process of projecting one vector onto another, resulting in a new vector that represents the component of the first vector in the direction of the second. This concept is crucial in understanding how vectors interact in terms of direction and magnitude, allowing for practical applications in physics, engineering, and geometry. By breaking down a vector into its components, one can analyze various properties such as forces, motion, and angles. 5 Must Know Facts For Your Next Test The formula for the projection of vector A onto vector B is given by: $$ ext{proj}_B A = \frac{A \cdot B}{B \cdot B} B$$, where $$A \cdot B$$ is the dot product. Vector projection helps in breaking down forces into components, which is essential for solving problems involving forces acting at angles. If the vectors being projected are orthogonal, then the projection of one onto the other will be zero, meaning they do not affect each other. The magnitude of the projection tells us how much of one vector acts in the direction of another, which can be crucial in optimizing designs or analyzing movements. In graphical terms, the projection can be visualized as dropping a perpendicular from one vector to another, creating a right triangle with sides corresponding to components. Review Questions How does vector projection relate to real-world applications such as physics and engineering? Vector projection is fundamental in physics and engineering because it allows for the analysis of forces acting in various directions. By projecting vectors onto axes or other vectors, engineers can determine how much force contributes to specific directions, which is essential for designing structures or analyzing mechanical systems. This method aids in breaking down complex problems into manageable parts, making it easier to apply Newton's laws and other physical principles. Explain how the dot product is used in calculating the projection of one vector onto another. The dot product plays a vital role in calculating vector projection by determining how much one vector extends in the direction of another. The formula for projecting vector A onto vector B involves calculating the dot product of A and B, which gives us a scalar that indicates their alignment. This scalar is then normalized by dividing by the dot product of B with itself, effectively scaling vector B to find the component of A that lies along B's direction. Evaluate the significance of understanding vector projection in multidimensional space and its implications for higher-level mathematics. Understanding vector projection in multidimensional space is significant because it extends concepts from basic geometry to higher-level mathematics and applications like computer graphics and machine learning. In these fields, projecting vectors allows for complex transformations and analyses, such as shading effects in 3D rendering or data reduction techniques in high-dimensional datasets. Mastering this concept enhances problem-solving skills and provides a deeper insight into how vectors interact within various dimensions, laying groundwork for advanced topics like linear algebra. Related terms Dot product:An operation that multiplies two vectors to produce a scalar, indicating the extent to which the two vectors point in the same direction. Unit vector:A vector with a magnitude of one, often used to indicate direction without regard to length. Orthogonal:Two vectors are orthogonal if they are perpendicular to each other, resulting in a dot product of zero. 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16324	https://artofproblemsolving.com/wiki/index.php/Trigonometric_identities?srsltid=AfmBOoqFjWrUp4xncq4FO0dlfVS-V9ArBRq9rs2Xr7CPCT7CUWie6Tna	Art of Problem Solving Trigonometric identities - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Trigonometric identities Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Trigonometric identities In trigonometry, trigonometric identities are equations involving trigonometric functions that are true for all input values. Trigonometric functions have an abundance of identities, of which only the most widely used are included in this article. Contents [hide] 1 Pythagorean identities 2 Angle addition identities 3 Double-angle identities 3.1 Cosine double-angle identity 4 Half-angle identities 5 Product-to-sum identities 6 Sum-to-product identities 7 Other identities 7.1 Triple-angle identities 7.2 Even-odd identities 7.3 Conversion identities 7.4 Euler's identity 7.5 Miscellaneous 8 Resources 9 See also Pythagorean identities The Pythagorean identities state that Using the unit circle definition of trigonometry, because the point is defined to be on the unit circle, it is a distance one away from the origin. Then by the distance formula, . To derive the other two Pythagorean identities, divide by either or and substitute the respective trigonometry in place of the ratios to obtain the desired result. Angle addition identities The trigonometric angle addition identities state the following identities: There are many proofs of these identities. For the sake of brevity, we list only one here. Euler's identity states that . We have that By looking at the real and imaginary parts, we derive the sine and cosine angle addition formulas. To derive the tangent addition formula, we reduce the problem to use sine and cosine, divide both numerator and denominator by , and simplify. as desired. Double-angle identities The trigonometric double-angle identities are easily derived from the angle addition formulas by just letting . Doing so yields: Cosine double-angle identity Here are two equally useful forms of the cosine double-angle identity. Both are derived via the Pythagorean identity on the cosine double-angle identity given above. In addition, the following identities are useful in integration and in deriving the half-angle identities. They are a simple rearrangement of the two above. Half-angle identities The trigonometric half-angle identities state the following equalities: The plus or minus does not mean that there are two answers, but that the sign of the expression depends on the quadrant in which the angle resides. Consider the two expressions listed in the cosine double-angle section for and , and substitute instead of . Taking the square root then yields the desired half-angle identities for sine and cosine. As for the tangent identity, divide the sine and cosine half-angle identities. Product-to-sum identities The product-to-sum identities are as follows: They can be derived by expanding out and or and , then combining them to isolate each term. Sum-to-product identities Substituting and into the product-to-sum identities yields the sum-to-product identities. Other identities Here are some identities that are less significant than those above, but still useful. Triple-angle identities All of these expansions can be proved using trick and perform the angle addition identities. Same for and for . Even-odd identities The functions , , , and are odd, while and are even. In other words, the six trigonometric functions satisfy the following equalities: These are derived by the unit circle definitions of trigonometry. Making any angle negative is the same as reflecting it across the x-axis. This keeps its x-coordinate the same, but makes its y-coordinate negative. Thus, and . Conversion identities The following identities are useful when converting trigonometric functions. All of these can be proven via the angle addition identities. Euler's identity Euler's identity is a formula in complex analysis that connects complex exponentiation with trigonometry. It states that for any real number , where is Euler's constant and is the imaginary unit. Euler's identity is fundamental to the study of complex numbers and is widely considered among the most beautiful formulas in math. Similar to the derivation of the product-to-sum identities, we can isolate sine and cosine by comparing and , which yields the following identities: They can also be derived by computing and . These expressions are occasionally used to define the trigonometric functions. Miscellaneous These are the identities that are not substantial enough to warrant a section of their own. Resources Table of trigonometric identities List of Trigonometric Identities See also Trigonometry Trigonometric substitution Proofs of trig identities Retrieved from " Category: Trigonometry Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
16325	https://genchem1.chem.okstate.edu/1314F00/Lecture/Chapter10/VSEPR2.html	VSEPR Supporting Evidence VSEPR Supporting Evidence The test of a good model, is how well it stands up to experimental data. How well does VSEPR work? We can evaluate several ways. One way we will use is related the polarity of compounds. That is, how well does VSEPR predict the polarity of molecules and how do experimental observations compare. What do I mean by polarity of molecules? A molecule is polar if as a result of unequal sharing of electrons, centers of positive and negative charge do not coincide. Recall in CHEM 1215 we discussed the polarity of chemical bonds in simple binary compounds. The criteria for determining whether a chemical bond is polar or nonpolar was the difference in the electronegativity of atoms involved in the chemical bond. If the two atoms differed in electronegativity the chemical bond was polar. If the electronegativity was the same the bond is nonpolar. We conclude that F 2 has a nonpolar bond because the electronegativity is the same for the two atoms sharing the pair of electrons in the bond. For HF the two atoms are different, Fluorine is more electronegative compared to hydrogen. The result is the electrons in the covalent bond spend more time on the fluorine atom than on the hydrogen atom. This produces a partial negative charge on the fluorine and a partial positive charge on the hydrogen. This separation of charge produces a small dipole making the molecular polar. The molecule can be thought of as a small magnet with a positive end and a negative end. Opposites attract so HF molecules are attracted to each other when in the liquid and solid phase (but more about that later). The dipole that is produced is represented as a vector with an arrow pointing towards the negatively charged end of the molecule and a positive symbol on the other end of the vector. (The magnitude of an electric dipole is reported as the electric dipole moment, in a unit called the debye (D).) So we conclude that HF is a polar molecule because the only bond in HF is polar. F 2 is a nonpolar molecule because it does not contain a polar bond. All heteronuclear diatomic molecules are polar and all homonuclear diatomic molecules are nonpolar. The presence or absence of an electric dipole can be determined experimentally if a sample of the compound of interest is placed in an electric field. If the molecule is polar alignment with an electric field will occur. The net result is the molecules effect the capacitance (the capacity of the plates to hold a charge) of the plates which maintain the electric field. By measuring the capacitance of the plates separated by different chemical substances allows the determination of dipole moments. If a molecule can be characterized as having polar bonds that do not cancel each other out the compound will have an experimentally measureable dipole moment. What happens in more complicated molecules? It turns out that, depending on the geometry, these individual polar bonds in molecule may cancel each other out, which results in a nonpolar compounds. If the polar bonds in a molecule do not cancel each other the molecule is polar and contains a permanent distribution of electron density which gives rise to a dipole moment. Lets consider several examples to demonstrate how the geometry of a molecule effects the dipole moment of a compound. We'll begin with CO 2 and H 2 O. Both molecule have a central atom, carbon or oxygen, surrounded by two terminal atoms. According to VSEPR CO 2 is linear and H 2 O is bent. If we consider CO 2 each C=O bond is polar, because of the difference in electronegativity of the atoms in the bond. However, because the geometry of the two polar bonds are exactly opposite one another and equal in magnitude the net effect is to cancel each other out so that the molecule has no dipole moment, CO 2 is a nonpolar compound. H 2 O, on the other hand, which also contains two polar bonds (O-H) has a molecular geometry that does not allow the contributions of the polar bonds to cancel out. Water has a permanent dipole moment and is a polar compound. Experimentally CO 2 has no measureable dipole moment, but H 2 O (1.87 x10–18 D) has a very large dipole moment. We will discuss how the polarity of water effects its physical properties in Chapter 14. We can generalize these two cases. Molecules containing central atoms with no lone pair electrons and bonding pairs of electrons to identical terminal atoms are nonpolar, Molecules containing central atoms with one or more lone pair electrons are polar. (Note: there are exceptions to this rule for central atoms with electron-pair geometries of trigonal bipyramidal and octahedral. But we will not discuss these exceptions.) Lets consider two additional examples to demonstrate how the geometry of a molecule effects the dipole moment of a compound. We will consider CCl 4 and CH 3 Cl. Both molecule have a central carbon atom, surrounded by four terminal atoms. According to VSEPR both CCl 4 and CH 3 Cl are tetrahedral. If we consider CCl 4 each C-Cl bond is polar, because of the difference in electronegativity of the atoms in the bond. However, because the geometry of the four polar bonds and that all four bonds are equally polar the net effect is to cancel each other out so that the molecule has no dipole moment, CCl 4 is a nonpolar compound. CH 3 Cl, on the other hand, which also contains polar bonds (C-H and C-Cl) is also tetrahedral, however because the terminal atoms are non identical the compound has a permanent dipole moment and is a polar compound. This introduces one more rule for predicting polar compounds, Anytime the molecule has a central atom with non identical terminal atoms the molecule will be polar. After all this if I give you the formula of a compound you should be able to draw the Lewis structure; then tell me the number of lone pair and bonding pair of electrons on the 'central' atom; the electron-pair and the molecular geometry; the bond angles around any 'central' atom; and if the molecule is polar or nonpolar. So lets try figuring out the polarity for BF 3 and NH 3. Go here to see what I came up with!
16326	https://web.mae.ufl.edu/uhk/TRIANGLE-LAWS.pdf	A LOOK AT THE PROPERTIES OF TRIANGLES The simplest two dimensional figure is the triangle. It is constructed by the intersection of three non-parallel lines. Students first run into this figure in early grade school and learn about its more esoteric properties by the time they graduate from high school. In college majors in mathematics, physics and engineering make extensive use of its properties. Our purpose here is to quickly derive all its properties using just elementary algebra and geometry. The order of the discussion will differ somewhat from the standard sequential methods used in most mathematics curricula. Area of Any Triangle: Our starting point is to draw a simple rectangle of width w and height h. We locate a point P along the top of this rectangle and from it draw two straight lines of length b and c to the opposite bottom corners of the rectangle as shown- This defines the shaded oblique triangle of side-lengths are a, b, and c and its three vertex angles designated by A, B, and C. A vertical blue line breaks the oblique triangle into two right triangles. The base length of the oblique triangle equals 2 1 w w a   , so that from the Pythagorean Theorem one has – 2 2 2 2 h b h c a     We see that the area of the oblique triangle is just the area of the rectangle minus the areas of the two right triangles lying outside the triangle. We thus have Area ObliqueTriangle= ) 2 ( ) )( 2 ( ) ( 2 1 2 1 ah w w h w w h     This is a well known result. We can eliminate the rectangle height h via the easily established identity- 2 2 2 2 2 ) ( ) 2 ( 2 1 a c b cb a h     This produces one of the more important properties of any triangle, namely that- 2 2 2 2 2 ) ( ) 2 ( 4 1 a c b cb Triangle Shaded Area     So, without resorting to any trigonometry and using only the Pythagorean Theorem we are able to express the area of any oblique triangle in terms of the length of its three sides. To test things out consider an equilateral triangle where a=b=c. It says the area will be- Area Equilateral Triangle= 2 4 3 a Also for a=3, b=4, and c=5 which produces a right triangle, we have – Area = 6 32 40 4 1 2 2   The above formula for the shaded oblique triangle is equivalent to Heron’s Formula which states that- Area Any Triangle ) )( )( ( c s b s a s s     , where s=(a+b+c)/2 is the semi-perimeter. I remember asking my math teacher in high school some 63 years ago how Heron was able to come up with his formula some 2000 years ago. She did not know. I later learned that it involves looking at a circle inscribed in the triangle. Properties of Oblique Triangles: Going back to the above figure, one can readily show that at point P the sum of the three angles around it and lying within the rectangle equal 180 degrees. Thus we have the triangle angle rule that- A+B+C=180 degrees=  radians. Note that the sum of the interior angles for the enclosing box is 4x90=360deg=2 radians. More generally the sum of the angles of any n sided polygon will equal (n-2) radians. The sum of the interior angles of a hexagon will be 720deg=4 radians. To relate the side lengths and the three corner angles to each other one needs to introduce the trigonometric functions. These are associated with right triangles having vertex angles ofA,B and /2 . The Pythagorean Theorem a2+b2=c2 is always valid for right triangles. The longest length c of a right triangle is referred to as its hypotenuse. The three most important of the trigonometric functions area- length base length opposite and hpotenuse length base hypotenuse length opposite    ) tan( ) cos( , ) sin(    From these definitions we have at once from the two right triangles inside the oblique triangle that- c h B and b h C   ) sin( ) sin( Eliminating h from these last two equations and recognizing that the concept of a hypotenuse is not applicable for oblique triangles, one arrives at the following Law of Sines - c C b B a A ) sin( ) sin( ) sin(   , which is valid for all triangles. Going back to the figure and applying the Pythagorean Theorem to the two right triangles within the oblique triangle we have- b w C b w h c h w a 2 2 2 2 2 2 2 2 2 ) cos( , , ) (       Eliminating h and w2 from these three equations produces the Law of Cosines- ) cos( 2 2 2 2 C ab b a c    When C=/2 radians this just reduces to the Pythagorean Theorem. For an equilateral triangle we have cos(/3)=cos(60deg)=0.5. According to this law one also knows that one of the acute angles of a 3-4-5 right triangle equals =arcos(3/5)=53.130 deg. The other acute angle is =arcos{4/5)=36.869deg . As expected , +=90deg=/2 rad . We could also have gotten  using the law of sines. There =arcsin(4/5). Some less known identities involving oblique triangles are the Mollweide Formulas. Mollweide lived from 1774 to 1825 and was a professor of mathematics at the University of Leipzig. It’s the same university where my father received his PhD back in 1933. Mollweide’s formulas are straight forward to derive using the Law of Sines. If one looks at the ratios of (a+b)/c and (a-b)/c and expresses the ratios in terms of angles, one gets- ) sin( )] sin( ) [sin( ) ( ) sin( )] sin( ) [sin( ) ( C B A c b a and C B A c b a       These are the two Mollweide Formulas. They can also be rewritten in terms of half angles, but this does not affect their ability to serve as an accuracy check for the triangles associated with geodetic measurements since they contain all side lengths and angles for any oblique triangle. Let us look at an example where the side-lengths are a=5, b=3 and c=7 . By the Law of Cosines this produces – cos(A)=11/14 , cos(B)=13/14 , and cos(C)=-1/2 from the first Mollweide Formula we then have- 3 7 13 14 11 14 7 ) 3 5 ( 2 2 2 2      which checks out. Inscribing and Circumscribing Triangles by Circles: We next ask what is the radius of the smallest circle which can circumscribe an oblique triangle and also what is the radius of the largest circle which can be placed inside the same triangle. We are looking at the following picture- We notice that the two circles are constructed by two different methods. The larger circle which circumscribes the oblique triangle ABC has its center determined by the intersection of three lines each intersecting the middle of the triangle sides at right angles. The radial distance from this circle center at [xc,yc] to the triangle vertexes have a constant value of R. The triangle side ‘a’ is kept horizontal with the coordinates at A, B , and C given by [x3,y3], [0,0], and [[a,0], respectively. After some manipulations one finds- c c c y R radius circle with y y a x x y a x      3 2 3 3 3 2 ] ) ( [ , 2 To test out this result let us look at an equilateral triangle of side-length a=b=c=1. Here x3=1/2 and y3=1/[2sqrt(3)]. This produces the circumscribing circle- 2 2 2 ) 3 ( )] 3 2 / 1 ( [ )] 2 / 1 ( [     y x The circle center lies at the centroid of the equilateral triangle as expected. The distance from the centroid to any of the three corners equals 1/sqrt(3). We next look at finding the largest circle which can be inscribed in any oblique triangle. We expect this circle to be centered at the intersection of the three lines which bisect the angles A, B, and C. Using the Pythagorean Theorem, the Law of Cosines, and the Law of Sines, one finds after some calculations that the inscribed circle is given by- 2 2 2 ) ( ) ( c c c y y y x x     , with-                 ) tan( ) tan( 1 ) tan( ) tan( ) tan( 1      a y and a x c c Here  and  are the half angles of B and C , respectively. They are given explicitly as-                     ab c b a ab and ac b c a ac 2 2 2 2 2 2 2 2 1 arccos 2 2 1 arccos   For the special case of an equilateral triangle with a=b=c=1 we get xc=1/2 and yc=1/[2sqrt(3)]. That is, the circle centers at [1/2, 1/2sqrt(3)] just like for the circumscribed circle. However this time the radius has the smaller value of- 3 2 1   c y R In general the circumscribing and inscribing circle of an oblique triangle will not coincide. Tiling with Oblique Triangles: The simplest form of tiling which leaves no gaps is accomplished by using rectangular or square plates. What is less often recognized is that any tile having the shape of an oblique triangle can also be used to tile a flat floor without leaving gaps. Let us demonstrate how this is done. One begins with the oblique triangle shown in the first figure above and draws two straight lines containing sides a and b to infinity as shown- Next equally spaced parallel lines are drawn to these to form a Rhomboidal grid. Each rhomboidal element contains two identical oblique triangles formed by drawing a diagonal line from A to B. The area of each rhomboidal element is equal to- Area Rhomboid=absinC 2 2 2 2 2 ) ( ) 2 ( 2 1 c b a ab     , so that the area of each identical oblique triangle will be just half of this. Although this result looks slightly different from the area formula we derived earlier in this article, the results yield the same value. An oblique triangle with a=5, b=3, and c=7 yields an area of (15/4)sqrt(3)=6.495… . One could also break up the existing triangle ABC into two sub-triangles to produce a gapless tiling with two different oblique triangle tiles. Alternatively one could generate two different oblique triangles by cutting a 1–sqrt(3)-2 right triangle with a single line emanating at C. A resultant two tile pattern follows- Here the tiles have the shape of an isosceles triangle and an equilateral triangle. Another type of tiling reminiscent of Penrose Tiling but differing from it by having a perfectly periodic pattern is the following- Here the kite and arrow sub-elements each contain two identical oblique triangles. Using Vectors to Define Triangles: A convenient way to handle many problems involving triangles is to employ vector forms for its sides. One can define three vectors as shown in the following figure- Here the absolute values of the vectors represent the side-length of the triangle so that a=abs(V1)=\|V1\| etc. .. The cosine of a particular angle is obtainable by using the dot product- \|] \|\| /[\| ] [ ) cos( 3 1 3 1 V V V V B   For an [a,b,c]=[5,3,7] triangle, we have- 2 3 2 3 2 3 2 3 ) ( ) 5 ( 9 ) ( ) ( 49 y x and y x      This means [x3,y3]=[13/2, sqrt(29)/2]. So we get- 14 / 13 35 / ]} ) 2 / 29 ( ) 2 / 13 [( ) 5 {( ) cos(     j i i B or B=21.7867 deg. We can verify this result by using the Law of Cosines. There we have- 14 / 13 ) 35 ( 2 ) 9 49 25 ( ) cos(     B Taking the absolute value of the cross product of vectors V1 and V3 one finds the area of a rhomboid with sides \|V1\| and \|V3\|. The area of the triangle formed is just half of this value. For the above case we get- Area Triangle= 4 29 5 0 2 / 29 2 / 13 0 0 5 2 1  k j i abs =6.7314.. For an equilateral triangle of side length a=b=c=1 we find its area to be sqrt(3)/4. Some Practical Applications of Triangles: Besides the mathematical aspects of triangles discussed above, triangles play important practical roles in connection with trusses and tripods. Trusses are used in railway bridges, towers, and buildings. A typical truss is composed of interlocking triangles as the following picture of typical roof trusses shows- A typical truss consists of triangular shaped sub-elements which will not distort when subjected to loads or moments. This is unlike what would happen to a rectangular sub-sections when subjected to a rotational torque without external constraints. Triangular truss components carry only compressive loads along their beams and hence do not require very strong gusset plates to hold its ends together. Some trusses have members which carry no load such as the two short pieces in the above Howe truss shown. Such zero force pieces have no functional purpose other than to keep the truss from warping out of its plane. The tripod is another practical device involving a triangle. The essence of a tripod are three legs which come together at a point where a camera or one’s eye level is located. The other ends of the legs form an oblique triangle at ground level. Here is a drawing of an early land surveyor (such as George Washington in his late teens)- The unique property of a tripod is that location of the three leg bottoms are sufficient to define a unique plane in space no matter how rough the ground . This observation stems from the fact that it takes only three points to define any plane.. Extra legs are not needed to define a plane and may even detract for finding a single plane to, for example, keep a four legged table from rocking. Many of you have probably had the experience when dining at an outdoor cafe of having your drink spill because you happen to be sitting at a table where one of the legs is not touching the uneven ground. Under such a condition one is dealing with different ground planes defined by just three legs at a time. My typical solution is to fold a napkin several times and then place it under the shortest leg relative to a plane defined by the three other legs. Someone should invent a little mechanical ratchet attachment capable of lengthening or shorting until all four leg bottoms find themselves in the same plane no matter how rough the ground. An alternative way of having a stable table would be to have just a single leg attached to a heavy round disc base and attach to the bottom of this disc three very short support points. In my wood shop I often find myself building bookshelves, chairs and tables with four legs not precisely of the same length. To solve this problem I turn the device upside down and then use a level lying on a rigid yard stick to mark the legs at the same height above the ground. Very often one can use self-sticking felt pads to adjust the height difference without needing to saw anything. February 2016
16327	https://www.britannica.com/science/coordination-compound/Isomerism	Coordination compound - Isomerism, Ligands, Complexes \| Britannica Search Britannica Click here to search Search Britannica Click here to search SUBSCRIBE SUBSCRIBE Login SUBSCRIBE Ask the ChatbotGames & QuizzesHistory & SocietyScience & TechBiographiesAnimals & NatureGeography & TravelArts & CultureProConMoneyVideos coordination compound Introduction Coordination compounds in nature Coordination compounds in industry History of coordination compounds Characteristics of coordination compounds Coordination number Ligands and chelates Mononuclear, monodentate Polydentate Polynuclear Nomenclature Structure and bonding of coordination compounds Geometry Isomerism Cis-trans isomerism Enantiomers and diastereomers Ionization isomerism Linkage isomerism Coordination isomerism Ligand isomerism Bonding theories Valence bond theory Crystal field theory Ligand field and molecular orbital theories Principal types of complexes Aqua complexes Halo complexes Carbonyl complexes Nitrosyl complexes Cyano and isocyano complexes Organometallic complexes Isopoly and heteropoly anions Important types of reactions of coordination compounds Acid-base Substitution Lability and inertness Isomerization Oxidation-reduction Synthesis of coordination compounds References & Edit HistoryQuick Facts & Related Topics Images For Students coordination compound summary Quick Summary Ask the Chatbot a Question ScienceChemistry Isomerism incoordination compoundin Structure and bonding of coordination compounds Quick Summary Ask the Chatbot Print Cite Share Feedback External Websites Also known as: complex compound, coordinate compound, coordination complex Written by Written by George B. Kauffman Professor of Chemistry, California State University, Fresno. Author of Inorganic Coordination Compounds and many others. George B. Kauffman , Jack Halpern Louis Block Distinguished Service Professor of Chemistry, University of Chicago. Author of papers on coordination compounds and reaction mechanisms. Jack Halpern•All Fact-checked by Fact-checked by The Editors of Encyclopaedia Britannica Encyclopaedia Britannica's editors oversee subject areas in which they have extensive knowledge, whether from years of experience gained by working on that content or via study for an advanced degree.... The Editors of Encyclopaedia Britannica Article History Table of Contents Table of Contents Quick Summary Ask the Chatbot a Question Coordination compounds often exist as isomers—i.e., as compounds with the same chemical composition but different structural formulas. Many different kinds of isomerism occur among coordination compounds. The following are some of the more common types. Cis-trans isomerism Cis-trans (geometric) isomers of coordination compounds differ from one another only in the manner in which the ligands are distributed spatially; for example, in the isomeric pair of diamminedichloroplatinum compounds the two ammonia molecules and the two chlorine atoms are situated next to one another in one isomer, called the cis (Latin for “on this side”) isomer, and across from one another in the other, the trans (Latin for “on the other side”) isomer. A similar relationship exists between the cis and trans forms of the tetraamminedichlorocobalt(1+) ion: Enantiomers and diastereomers So-called optical isomers (or enantiomers) have the ability to rotate plane-polarized light in opposite directions. Enantiomers exist when the molecules of the substances are mirror images but are not superimposable upon one another. In coordination compounds, enantiomers can arise either from the presence of an asymmetric ligand, such as one isomer of the amino acid, alanine (aminopropionic acid), or from an asymmetric arrangement of the ligands. Familiar examples of the latter variety are octahedral complexes carrying three didentate ligands, such as ethylenediamine, NH 2 CH 2 CH 2 NH 2. The two enantiomers corresponding to such a complex are depicted by the structures below. The ethylenediamine ligands above are indicated by a curved line between the symbols for the nitrogen atoms. Diastereomers, on the other hand, are not superimposable and also are not mirror images. Using AB as an example of a chelating ligand, in which the symbol AB implies that the two ends of the chelate are different, there are six possible isomers of a complex cis-[M(AB)2 X 2]. For example, AB might correspond to alanine [CH 3 CH(NH 2)C(O)O]−, where both N and O are attached to the metal. Alternatively, AB could represent a ligand such as propylenenediamine, [NH 2 CH 2 C(CH 3)HNH 2], where the two ends of the molecule are distinguished by the fact that one of the Hs on a C is substituted with a methyl (CH 3) group. Ionization isomerism Certain isomeric pairs occur that differ only in that two ionic groups exchange positions within (and without) the primary coordination sphere. These are called ionization isomers and are exemplified by the two compounds, pentaamminebromocobalt sulfate, [CoBr(NH 3)5]SO 4, and pentaamminesulfatocobalt bromide, [Co(SO 4)(NH 3)5]Br. In the former the bromide ion is coordinated to the cobalt(3+) ion, and the sulfate ion is outside the coordination sphere; in the latter the sulfate ion occurs within the coordination sphere, and the bromide ion is outside it. Linkage isomerism Isomerism also results when a given ligand is joined to the central atom through different atoms of the ligand. Such isomerism is called linkage isomerism. A pair of linkage isomers are the ions [Co(NO 2)(NH 3)5]2+and [Co(ONO)(NH 3)5]2+, in which the anionic ligand is joined to the cobalt atom through nitrogen or oxygen, as shown by designating it with the formulas NO 2−(nitro) and ONO−(nitrito), respectively. Another example of this variety of isomerism is given by the pair of ions [Co(CN)5(NCS)]3− and [Co(CN)5(SCN)]3−, in which an isothiocyanate (NCS)− and a thiocyanate group (SCN)− are bonded to the cobalt(3+) ion through a nitrogen or sulfur atom, respectively. Coordination isomerism Ionic coordination compounds that contain complex cations and anions can exist as isomers if the ligands associated with the two metal atoms are exchanged, as in the pair of compounds, hexaamminecobalt(3+) hexacyanochromate(3–), [Co(NH 3)6][Cr(CN)6], and hexaamminechromium(3+) hexacyanocobaltate(3–), [Cr(NH 3)6][Co(CN)6]. Such compounds are called coordination isomers, as are the isomeric pairs obtained by redistributing the ligands between the two metal atoms, as in the doubly coordinated pair, tetraammineplatinum(2+) hexachloroplatinate(2–), [Pt(NH 3)4][PtCl 6], and tetraamminedichloroplatinum(2+) tetrachloroplatinate(2–), [PtCl 2(NH 3)4][PtCl 4]. Ligand isomerism Isomeric coordination compounds are known in which the overall isomerism results from isomerism solely within the ligand groups. An example of such isomerism is shown by the ions, bis(1,3-diaminopropane)platinum(2+) and bis(1,2-diaminopropane)platinum(2+), Bonding theories Valence bond theory Several theories currently are used to interpret bonding in coordination compounds. In the valence bond (VB) theory, proposed in large part by the American scientists Linus Pauling and John C. Slater, bonding is accounted for in terms of hybridized orbitals of the metal ion, which is assumed to possess a particular number of vacant orbitals available for coordinate bonding that equals its coordination number. (See the article chemical bonding for a discussion of the theories of chemical bonding.) Each ligand donates an electron pair to form a coordinate-covalent bond, which is formed by the overlap of an unoccupied orbital of the metal ion and a filled orbital of a ligand. The configuration of the complex depends on the type and number of orbitals involved in the hybridization—e.g., s p (linear), s p 3 (tetrahedral), d s p 2 (square planar), and d 2 s p 3 (octahedral), in which the superscripts denote the number of orbitals of a particular type. In many cases, the number of unpaired electrons, as determined by magnetic susceptibility measurements, agrees with the theoretical prediction. The theory was modified in 1952 by the Canadian-born American Nobel chemistry laureate Henry Taube, who distinguished between inner orbital complexes (d 2 s p 3) and outer orbital complexes (s p 3 d 2) to account for discrepancies between octahedral complexes. The main defect of the simple VB theory lies in its failure to include the antibonding molecular orbitals produced during complex formation. Thus, it fails to offer an explanation for the striking colours of many complexes, which arise from their selective absorption of light of only certain wavelengths. From the early 1930s through the early 1950s, VB theory was used to interpret almost all coordination phenomena, for it gave simple answers to the questions of geometry and magnetic susceptibility with which chemists of that time were concerned. Crystal field theory Considerable success in understanding certain coordination compounds also has been achieved by treating them as examples of simple ionic or electrostatic bonding. The German theoretical physicist Walther Kossel’s ionic model of 1916 was revitalized and developed by the American physicists Hans Bethe and John H. Van Vleck into the crystal field theory (CFT) of coordination, used by physicists as early as the 1930s but not generally accepted by chemists until the 1950s. This view attributes the bonding in coordination compounds to electrostatic forces between the positively charged metal ions and negatively charged ligands—or, in the case of neutral ligands (e.g., water and ammonia), to charge separations (dipoles) that appear within the molecules. Although this approach meets with considerable success for complexes of metal ions with small electronegative ligands, such as fluoride or chloride ions or water molecules, it breaks down for ligands of low polarity (charge separation), such as carbon monoxide. It also requires modification to explain why the spectral (light-absorption) and magnetic properties of coordinated metal ions generally differ from those of the free ions and why, for a given metal ion, these properties depend on the nature of the ligands. Britannica Chatbot Chatbot answers are created from Britannica articles using AI. This is a beta feature. AI answers may contain errors. Please verify important information using Britannica articles. About Britannica AI. Load Next Page Feedback Corrections? Updates? Omissions? Let us know if you have suggestions to improve this article (requires login). Feedback Type Your Feedback Submit Feedback Thank you for your feedback Our editors will review what you’ve submitted and determine whether to revise the article. print Print Please select which sections you would like to print: [x] Table Of Contents [x] Introduction [x] Coordination compounds in nature [x] Coordination compounds in industry [x] History of coordination compounds [x] Characteristics of coordination compounds [x] Structure and bonding of coordination compounds [x] Principal types of complexes [x] Important types of reactions of coordination compounds [x] Synthesis of coordination compounds verified Cite While every effort has been made to follow citation style rules, there may be some discrepancies. Please refer to the appropriate style manual or other sources if you have any questions. Select Citation Style Kauffman, George B., Halpern, Jack. "coordination compound". Encyclopedia Britannica, 3 Aug. 2018, Accessed 29 September 2025. 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16328	https://artofproblemsolving.com/wiki/index.php/2012_USAMO_Problems/Problem_4?srsltid=AfmBOooTSZVmPfAoomqBiut6RPmXH5CuGm1YOExOxue1oSdrlKw5j4uQ	Art of Problem Solving 2012 USAMO Problems/Problem 4 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 2012 USAMO Problems/Problem 4 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2012 USAMO Problems/Problem 4 Contents [hide] 1 Problem 2 Solution 3 Summary 4 See Also Problem Find all functions (where is the set of positive integers) such that for all positive integers and such that divides for all distinct positive integers , . Solution By the first condition we have and , so or and similarly for . By the second condition, we have for all positive integers . Suppose that for some we have . We claim that for all . Indeed, from Equation (1) we have , and this is only possible if ; the claim follows by induction. We now divide into cases: Case 1: This gives always from the previous claim, which is a solution. Case 2: This implies for all , but this does not satisfy the initial conditions. Indeed, we would have and so , a contradiction. Case 3:, We claim always by induction. The base cases are and . Fix and suppose that . By Equation (1) we have that This implies (otherwise ). Also we have so . This gives the solutions and . The first case is obviously impossible, so , as desired. By induction, for all . This also satisfies the requirements. Case 4: We claim by a similar induction. Again if , then by (1) we have and so . Also note that and so . Then the only possible solution is . By induction, for all , and this satisfies all requirements. In summary, there are three solutions: . Summary Three functions: since , and , fixed points on the function. No elegant argument needed; so of course ! See Also 2012 USAMO (Problems • Resources) Preceded by Problem 3Followed by Problem 5 1•2•3•4•5•6 All USAMO Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Categories: Olympiad Algebra Problems Functional Equation Problems Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
16329	https://www.shaalaa.com/concept-notes/properties-of-a-square-property-the-diagonals-of-a-square-are-perpendicular-bisectors-of-each-other_12239	Properties of a Square - Property: The diagonals of a square are perpendicular bisectors of each other. Advertisements Topics Rational Numbers Linear Equations in One Variable Understanding Quadrilaterals Data Handling Practical Geometry Squares and Square Roots Cubes and Cube Roots Comparing Quantities Algebraic Expressions and Identities Mensuration Visualizing Solid Shapes Exponents and Powers Direct and Inverse Proportions Factorization Introduction to Graphs Playing with Numbers Theorem The diagonals of a square are perpendicular bisectors of each other. Given: ABCD is a square, where AC and BD is a diagonal bisect each other at a Point 'O'. To Prove: ∠AOD = ∠COD = 90°. Proof: ABCD is a square whose diagonals meet at O. ......(Given) OA = OC. ......(Since the square is a parallelogram)(1) In ΔAOD and ∆COD, OD = OD .........(Common side) OA = OC .........(From 1) AD = DC ..........(All the sides of square have equal length.) By SSS congruency condition, ∆AOD ≅ ∆COD Therefore, m∠ AOD = m∠ COD ......(C.A.C.T.) Since, m∠ AOD and m∠ COD are a linear pair, ∠AOD = ∠COD = 90°. Hence Proved. Shaalaa.com \| The diagonals of a square are perpendicular bisectors of each other. Shaalaa.com Next video Shaalaa.com Series: Property: The diagonals of a square are perpendicular bisectors of each other. Related QuestionsVIEW ALL In a parallelogram PQRS, T is any point on the diagonal PR. If the area of ΔPTQ is 18 square units find the area of ΔPTS. All rectangles are squares In a parallelogram PQRS, M and N are the midpoints of the sides PQ and PS respectively. If area of ΔPMN is 20 square units, find the area of the parallelogram PQRS. In square PQRS : (i) if PQ = 3x – 7 and QR = x + 3 ; find PS (ii) if PR = 5x and QS = 9x – 8. Find QS Prove that the bisectors of the interior angles of a rectangle form a square. Prove that the quadrilateral formed by joining the mid-points of a square is also a square. Identify all the quadrilaterals that have four right angles Explain how a square is a rhombus. Related concepts Advertisements Submit content Select a course
16330	https://www.timeanddate.com/news/time/russia-reintroduces-dst-2014.html	Will Russia revert to “winter time” in 2014? Published 15-Jan-2014. Changed 21-Jan-2014 For a third year in a row, there are reports that Russia will abandon permanent "summer time". Numerous Russian media outlets quote leading political figures saying that a decision to revert to standard time or re-introduce Daylight Saving Time (DST) has already been made but that its implementation will have to wait until after the Sochi Winter Olympics. Update June 11, 2014: The State Duma passed the bill on Tuesday, June 10, 2014. It is yet to be signed into law. Update January 21, 2014: The bill to change the current time zone rules in Russia has just been submitted to the State Duma. If adopted, clocks in Russia will be turned back to standard time (also known as “winter time”) on October 26, 2014 and no daylight saving time will be applied in future years. Time zone boundaries may also be changed under the bill to ensure that local times reflect solar time as much as possible. In addition, Moscow time may be changed from the current UTC+4 to UTC+3, and a tenth time zone may be re-introduced. The Grand Kremlin Palace in Moscow, Russia. ©thinkstockphoto.com Similar rumours had surfaced in 2012 and 2013, but no clock changes occurred then. DST may return after Olympics “The decision has been made. After the Olympics we will change to daylight savings time”, the Moscow Times cites Liberal Democratic Party chief, Vladimir Zhirinovsky. According to the head of the lower house healthcare committee, Sergey Kalashnikov, the bill will be submitted to the Russian parliament on January 20, 2014. According to Russia Today, Kalashnikov announced that the prepared draft would also “allow authorities in all of Russia's regions to decide what time zone they would prefer to live in.” Permanent “summer time” unpopular According to a survey by the Russia Public Opinion Research Center (VTSIOM), 43 percent of the Russian population support the re-introduction of Daylight Saving Time switches. Seasonal time change was abolished in 2011 by President Dmitry Medvedev. Time Zones Worldwide Time Zone Map Current local times around the world, including (DST) changes. Time Zone Tools Elsewhere on timeanddate.com Why Auroras Are About to Surge The northern and southern lights may be about to put on their best show in years. Here’s what makes this aurora season special for skywatchers. Skywatching Tips for September 2025 What’s up in the day and night sky in September 2025, including a Blood Moon and a partial lunar eclipse. DST 2025: Your Guide to the Clock Change Find out when Daylight Saving Time 2025 starts and ends, plus history, facts, and global observance details in this essential seasonal guide. What Is Daylight Saving Time (DST) About 40% of countries use Daylight Saving Time. Learn how it works, what makes it controversial, and why it’s not actually saving any daylight. Love Our Site? Become a Supporter © Time and Date AS 1995–2025 Company Legal Services Sites Follow Us © Time and Date AS 1995–2025. Privacy & Terms
16331	https://ocw.mit.edu/courses/18-01sc-single-variable-calculus-fall-2010/resources/clip-2-slope-of-tangent-to-circle-implicit/	Clip 2: Slope of Tangent to Circle: Implicit \| Single Variable Calculus \| Mathematics \| MIT OpenCourseWare Browse Course Material Syllabus 1. Differentiation Part A: Definition and Basic Rules Part B: Implicit Differentiation and Inverse Functions Exam 1 2. Applications of Differentiation Part A: Approximation and Curve Sketching Part B: Optimization, Related Rates and Newton's Method Part C: Mean Value Theorem, Antiderivatives and Differential Equa Exam 2 3. The Definite Integral and its Applications Part A: Definition of the Definite Integral and First Fundamental Part B: Second Fundamental Theorem, Areas, Volumes Part C: Average Value, Probability and Numerical Integration Exam 3 4. Techniques of Integration Part A: Trigonometric Powers, Trigonometric Substitution and Com Part B: Partial Fractions, Integration by Parts, Arc Length, and Part C: Parametric Equations and Polar Coordinates Exam 4 5. Exploring the Infinite Part A: L'Hospital's Rule and Improper Integrals Part B: Taylor Series Final Exam Course Info Instructor Prof. David Jerison Departments Mathematics As Taught In Fall 2010 Level Undergraduate Topics Mathematics Calculus Differential Equations Learning Resource Types grading Exams with Solutions notes Lecture Notes theaters Lecture Videos assignment_turned_in Problem Sets with Solutions laptop_windows Simulations theaters Problem-solving Videos Download Course menu search Give Now About OCW Help & Faqs Contact Us searchGIVE NOWabout ocwhelp & faqscontact us 18.01SC \| Fall 2010 \| Undergraduate Single Variable Calculus Menu More Info Syllabus 1. Differentiation Part A: Definition and Basic Rules Part B: Implicit Differentiation and Inverse Functions Exam 1 2. Applications of Differentiation Part A: Approximation and Curve Sketching Part B: Optimization, Related Rates and Newton's Method Part C: Mean Value Theorem, Antiderivatives and Differential Equa Exam 2 3. The Definite Integral and its Applications Part A: Definition of the Definite Integral and First Fundamental Part B: Second Fundamental Theorem, Areas, Volumes Part C: Average Value, Probability and Numerical Integration Exam 3 4. Techniques of Integration Part A: Trigonometric Powers, Trigonometric Substitution and Com Part B: Partial Fractions, Integration by Parts, Arc Length, and Part C: Parametric Equations and Polar Coordinates Exam 4 5. Exploring the Infinite Part A: L'Hospital's Rule and Improper Integrals Part B: Taylor Series Final Exam Session 14: Examples of Implicit Differentiation Clip 2: Slope of Tangent to Circle: Implicit » Accompanying Notes (PDF) From Lecture 5 of 18.01 Single Variable Calculus, Fall 2006 Video Player is loading. Play Video Play Mute Current Time 0:00 / Duration 49:01 Loaded: 0% 13:53 Stream Type LIVE Seek to live, currently behind live LIVE Remaining Time-35:08 1x Playback Rate Chapters Chapters Descriptions descriptions off, selected Captions captions settings, opens captions settings dialog captions off, selected English Captions Audio Track Picture-in-Picture download button Download video Fullscreen playback speed Playback speed 0.25 0.5 0.75 1, selected 1.25 1.5 This is a modal window. Beginning of dialog window. Escape will cancel and close the window. 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Transcript Download video Download transcript 0:00 The following content was created under a Creative 0:02 Commons License. 0:03 Your support will help MIT OpenCourseWare 0:05 continue to offer high quality educational resources for free. 0:09 To make a donation or to view additional materials 0:12 from hundreds of MIT courses, visit MIT OpenCourseWare 0:16 at ocw.mit.edu. 0:19 Professor: So, we're ready to begin the fifth lecture. 0:24 I'm glad to be back. 0:25 Thank you for entertaining my colleague, Haynes Miller. 0:33 So, today we're going to continue 0:35 where he started, namely what he talked about was the chain 0:40 rule, which is probably the most powerful technique 0:43 for extending the kinds of functions 0:45 that you can differentiate. 0:47 And we're going to use the chain rule in some rather clever 0:50 algebraic ways today. 0:54 So the topic for today is what's known 0:57 as implicit differentiation. 1:10 So implicit differentiation is a technique 1:15 that allows you to differentiate a lot of functions you didn't 1:17 even know how to find before. 1:20 And it's a technique - let's wait for a few people 1:24 to sit down here. 1:28 Physics, huh? 1:29 Okay, more Physics. 1:36 Let's take a break. 1:37 You can get those after class. 1:40 All right, so we're talking about implicit differentiation, 1:46 and I'm going to illustrate it by several examples. 1:53 So this is one of the most important and basic formulas 1:57 that we've already covered part way. 1:59 Namely, the derivative of x to a power is ax^(a-1). 2:06 Now, what we've got so far is the exponents, 0, plus or minus 2:15 1, plus or minus 2, etc. 2:19 You did the positive integer powers in the first lecture, 2:24 and then yesterday Professor Miller 2:30 told you about the negative powers. 2:32 So what we're going to do right now, 2:35 today, is we're going to consider 2:39 the exponents which are rational numbers, ratios of integers. 2:44 So a is m/n. 2:46 m and n are integers. 2:53 All right, so that's our goal for right now, 2:55 and we're going to use this method 2:56 of implicit differentiation. 2:58 In particular, it's important to realize that this 3:01 covers the case m = 1. 3:03 And those are the nth roots. 3:04 So when we take the one over n power, 3:07 we're going to cover that right now, 3:09 along with many other examples. 3:13 So this is our first example. 3:16 So how do we get started? 3:17 Well we just write down a formula for the function. 3:20 The function is y = x^(m/n). 3:24 That's what we're trying to deal with. 3:26 And now there's really only two steps. 3:30 The first step is to take this equation to the nth power, 3:38 so write it y^n = x^m. 3:42 Alright, so that's just the same equation re-written. 3:46 And now, what we're going to do is 3:50 we're going to differentiate. 3:52 So we're going to apply d/dx to the equation. 4:01 Now why is it that we can apply it to the second equation, not 4:05 the first equation? 4:06 So maybe I should call these equation 1 and equation 2. 4:10 So, the point is, we can apply it to equation 2. 4:13 Now, the reason is that we don't know how to differentiate 4:17 x^(m/n). 4:18 That's something we just don't know yet. 4:21 But we do know how to differentiate integer powers. 4:24 Those are the things that we took care of before. 4:29 So now we're in shape to be able to do the differentiation. 4:32 So I'm going to write it out explicitly 4:34 over here, without carrying it out just yet. 4:37 That's d/dx of y^n = d/dx of x^m. 4:46 And now you see this expression here 4:51 requires us to do something we couldn't do before yesterday. 4:55 Namely, this y is a function of x. 4:58 So we have to apply the chain rule here. 5:01 So this is the same as - this is by the chain rule now - 5:06 d/dy of y^n times dy/dx. 5:13 And then, on the right hand side, we can just carry it out. 5:15 We know the formula. 5:17 It's mx^(m-1). 5:21 Right, now this is our scheme. 5:24 And you'll see in a minute why we win with this. 5:29 So, first of all, there are two factors here. 5:32 One of them is unknown. 5:33 In fact, it's what we're looking for. 5:35 But the other one is going to be a known quantity, 5:38 because we know how to differentiate y 5:40 to the n with respect to y. 5:42 That's the same formula, although the letter 5:44 has been changed. 5:46 And so this is the same as - I'll write it underneath here - 5:53 n y^(n-1) dy/dx = m x^(m-1). 6:07 Okay, now comes, if you like, the non-calculus part 6:14 of the problem. 6:15 Remember the non-calculus part of the problem 6:17 is always the messier part of the problem. 6:20 So we want to figure out this formula. 6:22 This formula, the answer over here, 6:25 which maybe I'll put in a box now, 6:29 has this expressed much more simply, only in terms of x. 6:33 And what we have to do now is just solve for dy/dx 6:36 using algebra, and then solve all the way in terms of x. 6:39 So, first of all, we solve for dy/dx. 6:44 So I do that by dividing the factor on the left-hand side. 6:47 So I get here mx^(m-1) divided by ny^(n-1). 6:56 And now I'm going to plug in-- so I'll write this as m/n. 7:02 This is x^(m-1). 7:04 Now over here I'm going to put in for y, x^(m/n) times n-1. 7:15 So now we're almost done, but unfortunately we 7:18 have this mess of exponents that we have to work out. 7:22 I'm going to write it one more time. 7:25 So I already recognize the factor a out front. 7:28 That's not going to be a problem for me, 7:30 and that's what I'm aiming for here. 7:31 But now I have to encode all of these powers, 7:34 so let's just write it. 7:36 It's m-1, and then it's minus the quantity (n-1) m/n. 7:46 All right, so that's the law of exponents applied to this ratio 7:50 here. 7:50 And then we'll do the arithmetic over here on the next board. 7:58 So we have here m - 1 - (n-1) m/n = m - 1. 8:08 And if I multiply n by this, I get -m. 8:12 And if the second factor is minus minus, that's a plus. 8:15 And that's +m/n. 8:18 Altogether the two m's cancel. 8:21 I have here -1 + m/n. 8:23 And lo and behold that's the same thing as a - 1, 8:27 just what we wanted. 8:29 All right, so this equals a x^(n-1). 8:31 Okay, again just a bunch of arithmetic. 8:39 From this point forward, from this substitution 8:42 on, it's just the arithmetic of exponents. 8:51 All right, so we've done our first example here. 8:58 I want to give you a couple more examples, 9:00 so let's just continue. 9:04 The next example I'll keep relatively simple. 9:08 So we have example two, which is going to be the function x^2 + 9:15 y^2 = 1. 9:18 Well, that's not really a function. 9:21 It's a way of defining y as a function of x implicitly. 9:29 There's the idea that I could solve for y if I wanted to. 9:34 And indeed let's do that. 9:36 So if you solve for y here, what happens is you get y^2 = 1 - 9:42 x^2, and y is equal to plus or minus the square root of 1 - 9:47 x^2. 9:52 So this, if you like, is the implicit definition. 9:58 And here is the explicit function y, 10:00 which is a function of x. 10:04 And now just for my own convenience, 10:06 I'm just going to take the positive branch. 10:09 This is the function. 10:13 It's just really a circle in disguise. 10:15 And I'm just going to take the top part of the circle, 10:19 so we'll take that top hump here. 10:24 All right, so that means I'm erasing this minus sign. 10:27 I'm just taking the positive branch, just 10:35 for my convenience. 10:36 I could do it just as well with the negative branch. 10:40 Alright, so now I've taken the solution, 10:46 and I can differentiate with this. 10:49 So rather than using the dy/dx notation over here, 10:53 I'm going to switch notations over here, 10:55 because it's less writing. 10:56 I'm going to write y' and change notations. 10:59 Okay, so I want to take the derivative of this. 11:04 Well this is a somewhat complicated function here. 11:11 It's the square root of 1 - x^2, and the right way always 11:15 to look at functions like this is to rewrite them using 11:21 the fractional power notation. 11:26 That's the first step in computing 11:28 a derivative of a square root. 11:32 And then the second step here is what? 11:38 Does somebody want to tell me? 11:40 Chain rule, right. 11:43 That's it. 11:44 So we have two things. 11:45 We start with one, and then we do something else to it. 11:47 So whenever we do two things to something, 11:50 we need to apply the chain rule. 11:52 So 1 - x^2, square root. 11:55 All right, so how do we do that? 11:57 Well, the first factor I claim is 11:58 the derivative of this thing. 12:01 So this is 1/2 blah to the -1/2. 12:06 So I'm doing this kind of by the advanced method 12:09 now, because we've already graduated. 12:11 You already did the chain rule last time. 12:14 So what does this mean? 12:15 This is an abbreviation for the derivative with respect 12:20 to blah of blah ^ 1/2, whatever it is. 12:27 All right, so that's the first factor that we're going to use. 12:30 Rather than actually write out a variable for it 12:34 and pass through as I did previously 12:36 with this y and x variable here, I'm 12:39 just going to skip that step and let 12:41 you imagine it as being a placeholder for that variable 12:45 here. 12:45 So this variable is now parenthesis. 12:48 And then I have to multiply that by the rate of change of what's 12:52 inside with respect to x. 12:55 And that is going to be -2x. 12:58 The derivative of 1 - x^2 is -2x. 13:02 And now again, we couldn't have done this example two 13:09 before example one, because we needed 13:11 to know that the power rule worked not just for a integer 13:17 but also for a = 1/2. 13:19 We're using the case a = 1/2 right here. 13:22 It's 1/2 times, and this -1/2 here is a-1. - 13:29 So this is the case a = 1/2. 13:33 a-1 happens to be -1/2. 13:39 Okay, so I'm putting all those things together. 13:41 And you know within a week you have 13:44 to be doing this very automatically. 13:45 So we're going to do it at this speed now. 13:47 You want to do it even faster, ultimately. 13:49 Yes? 13:50 Student: [INAUDIBLE] 13:53 Professor: The question is could I have done it implicitly 13:56 without the square roots. 13:58 And the answer is yes. 13:59 That's what I'm about to do. 14:02 So this is an illustration of what's 14:04 called the explicit solution. 14:07 So this guy is what's called explicit. 14:13 And I want to contrast it with the method 14:17 that we're going to now use today. 14:18 So it involves a lot of complications. 14:20 It involves the chain rule. 14:21 And as we'll see it can get messier and messier. 14:23 And then there's the implicit method, 14:27 which I claim is easier. 14:29 So let's see what happens if you do it implicitly 14:36 The implicit method involves, instead of writing 14:41 the function in this relatively complicated way, 14:43 with the square root, it involves leaving it alone. 14:47 Don't do anything to it. 14:50 In this previous case, we were left with something which was 14:52 complicated, say x^(1/3) or x^(1/2) or something 14:56 complicated. 14:57 We had to simplify it. 14:59 We had an equation one, which was more complicated. 15:01 We simplified it then differentiated it. 15:03 And so that was a simpler case. 15:05 Well here, the simplest thing us to differentiate 15:09 is the one we started with, because squares are practically 15:13 the easiest thing after first powers, or maybe zeroth powers 15:16 to differentiate. 15:18 So we're leaving it alone. 15:19 This is the simplest possible form for it, 15:21 and now we're going to differentiate. 15:23 So what happens? 15:24 So again what's the method? 15:26 Let me remind you. 15:27 You're applying d/dx to the equation. 15:30 So you have to differentiate the left side of the equation, 15:33 and differentiate the right side of the equation. 15:35 So it's this, and what you get is 2x + 2yy' is equal to what? 15:51 0. 15:52 The derivative of 1 0. 15:56 So this is the chain rule again. 15:58 I did it a different way. 16:00 I'm trying to get you used to many different notations 16:02 at once. 16:04 Well really just two. 16:05 Just the prime notation and the dy/dx notation. 16:10 And this is what I get. 16:14 So now all I have to do is solve for y'. 16:19 So that y', if I put the 2x on the other side, is -2x, 16:24 and then divide by 2y, which is -x/y. 16:30 So let's compare our solutions, and I'll apologize, 16:34 I'm going to have to erase something to do that. 16:39 So let's compare our two solutions. 16:44 I'm going to put this underneath and simplify. 16:46 So what was our solution over here? 16:48 It was 1/2(1-x^2)^(-1/2) (-2x). 16:56 That was what we got over here. 17:02 And that is the same thing, if I cancel the 2's, and I change it 17:06 back to looking like a square root, 17:08 that's the same thing as -x divided by square root of 1 - 17:11 x^2. 17:13 So this is the formula for the derivative 17:18 when I do it the explicit way. 17:21 And I'll just compare them, these two expressions here. 17:29 And notice they are the same. 17:32 They're the same, because y is equal to square root of 1 - 17:37 x^2. 17:40 Yeah? 17:40 Question? 17:41 Student: [INAUDIBLE] 17:46 Professor: The question is why did the implicit method 17:48 not give the bottom half of the circle? 17:50 Very good question. 17:53 The answer to that is that it did. 17:57 I just didn't mention it. 17:59 Wait, I'll explain. 18:00 So suppose I stuck in a minus sign here. 18:05 I would have gotten this with the difference, so 18:08 with an extra minus sign. 18:10 But then when I compared it to what was over there, 18:12 I would have had to have another different minus sign over here. 18:15 So actually both places would get an extra minus sign. 18:19 And they would still coincide. 18:20 So actually the implicit method is a little better. 18:22 It doesn't even notice the difference 18:23 between the branches. 18:24 It does the job on both the top and bottom half. 18:28 Another way of saying that is that you're 18:31 calculating the slopes here. 18:33 So let's look at this picture. 18:35 Here's a slope. 18:36 Let's just take a look at a positive value 18:39 of x and just check the sign to see what's happening. 18:42 If you take a positive value of x over here, x is positive. 18:46 This denominator is positive. 18:48 The slope is negative. 18:49 You can see that it's tilting down. 18:52 So it's okay. 18:53 Now on the bottom side, it's going to be tilting up. 18:59 And similarly what's happening up here 19:01 is that both x and y are positive, and this x and this y 19:05 are positive. 19:06 And the slope is negative. 19:07 On the other hand, on the bottom side, x is still positive, 19:10 but y is negative. 19:11 And it's tilting up because the denominator is negative. 19:15 The numerator is positive, and this minus sign 19:17 has a positive slope. 19:19 So it matches perfectly in every category. 19:23 This complicated, however, and it's easier 19:26 just to keep track of one branch at a time, 19:30 even in advanced math. 19:32 Okay, so we only do it one branch at a time. 19:37 Other questions? 19:43 Okay, so now I want to give you a slightly more 19:47 complicated example here. 19:49 And indeed some of the-- so here's 19:52 a little more complicated example. 19:54 It's not going to be the most complicated example, 19:56 but you know it'll be a little tricky. 20:17 So this example, I'm going to give you a fourth order 20:22 equation. 20:23 So y^4 + xy^2 - 2 = 0. 20:31 Now it just so happens that there's 20:35 a trick to solving this equation, 20:38 so actually you can do both the explicit method 20:41 and the non-explicit method. 20:46 So the explicit method would say okay well, 20:50 I want to solve for this. 20:51 So I'm going to use the quadratic formula, but on y^2. 20:55 This is quadratic in y^2, because there's a fourth power 20:59 and a second power, and the first and third powers are 21:02 missing. 21:03 So this is y^2 is equal to -x plus or minus the square root 21:09 of x^2 - 4(-2) divided by 2. 21:19 And so this x is the b. 21:22 This -2 is the c, and a = 1 in the quadratic formula. 21:29 And so the formula for y is plus or minus the square root of -x 21:37 plus or minus the square root x^2 + 8 divided by 2. 21:45 So now you can see this problem of branches, 21:47 this happens actually in a lot of cases, 21:50 coming up in an elaborate way. 21:53 You have two choices for the sign here. 21:54 You have two choices for the sign here. 21:56 Conceivably as many as four roots for this equation, 21:59 because it's a fourth degree equation. 22:02 It's quite a mess. 22:02 You should have to check each branch separately. 22:06 And this really is that level of complexity, 22:09 and in general it's very difficult 22:11 to figure out the formulas for quartic equations. 22:17 But fortunately we're never going to use them. 22:21 That is, we're never going to need those formulas. 22:24 So the implicit method is far easier. 22:31 The implicit method just says okay I'll 22:35 leave the equation in its simplest form. 22:38 And now differentiate. 22:40 So when I differentiate, I get 4y^3 y' plus - 22:47 now here I have to apply the product rule. 22:50 So I differentiate the x and the y^2 separately. 22:56 First I differentiate with respect to x, so I get y^2. 22:59 Then I differentiate with respect to the other factor, 23:03 the y^2 factor. 23:04 And I get x(2 y y'). 23:08 And then the 0 gives me 0. 23:10 So minus 0 equals 0. 23:16 So there's the implicit differentiation step. 23:21 And now I just want to solve for y'. 23:26 So I'm going to factor out 4y^3 + 2xy. 23:32 That's the factor on y'. 23:35 And I'm going to put the y^2 on the other side. 23:39-y^2 over here. 23:43 And so the formula for y' is -y^2 divided by 4y^3 + 2xy. 23:55 So that's the formula for the solution. 24:01 For the slope. 24:06 You have a question? 24:07 Student: [INAUDIBLE] 24:16 Professor: So the question is for the y would 24:18 we have to put in what solved for in the explicit equation. 24:22 And the answer is absolutely yes. 24:24 That's exactly the point. 24:25 So this is not a complete solution to a problem. 24:30 We started with an implicit equation. 24:32 We differentiated. 24:33 And we got in the end, also an implicit equation. 24:36 It doesn't tell us what y is as a function of x. 24:39 You have to go back to this formula 24:43 to get the formula for x. 24:45 So for example, let me give you an example here. 24:49 So this hides a degree of complexity of the problem. 24:54 But it's a degree of complexity that we must live with. 24:58 So for example, at x = 1, you can see that y = 1 solves. 25:10 That happens to be-- solves y^4 + xy^2 - 2 = 0. 25:16 That's why I picked the 2 actually, 25:18 so it would be 1 + 1 - 2 = 0. 25:21 I just wanted to have a convenient solution there 25:23 to pull out of my hat at this point. 25:25 So I did that. 25:26 And so we now know that when x = 1, y = 1. 25:30 So at (1, 1) along the curve, the slope is equal to what? 25:41 Well, I have to plug in here, -1^2 / (41^3 + 211). 25:52 That's just plugging in that formula over there, 25:54 which turns out to be -1/6. 25:59 So I can get it. 26:00 On the other hand, at say x = 2, we're 26:13 stuck using this formula star here to find y. 26:32 Now, so let me just make two points 26:37 about this, which are just philosophical points for you 26:40 right now. 26:42 The first is, when I promised you 26:45 at the beginning of this class that we 26:47 were going to be able to differentiate 26:48 any function you know, I meant it very literally. 26:53 What I meant is if you know the function, 26:56 we'll be able give a formula for the derivative. 26:58 If you don't know how to find a function, 27:00 you'll have a lot of trouble finding the derivative. 27:02 So we didn't make any promises that if you 27:05 can't find the function you will be 27:06 able to find the derivative by some magic. 27:09 That will never happen. 27:10 And however complex the function is, 27:12 a root of a fourth degree polynomial 27:16 can be pretty complicated function of the coefficients, 27:20 we're stuck with this degree of complexity in the problem. 27:23 But the big advantage of his method, notice, 27:27 is that although we've had to find star, 27:29 we had to find this formula star, 27:31 and there are many other ways of doing these things numerically, 27:34 by the way, which we'll learn later, 27:36 so there's a good method for doing it numerically. 27:39 Although we had to find star, we never had to differentiate it. 27:43 We had a fast way of getting the slope. 27:46 So we had to know what x and y were. 27:48 But y' we got by an algebraic formula, 27:50 in terms of the values here. 27:54 So this is very fast, forgetting the slope, 27:57 once you know the point. yes? 28:02 Student: What's in the parentheses? 28:03 Professor: Sorry, this is-- Well let's see if I can manage this. 28:06 Is this the parentheses you're talking about? 28:16 Ah, "say". 28:17 That says "say". 28:17 Well, so maybe I should put commas around it. 28:19 But it was S A Y, comma comma, okay? 28:24 Well here was at x = 1. 28:28 I'm just throwing out a value. 28:33 Any other value. 28:34 Actually there is one value, my favorite value. 28:36 Well this is easy to evaluate right? x = 0, 28:39 I can do it there. 28:42 That's maybe the only one. 28:45 The others are a nuisance. 28:55 All right, other questions? 29:03 Now we have to do something more here. 29:06 So I claimed to you that we could differentiate 29:10 all the functions we know. 29:11 But really we can learn a tremendous 29:13 about functions which are really hard to get at. 29:17 So this implicit differentiation method 29:20 has one very, very important application 29:30 to finding inverse functions, or finding derivatives 29:38 of inverse functions. 29:40 So let's talk about that next. 29:51 So first, maybe we'll just illustrate by an example. 29:55 If you have the function y is equal to square root x, 29:59 for x positive, then of course this idea 30:04 is that we should simplify this equation 30:07 and we should square it so we get this somewhat simpler 30:10 equation here. 30:11 And then we have a notation for this. 30:14 If we call f(x) equal to square root of x, and g(y) = x, 30:21 this is the reversal of this. 30:25 Then the formula for g(y) is that it should be y^2. 30:33 And in general, if we start with any old y = f(x), 30:48 and we just write down, this is the defining relationship 30:52 for a function g, the property that we're saying is that 30:57 g(f(x)) has got to bring us back to x. 31:01 And we write that in a couple of different ways. 31:04 We call g the inverse of f. 31:08 And also we call f the inverse of g, 31:13 although I'm going to be silent about which variable 31:15 I want to use, because people mix them up a little bit, 31:19 as we'll be doing when we draw some pictures of this. 31:31 So let's see. 31:32 Let's draw pictures of both f and f inverse 31:42 on the same graph. 31:50 So first of all, I'm going to draw the graph of f(x) 32:02= square root of x. 32:06 That's some shape like this. 32:11 And now, in order to understand what g(y) is, 32:16 so let's do the analysis in general, 32:20 but then we'll draw it in this particular case. 32:23 If you have g(y) = x, that's really 32:31 just the same equation right? 32:34 This is the equation g(y) = x, that's y^2 = x. 32:37 This is y = square root of x, those are the same equations, 32:40 it's the same curve. 32:43 But suppose now that we wanted to write down what g(x) is. 32:49 In other words, we wanted to switch the variables, 32:51 so draw them as I said on the same graph with the same x, 32:55 and the same y axes. 32:59 Then that would be, in effect, trading the roles of x and y. 33:04 We have to rename every point on the graph which 33:07 is the ordered pair (x, y), and trade it for the opposite one. 33:12 And when you exchange x and y, so 33:15 to do this, exchange x and y, and when 33:23 you do that, graphically what that looks 33:27 like is the following: suppose you have a place here, 33:30 and this is the x and this is the y, 33:33 then you want to trade them. 33:35 So you want the y here right? 33:39 And the x up there. 33:41 It's sort of the opposite place over there. 33:44 And that is the place which is directly opposite this point 33:51 across the diagonal line x = y. 33:55 So you reflect across this or you flip across that. 33:58 You get this other shape that looks like that. 34:01 Maybe I'll draw it with a colored piece of chalk here. 34:10 So this guy here is y = f^(-1)(x). 34:24 And indeed, if you look at these graphs, 34:26 this one is the square root. 34:27 This one happens to be y = x^2. 34:34 If you take this one, and you turn it, 34:36 you reverse the roles of the x axis and the y axis, 34:39 and tilt it on its side. 34:43 So that's the picture of what an inverse function is, and now I 34:51 want to show you that the method of implicit differentiation 34:56 allows us to compute the derivatives 34:59 of inverse functions. 35:03 So let me just say it in general, 35:05 and then I'll carry it out in particular. 35:07 So implicit differentiation allows 35:16 us to find the derivative of any inverse function, 35:32 provided we know the derivative of the function. 35:53 So let's do that for what is an example, which 35:58 is truly complicated and a little subtle here. 36:02 It has a very pretty answer. 36:04 So we'll carry out an example here, 36:09 which is the function y is equal to the inverse tangent. 36:19 So again, for the inverse tangent 36:25 all of the things that we're going to do 36:30 are going to be based on simplifying 36:32 this equation by taking the tangent of both sides. 36:36 So, us let me remind you by the way, 36:38 the inverse tangent is what's also known as arctangent. 36:41 That's just another notation for the same thing. 36:45 And what we're going to use to describe 36:49 this function is the equation tan y = x. 36:55 That's what happens when you take 36:56 the tangent of this function. 36:59 This is how we're going to figure out 37:01 what the function looks like. 37:19 So first of all, I want to draw it, 37:23 and then we'll do the computation. 37:26 So let's make the diagram first. 37:32 So I want to do something which is 37:33 analogous to what I did over here with the square root 37:35 function. 37:38 So first of all, I remind you that the tangent function 37:43 is defined between two values here, which are pi/2 and -pi/2. 37:52 And it starts out at minus infinity 37:55 and curves up like this. 37:58 So that's the function tan x. 38:08 And so the one that we have to sketch 38:11 is this one which we get by reflecting this 38:14 across the axis. 38:21 Well not the axis, the diagonal. 38:25 This slope by the way, should be less - a little lower here so 38:33 that we can have it going down and up. 38:37 So let me show you what it looks like. 38:42 On the front, it's going to look a lot like this one. 38:44 So this one had curved down, and so the reflection 38:50 across the diagonal curved up. 38:52 Here this is curving up, so the reflection 38:54 is going to curve down. 38:56 It's going to look like this. 38:58 Maybe I should, sorry, let's use a different color, 39:02 because it's reversed from before. 39:04 I'll just call it green. 39:10 Now, the original curve in the first quadrant 39:15 eventually had an asymptote which was straight up. 39:17 So this one is going to have an asymptote which is horizontal. 39:24 And that level is what? 39:27 What's the highest? 39:29 It is just pi/2. 39:33 Now similarly, the other way, we're 39:35 going to do this: and this bottom level 39:40 is going to be -pi/2. 39:42 So there's the picture of this function. 39:47 It's defined for all x. 39:50 So this green guy is y = arctan x. 39:57 And it's defined all the way from minus infinity 39:59 to infinity. 40:05 And to use a notation that we had from limit notation 40:11 as x goes to infinity, let's say, x is equal to pi/2. 40:21 That's an example of one value that's of interest in addition 40:24 to the finite values. 40:28 Okay, so now the first ingredient 40:31 that we're going to need, is we're 40:34 going to need the derivative of the tangent function. 40:37 So I'm going to recall for you, and maybe you 40:40 haven't worked this out yet, but I hope that many of you have, 40:43 that if you take the derivative with respect to y of tan y. 40:48 So this you do by the quotient rule. 40:55 So this is of the form u/v, right? 40:59 You use the quotient rule. 41:00 So I'm going to get this. 41:06 But what you get in the end is some marvelous simplification 41:09 that comes out to cos^2 y. 41:12 1 over cosine squared. 41:14 You can recognize the cosine squared from the fact that you 41:17 should get v^2 in the denominator, 41:19 and somehow the numerators all cancel and simplifies to 1. 41:26 This is also known as secant squared y. 41:32 So that something that if you haven't done yet, 41:38 you're going to have to do this as an exercise. 41:48 So we need that ingredient, and now we're 41:50 just going to differentiate our equation. 41:59 And what do we get? 42:00 We get, again, (d/dy tan y) times dy/dx is equal to 1. 42:15 Or, if you like, 1 / cos^2 y times, in the other notation, 42:22 y', is equal to 1. 42:30 So I've just used the formulas that I just wrote down there. 42:35 Now all I have to do is solve for y'. 42:37 It's cos^2 y. 42:44 Unfortunately, this is not the form 42:47 that we ever want to leave these things in. 42:49 This is the same problem we had with that ugly square root 42:52 expression, or with any of the others. 42:54 We want to rewrite in terms of x. 42:58 Our original question was what is d/dx of arctan x. 43:05 Now so far we have the following answer to that question: 43:08 it's cos^2 (arctan x). 43:15 Now this is a correct answer, but way too complicated. 43:31 Now that doesn't mean that if you 43:33 took a random collection of functions, 43:35 you wouldn't end up with something this complicated. 43:37 But these particular functions, these beautiful circular 43:41 functions involved with trigonometry all 43:42 have very nice formulas associated with them. 43:45 And this simplifies tremendously. 43:48 So one of the skills that you need 43:50 to develop when you're dealing with trig functions 43:54 is to simplify this. 43:56 And so let's see now that expressions like this all 44:03 simplify. 44:07 So here we go. 44:10 There's only one formula, one ingredient 44:12 that we need to use to do this, and then we're 44:14 going to draw a diagram. 44:15 So the ingredient again, is the original defining relationship 44:18 that tan y = x. 44:22 So tan y = x can be encoded in a right triangle 44:27 in the following way: here's the right triangle and tan 44:34 y means that y should be represented as an angle. 44:38 And then, its tangent is the ratio 44:40 of this vertical to this horizontal side. 44:43 So I'm just going to pick two values that work, 44:46 namely x and 1. 44:48 Those are the simplest ones. 44:51 So I've encoded this equation in this picture. 44:57 And now all I have to do is figure out what the cosine of y 45:01 is in this right triangle here. 45:03 In order to do that, I need to figure out what the hypotenuse 45:06 is, but that's just square root of 1 + x^2. 45:13 And now I can read off what the cosine of y is. 45:18 So the cosine of y is 1 divided by the hypotenuse. 45:23 So it's 1 over square root, whoops, yeah, 1 + x^2. 45:32 And so cosine squared is just 1 / 1 + x^2. 45:39 And so our answer over here, the preferred answer which is way 45:43 simpler than what I wrote up there, 45:45 is that d/dx of tan inverse x is equal to 1 over 1 + x^2. 46:04 Maybe I'll stop here for one more question. 46:06 I have one more calculation which I can do even 46:10 in less than a minute. 46:11 So we have a whole minute for questions. 46:16 Yeah? 46:20 Student: [INAUDIBLE] 46:26 Professor: What happens to the inverse tangent? 46:34 The inverse tangent-- Okay, this inverse tangent 46:41 is the same as this y here. 46:44 Those are the same thing. 46:46 So what I did was I skipped this step here entirely. 46:50 I never wrote that down. 46:52 But the inverse tangent was that y. 46:54 The issue was what's a good formula 46:56 for cos y in terms of x? 47:01 So I am evaluating that, but I'm doing it using the letter y. 47:04 So in other words, what happened to the inverse 47:06 tangent is that I called it y, which 47:10 is what it's been all along. 47:15 Okay, so now I'm going to do the case 47:17 of the sine, the inverse sine. 47:20 And I'll show you how easy this is 47:22 if I don't fuss with-- because this one has an easy trig 47:27 identity associated with it. 47:29 So if y = sin^(-1) x, and sin y = x, 47:37 and now watch how simple it is when I do the differentiation. 47:40 I just differentiate. 47:42 I get (cos y) y' = 1. 47:50 And then, y', so that implies that = 1 / cos y, 48:00 and now to rewrite that in terms of x, 48:03 I have to just recognize that this is the same as this, 48:10 which is the same as 1 / square root of 1 - x^2. 48:14 So all told, the derivative with respect to x of the arcsine 48:19 function is 1 / square root of 1 - x^2. 48:30 So these implicit differentiations 48:32 are very convenient. 48:34 However, I warn you that you do have 48:38 to be careful about the range of applicability of these things. 48:42 You have to draw a picture like this one 48:44 to make sure you know where this makes sense. 48:47 In other words, you have to pick a branch for the sine function 48:50 to work that out, and there's something 48:52 like that on your problem set. 48:53 And it's also discussed in your text. 48:56 So we'll stop here. Course Info Instructor Prof. David Jerison Departments Mathematics As Taught In Fall 2010 Level Undergraduate Topics Mathematics Calculus Differential Equations Learning Resource Types grading Exams with Solutions notes Lecture Notes theaters Lecture Videos assignment_turned_in Problem Sets with Solutions laptop_windows Simulations theaters Problem-solving Videos Download Course Over 2,500 courses & materials Freely sharing knowledge with learners and educators around the world. Learn more © 2001–2025 Massachusetts Institute of Technology Accessibility Creative Commons License Terms and Conditions Proud member of: © 2001–2025 Massachusetts Institute of Technology You are leaving MIT OpenCourseWare close Please be advised that external sites may have terms and conditions, including license rights, that differ from ours. MIT OCW is not responsible for any content on third party sites, nor does a link suggest an endorsement of those sites and/or their content. Stay Here Continue
16332	https://math.stackexchange.com/questions/671748/show-that-if-ar-bs-1-for-some-r-and-s-then-a-and-b-are-relatively	elementary number theory - Show that if $ar + bs = 1$ for some $r$ and $s$ then $a$ and $b$ are relatively prime - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Show that if a r+b s=1 a r+b s=1 for some r r and s s then a a and b b are relatively prime Ask Question Asked 11 years, 7 months ago Modified9 years, 8 months ago Viewed 4k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. We were given the following problem as an assignment, we proved the converse in class for prime numbers; however, we can't use the assumption that gcd(a,b)=1 gcd⁡(a,b)=1. So I am struggling to think of my first possible move. We are given a r+b s=1 a r+b s=1, the only think I can think of is let d d be a common divisor or a r+b s a r+b s and then show that the d=1 d=1. However, I am not really sure how to do that. If anyone could just give me a hint or two to nudge me in the right direction I would be appreciative. elementary-number-theory divisibility gcd-and-lcm Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jan 30, 2016 at 13:27 Martin Sleziak 56.3k 20 20 gold badges 211 211 silver badges 391 391 bronze badges asked Feb 11, 2014 at 2:17 spitfireddspitfiredd 647 2 2 gold badges 10 10 silver badges 18 18 bronze badges 1 Some hint: Try some examples, using Euclidean algorithm. If the algorithm works, then we can determine some integers that work for a r+b s=1 a r+b s=1. Finally, work out the general case.Mr Mathster –Mr Mathster 2014-02-11 02:19:11 +00:00 Commented Feb 11, 2014 at 2:19 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 8 Save this answer. Show activity on this post. Hint: Suppose to the contrary that d>1 d>1 divides both a a and b b. Then d d divides a r a r and ……. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Feb 11, 2014 at 2:19 André NicolasAndré Nicolas 515k 47 47 gold badges 584 584 silver badges 1k 1k bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Hint Suppose d∣a,b d∣a,b is a common divisor of a,b.a,b. From the multiples a,b a,b of d d we may deduce further multiples of d d because the set of all multiples of d d enjoys a nice algebraic structure: it is closed under the operations of addition and subtraction, since j d±k d=(j±k)d.j d±k d=(j±k)d.. Further, it is closed under multiples m(k d)=(m k)d.m(k d)=(m k)d. Composing these two operations we deduce that the multiples of d d are closed under arbitrary integer linear combinations, i.e. if a,b a,b are multiples of d d then so too is j a+k b,j a+k b, for any integers j,k.j,k. In your case you happen to know a very small such linear combination which, being a multiple of d,d, greatly constrains the possible values of d.d. Remark This algebraic structure of (common) multiples plays a fundamental role in algebra - something that will become much clearer when one learns about ideals and modules. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Feb 11, 2014 at 3:03 Bill DubuqueBill Dubuque 284k 42 42 gold badges 339 339 silver badges 1k 1k bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions elementary-number-theory divisibility gcd-and-lcm See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 3If a r+b s=1 a r+b s=1, then gcd(a,s)=gcd(r,b)=gcd(r,s)=1 gcd(a,s)=gcd(r,b)=gcd(r,s)=1 0Bézout's Identity for Polynomial Ring 0How would we prove by contradiction that there do not exist integers x x and y y such that 15 x+1=21 y 15 x+1=21 y? 2Prove that ∀a, b, u, v ∈ Z − {0} ua + vb = 1 → gcd(a, b) = 1 0Proving a \| (b 2 b 2 + c 2 c 2) if a \| b and a \| c Related 19How does one show that two general numbers n!+1 n!+1 and (n+1)!+1(n+1)!+1 are relatively prime? 6Prove that any set of integers that are relatively prime in pairs are relatively prime 2Prove that if gcd(m,n)=1 gcd(m,n)=1 and m∣x m∣x and n∣x n∣x, then m n∣x m n∣x. 7Why is it true that if a x+b y=d a x+b y=d then gcd(a,b)gcd(a,b) divides d d? 5if p∣a p∣a and p∣b p∣b then p∣gcd(a,b)p∣gcd(a,b) 0Proof Verification, gcd(a,b)=p gcd(a,b)=p for prime p p implies gcd(a 2,b 2)=p 2 gcd(a 2,b 2)=p 2 0Common divisors of a,b a,b and a a mod b b 1Prove that among any five consecutive positive integers there is one integer which is relatively prime to the other four integers. 3Show that if gcd(a,b)>1 gcd(a,b)>1, then there can be at most one prime of the form p=a n+b p=a n+b. 1Solutions of a x+b y=c a x+b y=c satisfying gcd(a,y)=gcd(b,x)=1 gcd⁡(a,y)=gcd⁡(b,x)=1 Hot Network Questions With line sustain pedal markings, do I release the pedal at the beginning or end of the last note? 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16333	https://www.dreambox.com/math/skills/geometry/area-formula	Skip to content Author Michelle Griczika Published: Oct 2024 Key takeaways • Area is how we measure the space inside of a shape. Finding the area helps us understand the size of a bedroom or football field. • Different shapes have different area formulas. • There are many examples of calculating area in everyday life, like buying the right amount of paint or setting up a garden. The concept of area in math is more than just numbers and shapes; it’s about real-life applications. Picture this – choosing the right amount of paint for a wall or a room’s carpet size. That’s where area steps in. This simple yet essential math concept helps us navigate the world, making the abstract tangible. Definition of area in math What is area? Area in math is about counting squares. When we define area, it means how many square units fit inside a shape or figure. If you picture a square 3O units long on each side, the area is 9 square units. This is because 9 squares, each 1 unit by 1 unit, can fit inside. The concept of area math works the same way for all shapes, not just squares. From circles to triangles and more complex figures, the idea is to calculate how many square units fill up that shape. This area definition math principle helps us understand and describe the space around us. Real-life examples of calculating area Think about a farmer with a rectangular field. The length of his field is 200 feet, and the width is 100 feet. The farmer needs to know the total space of his field to buy the correct amount of seeds. He multiplies the length by the width and finds out his field is 20,000 square feet. In a different scenario, consider an architect who is designing a room. She has a perfect square room, with each side being 12 feet. She calculates the area to ensure the furniture will fit and leave enough space to move around. By multiplying the length and the width, she determines the room is 144 square feet. From farming to house design, the concept of area is essential. Table of contents What is it? Real life examples Formula Area vs perimeter Area vs volume Practice Get more practice with area and math with DreamBox! Start Free Trial Practice learning area more with DreamBox Math DREAMBOX MATH Get started for FREE today! Start Free Trial What is the formula for area? The formula to calculate area depends on the shape in question. For basic shapes such as a square or a rectangle, the area is the product of the length and the width. However, their area formula becomes more complicated as the shapes become complex. Let’s take a look at a variety of area formulas by shape: Square or Rectangle:In a square or rectangle, the formula for area is simply length times width (A = l x w). This basic formula is one of the most used in area calculation. If you have a square room that measures 10 feet on each side, the area of the room would be 10 feet (length) times 10 feet (width), which equals 100 square feet. For a rectangular space, you use the length and width of the rectangle similarly. Circle: The formula to calculate the area of a circle is pi times the radius squared (A = πr^2). Here, the radius is the distance from the circle’s center to its edge, and pi is a mathematical constant approximately equal to 3.14159. Triangle: For a triangle, the area formula is one-half the base times the height (A = ½ x b x h). This formula is used when you know the base and the height of the triangle. Parallelogram:The area of a parallelogram is calculated by multiplying the base by the height (A = b x h). Trapezoid:When dealing with a trapezoid, the area formula is the average of the lengths of the bases (the parallel sides) times the height (A = ½ (a + b)h)). Here’s a quick table of all the area formulas for reference: \| \| \| --- \| \| Shape \| Area Formula \| \| Square/Rectangle \| A = l x w \| \| Circle \| A = πr^2 \| \| Triangle \| A = ½ (b x h) \| \| Parallelogram \| A = b x h \| \| Trapezoid \| A = ½ (a + b)h \| The math program that drives results Get started today! DreamBox adapts to your child’s level and learning needs, ensuring they are appropriately challenged and get confidence-building wins. Try DreamBox for 14 Days Area vs perimeter How to measure area relates to the surface covered by a shape, while the perimeter speaks to the total boundary length around that shape. Both play crucial roles in geometry, but they’re distinct. Think of a football field. The entire space where the game is played that’s the area. Now consider the distance you’d travel if you ran around the edge of the field. That’s the perimeter. Area vs volume Volume measures the quantity of three-dimensional space an object inhabits. It’s like filling a box with small cubes and calculating how many fit inside. This measure considers not just the surface, as area does, but the entirety of the space within the object’s boundaries. So while area and volume are related, they offer distinct ways to understand and interact with space. If you think of area as the carpet on your living room floor, volume would be the space your living room occupies, from floor to ceiling. Practice Problems Click on the boxes below to see the answers! Calculate the area of a rectangle with length 8 units and width 5 units. The area of the rectangle is 40 square units. Calculate the area of a circle with radius 6 units. (Use Pi as 3.14) The area of the circle is approximately 113.04 square units. Calculate the area of a triangle with base 4 units and height 3 units. The area of the triangle is 6 square units. Calculate the area of a trapezoid with height 7 units and bases 6 units and 8 units. The area of the trapezoid is 49 square units. Calculate the area of a square with side 10 units. The area of the square is 100 square units. Calculate the area of a rectangle with length 15 units and width 7 units. The area of the rectangle is 105 square units. Calculate the area of a circle with radius 5 units. (Use Pi as 3.14) The area of the circle is approximately 78.5 square units. Calculate the area of a triangle with base 6 units and height 8 units. The area of the triangle is 24 square units. Calculate the area of a trapezoid with height 9 units and bases 5 units and 7 units. The area of the trapezoid is 54 square units. Calculate the area of a square with side 12 units. The area of the square is 144 square units. Area in Math Practice Problems Question 1 of 5 Q.1 What is the area of a rectangle with a length of 10 units and a width of 5 units? 30 50 100 15 Quiz Complete! You got 0 out of 5 questions correct. FAQs about area in math What is the formula for area? The formula varies depending on the shape. For rectangles and squares, A = length x width. The area of a circle is A = 𝜋r^2. Finally, the area of a triangle is A = ½ (base x height). Why do we calculate area? The area is calculated to determine the size of a surface or a two-dimensional space. This is crucial in real-world situations, like buying sufficient paint for a room or determining the number of tiles required to cover a floor. What is an example of area in real life? A practical example of area is when you need to paint a wall. You’d calculate the area of the wall to know how much paint you’ll need. What is the difference between area and perimeter? Area is just the inside space of a shape, counted in squares. Perimeter is the distance around the outside of the shape, counted in a straight line. Take at home math practice to the next level
16334	https://artofproblemsolving.com/wiki/index.php/Category:Sequences_and_series?srsltid=AfmBOorfgPZqx1tbOGmDUaYQHQI1yiaX7xvmnuayRS4YOGRedm7qb5gH	Art of Problem Solving Category:Sequences and series - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Category:Sequences and series Page CategoryDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Category:Sequences and series Pages in category "Sequences and series" The following 3 pages are in this category, out of 3 total. A Arithmetic sequence G Geometric sequence H Harmonic sequence Retrieved from " Category: Algebra Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
16335	https://www.varsitytutors.com/practice/subjects/gmat-quantitative/help/calculating-the-ratio-of-diameter-and-circumference	GMAT Quantitative - Calculating the ratio of diameter and circumference \| Practice Hub Skip to main content Practice Hub Search subjects AI TutorAI DiagnosticsAI FlashcardsAI WorksheetsAI SolverGamesProgress Sign In HomeGMAT QuantitativeLearn by ConceptCalculating the ratio of diameter and circumference Calculating the ratio of diameter and circumference Help Questions GMAT Quantitative › Calculating the ratio of diameter and circumference Questions 1 - 8 1 The circle with center , is inscribed in the square . What is the ratio of the diameter to the circumference of the circle given that the square has an area of ? Explanation To calculate the ratio of the diameter to the circumference of the square we should first get the diameter, which is the same as the length of a side of the sqaure. To do so we just need to take the square root of the area of the square, which is 4. Also we should remember that the circumference is given by , where is the length of the radius. Now we should notice that this formula can also be written . The ratio we are looking for is . Therefore, this ratio will always be and this is our final answer. 2 For any circle, what is the ratio of its circumference to its diameter? Explanation In order to calculate the ratio of circumference to diameter, we need an equation that involves both variables. The formula for circumference is as follows: We need to express the radius in terms of diameter. The radius of a circle is half of its diameter, so we can rewrite the formula as: If we divide both sides by the diameter, on the left side we will have , which is the ratio of circumference to diameter: So, for any circle, the ratio of its circumference to its diameter is equal to , which is actually the definition of this very important mathematical constant. 3 Find the circumference of a circle with a diameter measuring . Explanation The circumference of a circle is given by where We are told the diameter so we just need to plug in our value into the equation: 4 A given circle has a circumference of and a radius of . What is the ratio of the circle's circumference to its diameter? Explanation For a given circle of circumference and radius , . Since the radius of a circle is equal to half of the circle's diameter , we can then define as . Therefore, the ratio of the circumference to the diameter of this and all other circles is . 5 A given circle has a circumference of and a radius of . What is the ratio of the circle's circumference to its diameter? Explanation For a given circle of circumference and radius , . Since the radius of a circle is equal to half of the circle's diameter , we can then define as . Therefore, the ratio of the circumference to the diameter of this and all other circles is, . 6 A given circle has a circumference of and a radius of . What is the ratio of the circle's circumference to its diameter? Explanation For a given circle of circumference and radius , . Since the radius of a circle is equal to half of the circle's diameter , we can then define as, . Therefore, the ratio of the circumference to the diameter of this circles is, . 7 For any given circle, what is the ratio of its diameter to its circumference? Explanation To find the ratio between the diameter and the circumference of a circle, we need to use the formula for the circumference of a circle: We can see this formula is in terms of radius, so we need to express it in a way that the circumference is in terms of diameter. Using the knowledge that the radius is half of the diameter: Now that we have a simple formula involving the circumference and diameter, we can see that we will have the ratio of diameter to circumference if we divide both sides by the circumference. We then divide both sides by to isolate the ratio of diameter to circumference and find our solution: 8 Equilateral triangle is inscribed in a circle. The perimeter of the triangle is . What is the radius of the circle? Explanation The perimeter of the circle allows us to find the side of equilateral triangle ABC; or 2. From there, we can also find the length of the height of equilateral triangle ABC with the formula , which turns out to be . Since in an equilateral triangle, the center of gravity is at from any vertex, it follows that the radius of the circle is . Return to subject Powered by Varsity Tutors⋅ © 2025 All Rights Reserved
16336	https://farside.ph.utexas.edu/teaching/355/Surveyhtml/node209.html	Bulk Modulus Next:Sound WavesUp:Ideal GasPrevious:Adiabatic Atmosphere Bulk Modulus The bulk modulus of an ideal gas is a measure of its resistance to bulk compression, and is defined (5.146) In fact, an ideal gas possesses a number of different bulk moduli depending on what is held constant as the pressure is varied. The two most important bulk moduli are the isothermal bulk modulus, (5.147) and the isentropic bulk modulus, (5.148) The former describes situations in which the gas undergoes isothermal compression, whereas the latter describes situations in which the gas undergoes adiabatic compression. (Note that actually denotes entropy. However, a gas that undergoes compression at constant entropy is such that no heat is added to the gas during the compression.) According to the isothermal gas law, (5.114), (5.149) so (5.150) which implies that (5.151) According to the adiabatic gas law, (5.124), (5.152) (5.153) which implies that (5.154) Note that the isentropic bulk modulus of an ideal gas is greater than its isothermal bulk modulus (because ). In other words, an ideal gas resists adiabatic compression to a greater degree than it resists isothermal compression. This is the case because during adiabatic compression the work done on the gas causes its temperature to rise, leading to a greater increase in the pressure than would be obtained if the temperature were held constant. Next:Sound WavesUp:Ideal GasPrevious:Adiabatic Atmosphere
16337	https://sites.ualberta.ca/~rjia/Math418/Notes/chap3.pdf	Chapter 3. Absolutely Continuous Functions §1. Absolutely Continuous Functions A function f : [a, b] →I R is said to be absolutely continuous on [a, b] if, given ε > 0, there exists some δ > 0 such that n X i=1 \|f(yi) −f(xi)\| < ε, whenever {[xi, yi] : i = 1, . . . , n} is a ﬁnite collection of mutually disjoint subintervals of [a, b] with Pn i=1 \|yi −xi\| < δ. Clearly, an absolutely continuous function on [a, b] is uniformly continuous. Moreover, a Lipschitz continuous function on [a, b] is absolutely continuous. Let f and g be two absolutely continuous functions on [a, b]. Then f+g, f−g, and fg are absolutely continuous on [a, b]. If, in addition, there exists a constant C > 0 such that \|g(x)\| ≥C for all x ∈[a, b], then f/g is absolutely continuous on [a, b]. If f is integrable on [a, b], then the function F deﬁned by F(x) := Z x a f(t) dt, a ≤x ≤b, is absolutely continuous on [a, b]. Theorem 1.1. Let f be an absolutely continuous function on [a, b]. Then f is of bounded variation on [a, b]. Consequently, f ′(x) exists for almost every x ∈[a, b]. Proof. Since f is absolutely continuous on [a, b], there exists some δ > 0 such that Pn i=1 \|f(yi) −f(xi)\| < 1 whenever {[xi, yi] : i = 1, . . . , n} is a ﬁnite collection of mu-tually disjoint subintervals of [a, b] with Pn i=1 \|yi −xi\| < δ. Let N be the least inte-ger such that N > (b −a)/δ, and let aj := a + j(b −a)/N for j = 0, 1, . . . , N. Then aj −aj−1 = (b −a)/N < δ. Hence, ∨aj aj−1f < 1 for j = 0, 1, . . . , N. It follows that b _ a f = N X j=1 aj _ aj−1 f < N. This shows that f is of bounded variation on [a, b]. Consequently, f ′(x) exists for almost every x ∈[a, b]. 1 Theorem 1.2. If f is absolutely continuous on [a, b] and f ′(x) = 0 for almost every x ∈[a, b], then f is constant. Proof. We wish to show f(a) = f(c) for every c ∈[a, b]. Let E := {x ∈[a, c] : f ′(x) = 0}. For given ε > 0, there exists some δ > 0 such that Pn i=1 \|f(yi) −f(xi)\| < ε whenever {[xi, yi] : i = 1, . . . , n} is a ﬁnite collection of mutually disjoint subintervals of [a, b] with Pn i=1 \|yi −xi\| < δ. For each x ∈E, we have f ′(x) = 0; hence there exists an arbitrary small interval [ax, cx] such that x ∈[ax, cx] ⊆[a, c] and \|f(cx) −f(ax)\| < ε(cx −ax). By the Vitali covering theorem we can ﬁnd a ﬁnite collection {[xk, yk] : k = 1, . . . , n} of mutually disjoint intervals of this sort such that λ E \ ∪n k=1[xk, yk] < δ. Since λ([a, c] \ E) = 0, we have λ [a, c] \ ∪n k=1[xk, yk] = λ E \ ∪n k=1[xk, yk] < δ. Suppose a ≤x1 < y1 ≤x2 < · · · < yn ≤c. Let y0 := a and xn+1 := c. Then n X k=0 (xk+1 −yk) = λ [a, c] \ ∪n k=1[xk, yk] < δ. Consequently, n X k=0 \|f(xk+1) −f(yk)\| < ε. Furthermore, n X k=1 \|f(yk) −f(xk)\| < n X k=1 ε(yk −xk) ≤ε(c −a). It follows from the above inequalities that \|f(c) −f(a)\| ≤ n X k=0 \|f(xk+1) −f(yk)\| + n X k=1 \|f(yk) −f(xk)\| < ε(c −a + 1). This shows that \|f(c) −f(a)\| ≤ε(c −a + 1) for all ε > 0. Therefore, f(c) = f(a). 2 §2. The Fundamental Theorem of Calculus In this section we show that absolutely continuous functions are precisely those func-tions for which the fundamental theorem of calculus is valid. Theorem 2.1. If f is integrable on [a, b] and Z x a f(t) dt = 0 ∀x ∈[a, b], then f(t) = 0 for almost every t ∈[a, b]. Proof. By our assumption, Z d c f(t) dt = 0 for all c, d with a ≤c < d ≤b. If O is an open subset of [a, b], then O is a countable union of mutually disjoint open intervals (cn, dn) (n = 1, 2, . . .); hence, Z O f(t) dt = ∞ X n=1 Z dn cn f(t) dt = 0. It follows that for any closed subset K of [a, b], Z K f(t) dt = Z [a,b] f(t) dt − Z [a,b]\K f(t) dt = 0. Let E+ := {x ∈[a, b] : f(x) > 0} and E−:= {x ∈[a, b] : f(x) < 0}. We wish to show that λ(E+) = 0 and λ(E−) = 0. If λ(E+) > 0, then there exists some closed set K ⊆E+ such that λ(K) > 0. But R K f(t) dt = 0. It follows that f = 0 almost everywhere on K. This contradiction shows that λ(E+) = 0. Similarly, λ(E−) = 0. Therefore, f(t) = 0 for almost every t ∈[a, b]. Theorem 2.2. If f is integrable on [a, b], and if F is deﬁned by F(x) := Z x a f(t) dt, a ≤x ≤b, then F ′(x) = f(x) for almost every x in [a, b]. Proof. First, we assume that f is bounded and measurable on [a, b]. For n = 1, 2, . . ., let gn(x) := F(x + 1/n) −F(x) 1/n , x ∈[a, b]. 3 It follows that gn(x) = n Z x+1/n x f(t) dt, x ∈[a, b]. Suppose \|f(x)\| ≤K for all x ∈[a, b]. Then \|gn(x)\| ≤K for all x ∈[a, b] and n ∈I N. Since limn→∞gn(x) = F ′(x) for almost every x ∈[a, b], by the Lebesgue dominated convergence theorem, we see that for each c ∈[a, b], Z c a F ′(x) dx = lim n→∞ Z c a gn(x) dx. But F is continuous; hence, lim n→∞ Z c a gn(x) dx = lim n→∞n Z c+1/n c F(x) dx − Z a+1/n a F(x) dx = F(c) −F(a). Consequently, Z c a F ′(x) dx = lim n→∞ Z c a gn(x) dx = F(c) −F(a) = Z c a f(x) dx. It follows that Z c a [F ′(x) −f(x)] dx = 0 for every c ∈[a, b]. By Theorem 2.1, F ′(x) = f(x) for almost every x in [a, b]. Now let us assume that f is integrable on [a, b]. Without loss of any generality, we may assume that f ≥0. For n = 1, 2, . . ., let fn be the function deﬁned by fn(x) := f(x) if 0 ≤f(x) ≤n, 0 if f(x) > n. It is easily seen that F = Fn + Gn, where Fn(x) := Z x a fn(t) dt and Gn(x) := Z x a [f(t) −fn(t)] dt, a ≤x ≤b. Since f(t) −fn(t) ≥0 for all t ∈[a, b], Gn is an increasing function on [a, b]. Moreover, by what has been proved, F ′ n(x) = fn(x) for almost every x ∈[a, b]. Thus, we have F ′(x) = F ′ n(x) + G′ n(x) ≥F ′ n(x) = fn(x) for almost every x ∈[a, b]. Letting n →∞in the above inequality, we obtain F ′(x) ≥f(x) for almost every x ∈[a, b]. It follows that Z b a F ′(x) dx ≥ Z b a f(x) dx = F(b) −F(a). 4 On the other hand, Z b a F ′(x) dx ≤F(b) −F(a). Consequently, Z b a [F ′(x) −f(x)] dx = 0. But F ′(x) ≥f(x) for almost every x ∈[a, b]. Therefore, F ′(x) = f(x) for almost every x in [a, b]. Theorem 2.3. A function F on [a, b] is absolutely continuous if and only if F(x) = F(a) + Z x a f(t) dt for some integrable function f on [a, b]. Proof. The suﬃciency part has been established. To prove the necessity part, let F be an absolutely continuous function on [a, b]. Then F is diﬀerentiable almost everywhere and F ′ is integrable on [a, b]. Let G(x) := F(a) + Z x a F ′(t) dt, x ∈[a, b]. By Theorem 2.2, G′(x) = F ′(x) for almost every x ∈[a, b]. It follows that (F −G)′(x) = 0 for almost every x ∈[a, b]. By Theorem 1.2, F −G is constant. But F(a) = G(a). Therefore, F(x) = G(x) for all x ∈[a, b]. §3. Change of Variables for the Lebesgue Integral Let f be an absolutely continuous function on [c, d], and let u be an absolutely con-tinuous function on [a, b] such that u([a, b]) ⊆[c, d]. Then the composition f ◦u is not necessarily absolutely continuous. However, we have the following result. Theorem 3.1. Let f be a Lipschitz continuous function on [c, d], and let u be an absolutely continuous function on [a, b] such that u([a, b]) ⊆[c, d]. Then f ◦u is absolutely continuous. Moreover, (f ◦u)′(t) = f ′(u(t))u′(t) for almost every t ∈[a, b], where f ′(u(t))u′(t) is interpreted to be zero whenever u′(t) = 0 (even if f is not diﬀeren-tiable at u(t)). Proof. Since f is a Lipschitz continuous function on [c, d], there exists some M > 0 such that \|f(x)−f(y)\| ≤M\|x−y\| whenever x, y ∈[c, d]. Let ε > 0 be given. Since u is absolutely 5 continuous on [a, b], there exists some δ > 0 such that Pn i=1 \|u(ti)−u(si)\| < ε/M, whenever {[si, ti] : i = 1, . . . , n} is a ﬁnite collection of mutually disjoint subintervals of [a, b] with Pn i=1(ti −si) < δ. Consequently, n X i=1 \|(f ◦u)(ti) −(f ◦u)(si)\| = n X i=1 \|f(u(ti)) −f(u(si))\| ≤ n X i=1 M\|u(ti) −u(si)\| < ε. This shows that f ◦u is absolutely continuous on [a, b]. Since both u and f ◦u are absolutely continuous on [a, b], there exists a measurable subset E of [a, b] such that λ(E) = 0 and both u′(t) and (f ◦u)′(t) exist for all t ∈[a, b]\E. Suppose t0 ∈[a, b] \ E. If u′(t0) = 0, then for given ε > 0, there exists some h > 0 such that \|u(t) −u(t0)\| ≤ε\|t −t0\| whenever t ∈(t0 −h, t0 + h) ∩[a, b]. It follows that \|f ◦u(t) −f ◦u(t0)\| ≤M\|u(t) −u(t0)\| ≤Mε\|t −t0\| for all t ∈(t0 −h, t0 + h) ∩[a, b]. This shows that (f ◦u)′(t0) = 0 = f ′(u(t0))u′(t0). Now suppose t0 ∈[a, b] \ E and u′(t0) ̸= 0. Suppose u(t) ̸= u(t0). Then we have (f ◦u)(t) −(f ◦u)(t0) t −t0 = f(u(t)) −f(u(t0)) u(t) −u(t0) u(t) −u(t0) t −t0 . Since u′(t0) and (f ◦u)′(t0) exist, we obtain lim t→t0 (f ◦u)(t) −(f ◦u)(t0) t −t0 = (f ◦u)′(t0) and lim t→t0 u(t) −u(t0) t −t0 = u′(t0) ̸= 0. Consequently, lim t→t0 f(u(t)) −f(u(t0)) u(t) −u(t0) = (f ◦u)′(t0) u′(t0) . Let r := (f ◦u)′(t0)/u′(t0). For given ε > 0, there exists some δ > 0 such that r −ε < f(u(t)) −f(u(t0)) u(t) −u(t0) < r + ε ∀t ∈(t0 −δ, t0 + δ) ∩[a, b]. Since u′(t0) ̸= 0, there exists some η > 0 such that any x ∈(u(t0) −η, u(t0) + η) ∩[c, d] can be expressed as x = u(t) for some t ∈(t0 −δ, t0 + δ) ∩[a, b]. Therefore, r −ε < f(x) −f(u(t0)) x −u(t0) < r + ε ∀x ∈(u(t0) −η, u(t0) + η) ∩[c, d]. This shows that f ′(u(t0)) exists and f ′(u(t0)) = r = (f ◦u)′(t0)/u′(t0), as desired. 6 Theorem 3.2. Let g be a bounded and measurable function on [c, d], and let u be an absolutely continuous function on [a, b] such that u([a, b]) ⊆[c, d]. Then (g ◦u)u′ is integrable on [a, b]. Moreover, for any α, β ∈[a, b], Z u(β) u(α) g(x) dx = Z β α g(u(t))u′(t) dt. Proof. Let F(x) := Z x c g(t) dt, x ∈[c, d]. Since g is bounded, F is Lipschitz continuous. Moreover, F ′(x) = g(x) for almost every x ∈[a, b]. By Theorem 3.1, F ◦u is absolutely continuous on [a, b] and, for almost every t ∈[a, b], (F ◦u)′(t) = g(u(t))u′(t). Suppose α, β ∈[a, b] and α < β. By Theorem 2.3, we have (F ◦u)(β) −(F ◦u)(α) = F(u(β)) −F(u(α)) = Z u(β) u(α) F ′(x) dx = Z u(β) u(α) g(x) dx. On the other hand, (F ◦u)(β) −(F ◦u)(α) = Z β α (F ◦u)′(t) dt = Z β α g(u(t))u′(t) dt. This proves the desired formula for change of variables. Theorem 3.3. Let g be an integrable function on [c, d], and let u be an absolutely con-tinuous function on [a, b] such that u([a, b]) ⊆[c, d]. If (g ◦u)u′ is integrable on [a, b], then Z u(β) u(α) g(x) dx = Z β α g(u(t))u′(t) dt, α, β ∈[a, b]. Moreover, (g ◦u)u′ is integrable if, in addition, u is monotone. Proof. Suppose that g is integrable on [a, b]. Without loss of any generality, we may assume g ≥0. For n = 1, 2, . . ., let gn be the function deﬁned by gn(x) := g(x) if 0 ≤g(x) ≤n, 0 if g(x) > n. Then gn ≤gn+1 for all n ∈I N. Suppose α, β ∈[a, b] and α < β. By Theorem 3.2 we have Z u(β) u(α) gn(x) dx = Z β α gn(u(t))u′(t) dt. 7 If u is monotone, then u′(t) ≥0 for almost every t ∈[a, b]. Letting n →∞in the above equation, by the monotone convergence theorem we obtain Z u(β) u(α) g(x) dx = Z β α g(u(t))u′(t) dt. Since g is integrable on [c, d], it follows from the above equation that (g ◦u)u′ is integrable on [a, b]. More generally, we assume that (g ◦u)u′ is integrable on [a, b] but u is not necessarily monotone. Then \|gn(u(t))u′(t)\| ≤g(u(t))\|u′(t)\| for all n ∈I N and almost every t ∈[a, b]. Thus, an application of the Lebesgue dominated convergence theorem gives the desired formula for change of variables. 8
16338	https://www.quora.com/The-binomial-distribution-and-the-hypergeometric-distribution-both-determine-the-probability-of-having-k-successes-in-n-trials-why-are-they-different-How-can-you-tell-binomial-or-hypergeometric-distribution-is	Something went wrong. Wait a moment and try again. Probability (statistics) Statistical Analysis Hypergeometric Distributi... Probability Theory Discrete Probability Dist... Binomial Distribution Probability in Math Specific Probability Dist... 5 The binomial distribution and the hypergeometric distribution both determine the probability of having k successes in n trials.why are they different? How can you tell binomial or hypergeometric distribution is appropriate for particular experiment Justin Rising PhD in statistics · Author has 12.1K answers and 26.5M answer views · 4y The binomial distribution is what you get when you sample with replacement. The hypergeometric distribution, on the other hand, shows up in situations where you sample without replacement. The distinction between sampling with and without replacement is fundamental, and you must understand it. Related questions What is the difference between binomial and hypergeometric distribution? What is the critical value for a binomial distribution with a given probability? What are the difference and similiarities between Binomial and Hypergeometric Probability distribution? Define binomial distribution. How does it differ from Hypergeometric distribution? How do you describe probability distribution with special reference to binomial and hypergeometric distribution? Bo Jacoby · Author has 176 answers and 106K answer views · 3y From a hat containing K white pebbles and N-K black pebbles pick a sample of n pebbles and count the number k of white pebbles in the sample. This unknown number has the ‘hypergeometric’ distribution with mean value Kn/N. Introduce the probability p=K/N. Then the mean value is pn. The limiting case 1/N=0 gives a binomial distribution with mean value pn. If you do know N then use the hypergeometric distribution. If you only know that N is big then use the binomial distribution. Terry Moore PhD in statistics · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 16.6K answers and 29.3M answer views · 5y Related What is the difference between binomial and hypergeometric distribution? What is the difference between binomial and hypergeometric distribution? The short answer is that it’s the difference between sampling with replacement and sampling without replacement. In a binomial distribution the events are independent and have the same probability of success or failure. In the hypergeometric distribution the probabilities change at each draw and depend on the results of the previous draws. If the population is very large these are almost the same. Think of a Bernoulli distribution. This has two values and . The probability of is and the probability of is . We ca What is the difference between binomial and hypergeometric distribution? The short answer is that it’s the difference between sampling with replacement and sampling without replacement. In a binomial distribution the events are independent and have the same probability of success or failure. In the hypergeometric distribution the probabilities change at each draw and depend on the results of the previous draws. If the population is very large these are almost the same. Think of a Bernoulli distribution. This has two values and . The probability of is and the probability of is . We call observing a a success and observing a a failure. In independent Bernoulli trials the number of successes has a binomial distribution. If there are successes and failures the successes could have been observed in any of trials and the probability of successes in those trials and failures in all the others is . So the probability of exactly successes anywhere is A special case is drawing objects from a finite population with two types, e.g. drawing cards from a pile where we are interested in the number of spades. So spade is a success and any other card is a failure. If the card is replaced and the cards shuffled (completely randomised) after each draw the number of hearts has a binomial distribution. If you don’t replace the cards after each draw you get a hypergeometric distribution. Most books calculate the probability by counting the number of sequences of draws with exactly spades and dividing by the number of possible sequences of draws. Here’s a method that shows the similarity between the two distributions. As for the binomial, the number of places the successes could occur is . The probability that the successes occur in the first places and failures thereafter (provided that and , and, of course ) is . The denominators decrease by at each trial because there is one less object in the remaining population, the number of objects of the first type decreases by until we have observed of them, while the number of the other type starts at at draw and reduces by for each subsequent draw. Now here’s the crux: this is also the probability wherever the successes and the failures occur. To see this consider any particular set of positions for the successes. The probability that they occur in exactly those places is the same as the probability that they occur in the first places. Why? We want the marginal probability—the probability we would use if we didn’t know the results of the other draws. Or you can look at it another way, the probability calculation is just a rearrangement of the order of the numerators in the expression above. So the probability is . If is very large compared with then the factor is close to where and the factor is close to where and the probability is approximately binomial. Assistant Bot · 1y The binomial distribution and the hypergeometric distribution are both used to model the number of successes in a series of trials, but they apply to different scenarios based on the nature of the trials and the sampling method. Here’s a breakdown of their differences and how to determine which is appropriate for a particular experiment: Definitions and Key Differences Binomial Distribution: Trials: Fixed number of independent trials (n). Success Probability: The probability of success (p) is constant for each trial. Replacement: Each trial is independent, meaning the outcome of one trial does n The binomial distribution and the hypergeometric distribution are both used to model the number of successes in a series of trials, but they apply to different scenarios based on the nature of the trials and the sampling method. Here’s a breakdown of their differences and how to determine which is appropriate for a particular experiment: Definitions and Key Differences Binomial Distribution: Trials: Fixed number of independent trials (n). Success Probability: The probability of success (p) is constant for each trial. Replacement: Each trial is independent, meaning the outcome of one trial does not affect the others. This typically implies that sampling is done with replacement. Formula: Hypergeometric Distribution: Trials: Fixed number of draws (n) without replacement from a finite population. Success Probability: The probability of success changes on each draw because the population size decreases. Replacement: Each trial is dependent on the previous draws, as items are not replaced after being drawn. Formula: where: = total population size, = number of successes in the population, = number of draws, = number of observed successes. When to Use Each Distribution Use Binomial Distribution When: The number of trials is fixed. Each trial is independent. The probability of success remains constant across trials. Example: Flipping a fair coin 10 times and counting the number of heads. Use Hypergeometric Distribution When: The number of draws is fixed. The trials are dependent (sampling without replacement). The population is finite, and you are interested in the number of successes in that population. Example: Drawing 5 cards from a standard deck of 52 and counting how many are aces. Summary To decide which distribution to use, consider: - Independence vs. Dependence: If the trials are independent and have a constant probability of success, use the binomial distribution. If the trials are dependent and involve sampling without replacement, use the hypergeometric distribution. - Population Size: If you are sampling from a finite population without replacement, the hypergeometric distribution is appropriate. If the trials are thought of as being from an infinite or effectively infinite population, the binomial distribution is more suitable. Understanding these distinctions helps in selecting the correct statistical model for analyzing your data accurately. Related questions What is the formula for binomial distribution? What is the most likely value in a binomial distribution? How can I recognize a Binomial probability distribution problem? What is the Binomial Probability Formula? Does the default rate follow normal distribution or binomial distribution? Shivam Gupta BE in Computer Science, Thapar Institute of Engineering and Technology (Graduated 2021) · Author has 52 answers and 134.1K answer views · 6y Related How do we distinguish between hypergeometric, binomial, and Poisson distributions? How can you differentiate between binomial, hypergeometric, or Poisson? The basic distribution is binomial . If you change some assumptions of binomial distribution you will get hyper-geometric and Poisson distribution . So , lets start with binomial . Suppose you perform a experiment and result of that experiment is derived from some fixed number of trials , There are conditions applied on trials . Firstly they must be independent , meaning occurrence of a trial is not dependent on any other trial . Secondly , the probability of success in each trial is constant . Suppose you toss a fair coin for finite number , assume 100 . You are interested in number of head th The basic distribution is binomial . If you change some assumptions of binomial distribution you will get hyper-geometric and Poisson distribution . So , lets start with binomial . Suppose you perform a experiment and result of that experiment is derived from some fixed number of trials , There are conditions applied on trials . Firstly they must be independent , meaning occurrence of a trial is not dependent on any other trial . Secondly , the probability of success in each trial is constant . Suppose you toss a fair coin for finite number , assume 100 . You are interested in number of head that appear during the context of experiment . Your experiment is made of 100 trials . Each trial is independent( you get head in 67th toss , the outcome of any future toss is not dependent on the result of 67th toss) . Also , the probability of getting a head is exactly 0.5 in every trial . so this is a Bernoulli distribution .The parameters of binomial distribution are n(number of trials) and p(probability of success in individual trial). Lets start modifying the assumptions. Assume in binomial distribution we violate the assumption of finite trials . Now you can flip coin forever . This distribution now becomes Poisson . So out of infinite events , you are interested in certain finite number of successes . The events are still independent .Basically you write a the result using using binomial distribution and take limit of n equals infinity . You will see nice formula of Poisson distribution . The term np becomes lambda which intuitively is average number of successes out of infinite trials in a given time interval . We used a time interval here because we can not certainly keep on doing the experiment forever . Now disturb the independence of events conditions , you will get hyper-geometric . Suppose you have a jar of marbles and you have exactly w white marbles and b black marbles . You pick k marbles from the jar . We are interested in number of white marbles you pick . This is hyper-geometric . This may seem to be binomial but it is not . Lets see why… Assume in first trial you pick marble and it is white . Now it is slightly less likely that you will pick white marble again . Now the trials are no longer independent , picking some white marbles decreases the likelihood of picking white marbles again . The trials are dependent and hence distribution is not binomial , instead it is hyper-geometric . Now i tell you to sample with replacement , means i pick a marble and after noting its color put it back in jar . Now the trials are independent , the number of white marbles is same before each trial . Now the distribution will be binomial and not hyper-geometric . So summing up , if you are dealing with very large number of trials , use Poisson . When dealing with finite independent trials , go with binomial and with finite dependent trials , hyper-geometric is the answer . Last but not least , there’s a nice intuitive relation between between these three distributions where we will see how well can binomial approximate hyper-geometric , how well can Poisson approximate binomial etc . These things are slightly beyond the scope of asked question and i am too lazy to write it out . If you still want me to explain , write it out , i will explain . Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Mario Luis Iovaldi MD, Surgeon, not Statistician. Argentina · Author has 2.9K answers and 965.2K answer views · 4y Binomial: 2 mutually exclusive categories with replacement: you do not exclude the item selected, n is always the same for each trial Hypergeometric: the main difference is that it is without replacement, initial n for the 1st item, n - 1 for the next, and so on Filip Vander Stappen Mathematics and Physics teacher (2007–present) · Author has 161 answers and 220.1K answer views · 8y Related How do we distinguish between hypergeometric, binomial, and Poisson distributions? How can you differentiate between binomial, hypergeometric, or Poisson? I will give a general answer, as I assume you are not looking for a rigorous mathematical explanation but rather for a workeable way to decide which distribution is applicable. All distibutions apply when a statistical “experiment” is repeated. Now, for the binomial distribution to apply the following must hold: 1: the experiment is identical upon every repetition. 2: the outcome can be described in terms of succes versus failure. 3: All orders of succes versus failure are OK. To elucidate. Assume a wheel of fortune having four equally sized areas, one containing an A, one containing a B and two c I will give a general answer, as I assume you are not looking for a rigorous mathematical explanation but rather for a workeable way to decide which distribution is applicable. All distibutions apply when a statistical “experiment” is repeated. Now, for the binomial distribution to apply the following must hold: 1: the experiment is identical upon every repetition. 2: the outcome can be described in terms of succes versus failure. 3: All orders of succes versus failure are OK. To elucidate. Assume a wheel of fortune having four equally sized areas, one containing an A, one containing a B and two containing a C. Spin the wheel 20 times. The probability of getting 12 C’s is binomially distributed. 1: the chance of getting a C is the same upon every repetition. 2: success is getting a C, failure is getting any other letter. 3: all orders are OK. To elucidate rule 3. If one would repeat spinning until a C is spinned, that would violate this rule as now not all orders are OK, but rather only those where the last one is the first C apply. Spin the wheel 20 times. The probability of getting 12 C’s and 4 A’s is not binomially distributed. Why not? Because now it is not possible to define success versus failure. If success means getting a C then failure should mean : getting any other letter. But here the number of A’s is specified. So one needs 12 C’s, 4 A’s and 4 B’s violating rule number 2. Now the hypergeometrical distribution applies whenever an experiment is repeated and upon every repetition one of the options “disappears”. To elucidate: A vase contains 12 red balls, 6 blue balls and two white balls. Draw six balls without replacing them, what is the chance of getting 4 red balls? Now the binomial distribution does not apply, because rule 1 is violated. If e.g. one starts by drawing a red ball (with probability 12/20), the chance of getting another one in the second draw is no longer 12/20 but 11/19. The hypergeometrical distribution applies if: 1: One option disappears on every repetition. 2: All orders are OK. The Poisson distribution generally applies when: 1: The chance of success is independent of whatever has happend so far. 2: The chance of success is small. 3: The experiment is repeated often. Now, I know full well that “often” and “small” are not rigorously defined quantities but I am explaining in general terms. To elucidate rule 1, which is often violated when people use statistical models indiscriminately: Let X be the number of visitors of a bar. If X were normally distributed with mean - say - 1000 per day and on a specific day at 11 P.M. 1100 people had already visited one would assume a decreasing chance of more guests arriving as compared to a situation where only 950 had arrived. However using a Poisson model the probability of a guest arriving in the next - say - 1 minute would be completely independent of the fact that a relatively large number has already arrived. 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Bernoulli distribution This is a kind of distribution that will have only 2 possible outcomes-True or False. For example- Probability of getting a valid outcome while tossing a coin (Heads or Tails). Probability of getting a valid outcome while attending an exam (Pass or Fail). Probability of getting a valid outcome while attending a drug (effective or ineffective). Binomial distribution This is an extended version of Bernoulli distribution where multiple Bernoulli events are clubbed together. For example, If we want to find the probability of getting the head in tossing a coin exactly one time then it Bernoulli distribution This is a kind of distribution that will have only 2 possible outcomes-True or False. For example- Probability of getting a valid outcome while tossing a coin (Heads or Tails). Probability of getting a valid outcome while attending an exam (Pass or Fail). Probability of getting a valid outcome while attending a drug (effective or ineffective). Binomial distribution This is an extended version of Bernoulli distribution where multiple Bernoulli events are clubbed together. For example, If we want to find the probability of getting the head in tossing a coin exactly one time then it is a Bernoulli distribution. However, if we are tossing a coin more than one time (say 5 times) then what is the probability of getting exactly 3 heads? This is a classic example of the binomial distribution. Some more examples are- Probability of getting 3 heads in exactly 7 tosses. Probability of winning a game exactly 5 times when played for 10 times. Probability of getting passed in exactly 6 subjects out of 10 subjects in the final examinations. Geometric distribution Consider that we are standing in front of an election poll station. There are 2 candidates X and Y contesting in elections. Now we need to find the number of people who did not vote for X until the first vote was cast for him. Imagine that each person says for whom he voted after coming inside from the polling station. Person 1 said he voted for Y. Person 2 said he voted for Y. Person 3 said he voted for Y. Person 4 said he voted for Y. Person 5 said he voted for X. Here, 4 voters did not cast for X until he got the first vote. It means there were ’n’ failures before getting the first success. The geometric distribution is such a data distribution that measures the count of failures before the first success happens. There should be some conditions for geometric distributions to happen. Trials should be independent. The outcome should be binary. The probability of success should be equal in each trial. Negative binomial distribution This is an extension of geometric distribution where the number of success counts will be multiple instead of one. Here, we measure the number of failures before a specified number of successes gets occurs. For example, the number of voters who did not vote for X until X gets exactly 3 votes. Hypergeometric distribution This is a special kind of distribution, where the probability of a certain special characteristic of an object is measured when drawn from a big population. Consider that New York City has 1 million cats. We want to know the probability of the cat being a female if we select a random cat from the population. This is very difficult to determine with such a large population. Now, we randomly select 1000 cats from the total population and consider this as a “Sample”.Now, what is the probability that the selected cat from this random sample will be a female? This is nothing but a real-life example of the hypergeometric distribution. If we closely observe this theory, we can find that this is an extended version of the binomial distribution. Negative hypergeometric distribution This is similar to a negative binomial distribution with respect to the context of the hypergeometric distribution. Here, we need to measure the number of failed occurrences before the predefined number of successes. For the same real-world scenario explained above, what is the probability of getting exactly 120 female cats until we find 20 male cats if we count in the total cat population of New York (1 million)? This is nothing but an example of the negative hypergeometric distribution. Hope it helped you !!! Aditya N. Joshi Student of Probability · Author has 165 answers and 989.1K answer views · Updated 9y Related What is the binomial distribution? The Binomial Distribution is a probability distribution for a random variable [math]X[/math] which can take on only two discrete values. First, what is a random variable? A random variable is just a fancy way of associating a particular outcome with a variable. Let's illustrate this with an example. Let's define a random variable [math]X[/math] with respect to a coin flip. [math]X[/math] equals [math]1[/math] if heads and equals [math] 0 [/math] if tails. So tha The Binomial Distribution is a probability distribution for a random variable [math]X[/math] which can take on only two discrete values. First, what is a random variable? A random variable is just a fancy way of associating a particular outcome with a variable. Let's illustrate this with an example. Let's define a random variable [math]X[/math] with respect to a coin flip. [math]X[/math] equals [math]1[/math] if heads and equals [math] 0 [/math] if tails. So that illustrates the fact that the random variable [math]X[/math] has only two discrete possibilities - Heads or Tails. Now let's say we want to plot the Binomial Distribution for a fair coin that is flipped [math]n = 5[/math] times. Let [math]k[/math] be the number of heads observed after [math]n = 5[/math] flips. What is the probability of different [math]k[/math]'s? Our random variable [math]k[/math] now denotes the number of heads observed after [math]5[/math] flips. Well we can either get [math]0[/math], [math]1[/math], [math]2[/math], [math]3[/math], [math]4[/math] or [math]5[/math] heads. We find the probability of each of these cases and plot them on a graph. Since the coin is fair the probability of getting heads and getting tails is equal to [math]\frac {1}{2} [/math] The probability of getting [math]k[/math] heads is calculated with the formula illustrated by User-13259913648527583542 [ ] : [math]P(k= 0) = [/math] [math]{{5}\choose{0}} (0.5)^{0} (0.5)^{5} = 0.03125 (3.125\%) [/math] [math]P(k= 1) =[/math] [math]{{5}\choose{1}} (0.5)^{1} (0.5)^{4} = 0.15625 (15.625\%)[/math] [math]P(k= 2) =[/math] [math]{{5}\choose{2}} (0.5)^{2} (0.5)^{3} = 0.3125 (31.25\%)[/math] [math]P(k= 3) =[/math] [math]{{5}\choose{3}} (0.5)^{3} (0.5)^{2} = 0.3125 (31.25\%)[/math] [math]P(k= 4) =[/math] [math]{{5}\choose{4}} (0.5)^{4} (0.5)^{1} = 0.15625 (15.625\%)[/math] [... Promoted by Spokeo Spokeo - People Search \| Dating Safety Tool Dating Safety and Cheater Buster Tool · Apr 16 Is there a way to check if someone has a dating profile? Originally Answered: Is there a way to check if someone has a dating profile? 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Dating Safety Check – Review your date’s background to help keep you safe. Terry Moore PhD in statistics · Author has 16.6K answers and 29.3M answer views · Updated 6y Related How do we distinguish between hypergeometric, binomial, and Poisson distributions? How can you differentiate between binomial, hypergeometric, or Poisson? Well, the formulae for the probabilities are different for the binomial, hypergeometric and Poisson distributions. Apart from that, one obvious distinguishing factor for the Poisson distribution is its domain, the others have a finite number of possible values, the Poisson has an infinite number of possibilities. (So in practice it doesn’t exist, nothing in the real world is infinite, but the probability of values so large that they are unobservable is negligible under the Poisson, so the model can work well in practice.) If you mean how can you decide which model to use, you need to know how t Well, the formulae for the probabilities are different for the binomial, hypergeometric and Poisson distributions. Apart from that, one obvious distinguishing factor for the Poisson distribution is its domain, the others have a finite number of possible values, the Poisson has an infinite number of possibilities. (So in practice it doesn’t exist, nothing in the real world is infinite, but the probability of values so large that they are unobservable is negligible under the Poisson, so the model can work well in practice.) If you mean how can you decide which model to use, you need to know how they arise in practice. The binomial applies to a fixed number of independent binary events (i.e. two possibilities which are often called success or failure) with success having the same probability at each trial. If you can recognise that these assumptions hold and you want the probability distribution of the number of successes then you have a binomial distribution. The hypergeometric applies to a similar situation to the binomial except that the success probability at each trial changes and the events are not independent. Specifically, if you are selecting people randomly from a population and you classify the people into two categories, then at each draw there is one less person, and one less person of the category observed at the previous draw. If the population is large and you only take a small proportion of the population, the distribution is approximately binomial, but when sampling from a small population you need to use the hypergeometric distribution. The Poisson distribution also applies to independent events, but there is no a fixed population. It can be used for accidents that occur at random times and we count the number that occur over a fixed time interval. It can be derived from the binomial by letting the number of trials increase indefinitely. Martin Jansche biased estimator · Author has 3.7K answers and 3.7M answer views · Updated 1y Related How do I formulate the likelihood of a binomial mixture? I know the number of successes (k) and success probability (p) in a set of experiments but not the number of trials (n) in each experiment (n varies between experiments, p is constant)? [Rewritten completely based on a long discussion in the comments.] The scenario in question involves two distinct uses of binomial variables. We’ll illustrate this by describing the assumed data generation process. This is couched in terms of latent experiment results which are assumed to have been drawn from three distinct binomial distributions, with different parameters [math]n\in{1,2,3}[/math] and a known, shared parameter [math]p[/math]. We’ll denote the latent experiment results by a sequence of random variables [math]X_i[/math]. These come from [math]N_1[/math] iid experiments with [math]n=1[/math]; and from [math]N_2[/math] iid experiments with [math]n=2[/math]; and from [math]N_3[/math] [Rewritten completely based on a long discussion in the comments.] The scenario in question involves two distinct uses of binomial variables. We’ll illustrate this by describing the assumed data generation process. This is couched in terms of latent experiment results which are assumed to have been drawn from three distinct binomial distributions, with different parameters [math]n\in{1,2,3}[/math] and a known, shared parameter [math]p[/math]. We’ll denote the latent experiment results by a sequence of random variables [math]X_i[/math]. These come from [math]N_1[/math] iid experiments with [math]n=1[/math]; and from [math]N_2[/math] iid experiments with [math]n=2[/math]; and from [math]N_3[/math] iid experiments with [math]n=3[/math]: [math]X_1, \dots, X_{N_1} \sim \mathrm{binom}(1, p)[/math] [math]X_{N_1+1}, \dots, X_{N_1+N_2} \sim \mathrm{binom}(2, p)[/math] [math]X_{N_1+N_2+1}, \dots, X_{N_1+N_2+N_3} \sim \mathrm{binom}(3, p)[/math] We do not observe the [math]X_i[/math] directly. Instead we observe how often they take on the values 1, 2, and 3. Using Iverson bracket notation: [math]\displaystyle K_1 = \sum_i [X_i = 1][/math] [math]\displaystyle K_2 = \sum_i [X_i = 2][/math] [math]\displaystyle K_3 = \sum_i [X_i = 3][/math] Note that each [math]K_j[/math] is the sum of [math]4-j[/math] binomial random variables. We can see this by writing e.g. [math]\displaystyle K_2 = K_{2,2} + K_{2,3} = \sum_{i=N_1+1}^{N_1+N_2} [X_i = 2] + \sum_{i=N_1+N_2+1}^{N_1+N_2+N_3} [X_i = 2][/math] Then we find that [math]K_{2,2} \sim \mathrm{binom}(N_2, f(2 \mid 2, p))[/math] [math]K_{2,3} \sim \mathrm{binom}(N_3, f(2 \mid 3, p))[/math] where [math]f[/math] is the binomial probability mass function. Here [math]K_{2,3}[/math] is the count of experiments with outcome 2 that arose from the experiments conducted with [math]n=3[/math]. The challenge is that [math]K_2 = K_{2,2} + K_{2,3}[/math] is the sum of two binomial RVs, but is generally not a binomial RV itself. We’ll return to this issue below. We are faced with the following problem: Given [math]K_1[/math], [math]K_2[/math], [math]K_3[/math], and [math]p[/math], what can we say about [math]N_1[/math], [math]N_2[/math], and [math]N_3[/math]? I would approach this problems in a Bayesian setting, by putting independent priors on [math]N_1[/math], [math]N_2[/math], and [math]N_3[/math], and performing posterior inference by simulation. Concretely, using PyMC: import arviz import numpy import pymc as pm import scipy p = 0.75 # Sample from the assumed data generating process:experiments = numpy.concatenate([ scipy.stats.binom(1, p).rvs(10), scipy.stats.binom(2, p).rvs(50), scipy.stats.binom(3, p).rvs(100),]) # This gets us the following observations:k1 = numpy.sum(experiments == 1)k2 = numpy.sum(experiments == 2)k3 = numpy.sum(experiments == 3) def f(k, n): return scipy.stats.binom.pmf(k, n, p) with pm.Model() as model: # Uniform priors on the three experiment sizes: n1 = pm.DiscreteUniform("n1", lower=0, upper=200) n2 = pm.DiscreteUniform("n2", lower=0, upper=200) n3 = pm.DiscreteUniform("n3", lower=0, upper=200) # The observed count of threes can only have come from # the experiment with n=3. We can model it directly as a # binomial variable: pm.Binomial("k3", n=n3, p=f(3, 3), observed=k3) # The other observed counts arise as sums of binomial # RVs with unequal success probabilities. # Let's approximate those with normal variables: m11 = n1 f(1, 1) m12 = n2 f(1, 2) m13 = n3 f(1, 3) v11 = m11 (1 - f(1, 1)) v12 = m12 (1 - f(1, 2)) v13 = m13 (1 - f(1, 3)) mean1 = m11 + m12 + m13 var1 = v11 + v12 + v13 pm.Normal("k1", mu=mean1, sigma=var10.5, observed=k1) m22 = n2 f(2, 2) m23 = n3 f(2, 3) v22 = m22 (1 - f(2, 2)) v23 = m23 (1 - f(2, 3)) mean2 = m22 + m23 var2 = v22 + v23 pm.Normal("k2", mu=mean2, sigma=var20.5, observed=k2) with model: idata = pm.sample(10_000) arviz.plot_posterior(idata, show=True) This resulted in the following posterior distributions for me: The highest density intervals of these posteriors all include the true values from which the data were generated in the first place. This works well enough for relatively large values of [math]p[/math], which means we see a lot of 3s, which in turn reveal a lot of information about [math]N_3[/math]. For small values of [math]p[/math], however, there is not enough information in the observations to draw very reliable inferences. We’ll predominantly see 0s and 1s, which could have plausibly come from any of the underlying experiment types. For very small [math]p[/math] the normal approximation is increasingly inappropriate, but I doubt that this is the main problem. Still, one refinement would be to replace the normal approximation of the distribution of [math]K_1[/math] and [math]K_2[/math] with the proper convolution corresponding to the sum of two or three binomial variables. One would need to work out at least the unnormalized log-likelihood as part of a custom distribution. I’ll leave this as an exercise for a later time. Edited to add: On further reflection, here is an additional perspective. What I said above captures the marginal distributions of the observed counts, but does not fully reflect the data generation process (because it does not enforce the correct marginal totals and because it approximates the convolution of the marginals with a normal distribution). Let’s fix that. We first need a matrix of the relevant binomial probabilities. (I’m going to make it symmetric so that I don’t confuse myself about the indices.) In Python: import numpy as npfrom scipy.stats import binom def binomial_probabilities(p): return np.array([ [binom.pmf(k, n, p) for k in range(4)] for n in range(4) ]) Then for example: P = binomial_probabilities(0.8)>>> print(P) The observations can be viewed as having been generated as follows. Let’s say for the moment that we know the true underlying experiment counts [math]n_1[/math], [math]n_2[/math], and [math]n_3[/math]: n1, n2, n3 = 10, 50, 100n = np.array([0, n1, n2, n3]) The experiment with size 3 generates counts according to the following multinomial distribution: from scipy.stats import multinomial np.random.seed(1923)c = multinomial.rvs(n, P) >>> print(c)[ 0 8 36 56] As you can see, this correctly sums to [math]n_3=100[/math] and we have [math]c_k\sim\mathrm{binom}(n_3, f(k\mid 3, 0.8))[/math] where [math]f[/math] is the binomial pmf. The complete latent data are generated as follows: K = np.array([ multinomial.rvs(n[i], P[i]) for i in range(4)]) >>> print(K) The sum along the rows of [math]K[/math] gives us the underlying experiment counts: ``` print(K.sum(1))[ 0 10 50 100] ``` The sum along the columns of [math]K[/math] gives us the observed data math[/math], ignoring the fact that we don’t observe [math]k_0[/math]: ``` print(K.sum(0))[ 5 34 70 51] ``` The original formulation of the inference problem was this: Given a known [math]p[/math] and observations math[/math], find the underlying math[/math]. We now modify this as follows: Given a known [math]p[/math] and observations math[/math], find the complete latent data [math]K[/math]. The matrix [math]K[/math] contains the complete information about the setup that is assumed to have generated the data. Not only does it tell us the marginal counts, it in fact tells us e.g. how many of the observations with label “2” came from the experiment with parameter value 3: that count is [math]K_{3,2}[/math]. The likelihood for [math]K[/math] can be combined with flat independent priors over the [math]n_i[/math] to give us a posterior distribution. Independence of the [math]n_i[/math] means the rows of [math]K[/math] are conditionally independent, hence the log-likelihood of [math]K[/math] is simply: def log_likelihood(K, P): return multinomial.logpmf(K, K.sum(1), P).sum() >>> print(log_likelihood(K, P)) -11.253706578210695 If necessary, the likelihood of math[/math] could be computed by summing the likelihoods of all [math]K[/math] matrices with fixed row-wise and column-wise margins. For example has the same margin totals as the example of [math]K[/math] above. With the margins fixed, there are finitely many such matrices, and they can be enumerated straightforwardly. That said, I don’t think this is strictly necessary. No matter whether we look at the likelihood of [math]K[/math] or the likelihood of its marginal totals math[/math], we are facing a similar discrete optimization problem. A simple way of approaching this is via random-walk Metropolis-Hastings, in which case it doesn’t seem hugely important whether we carry out that random walk on math[/math] or directly on [math]K[/math]. (To be continued.) Steven Smith Earned 98% or higher in all my math classes at UCMO. · Author has 3.4K answers and 9.1M answer views · 7y Related How can I differ binomial distribution, geometric distribution, and hypergeometric distribution in a test? What are some examples? Both the binomial distribution and geometric distribution deal with independent trials with equal probability. Some common examples for that include rolling dice, drawing items WITH REPLACEMENT, getting item drops in a videogame like WoW, flipping coins, etc… In all of those examples, there’s no relation between the result of one attempt and the result of another attempt. If this is satisfied, then you have to know what information you are trying to determine. In the binomial distribution, you want to know the probability of getting a certain number of successes in an exact number of attempts. For Both the binomial distribution and geometric distribution deal with independent trials with equal probability. Some common examples for that include rolling dice, drawing items WITH REPLACEMENT, getting item drops in a videogame like WoW, flipping coins, etc… In all of those examples, there’s no relation between the result of one attempt and the result of another attempt. If this is satisfied, then you have to know what information you are trying to determine. In the binomial distribution, you want to know the probability of getting a certain number of successes in an exact number of attempts. For example, the probability of seeing a 6 3 times when you roll a die 10 times. In the geometric distribution, you want to know the probability of your VERY FIRST success happening with a certain number of attempts. For example, the probability that your first heads will happen on the 5th flip of a coin. The hypergeometric distribution is different. This does not deal with independence, this is commonly used to calculate probabilities when you draw a certain number of items WITHOUT REPLACEMENT. Some examples of where this could be applied are drawing items to inspect for defective items, your tickets being drawn in a raffle, or drawing cards from a deck of cards. For example, if you draw 10 cards from a deck of cards, and you want to know the probability that 2 of those are aces. Terry Moore PhD in statistics · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 16.6K answers and 29.3M answer views · 8y Related What is the difference between Poisson distribution, geometric distribution and hypergeometric distribution? The Poisson distribution, Geometric distribution and Hypergeometric distributions are all discrete and take all positive integer values. The Poisson and hyoergeometric distributions also take the value 0. The geometric distribution doesn’t, but a simple modification of it does. The geometric distribution is given by the number of trials before the first failure in a sequence of independent Bernoulli trials. This number-1 is the number of successes before the first failure and can be zero if a failure occurs at the first trial. So it’s just a custom that the geometric distribution can’t take the The Poisson distribution, Geometric distribution and Hypergeometric distributions are all discrete and take all positive integer values. The Poisson and hyoergeometric distributions also take the value 0. The geometric distribution doesn’t, but a simple modification of it does. The geometric distribution is given by the number of trials before the first failure in a sequence of independent Bernoulli trials. This number-1 is the number of successes before the first failure and can be zero if a failure occurs at the first trial. So it’s just a custom that the geometric distribution can’t take the value zero. (Some writers swap the words success and failure in this definition.) The negative binomial occurs in a similar way as the number of trials before n failures (or the number of successes before n failures—this is better as it has a version with non-integer n, but having nothing to do with Bernoulli trials). The hypergeometric is a generalisation of the binomial distribution. The binomial is the number of successes in n independent Bernoulli trials. In the hypergeometric the trials are not independent. Think of balls of two colours drawn from an urn without replacement. If the balls are replaced the number of one colour has a binomial distribution. If they are not replaced you get a hypergeometric distribution. The Poisson distribution arises from a continuous model of events occurring independently and randomly in time or space. The number of events in a given time period (or the number of weeds in a given area) has a Poisson distribution. (Well, no model is perfect: a weed will inhibit growth of nearby weeds.) It would be a nice exercise for you to find a formula for each of the distributions mentioned above. Maybe the Poisson is a little harder than the others. One way to get it is to let p in the binomial distribution tend to zero and the number of trials tend to infinity while k = np remains constant. You need to know that (1+x/n)^n tends to e^n as n tends to infinity. Related questions What is the difference between binomial and hypergeometric distribution? What is the critical value for a binomial distribution with a given probability? What are the difference and similiarities between Binomial and Hypergeometric Probability distribution? Define binomial distribution. How does it differ from Hypergeometric distribution? How do you describe probability distribution with special reference to binomial and hypergeometric distribution? What is the formula for binomial distribution? What is the most likely value in a binomial distribution? How can I recognize a Binomial probability distribution problem? What is the Binomial Probability Formula? Does the default rate follow normal distribution or binomial distribution? 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16339	https://www.feinberg.northwestern.edu/sites/firstdailylife/docs/Microscopic_Polyangiitis.pdf	2022 American College of Rheumatology/European Alliance of Associations for Rheumatology Classiﬁcation Criteria for Microscopic Polyangiitis Ravi Suppiah,1 Joanna C. Robson,2 Peter C. Grayson,3 Cristina Ponte,4 Anthea Craven,5 Sara Khalid,5 Andrew Judge,6 Andrew Hutchings,7 Peter A. Merkel,8 Raashid A. Luqmani,5 and Richard A. Watts9 Objective. To develop and validate classiﬁcation criteria for microscopic polyangiitis (MPA). Methods. Patients with vasculitis or comparator diseases were recruited into an international cohort. The study proceeded in 5 phases: 1) identiﬁcation of candidate items using consensus methodology, 2) prospective collection of candidate items present at the time of diagnosis, 3) data-driven reduction of the number of candidate items, 4) expert panel review of cases to deﬁne the reference diagnosis, and 5) derivation of a points-based risk score for disease clas-siﬁcation in a development set using least absolute shrinkage and selection operator logistic regression, with subse-quent validation of performance characteristics in an independent set of cases and comparators. Results. The development set for MPA consisted of 149 cases of MPA and 408 comparators. The validation set consisted of an additional 142 cases of MPA and 414 comparators. From 91 candidate items, regression analysis iden-tiﬁed 10 items for MPA, 6 of which were retained. The ﬁnal criteria and their weights were as follows: perinuclear antineutrophil cytoplasmic antibody (ANCA) or anti–myeloperoxidase-ANCA positivity (+6), pauci-immune glomerulo-nephritis (+3), lung ﬁbrosis or interstitial lung disease (+3), sino-nasal symptoms or signs (3), cytoplasmic ANCA or anti–proteinase 3 ANCA positivity (1), and eosinophil count ≥1 109/liter (4). After excluding mimics of vasculitis, a patient with a diagnosis of small- or medium-vessel vasculitis could be classiﬁed as having MPA with a cumulative score of ≥5 points. When these criteria were tested in the validation data set, the sensitivity was 91% (95% conﬁdence interval [95% CI] 85–95%) and the speciﬁcity was 94% (95% CI 92–96%). Conclusion. The 2022 American College of Rheumatology/European Alliance of Associations for Rheumatology classiﬁcation criteria for MPA are now validated for use in clinical research. This article is published simultaneously in the March 2022 issue of Annals of the Rheumatic Diseases. The Diagnostic and Classiﬁcation Criteria in Vasculitis (DCVAS) study, of which the development of these classiﬁcation criteria was a part, was funded by grants from the American College of Rheumatology (ACR), the European Alliance of Associations for Rheumatology (EULAR), the Vasculitis Foundation, and the University of Pennsylvania Vasculitis Center. 1Ravi Suppiah, MD: Auckland District Health Board, Auckland, New Zealand; 2Joanna C. Robson, MBBS, PhD: Centre for Health and Clinical Research, University of the West of England, and University Hospitals and Weston NHS Foundation Trust, Bristol, UK; 3Peter C. Grayson, MD, MSc: National Institute of Arthritis and Musculoskeletal and Skin Diseases, NIH, Bethesda, Maryland; 4Cristina Ponte, MD, PhD: Hospital de Santa Maria, Centro Hospitalar Univer-sit ario Lisboa Norte, Universidade de Lisboa, and Centro Académico de Medicina de Lisboa, Lisbon, Portugal; 5Anthea Craven, BSc, Sara Khalid, DPhil, Raashid A. Luqmani, DM, FRCP: Oxford NIHR Biomedical Research Centre and University of Oxford, Oxford, UK; 6Andrew Judge, PhD: Oxford NIHR Biomedical Research Centre and University of Oxford, Oxford, UK, and Bristol NIHR Biomedical This criteria set has been approved by the American College of Rheumatology (ACR) Board of Directors and the European Alliance of Associations for Rheumatology (EULAR) Executive Committee. This signiﬁes that the criteria set has been quantitatively validated using patient data, and it has undergone validation based on an independent data set. All ACR/EULAR-approved criteria sets are expected to undergo intermittent updates. The ACR is an independent, professional, medical and scientiﬁc society that does not guarantee, warrant, or endorse any commercial product or service. 1 Arthritis & Rheumatology Vol. 0, No. 0, Month 2022, pp 1–7 DOI 10.1002/art.41983 © 2022, American College of Rheumatology INTRODUCTION The ﬁrst description of “periarteritis nodosa” was made by Kussmaul and Maier in 1866 (1). In 1948, Davson et al described 14 cases at autopsy that ﬁtted the clinical description of periarter-itis nodosa (2). They divided the cases into 2 groups based on the histologic ﬁndings in the kidneys. The clinical presentations of both groups were similar, but their pathologic features differed: 9 patients showed a distinctive pattern of necrotizing glomerulo-nephritis with no arterial aneurysms, whereas the other 5 patients showed no glomerular lesions in the kidney but had widespread renal arterial aneurysms and renal infarcts. This is the ﬁrst time that a clear distinction was made between the microscopic form of polyarteritis nodosa (now called microscopic polyangiitis [MPA]) and classic polyarteritis nodosa (PAN). The 1990 American Col-lege of Rheumatology (ACR) criteria for the classiﬁcation of vascu-litis did not make this distinction; instead both entities were included under the term “polyarteritis nodosa” (3) or possibly “granulomatosis with polyangiitis” (then called Wegener’s granulomatosis). The publication that resulted from the 1994 Chapel Hill Consensus Conference (CHCC) aimed to standardize the nomenclature and commented that “different names are being used for the same disease and the same name is being used for different diseases” (4). The distinction between MPA and PAN is recognized in the CHCC deﬁnitions. The main discrimi-nating feature between MPA and PAN is the presence in MPA of pauci-immune vasculitis in arterioles, venules, or capillaries. PAN is restricted to a medium-vessel disease, and MPA is a pre-dominantly small-vessel vasculitis that can also involve medium-sized vessels. The resulting inconsistency between disease deﬁnitions and existing classiﬁcation criteria highlights an important need to update the classiﬁcation criteria and to include MPA as its own entity. Additionally, over time there have been improvements in our understanding of the different forms of vasculitis, which have been informed in part by routine testing for antineutrophil cyto-plasmic antibody (ANCA) in patients with vasculitis and increased utilization of cross-sectional imaging, both of which have occurred since the 1990 ACR criteria were published. Indeed, most investigators regard MPA as part of the group of small-vessel vasculitides related to the presence of ANCA. This article outlines the development and validation of the new ACR/ European Alliance of Associations for Rheumatology (EULAR)– endorsed classiﬁcation criteria for MPA. METHODS A detailed and complete description of the methods involved in the development and validation of the classiﬁcation criteria for MPA is provided in Supplementary Appendix 1, available on the Arthritis & Rheumatology website at com/doi/10.1002/art.41983/abstract. Brieﬂy, an international Steering Committee comprising clinician investigators with exper-tise in vasculitis, statisticians, and data managers was established to oversee the overall Diagnostic and Classiﬁcation Criteria in Vas-culitis (DCVAS) project. The Steering Committee established a 5-stage plan using data-driven and consensus methodology to develop the criteria for each of 6 forms of vasculitis. Stage 1: generation of candidate classiﬁcation items for the systemic vasculitides. Candidate classiﬁcation items were generated by expert opinion and reviewed by a group of vasculitis experts across a range of specialties using a nominal group technique. Stage 2: DCVAS prospective observational study. A prospective, international multisite observational study was con-ducted (see Appendix A for study investigators and sites). Ethical approval was obtained from national and local ethics committees. Consecutive patients representing the full spectrum of disease were recruited from academic and community practices. Patients were included if they were 18 years or older and had a diagnosis of vas-culitis or a condition that mimics vasculitis. Patients with ANCA-associated vasculitis (AAV) could only be enrolled within 2 years of diagnosis. Only data present at diagnosis were recorded. Stage 3: reﬁnement of candidate items speciﬁcally for AAV. The Steering Committee conducted a data-driven pro-cess to reduce the number of candidate items of relevance to cases and comparators for AAV. Items were selected for exclu-sion if they had a prevalence of <5% within the data set and/or they were not clinically relevant for classiﬁcation criteria (e.g., related to infection, malignancy, or demographic characteristics). Low-frequency items of clinical importance could be combined, when appropriate. Stage 4: expert review to derive a gold standard– deﬁned set of cases of AAV. Experts in vasculitis from a wide range of geographic locations and specialties reviewed all submit-ted cases of vasculitis and a random selection of mimics of Research Centre and University of Bristol, Bristol, UK; 7Andrew Hutchings, MSc: London School of Hygiene and Tropical Medicine, London, UK; 8Peter A. Merkel, MD, MPH: University of Pennsylvania, Philadelphia; 9Richard A. Watts, MD: Oxford NIHR Biomedical Research Centre and University of Oxford, Oxford, UK, and Uni-versity of East Anglia, Norwich, UK. Author disclosures are available at downloadSupplement?doi=10.1002%2Fart.41983&ﬁle=art41983-sup-0001-Disclosureform.pdf. Address correspondence to Peter A. Merkel, MD, MPH, Chief, Division of Rheumatology, University of Pennsylvania, White Building, Fifth Floor, 3400 Spruce Street, Philadelphia, PA 19104 (email: pmerkel@upenn.edu); or to Raashid A. Luqmani, DM, FRCP, Rheumatology Department, Nufﬁeld Ortho-paedic Centre, University of Oxford, Windmill Road, Oxford OX3 7LD, UK (email: raashid.luqmani@ndorms.ox.ac.uk). Submitted for publication February 28, 2020; accepted in revised form September 21, 2021. SUPPIAH ET AL 2 vasculitis. Each reviewer was asked to review ~50 submitted cases to conﬁrm the diagnosis and to specify the certainty of their diagnosis as follows: very certain, moderately certain, uncertain, or very uncertain. Only cases agreed upon with at least moderate certainty were retained for further analysis. Stage 5: derivation and validation of the ﬁnal classi-ﬁcation criteria for MPA. The DCVAS AAV data set was ran-domly split into development (50%) and validation (50%) sets. Comparisons were performed between cases of MPA and a com-parator group randomly selected from the DCVAS cohort in the fol-lowing proportions: another type of AAV (including granulomatosis with polyangiitis [GPA] and eosinophilic granulomatosis with polyangiitis [EGPA]), 60%; another form of small-vessel vasculitis (e.g., cryoglobulinemic vasculitis) or medium-vessel vasculitis (e.g., PAN), 40%. Least absolute shrinkage and selection operator (lasso) logistic regression was used to identify items from the data set and create a parsimonious model including only the most important items. The ﬁnal items in the model were formulated into a clinical risk-scoring tool with each factor assigned a weight based on its respective regression coefﬁcient. A threshold that best bal-anced sensitivity and speciﬁcity was identiﬁed for classiﬁcation. In sensitivity analyses, the ﬁnal classiﬁcation criteria were applied to an unselected population of cases and comparators from the DCVAS data set based on the submitting physician diagnosis. RESULTS Generation of candidate classiﬁcation items for the systemic vasculitides. The Steering Committee identiﬁed >1,000 candidate items for the DCVAS case report form (see Supplementary Appendix 2, available on the Arthritis & Rheuma-tology website at 41983/abstract). DCVAS prospective observational study. Between January 2011 and December 2017, the DCVAS study recruited 6,991 participants from 136 sites in 32 countries. Information on the DCVAS sites, investigators, and participants is listed in Supplementary Appendices 3, 4, and 5, available on the Arthritis & Rheumatology website at 1002/art.41983/abstract. Reﬁnement of candidate items speciﬁcally for AAV. Following a data-driven and expert consensus process, 91 items from the DCVAS case report form were retained for regression analysis, including 45 clinical (14 composite), 18 laboratory (2 composite), 12 imaging (all composite), and 16 biopsy (1 com-posite) items. Some clinical items were removed in favor of similar but more speciﬁc pathophysiologic descriptors. For example, “Hearing loss or reduction” was removed, and the composite item “Conductive hearing loss/sensorineural hearing loss” was retained. See Supplementary Appendix 6, available on the Arthri-tis & Rheumatology website at 10.1002/art.41983/abstract, for the ﬁnal candidate items used in the derivation of the classiﬁcation criteria for GPA, MPA, and EGPA. Expert review to derive a gold standard–deﬁned ﬁnal set of cases of AAV. Fifty-ﬁve independent experts reviewed vignettes derived from the case report forms for 2,871 cases submitted with a diagnosis of either small-vessel vasculitis (90% of case report forms) or another type of vasculitis or a mimic of vasculitis (10% of case report forms). The characteristics of the expert reviewers are shown in Supplementary Appendix 7, available on the Arthritis & Rheumatology website at http:// onlinelibrary.wiley.com/doi/10.1002/art.41983/abstract. A ﬂow chart showing the results of the expert review process is shown in Supplementary Appendix 8, available on the Arthritis & Table 1. Demographic and disease features of cases of MPA and comparators MPA (n = 291) Comparators (n = 822)† P Age, mean SD years 65.5 13.2 52.0 16.9 <0.001 Sex, no. (%) female 164 (56.4) 394 (47.9) 0.016 Maximum serum creatinine, mean <0.001 μmoles/liter 126.4 185.2 mg/dl 1.4 2.1 cANCA positive, no. (%) 11 (3.8) 257 (31.3) <0.001 pANCA positive, no. (%) 236 (81.1) 136 (16.5) <0.001 Anti–PR3-ANCA positive, no. (%) 6 (2.1) 265 (32.2) <0.001 Anti–MPO-ANCA positive, no. (%) 279 (95.9) 142 (17.3) <0.001 Maximum eosinophil count ≥1 109/liter, no. (%) 15 (5.2) 244 (29.7) <0.001 cANCA = cytoplasmic antineutrophil cytoplasmic antibody; pANCA = perinuclear ANCA; anti–PR3-ANCA = anti–proteinase 3–ANCA; anti–MPO-ANCA = anti–myeloperoxidase-ANCA. † Diagnoses of comparators for the classiﬁcation criteria for microscopic polyangiitis (MPA) included granulo-matosis with polyangiitis (n = 300), eosinophilic granulomatosis with polyangiitis (n = 226), polyarteritis nodosa (n = 51), non–ANCA-associated small-vessel vasculitis that could not be subtyped (n = 51), Behçet’s dis-ease (n = 50), IgA vasculitis (n = 50), cryoglobulinemic vasculitis (n = 34), ANCA-associated vasculitis that could not be subtyped (n = 25), primary central nervous system vasculitis (n = 19), and anti–glomerular basement membrane disease (n = 16). ACR/EULAR CLASSIFICATION CRITERIA FOR MPA 3 Rheumatology website at 1002/art.41983/abstract. A total of 2,072 cases (72%) passed the process and were designated as cases of vasculitis; these cases were used for the stage 5 analyses. After expert panel review by 55 investigators, 269 of 404 of the cases retained the submitting physician diagnosis of MPA, and 22 additional cases were reclassiﬁed as having MPA by consensus of 2 expert reviewers. Compared to the 291 patients with a reference diagnosis of MPA, the 135 cases that were excluded had lower rates of perinuclear ANCA (pANCA) or anti–myeloperoxidase-ANCA (anti–MPO-ANCA) positivity (76% versus 98%; P < 0.01), were less likely to have pauci-immune glomerulonephritis (16% versus 49%; P < 0.01), were more likely to have maximum eosinophil counts ≥1 109/liter (12% versus 6%; P = 0.02), and were more likely to be cytoplasmic ANCA– or proteinase 3–ANCA–positive (20% versus 4%; P < 0.01). There were 822 comparators randomly selected for analysis. Table 1 shows the demographic and disease features of the 1,113 cases included in this analysis (291 patients with MPA and 822 comparators), of which 557 (50%; 149 patients with MPA and 408 comparators) were in the development set, and 556 (50%; 142 patients with MPA and 414 comparators) were in the validation set. Derivation and validation of the ﬁnal classiﬁcation criteria for MPA. Lasso regression of the previously selected 91 items yielded 10 independent items for MPA (see Supplementary Appendix 9C, available on the Arthritis & Rheuma-tology website at 41983/abstract). Each item was then adjudicated by the DCVAS Steering Committee for inclusion based on clinical relevance and speciﬁcity to MPA, resulting in 6 ﬁnal items. Weighting of an indi-vidual criterion was based on logistic regression ﬁtted to the 6 selected items (see Supplementary Appendix 10C, available on the Arthritis & Rheumatology website at wiley.com/doi/10.1002/art.41983/abstract). Figure 1. 2022 American College of Rheumatology/European Alliance of Associations for Rheumatology classiﬁcation criteria for microscopic polyangiitis. SUPPIAH ET AL 4 Model performance. Use of a cutoff of ≥5 in total risk score (see Supplementary Appendix 11C, available on the Arthritis & Rheumatology website at 1002/art.41983/abstract, for different cut points) yielded a sensitiv-ity of 90.8% (95% conﬁdence interval [95% CI] 84.9–95.0%) and a speciﬁcity of 94.2% (95% CI 91.5–96.3%) in the validation set. The area under the curve for the model was 0.98 (95% CI 0.97–0.99) in the development set and 0.97 (95% CI 0.95–0.98) in the validation set for the ﬁnal MPA classiﬁcation criteria (Supplementary Appendix 12C, available on the Arthritis & Rheumatology website at http:// onlinelibrary.wiley.com/doi/10.1002/art.41983/abstract). The ﬁnal classiﬁcation criteria for MPA are shown in Figure 1 (for the slide presentation version, see Supplementary Figure 1, available on the Arthritis & Rheumatology website at com/doi/10.1002/art.41983/abstract). Sensitivity analysis. The classiﬁcation criteria for MPA were applied to 2,871 patients in the DCVAS database using the original physician-submitted diagnosis (n = 404 cases of MPA and 2,467 randomly selected comparators). Use of the same cut point of ≥5 points for the classiﬁcation for MPA yielded a similar speciﬁcity of 92.5% but a lower sensitivity of 82.4%. This is con-sistent with the a priori hypothesis that speciﬁcity would remain unchanged but sensitivity would be reduced in a population with fewer clearcut diagnoses of MPA (i.e., cases that did not pass expert panel review). DISCUSSION Presented here are the 2022 ACR/EULAR MPA classiﬁcation criteria. These are the ﬁrst formal criteria for MPA. A 5-stage approach has been used, underpinned by data from the multina-tional prospective DCVAS study and informed by expert review and consensus at each stage. The comparator group for devel-oping and validating the criteria were predominantly patients with other forms of AAV and other small- and medium-vessel vasculit-ides, the clinical entities where discrimination from MPA is difﬁcult, but important. The new criteria for MPA have excellent sensitivity and speciﬁcity and incorporate ANCA testing and modern imag-ing techniques. The criteria were designed to have face and con-tent validity for use in clinical trials and other research studies. These criteria are validated and intended for the purpose of classiﬁcation of vasculitis and are not appropriate for use in estab-lishing a diagnosis of vasculitis. The aim of the classiﬁcation cri-teria is to differentiate cases of MPA from similar types of vasculitis in research settings. Therefore, the criteria should only be applied when a diagnosis of small- or medium-vessel vasculitis has been made and all potential “vasculitis mimics” have been excluded. The exclusion of mimics is a key aspect of many classi-ﬁcation criteria, including those for Sjögren’s syndrome (5) and rheumatoid arthritis (6). The 1990 ACR classiﬁcation criteria for vasculitis perform poorly when used for diagnosis (i.e., when used to differentiate between cases of vasculitis versus mimics without vasculitis) (7), and it is expected that the 2022 criteria would also perform poorly if used inappropriately as diagnostic criteria in people in whom alternative diagnoses, such as infection or other non-vasculitis inﬂammatory diseases, are still being considered. The relatively low weight assigned to glomerulonephritis in these classiﬁcation criteria highlights the distinction between classiﬁca-tion and diagnostic criteria. While detection of kidney disease is important to diagnose MPA, glomerulonephritis is common among patients with either GPA or MPA and thus does not func-tion as a strong classiﬁer between these conditions. MPA was not recognized as a separate entity in the 1990 ACR classiﬁcation criteria for vasculitis, although the disease was recognized as pathologically distinct from PAN over 40 years earlier. This omission of MPA caused difﬁculties in deﬁning clear homogeneous populations for research; thus, over the last 2 decades, investigators have often relied on the disease deﬁni-tions of the CHCC nomenclature for eligibility criteria when enroll-ing patients with MPA into clinical trials (4,8–11). This approach resulted in heterogeneity between patients enrolled in therapeutic trials and epidemiologic studies (12). Due to inconsistent methods employed by researchers when applying the 1990 ACR criteria and the CHCC deﬁnitions in parallel, the European Medicines Agency (EMA) convened meetings to develop a consensus on how to utilize the 2 systems, leading to the publication of the EMA algorithm in 2007 (13). The algorithm works by ﬁrst exclud-ing EGPA and GPA, and then relying on the CHCC histologic descriptions to discriminate between MPA and PAN. The new 2022 ACR/EULAR classiﬁcation criteria for MPA and other vascu-litides provide validated criteria that can replace the EMA interim solution and should harmonize future research studies. A potential limitation of these new criteria is that, through the expert panel consensus methodology, only the most deﬁnite cases were included in the analyses. However, the purpose of these criteria is to enable homogeneous groupings so that individ-ual diseases can be studied. Overall, the use of more deﬁnitive cases is consistent with the purpose of classiﬁcation criteria. Additionally, positive testing for MPO-ANCA is weighted heavily in the criteria, and it is theoretically possible to classify a patient as having MPA on the basis of a positive test for MPO-ANCA only. However, the criteria are intended to only be applied to patients with an established diagnosis of small- or medium-vessel vasculi-tis; in this setting, the criteria sets should result in a reduction of the score away from a classiﬁcation of MPA if the patient has fea-tures of another form of AAV. When criteria were tested in a much less clearly deﬁned population using the submitting physician diagnosis as the gold standard, the sensitivity of the criteria fell substantially despite 91% of this group being pANCA- or MPO-ANCA positive, which supports the contention that ANCA positiv-ity is not overly dominant for the classiﬁcation. Nonetheless, ANCA testing is obviously a key discriminator between the different forms of AAV and other small- and medium-vessel vasculitides. ACR/EULAR CLASSIFICATION CRITERIA FOR MPA 5 There are some additional study limitations to consider. Although this was the largest international study ever conducted in vasculitis, most patients were recruited from Europe, Asia, and North America. The performance characteristics of the cri-teria should be further tested in African and South American populations, which may have different clinical presentations of vasculitis. These criteria were developed using data collected from adult patients with vasculitis. Although the clinical character-istics of MPA and the other vasculitides which these criteria were tested against are not known to differ substantially between adults and children, these criteria should be applied to children with some caution. The scope of the criteria is intentionally narrow and applies only to patients who have been diagnosed as having vasculitis. Diagnostic criteria are not speciﬁed. The criteria are intended to identify homogeneous populations of disease and, therefore, may not be appropriate for studies focused on the full spectrum of clinical heterogeneity in these conditions. To maxi-mize relevance and face validity of the new criteria, study sites and expert reviewers were recruited from a broad range of coun-tries and different medical specialties. Nonetheless, the majority of patients were recruited from academic rheumatology or nephrol-ogy units, which could have introduced referral bias. The 2022 ACR/EULAR classiﬁcation criteria for MPA are the product of a rigorous methodologic process that utilized an extensive data set generated by the work of a remarkable inter-national group of collaborators. These are the ﬁrst classiﬁcation criteria for this disease. The criteria can now be applied to patients who have been diagnosed as having a small- or medium-vessel vasculitis. These criteria have been endorsed by the ACR and EULAR and are now ready for use to differentiate one type of vasculitis from another to deﬁne populations in research studies. ACKNOWLEDGMENTS We acknowledge the patients and clinicians who provided data to the DCVAS project. AUTHOR CONTRIBUTIONS All authors were involved in drafting the article or revising it critically for important intellectual content, and all authors approved the ﬁnal ver-sion to be published. Dr. Merkel had full access to all of the data in the study and takes responsibility for the integrity of the data and the accu-racy of the data analysis. Study conception and design. Suppiah, Robson, Grayson, Ponte, Cra-ven, Judge, Hutchings, Merkel, Luqmani, Watts. Acquisition of data. Suppiah, Robson, Grayson, Ponte, Craven, Merkel, Luqmani, Watts. Analysis and interpretation of data. Suppiah, Robson, Grayson, Ponte, Craven, Khalid, Judge, Hutchings, Merkel, Luqmani, Watts. REFERENCES 1. Kussmaul A, Maier R. 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APPENDIX A: THE DCVAS INVESTIGATORS The DCVAS study investigators are as follows: Paul Gatenby (ANU Medical Centre, Canberra, Australia); Catherine Hill (Central Adelaide Local Health Network: The Queen Elizabeth Hospital, Australia); Dwarakanathan Ranganathan (Royal Brisbane and Women’s Hospital, Australia); Andreas Kronbichler (Medical University Innsbruck, Austria); Daniel Blockmans (University Hospitals Leuven, Belgium); Lillian Barra (Lawson Health Research Institute, London, Ontario, Canada); Simon Carette, Christian Pagnoux (Mount Sinai Hospital, Toronto, Canada); Navjot Dhindsa (University of Manitoba, Winnipeg, Canada); Aurore Fiﬁ-Mah (University of Calgary, Alberta, Canada); Nader Khalidi (St Joseph’s Healthcare, Hamil-ton, Ontario, Canada); Patrick Liang (Sherbrooke University Hospital Cen-tre, Canada); Nataliya Milman (University of Ottawa, Canada); Christian SUPPIAH ET AL 6 Pineau (McGill University, Canada); Xinping Tian (Peking Union Medical College Hospital, Beijing, China); Guochun Wang (China-Japan Friendship Hospital, Beijing, China); Tian Wang (Anzhen Hospital, Capital Medical University, China); Ming-hui Zhao (Peking University First Hospital, China); Vladimir Tesar (General University Hospital, Prague, Czech Republic); Bo Baslund (University Hospital, Copenhagen [Rigshos-pitalet], Denmark); Nevin Hammam (Assiut University, Egypt); Amira Sha-hin (Cairo University, Egypt); Laura Pirila (Turku University Hospital, Finland); Jukka Putaala (Helsinki University Central Hospital, Finland); Bernhard Hellmich (Kreiskliniken Esslingen, Germany); Jörg Henes (Universitätsklinikum Tübingen, Germany); Peter Lamprecht (Klinikum Bad Bramstedt, Germany); Thomas Neumann (Universitätsklinikum Jena, Germany); Wolfgang Schmidt (Immanuel Krankenhaus Berlin, Germany); Cord Sunderkoetter (Universitätsklinikum Müenster, Germany); Zoltan Szekanecz (University of Debrecen Medical and Health Science Center, Hungary); Debashish Danda (Christian Medical College & Hospital, Vellore, India); Siddharth Das (Chatrapathi Shahuji Maharaj Medical Center, Luck-now [IP], India); Rajiva Gupta (Medanta, Delhi, India); Liza Rajasekhar (NIMS, Hyderabad, India); Aman Sharma (Postgraduate Institute of Medi-cal Education and Research, Chandigarh, India); Shrikant Wagh (Jehangir Clinical Development Centre, Pune [IP], India); Michael Clarkson (Cork University Hospital, Ireland); Eamonn Molloy (St. Vincent’s University Hospital, Dublin, Ireland); Carlo Salvarani (Santa Maria Nuova Hospital, Reggio Emilia, Italy); Franco Schiavon (L’Azienda Ospedaliera of University of Padua, Italy); Enrico Tombetti (Università Vita-Salute San Raffaele Milano, Italy); Augusto Vaglio (University of Parma, Italy); Koichi Amano (Saitama Medical University, Japan); Yoshihiro Arimura (Kyorin University Hospital, Japan); Hiroaki Dobashi (Kagawa University Hospital, Japan); Shouichi Fujimoto (Miyazaki University Hospital [HUB], Japan); Masayoshi Harigai, Fumio Hirano (Tokyo Medical and Dental University Hospital, Japan); Junichi Hirahashi (University Tokyo Hospital, Japan); Sakae Honma (Toho University Hospital, Japan); Tamihiro Kawakami (St. Marianna University Hospital Dermatology, Japan); Shigeto Kobayashi (Juntendo University Koshigaya Hospital, Japan); Hajime Kono (Teikyo University, Japan); Hirofumi Makino (Okayama University Hospital, Japan); Kazuo Matsui (Kameda Medical Centre, Kamogawa, Japan); Eri Muso (Kitano Hospital, Japan); Kazuo Suzuki, Kei Ikeda (Chiba University Hospital, Japan); Tsutomu Takeuchi (Keio University Hospital, Japan); Tat-suo Tsukamoto (Kyoto University Hospital, Japan); Shunya Uchida (Teikyo University Hospital, Japan); Takashi Wada (Kanazawa University Hospital, Japan); Hidehiro Yamada (St. Marianna University Hospital Internal Medi-cine, Japan); Kunihiro Yamagata (Tsukuba University Hospital, Japan); Wako Yumura (IUHW Hospital [Jichi Medical University Hospital], Japan); Kan Sow Lai (Penang General Hospital, Malaysia); Luis Felipe Flores-Suarez (Instituto Nacional de Enfermedades Respiratorias, Mexico City, Mexico); Andrea Hinojosa (Instituto Nacional de Ciencias Médicas y Nutrici on Salvador Zubir an, Mexico City, Mexico); Bram Rutgers (University Hospital Groningen, Netherlands); Paul-Peter Tak (Academic Medical Centre, University of Amsterdam, Netherlands); Rebecca Grain-ger (Wellington, Otago, New Zealand); Vicki Quincey (Waikato District Health Board, New Zealand); Lisa Stamp (University of Otago, Christ-church, New Zealand); Ravi Suppiah (Auckland District Health Board, New Zealand); Emilio Besada (Tromsø, Northern Norway, Norway); Andreas Diamantopoulos (Hospital of Southern Norway, Kristiansand, Norway); Jan Sznajd (University of Jagiellonian, Poland); Elsa Azevedo (Centro Hospitalar de S~ ao Jo~ ao, Porto, Portugal); Ruth Geraldes (Hospital de Santa Maria, Lisbon, Portugal); Miguel Rodrigues (Hospital Garcia de Orta, Almada, Portugal); Ernestina Santos (Hospital Santo Anto-nio, Porto, Portugal); Yeong-Wook Song (Seoul National University Hospi-tal, Republic of Korea); Sergey Moiseev (First Moscow State Medical University, Russia); Alojzija Hoc ˇevar (University Medical Centre Ljubljana, Slovenia); Maria Cinta Cid (Hospital Clinic de Barcelona, Spain); Xavier Solanich Moreno (Hospital de Bellvitge-Idibell, Spain); Inoshi Atukorala (University of Colombo, Sri Lanka); Ewa Berglin (Umeå University Hospital, Sweden); Aladdin Mohammed (Lund-Malmo University, Sweden); Mårten Segelmark (Linköping University, Sweden); Thomas Daikeler (University Hospital Basel, Switzerland); Haner Direskeneli (Marmara University Med-ical School, Turkey); Gulen Hatemi (Istanbul University, Cerrahpasa Medi-cal School, Turkey); Sevil Kamali (Istanbul University, Istanbul Medical School, Turkey); Ömer Karada g (Hacettepe University, Turkey); Seval Pehlevan (Fatih University Medical Faculty, Turkey); Matthew Adler (Frimley Health NHS Foundation Trust, Wexham Park Hospital, UK); Neil Basu (NHS Grampian, Aberdeen Royal Inﬁrmary, UK); Iain Bruce (Manchester University Hospitals NHS Foundation Trust, UK); Kuntal Chakravarty (Barking, Havering and Redbridge University Hospitals NHS Trust, UK); Bhaskar Dasgupta (Southend University Hospital NHS Foun-dation Trust, UK); Oliver Flossmann (Royal Berkshire NHS Foundation Trust, UK); Nagui Gendi (Basildon and Thurrock University Hospitals NHS Foundation Trust, UK); Alaa Hassan (North Cumbria University Hos-pitals, UK); Rachel Hoyles (Oxford University Hospitals NHS Foundation Trust, UK); David Jayne (Cambridge University Hospitals NHS Foundation Trust, UK); Colin Jones (York Teaching Hospitals NHS Foundation Trust, UK); Rainer Klocke (The Dudley Group NHS Foundation Trust, UK); Peter Lanyon (Nottingham University Hospitals NHS Trust, UK); Cathy Laver-such (Taunton & Somerset NHS Foundation Trust, Musgrove Park Hospi-tal, UK); Raashid Luqmani, Joanna Robson (Nufﬁeld Orthopaedic Centre, Oxford, UK); Malgorzata Magliano (Buckinghamshire Healthcare NHS Trust, UK); Justin Mason (Imperial College Healthcare NHS Trust, UK); Win Win Maw (Mid Essex Hospital Services NHS Trust, UK); Iain McInnes (NHS Greater Glasgow & Clyde, Gartnavel Hospital & GRI, UK); John Mclaren (NHS Fife, Whyteman’s Brae Hospital, UK); Matthew Morgan (University Hospitals Birmingham NHS Foundation Trust, Queen Elizabeth Hospital, UK); Ann Morgan (Leeds Teaching Hospitals NHS Trust, UK); Chetan Mukhtyar (Norfolk and Norwich University Hospitals NHS Founda-tion Trust, UK); Edmond O’Riordan (Salford Royal NHS Foundation Trust, UK); Sanjeev Patel (Epsom and St Helier University Hospitals NHS Trust, UK); Adrian Peall (Wye Valley NHS Trust, Hereford County Hospital, UK); Joanna Robson (University Hospitals Bristol NHS Foundation Trust, UK); Srinivasan Venkatachalam (The Royal Wolverhampton NHS Trust, UK); Erin Vermaak, Ajit Menon (Staffordshire & Stoke on Trent Partnership NHS Trust, Haywood Hospital, UK); Richard Watts (East Suffolk and North Essex NHS Foundation Trust, UK); Chee-Seng Yee (Doncaster and Bassetlaw Hospitals NHS Foundation Trust, UK); Daniel Albert (Dartmouth-Hitchcock Medical Center, US); Leonard Calabrese (Cleveland Clinic Foundation, US); Sharon Chung (University of California, San Francisco, US); Lindsy Forbess (Cedars-Sinai Medical Center, US); Angelo Gaffo (University of Alabama at Birmingham, US); Ora Gewurz-Singer (University of Michigan, US); Peter Grayson (Boston University School of Medicine, US); Kimberly Liang (University of Pittsburgh, US); Eric Matteson (Mayo Clinic, US); Peter A. Merkel (University of Pennsylvania, US); Jason Springer (University of Kansas Medical Center Research Insti-tute, US); and Antoine Sreih (Rush University Medical Center, US). ACR/EULAR CLASSIFICATION CRITERIA FOR MPA 7
16340	https://www.jbc.org/article/S0021-9258(19)33722-6/fulltext	Passive and Facilitated Transport in Nuclear Pore Complexes Is Largely Uncoupled - Journal of Biological Chemistry Skip to Main ContentSkip to Main Menu Login to your account Email/Username Your email address is a required field. E.g., j.smith@mail.com Password Show Your password is a required field. Forgot password? [x] Remember me Don’t have an account? Create a Free Account If you don't remember your password, you can reset it by entering your email address and clicking the Reset Password button. 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Ok Membrane Transport, Structure, Function, and BiogenesisVolume 282, Issue 6p3881-3888 February 2007 Open access Download Full Issue Download started Ok Passive and Facilitated Transport in Nuclear Pore Complexes Is Largely Uncoupled Bracha Naim Bracha Naim Affiliations Department of Biological Chemistry, Weizmann Institute of Science, Rehovot 76100, Israel Search for articles by this author ‡ ∙ Vlad Brumfeld Vlad Brumfeld Affiliations Department of Plant Sciences, Weizmann Institute of Science, Rehovot 76100, Israel Search for articles by this author § ∙ Ruti Kapon Ruti Kapon Affiliations Department of Biological Chemistry, Weizmann Institute of Science, Rehovot 76100, Israel Search for articles by this author ‡ ∙ Vladimir Kiss Vladimir Kiss Affiliations Department of Biological Chemistry, Weizmann Institute of Science, Rehovot 76100, Israel Search for articles by this author ‡ ∙ Reinat Nevo Reinat Nevo Affiliations Department of Biological Chemistry, Weizmann Institute of Science, Rehovot 76100, Israel Search for articles by this author ‡ ∙ Ziv Reich Ziv Reich Correspondence To whom correspondence should be addressed. Tel.: 972-8-934-2982; Fax: 972-8-934-6010; ziv.reich@weizmann.ac.il Affiliations Department of Biological Chemistry, Weizmann Institute of Science, Rehovot 76100, Israel Search for articles by this author ‡ziv.reich@weizmann.ac.il Affiliations & Notes Article Info ‡Department of Biological Chemistry, Weizmann Institute of Science, Rehovot 76100, Israel §Department of Plant Sciences, Weizmann Institute of Science, Rehovot 76100, Israel This work was supported by grants from the Human Frontier Science Program and from the Minerva Foundation with funding from the federal German ministry for education and research. The costs of publication of this article were defrayed in part by the payment of page charges. This article must therefore be hereby marked “advertisement” in accordance with 18 U.S.C. Section 1734 solely to indicate this fact. Publication History: Received August 31, 2006; Revised November 21, 2006; Published online December 12, 2006 DOI: 10.1074/jbc.M608329200 External LinkAlso available on ScienceDirect External Link Copyright: © 2007 ASBMB. Currently published by Elsevier Inc; originally published by American Society for Biochemistry and Molecular Biology. User License: Creative Commons Attribution (CC BY 4.0) \| Elsevier's open access license policy Download PDF Download PDF Outline Outline Abstract EXPERIMENTAL PROCEDURES RESULTS DISCUSSION Acknowledgments References Article metrics Related Articles Share Share Share on Email X Facebook LinkedIn Sina Weibo Add to Mendeley bluesky Add to my reading list More More Download PDF Download PDF Cite Share Share Share on Email X Facebook LinkedIn Sina Weibo Add to Mendeley Bluesky Add to my reading list Set Alert Get Rights Reprints Download Full Issue Download started Ok Previous articleNext article Show Outline Hide Outline Abstract EXPERIMENTAL PROCEDURES RESULTS DISCUSSION Acknowledgments References Article metrics Related Articles Abstract Nuclear pore complexes provide the sole gateway for the exchange of material between nucleus and cytoplasm of interphase eukaryotic cells. They support two modes of transport: passive diffusion of ions, metabolites, and intermediate-sized macromolecules and facilitated, receptor-mediated translocation of proteins, RNA, and ribonucleoprotein complexes. It is generally assumed that both modes of transport occur through a single diffusion channel located within the central pore of the nuclear pore complex. To test this hypothesis, we studied the mutual effects between transporting molecules utilizing either the same or different modes of translocation. We find that the two modes of transport do not interfere with each other, but molecules utilizing a particular mode of transport do hinder motion of others utilizing the same pathway. We therefore conclude that the two modes of transport are largely segregated. Eukaryotic cell nuclei are separated from the cytoplasm by a double lipid bilayer system known as the nuclear envelope (NE).2 Exchange of material between the two compartments proceeds through nuclear pore complexes (NPCs), large protein assemblies that span the NE and provide the sole medium for exchange. The vertebrate NPC has a molecular mass of ∼125 MDa (1 1. Reichelt, R. ∙ Holzenburg, A. ∙ Buhle, Jr., E.L. ... J. Cell Biol. 1990; 110:883-894 Crossref Scopus (376) PubMed Google Scholar ) and is made up of ∼30 different proteins, called nucleoporins, most of which are present in multiples of eight (2 2. Cronshaw, J.M. ∙ Krutchinsky, A.N. ∙ Zhang, W. ... J. Cell Biol. 2002; 158:915-927 Crossref Scopus (800) PubMed Google Scholar , 3 3. Rout, M.P. ∙ Aitchison, J.D. ∙ Suprapto, A. ... J. Cell Biol. 2000; 148:635-651 Crossref Scopus (1188) PubMed Google Scholar ). The core of the NPC consists of a symmetrical framework, measuring ∼120 × 90 nm, which is made of two coaxial rings sandwiching a wheel-like array of eight spoke-shaped domains. The spoke-ring assembly encircles the central pore channel, which resembles an hourglass 45–50 nm wide at its waist (4–7 4. Hinshaw, J.E. ∙ Carragher, B.O. ∙ Milligan, R.A. Cell. 1992; 69:1133-1141 Abstract Full Text (PDF) Scopus (387) PubMed Google Scholar 5. Akey, C.W. ∙ Radermacher, M. J. Cell Biol. 1993; 122:1-19 Crossref Scopus (357) PubMed Google Scholar 6. Stoffler, D. ∙ Feja, B. ∙ Fahrenkrog, B. ... J. Mol. Biol. 2003; 328:119-130 Crossref Scopus (194) PubMed Google Scholar 7. Beck, M. ∙ Forster, F. ∙ Ecke, M. ... Science. 2004; 306:1387-1390 Crossref Scopus (416) PubMed Google Scholar ). In addition to the central framework, NPCs contain peripheral structures, which are anchored to the ring moieties of the spoke-ring assembly and serve as docking sites for nuclear transport receptors and effectors. These structures include eight short (∼50 nm) filaments that protrude toward the cytoplasm and a massive, fish trap-like structure, termed the nuclear basket, which extends into the nucleus (4–11 4. Hinshaw, J.E. ∙ Carragher, B.O. ∙ Milligan, R.A. Cell. 1992; 69:1133-1141 Abstract Full Text (PDF) Scopus (387) PubMed Google Scholar 5. Akey, C.W. ∙ Radermacher, M. J. Cell Biol. 1993; 122:1-19 Crossref Scopus (357) PubMed Google Scholar 6. Stoffler, D. ∙ Feja, B. ∙ Fahrenkrog, B. ... J. Mol. Biol. 2003; 328:119-130 Crossref Scopus (194) PubMed Google Scholar 7. Beck, M. ∙ Forster, F. ∙ Ecke, M. ... Science. 2004; 306:1387-1390 Crossref Scopus (416) PubMed Google Scholar 8. Goldberg, M.W. ∙ Allen, T.D. J. Cell Sci. 1993; 106:261-274 PubMed Google Scholar 9. Ris, H. EMSA Bull. 1991; 21:54-56 Google Scholar 10. Nevo, R. ∙ Markiewicz, P. ∙ Kapon, R. ... Single Molecules. 2000; 1:109-114 Crossref Scopus (10) Google Scholar 11. Kapon, R. ∙ Nevo, R. ∙ Shuhmaher, N. ... Single Molecules. 2001; 2:86-89 Crossref Google Scholar ). Yeast NPCs have an overall similar architecture but are smaller (3 3. Rout, M.P. ∙ Aitchison, J.D. ∙ Suprapto, A. ... J. Cell Biol. 2000; 148:635-651 Crossref Scopus (1188) PubMed Google Scholar , 12–14 12. Rout, M.P. ∙ Blobel, G. J. Cell Biol. 1993; 123:771-783 Crossref Scopus (221) PubMed Google Scholar 13. Fahrenkrog, B. ∙ Hurt, E.C. ∙ Aebi, U. ... J. Cell Biol. 1998; 143:577-588 Crossref Scopus (95) PubMed Google Scholar 14. Yang, Q. ∙ Rout, M.P. ∙ Akey, C.W. Mol. Cell. 1998; 1:223-234 Full Text Full Text (PDF) Scopus (291) PubMed Google Scholar ). Transport across the NPC has been reviewed in detail (15–22 15. Corbett, A.H. ∙ Silver, P.A. Microbiol. Mol. Biol. Rev. 1997; 61:193-211 Crossref Scopus (170) PubMed Google Scholar 16. Mattaj, I.W. ∙ Englmeier, L. Annu. Rev. Biochem. 1998; 67:265-306 Crossref Scopus (1028) PubMed Google Scholar 17. Gorlich, D. ∙ Kutay, U. Annu. Rev. Cell Dev. Biol. 1999; 15:607-660 Crossref Scopus (1727) PubMed Google Scholar 18. Yoneda, Y. Genes Cells. 2000; 5:777-787 Crossref Scopus (173) PubMed Google Scholar 19. Fried, H. ∙ Kutay, U. Cell Mol. Life Sci. 2003; 60:1659-1688 Crossref Scopus (413) PubMed Google Scholar 20. Weis, K. Cell. 2003; 112:441-451 Full Text Full Text (PDF) Scopus (595) PubMed Google Scholar 21. Suntharalingam, M. ∙ Wente, S.R. Dev. Cell. 2003; 4:775-789 Full Text Full Text (PDF) Scopus (372) PubMed Google Scholar 22. Pemberton, L.F. ∙ Paschal, B.M. Traffic. 2005; 6:187-198 Crossref Scopus (591) PubMed Google Scholar ) and can be divided into two modes. Small molecules, such as ions, metabolites, and intermediate-sized macromolecules, can pass unassisted by diffusion which becomes increasingly restricted as the particle approaches a size limit of ∼10 nm in diameter (23 23. Paine, P.L. ∙ Moore, L.C. ∙ Horowitz, S.B. Nature. 1975; 254:109-114 Crossref Scopus (622) PubMed Google Scholar , 24 24. Keminer, O. ∙ Peters, R. Biophys. J. 1999; 77:217-228 Full Text Full Text (PDF) Scopus (195) PubMed Google Scholar ). In contrast, proteins, RNAs, and their complexes are ushered selectively by dedicated soluble transport receptors, which recognize specific import (NLS) or nuclear export signal (NES) peptides displayed by the cargo. In most cases, assembly and disassembly of transport complexes in the appropriate cellular compartment are coordinated by the small GTPase Ran, which performs these activities by binding to transport receptors in its GTP-bound form. Facilitated translocation is often coupled to an input of metabolic energy that permits transport against concentration gradients and can accommodate objects with a diameter of up to ∼40 nm (25 25. Pante, N. ∙ Kann, M. Mol. Biol. Cell. 2002; 13:425-434 Crossref Scopus (672) PubMed Google Scholar ). It has been estimated that in proliferating HeLa cells a single NPC supports the facilitated translocation of ≥10–20 MDa of macromolecules and macromolecular complexes/s (26 26. Ribbeck, K. ∙ Gorlich, D. EMBO J. 2001; 20:1320-1330 Crossref Scopus (605) PubMed Google Scholar ). At the same time, each NPC most likely has to accommodate comparable flows of small molecules that cross it by diffusion. It is generally assumed that both modes of transport share a single diffusion channel located within the central pore of the NPC (see, e.g. Refs. 23 23. Paine, P.L. ∙ Moore, L.C. ∙ Horowitz, S.B. Nature. 1975; 254:109-114 Crossref Scopus (622) PubMed Google Scholar and 27–30 27. Stevens, B.J. ∙ Swift, H. J. Cell Biol. 1966; 31:55-77 Crossref Scopus (258) PubMed Google Scholar 28. Peters, R. EMBO J. 1984; 3:1831-1836 Crossref Scopus (120) PubMed Google Scholar 29. Dworetzky, S.I. ∙ Lanford, R.E. ∙ Feldherr, C.M. J. Cell Biol. 1988; 107:1279-1287 Crossref Scopus (230) PubMed Google Scholar 30. Fahrenkrog, B. ∙ Aebi, U. Nat. Rev. Mol. Cell. Biol. 2003; 4:757-766 Crossref Scopus (348) PubMed Google Scholar ). However, up to date, no study had directly tested for concurrent transport of cargoes utilizing different modes. The issue of whether or not a single diffusion channel mediates both modes of transport therefore remains unsettled, despite the fact that it pertains to the very basic function of the NPC. In the present work, we addressed this issue by testing for mutual effects of molecules utilizing either the same or different modes of transport when crossing NPCs. We found that molecules transporting in different modes did not hinder the passage of each other. In contrast, molecules that utilize a given mode of transport for translocation sterically interfered with the passage of other molecules that make use of the same mode. These results indicate that passive and facilitated transport through NPCs proceed along largely segregated routes. EXPERIMENTAL PROCEDURES Transport Cargoes—FITC-labeled 9.5- and 19.5-kDa dextrans and RITC-labeled 17.2-kDa dextran were purchased from Sigma. TRITC-labeled bovine serum albumin (BSA) and TRITC-labeled BSA covalently linked to ∼10 SV40 large T-antigen NLS peptides (CTPPKKKRKV; the number of peptides attached was determined by mass spectrometry) were a kind gift from Michael Elbaum (Weizmann Institute of Science). A chimera consisting of IgM conjugated to green fluorescent protein (GFP) fused to the naturally NLS-carrying protein, nucleoplasmin, was prepared as follows. Nucleoplasmin-GFP (1 mg in PBS; also a gift from M. Elbaum) was reacted with 92 nmol of m-maleimidobenzoyl-N-hydroxysuccinimide ester (Pierce) for 1 h at room temperature. The IgM (Jackson ImmunoResearch, 0.9 mg dissolved in 200 μl of 0.01 m sodium phosphate buffer (pH 7.6) and 0.5 m NaCl) was incubated with 46 nmol N-succinimidyl 3-(2-pyridyl-dithio)propionate (Pierce) for 30 min (room temperature). Desalting of both mixtures was carried out using Sephadex G-25 spin columns equilibrated with PBS. Dithiothreitol was added to the IgM-SPDP mixture to a final concentration of 1 m m, and the solution was immediately passed through a PD-10 desalting column (Amersham Biosciences), equilibrated with PBS. Fractions containing the IgM-SPDP without dithiothreitol (verified with 5,5′-dithiobis (2-nitrobenzoic acid), Ellman's reagent) were then added to the solution containing Nup-GFP-m-maleimidobenzoyl-N-hydroxysuccinimide ester. Following 30 min of incubation at room temperature, the mixture was concentrated (Vivaspin 6, MWCO 30,000; Vivascience) and separated on Superdex 200 column (Amersham Biosciences). The species used in the experiments had one molecule of nucleoplasmin-GFP per IgM. A plasmid encoding for transportin (human) fused to glutathione S-transferase (GST) was kindly provided by Yuh Min Chook (University of Texas Southwestern Medical Center). The fusion protein was expressed and purified as described (31 31. Chook, Y.M. ∙ Blobel, G. Nature. 1999; 399:230-237 Crossref Scopus (302) PubMed Google Scholar ). The protein was labeled with AlexaFluor 633 succinimidyl ester carboxylic acid by incubation (in PBS, pH 8.4) with 5-fold molar excess of the dye at 4 °C for 14 h. Free dye was removed using PD-10 desalting column (Amersham Biosciences) equilibrated with transport buffer (TB; 30 m m sodium chloride, 90 m m potassium acetate, 5 m m magnesium acetate, 1 m m EDTA, 2 m m dithiothreitol, 250 m m sucrose, and 20 m m Tris-HCl, pH 7.4). Following labeling, the protein was concentrated using Vivaspin 6 (MWCO 30,000; Vivascience) and stored at -70 °C in TB containing 20% (v/v) glycerol. GST sometimes tends to dimerize (24 24. Keminer, O. ∙ Peters, R. Biophys. J. 1999; 77:217-228 Full Text Full Text (PDF) Scopus (195) PubMed Google Scholar ). Mass spectrometry analyses revealed, however, that the fusion is monomeric. Cell Culture—HeLa cells (ATCC CCL-2; up to five passages) were cultured in Dulbecco's modified Eagle's medium with 10% fetal calf serum/antibiotics and were used 1 day after plating, at which time they were 50–70% confluent. Media, antibiotics, and fetal calf serum were purchased from Invitrogen. Transport Assays and Microinjection—Microinjections were performed using a semi-automatic injection system (Eppendorf 5246 Transjector attached to an Eppendorf 5171 micromanipulator) with a Z-depth limit option, using glass micropipettes (Femtotips; Eppendorf). Air was administrated at 100–250 hPa for 0.2–0.3 s. The cells were kept in Dulbecco's modified Eagle's medium, 25 m m Hepes (pH 7.3) at 37 °C in an open perfusion micro-incubator (PDMI II, Harvard Apparatus) mounted on the stage of the microscope used for the recordings. All of the samples were injected with Cy5-labeled secondary anti-goat antibody (IgG, 150 kDa; Jackson ImmunoResearch), which served as a marker for nuclear envelope integrity. A rough approximation of the initial cytosolic concentration of the injected material was made based on the injected volume, using a spheroid model of the cell. The cells were viewed with an IX70-based Olympus FluoView 500 confocal laser-scanning microscope, using 0.85-numerical aperture 40× (UplanApo) objective; the focal plane was set such that it traversed through the mid-section of the nuclei. The images were collected every 1.6 s before, during, and following injection, until a steady state had been reached. FITC-dextran, GFP, and IgM-nucleoplasmin-GFP were visualized using an argon laser (λ ex = 488 nm) and a 505–525-nm band pass filter. TRITC-BSA and RITC-dextran were excited at 543 nm, using a helium-neon laser, and the emitted fluorescence was collected between 560 and 600 nm. Visualization of the anti-goat Cy5-labeled antibody was made using a helium-neon laser (λ ex = 633 nm) and a >660-nm high-pass filter. The images were acquired sequentially to prevent spectral bleed-through of the fluorescence emission and were processed using the ImagePro Plus (v4.5) software. In all cases, the time-dependent rise in the ratio between (background-corrected) nuclear and cytoplasmic mean fluorescence intensity, F(t), fitted well to a single-exponential curve of the form, F(t)=F max(1-e-k t) (Eq. 1) where t is the time post injection, and k is the first order rate constant. Translocation rates (particles/NPC/second) were estimated using the values reported by Ribbeck and Görlich (26 26. Ribbeck, K. ∙ Gorlich, D. EMBO J. 2001; 20:1320-1330 Crossref Scopus (605) PubMed Google Scholar ) for the volume and NPC density of HeLa cell nuclei. Fluorescence recovery after photobleaching. HeLa cells stably expressing GFP were obtained by transfection with pCEFL and selection under 700 μg/ml G418 for 3 weeks. The cells were serum-starved (0.15% fetal calf serum) for 14 h. An hour and a half before the measurements, the medium was replaced by a medium containing 10% fetal calf serum, which was used throughout the experiments. The measurements were made at 37 °C with the microscope set-up described above, using a 60× PlanApo oil-immersion objective (NA 1.4). Prior to bleaching, the target cell was imaged using 0.2% of the maximal laser intensity. Photobleaching of nuclear GFP was achieved by repetitive scans of a 72 × 24 pixel area inside the nucleus (covering most of its area) for 3–5 s, using full illumination intensity (∼0.3 milliwatt at the focal point). Following bleaching, the laser intensity was brought back to its initial value, and time lapse image sequences were acquired at 2 Hz. The relatively slow acquisition rate and the fact that bleaching often extended to cytosolic domains rendered the rate constants derived from the measurements only approximations. However, we were interested only in relative rates, for which the analysis fully sufficed. Fluorescence recovery inside the nucleus was fitted using the following equation, F(t)F∞=F 0 F∞+1-F 0 F∞(1-e-k t) (Eq. 2) where F 0 and F° denote the mean nuclear fluorescence intensity ratio immediately after bleaching and at the end of recovery. Permeabilized Cells—HeLa cells were washed with cold TB (see above) and incubated for 5 min with 8 μg/ml digitonin (in TB) on ice, after which they were rinsed five times with cold TB. The measurements were made at 37 °C as described for the microinjection assays, using RITC-labeled 77-kDa dextran as nuclear integrity marker. The images were collected every 1.4 s before, during, and following the addition of the transport cargoes (FITC-labeled 9.5-kDa dextran and AlexaFluor 633-GST-transportin). RESULTS Microinjection Experiments—The first set of experiments consisted of competition assays in intact cells. Fluorescently labeled substrates, capable of crossing NPCs either by passive diffusion or in a signal-dependent, receptor-mediated fashion, were injected into the cytoplasm of HeLa cells, either separately or together, and their nuclear import kinetics was followed by confocal microscopy. Passively transported cargoes included 9.5-, 17.2-, and 19.5-kDa dextran molecules (Table 1). Receptor mediated transport was represented by BSA covalently linked to ∼10 SV40 large T-antigen NLS peptides, which serve as substrates for importin α/β transport receptor complexes. Within the time scale of our measurements, BSA devoid of NLS was virtually excluded from the nuclei and thus can be considered as exclusively dependent on receptor mediation for its transport. To verify that the NE of the injected cells remains intact during the experiments and, hence, that nuclear entry of the probes is the result of authentic translocation events, fluorescently labeled anti-goat secondary antibody (∼150 kDa) was co-injected with the transport substrates. Only cells capable of completely excluding the antibody from the nucleus throughout the entire measurement were used in the analysis. Injections and recordings were performed at 37 °C, using an environmental chamber that was mounted on the microscope stage. \| Probe/substrate \| Hydrodynamic radii \| --- \| \| \| nm \| \| FITC-dextran (19.5 kDa) \| 3.4a \| \| RITC-dextran (17.2 kDa) \| 3.3b \| \| FITC-dextran (9.5 kDa) \| 2.8b \| \| GFP (27 kDa) \| 2.4c \| \| TRITC-BSA-NLS (79 kDa) \| 3.6a \| \| IgM-nucleoplasmin-GFP (1.1 MDa) \| ∼35–40d \| \| Alexa fluor 633-GST-transportin (130.6 kDa) \| ∼9d \| TABLE 1 Hydrodynamic radii of molecules used in this study a From Ref. 59 59. Peters, R. Biochim. Biophys. Acta. 1986; 864:305-359 Crossref Scopus (219) PubMed Google Scholar b From Ref. 24 24. Keminer, O. ∙ Peters, R. Biophys. J. 1999; 77:217-228 Full Text Full Text (PDF) Scopus (195) PubMed Google Scholar c From Ref. 26 26. Ribbeck, K. ∙ Gorlich, D. EMBO J. 2001; 20:1320-1330 Crossref Scopus (605) PubMed Google Scholar d Estimated from crystal structures Open table in a new tab To establish the validity of the experimental system, we injected cells with a mixture containing FITC-labeled 19.5-kDA dextran (green), TRITC-labeled BSA-NLS (red), and the NE integrity marker antibody (Cy5-labeled, magenta). Time series images in Fig. 1 A show a homogenous fluorescence of both dextran and BSA inside the cytoplasm within 5 s post-injection. Nuclear import of the two substrates ensued almost immediately after injection and reached a steady state a few minutes later. At this stage, the dextran molecules were distributed more or less evenly between the two cellular compartments, whereas the NLS-carrying BSA markedly accumulated inside the nucleus. Figure viewer FIGURE 1Nuclear import of passively and receptor-mediated transporting cargoes following cytoplasmic injections. FITC-labeled dextran (green) and TRITC-labeled NLS-carrying BSA (red) were injected into the cytoplasm of HeLa cells, separately or together, in the presence of Cy5-labeled secondary antibody (magenta), which served as a marker for NE integrity. Double or triple fluorescence images were acquired in a sequential mode every 1.6 s, before, during, and following injection. A, time series images obtained for a cell injected with a mixture containing 19.5-kDa dextran and BSA-NLS at a molar ratio of 1.5 (dextran per BSA-NLS), as well as the NE integrity marker. B, nuclear accumulation versus time recorded for cells injected with either 19.5-kDa dextran (mixed with BSA lacking NLS) or BSA-NLS. Shown are fits to the averaged normalized data sets. In each case, the data points fit well (solid lines) to a single-exponential of the form F(t) = F max (1 - e-kt), suggesting that nuclear entry of both species follows first order kinetics. Inset, single-cell traces. In all the experiments, BSA and BSA-NLS were injected at 52 μ m; their cytoplasmic concentrations following injection are estimated to be ∼2 μ m. Dextran concentration in the cytoplasm was likewise estimated to be 3 μ m. Within the time scale of the experiments, BSA lacking NLS and dextran molecules ≥40 kDa failed to enter the nuclei to a detectable level (data not shown). We next set out to obtain the base-line rate constants characteristic of each mode of transport on its own. These rates will serve for comparison in the competition experiment. The inset in Fig. 1 B shows single-cell traces of the change in the nuclear/cytoplasmic mean fluorescence intensity ratio, F(t), of either BSA-NLS or 19.5-kDa dextran, plotted against time. Fits to the averaged normalized data sets are shown in the main figure. Both traces fit well to a single exponent F(t) = F max(1 - e-kt), akin to first order kinetics. BSA-NLS imported into the nuclei with an apparent rate constant of 7.3 × 10-3 s-1. When the concentration difference between cytoplasm and nucleus is set to 1 μ m, this rate leads to a calculated initial influx of ∼5000 molecules of BSA-NLS/s or to a translocation rate of approximately two molecules/NPC/s. The rate constants derived for diffusing 9.5- and 19.5-kDa dextran molecules were 27.7 × 10-3 and 18.5 × 10-3 s-1, respectively. These rate constants are 3.8 and 2.5 times higher than that determined for BSA-NLS. Normalized by size, however, the flow rate of the two dextrans through the pores is actually similar to or even slower than that of BSA-NLS bound to its transport receptor(s). This is despite the fact that transport of BSA-NLS involves binding to importin α/β, in the cytoplasm, as well as binding of RanGTP to importin β, at the nucleoplasmic face of the pores. Importin β-mediated translocation of BSA-NLS across the NPC thus appears to proceed in a facilitated manner. Similar observations have been made before on the passage of other transport receptors (26 26. Ribbeck, K. ∙ Gorlich, D. EMBO J. 2001; 20:1320-1330 Crossref Scopus (605) PubMed Google Scholar , 32 32. Siebrasse, J.P. ∙ Peters, R. EMBO Rep. 2002; 3:887-892 Crossref Scopus (41) PubMed Google Scholar , 33 33. Kubitscheck, U. ∙ Grunwald, D. ∙ Hoekstra, A. ... J. Cell Biol. 2005; 168:233-243 Crossref Scopus (208) PubMed Google Scholar ). Having established the validity of the experimental system and having obtained the base-line rate constants for the substances used, we proceeded with the competition experiments. First, we examined the effect of passive transport of dextran, on receptor-mediated traffic of BSA-NLS (Fig. 2). Fig. 2 A shows the import rate constants derived for BSA-NLS in the presence of 9.5- or 19.5-kDa dextran. The molar ratio between the two species was 1.5 dextran molecules/BSA-NLS. Under these conditions, neither the 9.5-kDa nor the 19.5-kDa dextran had a significant effect on the rate of BSA-NLS import into the nucleus. In principle, the lack of BSA-NLS sensitivity to the application of dextran can be simply the result of too low a concentration of dextran competitors. To ascertain that this is not the case, we conducted two more experiments: one in which differently labeled BSA-BLS substrates competed against each other at a low molar ratio and one at which the ratio of dextran to BSA-NLS was raised dramatically. For the first experiment TRITC-BSA-NLS at a concentration of 2 μ m was introduced alone or together with FITC-BSA-NLS at the same concentration. Even at this low molar ratio of 1:1, a significant reduction of ∼1.6 in the transport rate of TRITC-BSA-NLS was observed. Figure viewer FIGURE 2Import rate constants of inert and NLS-carrying cargoes injected alone or together. In all experiments BSA-NLS concentration was kept constant at 2 μ m, and dextran concentration was varied from 3 to 100 μ m to achieve different molar ratios. A, rate constants of BSA-NLS injected into the cell cytoplasm alone or in the presence of 9.5- or 19.5-kDa dextran. No attenuation in import rate of BSA-NLS is observed up to 50-fold molar excess of dextran. In the latter case, import is actually facilitated, probably because of a crowding effect, which promotes binding of BSA-NLS to its transport receptor. Consistent with this, 40-kDa dextran, which is too big to cross the pores unassisted, also speeded up import of BSA-NLS into the nucleus. B, rate constants of dextran molecules coinjected with BSA or NLS-carrying BSA at molar ratio of 0.7 BSA per dextran or with 10- or 50-fold molar excess of 17.2-kDa dextran. The presence of BSA-NLS in the injected samples did not cause significant changes in the transport rate of the dextran. In contrast, 17.2-kDa dextran (at 30 and 150 μ m) effectively attenuated the rate of nuclear entry of the 19.5-kDa dextran probe (kept at 3 μ m). The data are presented as the means ± 2 S.E. of determinations involving between 8 and 13 cells for each column. n.s., no significant difference (t test; α = 0.95). Next, we increased the molar ratio between dextran (19.5 kDa) and BSA-NLS to 50 (by increasing the concentration of the former). At this concentration (∼100 μ m), the dextran molecules imposed an additional mass flow of roughly 6 MDa (∼300 molecules)/s on each NPC, ∼25–50% of the estimated flow of material crossing in a facilitated manner. Notably, this additional load not only failed to slow down the transport of BSA-NLS but, in fact, exerted an opposite effect, increasing the rate constant of the latter by ∼2.5-fold. We hypothesized that this latter effect might be due to depletion interactions (molecular crowding), affected by the high concentration of dextran in the cytoplasm, which facilitated association of BSA-NLS with its transport receptors. To test this, we substituted the 19.5-kDa dextran with 40-kDa dextran, which is too big to diffuse through the pores, thus separating possible effects in the cytoplasm and within the pores. As shown, the 40-kDa dextran also speeded the transport of BSA-NLS into the nucleus. The smaller effect produced by this dextran as compared with that exerted by the 20-kDa dextran is expected because it is larger and, because of solubility limits, was applied at lower molar excess (30-fold versus 50-fold). Next, we conducted the reverse experiment, where the effect of BSA-NLS on the nuclear import of dextran was studied (Fig. 2 B). Here too, no statistically significant change could be observed in the rate constant of the dextran probes when BSA-NLS was co-injected at a molar ratio of 0.7 BSA-NLS/dextran. Substituting dextran with GFP led to similar results.3 In this set of experiments, however, we could not significantly increase the ratio between the two species (in favor of BSA-NLS) because it was impossible to concentrate the protein much further or, alternatively, to sufficiently dilute the dextran (or GFP) solution and still retain a reliable signal. We therefore took another approach where the pores were plugged by chimeric molecules consisting of IgM molecules conjugated to NLS (5x)-carrying nucleoplasmin fused to GFP. This construct has an estimated diameter of 35–40 nm and thus should occupy a very large fraction of the central channel of the pores, at least at its waist. To ensure that potential effects on facilitated transport are not due to competition over available transport receptors, the IgM-nucleoplasmin-GFP chimeras were injected into the cells (cytoplasm) at very low concentrations such that their intracellular concentration (1–10 n m) was well below that of their transport receptors ∼5–10 μ m; (34 34. Ribbeck, K. ∙ Lipowsky, G. ∙ Kent, H.M. ... EMBO J. 1998; 17:6587-6598 Crossref Scopus (368) PubMed Google Scholar ). Because of its size, the chimera transports into the nucleus very slowly with a half-life of 2 h (data not shown). The cells were therefore let to incubate for ∼2 h (37 °C), during which the chimeras reached the NE and began translocating through the pores (Fig. 3 A). Following incubation, the cells containing the IgM chimeras were injected again with either BSA-NLS or 17.2-kDa dextran, and the nuclear import kinetics of either species was monitored. The presence of the IgM conjugates inside the cells had no effect on the transport of the 17.2-kDa dextran (Fig. 3 B). On the other hand, the presence of the chimera molecules led to a 2-fold decrease in the transport rate constant of BSA-NLS (Fig. 3 C). The relatively small inhibitory effect probably reflects that only a small fraction of pores contained IgM-nucleoplasmin-GFP molecules at the time of the second injection (Fig. 3 A). Figure viewer FIGURE 3Effect of translocating IgM-nucleoplasmin-GFP chimeras on nuclear import of inert and NLS-carrying cargo.A, IgM conjugated to NLS-carrying nucleoplasmin-GFP fusion (green) was injected into the cell cytoplasm together with Cy5-labeled secondary antibody (magenta). ∼2 h later, the cells were injected again with either 17.2-kDa dextran (B) or BSA-NLS (C), and nuclear import was followed by confocal imaging. The IgM conjugates noticeably interfered with the import of BSA-NLS, leading to a ∼2-fold decrease in its apparent transport rate constant but had no effect on the passage of the dextran molecules. The data are presented as the means ± 2 S.E. of determinations involving between 7 and 15 cells for each column. n.s., no significant difference (t test, α = 0.95). The above set of experiments demonstrates that passively diffusing molecules do not slow down the translocation of receptor-cargo complexes through NPCs and vice versa. In principle, the observed lack of inhibition may be attributed to the use of insufficiently high loads. However, in the case of dextran, very high intracellular concentrations could be achieved. Moreover, the results obtained from the last experiment show that IgM chimeras, although not affecting the transport of passively diffusing dextran, significantly attenuated the passage of BSA-NLS molecules through the pores. If this differential effect of receptor-mediated transport cargoes on other complexes is due to a steric segregation of paths, then the same should hold true for passively diffusing cargoes. We have already shown that these do not interfere with their receptor-mediated counterparts. It thus remains to be shown that they do interfere with other passively diffusing cargoes. To this end, we studied the effect of 17.2-kDa dextran on the nuclear import of 19.5-kDa dextran, which served as the test probe. The concentrations and molar ratios between the two dextrans were almost identical to those used in the experiments involving dextran and BSA-NLS, ensuring that similar loads are applied. Applied at a molar excess of 10, the 17.2-kDa dextran molecules led to an almost 2-fold decrease in the import rate constant of the 19.5-kDa dextran probe. Increasing the ratio between the two species to 50 led to a further diminution, decreasing the rate by a factor of ∼3 (Fig. 2 B). Thus, both inert and signal-carrying molecules can selectively interfere with the passage of other molecules that utilize the same pathway but have no significant effect on the passage of molecules transporting in a different mode. FRAP Experiments—Another strategy we employed to study the effects produced by receptor-bound translocating material on the passive diffusion of inert molecules relied on the fact that nuclear export of RNA molecules and their protein complexes constitutes a major fraction of the total mass flow through NPCs. In addition, ribonucleoprotein complexes can be very large and thus can plug the pores for relatively long times during their translocation. As inert cargo, capable of freely diffusing through the pores, we used GFP that was stably expressed in the (HeLa) cells. In the experiments, the nuclear import kinetics of GFP was followed by semi-quantitative FRAP (see “Experimental Procedures”; Fig. 4 A) in control cells and in cells treated with actinomycin D, which blocks synthesis of all RNA types and inhibits nuclear import of pre-mRNA-binding proteins (35 35. Pinol-Roma, S. ∙ Dreyfuss, G. Nature. 1992; 355:730-732 Crossref Scopus (775) PubMed Google Scholar ). Figure viewer FIGURE 4Effect of RNA export on nuclear import of endogenously expressed GFP studied by FRAP. GFP was stably expressed in HeLa cells, and its nuclear import was followed by FRAP in control cells or in cells treated with 5 μg/ml actinomycin D. Treated cells were incubated with the drug for different times before measurements (70–180 min). Recordings were taken at 10-min intervals. However, because no significant changes were observed throughout the entire time range, the data were clustered and analyzed collectively. A, images of an untreated cell before and after photobleaching of nuclear GFP. B, fluorescence recovery recorded in control cells and in cells treated with actinomycin D. For clarity, standard errors are shown only for control cells; similar values of S.E. were observed for treated cells. Inset, corresponding transport rate constants. Blocking RNA synthesis (and, consequently, its export) had no statistically significant effect on the nuclear import rate of GFP. The data are presented as the means ± 2 S.E. of determinations involving between 9 and 28 cells for each column. n.s., no significant difference (one-way analysis of variance; α = 0.95). An averaged trace of the time-dependent recovery of GFP fluorescence inside the nuclei of untreated cells is shown in Fig. 4 B (black dots). Consistent with the results obtained from the microinjection experiments, nuclear import of GFP followed apparent first order kinetics with a half-time of 34 s. The transport rate constant derived from the fit (red, using Equation 2 under “Experimental Procedures”) is 22.6 × 10-3 s-1, close to the 27.7 × 10-3 s-1 value we measured for the similarly sized 9.5-kDa dextran. To study the effect of RNA synthesis blockage and, consequently, lack of its nuclear export on GFP transport through the pores, we subjected cells to 5 μg/ml actinomycin D. To cover most of the activation time range of the drug (36 36. Andersen, J.S. ∙ Lam, Y.W. ∙ Leung, A.K. ... Nature. 2005; 433:77-83 Crossref Scopus (983) PubMed Google Scholar ), we took multiple measurements at 10-min intervals, starting 70 min and ending 180 min after addition of the reagent. Because no significant change in import kinetics was observed throughout the entire time range, the data were clustered and analyzed collectively. The global fit of the data, shown in Fig. 4 B (green line), is statistically similar to the curve obtained for the untreated cells. The rate constant derived from the fit is 25 × 10-3 s-1, statistically indistinguishable from the value determined in the untreated cells (Fig. 2 B, inset). Competition Assays in Permeabilized Cells—A problem with the above set of experiments is that the load exerted on the pores by the transporting RNA molecules, although probably substantial, is essentially unknown. To overcome this limitation, we turned to permeabilized cells (37 37. Adam, S.A. ∙ Marr, R.S. ∙ Gerace, L. J. Cell Biol. 1990; 111:807-816 Crossref Scopus (822) PubMed Google Scholar ). Such cells allow the introduction of very large amounts of transporting substrates and are largely devoid of endogenous nucleocytoplasmic exchange, leaving the majority of NPCs in an “empty” state. As a substrate for facilitated transport, we chose the protein import receptor transportin, which can be applied at high concentrations (i.e.>50 μ m) and imports efficiently and irreversibly into the nucleus in the absence of RanGTP (26 26. Ribbeck, K. ∙ Gorlich, D. EMBO J. 2001; 20:1320-1330 Crossref Scopus (605) PubMed Google Scholar ). The combination of permeabilized cells with exogenously applied transportin thus provides an efficient, simple translocation system where the competition between substrates transporting in a passive or facilitated manner can be directly assessed. In addition, transportin translocates through NPCs by a pathway that is distinct from that utilized by importin β (which mediates the passage of BSA-NLS used in the microinjection assays), extending the scope of potentially competing pathways. For the experiments, we used recombinant (human) transportin fused to GST. The fused protein entered the nuclei of permeabilized HeLa cells with a rate constant of 8.6 × 10-3 s-1 (Fig. 5, inset). At the highest concentration we used (100 μ m), this corresponds to an initial translocation rate of ∼200 molecules·NPC-1·s-1; we note that it has previously been shown that nuclear entry of transportin deviates from ideal Michaelis-Menten kinetics at concentrations above 5–10 μ m (26 26. Ribbeck, K. ∙ Gorlich, D. EMBO J. 2001; 20:1320-1330 Crossref Scopus (605) PubMed Google Scholar ), indicating that above this limit, the passage of transporting molecules is influenced by the presence of others. Figure viewer FIGURE 5Effect of increasing concentrations of transportin on the nuclear entry of passively diffusing dextran studied in permeabilized cells. Nuclear import of 9.5-kDa dextran (Δ C = 2 μ m) was studied in permeabilized HeLa cells in the absence and presence of increasing concentrations of recombinant (human) transportin fused to GST. With the exception of 60 μ m, no significant change in the rate of import of dextran into the nuclei was observed when the concentration of the protein was varied between 2 and 100 μ m. The data are presented as the means ± 2 S.E. of determinations involving between 12 and 50 cells for each column. n.s., no significant difference (one-way analysis of variance; α = 0.95). Inset, averaged mean fluorescence intensity versus time measured for GST-transportin (Δ C = 60 μ m) in permeabilized HeLa cells. The fusion protein entered the nuclei with a (first order) rate constant of 8.6 × 10-3 s-1. Fig. 5 summarizes the effect produced by increasing concentrations of GST-transportin on the nuclear import of 9.5-kDa dextran applied at a constant concentration of 2 μ m. No significant changes in the nuclear entry rate constant of the dextran molecules were observed when GST-transportin was added up to a final concentration of 50 μ m (Fig. 5). When present at a concentration of 60 μ m, GST-transportin led to a small but statistically significant drop (∼15%) in the transport rate constant of the dextran, but a further increase of the concentration of the protein to 100 μ m had, again, no significant effect on this constant. Overall, there is no apparent trend in the dependence of the rate constant of nuclear import of the dextran probe. Thus, even when competed by heavy loads of cargoes translocating in a facilitated manner, diffusion of molecules through the pores remains unperturbed. DISCUSSION Traffic through NPCs can be considered as consisting of two components. The first is comprised of a population of small, passively diffusing molecules, and the second includes protein, RNA, and ribonucleoprotein complexes requiring interaction with receptors to be transported. Maintaining these two components unperturbed is essential for cell viability and homeostasis, requiring the NPCs to accommodate huge fluxes of both populations at the same time. Mechanistically, as well physiologically, it makes sense that the two components should traffic along spatially separated routes. Such a separation would minimize the risks of steric hindrances that may compromise flows and should simplify control over transport occurring in either mode. Performing single-particle three-dimensional reconstructions on NPCs released from Xenopus oocytes NEs by detergent treatment, Hinshaw et al. (4 4. Hinshaw, J.E. ∙ Carragher, B.O. ∙ Milligan, R.A. Cell. 1992; 69:1133-1141 Abstract Full Text (PDF) Scopus (387) PubMed Google Scholar ) showed that the central pore channel is surrounded by eight 10-nm-wide channels that perforate the central framework of the NPC. Morphologically similar channels (although positioned at smaller radii) were observed in three-dimensional reconstructions of detergent-released frozen-hydrated NPCs from Xenopus (5 5. Akey, C.W. ∙ Radermacher, M. J. Cell Biol. 1993; 122:1-19 Crossref Scopus (357) PubMed Google Scholar ) and yeast (14 14. Yang, Q. ∙ Rout, M.P. ∙ Akey, C.W. Mol. Cell. 1998; 1:223-234 Full Text Full Text (PDF) Scopus (291) PubMed Google Scholar ). A recent cryo-electron tomography study (6 6. Stoffler, D. ∙ Feja, B. ∙ Fahrenkrog, B. ... J. Mol. Biol. 2003; 328:119-130 Crossref Scopus (194) PubMed Google Scholar ) of native, NE-associated NPCs, also revealed the presence of eight channels roughly similar in shape, size, and radial position to those identified by Hinshaw et al. These channels have been implicated in trafficking of integral membrane proteins to the inner nuclear membrane (38 38. Soullam, B. ∙ Worman, H.J. J. Cell Biol. 1995; 130:15-27 Crossref Scopus (183) PubMed Google Scholar ) and in maintenance of NE electrical conductance (39–42 39. Shahin, V. ∙ Danker, T. ∙ Enss, K. ... FASEB J. 2001; 15:1895-1901 Crossref Scopus (66) PubMed Google Scholar 40. Danker, T. ∙ Schillers, H. ∙ Storck, J. ... Proc. Natl. Acad. Sci. U. S. A. 1999; 96:13530-13535 Crossref Scopus (36) PubMed Google Scholar 41. Enss, K. ∙ Danker, T. ∙ Schlune, A. ... J. Membr. Biol. 2003; 196:147-155 Crossref Scopus (14) PubMed Google Scholar 42. Mazzanti, M. ∙ Bustamante, J.O. ∙ Oberleithner, H. Physiol. Rev. 2001; 81:1-19 Crossref Scopus (138) PubMed Google Scholar ). However, they may also serve as an additional conduit for passively diffusing molecules (4 4. Hinshaw, J.E. ∙ Carragher, B.O. ∙ Milligan, R.A. Cell. 1992; 69:1133-1141 Abstract Full Text (PDF) Scopus (387) PubMed Google Scholar , 43 43. Feldherr, C.M. ∙ Akin, D. J. Cell Sci. 1997; 110:3065-3070 Crossref PubMed Google Scholar ). In a meticulous assay, Feldherr and Akin (43 43. Feldherr, C.M. ∙ Akin, D. J. Cell Sci. 1997; 110:3065-3070 Crossref PubMed Google Scholar ) studied the distribution of gold colloids following microinjection into cytoplasm or nucleoplasm of Xenopus oocytes. They found that particles smaller than 10 nm could be found at sites that correspond to peripheral channels in addition to those found in the central pore channel. Larger particles were not observed at these sites. These studies indicate that facilitated transport occurs solely in the central channel, whereas passive diffusion of small particles may occur in both the central and the peripheral channels. On the other hand, a transport assay on isolated nuclear patches (24 24. Keminer, O. ∙ Peters, R. Biophys. J. 1999; 77:217-228 Full Text Full Text (PDF) Scopus (195) PubMed Google Scholar ) suggested that passive transport occurs only through the central channel of the NPC. Using optical single transporter recording and statistical analysis of the transport of 4–20-kDa dextran molecules, the authors concluded that NPCs contain a single diffusion conduit, located within the central pore channel, which mediates both passive and receptor-mediated transport. However, a recent model published by this group suggests that transport of small molecules occurs within a dedicated tube located in the center of the central pore channel (44 44. Peters, R. Traffic. 2005; 6:421-427 Crossref Scopus (182) PubMed Google Scholar ), whereas macromolecules traffic along the walls of the NPC. In the present study, we tackled the question of how the two components of traffic interact by presenting the NPC with cargoes of both types, at various ratios, concomitantly. If both components utilize the same conduit exclusively, one would expect that, at high enough loads, some inhibitory effect would be observed. However, in our experiments, such an inhibitory effect was observed only for species belonging to the same component. At all molar ratios used in our experiments, up to 50, and with applied loads greater than 100 μ m, no inhibitory effect was observed for passive transport on facilitated transport, or vice versa. One could argue that the molar ratios used in our experiments are not sufficient to observe hindrance. However, because the same ratios were sufficient to impede members of the same group, the reason for lack of mutual effects of the two traffic components must lay elsewhere. The most parsimonious explanation for this indifference is that they proceed, at least partly, along physically segregated routes. Our data do not preclude that passive diffusion can also take place across the central pore of the NPC; indeed there is no reason to believe that small molecules do not take advantage of this massive conduit as observed by Feldherr (43 43. Feldherr, C.M. ∙ Akin, D. J. Cell Sci. 1997; 110:3065-3070 Crossref PubMed Google Scholar ). However, our results indicate that a significant fraction of the molecules that cross NPCs by diffusion utilize alternative routes that are sterically not overlapping with the central pore channel. This conclusion is in accordance with observations that passive diffusion of small inorganic ions is largely uncoupled from receptor-mediated, facilitated transport (39–42 39. Shahin, V. ∙ Danker, T. ∙ Enss, K. ... FASEB J. 2001; 15:1895-1901 Crossref Scopus (66) PubMed Google Scholar 40. Danker, T. ∙ Schillers, H. ∙ Storck, J. ... Proc. Natl. Acad. Sci. U. S. A. 1999; 96:13530-13535 Crossref Scopus (36) PubMed Google Scholar 41. Enss, K. ∙ Danker, T. ∙ Schlune, A. ... J. Membr. Biol. 2003; 196:147-155 Crossref Scopus (14) PubMed Google Scholar 42. Mazzanti, M. ∙ Bustamante, J.O. ∙ Oberleithner, H. Physiol. Rev. 2001; 81:1-19 Crossref Scopus (138) PubMed Google Scholar ). It also rationalizes early results reported on the effect of the lectin wheat germ agglutinin on passive and signal-dependent transport. This substance, which binds to O-linked N-acetyl-glucosamine moieties present on certain nucleoporins (45–47 45. Davis, L.I. ∙ Blobel, G. Proc. Natl. Acad. Sci. U. S. A. 1987; 84:7552-7556 Crossref Scopus (295) PubMed Google Scholar 46. Holt, G.D. ∙ Snow, C.M. ∙ Senior, A. ... J. Cell Biol. 1987; 104:1157-1164 Crossref Scopus (361) PubMed Google Scholar 47. Snow, C.M. ∙ Senior, A. ∙ Gerace, L. J. Cell Biol. 1987; 104:1143-1156 Crossref Scopus (440) PubMed Google Scholar ), effectively blocks receptor-mediated translocation (48–52 48. Newmeyer, D.D. ∙ Forbes, D.J. Cell. 1988; 52:641-653 Abstract Full Text (PDF) Scopus (452) PubMed Google Scholar 49. Finlay, D.R. ∙ Newmeyer, D.D. ∙ Price, T.M. ... J. Cell Biol. 1987; 104:189-200 Crossref Scopus (432) PubMed Google Scholar 50. Yoneda, Y. ∙ Imamoto-Sonobe, N. ∙ Yamaizumi, M. ... Exp. Cell Res. 1987; 173:586-595 Crossref Scopus (201) PubMed Google Scholar 51. Adam, E.J. ∙ Adam, S.A. J. Cell Biol. 1994; 125:547-555 Crossref Scopus (277) PubMed Google Scholar 52. Rutherford, S.A. ∙ Goldberg, M.W. ∙ Allen, T.D. Exp. Cell Res. 1997; 232:146-160 Crossref Scopus (46) PubMed Google Scholar ) but has little effect on the diffusion of intermediate-sized polymers and proteins (49 49. Finlay, D.R. ∙ Newmeyer, D.D. ∙ Price, T.M. ... J. Cell Biol. 1987; 104:189-200 Crossref Scopus (432) PubMed Google Scholar , 50 50. Yoneda, Y. ∙ Imamoto-Sonobe, N. ∙ Yamaizumi, M. ... Exp. Cell Res. 1987; 173:586-595 Crossref Scopus (201) PubMed Google Scholar , 53 53. Adachi, M. ∙ Fukuda, M. ∙ Nishida, E. EMBO J. 1999; 18:5347-5358 Crossref Scopus (243) PubMed Google Scholar ). Electron microscopy studies indicate that wheat germ agglutinin accumulates mostly, although not exclusively, in the central region of the pores (52 52. Rutherford, S.A. ∙ Goldberg, M.W. ∙ Allen, T.D. Exp. Cell Res. 1997; 232:146-160 Crossref Scopus (46) PubMed Google Scholar , 54 54. Akey, C.W. J. Cell Biol. 1989; 109:955-970 Crossref Scopus (174) PubMed Google Scholar , 55 55. Wilken, N. ∙ Kossner, U. ∙ Senecal, J.L. ... J. Cell Biol. 1993; 123:1345-1354 Crossref Scopus (48) PubMed Google Scholar ) and does not prevent binding of NLS-carrying cargoes to the surface of the pores (48 48. Newmeyer, D.D. ∙ Forbes, D.J. Cell. 1988; 52:641-653 Abstract Full Text (PDF) Scopus (452) PubMed Google Scholar , 52 52. Rutherford, S.A. ∙ Goldberg, M.W. ∙ Allen, T.D. Exp. Cell Res. 1997; 232:146-160 Crossref Scopus (46) PubMed Google Scholar , 56 56. Dabauvalle, M.C. ∙ Schulz, B. ∙ Scheer, U. ... Exp. Cell Res. 1988; 174:291-296 Crossref Scopus (126) PubMed Google Scholar ). The different effects exerted by wheat germ agglutinin on passive and facilitated transport through NPCs are naturally explained by the existence of separate routes of transport for each mode. Based on the above, we conclude that passive and facilitated transport across NPCs proceed, to a large extent, through routes that are sterically not overlapping with each other. Segregation of the two modes of transport may follow the configuration proposed by Peters (44 44. Peters, R. Traffic. 2005; 6:421-427 Crossref Scopus (182) PubMed Google Scholar ), in which facilitated transport occurs along the walls of the central pore channel and passive diffusion occurs through a narrow diffusion tube located at the pore center. More probable to our opinion, the routes mediating passive diffusion of small molecules are provided by the peripheral channels that surround the central pore, as was originally proposed by Hinshaw et al. (4 4. Hinshaw, J.E. ∙ Carragher, B.O. ∙ Milligan, R.A. Cell. 1992; 69:1133-1141 Abstract Full Text (PDF) Scopus (387) PubMed Google Scholar ). In contrast to the central pore that possesses a high degree of hydrophobicity, which is believed to promote passage of nuclear transport receptors (34 34. Ribbeck, K. ∙ Lipowsky, G. ∙ Kent, H.M. ... EMBO J. 1998; 17:6587-6598 Crossref Scopus (368) PubMed Google Scholar , 44 44. Peters, R. Traffic. 2005; 6:421-427 Crossref Scopus (182) PubMed Google Scholar , 57 57. Ribbeck, K. ∙ Kutay, U. ∙ Paraskeva, E. ... Curr. Biol. 1999; 9:47-50 Full Text Full Text (PDF) Scopus (109) PubMed Google Scholar , 58 58. Macara, I.G. Microbiol. Mol. Biol. Rev. 2001; 65:570-594 Crossref Scopus (774) PubMed Google Scholar ), the peripheral channels (or the central diffusion channel in the model proposed by Peters) are hydrophilic in nature. The difference in local dielectric may then act to sort incoming transport substrates into the different translocation paths according to their surface hydrophobicity/hydrophilicity. 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16341	https://blog.csdn.net/weixin_42292229/article/details/89945567	枚举法用于逻辑问题的处理_逻辑枚举法-CSDN博客博客下载学习社区 GitCode InsCodeAI 会议搜索 AI 搜索登录登录后您可以：复制代码和一键运行与博主大V深度互动解锁海量精选资源获取前沿技术资讯立即登录会员·新人礼包消息历史创作中心创作枚举法用于逻辑问题的处理 VeggieOrz于 2019-05-08 12:37:42 发布阅读量5.9k收藏 3 点赞数 3 CC 4.0 BY-SA版权文章标签：枚举逻辑问题版权声明：本文为博主原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接和本声明。本文链接：博客介绍了用计算机枚举解决逻辑问题的方法。通过警察局抓小偷、老师预测竞赛名次、公安审问嫌疑犯等实例，将问题转化为逻辑表达式，枚举所有可能情况，找出满足条件的解，避免了人工枚举的繁琐。前言遇到一些逻辑问题的时候，因为数据量不大的关系，我们通常只是需要人工枚举出所有的情况就可以。今天发现了怎么用计算机去枚举，况且记录一波。逻辑枚举问题 1：警察局抓了a,b,c,d四名偷窃嫌疑犯，其中有一人是小偷。审问中： a说：“我不是小偷”。 b说：“c是小偷”。 c说：“小偷肯定是d”。 d说：“c冤枉人”。现在已经知道四人中三人说的是真话，一人说的是假话。问到底谁是小偷？分析问题只有4中情况，而且他们4个人说的话只有4句为真。假设小偷为x，他们四个人说的话可以转换为以下四条逻辑表达式： a的话：x !=a b的话：x = c c的话：x = d d的话：x != d 接下来就只需要枚举x可取的四个值，满足上面三条逻辑表达式有三条为真的情况就得到了正解代码如下 ```c include include using namespace std; int main() { int x; for (x = 'a'; x <= 'd'; x++) if (((x != 'a') + (x == 'c') + (x == 'd') + (x !=' d')) == 3) printf("%c",x); return 0; } ``` AI写代码 c 运行 1 2 3 4 5 6 7 8 9 10 11 问题 2： 3位老师对某次数学竞赛进行预测，他们的预测为: 甲说：学生A得第一名，学生B得第三名乙说：学生C得第一名，学生D得第四名丙说：学生D得第二名，学生A得第三名。竞赛结果表明，他们都说对一半，说错一半，且无并列名次，试编写程序a,b,c,d各自的名称。分析如果不除去并列名次，则有 444^4 4 4 种情况，人工枚举会稍显麻烦。我们还是可以按照上一题的思想进行枚举。假设a,b,c,d分别表示学生A，学生B，学生C和学生D所获得的名次，那么三位教练说的话可以可以用如下逻辑表达式表示：甲说的话：a = 1 ， b = 3 乙说的话：c = 1 ， d = 4 丙说的话：d = 2 ， a = 3 由于他们说的话都只有一半是对的，所以我们可以得到以下条件句： (a=1) + (b=3) = 1 (c=1) + (d=4) = 1 (d=2) + (a=3) = 1 代码如下 ```c include include using namespace std; int main() { int a,b,c,d; for(a=1;a<=4;a++){ for(b=1;b<=4;b++)if(a!=b){//去除重复名次 for(c=1;c<=4;c++)if(a!=c&&b!=c){ d=10-a-b-c; //根据a，b，c的名次计算d的名次 if(a!=d&&b!=d&&c!=d){ if((a==1)+(b==3)==1&&(c==1)+(d==4)==1&&(d==2)+(a==3)==1) printf("a=%d, b=%d, c=%d, d=%d",a,b,c,d); } } } } return 0; } ``` AI写代码 c 运行 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 实战演练问题描述公安人员审问甲、乙、丙、丁四个嫌疑犯，已确知，这四个人当中仅有一人是偷窃者，还知道这四个人的答话，要么完全诚实，要么完全说谎。在回答公安人员的问话中：甲说：“乙没有偷，是丁偷的。” 乙说：“我没有偷，是丙偷的。” 丙说：“甲没有偷，是乙偷的。” 丁说：“我没有偷，我用的那东西是我家里的。” 请根据上述四人答话，判断谁是偷窃者。输入格式无输出格式输出一个字符，表示偷窃者是谁，A表示甲，B表示乙，C表示丙，D表示丁。分析这对绝大多数人来说，都只是一道水题。虽然情况不少，我们只要通过人工的逻辑推理很快可以得到正确的答案。下面我们继续用上面讲解到的思路进行求解，我们假设凶手为x，可以得到他们说的话对应的逻辑表示式：甲说的话表示x != B , x = D 乙说的话表示x != B , x = C 丙说的话表示x != A , x = B 丁说的话表示x != D 因为甲乙丙说的话都包含两个部分，而且他们说的话要么全为真，要么全为假。（至于丁说的话对结果有没有影响，我还有点疑问）就可得以下条件句： (x!=‘B’)+(x=‘D’)!=1 (x!=‘B’)+(x=‘C’)!=1 (x!=‘A’)+(x=‘B’)!=1 代码如下 ```c include include define x1 (x!='B')+(x=='D') define x2 (x!='B')+(x=='C') define x3 (x!='A')+(x=='B') using namespace std; int main() { int x; for(x='A';x<='D';x++) if((x1!=1)&&(x2!=1)&&(x3!=1)) printf("%c",x); return 0; } ``` AI写代码 c 运行 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 确定要放弃本次机会？福利倒计时 : : 立减 ¥ 普通VIP年卡可用立即使用 VeggieOrz 关注关注 3点赞踩 3 收藏觉得还不错? 一键收藏 0评论分享复制链接分享到 QQ 分享到新浪微博扫一扫举报举报参与评论您还未登录，请先登录后发表或查看评论关于我们招贤纳士商务合作寻求报道 400-660-0108 kefu@csdn.net 在线客服工作时间 8:30-22:00 公安备案号11010502030143 京ICP备19004658号京网文〔2020〕1039-165号经营性网站备案信息北京互联网违法和不良信息举报中心家长监护网络110报警服务中国互联网举报中心 Chrome商店下载账号管理规范版权与免责声明版权申诉出版物许可证营业执照 ©1999-2025北京创新乐知网络技术有限公司 VeggieOrz 博客等级码龄7年 138 原创1177 点赞 4821 收藏 723 粉丝关注私信 🆓注册即刻获得高达 $200 抵扣金☁️开始您的云计算之旅同时获得30余项永久免费服务广告热门文章 Spring Boot实现简单的用户权限管理（超详细版） 74900 面试官问你HTTP状态码，你敢答吗？ 48497 iOS 准确获取 iPhone 状态栏、导航栏、TabBar高度，看这篇就够了 42781 剑指Offer #07 斐波那契数列（四种解法）\| 图文详解 35521 wdCP面板升级MySQL版本为5.6 34910 分类专栏剑指Offer14篇 Debug笔记6篇 iOS 开发21篇 CSP认证8篇蓝桥杯题解8篇上一篇： C++描述的位运算总结下一篇： P1464 Function（递归式的记忆化搜索）最新评论 iOS - 沙盒文件操作指南胖虎1:感谢博主分享，文章内容丰富，感谢分享 iOS - 在 UITextView 的光标处插入文字（详细版）孚亭:代码中let addText = "new"需要修改成let addText: Character = "#"吧，不然会报错解决 VSCode 配置 tab 空格数 Dart 语言无效的问题 Krorainas:最后四个设置成功了嘛。代码太密集了看的很麻烦解决 VSCode 编写 C++11 代码报红问题鲤鱼_1:打开settings.json后发现里面已经存在如下内容： { "files.associations": { "iostream": "cpp", "stack": "cpp" }, "editor.suggest.snippetsPreventQuickSuggestions": false, "aiXcoder.showTrayIcon": true, "C_Cpp.default.compilerPath": "" } 请问此时该如何添加您最后给出的代码？ jsp uri=" 代不了一点码:请问 Web项目的WEB-INF 在哪大家在看 CSS基础入门指南 69 对象创建流程大模型技术精要：预训练、适配微调、提示学习与知识增强，收藏学习！ CTFHub命令注入绕过过滤第四部分：VTK常用类详解（第120章 vtkWarpTo变形到类）最新文章 Swift 高阶函数详解（forEach、filter、map、flatMap、compactMap、reduce、sorted）浅析 C 语言的共用体、枚举和位域如果你想知道 Swift 的 KeyPath 特性，不妨看看这个 2023年 4篇 2022年 7篇 2021年 22篇 2020年 49篇 2019年 48篇 2018年 8篇目录前言逻辑枚举实战演练展开全部收起目录前言逻辑枚举实战演练展开全部收起 🆓注册即刻获得高达 $200 抵扣金☁️开始您的云计算之旅同时获得30余项永久免费服务广告上一篇： C++描述的位运算总结下一篇： P1464 Function（递归式的记忆化搜索）分类专栏剑指Offer14篇 Debug笔记6篇 iOS 开发21篇 CSP认证8篇蓝桥杯题解8篇展开全部收起登录后您可以享受以下权益：免费复制代码和博主大V互动下载海量资源发动态/写文章/加入社区 ×立即登录评论被折叠的条评论为什么被折叠?到【灌水乐园】发言查看更多评论添加红包祝福语请填写红包祝福语或标题红包数量个红包个数最小为10个红包总金额元红包金额最低5元余额支付当前余额 3.43 元前往充值 > 需支付：10.00 元取消确定成就一亿技术人! 领取后你会自动成为博主和红包主的粉丝规则 hope_wisdom 发出的红包实付元使用余额支付点击重新获取扫码支付钱包余额 0 抵扣说明： 1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。 2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。余额充值确定取消举报选择你想要举报的内容（必选）内容涉黄政治相关内容抄袭涉嫌广告内容侵权侮辱谩骂样式问题其他原文链接（必填）请选择具体原因（必选）包含不实信息涉及个人隐私请选择具体原因（必选）侮辱谩骂诽谤请选择具体原因（必选）搬家样式博文样式补充说明（选填）取消确定 AI助手 AI 搜索智能体 AI 编程 AI 作业助手下载APP 程序员都在用的中文IT技术交流社区公众号专业的中文 IT 技术社区，与千万技术人共成长视频号关注【CSDN】视频号，行业资讯、技术分享精彩不断，直播好礼送不停！客服返回顶部
16342	https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_14?srsltid=AfmBOoruLpK1dgaBQfo1JHnTQAHbgR52SDQZRBIiU7upSqseShqr3hMs	Art of Problem Solving 2021 AMC 12A Problems/Problem 14 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 2021 AMC 12A Problems/Problem 14 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2021 AMC 12A Problems/Problem 14 Contents [hide] 1 Problem 2 Solution 1 (Properties of Logarithms) 3 Solution 2 (Properties of Logarithms) 4 Solution 3 (Estimations and Answer Choices) 5 Solution 4 (Properties of Logarithms) 6 Video Solution by Punxsutawney Phil 7 Video Solution by Hawk Math 8 Video Solution by OmegaLearn (Using Logarithmic Manipulations) 9 Video Solution by TheBeautyofMath (Using Magical Ability) 10 Video Solution by The Power of Logic 11 Video Solution (Logic and Simplification) 12 See also Problem What is the value of Solution 1 (Properties of Logarithms) We will apply the following logarithmic identity: which can be proven by the Change of Base Formula: Now, we simplify the expressions inside the summations: and Using these results, we evaluate the original expression: ~MRENTHUSIASM (Solution) ~JHawk0224 (Proposal) Solution 2 (Properties of Logarithms) First, we can get rid of the exponents using properties of logarithms: (Leaving the single in the exponent will come in handy later). Similarly, Then, evaluating the first few terms in each parentheses, we can find the simplified expanded forms of each sum using the additive property of logarithms: In we use the triangular numbers equation: Finally, multiplying the two logarithms together, we can use the chain rule property of logarithms to simplify: Thus, ~Joeya (Solution) ~MRENTHUSIASM (Reformatting) Solution 3 (Estimations and Answer Choices) In note that the addends are greater than for all In note that the addends are greater than for all We have the inequality which eliminates choices and We get the answer by either an educated guess or a continued approximation: Observe that and Therefore, we obtain the following rough underestimation: From here, it should be safe to guess that the answer is ~MRENTHUSIASM Solution 4 (Properties of Logarithms) Using the identity simplify and . Now we have the product: With the reciprocal rule the logarithms cancel out leaving: ~PowerQualimit Video Solution by Punxsutawney Phil Video Solution by Hawk Math Video Solution by OmegaLearn (Using Logarithmic Manipulations) Video Solution by TheBeautyofMath (Using Magical Ability) ~IceMatrix Video Solution by The Power of Logic Video Solution (Logic and Simplification) ~Education, the Study of Everything See also 2021 AMC 12A (Problems • Answer Key • Resources) Preceded by Problem 13Followed by Problem 15 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25 All AMC 12 Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
16343	https://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-geometry/cc-8th-pythagorean-theorem/e/use-area-of-squares-to-visualize-pythagorean-theorem	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
16344	https://www.math.utah.edu/~palais/AMissingPiece/ConicSections.html	\| \| \| A Missing Piece: Early Elementary Plane Rotations by Bob Palais Notices of the AMS February 2014, Doceamus, Web Supplement \| \| The x-y equations of conic sections are often derived by intersecting tilted planes with the standard right circular cone x2 +y2 =z2. The standard form is messy, and neglects the fact that x-y-coordinates in the tilted plane will not be the same as the x-y-coordinates of the original x-y-z space due to the tilt. The equations can be derived more easily, correctly, and in a cleaner standard form by instead rotating the cone using the rotation formula, shifting the vertex along the rotated axis, then intersecting with the x-y-plane, z=0, in other words, by omitting any terms involving z! The details are carried out here, and an animation resulting from the process is shown below. The elementary rotation formula in the x-y-plane makes it possible to treat quadratics with a cross term ax2+2bxy+cy2=1, and connect the type of conic section with the sign of the discriminant of ax2+2bx+c, with the determinant of the corresponding matrix, and eventually, with determining whether a critical point of a function f(x,y) is an extreme or saddle point. \| \| \| \| \| \| \| \| Acknowledgements \| \| Cosine Subtraction \| \| Conic Sections \| \| \| \| \| \| \| \| \|
16345	https://artofproblemsolving.com/wiki/index.php/2006_iTest_Problems?srsltid=AfmBOoqiq2hBESjozl7BRD9KtmUO4kxUq_mVyEHgdgVx9c472dYr9PDh	Art of Problem Solving 2006 iTest Problems - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 2006 iTest Problems Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2006 iTest Problems Contents [hide] 1 Multiple Choice Section 1.1 Problem 1 1.2 Problem 2 1.3 Problem 3 1.4 Problem 4 1.5 Problem 5 1.6 Problem 6 1.7 Problem 7 1.8 Problem 8 1.9 Problem 9 1.10 Problem 10 1.11 Problem 11 1.12 Problem 12 1.13 Problem 13 1.14 Problem 14 1.15 Problem 15 1.16 Problem 16 1.17 Problem 17 1.18 Problem 18 1.19 Problem 19 1.20 Problem 20 2 Short Answer Section 2.1 Problem 21 2.2 Problem 22 2.3 Problem 23 2.4 Problem 24 2.5 Problem 25 2.6 Problem 26 2.7 Problem 27 2.8 Problem 28 2.9 Problem 29 2.10 Problem 30 2.11 Problem 31 2.12 Problem 32 2.13 Problem 33 2.14 Problem 34 2.15 Problem 35 2.16 Problem 36 2.17 Problem 37 2.18 Problem 38 2.19 Problem 39 2.20 Problem 40 3 Ultimate Question 3.1 Problem 41 3.1.1 Problem U1 3.1.2 Problem U2 3.1.3 Problem U3 3.2 Problem 42 3.2.1 Problem U4 3.2.2 Problem U5 3.2.3 Problem U6 3.2.4 Problem U7 3.3 Problem 43 3.3.1 Problem U8 3.3.2 Problem U9 3.3.3 Problem U10 4 See Also Multiple Choice Section Problem 1 Find the number of positive integral divisors of 2006. Solution Problem 2 Find the harmonic mean of 10 and 20. Solution Problem 3 Let be distinct positive integers such that the product . What is the largest possible value of the sum ? Solution Problem 4 Four couples go ballroom dancing one evening. Their first names are Henry, Peter, Louis, Roger, Elizabeth, Jeanne, Mary, and Anne. If Henry's wife is not dancing with her husband (but with Elizabeth's husband), Roger and Anne are not dancing, Peter is playing the trumpet, and Mary is playing the piano, and Anne's husband is not Peter, who is Roger's wife? Solution Problem 5 A line has y-intercept and forms a right angle to the line . Find the x-intercept of the line. Solution Problem 6 What is the remainder when is divided by 7? Solution Problem 7 The sum of consecutive integers is . Find the second largest integer. Solution Problem 8 The point is a point on a circle with center . Perpendicular lines are drawn from to perpendicular diameters, and , meeting them at points and , respectively. If the diameter of the circle is , what is the length of ? Solution Problem 9 If and is in the third quadrant, what is the absolute value of ? Solution Problem 10 Find the number of elements in the first rows of Pascal's Triangle that are divisible by . Solution Problem 11 Find the radius of the inscribed circle of a triangle with sides of length , , and . Solution Problem 12 What is the highest possible probability of getting of these multiple choice questions correct, given that you don't know how to work any of them and are forced to blindly guess on each one? Solution Problem 13 Suppose that are three distinct prime numbers such that . Find the maximum possible value for the product . Solution Problem 14 Find , where is the smallest positive integer such that leaves a remainder of when divided by , , and . Solution Problem 15 How many integers between and , inclusive, are perfect squares? Solution Problem 16 The Minnesota Twins face the New York Mets in the 2006 World Series. Assuming the two teams are evenly matched (each has a probability of winning any game) what is the probability that the World Series (a best of 7 series of games which lasts until one team wins four games) will require the full seven games to determine a winner? Solution Problem 17 Let . Find the numerical value of . Solution Problem 18 Every even number greater than 2 can be expressed as the sum of two prime numbers.' Name the mathematician for which this theorem was named, and then name the mathematician to whom he transmitted this theorem via letter in 1742. Solution Problem 19 Questions 19 and 20 are Sudoku-related questions. Sudoku is a puzzle game that has one and only one solution for each puzzle. Digits from 1 to 9 must go into each space on the grid such that every row, column, and square contains one and only one of each digit. Find the sum of by solving the Sudoku puzzle below. 1 _ _ \| 3 5 8 \| _ _ 6 4 _ _ \| _ _ _ \| _ x 8 _ _ 9 \| _ 1 _ \| 7 _ _ _ z _ \| 1 _ _ \| _ 5 _ _ _ 3 \| 2 _ 4 \| 8 _ _ _ 2 _ \| w _ 9 \| _ _ _ _ _ 6 \| _ 2 _ \| 9 _ _ 3 _ _ \| _ y _ \| _ _ 1 2 _ _ \| 8 4 3 \| _ _ 7 Solution Problem 20 Sudoku is a puzzle game that has one and only one solution for each puzzle. Digits from 1 to 9 must go into each space on the grid such that every row, column, and square contains one and only one of each digit. Find the sum of by solving the Sudoku puzzle below. _ _ _ \| _ 4 _ \| _ z _ 1 _ 6 \| _ _ _ \| 7 _ 3 5 _ _ \| 9 _ _ \| _ _ 2 _ 8 3 \| w 2 _ \| 5 _ _ 2 _ _ \| 5 _ 9 \| _ _ 7 _ _ 7 \| _ 8 _ \| 9 2 _ 3 _ _ \| _ _ 1 \| _ _ 6 8 _ 9 \| x _ _ \| 3 _ 5 _ y _ \| _ 3 _ \| _ _ _ Solution Short Answer Section Problem 21 What is the last (rightmost) digit of ? Solution Problem 22 Triangle has sidelengths , , and . Point is the foot of the altitude from , and lies on segment such that . Find the area of the triangle . Solution Problem 23 Jack and Jill are playing a chance game. They take turns alternately rolling a fair six sided die labeled with the integers 1 through 6 as usual (fair meaning the numbers appear with equal probability.) Jack wins if a prime number appears when he rolls, while Jill wins if when she rolls a number greater than 1 appears. The game terminates as soon as one of them has won. If Jack rolls first in a game, then the probability of that Jill wins the game can be expressed as where and are relatively prime positive integers. Compute . Solution Problem 24 Points and are chosen on side of triangle such that is between and and , . If , the area of can be expressed as , where and are relatively prime positive integers and is a positive integer not divisible by the square of any prime. Compute . Solution Problem 25 The expression reduces to , where and are relatively prime positive integers. Find . Solution Problem 26 A rectangle has area and perimeter . The largest possible value of can be expressed as , where and are relatively prime positive integers. Compute . Solution Problem 27 Line passes through and into the interior of the equilateral triangle . and are the orthogonal projections of and onto respectively. If and , then the area of can be expressed as , where and are positive integers and is not divisible by the square of any prime. Determine . Solution Problem 28 The largest prime factor of is greater than . Determine the remainder obtained when this prime factor is divided by . Solution Problem 29 The altitudes in triangle have lengths 10, 12, and 15. The area of can be expressed as , where and are relatively prime positive integers and is a positive integer not divisible by the square of any prime. Find . Solution Problem 30 Triangle is equilateral. Points and are the midpoints of segments and respectively. is the point on segment such that . Let denote the intersection of and , The value of can be expressed as where and are relatively prime positive integers. Find . Solution Problem 31 The value of the infinite series can be expressed as where and are relatively prime positive numbers. Compute . Solution Problem 32 Triangle is scalene. Points and are on segment with between and such that , , and . If and trisect , then can be written uniquely as , where and are relatively prime positive integers and is a positive integer not divisible by the square of any prime. Determine . Solution Problem 33 Six students sit in a group and chat during a complicated mathematical lecture. The professor, annoyed by the chatter, splits the group into two or more smaller groups. However, the smaller groups with at least two members continue to produce chatter, so the professor again chooses one noisy group and splits it into smaller groups. This process continues until the professor achieves the silence he needs to teach Algebraic Combinatorics. Suppose the procedure can be carried out in ways, where the order of group breaking matters (if A and B are disjoint groups, then breaking up group A and then B is considered different form breaking up group B and then A even if the resulting partitions are identical) and where a group of students is treated as an unordered set of people. Compute the remainder obtained when is divided by . Solution Problem 34 For each positive integer let denote the set of positive integers such that is divisible by . Define the function by the rule Let be the least upper bound of and let be the number of integers such that and . Compute the value of . Solution Problem 35 Compute the of ordered quadruples of complex numbers (not necessarily nonreal) such that the following system is satisfied: Solution Problem 36 Let denote . The recursive sequence satisfies and, for all positive integers , Suppose that the series can be expressed uniquely as , where and are coprime positive integers and is not divisible by the square of any prime. Find the value of . Solution Problem 37 The positive reals , , satisfy the relations The value of can be expressed uniquely as , where , , , are positive integers such that is not divisible by the square of any prime and no prime dividing divides both and . Compute . Solution Problem 38 Segment is a diameter of circle . Point lies in the interior of segment such that , and is a point on such that . Segment is a diameter of the circle . A third circle, , is drawn internally tangent to , externally tangent to , and tangent to segment . If is centered on the opposite side of as , then the radius of can be expressed as , where and are relatively prime positive integers. Compute . Solution Problem 39 is a regular dodecagon. The number 1 is written at the vertex A, and 0's are written at each of the other vertices. Suddenly and simultaneously, the number at each vertex is replaced by the arithmetic mean of the two numbers appearing at the adjacent vertices. If this procedure is repeated a total of times, then the resulting number at A can be expressed as , where and are relatively prime positive integers. Compute the remainder obtained when is divided by . Solution Problem 40 Acute triangle satisfies and . Tetrahedron is formed by choosing points , , and on the segments , , and (respectively) and folding , , , over , , and (respectively) to the common point . Let denote the circumradius of . Compute the smallest positive integer for which we can be certain that . It may be helpful to use . Solution Ultimate Question In the next 2 problems, the problem after will require the answer of the current problem. TNFTPP stands for the number from the previous problem. Problem 41 requires the answer to the third problem. Problem 42 requires the answer to the seventh problem. Problem 43, however, requires the sum of the answers to all ten questions. For those who want to try these problems without having to find the T-values of the previous problem, a link will be here. Also, all solutions will have the T-values substituted. Problem 41 Problem U1 Find the real number such that Solution Problem U2 Let . Points and lie on a circle centered at such that is right. Points and lie on radii and respectively such that , , and . Determine the area of quadrilateral . Solution Problem U3 Let . When properly sorted, math books on a shelf are arranged in alphabetical order from left to right. An eager student checked out and read all of them. Unfortunately, the student did not realize how the books were sorted, and so after finishing the student put the books back on the shelf in a random order. If all arrangements are equally likely, the probability that exactly of the books were returned to their correct (original) position can be expressed as , where and are relatively prime positive integers. Compute . Solution Problem 42 Problem U4 Let . As ranges over the integers, the expression evaluates to just one prime number. Find this prime. Solution Problem U5 Let , and let be the sum of the digits of . In triangle , points , , and are the feet of the angle bisectors of , , respectively. Let point be the intersection of segments and , and let denote the perimeter of . If , , and , then the value of can be expressed uniquely as where and are positive integers such that is not divisible by the square of any prime. Find . Solution Problem U6 Let . and are nonzero real numbers such that The smallest possible value of is equal to where and are relatively prime positive integers. Find . Solution Problem U7 Let . Triangle has integer side lengths, including , and a right angle, . Let and denote the inradius and semiperimeter of respectively. Find the perimeter of the triangle ABC which minimizes . Solution Problem 43 Problem U8 Let , and let be the sum of the digits of . Cyclic quadrilateral has side lengths , , , and . Let and be the midpoints of sides and . The diagonals and intersect at and respectively. can be expressed as where and are relatively prime positive integers. Determine . Solution Problem U9 Let . Determine the number of 5 element subsets of such that the sum of the elements of is divisible by 5. Solution Problem U10 Let and let be the sum of the digits of . Point in the interior of triangle satisfies , , and . If the sides of ABC satisfy then the area of triangle can be expressed as , where and are relatively prime positive integers. Compute the remainder obtained when is divided by . Recall that you are turning in the sum of all ten answers, NOT the answer to this problem. Solution See Also iTest Problems and Solutions 2006 iTest(Problems, Answer Key) Preceded by: 2005 iTestFollowed by: 2007 iTest 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25•26•27•28•29•30•31•32•33•34•35•36•37•38•39•40•U1•U2•U3•U4•U5•U6•U7•U8•U9•U10 Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
16346	https://math.stackexchange.com/questions/1417204/general-equation-of-line-that-goes-through-center-of-a-circle-and-a-point	geometry - General equation of line that goes through center of a circle and a point - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more General equation of line that goes through center of a circle and a point Ask Question Asked 10 years, 1 month ago Modified10 years, 1 month ago Viewed 2k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Given an arbitrary point P P, at (x 1,y 1)(x 1,y 1), is there a general expression of a line that goes through a circle of radius r r centered at the origin? I know there are infinite number of such lines/diameters when P=0 P=0. If such an expression exists, is it possible to algebraically, or otherwise, find the general equation for all intersection points between the line and the circle? geometry analytic-geometry Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Sep 1, 2015 at 18:45 KReiser 75.6k 16 16 gold badges 65 65 silver badges 118 118 bronze badges asked Sep 1, 2015 at 15:46 RazorlanceRazorlance 275 1 1 silver badge 10 10 bronze badges 1 you can wright equation of aline when you have two points of it. just google it. if p isnot equal to center then it has two point intersectin with circle.(excuse me for weak english)R.N –R.N 2015-09-01 15:52:09 +00:00 Commented Sep 1, 2015 at 15:52 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. An arbitrary line is given by the equation a x+b y=c(∗)a x+b y=c(∗) The circle through the origin with radius r r has the equation x 2+y 2=r 2 x 2+y 2=r 2 The requirements on the line is to go through the origin (0,0)(0,0) which happens to be the center of the circle and some point P=(x 1,y 1)P=(x 1,y 1). This line can be written in parametric form as x=(1−t)0+t x 1=t x 1 y=(1−t)0+t y 1=t y 1 x=(1−t)0+t x 1=t x 1 y=(1−t)0+t y 1=t y 1 for t∈R t∈R. Using (∗)(∗) we insert the origin and get a 0+b 0=c⇒c=0 a 0+b 0=c⇒c=0 Then we insert P P and get a x 1+b y 1=0 a x 1+b y 1=0 If P P is different from the origin, this gives another condition. E.g. if x 1≠0 x 1≠0 we have a=−y 1 x 1 b a=−y 1 x 1 b and −y 1 x 1 b x+b y=0−y 1 x 1 b x+b y=0 which means b=0 b=0 or y=(y 1/x 1)x y=(y 1/x 1)x. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Sep 1, 2015 at 16:10 answered Sep 1, 2015 at 15:59 mvwmvw 35.2k 2 2 gold badges 34 34 silver badges 65 65 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. A line is often written in standard form as y=m x+b y=m x+b where m m is the slope and b b is some constant. We can solve for m m easily: m=x point−x center of the circle y point−y center of the circle m=x point−x center of the circle y point−y center of the circle Plugging this back in, we have y=x point−x center of the circle y point−y center of the circle x+b y=x point−x center of the circle y point−y center of the circle x+b Then, we can once again use the coordinates of P P to find b b: y center of the circle−x point−x center of the circle y point−y center of the circle x center of the circle=b y center of the circle−x point−x center of the circle y point−y center of the circle x center of the circle=b Therefore, we have the equation y=x point−x center of the circle y point−y center of the circle x+(y center of the circle−x point−x center of the circle y point−y center of the circle x center of the circle)y=x point−x center of the circle y point−y center of the circle x+(y center of the circle−x point−x center of the circle y point−y center of the circle x center of the circle) Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Sep 1, 2015 at 15:58 HDE 226868HDE 226868 2,392 1 1 gold badge 21 21 silver badges 36 36 bronze badges Add a comment\| You must log in to answer this question. 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16347	https://www.scribd.com/document/660266631/Practice-Problems-on-Kinetic-vs-Thermodynamic-Control	Practice Problems On Kinetic Vs Thermodynamic Control \| PDF \| Chemistry \| Chemical Substances Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 409 views 2 pages Practice Problems On Kinetic Vs Thermodynamic Control This document discusses kinetic vs thermodynamic control in chemical reactions. It provides examples of reactions adding hydrogen halides to alkenes and asks to draw the major product(s) tha… Full description Uploaded by Denis Coromoto Uzcátegui AI-enhanced title and description Go to previous items Go to next items Download Save Save Practice Problems on Kinetic vs Thermodynamic Cont... For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Translate Ask AI Report Download Save Practice Problems on Kinetic vs Thermodynamic Cont... For Later You are on page 1/ 2 Search Fullscreen !"#$%&$' !")+',- ). /&.'%&$ 0- 12'",)34.#,&$ 5).%")+ 67 8"#9 %2' -%":$%:"'- ); %2' ,#<)" =")3:$%- ); '#$2 ); %2' ;)++)9&.>"'#$%&).-7 6; %2' ,#<)" =")3:$% &- ).' -='$&;&$ -%'"')&-),'"? ' -:"' %) 3"#9 %2' $)""'$% ).'7 6; %2' ,#<)" =")3:$% &- # ,&@%:"' ); -%'"')&-),'"- &. #==")@&,#%'+4 'A:#+ #,):.%- $+'#"+4 3"#9 %2',7 a)HBr (1 equiv.)(high temp.thermodynamic control)b)HBr (1 equiv.)(low temp.kinetic control)c)d)Br 2 Kinetic Control e)Br 2 Thermodynamic Control f)HCl Kinetic Control HCl Thermodynamic Control adDownload to read ad-free 667 B)" '#$2 ); %2' ;)++)9&.> #33&%&). ); 243")>'. 2#+&3'- %) $).<:>#%'3 3&'.'- 3"#9 &. %2' )@ %2' $)""'-=).3&.> =")3:$%C-D7 HBr a)+Kinetic Control Thermodynamic Control b)HCl+Kinetic Control Thermodynamic Control c)HCl+Kinetic Control Thermodynamic Control d)HBr+Kinetic Control Thermodynamic Control e)HCl+Kinetic Control Thermodynamic Control Read this document in other languages Français Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like Unit 4 AoS 1 SAC Revision 2024 Solutions No ratings yet Unit 4 AoS 1 SAC Revision 2024 Solutions 25 pages CH 16 Problem Set (MCM) - 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16348	https://math.stackexchange.com/questions/1627986/line-equation-through-point-parallel-to-plane-and-intersecting-line	algebra precalculus - Line equation through point, parallel to plane and intersecting line - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Line equation through point, parallel to plane and intersecting line Ask Question Asked 9 years, 8 months ago Modified6 years, 6 months ago Viewed 5k times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. Write the equations of the line that passes through point M(1,0,7)M(1,0,7), is parallel with the plane 3 x−y+2 z−15=0 3 x−y+2 z−15=0 and intersects line x−1 4=y−3 2=z 1 x−1 4=y−3 2=z 1 Alright, so from what I know, if a line is parallel to a plane, that means that the plane's normal vector is perpendicular to the line's direction vector, so their dot product must be 0. The plane's normal vector is v=⟨3,−1,2⟩v=⟨3,−1,2⟩ from what I can tell. What other conditions are there for a line to intersect another line, in 3D? Also, can I get an equation from the given point as well? EDIT: To get the point of intersection between my given line and the plane, I write down the parametric equations for my line. I have z=t,y=2 t+3,x=4 t+1 z=t,y=2 t+3,x=4 t+1. I substitute these in the plane equation that you've given me(in which I have my line) and I get the point of intersection between the plane and the line t=17 13 t=17 13. That point should be on the line I'm looking for as well. I get point B(68 13,37 13,−74 13)B(68 13,37 13,−74 13). Then, when I try to compute the line equation from the two points I know are on it, I get x=81 13+68 13 t,y=37 13+37 13 t,z=17 13−74 13 t x=81 13+68 13 t,y=37 13+37 13 t,z=17 13−74 13 t. The thing is, when I try to compute the dot product between the line's direction vector and the normal vector of the plane to which it is parallel to, I don't get 0 0. I ran this many times, and I get the same result every time. Am I doing something wrong? I don't think it's a computation error. algebra-precalculus geometry analytic-geometry solid-geometry Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jun 12, 2020 at 10:38 CommunityBot 1 asked Jan 26, 2016 at 17:22 MikhaelMMikhaelM 1,479 1 1 gold badge 15 15 silver badges 24 24 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. Another condition "for a line to intersect another line, in 3D" is that there is a point that is on both lines. That sounds simplistic but that really is the condition. Here is one way to solve your problem. The line that you want is in a plane that is parallel to the plane 3 x−y+2 z−15=0 3 x−y+2 z−15=0. Any such plane has the equation 3 x−y+2 z+D=0 3 x−y+2 z+D=0 for some constant D D. That plane must also go through the point M(1,0,7)M(1,0,7), so we can substitute to find the value of D D: 3⋅1−0+2⋅7+D=0 3⋅1−0+2⋅7+D=0 So D=−17 D=−17 and the plane is 3 x−y+3 z−17=0 3 x−y+3 z−17=0. Combining that with your equations x−1 4=y−3 2=z 1 x−1 4=y−3 2=z 1 you can find the single point that is the intersection of that plane and your given line. The line you want goes through those two points. You should be able to finish from here. There are multiple ways to find the intersection of the plane and of the given line. You showed one way, by giving a parameterization of the line and solving for the parameter. You did make a computation error: the equation you get is 3(4 t+1)−(2 t+3)+2(t)−17=0 3(4 t+1)−(2 t+3)+2(t)−17=0 and the solution is t=17 12 t=17 12 which makes x=20 3,y=35 6,z=17 12 x=20 3,y=35 6,z=17 12. Substituting this solution into all the equations checks, so this is the right answer. Thus the intersection point is Q(20 3,35 6,17 12)Q(20 3,35 6,17 12) I got the intersection point by different means, by setting up and solving these simultaneous linear equations: 1 x+0 y−4 z y−2 z 3 x−y+2 z=1=3=17 1 x+0 y−4 z=1 y−2 z=3 3 x−y+2 z=17 I got the same intersection point. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jan 26, 2016 at 19:00 answered Jan 26, 2016 at 18:03 Rory DaultonRory Daulton 32.9k 6 6 gold badges 48 48 silver badges 65 65 bronze badges 3 Thanks! I'm still having a bit of an issue with the final result I get. I edited my initial post to outline the steps I've taken.MikhaelM –MikhaelM 2016-01-26 18:38:46 +00:00 Commented Jan 26, 2016 at 18:38 @MikhaelM: See the addition to my answer. Let me know if you can't finish from there.Rory Daulton –Rory Daulton 2016-01-26 18:54:33 +00:00 Commented Jan 26, 2016 at 18:54 Thank you! Seems I can't add >.<MikhaelM –MikhaelM 2016-01-26 19:41:51 +00:00 Commented Jan 26, 2016 at 19:41 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algebra-precalculus geometry analytic-geometry solid-geometry See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 8Equation for a line through a plane in homogeneous coordinates. 1Find the equation of a plane that is perpendicular to another plane, parallel to a line and goes through a point 1Equation of a plane through line and point 2find a plane that passes through one line and is parallel to another 2Finding an equation of a plane through the origin that is parallel to a given plane and parallel to a line. 2Find the equation of a plane given point, parallel line and angle between plane and line 3Find equation of a plane, that contains a point, is perpendicular to plane and parallel to a line. 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16349	https://www.vedantu.com/question-answer/a-plant-cell-becomes-turgid-due-to-aplasmolysis-class-11-biology-cbse-5fd2669e164c3d4fbe5271a8	Talk to our experts 1800-120-456-456 A plant cell becomes turgid due to A)Plasmolysis B)Exosmosis C)Endosmosis D)Electrolysis Repeaters Course for NEET 2022 - 23 © 2025.Vedantu.com. All rights reserved
16350	https://statkat.com/find-critical-value/significance-level-alpha/t.php	You can easily find the critical t value given the significance level alpha with our . If you want to find the critical t value by using a table with critical t values, instructions are given below. Finding the critical value $t^$ given a significance level $\alpha$, using the table with critical $t$ values Assuming a table with a row per degrees of freedom and a column per upper tail probability \| \| \| --- \| \| Two sided \| Reject the null hypothesis if the observed $t$ falls in one of the two most extreme $\alpha$ / 2 areas of the $t$ distribution. In order to find the critical values $t^$ and $-t^$ that correspond to these tail areas: 1. Find the row with the appropriate number of degrees of freedom (df) 2. Find the column for the upper tail probability equal to $\alpha / 2$ 3. At the intersection point of this row and column, you find the positive critical value $t^$. Observed $t$ values that are at least as extreme as $t^$ (positive or negative), are significant. In other words, observed $t$ values that are equal to or larger than $t^$, and observed $t$ values that are equal to or smaller than $-t^$, lead to rejection of the null hypothesis \| \| Right sided \| Reject the null hypothesis if the observed $t$ falls in the highest $\alpha$ area of the $t$ distribution. In order to find the critical value $t^$ that corresponds to this upper tail area: 1. Find the row with the appropriate number of degrees of freedom (df) 2. Find the column for the upper tail probability equal to $\alpha$ 3. At the intersection point of this row and column, you find the critical value $t^$. Observed $t$ values that are equal to or larger than $t^$, lead to rejection of the null hypothesis \| \| Left sided \| Reject the null hypothesis if the observed $t$ falls in the lowest $\alpha$ area of the $t$ distribution. In order to find the critical value $t^$ that corresponds to this lower tail area: 1. Find the row with the appropriate number of degrees of freedom (df) 2. Find the column for the upper tail probability equal to $\alpha$ 3. Multiply the value at the intersection point of this row and column, by -1. The resulting value is the critical value $t^$. Observed $t$ values that are equal to or smaller than $t^$, lead to rejection of the null hypothesis \|
16351	https://www.youtube.com/watch?v=l3R6wdHs9n8	The Counting by Hundreds Song \| Counting Songs \| Scratch Garden Scratch Garden 1890000 subscribers 6770 likes Description 4981353 views Posted: 23 Mar 2020 Learn to count by one hundreds with Scratch Garden's fun Counting by Hundreds Song! This Skip Counting by 100s video is great for learning to count in the classroom, practicing multiplication, and practicing skip counting by 100 to 2000! Like our videos? Help support us on Patreon (and access over 50 member-only videos!) ►► ◄◄ The Counting by 100s Song teaches skip counting by 100. If you teaching math and early years numeracy, check out the Counting by One Hundreds Song by Scratch Garden! Watch more Counting Videos from Scratch Garden: Watch many more fun Scratch Garden Educational Videos here: Scratch Garden makes entertaining educational videos for people that like to laugh and learn! Please Subscribe to see more great fun learning videos from Scratch Garden! Website: Instagram: Facebook: countingbyhundreds #skipcountingsongs #scratchgardensongs Transcript: Counting by 100s Counting by 100s Now you start with 100, and as you go along Jump 100 numbers and practice this song Here’s a little hint to help as you go You kind of count up by one and then add two zeros! 100, 200, 300, 400, 500 Counting by 100s Counting by 100s Let’s do it again but let's go higher 100, 200, 300, 400, 500, 600, 700, 800, 900, 1000 Counting by 100s Counting by 100s Whoa that is high, this sure is fun Let’s get to 2000 and maybe we’re done 100, 200, 300, 400, 500, 600, 700, 800, 900, 1000, 1100 1200, 1300, 1400, 1500, 1600, 1700, 1800, 1900, 2000 Counting by 100s Counting by 100s This number here, it’s one thousand one hundred you’ll say But let’s count up again in a slightly different way Let’s call it eleven hundred, yeah it means the same thing So don’t say the word “thousand” this time when you sing 100, 200, 300, 400, 500, 600, 700, 800, 900, 1000, eleven hundred, twelve hundred, fourteen hundred, fifteen hundred, sixteen hundred, seventeen hundred, eighteen hundred, nineteen hundred, 2000 Counting by 100s Counting by 100s Counting by 100s Counting by 100s
16352	https://openlibrary.org/books/OL7586033M/The_History_of_Money	The History of Money by Jack Weatherford \| Open Library It looks like you're offline. Donate ♥ English (en) Čeština (cs) Deutsch (de) English (en) Español (es) Français (fr) हिंदी (hi) Hrvatski (hr) Italiano (it) Português (pt) Română (ro) Sardu (sc) తెలుగు (te) Українська (uk) 中文 (zh) My Books Browse Menu Subjects Trending Library Explorer Lists Collections K-12 Student Library Book Talks Random Book Advanced Search All [x] Log In Sign Up My Open Library Log InSign Up Browse Subjects Trending Library Explorer Lists Collections K-12 Student Library Book Talks Random Book Advanced Search Contribute Add a Book Recent Community Edits Resources Help & Support Developer Center Librarians Portal My Books Browse Menu Subjects Trending Library Explorer Lists Collections K-12 Student Library Book Talks Random Book Advanced Search Overview View 4 Editions Details Reviews Lists Related Books An edition of The History of Money (1998) The History of Money by Jack Weatherford ★★2.0 (1 rating) 5 Want to read 2 Have read Share ×Close Facebook Twitter Pinterest Embed Copy URL QR Code The History of Money Jack Weatherford, Jack Weather ... The History of Money ×Close Locate ✓Want to Read Remove From Shelf Want to Read Currently Reading Already Read My Reading Lists: [x] Use this Work Create a new list Create a new list ×Close Name: Description: Create new listCancel 1 2 3 4 5 Clear my rating Review Notes My Book Notes ×Close My private notes about this edition: Delete Note Save Note Share ×Close Facebook Twitter Pinterest Embed Copy URL QR Code ★★2.0 (1 rating) 5 Want to read 2 Have read Review 1 2 3 4 5 Clear my rating Notes My Book Notes ×Close My private notes about this edition: Delete Note Save Note Share ×Close Facebook Twitter Pinterest Embed Copy URL QR Code Check nearby libraries WorldCat Buy this book Better World Books $8.02 (used) - includes shipping Amazon $9.56 More Bookshop.org When you buy books using these links the Internet Archive may earn a small commission. Overview View 4 Editions Details Reviews Lists Related Books Last edited by ImportBot March 28, 2025 \| History Edit An edition of The History of Money (1998) The History of Money by Jack Weatherford ★★2.0 (1 rating) 5 Want to read 2 Have read The History of Money Edit In his most widely appealing book yet, one of today's leading authors of popular anthropology looks at the intriguing history and peculiar nature of money, tracing our relationship with it from the time when primitive men exchanged cowrie shells to the imminent arrival of the all-purpose electronic cash card. 320 pp. Author tour. National radio publicity. 25,000 print.From the Hardcover edition. Read more ▾Read less ▲ Publish Date March 10, 1998 Publisher Three Rivers Press Language English Pages 304 Check nearby libraries WorldCat Buy this book Better World Books $8.02 (used) - includes shipping Amazon $9.56 More Bookshop.org When you buy books using these links the Internet Archive may earn a small commission. Subjects Business, Nonfiction, Money, History, Money, history Show entries Search: \| Edition \| Availability↑ \| --- \| \| 4 The History of Money March 10, 1998, Three Rivers Press in English 0609801724 9780609801727 \| aaaa Locate \| \| 1 The History of Money Feb 25, 2014, Brilliance Audio mp3 cd 1480590029 9781480590021 \| zzzz Locate \| \| 2 The History of Money Feb 25, 2014, Brilliance Audio audio cd 1480591890 9781480591899 \| zzzz Locate \| Showing 1 to 3 of 4 entries FirstPrevious12NextLast Add another edition? Book Details First Sentence "IN THE CENTER OF THE AZTEC IMPERIAL CAPITAL OF Tenochtitlan, the priests performed their daily sacrifices." 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16353	https://pmc.ncbi.nlm.nih.gov/articles/PMC10690513/	Team Cohesion Profiles: Influence on the Development of Mental Skills and Stress Management - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. PMC Search Update PMC Beta search will replace the current PMC search the week of September 7, 2025. Try out PMC Beta search now and give us your feedback. Learn more Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide New Try this search in PMC Beta Search View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice J Sports Sci Med . 2023 Dec 1;22(4):637–644. doi: 10.52082/jssm.2023.637 Search in PMC Search in PubMed View in NLM Catalog Add to search Team Cohesion Profiles: Influence on the Development of Mental Skills and Stress Management Amaia Ramírez Muñoz Amaia Ramírez Muñoz 1 Faculty of Education at Universidad Internacional de La Rioja (UNIR). TECNODEF Research Group. La Rioja, Spain Find articles by Amaia Ramírez Muñoz 1, Marta Vega-Díaz Marta Vega-Díaz 2 Faculty of Education at the University of Alfonxo X el Sabio, Madrid, Spain Find articles by Marta Vega-Díaz 2,✉, Higinio González-García Higinio González-García 1 Faculty of Education at Universidad Internacional de La Rioja (UNIR). TECNODEF Research Group. La Rioja, Spain Find articles by Higinio González-García 1 Author information Article notes Copyright and License information 1 Faculty of Education at Universidad Internacional de La Rioja (UNIR). TECNODEF Research Group. La Rioja, Spain 2 Faculty of Education at the University of Alfonxo X el Sabio, Madrid, Spain ✉ Matías López, Nº26-30. CP:27600. Galicia, Spain Received 2022 Sep 19; Accepted 2023 Sep 23; Collection date 2023 Dec. © Journal of Sports Science and Medicine PMC Copyright notice PMCID: PMC10690513 PMID: 38045748 Abstract High-level sports competitions involve facing highly challenging situations. Athletes must maintain strong team cohesion with peers, have specific mental abilities, and high-stress control to overcome adversity and report high sports performance. This research aimed to identify team cohesion profiles and examine whether participants differed significantly in their mental abilities and stress management. The sample consisted of 146 promising and talented athletes from the Sports Talent Development of the Provincial Council of Guipúzcoa (Spain) (M age = 20.08; SD = 4.68), who completed the questionnaire on Psychological Characteristics Related to Sports Performance (CPRD). Cluster analyzes revealed three profiles; (a) profile with low team cohesion; (b) profile with average team cohesion; (c) profile with high team cohesion. Results showed significant differences in mental abilities (i.e., positive self-talk), and marginally significant differences in self-confidence, between the profiles. The best scores were reported in profile (b). In conclusion, the combination of low individualism, high social cohesion, and medium team spirit seems to be the most recommendable for promoting mental abilities and self-confidence in athletes' samples. As practical implications, the programs that train the mental abilities of athletes and control management should consider the importance of team cohesion to obtain improvements in the results of the competitions. Key points. High-level sports competitions involve facing challenging situations in which it is essential for athletes to feel cohesion with other team members to maintain high performance. Identifying functional and dysfunctional profiles of team cohesion can help athletes find an explanation for their mental abilities during competitions. Programs that train athletes must strive to ensure that athletes maintain a good group spirit, are not individualistic, and maintain strong bonds with their teammates since their ability to control stress in the face of adversity depends on it. Key words: Mental ability, group analysis, sport, team Introduction The fundamental objective of competitive sports is that athletes perform to the maximum of their possibilities to achieve outstanding success when competing with other athletes or sports teams (Fernando and Pérez-Llantada, 2007). However, the stressful situations that must be faced during sports competitions can imply an alteration in the psychological functioning of athletes, which hinders their probability of success. As a result, athletes may experience a loss of attentional focus, and increased stress, among others (Brown and Fletcher, 2016; Guerrero et al., 2017). Regarding these problems, teams with high team cohesion, manage to respond as a closed unit to the adversities of competition, which is why they usually obtain favorable sporting results (Palmi, 1994). In addition to team cohesion, other variables facilitate sporting success: mental abilities and stress management (Meyers et al., 1979). Considering the aforementioned implications of team cohesion, mental abilities, and stress management on success in competitive sports, it is interesting to investigate these variables to maximize athletes' achievements simultaneously. Previous literature has examined team cohesion within sports field (Iturbide et al., 2010; Pescosolido and Saavedra, 2012; Sezer and Kocaeksi, 2018). Team cohesion is the tendency of a group not to separate and to remain together in pursuit of goals and objectives and in satisfying the affective needs of its members (Carron et al., 1998). According to Gimeno et al. (2001) team cohesion is divided into: individualism vs collectivism, social cohesion and team spirit. To assess team cohesion, most sport-related literature in the Spanish language has adopted Gimeno et al.’s measure (2001). Individualism implies that the person prefers to work individually, and he/she believes that the team success depends exclusively on its performance (Hofstede, 1980). Thus, those high in individualism can be highly motivated by competition, individual rewards, and recognition (Hadjiyankova and Iancheva, 2021). Moreover, individualistic athletes often work and invest efforts to achieve individual goals. Oppositely, athletes who tend to be collectivist prefer working in a group (Tan et al., 1998), and present goals aligned with teammates (Triandis and Gelfand, 1998; Yamaguchi 1994). Social cohesion refers to the extent that the team is consolidated in social life (Eys and Brawley, 2018). In the same way, Richardson (2013) added that social cohesion is the degree to which team members create a conciliatory environment to interact, and perceive liking towards their teammates. Therefore, teams with social cohesion will improve together regardless of sporting results (Erikstad et al., 2018; Silveira and Oliveira, 2017). Finally, team spirit is the feeling of respect and shared responsibility for the successes and failures of the team, and the belief in the group's ideology (Salas et al., 2015). Previous studies examined the relationship between team cohesion and mental abilities (Gu and Xue, 2022; Villegas, 2019). Within the sports field, Loehr (1986) described mental abilities as psychological skills that allow staying focused on tasks and maintaining confidence to face challenges. Mental abilities are proposed to include: practice in imagination, goal setting, positive self-talk, objective analysis of own performance and tension control (Gimeno et al., 2001; Heydari et al., 2018; Jeon et al., 2021; Lago, 2008; Riera et al., 2017). Practice in imagination involves predicting what will happen in the competition and mentally rehearsing what to do to correct mistakes (Martin, 1999; Riera et al., 2017). Goal setting directs people to focus their efforts on actions related to their goals and ignore irrelevant activities (Jeon et al., 2021). In addition, goal setting energizes people toward the most difficult and effort-intensive goals. Positive self-talk is the internal conversation that a person can do aloud or silently (before, during, or after the competition) and through which the person strengthens him/herself (Calvete and Cardeñoso, 2002). The objective analysis of own performance means that each person must identify strengths and weaknesses that may affect performance in competition (Lago, 2008). Tension control means that an athlete perceives that he/she is ready or able to face a competition (Lawther, 1998). Although there seems not to be much sports literature that relates each dimension of team cohesion with mental abilities, in the work of Fitzgerald (2019), it is stated that cohesive teams are constantly looking for a competitive advantage. As such, it implies the display of mental abilities to achieve success, such as the practice of imagination. On the other hand, team cohesion encourages athletes to fight for common goals (Miçoogullari, 2013), which could be related to goal setting, and favor positive self-talk (Cardeñoso et al., 2007; Villa, 2005). More specifically, referring to the sub-dimensions that constitute team cohesion, it is known that the objective analysis of own performance allows each person to reflect to identify strengths that can be maximized, and weaknesses that can be improved. Individualistic athletes believe the team's success depends exclusively on its performance (Hofstede, 1980). Thus, they could spend more time reflecting on their performance in the competition (against collectivists who believe in the importance of the group). Regarding tension control, it is known that individualism is less effective than collectivism in regulating this variable within the competitive field (Csikszentmihalyi et al., 1993). On the other hand, social cohesion and team spirit positively help to tension control (Csikszentmihalyi et al., 1993). Previous works revealed a connection between team cohesion and stress (Driskell et al., 2015), self-confidence (Chicau et al., 2012; Prapavessis and Carron, 1996), and attention (Salas et al., 2015); variables influencing athletes' stress management abilities. Stress can be described as a state of physical and psychological activation in response to external demands that exceed one’s ability to cope and requires a person to adapt or change behavior (Dos Santos et al., 2020). Self-confidence is the belief that athletes have about their ability to be successful in sport (Vealey, 1986; Vealey and Chase, 2008). Attention refers to an individual's effort to mentally focus on a stimulus and eliminate distracting sources (Hill et al., 2019). Some studies have identified some beneficial outcomes for athlete stress on sport physical achievement (Galli and Reel, 2012; Howells et al., 2015). This happens because every athlete has a certain stress level that is needed to optimize his or her game (Bali, 2015), and because stress prepares the body with greater motivation and enthusiasm to face the sporting demands (Ferreira et al., 2002). However, other research has shown negative consequences of stress on performance (Olmedilla et al., 2021). Regarding self-confidence, some studies reveal significant benefits of this variable in the sports performance of athletes (Draper et al., 2011; Hassmén et al., 2004; Terry and Slade, 1995) and others do not indicate any benefit (Bejek and Hagtvet, 1996; McAuley, 1985; Jerome and Williams, 2000). Finally, attention allows changing the attentional focus to the different tasks that must be carried out during the competitions (García et al., 2011). Because of this, attention is often associated with high athletic performance. Regarding the action of team cohesion on stress, self-confidence, and attention, it is known that individualims does not offer the possibility of cooperating to face tasks where low efficacy and stress are perceived. Collectivist athletes cooperate and merge with the group during competitions (Hadjiyankova and Iancheva, 2021). Therefore, they will be able to overcome situations perceived as stressful to a better extent than individualistic athletes. On the other hand, subjects with low social cohesion tend to perceive unpleasantness when being with other people (Richardson, 2013) and are related to low self-confidence. Finally, some researchers consider that sharing the group's team spirit improves the psychological state of athletes and decreases their perceptions of stress (Prapavessis and Carron, 1996). The aim of this study is to assess the reported mental abilities and stress management (stress, self-confidence and attention) across different team cohesion profiles. The findings of which may influence training programs that focus on the team cohesion of athletes. In relation to previous studies that found differences in team cohesion and mental abilities (Cardeñoso et al., 2007; Csikszentmihalyi et al., 1993; Villa, 2005) and skills that influence stress management (Prapavessis and Carron, 1996), the established hypotheses were: (a) Participants who perceive high team cohesion will report higher scores in mental abilities and stress management; (b) Participants who perceived low team cohesion will report lower scores in the mental abilities and stress management. Methods Design The study followed a cross-sectional non-probabilistic design in which researchers tried to collect participants that trained in the “Development for talent in sports program of the Provincial Council of Guipúzcoa” in the season 2017/2018. The inclusion criteria of the sample were: be a promising athlete and meet the criteria specified in the call. These inclusion criteria were added to ensure that athletes that participated in the study were competitors and they were training in groups. Participants The sample was made up of 146 promising and talented athletes from the Development for talent in sports program of the Provincial Council of Guipúzcoa (M age = 20.08; SD = 4.68). Particularly, 71 were women (M age = 21.09; SD = 4.70), 48.6% and 76 were men (M age = 21.02; SD = 4.73), 51.4%, aged between 12 and 20 years old. From the total sample, 124 were designated promising athletes, and 22 talented athletes. The modalities included in the sample were: athletics (16.9%), canoeing (10.1%), cycling (10.1%), rowing (8.8%), hockey (5.4%), handball (5.4%), climbing (4.7%), surfing (2.7%), badminton (2.7%), table tennis (2%), others (33.9%). Among women, the sport most practiced was athletics (11.26%), canoeing (9.85%) and rowing (9.85%). Among men, the most practiced sports were canoeing (10.52%), rowing (6.57%) and athletics (6.57%). Instruments To examine the psychological characteristics related to performance in sport, the "Psychological Characteristics Related to Sports Performance" (CPRD) questionnaire in the Spanish version was used (Gimeno et al., 2001). This questionnaire assesses the psychological aptitudes for sports performance and this is the most widely used instrument in Spain for assessing the psychological abilities related to performance. The questionnaire is made up of 55 items in a Likert-type response format with five points. The CPRD is made up of five scales: a) Stress management (CE), which covers the characteristics of the athlete's response concerning the demands of training and competition (e.g., concentration) and potentially stressful situations that can cause stress and where the presence of control is necessary; b) Influence of Performance Evaluation (IER), which covers the characteristics of the athlete's response to situations in which he or others evaluate his/her performance (e.g., losing concentration); c) Motivation (M), which refers to the motivation to improve day by day, the establishment of goals and the importance of sport about other activities; d) Mental Skill (HM), which includes psychological skills that favor sports performance (e.g., goal setting); and e) Team Cohesion (CEQ), which considers the integration of the athlete in his/her team or sports group. The CPRD has reported good psychometric properties of reliability and validity in several studies, as shown by the compilation of Gimeno and Pérez-Llantada (2010) and other specific reviews (Gimeno and Pérez-Llantada, 2010; López-López et al., 2013). Regarding the reliability of the subscales used in this study, the following Cronbach alphas and the Joreskog rhô coefficient were reported: team cohesion (α = 0.72; ω = 0.76), mental abilities (α = 0.70; ω = 0.89), and stress management (α = 0.91; ω = 0.90). Moreover, the CPRD factors may be divided into little subfactors to provide further knowledge of each variable (Gimeno and Pérez-Llantada, 2010; López-López et al., 2013). In this study, team cohesion was subidivided into: Individualism vs collectivism (2 items; r = 0.27), social cohesion (2 items; r = 0.36) and team spirit (2 items; r = 0.73). Mental abilities were subdivided into: practice in imagination (4 items; r = 0.26), goal setting (3 items; r = 0.24), positive self-talk (1 item), objective analysis of own performance (1 item), skill deficit (1 item; r = 0.29), tension control (1 item) and others (4 items; r = 0.30). In addition, stress (5 items; α = 0.71), attention (5 items; α = 0.77) and self-confidence (10 items; α = 0.90) were assessed. As Cronbach alpha increases with the number of items of the scale (Clark and Watson, 1995), the average inter-item correlation was taken in the subscales in which there were factors with few items. Particularly, some scholars have shown the reliability of the average mean inter-item correlation as an internal consistency marker (Clark and Watson, 1995). Finally, all Cronbach alphas and the Joreskog rhô coefficient were suitable and the as they ranged higher than 0.70 as well as the inter-item correlation as they ranged higher than 0.15. Procedure The study complied with the ethical guidelines established by the American Association of Psychology in its seventh edition (APA 7) (American Psychological Association, 2020), the anonymity was preserved, and the research followed the principles of the Declaration of Helsinki 2013. The research was carried out and approved by the Development for talent in the sports program of the Provincial Council of Guipúzcoa. As a participation requirement, the government should consider athletes as promising or sports talent. This criterion was inquired to ensure the participants had a certain degree of performance to meet the study purposes. First, psychology practitioners contacted the parents of the athletes. Then, the parents of the interested participants completed informed consent and voluntarily agreed to participate in this program. In the consent, it was stated the type of questionnaires, conditions, information and purposes of the study. This resulted in a total of 124 promising athletes and 22 sports talents. Once the appointment was concluded, the CPRD questionnaire was carried out online through Google forms platform by the practitioners of the Development for talent in the sports program. The approximate time to carry out the questionnaire was 15 minutes, all the questions were mandatory and for this purpose, some keys and a web page were enabled for the athletes to complete the survey. There was no IP control in fulfilling the questionnaire and there were no funds for completing the survey. Data analysis The analyses were performed using SPSS version 20 software. First, the data were filtered for multivariate outliers, data screening and multicollinearity of scales. Second, to increase confidence in the stability of the cluster solution, a two-step approach was performed that included both hierarchical and non-hierarchical cluster analyzes using standardized CPRD scores (Gimeno et al., 2001). In particular, hierarchical group analysis was performed to identify the number of groups (team cohesion) (Ward's method of linking with the squared Euclidean distance). Then, a group analysis of k means was performed, using the most appropriate group solution identified in step one. Third, group analysis was performed on this variable to examine the differences between the groups in team cohesion. A multivariate analysis of variance (MANOVA) was carried out with the outcome variables of the team cohesion profiles entered as dependent variables. The partial eta squared (η 2) was evaluated to provide an effect size index. Finally, a MANOVA was performed with quantitative demographic variables to explore the possible confusion of demographic groups. In addition, a series of chi-square tests were carried out with demographic variables such as gender and age. Results Team cohesion perceived by athletes A MANOVA analysis was performed to detect significant multivariate effects between the groups in the dimensions of team cohesion perceived by the athletes (Wilk's Lambda = 0.18, F (6) = 63.76, p< 0.001, η 2 = 0.57). Subsequently, the ANOVAs indicated significant differences (p< 0.001) in all dimensions of perceived team cohesion, which provides evidence of the sustainability of the cluster solution (Table 1). Table 1. Perceived team cohesion scores among athletes. \| \| Participants with low team cohesion (n = 5) \| Participants with average team cohesion (n = 26) \| Participants with high team cohesion (n = 115) \| F \| P \| Eta 2 \| :---: :---: :---: \| M (SD) \| M (SD) \| M (SD) \| \| Individualism vs collectivism \| 3.20 (2.16) \| 1.08 (1.29) \| 5.33 (1.70) \| 71.55 \| 0.00 \| 0.70 \| \| Social cohesion \| 1.60 (2.19) \| 6.58 (1.72) \| 7.59 (.84) \| 75.23 \| 0.00 \| 0.71 \| \| Team spirit \| 3.00 (3.31) \| 3.69 (2.65) \| 7.65 (.70) \| 105.52 \| 0.00 \| 0.72 \| Open in a new tab p < .001 The different scores in individualism vs collectivism, social cohesion and team spirit are detailed in Figure 1. The descriptive labels for these groups are: (a) low team cohesion; which includes athletes with average scores in individualism vs collectivism, low scores in social cohesion, and average scores in team spirit; (b) average team cohesion; which includes athletes with low scores in individualism vs collectivism, high scores in social cohesion, and average scores in team spirit; (c) high team cohesion; which includes athletes with high scores in individualism vs collectivism, social cohesion, and team spirit. Figure 1. Open in a new tab Clusters team cohesion scores. Differences between cluster groups in perceived team cohesion and mental abilities The differences between the cluster groups in team cohesion and mental abilities (Wilk's Lambda = 0.87, F (14) = 1.30, p< 0.001, η 2 = .20) revealed that there are significant differences (p< 0.05) between the groups in the results of this variables. In Table 2, the ANOVA results show that the participants of profile (b) reported higher scores in positive self-talk than those of the profiles (a) and (c). Table 2. Scores of mental abilities developed by athletes. \| \| Participants with low team cohesion (n = 5) \| Participants with average team cohesion (n = 26) \| Participants with high scores in team cohesion (n = 115) \| F \| P \| Eta 2 \| :---: :---: :---: \| Practice in imagination \| 10.00 (0.70) \| 11.54 (2.19) \| 10.81 (2.16) \| 1.69 \| 0.18 \| 0.15 \| \| Goal setting \| 9.20 (2.95) \| 8.38 (1.92) \| 8.67 (1.90) \| .44 \| 0.64 \| 0.07 \| \| Positive self-talk \| 2.20 (1.64) \| 3.42 (0.85) \| 3.27 (0.94) \| 3.48 \| 0.03 \| 0.21 \| \| Objective analysis of own performance \| 3.00 (0.70) \| 2.96 (0.99) \| 3.03 (1.00) \| .04 \| 0.95 \| 0.02 \| \| Skill deficit \| 7.00 (1.73) \| 8.58 (2.30) \| 8.24 (2.46) \| .90 \| 0.40 \| 0.11 \| \| Tension control \| 1.80 (0.83) \| 2.77 (0.86) \| 2.74 (0.99) \| 2.34 \| 0.10 \| 0.17 \| \| Others \| 10.40 (2.30) \| 11.96 (2.01) \| 11.96 (2.68) \| .88 \| 0.41 \| 0.11 \| Open in a new tab Differences between cluster groups in team cohesion and stress management The differences between the cluster groups in team cohesion and stress management (Wilk's Lambda = 0.78, F (6) = 1.82, p< 0.001, η 2 = 0.93) revealed that there are marginal differences (p< .07) between the groups in the results of this variables. In Table 3, the ANOVA results show that the participants of profile (b) reported higher scores in stress control than those of the profiles (a) and (c). Table 3. Stress scores experienced by athletes. \| \| Participants with low team cohesion (n = 5) \| Participants with average team cohesion (n = 26) \| Participants with high team cohesion (n = 115) \| F \| P \| Eta 2 \| :---: :---: :---: \| M (SD) \| M (SD) \| M (SD) \| \| Stress \| 11. 00 (.70) \| 13.92 (3.50) \| 13.43 (3.60) \| 1.43 \| 0.24 \| 0.39 \| \| Self-confidence \| 20.60 (8.20) \| 29.62 (6.94) \| 27.28 (8.35) \| 2.70 \| 0.07+ \| 0.19 \| \| Attention \| 12.80 (4.43) \| 14.69 (3.23) \| 25.50 (3.55) \| 1.82 \| 0.16 \| 0.15 \| Open in a new tab p < 0.07 (marginally significant) Covariation between cluster groups and sociodemographic variables The results of the chi-square tests did not show significant differences (p> 0.05) between gender (χ 2 (2) = 3.58). In particular, the highest number of women was included in profile (c) high team cohesion. On the other hand, men mostly belonged to profile (c) high team cohesion. Furthermore, no significant differences were found in the age of the participants (p> 0.05; χ 2 (16) = 21.12). In particular, the participants with the best scores in team cohesion and team spirit were associated with the profile (c) high team cohesion and with a mean age of 26 years old. However, the worst scores in team cohesion and team spirit were associated with the participants of the profile (a) low team cohesion, with a mean age of 23 years old. Discussion This research aimed to identify team cohesion profiles and examine whether participants differed significantly in their mental abilities and stress management. The results of the study have managed to increase the knowledge available in the scientific literature on the subject of team cohesion and their action on mental abilities and stress management but through the methodology of profiles. Within the sample of this project, three team cohesion profiles were identified: (a) characterized by low scores in team cohesion; which includes athletes with average scores in individualism vs collectivism, low scores in social cohesion, and average scores in team spirit; (b) average team cohesion; which includes athletes with low scores in individualism vs collectivism, high scores in social cohesion, and average scores in team spirit; (c) high team cohesion; which includes athletes with high scores in individualism vs collectivism, social cohesion, and team spirit. Firstly, participants from profile (a) [low scores in team cohesion] are defined by athletes who have an intermediate interest in cooperating with other group members (Hofstede, 1980), and their goals partially coincide with those of their peers. On the other hand, these athletes do not perceive liking towards their teammates (Richardson, 2013), and partially share the group ideology and their responsibilities within the team (Filho et al., 2014). Regarding participants from profile (b) [average team cohesion], this profile is defined by athletes who prefer to work in a group (Tan et al., 1998) and share goals with team members (Triandis and Gelfand, 1998). These participants perceive a conciliatory environment with their peers and enjoy sharing time with them (Richardson, 2013). However, these athletes do not fully share the ideology of their competition group (Richardson, 2013). Finally, the athletes of the profile (c) [high team cohesion] are defined by participants who prefer to work individually, understand successes as a result of their individual performance (Hofstede, 1980) and do not share goals with their peers (Stone-Romero and Stone, 2002; Yamaguchi 1994). However, these participants perceive enjoyment in sharing experiences with their peers (Triandis and Gelfand, 1998) and they fully agree with the group's philosophy (Richardson, 2013). Secondly, results revealed that athletes from profile (b) reported higher positive self-talk than profiles (a) and (c). It is widely recognized that people need to perceive affiliation need satisfaction (McClelland, 1987), a requirement that seems to be satisfied in the profile (b) participants with average team cohesion (given his/her low individualism vs collectivism and its high social cohesion). People who do not satisfy this affiliation often present depression and sadness (Baumeister and Leary, 1995). Moreover, these people often use self-talk due to the lack of social relationships (Jonanson et al., 2008), but the self-talk is usually negative (Brinthaupt and Dove, 2012). This would explain why, in the profile (b), participants with average team cohesion revealed higher levels of self-talk. However, due to a scarcity of previous research that has unravelled this relationship, it seems that maybe there is another variable that may influence this results (Santos-Rosa et al., 2022). Perhaps, there is another contextual, personal or situational variable that may modify the self-talk of this group and which is not been measured. As such, it would be interesting to further explain this connection in future research, and to examine if profiles behave in the same way of the current study. Nevertheless, the results indicate that the profile with average scores in team cohesion maybe more adapted in terms of creating positive self-talk in athletes which may advert to comprehend team cohesion from a multivariate approach rather than from a bivariate one. This means that the combination of profiles is precise in detecting those dysfunctional variables. On the other hand, in this research found marginally significant differences in self-confidence across profiles. Particularly, participants in profile (b) reported the highest scores in this variable (these athletes strongly believe in their chances of sporting success) (Vealey, 1986; Vealey and Chase, 2008). The aforementioned profile (b) is defined by participants with a quite high social cohesion. Previously, it was found that athletes with low social cohesion tend to perceive unpleasantness when being with other people (Richardson, 2013). Furthermore, in the research of Chicau et al. (2012) group cohesion was related to good self-confidence scores. However, this would lead to think that those in profile (c) should have more self-confidence than those in profile (b), because profile (c) reported higher levels of social cohesion. Despite this, it cannot be ignored that those in profile (c) recorded the highest scores in individualism (athletes are highly motivated by competition, individual rewards, and recognition) (Hadjiyankova and Iancheva, 2021). Therefore, athletes do not believe in the support of the group to achieve certain objectives, which could lead them to sporting failures and loss of confidence in themselves. A limitation of the cluster analysis methodology is that it is based on data collected from self-report measures. In addition, in this research only the point of view of the athletes has been considered. On the other hand, the few sample obtained in this study was due to the difficulty to find athletes in training settings because of their strict routines of training and competition. Therefore, it is suggested that future research attempts to replicate the current study with other objective points of view that could incorporate information from parents of athletes or coaches. In addition, it would be advisable to include athletes from other cultures to check if the results are generalizable to them. Moreover, it was a limitation to do not measure the influence of peer or other contextual variables in mental skills which maybe influencing in the development of athletes’ mental skills. Despite the detailed limitations, this work has proposed an approach focused on the perception of athletes that can be useful to examine the dimensions that influence the team cohesion. In addition, this work allows to know how the combination of team cohesion variables could be related to mental abilities and stress management, which is essential to maximize the performance of athletes. As future lines of research, the study could include the variables of satisfaction of basic psychological needs and mental toughness perceived by Spanish athletes. A priori, the satisfaction of basic psychological needs should have a positive impact on team cohesion, as found in Brazilian culture (Andrade et al., 2019). Also, strong team cohesion should have a positive effect on the mental toughness of athletes, as found by Gu and Xue (2022) in Chinese culture. As practical implications, this work has made it possible to verify the coexistence of different variables of the team cohesion construct in Spanish athletes and see how they are related with reported mental abilities and self-confidence during competitions. Therefore, these athletes will be able to find an explanation for the psychological perceptions they experience during sports-competitive practice and be aware of the importance of team cohesion. Taking into account that team cohesion is modifiable, programs that train sports competences to achieve high performance in competitions must pay attention to individualism/collectivism, social cohesion, and team spirit to enhance the mental abilities and confidence in athletes, that are associated with increased chances of success. This will be achieved as long as athletes are helped to replace their preferences for individual work with the collective, bonds between teammates are strengthened, and athletes are fought to believe firmly in a spirit of group. Conclusion In conclusion, the optimal perception of team cohesion could be related with the development of mental abilities such as positive self-talk, and also with self-confidence. The profile of athletes where perceptions of team cohesion based on low scores in individualism, high social cohesion, and average team spirit are combined seems to be the most functional for the development of psychological abilities that favor sports performance. Athletes who perceive low team cohesion should attend programs where competences are trained that allow their participants to learn to enjoy working in groups, understand the importance of peer bons and fight for a shared vision. Thanks to this, more opportunities will open up for athletes to grow professionally, achieve success and maintain their performance in the long term. Acknowledgements There is no conflict of interest. The present study complies with the current laws of the country in which it was performed. The datasets generated and analyzed during the current study are not publicly available but are available from the corresponding author, who was an organizer of the study. The authors acknowledge Universidad Internacional de La Rioja (UNIR) for providing funds in this research. Biographies Amaia RAMÍREZ MUÑOZ Employment Professor at Universidad Internacional de La Rioja (UNIR), La Rioja, Spain. Degree PhD Research interests Coach leadership and performance in sport; Psychological abilities in high performance; Development of sports talent. E-mail:amaia.ramirez@unir.net Marta VEGA-DÍAZ Employment Professor in University of Alfonso X el Sabio (UAX), Madrid, Spain Degree Master in Attention to Diversity and Inclusive Education. Specialist Teacher in Early Childhood Education with Mention in Physical Education, Graduated in Biology. Research interests Parental educational styles and physical and psychological health E-mail:mdiazveg@uax.es Higinio GONZÁLEZ-GARCÍA Employment Professor at Universidad Internacional de La Rioja (UNIR), La Rioja, Spain. 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16355	https://brainly.com/question/31289914	[FREE] What is the sum of the two solutions to the equation 54 - 15x - x^2 = 0? - brainly.com 4 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +48k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +44,6k Ace exams faster, with practice that adapts to you Practice Worksheets +8,5k Guided help for every grade, topic or textbook Complete See more / Mathematics Expert-Verified Expert-Verified What is the sum of the two solutions to the equation 54−15 x−x 2=0? 2 See answers Explain with Learning Companion NEW Asked by chengyue368238 • 03/25/2023 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 68554733 people 68M 5.0 0 Upload your school material for a more relevant answer 15 Given quadratic equation: 54 - 15x - x² = 0 Write it in standard form of ax² + bx + c = 0: x² + 15x - 54 = 0 According to Vieta's theorem, sum of the two roots is - b/a, that is: x₁ + x₂ = - 15/1 = - 15 Answered by mhanifa •13.6K answers•68.6M people helped Thanks 0 5.0 (1 vote) Expert-Verified⬈(opens in a new tab) This answer helped 68554733 people 68M 5.0 0 Upload your school material for a more relevant answer The sum of the two solutions to the equation 54−15 x−x 2=0 is −15. This is determined using Vieta's theorem, which establishes that the sum of the roots is −b/a. In this case, with b=15 and a=1, the sum is −15. Explanation To find the sum of the two solutions to the equation 54−15 x−x 2=0, we first need to rewrite this equation in standard quadratic form, which is a x 2+b x+c=0. The given equation can be rearranged as follows: −x 2−15 x+54=0 Multiplying the entire equation by -1 to make the coefficient of x 2 positive gives us: x 2+15 x−54=0 Next, we can apply Vieta's theorem, which states that for a quadratic equation in the form a x 2+b x+c=0, the sum of the roots, denoted as x 1+x 2, is given by −a b. In our equation, a=1 and b=15. Therefore, the sum of the solutions is: x 1+x 2=−1 15=−15 Thus, the sum of the two solutions to the equation 54−15 x−x 2=0 is −15. Examples & Evidence For example, if we had a different quadratic equation, like x 2+6 x+9=0, we could find the sum of the solutions using Vieta's theorem as well. Here, the sum would be −1 6=−6. Vieta's theorem is a well-established principle in algebra, widely taught in mathematics courses. The approach to converting the equation to standard form and applying the theorem follows standard methods used in solving quadratic equations. Thanks 0 5.0 (1 vote) Advertisement Community Answer This answer helped 8944701 people 8M 0.0 0 Explanation x^2 -15x + 54 = 0 x^2 + 15x - 54 = 0 the two solutions will add up to - b/a = - 15 Or you can factor it ( -x^2 -15x+54) = 0 (x+18)(-x+3) = 0 shows x = -18 or 3 sum = - 15 Answered by jsimpson11000 •14.4K answers•8.9M people helped Thanks 0 0.0 (0 votes) Advertisement ### Free Mathematics solutions and answers Community Answer 3 Wendy has 180 feet of fencing. She needs to enclose a rectangular space with an area that is ten times its perimeter. If she uses up all her fencing material, how many feet is the largest side of the enclosure? Community Answer 5.0 4 If $(ax+b)(2x+3)=20x^2+44x+21$, where $a$ and $b$ are two distinct integers, what is the value of the sum $a+b$? Community Answer 5.0 If x+y=7/13 and x-y=1/91, what is the value of x²-y²? Express your answer as a common fraction. Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? New questions in Mathematics Determine which function produces the same graph as f(x)=(8 3 2x)(1 6 2 1x).A. f(x)=4 x B. f(x)=4 2 x C. f(x)=8 3 x D. f(x)=1 6 2 x For each conjecture, test the conjecture with several more examples or find a counterexample to disprove it. a. For every integer n, the value of n² is positive. b. A number is divisible by 4 if the last two digits are divisible by 4. A table represents the possibility of an association between hair color and eye color. \| \| Eye Color \| \| \| :--- --- \| \| Hair Color \| Blue \| Green \| Brown \| \| Blonde \| 25 \| 27 \| 31 \| \| Brown \| 26 \| 18 \| 22 \| In order to determine if blue eye color and blonde hair color differ significantly, a chi-square test for homogeneity should be performed. What is the expected frequency of blonde hair and blue eyes (answer choices are rounded to the nearest hundredth)? The function f(x)=x+15 is one-to-one. a. Find an equation for f−1(x), the inverse function. b. Verify that your equation is correct by showing that f(f−1(x))=x and $f^{-1}(f(x))=x. A. f−1(x)=□ , for x=□ B. f−1(x)=□ , for x≤□ C. f−1(x)=□ , for all x D. f−1(x)=□ for x≥□ The hypotenuse of a 4 5∘−4 5∘−9 0∘ triangle measures 4 cm. What is the length of one leg of the triangle? 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16356	https://symbolica.io/docs/pattern_matching.html	Pattern matching Introduction Pattern matching is an integral part of a computation toolkit. It allows to isolate and modify parts of an expression with replacement rules. A properly broad pattern matcher is effectively its own Turing complete programming language! For example, Conway’s game of life can be completely emulated by pattern matching, where each cell together with its surrounding cells determined the future value of the cell. Other common uses of pattern matching is in regular expression (regex), widely used for matching patterns in text. The following pattern ^[a-zA-Z0-9]+@gmail\.com$ matches e-mail addresses with at least one alphanumerical character and ending in @gmail.com. To faithfully parse an e-mail address, a horrendous pattern match is required! Pattern matching is also ubiquitous in mathematics. For example, the recursive definition of the factorial function [ \begin{align} f(0) &= 1\ f(n) &= n f(n - 1);\quad n > 0 \end{align} ] is a pattern match. First, the pattern (f(0)) is tested. If it is matched, it is replaced by 1. Then the pattern (f(n)) is matched, where (n) needs to be positive and it is replaced by (n f(n-1)), where the (n) takes the value of the (n) matched in the pattern. Symbolica has rich support for pattern matching. Let us see some examples where we match and replace all literal parts of the expression. In this part we will mostly use the Python API, with shorthand notation. We match (x) in the expression (x+f(x) + 5) and replace it by 6, using use replace. Python Rust from symbolica import from importx, f = S('x', 'f') = 'x' 'f'e = x + f(x) + 5 = + + 5e = e.replace(x, 6) = 6print(e) print use symbolica::{usesymbolica::{ atom::{Atom, AtomCore},atom::{,}, function, symbol,,,};};fn main() {fn{ let (x, f) = symbol!("x", "f"); let, =symbol! "x", "f"; let mut e = function!(f, x) + x + 5; let mut =function!, + + 5; e = e.replace(x).with(Atom::new_num(6)); =..Atom:: 6; println!("{}", e);println!"{}",;}} We get 11 + f(6) To replace f(x) by 6, we change the pattern: e = e.replace(f(x), 6) = 6 which yields 11 + x. To replace multiple patterns at the same time, use replace_multiple. For example: x, y, f = S('x', 'y', 'f') = 'x' 'y' 'f'k = f(x, y).replace_multiple([Replacement(x, y), Replacement(y, x)]) =print(k) print replaces f(x,y) by f(y,x). Wildcards Let us now try to replicate the factorial function pattern match mentioned above. For that we need to be able to match to any numerical value in the function f and we do not know that value in advance. Writing out an explicit rule for every desired number, e.g., f(40) = 40f(39) is possible, but tedious. Instead, we use a wildcard that can match any subexpression. The specific rules will be explained later. A wildcard is defined as a variable ending in a _ (note that the name of the Python variable does not matter). We define a wildcard n_ and write the rule: from symbolica import from importn_, f = S('n_', 'f') = 'n_' 'f'e = f(4) = 4e = e.replace(f(0), 1) = 0 1e = e.replace(f(n_), n_f(n_-1)) = - 1print(e) print This is almost perfect; the pattern still matches to negative numbers, and non-numbers. We can restrict the pattern by adding a third argument to replace: e = e.replace(f(n_), n_f(n_-1), n_ > 0) = - 1> 0 Rules We now set out all the rules for when patterns match. Syntactic matching Patterns match syntactically / structurally, which means that the patten must appear explicitly in the tree representation of the expression. For example, the pattern (x^2) is not found in the expression ((x+1)^2), even though it appears in the mathematically equivalent (but structurally different) expression (x^2+2x+1). The pattern does not necessarily have to appear verbatim in the tree structure though. Symbolica uses mathematical properties such as the symmetry of the multiplication and addition operator, and of symmetric functions, to match atoms in unordered fashion. For example (x + z) matches (x+y+z), even though (y) is “in between” in the normal form. Antisymmetric functions Patterns are automatically written in canonical form, also if they contain wildcards. For an antisymmetric function f this means that a user-written pattern f(2,1) will be turned into -f(1,2). This pattern will not match f(1,2) as the minus sign is not present. Limiting options Since matches on products, sums and symmetric functions are unordered, in order to establish that there is no match, in the worst case all permutations or subsets have to be tested. This could be expensive. Providing constraints on the pattern, especially length constraints, will help reduce the cost. Wildcards Wildcards are variables ending in one or more underscores _. The number of underscores signify various restrictions on the wildcard length. Wildcards can match any subexpression and will always include the operator it matches. For example, matching x___ (no length restrictions) on abc yields the following options: Note that the empty set is a valid option. Because of the rule of syntactic matching, the pattern y + x___ is not present in the expression y even though x___ can be the empty set. The cases where one or zero sub-atoms are matched is subject to downcasting. For example, is transformed to which means that the atom type of x___ is no longer Mul but the type of a. The atom type of the empty match is Empty. Downcasting ensures that repeated patterns such as x_f(x_) match to xf(x) even though the first x_ matches Mul Var x and the second one Arg Var x. Function arguments behave analogously to the case of Mul described above, except that the order is preserved if the function is non-symmetric. The pattern matcher interprets function arguments as an operator of type Arg. For example, when matching f(x___) to f(a,b,c) the pattern matcher views the arguments as When a wildcard that matches multiple arguments is used outside of a function context on the right-hand side of a replacement, the matched arguments will be wrapped in the internal function arg. For example: e = f(1,2).replace(f(x___), x___) = 1 2 yields arg(1,2). An arg function that is placed into a function context again, is flattened during normalization, e.g., f(arg(1,2)) = f(1,2). Repeated wildcards A repeated wildcard means that the matched value must evaluate to exactly the same atoms. For example: e = xyf(xy) =e = e.replace(x___f(x___), 1) = 1 matches because x___ matches to xy outside of the function and inside the function. See Examples for more complicated examples. Precedence rules The expression that is tested first is the root node with all its children. If that does not match, the subset of the root node with one sub-atoms removed is tested. If no match is found, two sub-atoms are removed, etc. Once all subsets are tested, the pattern matcher repeats the process for each of the sub-atoms as the new root. For example, the pattern x_ will match any subexpression in any expression, so it will match the root first: e = xyz =e = e.replace(x_, 1) = 1 will yield 1, since x_ matches xyz. In the following case the pattern matches at the root level, so a match is returned, even though the child would have matched too: e = f(f(x)) =e = e.replace(f(x_), f(1)) = 1 will yield f(1), since x_ matches f(x). In the following case the root did not match, so the sub-atom is matched: e = g(f(x)) =e = e.replace(f(x_), f(1)) = 1 will yield g(f(1)), since x_ matches f(x). replace will always replace the first match found for every non-overlapping match. To iterate matches, see Match iteration. Ordering A wildcard will match to a given set of atoms only once and it will have the order of the normal form. This means that for a symmetric function f(1,2,3), f(x___) will generate one match, x___ = 1,2,3, and it will never permute through all options (e.g. x___ = 2,3,1). Consequently, f = S('f') = 'f'f_sym = S('fsym', is_symmetric=True) = 'fsym' = Truee = f(3,2,1)fsym(1,2,3) = 3 2 1 1 2 3e = e.replace(f(x___)fsym(x___), f(1)) = 1 will not match. Conditions Wildcards can be restricted based on arbitrary conditions. Below are some built-in options. Comparison and condition functions Wildcards can be restricted by any Condition. A Condition is an object that yields True, False, or Inconclusive when evaluated. There are several functions on expressions that yield conditions, such as contains or matches: g, a, x_ = S('g', 'a', 'x_') = 'g' 'a' 'x_'g(a).replace(x_, 1, x_.contains(E('a'))) # yields 1 as a is contained in g(a) 1 'a'# yields 1 as a is contained in g(a)g(2).replace(x_, 1, x_.contains(E('a'))) # no match 2 1 'a' # no match A comparison of a wildcard with any other expression also yields a condition. For example x_ > 1 in: g, x, x_ = S('g', 'x', 'x_') = 'g' 'x' 'x_'e = g(10).replace(g(x_), g(x_-1) + g(x_-2), x_ > 1, repeat=True) = 10 - 1 + - 2> 1 = True which keeps on applying the replacement if x > 1 in g(x). Note that a condition is not directly evaluated when it is created, thus allowing the substitution of x_ with its matched value before evaluating. Wildcard length It is possible to filter on the length of the wildcard before downcasting. The length is defined as the number of sub-atoms for Add, Mul, and Arg and is 1 for any other atom. Note A function has a length of 1 (it counts as one atom) regardless of the number of arguments. This way x_ with a wildcard length of 1 matches f(...) regardless of the number arguments. The number of underscores at the end of the wildcard name determines pre-set length restrictions. We have: x_: must match a single atom x__: must match one or more atoms x___: must match zero or more atoms The length of x__ and x___ can be further restricted. For example, let us restrict the wildcard x__ to a length between 2 and 3 e1 = f(1).replace(f(x__), 1, x__.req_len(2, 3)) = 1 1 2 3e2 = f(1,2).replace(f(x__), 1, x__.req_len(2, 3)) = 1 2 1 2 3e3 = f(1,2,3).replace(f(x__), 1, x__.req_len(2, 3)) = 1 2 3 1 2 3 then only e2 and e3 will yield 1. Even though the wildcard length of x_ is restricted to 1 and xy has a length of 2, e1 = f(xy).replace(f(x_, 1)) = 1 we still get the match x_ = xy. This is because the length constraint is applied before downcasting. A wildcard length restriction check is applied at any point that a wildcard is used and therefore xyf(xy).replace(x__f(x__), 1, x__.req_len(2, 2)) 1 2 2 does not match, since x__ has wildcard length 1 inside the function argument. To restrict the length of the atom matched to x__ after downcasting, use a filter function. Type restriction A wildcard can be restricted based on the atom type. For example, x_ can only match a variable: e1 = f(x).replace(f(x_), 1, x_.req_type(AtomType.Var)) = 1 Filter function Custom filters can be added to restrict any wildcard. This is a function receives the wildcard name and its match expression as an argument and should return a boolean. If true, the match is accepted. from symbolica import from importx_, f = S('x_', 'f') = 'x_' 'f'e = f(1)f(2)f(3) = 1 2 3e = e.replace(f(x_), 1, x_.req(lambda m: m == 2 or m == 3)) = 1 lambda == 2 or == 3 Tip Since a Python function is called, this operation has significant overhead. For high-performance code, consider implementing the filter in Rust, or use one of the built-in filters, such as req_gt and reg_cmp_gt that does numerical comparisons in Rust. Comparison function A more complicated class of custom filters is based on using the values of two wildcards. from symbolica import from importx_, y_, f = S('x_', 'y_', 'f') = 'x_' 'y_' 'f'e = f(1)f(2)f(3) = 1 2 3e = e.replace(f(x_)f(y_), 1, x_.req_cmp(y_, lambda m1, m2: m1 + m2 == 4)) = 1 lambda + == 4 Match stack filter It is also possible to filter patterns based on the currently matched wildcards. For example, here is a filter for an ascending order of 3 variables: f, x_, y_, z_ = S('f', 'x_', 'y_', 'z_') = 'f' 'x_' 'y_' 'z_'def filter(m: dict[Expression, Expression]) -> int: def filter dict -> int if x_ in m and y_ in m: if in and in if m[x_] > m[y_]: if> return -1 # no match return - 1 # no match if z_ in m: if in if m[y_] > m[z_]: if> return -1 return - 1 return 1 # match return 1 # match return 0 # inconclusive return 0 # inconclusivee = f(1, 2, 3).replace(f(x_, y_, z_), 1, = 1 2 3 1 PatternRestriction.req_matches(filter)) filter The filter function is called after a change occurs to the currently matched wildcards, so the user has to treat cases where wildcards they want to act on are not yet matched. The function must return an int instead of a bool, as it needs to distinguish the cases of certainly no match (<0), potentially a match (0), or certainly a match (>0). Failing to properly account for the inconclusive case may lead to unexpected pattern matches. Logical conditions Multiply wildcard conditions can be chained using the and (&), or (\|) and not (~ in Python and ! in Rust) operators. For example: x, x___, x___, y___ = S('x', 'x_', 'x___', 'y___') = 'x' 'x_' 'x___' 'y___'f = S('f') = 'f'e = f(3, x, 2).replace(f(x___, x___, y___), = 3 2 f(x___), x___.req_lt(3) \| x___.req_type(AtomType.Var)) 3 \| matches if x___ is smaller than 3 or is a variable. The example yields f(2). Level restrictions Sometimes a pattern should only match in a function or at a certain depth of the expression tree. This can be set with level_range, which defines the inclusive range for which the pattern should match. For example: x, f = S('x', 'f') = 'x' 'f'r = x+x2f(x2, x, f(x)) = + 2 2print(r.replace(x, 1, level_range=(1, 1))) print 1 = 1 1 yields x+x^2f(1,1,f(x)). By default, the level signifies the function depth. It can be set to signify the depth in the expression tree by the optional argument level_is_tree_depth=True. Additional settings Wildcards are greedy by default, i.e. they try to be as large as possible. In the example of the previous section we see that the result is f(2) instead of f(x), because x___ is greedy. By passing the setting non_greedy_wildcards we can change this behaviour: e = f(3, x, 2).replace(f(x___, x_, y___), = 3 2 f(x_), x_.req_lt(3) \| x_.req_type(AtomType.Var), 3 \| non_greedy_wildcards=[x___]) = This code yields f(x). Match iteration It is possible to iterate through all matches without modifying the expression. For example, here f(x_) is matched: from symbolica import from importx_, f = S('x_', 'f') = 'x_' 'f'e = f(1)+f(2)+f(3) = 1 + 2 + 3for m in e.match(f(x_)): for in print(m[x_]) print which yields 1 2 3 The match function returns a dictionary with the wildcard as keys. The match function also yields nested matches: from symbolica import from importx_, f = S('x_', 'f') = 'x_' 'f'e = f(f(1)) = 1for m in e.match(f(x_)): for in print(m[x_]) print which yields f(1) 1 Replacements We have already seen many examples of replacing all occurrences of a pattern in this section. An iterator over single replacements can be generated as well: from symbolica import from importx_, f = S('x_', 'f') = 'x_' 'f'e = f(1)f(2)f(3) = 1 2 3for r in e.replace_iter(f(x_), f(x_ + 1)): for in + 1 print(r) print which yields f(2)f(2)f(3) f(1)f(3)f(3) f(1)f(2)f(4) Transformers Often the goal of a pattern match and replacement is to isolate a part of the expression and to modify it to something else. Rather than simply allowing one atom to be replaced by another, functions can be executed for the replacement instead. These are called transformers (see here for more information). A transformer takes a pattern with wildcards as an argument and executes an operation on it after the wildcards have been substituted. For example: from symbolica import from importx, x_, f = S('x', 'x_', 'f') = 'x' 'x_' 'f'e = f((x+1)2) = + 1 2print(e.replace(f(x_), f(x_.transform().expand()))) print f(x^2+2x+1) Replace function The right-hand-side of a replacement can also be a function that maps wildcards to any expression. This can be used to do transformations that are not possible by just using Expression, for example string manipulations. Here is an example: k, p, id1_, v1_, idx1_, id2_, v2_ = S('k', 'p', 'id1_', 'v1_', 'idx1_', 'id2_', 'v2_') = 'k' 'p' 'id1_' 'v1_' 'idx1_' 'id2_' 'v2_'r = (k(1, 2)p(3, 2)).replace(id1_(v1_, idx1_)id2_(v2_, idx1_), = 1 2 3 2 lambda m: E("{}(mu{}){}(mu{})".format(m[id1_], m[v1_], m[id2_], m[v2_]))) lambda "{}(mu{}){}(mu{})" format This maps k(1, 2)p(3, 2) into k(mu1)p(mu3) Examples Multiple matches from symbolica import from importx_, y_, z_, w_, a_, f, g = S('x__', 'y__', 'z__', 'w__', 'a__', 'f', 'g') = 'x__' 'y__' 'z__' 'w__' 'a__' 'f' 'g'e = f(1,2,3,4)g(5,2,3,6,7) = 1 2 3 4 5 2 3 6 7for m in e.match(f(x_,y_,z_)g(w_,y_,a_)): for in print('{}\t{}\t{}\t{}\t{}'.format(m[x_], m[y_], m[z_], m[w_], m[a_])) print '{} \t{} \t{} \t{} \t{} ' format 1 2 arg(3,4) 5 arg(3,6,7) 1 arg(2,3) 4 5 arg(6,7) arg(1,2) 3 4 arg(5,2) arg(6,7) Coefficient extraction The coefficients of the polynomial x^2(y+z) + x^3(y+z^2) in x are matched: from symbolica import from importx, y, z, n_, x_ = S('x', 'y', 'z', 'n_', 'x_') = 'x' 'y' 'z' 'n_' 'x_'e = x2(y+z) + x3(y+z2) = 2 + + 3 + 2 # match returns a dictionary that maps every wildcard to its value # match returns a dictionary that maps every wildcard to its valuecoeff_list = [(m[n_], m[x_]) for m in e.match(xn_x_)] = for infor (pow, content) in coeff_list: for pow in print(xpow, content) print pow Unbound wildcards on the right-hand side Symbolica supports wildcards appearing on the right-hand side, that do not appear in the pattern. This can be used to construct new patterns that depend on – for example – structural information in the expression. In the example below, we want to find all automorphisms of a small graph, with vertices v and edges e_i. We replace all edges e[i] with a unique wildcard e[i]_ and apply the new pattern to the graph we started out with. v = S('v', is_symmetric=True) = 'v' = Truee = S('e0', 'e1', 'e2', 'e3', 'e4', 'e5', 'e6') = 'e0' 'e1' 'e2' 'e3' 'e4' 'e5' 'e6'e_ = S('e0_', 'e1_', 'e2_', 'e3_', 'e4_', 'e5_', 'e6_') = 'e0_' 'e1_' 'e2_' 'e3_' 'e4_' 'e5_' 'e6_'x_ = S('x_') = 'x_'graph = v(e, e, e)v(e, e, e, e) \ = 0 1 4 1 2 5 6 \ v(e, e, e)v(e, e, e) 0 2 3 3 4 5graph_pat = graph =for edge, edge_pat in zip(e, e_): for in zip graph_pat = graph_pat.replace(= edge, edge_pat, allow_new_wildcards_on_rhs=True) = Trueprint('Pattern: ', graph_pat) print'Pattern: 'for x in graph.match(graph_pat): for in for ee in e_: for in print('{}={}'.format(ee, x[ee]), end=' ') print '{} ={} ' format = ' ' print() print We obtain 6 automorphisms: Pattern: v(e0_,e1_,e4_)v(e0_,e2_,e3_) v(e3_,e4_,e5_)v(e1_,e2_,e5_,e6_) e0_=e0 e1_=e1 e2_=e2 e3_=e3 e4_=e4 e5_=e5 e6_=e6 e0_=e4 e1_=e1 e2_=e5 e3_=e3 e4_=e0 e5_=e2 e6_=e6 e0_=e0 e1_=e2 e2_=e1 e3_=e4 e4_=e3 e5_=e5 e6_=e6 e0_=e3 e1_=e2 e2_=e5 e3_=e4 e4_=e0 e5_=e1 e6_=e6 e0_=e3 e1_=e5 e2_=e2 e3_=e0 e4_=e4 e5_=e1 e6_=e6 e0_=e4 e1_=e5 e2_=e1 e3_=e0 e4_=e3 e5_=e2 e6_=e6
16357	https://en.wikipedia.org/wiki/Complex_analysis	Jump to content Search Contents (Top) 1 History 2 Complex functions 3 Holomorphic functions 4 Conformal map 5 Major results 6 See also 7 References 8 Sources 9 External links Complex analysis العربية Asturianu Azərbaycanca Башҡортса Беларуская Български Català Чӑвашла Čeština Cymraeg Dansk Deutsch Eesti Ελληνικά Español Esperanto Euskara فارسی Français Galego 한국어 Հայերեն हिन्दी Bahasa Indonesia Ирон Íslenska Italiano עברית ქართული Lombard Magyar Македонски Malti მარგალური Nederlands 日本語 Norsk bokmål Polski Português Romnă Русский Scots Shqip සිංහල Simple English Slovenčina کوردی Српски / srpski Suomi Svenska ไทย Türkçe Українська Tiếng Việt 吴语粵語中文 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikimedia Commons Wikibooks Wikiquote Wikidata item Appearance From Wikipedia, the free encyclopedia Branch of mathematics studying functions of a complex variable Not to be confused with Complexity theory. \| \| \| This article includes a list of general references, but it lacks sufficient corresponding inline citations. Please help to improve this article by introducing more precise citations. (March 2021) (Learn how and when to remove this message) \| \| \| \| Mathematical analysis → Complex analysis \| \| Complex analysis \| \| Complex numbers \| \| Real number Imaginary number Complex plane Complex conjugate Euler's formula \| \| Basic theory \| \| Argument principle Residue Essential singularity Isolated singularity Removable singularity Zeros and poles \| \| Complex functions \| \| Complex-valued function Antiderivative Analytic function Entire function Holomorphic function Meromorphic function Cauchy–Riemann equations Formal power series Laurent series \| \| Theorems \| \| Analyticity of holomorphic functions Cauchy's integral theorem Cauchy's integral formula Residue theorem Liouville's theorem Picard theorem Weierstrass factorization theorem \| \| Geometric function theory \| \| Complex manifold Conformal map Quasiconformal mapping Riemann surface Schwarz lemma Winding number Analytic continuation Riemann's mapping theorem \| \| People \| \| Augustin-Louis Cauchy Leonhard Euler Carl Friedrich Gauss Bernhard Riemann Karl Weierstrass \| \| Mathematics portal \| \| v t e \| Complex analysis, traditionally known as the theory of functions of a complex variable, is the branch of mathematical analysis that investigates functions of complex numbers. It is helpful in many branches of mathematics, including algebraic geometry, number theory, analytic combinatorics, and applied mathematics, as well as in physics, including the branches of hydrodynamics, thermodynamics, quantum mechanics, and twistor theory. By extension, use of complex analysis also has applications in engineering fields such as nuclear, aerospace, mechanical and electrical engineering. As a differentiable function of a complex variable is equal to the sum function given by its Taylor series (that is, it is analytic), complex analysis is particularly concerned with analytic functions of a complex variable, that is, holomorphic functions. The concept can be extended to functions of several complex variables. Complex analysis is contrasted with real analysis, which deals with the study of real numbers and functions of a real variable. History [edit] Complex analysis is one of the classical branches in mathematics, with roots in the 18th century and just prior. Important mathematicians associated with complex numbers include Euler, Gauss, Riemann, Cauchy, Weierstrass, and many more in the 20th century. Complex analysis, in particular the theory of conformal mappings, has many physical applications and is also used throughout analytic number theory. In modern times, it has become very popular through a new boost from complex dynamics and the pictures of fractals produced by iterating holomorphic functions. Another important application of complex analysis is in string theory which examines conformal invariants in quantum field theory. Complex functions [edit] A complex function is a function from complex numbers to complex numbers. In other words, it is a function that has a (not necessarily proper) subset of the complex numbers as a domain and the complex numbers as a codomain. Complex functions are generally assumed to have a domain that contains a nonempty open subset of the complex plane. For any complex function, the values from the domain and their images in the range may be separated into real and imaginary parts: where are all real-valued. In other words, a complex function may be decomposed into : and i.e., into two real-valued functions (, ) of two real variables (, ). Similarly, any complex-valued function f on an arbitrary set X (is isomorphic to, and therefore, in that sense, it) can be considered as an ordered pair of two real-valued functions: (Re f, Im f) or, alternatively, as a vector-valued function from X into Some properties of complex-valued functions (such as continuity) are nothing more than the corresponding properties of vector valued functions of two real variables. Other concepts of complex analysis, such as differentiability, are direct generalizations of the similar concepts for real functions, but may have very different properties. In particular, every differentiable complex function is analytic (see next section), and two differentiable functions that are equal in a neighborhood of a point are equal on the intersection of their domain (if the domains are connected). The latter property is the basis of the principle of analytic continuation which allows extending every real analytic function in a unique way for getting a complex analytic function whose domain is the whole complex plane with a finite number of curve arcs removed. Many basic and special complex functions are defined in this way, including the complex exponential function, complex logarithm functions, and trigonometric functions. Holomorphic functions [edit] Main article: Holomorphic function Complex functions that are differentiable at every point of an open subset of the complex plane are said to be holomorphic on . In the context of complex analysis, the derivative of at is defined to be Superficially, this definition is formally analogous to that of the derivative of a real function. However, complex derivatives and differentiable functions behave in significantly different ways compared to their real counterparts. In particular, for this limit to exist, the value of the difference quotient must approach the same complex number, regardless of the manner in which we approach in the complex plane. Consequently, complex differentiability has much stronger implications than real differentiability. For instance, holomorphic functions are infinitely differentiable, whereas the existence of the nth derivative need not imply the existence of the (n + 1)th derivative for real functions. Furthermore, all holomorphic functions satisfy the stronger condition of analyticity, meaning that the function is, at every point in its domain, locally given by a convergent power series. In essence, this means that functions holomorphic on can be approximated arbitrarily well by polynomials in some neighborhood of every point in . This stands in sharp contrast to differentiable real functions; there are infinitely differentiable real functions that are nowhere analytic; see Non-analytic smooth function § A smooth function which is nowhere real analytic. Most elementary functions, including the exponential function, the trigonometric functions, and all polynomial functions, extended appropriately to complex arguments as functions , are holomorphic over the entire complex plane, making them entire functions, while rational functions , where p and q are polynomials, are holomorphic on domains that exclude points where q is zero. Such functions that are holomorphic everywhere except a set of isolated points are known as meromorphic functions. On the other hand, the functions , , and are not holomorphic anywhere on the complex plane, as can be shown by their failure to satisfy the Cauchy–Riemann conditions (see below). An important property of holomorphic functions is the relationship between the partial derivatives of their real and imaginary components, known as the Cauchy–Riemann conditions. If , defined by , where , is holomorphic on a region , then for all , In terms of the real and imaginary parts of the function, u and v, this is equivalent to the pair of equations and , where the subscripts indicate partial differentiation. However, the Cauchy–Riemann conditions do not characterize holomorphic functions, without additional continuity conditions (see Looman–Menchoff theorem). Holomorphic functions exhibit some remarkable features. For instance, Picard's theorem asserts that the range of an entire function can take only three possible forms: , , or for some . In other words, if two distinct complex numbers and are not in the range of an entire function , then is a constant function. Moreover, a holomorphic function on a connected open set is determined by its restriction to any nonempty open subset. Conformal map [edit] This section is an excerpt from Conformal map.[edit] In mathematics, a conformal map is a function that locally preserves angles, but not necessarily lengths. More formally, let and be open subsets of . A function is called conformal (or angle-preserving) at a point if it preserves angles between directed curves through , as well as preserving orientation. Conformal maps preserve both angles and the shapes of infinitesimally small figures, but not necessarily their size or curvature. The conformal property may be described in terms of the Jacobian derivative matrix of a coordinate transformation. The transformation is conformal whenever the Jacobian at each point is a positive scalar times a rotation matrix (orthogonal with determinant one). Some authors define conformality to include orientation-reversing mappings whose Jacobians can be written as any scalar times any orthogonal matrix. For mappings in two dimensions, the (orientation-preserving) conformal mappings are precisely the locally invertible complex analytic functions. In three and higher dimensions, Liouville's theorem sharply limits the conformal mappings to a few types. The notion of conformality generalizes in a natural way to maps between Riemannian or semi-Riemannian manifolds. Major results [edit] One of the central tools in complex analysis is the line integral. The line integral around a closed path of a function that is holomorphic everywhere inside the area bounded by the closed path is always zero, as is stated by the Cauchy integral theorem. The values of such a holomorphic function inside a disk can be computed by a path integral on the disk's boundary (as shown in Cauchy's integral formula). Path integrals in the complex plane are often used to determine complicated real integrals, and here the theory of residues among others is applicable (see methods of contour integration). A "pole" (or isolated singularity) of a function is a point where the function's value becomes unbounded, or "blows up". If a function has such a pole, then one can compute the function's residue there, which can be used to compute path integrals involving the function; this is the content of the powerful residue theorem. The remarkable behavior of holomorphic functions near essential singularities is described by Picard's theorem. Functions that have only poles but no essential singularities are called meromorphic. Laurent series are the complex-valued equivalent to Taylor series, but can be used to study the behavior of functions near singularities through infinite sums of more well understood functions, such as polynomials. A bounded function that is holomorphic in the entire complex plane must be constant; this is Liouville's theorem. It can be used to provide a natural and short proof for the fundamental theorem of algebra which states that the field of complex numbers is algebraically closed. If a function is holomorphic throughout a connected domain then its values are fully determined by its values on any smaller subdomain. The function on the larger domain is said to be analytically continued from its values on the smaller domain. This allows the extension of the definition of functions, such as the Riemann zeta function, which are initially defined in terms of infinite sums that converge only on limited domains to almost the entire complex plane. Sometimes, as in the case of the natural logarithm, it is impossible to analytically continue a holomorphic function to a non-simply connected domain in the complex plane but it is possible to extend it to a holomorphic function on a closely related surface known as a Riemann surface. All this refers to complex analysis in one variable. There is also a very rich theory of complex analysis in more than one complex dimension in which the analytic properties such as power series expansion carry over whereas most of the geometric properties of holomorphic functions in one complex dimension (such as conformality) do not carry over. The Riemann mapping theorem about the conformal relationship of certain domains in the complex plane, which may be the most important result in the one-dimensional theory, fails dramatically in higher dimensions. A major application of certain complex spaces is in quantum mechanics as wave functions. See also [edit] Complex geometry Hypercomplex analysis List of complex analysis topics Monodromy theorem Riemann–Roch theorem Runge's theorem Vector calculus References [edit] ^ "Industrial Applications of Complex Analysis". Newton Gateway to Mathematics. October 30, 2019. Retrieved November 20, 2023. ^ Rudin, Walter (1987). Real and Complex Analysis (PDF). McGraw-Hill Education. p. 197. ISBN 978-0-07-054234-1. ^ Blair, David (2000-08-17). Inversion Theory and Conformal Mapping. The Student Mathematical Library. Vol. 9. Providence, Rhode Island: American Mathematical Society. doi:10.1090/stml/009. ISBN 978-0-8218-2636-2. S2CID 118752074. Sources [edit] Ablowitz, M. J. & A. S. Fokas, Complex Variables: Introduction and Applications (Cambridge, 2003). Ahlfors, L., Complex Analysis (McGraw-Hill, 1953). Cartan, H., Théorie élémentaire des fonctions analytiques d'une ou plusieurs variables complexes. (Hermann, 1961). English translation, Elementary Theory of Analytic Functions of One or Several Complex Variables. (Addison-Wesley, 1963). Carathéodory, C., Funktionentheorie. (Birkhäuser, 1950). English translation, Theory of Functions of a Complex Variable (Chelsea, 1954). [2 volumes.] Carrier, G. F., M. Krook, & C. E. Pearson, Functions of a Complex Variable: Theory and Technique. (McGraw-Hill, 1966). Conway, J. B., Functions of One Complex Variable. (Springer, 1973). Fisher, S., Complex Variables. (Wadsworth & Brooks/Cole, 1990). Forsyth, A., Theory of Functions of a Complex Variable (Cambridge, 1893). Freitag, E. & R. Busam, Funktionentheorie. (Springer, 1995). English translation, Complex Analysis. (Springer, 2005). Goursat, E., Cours d'analyse mathématique, tome 2. (Gauthier-Villars, 1905). English translation, A course of mathematical analysis, vol. 2, part 1: Functions of a complex variable. (Ginn, 1916). Henrici, P., Applied and Computational Complex Analysis (Wiley). [Three volumes: 1974, 1977, 1986.] Kreyszig, E., Advanced Engineering Mathematics. (Wiley, 1962). Lavrentyev, M. & B. Shabat, Методы теории функций комплексного переменного. (Methods of the Theory of Functions of a Complex Variable). (1951, in Russian). Markushevich, A. I., Theory of Functions of a Complex Variable, (Prentice-Hall, 1965). [Three volumes.] Marsden & Hoffman, Basic Complex Analysis. (Freeman, 1973). Needham, T., Visual Complex Analysis. (Oxford, 1997). Remmert, R., Theory of Complex Functions. (Springer, 1990). Rudin, W., Real and Complex Analysis. (McGraw-Hill, 1966). Shaw, W. T., Complex Analysis with Mathematica (Cambridge, 2006). Stein, E. & R. Shakarchi, Complex Analysis. (Princeton, 2003). Sveshnikov, A. G. & A. N. Tikhonov, Теория функций комплексной переменной. (Nauka, 1967). English translation, The Theory Of Functions Of A Complex Variable (MIR, 1978). Titchmarsh, E. C., The Theory of Functions. (Oxford, 1932). Wegert, E., Visual Complex Functions. (Birkhäuser, 2012). Whittaker, E. T. & G. N. Watson, A Course of Modern Analysis. (Cambridge, 1902). 3rd ed. (1920) External links [edit] Complex analysis at Wikipedia's sister projects Definitions from Wiktionary Media from Commons Quotations from Wikiquote Wolfram Research's MathWorld Complex Analysis Page \| v t e Major topics in mathematical analysis \| \| Calculus: Integration Differentiation Differential equations + ordinary + partial + stochastic Fundamental theorem of calculus Calculus of variations Vector calculus Tensor calculus Matrix calculus Lists of integrals Table of derivatives \| \| Real analysis Complex analysis Hypercomplex analysis (quaternionic analysis) Functional analysis Fourier analysis Least-squares spectral analysis Harmonic analysis P-adic analysis (P-adic numbers) Measure theory Representation theory Functions Continuous function Special functions Limit Series Infinity \| \| Mathematics portal \| \| Authority control databases \| \| International \| \| \| National \| United States Czech Republic Israel \| \| Other \| Yale LUX \| Retrieved from " Categories: Complex analysis Complex numbers Hidden categories: Articles with short description Short description matches Wikidata Articles lacking in-text citations from March 2021 All articles lacking in-text citations Articles with excerpts Pages using Sister project links with hidden wikidata Complex analysis Add topic
16358	https://math.stackexchange.com/questions/4506055/number-of-3-digit-numbers-with-digits-in-descending-order	combinatorics - Number of $3$-digit numbers with digits in descending order - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Number of 3 3-digit numbers with digits in descending order Ask Question Asked 3 years, 1 month ago Modified2 years, 1 month ago Viewed 3k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. Question: How many 3 3-digit numbers are possible in which the digits are in descending order? Now, I did: 9×9×8 6 9×9×8 6 while the video did: 10×9×8 6 10×9×8 6. As apparent, the video considered all possible digits that can come in 3 3 places and then dividing by 6 6 will leave only those 3 3-digit numbers with their digits in descending order. Now I also reasoned exactly the same but naturally, hundred's digit can't have 0 0 so didn't include it but in my mental imagery all the numbers were resulting in what was asked by the question. I know I'm wrong. But I can't "see" (mentally) how/why? Please help. Note: I'm not really interested in solving the question by some other method though I'll be glad to know them. My main concern is to check and correct my logical fallacy so I can tackle similar problems in the future without repeating the same mistake. Edit: Further explanation of the thought process involved into 9×9×8 6 9×9×8 6: The 100 100 s' digit may contain digits from 1 1 to 9 9, then 10 10 s' digit may contain numbers from 0 0 to 9 9 except the digit now 100 100's place is using. Finally, unit's place may also contain digits form 0 0 to 9 9 except the digits now 100 100 s' and 10 10 s' place are using. Now I multiply them: 9×9×8 9×9×8. Now to select the ones with specific descending order, I divide by 6 6. combinatorics Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Aug 4, 2022 at 16:41 VoidGawdVoidGawd asked Aug 4, 2022 at 15:34 VoidGawdVoidGawd 3,110 2 2 gold badges 15 15 silver badges 44 44 bronze badges 7 The first digit can't be one but the last digit can be. He's not picking the digits in order of their place values. He's just picking the three different digits. And in decreasing order the 0 0 will not be in first so we don't need to worry about making a case for it.fleablood –fleablood 2022-08-04 15:47:18 +00:00 Commented Aug 4, 2022 at 15:47 It is a matter of choosing exactly 3 3 distinct digits and after rhat ordering them in descending order. Because the order is descending it will be automatically that the first digit will not be zero. Things would be different by ascending order.drhab –drhab 2022-08-04 15:47:33 +00:00 Commented Aug 4, 2022 at 15:47 1 Why do you have 9×9×8 9×9×8 as numerator? I suspect the first 9 9 corresponds with the first digit (no 0 0). However there are only 8 8 options for the first digit. This because of the descending order.drhab –drhab 2022-08-04 16:28:48 +00:00 Commented Aug 4, 2022 at 16:28 1 @drhab Because 100s' digit may contain digits from 1 to 9, then 10s' digit may contain numbers from 0 to 9 except the digit now 100's place is using. Finally, unit's place may also contain digits form 0 to 9 except the digits now 100s' and 10s' place are using. Now I multiply them: 9x9x8. Now to select the ones with specific descending order, I divide by 6.VoidGawd –VoidGawd 2022-08-04 16:33:12 +00:00 Commented Aug 4, 2022 at 16:33 1 To get more grip on intuition solve the following: we only allow the digits 0 0 and 1 1. "How many 2 2-digit numbers are there that have digits in strictly descending order?" It is evident that the correct answer is 1 1 (only 10 10 works) but according to your reasoning we get 1×1 2=0.5 1×1 2=0.5 which is absurd.drhab –drhab 2022-08-04 17:00:42 +00:00 Commented Aug 4, 2022 at 17:00 \|Show 2 more comments 6 Answers 6 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. As the other answers have explained, 10⋅9⋅8 6=8⋅3⋅5=120 10⋅9⋅8 6=8⋅3⋅5=120 is the correct answer. To show why your answer is wrong, I'm first going to make an educated guess about how you came to your solution. For a 3 digit number, there are 9 9 choices for the first digit, 9 9 choices for the second, and 8 8 for the third. Thus there are 9⋅9⋅8 9⋅9⋅8 three-digit numbers with unique digits. Now for every set of 3 numbers, there are 6 three-digit numbers there are 6 three-digit numbers, only one of which is sorted in decreasing order. Thus the number of descending three-digit numbers is one-sixth of the total amount of three-digit numbers with distinct digits. Thus there are 9⋅9⋅8 6 9⋅9⋅8 6 three-digit numbers with digits sorted in decreasing order. Note the expression in red. It is false. This is because, when excluding numbers that start with zero, you've made it so that sets of three numbers which contain zero only appear 4 times among the 3-digit numbers. With this knowledge, you can correct your solution by tracking the numbers with zero separately. There are 9⋅8⋅7 9⋅8⋅7 numbers with no zeros, 9⋅1⋅8 9⋅1⋅8 numbers with a zero in the tens place, and 9⋅8⋅1 9⋅8⋅1 numbers with a zero in the ones place. This gives a total of 9⋅8⋅2 9⋅8⋅2 numbers with a zero. From the numbers with no zeros, there are six numbers per set of digits with one decreasing. Thus we get 9⋅8⋅7 6=4⋅3⋅7 9⋅8⋅7 6=4⋅3⋅7 decreasing numbers from here. From the numbers with a zero, there are four numbers per set of digits, with one decreasing. Thus we get 9⋅8⋅2 4=4⋅9 9⋅8⋅2 4=4⋅9 decreasing numbers. Adding these totals together, we get (4⋅3⋅7)+(4⋅9)=12⋅(7+3)=120(4⋅3⋅7)+(4⋅9)=12⋅(7+3)=120 numbers with decreasing digits. Which is the amount given by the solutions. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Aug 4, 2022 at 16:59 AngelicaAngelica 3,049 11 11 silver badges 23 23 bronze badges 3 Thanks a lot. Greatly explained. Now I get it. Upvoted. Also, could you please address this too: How to not make such a mistake? Should I never treat a 3-digit number as a 3 digit number but as an array of length 3 with only digits allowed, for probability and combinatorics purposes? Will this change help me or do you have an even better suggestion?VoidGawd –VoidGawd 2022-08-04 17:12:13 +00:00 Commented Aug 4, 2022 at 17:12 1 @InanimateBeing I think just being aware of (or going back though and listing) what assumptions and generalizations you're making would go a long way. In a counting problem, consider what effect any exclusion has on the remainder of your argument. For a more specific tip, no exclusions are "more important" than others, and even if they were they don't have to be first--in this case there's no reason to exclude 063 063 in an earlier step than 482 482.Angelica –Angelica 2022-08-04 17:20:52 +00:00 Commented Aug 4, 2022 at 17:20 Thanks a lot! That's what I wanted. Amazing. :-)VoidGawd –VoidGawd 2022-08-04 17:47:35 +00:00 Commented Aug 4, 2022 at 17:47 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Apparently, though not explicitly written in the question, what is wanted is three digit numbers in strictly descending order. Anyway, don't visualize it as hundreds digit and so on, just a block of three different digits. When arranged in strictly descending order, a 0 0, if it is used at all, must necessarily be in the units position. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Aug 4, 2022 at 16:17 true blue aniltrue blue anil 49.2k 4 4 gold badges 31 31 silver badges 64 64 bronze badges 3 I'm not able to convey my problem yet I think. I want to know why 9×9×8 6 9×9×8 6 is not working?VoidGawd –VoidGawd 2022-08-04 16:24:49 +00:00 Commented Aug 4, 2022 at 16:24 1 If 0's were not allowed at all, then there would be only 9×8×7 6=84 9×8×7 6=84 solutions. Allowing a 0 in the units place lets us choose 9×8 2=36 9×8 2=36 permutations of the first two places. Thus, total number of solutions is 84 + 36 = 120.Dan –Dan 2022-08-04 16:52:32 +00:00 Commented Aug 4, 2022 at 16:52 1 @Dan got it, thanks. This is a nice way to deal with 0 too.VoidGawd –VoidGawd 2022-08-05 03:43:24 +00:00 Commented Aug 5, 2022 at 3:43 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. To get more grip on intuition solve the following: If we only allow the digits 0 0 and 1 1 and also in this case the first digit is not allowed to be 0 0 then: "How many 2 2-digit numbers are there that have digits in strictly descending order?" It is evident that the correct answer is 1 1 (only 10 10 works) but according to your reasoning we get 1×1 2=0.5 1×1 2=0.5 which is absurd. I advice you to think about this. It is the same problem but made "smaller". That can be healthy if your intuition has some blind spot. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Aug 4, 2022 at 17:06 drhabdrhab 154k 11 11 gold badges 87 87 silver badges 222 222 bronze badges 2 Thanks a lot. Great example! It really helps understand the fallacy in our logic and that's always the first step towards solution.VoidGawd –VoidGawd 2022-08-04 17:14:22 +00:00 Commented Aug 4, 2022 at 17:14 Glad to hear that.drhab –drhab 2022-08-04 17:17:12 +00:00 Commented Aug 4, 2022 at 17:17 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. It depends on whether you're allowed to have duplicated digits. If they must be three distinct digits, then there are (10 3)=120(10 3)=120 possible combinations of digits, each with exactly one way to arrange those digits in descending order. However, with duplicate digits allowed, there are 99 additional possibilities: 9 with all three digits the same (111, 222, 333, ..., 999), 45 (=9+8+7+...+1) with duplicated first/second digits, and 45 with duplicated second/third digits. This raises the total number of descending-digit numbers to 219. The video's answer of 10×9×8 6=120 10×9×8 6=120 indicates the assumption of distinct digits. I don't know where you got 9×9×8 6=108 9×9×8 6=108 from. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Aug 4, 2022 at 16:44 DanDan 18.5k 3 3 gold badges 29 29 silver badges 53 53 bronze badges 2 I have added an "edit" in my question, please check.VoidGawd –VoidGawd 2022-08-04 16:48:32 +00:00 Commented Aug 4, 2022 at 16:48 Thanks a lot for your efforts. Upvoted.VoidGawd –VoidGawd 2022-08-04 17:48:40 +00:00 Commented Aug 4, 2022 at 17:48 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. I already know there're (9 3)(9 3) = 84 numbers of three distinct digits in increasing order with no zeros in them (because there's no way a 0 can be in a increasing order number between 100 and 999). Check: Number of 3 3-digit numbers with strictly increasing digits, URL (version: 2023-08-22): This implies there are 84 numbers (with no zero) in decreasing order, which are the reversals of the ones in increasing order. E.g: 321 is the reversal of 123. Then you have to find how many numbers in decreasing order are there with a 0 as its third digit. Because there's no way it can be the first or middle digit!. This is a problem that can be solved thinking in how many numbers of two digits in decreasing order are there. This is (9 2)(9 2) numbers. Then there're (9 2)+(9 3)=36+84=120(9 2)+(9 3)=36+84=120 numbers of three distinct digits in decreasing order. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Aug 22, 2023 at 4:35 answered Aug 22, 2023 at 4:30 itsMontitsMont 131 5 5 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. He's not picking the digits in order. His picking the digits. period. There are three digits and they can be any combination of three numbers. There are (10 3)(10 3) ways to choose them. For each combination there is exactly one way to arrange them in descending order. And if there is a 0 0 it will be at the end so that's acceptable. Had the problem been how many in ascending order then you can not have 0 0 in your three digits at all (else it would have to begin with 0 0) and the answer would be (9 3)=9⋅8⋅7 6(9 3)=9⋅8⋅7 6 Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Aug 4, 2022 at 15:53 fleabloodfleablood 132k 5 5 gold badges 52 52 silver badges 142 142 bronze badges 1 I get the logical plausibility of the video's approach as you have now explained too. My confusion is that I too see the numbers rearranging themselves in descending order despite not including 0 as a candidate for 100s' place.VoidGawd –VoidGawd 2022-08-04 15:58:36 +00:00 Commented Aug 4, 2022 at 15:58 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions combinatorics See similar questions with these tags. 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16359	https://www.youtube.com/watch?v=QEX2mE8zviY	HPC, 7.3.2, Solving Inconsistent and Consistent-Dependent 3x3 linear systems mattemath 2390 subscribers 23 likes Description 7811 views Posted: 25 Jan 2013 In this video I cover how to solve and recognize algebraically inconsistent and consistent-dependent systems. I go over what things to look for. I also cover how to define a parameter and define all infinitely many solutions to a consistent-dependent system. 5 comments Transcript: hello youtube in this video here we're gonna be solving a couple of three by three linear systems but uh that produce uh two very special cases and so that would be the inconsistent system that would be where three planes uh never intersect say at a common point or at common points and then a consistent dependent system in which all three of our planes say maybe intersect along a particular line or maybe just all lie flat on top of each other it's you know it's all ordered triples you could ever think of satisfy each one so let's go ahead and start by taking a look at the the inconsistent system so first things first when we say inconsistent we want to use this word here inconsistent okay this is a fancy way of saying like no solutions there would be no ordered triple x y z that would satisfy all three of these equations at the same time so we say it does not exist does not exist in xyz it is a solution all three of these but how to recognize it so let's go ahead and take a look at this system here starting with this we're going to use of course our elementary row operations to put this in what we refer to as row echelon form that is that that stair step pattern in which the leading coefficient of each row so we'd say it's like 1 so 1x we want a 1y and a 1z and so let's start with this i kind of like to switch my row 1 and my row 2 around because because this already has a leading coefficient of 1. so um we're going to go ahead and rewrite the system into a new row equivalent system in which that second row is now our first row because the leading coefficient of one and let's say we switch it with our i don't know our first row so we say row one and row two if i'm going to show my work here row one row two uh we're switching them around so there's my work off to the right side so we say row one row two we say three x minus two y plus four z was was our first row now our second row and then two x minus three y plus 6z is equal to 8. so now this being said we need to eliminate anything that is underneath this leading 1 up here in the top left so showing our work we say well since it's a 1 you know we could just scale it up to be a three and a two respectively and we want it to be negative in each case to cancel out with our positive three and our positive two respectively so here's how we'll show our work we'll say negative three times row one dumped on to row two and negative two times that first row dumped on to row three so what this will do is it'll produce produce excuse me our new row equivalent system that has the same exact solutions as the original system but now looks like this our x's would be gone in these second equations here but let's talk about our second equation if we were to negative triple our first equation and add it to row 2 so then we'd have this we'd have negative 3 times y is negative 3y negative 3y plus a negative 2y is negative 5y so negative 5y now in our second row negative 3 times negative 2z is positive 6z and 6z plus 4z is 10z so plus z and now our constants we say negative 3 times positive 3 here is negative 9 negative 9 plus 1 is negative 8. so there's our new second equation now our third equation in which we doubled and negated the first row and added on to the third equation our x is now gone we say negative 2 times y is negative 2y negative 2y plus a negative 3y is negative 5y so negative 5y let's talk about our z's now we say negative 2 times negative 2 z is positive 4z positive 4z plus a6z is 10z and then last but not least we have negative 2 times 3 here is negative 6 and negative 6 plus our original 8 gives us 2. so now at this point i would say well let's get let's get this to be a 1 here you know we could take the second row times you know negative 1 5 or the reciprocal of this leading coefficient but what i want us to recognize at this point is this we have two sys you know we get a it's almost like a mini system down here a two by two system but you'll notice that they're nearly identical and in this instance i can say well what if we were to just go ahead and make things simple on ourselves what if we were to just like take negative row 2 plus row 3. just go ahead and dump it onto row 3. what we get is this new system it would look like this would say x plus y minus 2 z is equal to 3 our second equation would still be negative 5 y plus 10 z and it would be equal to negative 8. but since we negated and added it we'd say well then it'd be a positive 5y plus a negative 5y which would cancel out our y's here we'd no longer have any y's here but since we negated our 10 z in our second row as well and added it to our positive 10z down here it also cancels out our z's so we say like those zero y's plus zero z's and this is equal to now negative negative 8 in our second row added on to our 2 would be 8 plus 2 we get 10. in other words we get a third equation that makes this bold statement right here but it says 0 equals 10 which of course we know is not valid this is not a valid statement when is it valid never so we we'd say whenever you end up with an invalid statement where all your variables are gone but you're left with an untrue statement okay always false we say we have an inconsistent system and so you know again things to look for are these we say well if you had two equations that were nearly identical like like all the variables were pretty much the same thing but the constants are not are not then we say okay well then you're going to have an inconsistent system so let's talk about a consistent dependent system a little bit different uh animals so we have our uh system here three by three three equations three variables you'll notice that our top left x here is already a one that is lovely we're going to try to get everything underneath it to be nothing we want to get rid of those to put it in our stair step fashion so showing our work we have this we have negative 2 times row 1 we're going to add that to row 2. that'll get rid of our 2x there and we say negative 3 times that first x there in row 1 added on to row 3 that's going to eliminate the 3x in the second or the third row so we have x plus 2y minus 7 z is equal to negative 4. second equation now our 2x is gone we say okay so negative 3 times the original 2y in the original equation we say is negative 6y negative 6y plus positive 3y is negative 3y we say okay so negative oh oh one moment we were taking negative 2 times row 1. i got to be careful here so negative 2 times our 2y here is negative 4y negative 4y plus a positive 3y is negative y so we say negative 2 times this negative 7z i'll be positive 14z and positive 14z plus our original z gives us 15 z's and we say negative 2 times our original negative 4 here is 8 and 8 plus 5 is 13. okay so now our third equation here we'd say we're tripling negative our first row adding it to our third our x's would be gone triple this negative 2 would be negative 6 negative 6 y plus a positive y there would be just y or positive 7y would be just a y if we say negative 3 times negative 7z is 21z and 21z positive added to negative 36z would be i believe negative 15 z and you can see this pattern popping up again here and we say okay so negative 3 now times this original negative 4 would be 12 and 12 plus a negative 25 gives us negative 13. so again we're seeing this situation here in which you know i i should continue i mean i should keep going actually i will i want to i want to switch row two and row three and everybody watching this video you can roll your eyes i just want this positive coefficient of y to be here it's like it's weird man when i go places and i get like a beverage and those little tabs you push down you gotta push them down i have to push them down it like drives me nuts so y minus 15z equals negative 13 is now our second row we say negative y plus 15 z equals positive 13. if if now we were to add row 2 onto row 3 it would cancel out your y's okay so you say you get this new system over here x plus 2y minus 7z minus 7z equals negative 4. uh we'd say y minus 15z equals negative 13 is still our second row but if we dumped our second row onto our third row what we end up with is this we get zero is zero and this is something that is totally consistent what does this mean it means it's true so it gives us at least one solution but it's consistent dependent why because we say we have a situation in which there are several points that satisfy this not just one but several consistent dependent so here's the deal we actually a lot of times in math when you have three by three systems or higher we want to develop what we call a parameter and you know later in the year i know we typically use the parameter as time okay but in this case we're going to say a parameter is just a certain value that we're going to allow and allow what typically z to be equal to so just follow my lead here but basically what we're going to do is this we're going to pick a variable we typically go with z i typically go with c but we say let's allow z to equal a i'm going to use the word let here let z equal a what this means is as far as our solution goes this ordered triple we're saying that the last part's this parameter a that's allowed to wander they say well okay so what about y and what about x well this is where that backsol thing comes in if z were equal to a we could back solve by substituting that in for z in the second equation and so if we do this we get now this new equation y minus 15 times blank equals negative 13 where we're plugging in a for a z so now we have this y minus 15 a equals negative 13 so y actually equals 15 a if we add that 15a to the right side minus 13. and so now this is what y equals we say 15 a's minus 13. so you might be saying to yourself self now that i know z and i know y how can i find x we're going to take these now both we'll switch our colors here so we say this is z equal to a we're going to put it into our first equation we're going to take this y value now that we just found we're going to stick it into our second or second y there but in our first equation we end up with this we say x plus two times i leave a big blank here y minus 7 z equals negative 4 in which z we said it was a and we say this is our y value here so we're going to plug this in right here we get 15 a's minus 13 constants so we've got a little cleaning to do but we get this we have x plus 30a minus 26 minus 7a is negative four and now we're going to isolate our x so we'd say okay well these thirty a's here and these negative seven a's first of all those produce 23 a's minus 26 constants negative four and if we were to now add 26 to both sides and subtract 23 a's you know over to the other side we get this we get that x equals negative 23a and then 26 constants plus negative 4 constants would give us 22. so now we have x in terms of a so negative 23 a plus 22. so i know what you might be thinking guys ugly i don't like this at all but these are our infinitely many solutions right here if we plugged in a with 0 you know we'd get say 22 comma negative 13 comma 0. if we plugged in a as 1 you know one for a we'd say well then we get negative 23 plus positive 22 is negative one and uh if we plugged in one for a for our y here 15 a minus 13 would be 15 times 1 is 15 minus 13 is uh 2. both of these solutions if you plug them back in the original system you see they work they're infinitely many solutions depending upon what this last a value equals or our z value so the rule of thumb is this if you ever get a consistent dependent system just say z equals a and then back solve by plugging that back into your second and first equations respectively to find x and y enjoy
16360	https://www.youtube.com/watch?v=VozE0v5_iMw	14.4 Standing Waves \| General Physics Chad's Prep 170000 subscribers 326 likes Description 13106 views Posted: 15 Jan 2024 In this lesson Chad provides a lesson on Standing Waves. An introduction first explains the condition of resonance that leads to standing waves along with the identification of both nodes and antinodes in standing waves. Standing waves on a string fixed at both ends are covered first along with the derivation of the equations for both wavelength and frequency. The first few standing waves are drawn. The longest standing wave is referred to as the fundamental frequency or the first harmonic, while each successive standing wave is referred to as the second harmonic, third harmonic, etc. The lesson moves next to waves on a string fixed at one end and the derivation of the equations for both wavelength and frequency. It is also shown how the equations differ from the waves on a string fixed at both ends, and how only the odd harmonics exist (1st harmonic, 3rd harmonic, 5th harmonic, etc.). The lesson moves on to a pipe open at both ends and a pipe closed at one end. Chad shows how the equations for the wavelengths and frequencies for a pipe open at both ends are the same as for the string fixed at both ends, and how the equations for a pipe closed at one end are the same as for a string fixed at one end. The lesson is concluded with the solving of a few physics standing waves practice problems. 00:00 Lesson Introduction 00:40 Introduction to Standing Waves 03:08 Standing Waves on a String Fixed at Both Ends 10:08 Standing Waves on a String Fixed at One End 13:45 Standing Waves in a Pipe Open at Both Ends or Closed at One End 15:46 Standing Waves Practice Problems Check out Chad's General Physics Master Course: 29 comments Transcript: Lesson Introduction standing wave is going to be the topic of this lesson in my new General Physics playlist which when complete will cover a full year of University algebra based physics now in this lesson we're going to talk about waves on a string that's either fixed at one end or both ends and then wav sound waves in a pipe uh again that's either open at both ends or only open at one end my name is Chad and welcome to Chad's prep where my goal is to take the stress out of learning science now if you're new to the channel we've got comprehensive playlists for General chemistry organic chemistry General Physics and high school chemistry and on Chads prep.com you'll find premium Master courses for the same that include study guides and a ton of practice you'll also find comprehensive prep courses for the DAT the MCAT and the oat and we're going to introduce the Introduction to Standing Waves idea of standing waves with a string that is fixed at one end and where uh a string does not move where there's no displacement of the string we end up calling that a node and since it's fixed at one end like in this case attached to a wall we'll call that a node and the other end we're going to oscillate it I'm just going to oscillate it once here I'm just going to give it a quick little JW musle and what that's going to do by oscillating it in an upward Direction that's going to introduce an upward wave so to speak so and as the string pulls upward it pulls the next segment of the string upward as well which then pulls the next one upward and that's going to make it look like the wave is traveling down the string towards the wall now when it gets to the wall the end the very end of the string doesn't have any other string to pull upward on so it's going to attempt to pull upward on the wall but the Wall's immovable so in the same way if I try to push this wall if I try to push this wall rather than the wall moving that way I end up moving out off the wall from that equal and opposite Force according to Newton's third law well same thing for the string here as it attempts to pull upward on the wall the wall is going to exert a downward force and the wall wins because the wall is not moving and as a result what we see is an inversion in the wave that's reflected back and it'll travel back all right now as long as all we're doing is a ating or jostling this string once one wave travels to the wall it becomes inverted and travels back and that's it but what if we just keep oscillating this every so often and we can you know make it faster or slower so on and so forth well it turns out that you're going to be sending multiple waves toward the wall and then you're going to be having multiple waves reflected back so and depending on how fast you uh oscillate this and things of this sort you can get a condition of what's called a standing waves and it turns out it happens when we hit what's called a resonance frequen queny or the condition of resonance is met so in this case you're going to end up with regions not just necessarily the fixed end right there but regions of the string that are not being displaced whatsoever and we might end up having some additional nodes now again we're going to have this node always at the end the question is are we going to have any additional nodes internal to the string here in this case in which case we'd call them internal nodes and it turns out there are multiple of these resonant frequencies and we have to drive some equations for how we come up with these uh the corresponding wavelengths and frequencies that correspond to these standing waves these resonant frequencies all right so if we take a Standing Waves on a String Fixed at Both Ends look uh we started with a wave on a string that's fixed at one end but we classically are going to begin with equations for a wave on a string that's fixed at both ends so we got the wave that's fixed at both ends and by definition therefore it's fixed at those ends they're not going to be experiencing any displacement that's going to be a node so when it turns out you're going to start with your longest possible wavelength and get progressively shorter and your longest possible wavelength is going to happen when you don't have any internal nodes and in this case we obviously can't you know oscillate it from one end or the other this would be like a string and a guitar where you pluck it instead so and in this case we might pluck it and the longest possible wave we could get is one where there's no additional internal nodes no additional nodes whatsoever and then the reflected wave would go back now we've talked about the nodes the the other thing we want to talk about what called the anti noes and here the antinode is the point where you get the maximum displacement which is right in the middle right here okay now this is one possibility an additional possibility for this string if we pluck it just right again we're always going to have nodes at the ends but can we get one additional node in this case the nodes tend to be symmetrically distributed and so could we get a node right in the middle of the string in this case if we plucked it just right we could get that lovely wave right there and then the reflected wave would pass through the very same node so it end up right back at the same place at a node at both ends so we're constrained for this particular wavelength having nodes at the end and a node right in the middle it's possible all right let's take this one step further and let's add one additional node so again we've got still the nodes at the end so but now if we want to put two nodes in there if we want to make them symmetrically distributed got to kind of divide this string up into thirds there's 1/3 there's 1/3 there's 1/3 so and then our wavelength would look just like so and then the reflected wave would look just like so so now we have two internal nodes we went from zero internal nodes to One internal node to two internal nodes we went from one anti- node there to two anti noes and now to three anti- no so each successive standing wave is going to have an additional node and an addition antinode as well now what we want to do is come up with an equation for the wavelength of each of these waves and we're going to start with the middle one here because this is the easiest because we can see that over the length of the string we complete one full wavelength and you could just simply say that the wavelength is equal to l the length of the string okay so if we look at the first one we haven't actually even completed a full wavelength so we'd have to keep going this is the first half of the wavelength and then the second half to finish it off and so a full wavelength would actually be twice as long as the string as it turns out and so Lambda would equal 2 L well in this last one here now we've actually completed one full wavelength by the time we've crossed 2/3 of the string and then we get another half by the time we reach the end and stuff like that so how long is the wavelength well it's just 2/3 of the string and so here we'd say that Lambda equal 2/3 the length of the string for a full wavelength okay now we've got to come with you know some equation that we can apply that describes each of these and it turns out may not be intuitive but it turns out the equation is going to be Lambda = 2 l/ N where n is equal to any integer there's the equation for the wavelength and it turns out n in this case for this example would could you could look at it as corresponding to the anti noes or you could just say that the longest wavelength has n equals 1 and then the next one with one additional internal node is n equal 2 and so on and so forth however you want to remember it so but here we've got n equals 1 and if n equals 1 then Lambda equal 2 L over 1 or just simply 2 L so for the second one we have a so Lambda equal 2 L over 2 and 2 L over2 is just Lambda equal L for the next one it' be Lambda = 2 L over 3 which is the same thing as saying Lambda = 2/3 L for the fourth one n = 4 be Lambda = 2 L over 4 or Lambda = 12 lamb or 12 l so on and so for now it turns out we talk about this original wavelength here so as being the longest possible one well we want to look at the corresponding frequencies and the frequency that corresponds to that longest wavelength is what we call the fundamental frequency or the first harmonic now we want to derive an expression for these frequencies and you might recall that the velocity of a wave is equal to its frequency times its wavelength and if you rearrange this and solve for frequency you'd get velocity over wavelength well in this this case that would be velocity all over the wavelength which is 2 l/ N and dividing by a fraction you might recall is multiplying by the reciprocal and so this would just equal n V over 2L and that is the expression we want to derive for this frequency as well so it's n V over 2L and sometimes you you'll commonly see this written where we'll say Lambda n or frequency n so and it just tells you that these describ a series of wavelengths and a series of frequencies where n is any integer okay so a couple things to look at here we talk about this frequency now when n equals 1 that again corresponds to the fundamental frequency and if we kind of look at this just a little bit different so FN = n v/ 2L so when again n equals 1 you get your fundamental frequency and so your fundamental frequency is just V over 2L and notice every other additional frequency is as possible is just going to be multiplying that fundamental frequency v over2 l by a different integer now again when n equals 1 that's your fundamental frequency or first harmonic so and then the additional ones when n equals 2 that's your second harmonic when n equals 3 your third harmonic when n equals 4 your fourth harmonic so on and so forth and all of those harmonics are just equal to a multiple of that first harmonic that fundamental frequency so again when n equals 1 the frequency is just v/2 well and if I told you that the fundamental frequency was 100 Herz and then I said what's the second harmonic well it' be two times that fundamental frequency would be 200 htz what would be the third harmonic well it be three times that fundamental frequency it would be 300 htz what's the value of the fourth harmonic what would be four times that fundamental frequency it would be 400 HZ so on and so forth so all the successive harmonics are just a multiple of that first harmonic of that First Fundamental frequency okay so this is kind of how works out for uh a wave on a string that's fixed at both ends now we want to go back and take a further look at a string that's fixed at just one end all Standing Waves on a String Fixed at One End right for a string that is fixed now at just one end again on that one end you're going to have a note so but now we're going to be oscillating it from the other end and so it's going to experience a point of Maximum displacement at that other end by definition because that's where we're oscillating it from and so that has to be an anti Noe and so we have to see okay how can we come up with some standing waves here so not only does the wave have to start on an antinode and then hit a node but the reflected wave then have to exit on an antinode as well and we see that we can indeed do this without introducing any new internal nodes so your wave coming in hits a node and then reflected right back out so this is when Nal 1 it turns out and if we try to come up with an expression for the wavelength the wavelength is not equal to 2L anymore so again if you look back over here what was fixed at both ends over the length of the string we completed a whole half of a wavelength well here it looks like we've only completed like the second half of that half of a wavelength this is only a fourth of a wavelength so as a result if I wanted a full wavelength I would need something four times the length of the string here and so Lambda would equal 4 L cool and you might be like well hey we just kind of took that and said for the fundamental here that longest wavelength and we just said divide it by n and life is good and then that's pretty much what we're going to do almost so Lambda n here is going to equal 4 L over n and then we can derve the corresponding formula for the frequency but we run into a problem and we try to draw the next standing wave if we go to draw the next standing wave we might be tempted oh you just put a node an internal node right in the middle of the string if you put a node right in the middle of the string there is no way to have a wave going towards the wall and reflected wave both travel through that node both start on an antinode so terminate on a node and both travel through the internal node it's impossible and it turns out there's no corresponding wavelength for n equals 2 when you go to n equals 3 it turns out you can put an internal node 13 of the way in so 2/3 from from the wall and this works and so you start on an anti-node yet again terminate on a node at the wall and there's your wave going toward the wall the reflected waves [Music] inverted and perfectly goes right back through the node and ter or exits the string if you will I say exits so ends the other end of the string with an anti- Noe and it works but that's for n equals 3 and it turns out for a string that's fixed at one end not all values of n are possible only the odd values of n actually give you a standing wave and a resonant frequency if you will will so in this case we have to say that n = 1 comma 3 comma 5 so with the string that was fixed at two ends any integer value of n but for a string that's fixed at one end just the odd values of n are possible and so when n equals 1 we still call that the fundamental frequency or the first harmonic but there is no second harmonic there is no fourth harmonic there's just a third harmonic and a fifth harmonic and a seventh harmonic so on and so forth for a string that's only fixed at one end all right so if we come up with the equation for the frequency so it' be the same kind of a thing and we'd end up with frequency is equal to NV over 4 L where once again n has to be one of the odd numbers so we've derived our equations Standing Waves in a Pipe Open at Both Ends or Closed at One End for frequency and wavelength here for standing waves either on a string that's fixed at both ends or a string that is fixed at one end but it turns out we use these exact same equations for talking about sound waves in a pipe and it's a pipe that's either open at both ends or a pipe that's open at one end and if we kind of look you can kind of see why it works out this way so if we have a pipe that's open at both ends versus over here where we have a pipe that's open at one end and so sound's going to come in let's let's look at the one end first sound's going to come into the pipe so it's going to bounce off the the far wall and then reflect back out of the pipe and where the sound enters and exits it's always on an anti Noe and so we could see that this would work out so the long long EST wavelength possible would start at an anti Noe terminate on a node at the wall and then reflect back so at the other end on an antinode and that works so it turns out you end up with exactly analogous situation like we saw here so and so on and so forth where only odd values of n actually result in standing waves now what's different with the pipe here though is that uh again where the sound enters the pipe is going to be an anti Noe and where it exits the pipe is going to be an anti-node as well and not all the sound actually exits here some of it will be reflected back and there will be a reflected wave for the standing wave here but it's the kind of the inverse of what we saw with the string whereas the string we had nodes at either end so here we have anti noes at either end but ultimately the equations work out exactly the same where all integer values of n are possible so that your longest wavelength possible is still going to be double the length of in this case your pipe not your string so so on and so forth but the same equations apply and so now we know how to treat either a string that's fixed at both ends or a pipe that's open at both ends we know how to treat a string that's fixed at one end or a pipe that's only open at one end we know which set of equations to use and I think we're ready to do some plugin and chugging all right Standing Waves Practice Problems for the first question we're going to tackle here it says what is the wavelength of the fundamental frequency for a025 M long string that is fixed at both ends if the speed of sound and air is 340 m second what is the fundamental frequency so a couple parts to this question so in the first part is just what is the wavelength of that fundamental frequency and we're told the string is a length of 0.25 M and if you recall our fundamental frequency the corresponding wavelength spans half a wavelength over that so that the full wavelength for that longest wavelength would be double the length or you could just go over here and say it's 2 L over one for that fundamental wavelength or at least the wavelength corresponding to the fundamental frequency see and 2 L over 1 is 2 L and 2 L here would equal 2 0.25 M which is going to equal 0.5 M okay so there's the first half of the question second half says if the speed of sound in air is 340 m/ second what is the fundamental frequency and so again here fundamental frequency is fub1 where n equal 1 and again it's going to equal NV v/ 2 L that's going to be 1 times the V we're given 340 m/s all over 2 L 2 0.25 met all right 2 .25 is 0.5 dividing by .5 which is a half dividing by half same thing as multiplying by 2 and 340 2 is 680 and this has units of Hertz or per seconds if you notice the meters part would cancel and it's just per seconds or Hertz cool and that's all there is to it so we were told it was string fixed at both ends so we knew we were supposed to use this set of equations not this set of equations all right let's take a look at the next one here next one says what is the wavelength of the fundamental frequency for a 0.25 meter long string that is fixed now at one end if the speed of sound in air is 340 m/ second what is the fun Al frequency so it's really the same question but now dealing with a string that's only fixed at one end and so we've got to use the other set of equations instead and this one's not as easy to picture so but again the key is that uh for the string that's fixed at both ends the fundamental wavelength is going to be double the length of the string now it's going to be four times the length of the string instead all right so Lambda n = 4 L / n which in this case is going to be 4 0.25 M all over Nal 1 for the fundamental which comes out to 1.0 M okay and then the fundamental wavelength which is NV over 4 L is going to equal 1 told to use 340 m/s all over 4 L which again is 4 0.25 M 4 .25 is 1 3 340 ID 1 is still just 340 and again units here are Hertz The Meters Parts cancel and so again same question just dealing with string fixed at one and instead of two and the key is just again knowing which set of equations you need to use based on which kind of a string or which kind of a pipe you've got all right last one here says the fundamental frequency of a guitar string is 83 Hertz what is the frequency of the third harmonic and if the speed of sound in air is 340 m/s what is the wavelength of the third harmonic all right so first off we're told the fundamental frequency which is F1 is 83 Hertz and the question is what is the frequency of the third harmonic let's go back here so again we're dealing with a guitar a guitar is a string that's fixed at both ends so it's not super critical we'll find out but string is fixed at two ends uh and in this case that means that your frequency is NV 2L well again when n equals 1 that's your fundamental frequency it's just V over 2 L so the third harmonic is going to happen when we plug in Nal 3 instead and it's just going to equal the N /2l the fundamental frequency times three and so if you realize that you could just say it's 3 times that 83 Hertz so in this case 3 80 is 240 and 3 3 is 9 so it's going to be 249 Hertz for that third harmonic and then finally the question says if the speed of sound in air is 340 meters per second what is the wavelength of the third harmonic and in this case I've got a couple different approaches I can take but I don't actually know the length of the guitar string here but I do know that for any sound wave or any wave for that matter wavelength times frequency equals the speed of the wave so and I want that wavelength so it's going to equal the speed divided by the frequency we're told the speed of sound in this case is 340 m/s we can see that the wave I'm sorry with the frequency was 249 Hertz which I'll write as a per second and we'll let our calculator do the rest of the heavy lifting here but we should get 1 point something meters for that wavelength here so 340 / 249 equals 1365 and in this case I want to round that to two SigFig so 1.4 M and that was that third the wavelength of the third harmonic so again 1.4 M and that's all there is to it if you have found this lesson helpful consider giving it a like happy studying
16361	https://dspace.mit.edu/bitstream/handle/1721.1/104427/6-042j-spring-2005/contents/recitations/rec12.pdf	6.042/18.062J Mathematics for Computer Science March 18, 2005 Srini Devadas and Eric Lehman Notes for Recitation 12 1 Solving linear recurrences Guessing a particular solution. Recall that a general linear recurrence has the form: f (n) = a1f (n −1) + a2f ( + adf (n −d) + g(n) n −2) + · · · As explained in lecture, one step in solving this recurrence is ﬁnding a particular solu tion; i.e., a function f (n) that satisﬁes the recurrence, but may not be consistent with the boundary conditions. Here’s a recipe to help you guess a particular solution: • If g(n) is a constant, guess that f (n) is some constant c. Plug this into the recurrence and see if any constant actually works. If not, try f (n) = bn + c, then f (n) = an2 + bn + c, etc. • More generally, if g(n) is a polynomial, try a polynomial of the same degree. If that fails, try a polynomial of degree one higher, then two higher, etc. For example, if 2 g(n) = n, then try f (n) = bn + c and then f (n) = an + bn + c. • If g(n) is an exponential, such as 3n, then ﬁrst guess that f (n) = c3n. Failing that, try f (n) = bn3n + c3n and then an23n + bn3n + c3n, etc. In practice, your ﬁrst or second guess will almost always work. Dealing with repeated roots. In lecture we saw that the solutions to a linear recurrence are determined by the roots of the characteristic equation: For each root r of the equation, the function rn is a solution to the recurrence. Taking a linear combination of these solutions, we can move on to ﬁnd the coefﬁcients. The situation is a little more complicated when r is a repeated root of the characteristic equation: if its multiplicity is k, then (not only rn, but) n n 2 n each of the functions r , nr , n r , . . . , nk−1rn is a solution to the recurrence, so that our linear combination must use all of them. Recitation 12 2 2 MiniTetris (again) Remember MiniTetris from Recitation 4? Here is an overview: A winning conﬁguration in the game is a complete tiling of a 2 × n board using only the three shapes shown below: For example, the several possible winning conﬁgurations on a 2 × 5 board include: In that past recitation, we had deﬁned Tn to be the number of different winning conﬁg urations on a 2 × n board. Then we had to inductively prove Tn equals some particular closed form expression. Remember that expression? Probably not. But no damage, now you can ﬁnd it on your own. (a) Determine the values of T1, T2, and T3. Solution. T1 = 1, T2 = 3, and T3 = 5. (b) Find a recurrence equation that expresses Tn in terms of Tn−1 and Tn−2. Solution. Every winning conﬁguration on a 2 × n board is of one three types, dis tinguished by the arrangment of pieces at the top of the board. n − 1 n − 2 n − 2 There are Tn−1 winning conﬁgurations of the ﬁrst type, and there are Tn−2 winning conﬁgurations of each of the second and third types. Overall, the number of win ning conﬁgurations on a 2 × n board is: Tn = Tn−1 + 2Tn−2 Recitation 12 3 (c) Find a closedform expression for Tn. Solution. The characteristic polynomial is r2 −r −2 = (r −2)(r +1), so the solution n is of the form A2n + B(−1) . Setting n = 1, we have 1 = T1 = 2A −B. Setting 2 n = 2, we have 3 = T2 = A22 + B(−1) = 4A + B. Solving these two equations, we conclude A = 2/3 and B = 1/3. That is, the closed form expression for Tn is n 2 1 Tn = 2n + (−1)n = 2n+1 + (−1) . 3 3 3 Remember it now? 3 Inhomogeneous linear recurrences Find a closedform solution to the following linear recurrence. T0 = 0 T1 = 1 Tn = Tn−1 + Tn−2 + 1 () (a) First ﬁnd the general solution to the corresponding homogenous recurrence. 2 Solution. The characteristic equation is r −r −1 = 0. The roots of this equation are: 1 + √ 5 r1 = 2 1 − √ 5 r2 = 2 Therefore, the solution to the homogenous recurrence is of the form 1 − √ 5 ! n n 1 + √ 5 ! Tn = A + B . 2 2 (b) Now ﬁnd a particular solution to the inhomogenous recurrence. Solution. Since the inhomogenous term is constant, we guess a constant solution, c. So replacing the T terms in () by c, we require c = c + c + 1, namely, c = −1. That is, Tn ≡−1 is a particular solution to (). Recitation 12 4 (c) The complete solution to the recurrence is the homogenous solution plus the partic ular solution. Use the initial conditions to ﬁnd the coefﬁcients. Solution. 1 − √ 5 ! n n 1 + √ 5 ! Tn = A + B −1 2 2 All that remains is to ﬁnd the constants A and B. Substituting the initial conditions gives a system of linear equations. 0 = A + B −1 1 − √ 5 ! 1 + √ 5 ! 1 = A + B −1 2 2 The solution to this linear system is: 5 + 3 √ 5 A = 10 5 −3 √ 5 B = 10 (d) Therefore, the complete solution to the recurrence is: Solution. ! 5 −3 √ 5 ! 1 − √ 5 ! n n 5 + 3 √ 5 ! 1 + √ 5 Tn = + −1. 10 · 2 10 · 2 4 Back to homogeneous ones Let’s get back to homogeneous linear recurrences. Find a closedform solution to this one. S0 = 0 S1 = 1 Sn = 6Sn−1 −9Sn−2 Anything strange? 2 Solution. The characteristic polynomial is r −6r + 9 = (r −3)2, so we have a repeated root: r = 3, with multiplicity 2. The solution is of the form A3n + Bn3n for some constants A and B. Setting n = 0, we have 0 = S0 = A30 + B · 30 = A. Setting n = 1, we have 0 · 1 = S1 = A31 + B · 31 = 3B, so B = 1/3. That is, 1 · 1 n3n Sn = 0 · 3n + = n3n−1 . 3 · 5 Recitation 12 Short Guide to Solving Linear Recurrences A linear recurrence is an equation f(n) = a1f(n − 1) + a2f(n − 2) + . . . + adf(n − d) +g(n) \| {z } \| {z } homogeneous part inhomogeneous part together with boundary conditions such as f(0) = b0, f(1) = b1, etc. 1. Find the roots of the characteristic equation: n x = a1x n−1 + a2x n−2 + . . . + ak 2. Write down the homogeneous solution. Each root generates one term and the homoge n neous solution is the sum of these terms. A nonrepeated root r generates the term cr r , where cr is a constant to be determined later. A root r with multiplicity k generates the terms: n n 2 n n cr1 r , cr2 nr , cr3 n r , . . . , crk n k−1 r where cr1 , . . . , c are constants to be determined later. rk 3. Find a particular solution. This is a solution to the full recurrence that need not be con sistent with the boundary conditions. Use guess and verify. If g(n) is a polynomial, try a polynomial of the same degree, then a polynomial of degree one higher, then two higher, etc. For example, if g(n) = n, then try f(n) = bn+c and then f(n) = an2 +bn+c. If g(n) is an exponential, such as 3n, then ﬁrst guess that f(n) = c3n . Failing that, try f(n) = bn3n + c3n and then an23n + bn3n + c3n, etc. 4. Form the general solution, which is the sum of the homogeneous solution and the partic ular solution. Here is a typical general solution: n + f(n) = c2n + d(−1) 3n + 1 \| {z } \| {z } homogeneous solution particular solution 5. Substitute the boundary conditions into the general solution. Each boundary condition gives a linear equation in the unknown constants. For example, substituting f(1) = 2 into the general solution above gives: 2 = c · 21 + d · (−1)1 + 3 · 1 + 1 ⇒ −2 = 2c − d Determine the values of these constants by solving the resulting system of linear equa tions.
16362	https://eprints.whiterose.ac.uk/id/eprint/140116/3/latest%20revision%20SB_PB2_corrections%281%29.pdf	This is a repository copy of Biallelic Mutations in LRRC56, Encoding a Protein Associated with Intraflagellar Transport, Cause Mucociliary Clearance and Laterality Defects. White Rose Research Online URL for this paper: Version: Accepted Version Article: Bonnefoy, S, Watson, CM, Kernohan, KD et al. (22 more authors) (2018) Biallelic Mutations in LRRC56, Encoding a Protein Associated with Intraflagellar Transport, Cause Mucociliary Clearance and Laterality Defects. American Journal of Human Genetics, 103 (5). pp. 727-739. ISSN 0002-9297 © 2018. Licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License ( eprints@whiterose.ac.uk Reuse This article is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivs (CC BY-NC-ND) licence. This licence only allows you to download this work and share it with others as long as you credit the authors, but you can’t change the article in any way or use it commercially. More information and the full terms of the licence here: Takedown If you consider content in White Rose Research Online to be in breach of UK law, please notify us by emailing eprints@whiterose.ac.uk including the URL of the record and the reason for the withdrawal request. 1 TITLE Biallelic Mutations in LRRC56 encoding a protein associated with intraflagellar transport, cause mucociliary clearance and laterality defects AUTHORS Serge Bonnefoy,1,10 Christopher M. Watson,2,3,10 Kristin D. Kernohan,4 Moara Lemos,1 Sebastian Hutchinson1, James A. Poulter,3 Laura A. Crinnion,2,3 Ian Berry,2 Jennifer Simmonds,2 Pradeep Vasudevan,5 Chris O'Callaghan,5,6 Robert A. Hirst,5 Andrew Rutman,5 Lijia Huang,4 Taila Hartley,4 David Grynspan,7 Eduardo Moya,8 Chunmei Li,9 Ian M. Carr,3 David T. Bonthron,2,3 Michel Leroux,9 Care4Rare Canada Consortium,4 Kym M. Boycott,4 Philippe Bastin,1,10, and Eamonn G. Sheridan2,3,10, AFFILIATIONS 1: Trypanosome Cell Biology Unit & INSERM U1201, Institut Pasteur, 25, rue du Docteur Roux, 75015 Paris, France ʹǣ Yorkshire Regional Genetics Serviceǡ StǤ Jamesǯs University (ospitalǡ Leedsǡ LSͻ ͹TFǡ United Kingdom ͵ǣ School of Medicineǡ University of Leedsǡ StǤ Jamesǯs University Hospital, Leeds, LS9 7TF, United Kingdom Ͷǣ Childrenǯs (ospital of Eastern Ontario Research )nstituteǡ University of Ottawaǡ Ottawa, K1H 8L1, Canada 5: Centre for PCD Diagnosis and Research, Department of Infection, Immunity and Inflammation, RKCSB, University of Leicester, Leicester, LE2 7LX, United Kingdom 2 6: Respiratory, Critical Care & Anaesthesia, Institute of Child Health, University College London Ƭ Great Ormond Street Childrenǯs (ospitalǡ ͵Ͳ Guilford Streetǡ London WCͳN 1EH, United Kingdom 7: Department of Pathologyǡ Childrenǯs (ospital of Eastern Ontarioǡ ͶͲͳ Smyth Roadǡ Ottawa, K1H 8L1, Canada 8: Bradford Royal Infirmary, Bradford, West Yorkshire, BD9 6R, United Kingdom 9: Department of Molecular Biology and Biochemistry, and Centre for Cell Biology, Development and Disease, Simon Fraser University, Burnaby, Canada 10: These authors contributed equally to this work CORRESPONDENCE EMAILS Correspondence: e.sheridan@leeds.ac.uk (E.G.S.), philippe.bastin@pasteur.fr (P.B) 3 ABSTRACT Primary defects in motile cilia result in dysfunction of the apparatus responsible for generating fluid flows. Defects in these mechanisms underlie disorders characterised by poor mucus clearance , resulting in susceptibility to chronic recurrent respiratory infections, often associated with infertility; laterality defects occur in about 50 % of such individuals. Here we report biallelic variants in LRRC56 (known as oda8 in Chlamydomonas) identified in two unrelated consanguineous families. The phenotype comprises laterality defects and chronic pulmonary infections. High speed video microscopy of cultured epithelial cells from an affected individual showed severely dyskinetic cilia, but no obvious ultra-structural abnormalities on routine transmission electron microscopy (TEM). Further investigation revealed that LRRC56 interacts with the intraflagellar transport (IFT) protein IFT88. The link with IFT was interrogated in Trypanosoma brucei. In this protist, LRRC56 is recruited to the cilium during axoneme construction, where it co-localises with IFT trains and is required for the addition of dynein arms to the distal end of the flagellum. In T. brucei carrying LRRC56 null mutations, or a variant resulting in the p.Leu259Pro susbtitution corresponding to the p.Leu140Pro variant seen in one of the affected families, we observed abnormal ciliary beat patterns and an absence of outer dynein arms restricted to the distal portion of the axoneme. Together, our findings confirm that deleterious variants in LRRC56 result in a human disease, and suggest this protein has a likely role in dynein transport during cilia assembly that is evolutionarily important for cilia motility. 4 MAIN TEXT INTRODUCTION Cilia are highly conserved eukaryotic organelles that are classified into motile and non-motile forms. Motile cilia and flagella, a hallmark of eukaryotes, display remarkable structural and molecular conservation1. Most motile cilia exhibit a 9+2 configurationȄa pair of single microtubules surrounded by 9 peripheral doublets. Connected to each doublet of peripheral microtubules is an inner and outer dynein arm, consisting of multiple dynein chains that provide ATPase-mediated energy for movement2. Dynein arms are preassembled in the cytosol and are transported to an assembly site at the distal end of the growing axoneme. This requires intraflagellar transport (IFT)3,4 5, an evolutionary conserved bidirectional transport system that delivers axoneme building blocks6,7 such as tubulin, to the flagellar tip8. Defective motile cilia underlie the pathophysiology of individuals with impaired mucociliary clearance, which increases susceptibility to respiratory complications, including sinusitis, bronchitis, pneumonia, and otitis media2. Chronic infections frequently lead to progressive pulmonary damage and bronchiectasis. Spermatozoal dysmotility in affected men causes infertility2. These clinical features characterise the hallmark disease of motile cilia, Primary Ciliary Dyskinesia ( [MIM: PS244400]), guidelines for the diagnosis of PCD have recently been published9. Approximately half of all individuals with PCD display laterality defects varying from partial to complete situs inversus, a consequence of dysfunctional embryonic nodal cilia2. At least 36 genes currently account for the heterogeneous genetic disorder PCD, which displays mainly autosomal recessive inheritance and is characterised by cilia dyskinesis and structural 5 defects observed by TEM of a nasal biopsy. However this is not always the case. Variants in CCNO ([MIM: 607752]) result in a congenital mucociliary clearance disorder with reduced generation of motile cilia ([CILD29 MIM:615872]), which is not associated with laterality defects10. Mutations in DNAH11 ([MIM: 603339]) result in CILD7 ([MIM: 611884]) due to an abnormally rapid ciliary beat frequency without a discernible structural defect11. These data highlight the molecular complexity underlying the formation and function of cilia2. Here, we report biallelic variants in LRRC56 which result in a disease entity within the group of mucociliary clearance disorders, possibly distinct from PCD. It is associated with bronchiectasis and laterality defects. In humans, defective ciliary function was only detected by HSVA of cultured material. We show human LRRC56 interacts with the IFT subunit, IFT88. Functional studies using Trypanosoma brucei reveal that LRRC56 is recruited during axoneme construction in an IFT dependent manner, and is required for the addition of dynein arms to the distal segment of the flagellum. Our findings add LRRC56 to an expanding list of genes whose disruption results in an atypical ciliary phenotype and reveal a mechanism whereby disruption of LRRC56 leads to defective IFT-dependent targeting of dynein to cilia, and loss of ciliary motility. 6 MATERIAL AND METHODS Subject evaluation Two unrelated families were independently ascertained with features suggestive of a ciliopathy (Figure 1A). Family 1 consisted of a single female whose parents are first cousins of UK Pakistani ethnicity. She presented with chronic chest infections; nasal biopsy was obtained and respiratory epithelial cultures prepared. These were investigated by transmission electron microscopy (TEM) and high-speed video microscopy. Family 2 consisted of two affected individuals, the offspring of first cousin consanguineous parents from Kuwait, ascertained during pregnancy to have lethal congenital heart disease. Both pregnancies were terminated and post mortem pathological investigations were performed. The families provided signed informed consent to participate in studies approved by the Leeds East Research Ethics Committee ȋͲ͹Ȁ(ͳ͵Ͳ͸Ȁͳͳ͵Ǣ Family ͳȌ and the Review Ethics Board of the Childrenǯs (ospital of Eastern Ontario (11/04E; Family2). Review of clinical samples previously investigated by targeted next generation sequencing subsequently identified a further case with biallelic variants (Family 3.) Genetic Analysis In view of the consanguinity in both families, genetic analysis was performed under an autosomal recessive model. Whole exome sequencing of genomic DNA was performed at the University of Leeds and the Childrenǯs (ospital of Eastern OntarioǤ Target enrichment was performedǡ following manufacturerǯs protocolsǡ using SureSelect hybridization capture reagents with v5 and v4 Human All Exon probes for Family 1 and 2, respectively (Agilent Technologies). Enriched library preparations were sequenced on either HiSeq 2000/2500 platforms (Illumina) using paired end 100-bp reads. 7 Bioinformatic data processing was performed as previously described, with all variants being reported against human reference genome build hg1912,13. Genomic regions corresponding to runs of autozygosity were identified from pipeline-produced variant call format (VCF) files using the tool AgileMultiIdeogram (see Web Resources). These were subsequently used to filter Alamut Batch-annotated variant reports. Additional filtering criteria included the exclusion of common variants (those with a minor allele frequency ηͳΨȌ represented in the N(LB) Exome Variant Server (EVS) or an in-house cohort of >1,500 control exomes and the exclusion of genes with biallelic functional variants reported to the Exome Aggregation Consortium (ExAC). Pathogenic variants and segregation in the families were confirmed using Sanger sequencing with an ABI3130xl. Primer sequences and thermocycling conditions are available upon request. RNA splicing assay Total RNA was extracted from peripheral blood of the affected proband in Family 1 using the QIAamp RNA blood mini kit (Qiagen). A gene-specific primer spanning the boundary of LRRC56 exons 10 and 11 (NM_198075.3) (dCTTGGCCAGCACCATGGGTGAG) was used to perform first-strand cDNA synthesis with a SuperScriptTM II RT kit (Life Technologies). Two PCR amplicons were generated from the resulting cDNA using an exon 6 forward primer (dCAACCTGGACCAACTGAAGC) combined with either an exon 8 (dCCTCCAGGGTGAGCATGG) or exon 10 (dCCAGGTCCTCAGAAAGCAGG) reverse primer. PCR products were used to create Illumina-compatible sequencing libraries with NEBNext® UltraTM reagents (New England Biolabs) and sequenced on an Illumina MiSeq using a paired-end 150bp read length configuration. Co-IP experiments 8 Human LRRC56 was amplified from an untagged image clone (SC123392, Origene Technologies), inserted into the pDONR201 Gateway cloning vector (Invitrogen) and subsequently pDEST40 destination vector according to the manufacturerǯs instructionsǤ The human IFT88-eYFP construct was provided by Professor Colin Johnson14. Both plasmids were sequenced and maxi-prepped prior to transfection. HEK293 cells were co-transfected with 1.5µg of each plasmid using Lipofectamine 2000 (Invitrogen). Cells were lysed 48 h post transfection,using NP40 cell lysis buffer, containing 1x protease and phosphatase inhibitors, and protein extracted as per standard protocols. Immunoprecipitation (IP) of eYFP was performed with 1000 µg of both transfected and untransfected cell extracts using the GFP-Trap®_M kit (Chromotek) according to the manufacturerǯs instructionsǤ Whole cell extracts ȋWCEȌ and immunoprecipitates ȋ)PȌ were blotted using mouse anti-V5 (Invitrogen, 1/2,000) or mouse anti-GFP that detects YFP (Invitrogen, 1/1,000). A secondary goat anti-mouse HRP (Dako) was used at 1/10,000 and detected using the SuperSignal West Femto kit (Thermo Fisher Scientific). Ciliary function measurements In IV:1, Family 1, ciliary beat frequency (CBF) was measured using a digital high-speed video imaging system, as described previously15,16. Ciliary beat pattern was investigated in three different planes: a sideways profile, beating directly toward the observer and from directly above. Dyskinesia was defined as an abnormal beat pattern that included reduced beat amplitude, stiff beat pattern, flickering motion or a twitching motion. Ciliary beat pattern is associated with specific ultrastructural defects in primary ciliary dyskinesia. Dyskinesia index was calculated as the percentage of dyskinetic cilia within the sample (number of dyskinetic readings/total number of readings for sample ×100). All measurements were taken with the solution temperature between 36.5 and 37.5° C 9 and the pH between 7.35 and 7.45. Normal ranges for CBF and percentage dyskinetic cells are 8-17Hz and 4-49% respectively as previously reported17. Air liquid interface cell culture The modified method we used has previously been described18. Briefly, nasal brush biopsy samples were grown on collagen (0.1%, Vitrogen, Netherlands) coated tissue culture trays (12 well) in Bronchial Epithelial Growth Media (BEGM, Lonza, USA) for 2-7 days. The confluent unciliated basal cells were expanded into collagen-coated 80 cm2 flasks and the BEGM was replaced every 2-3 days. The basal cells were then seeded at approximately 1-3 x 105 cells per cm2 on a collagen coated 12mm diameter transwell clear insert (Costar, Corning, USA) under BEGM for 2 days. After confluency, the basal cell monolayer was fed on the basolateral side only with ALI-media (50% BEGM and 50% Hi-glucose minimal essential medium containing 100 nM retinoic acid). The media was exchanged every 2-3 days and the apical surface liquid was removed by gentle washing with phosphate buffered saline when required. When cilia were observed on the cultures they were physically removed from the transwell insert by gentle scraping with a spatula and washing with 1ml of HEPES (20 mM) buffered medium 199 containing penicillin (50 µg/ml), streptomycin (50 µg/ml) and Fungizone (1 µg/ml). The recovered ciliated epithelium was then dissociated by gentle pipetting and 100µl of the cell suspension was examined inside a chamber slide and ciliary beat frequency and pattern assessed as described above 15,16. The remaining 900 µl was fixed in 4% glutaraldehyde for transmission electron microscopy (TEM) analysis of cilia and ciliated epithelium. TEM of Human Samples 10 TEM was performed as previously described19, using a Jeol 1200 instrument. For TEM, the ciliated cultures were fixed with glutaraldehyde (4%w/v) in Sorensenǯs phosphate buffered (pH 7.4). After post-fixation in osmium tetroxide (1%w/v), samples were dehydrated through graded ethanol series and immersed in hexamethyldisilazane. Sections were cut at 90nm, with cross-sections categorised for height in the cilium using histological parameters. The bottom (2µm) of the axoneme cross-sections were located because they were associated with microvilli. The middle (2µm) cross sections were identified by wide cross-sections with the outer dynein (ODA) away from the axoneme membrane (with no microvilli present). The top 1um of the cilia was represented by cross-sections which have a slightly smaller diameter compared to the middle sections. In addition, the axonemal membrane was tightly wrapped to the microtubules and the ODA. The tips (0.6µm) appear as thin cross sections and have no dynein arms. Trypanosome cell culture All cloned cell lines used for this work were derivatives of T. brucei strain Lister 427 and were cultured in SDM79 medium supplemented with hemin and 10 % fetal calf serum20. Cell lines IFT88RNAi (targeting an essential protein for anterograde IFT)21, IFT140RNAi (essential protein for retrograde IFT)22, DNAI1RNAi (component of the dynein arm)23, and ODA7RNAi (cytoplasmic assembly machinery of the dynein arm)24 have previously been described. They all express double-stranded RNA under the control of tetracycline-inducible T7 promoters25,26. Expression of endogenous LRRC56 fused to YFP 11 Endogenous tagging of the T. brucei Tb427.10.15260LRRC56 was carried out by direct integration into LRRC56 using the p2675TbLRRC56 plasmid. This vector is derived from the p2675 plasmid and contains a 1495bp fragment of the trypanosome LRRC56 sequence (1-1495) downstream of YFP27. Transfection was achieved by nucleofection of T. brucei cells using program X-014 of the AMAXA Nucleofector® apparatus (Lonza) as previously described 28, with 10 µg plasmid linearized with AfeI in the target LRRC56 sequence, for homologous recombination with the target allele. The mutant allele was obtained following T776/C base substitution (Genecust Europe, Luxembourg) to substitute leucine 259 for proline in the p2675TbLRRC56. The resulting p2675TbLRRC56L259P plasmid was linearized with NheI prior to nucleofection. As a result, the expression of the YFP fluorescent fusion protein is under the control of the endogenous ͵Ԣ untranslated region of the LRRC56 resulting in a similar expression level as the wild-type allele since most gene expression regulation is determined by ͵ԢUTR sequences29. Transgenic cell lines were obtained following puromycin selection and cloning by serial dilution. LRRC56 deletion or replacement Sequential LRRC56 replacement in T. brucei was used to obtain lrrc56-/- cells. Sequences containing either the puromycin (PURO) or the blasticidin (BLA) drug resistance gene flanked by 300 bp long upstream and downstream regions of the LRRC56 were synthesized and cloned in a pUC57 plasmid (Genecust Europe, Luxembourg). Amplicons were generated by PCR with primers P1: dTTTGAAGGTGCTGTGTGAGGG and P2 dAGGTAGAGGGAGGCGTTGAG (Eurogentec) which anneal to the sequences 300 bp upstream of the LRRC56 start codon and downstream of LRRC56 stop codon respectively. Prior to nucleofection PCR fragments 12 containing either the blasticidin (BLA) resistance gene or the G418 neomycin (NEO) resistance gene were purified using Nucleospin gel and PCR clean-up kit (Macherey Nagel). Single allele knockout cells were obtained following nucleofection with the BLA amplicon containing LRRC56 flanking sequences, blasticidin selection and cloning. LRRC56 allele deletion was confirmed using PCR with ͷԢUTR LRRC56-specific primer P3 dTCACCATCACGCCCTTTTGT and BLA-specific primer P4 dCTGGCGACGCTGTAGTCTT. Replacement of the remaining LRRC56 allele was performed with either the YFP::LRRC56 or the mutated YFP::LRRC56L259P in this single knockout cell line using linearized p2675TbLRRC56 or p2675TbLRRC56L259P plasmid and puromycin selection as described above. Double LRRC56 knockout was achieved following nucleofection of the cell line with a single allele knockout carrying the mutated YFP::LRRC56L259P with NEO amplicon containing LRRC56 flanking sequences and validation obtained following PCR analysis of G418-resistant cells with LRRC56-specific primers P5 dCCGTAGCATCATCCGAGACC and P6 dACTATTTGCGTCAGGTGGCA. Primers amplifying a 1511-bp region of the unrelated aquaporin 1 gene served as positive controls. For whole-genome sequencing, genomic DNA was extracted using the Qiagen DNeasy kit. Short insert Illumina sequencing libraries were constructed and sequenced at the Beijing Genomics Institute, generating approximately 6.6 million 100-bp paired end reads. Reads were aligned to the T. brucei TREU927 genome (TriTrypDB release 35)30 using bowtie2 in very-sensitive-local alignment mode, with an 88.6% alignment rate30,31. Alignment files were sorted, merged and indexed using SAMtools32. Aligned reads were visualised and analysed, including counting reads per CDS using the Artemis pathogen sequence browser33. Indirect immunofluorescence assay (IFA) 13 Cultured trypanosomes were washed twice in SDM79 medium without serum and spread directly on poly-L-lysine coated slides (Thermoscientific, Menzel-Gläser) before fixation. For methanol fixation, cells were air-dried and fixed in methanol at -20°C for 5 min followed by a rehydration step in PBS for 15 min. For paraformaldehyde-methanol fixation, cells were left for 10 min to settle on poly-L-lysine coated slides. Adhered cells were washed briefly in PBS before being incubated for 5 min at room temperature with a 4% PFA solution in PBS at pH 7 and fixed with methanol at -20°C for an additional 5 min followed by a rehydration step in PBS for 15 min. For immunodetection, slides were incubated for 1 h at 37°C with the appropriate dilution of the first antibody in 0.1% BSA in PBS; mAb25 recognises the axonemal protein TbSAXO134; a monoclonal antibody against the IFT-B protein IFT17222, and a mouse polyclonal antiserum against DNAI1 24. YFP::LRRC56 was observed upon fixation by immunofluorescence using a 1:500 dilution of a rabbit anti-GFP antibody that detects YFP (Life Technologies). After several 5 min washes in PBS, species and subclass-specific secondary antibodies coupled to the appropriate fluorochrome (Alexa 488, Cy3 or Cy5, Jackson ImmunoResearch) were diluted 1/400 in PBS containing 0.1% BSA and were applied for 1 h at 37°C. After washing in PBS as indicated above, cells were stained with a 1 µg/ml solution of the DNA-dye DAPI (Roche) and mounted with Slowfade antifade reagent (Invitrogen). Slides were either stored at -20°C or immediately observed with a DMI4000 microscope (Leica) with a 100X objective (NA 1.4) using a Hamamatsu ORCA-03G camera with an EL6000 (Leica) as light source. Image acquisition was performed using Micro-manager software and images were analysed using ImageJ (National Institutes of Health, Bethesda, MD) Motility analyses 14 Volumes of 250µl at 5x106 trypanosomes/ml in warm medium were distributed in 24 well plates. Samples were analysed under 10X objective of a DM IL LED microscope (Leica) coupled to a DFC3000G camera (Leica). Movies were recorded (200 frames, 50 ms of exposure) using LASX Leica software, converted to .avi files and analysed with the medeaLAB CASA Tracking v.5.5 software (medea AV GmbH) 35. TEM analysis of trypanosome samples Cells were fixed with 2.5% glutaraldehyde directly in the suspension culture for 10 min and then treated with 4% paraformaldehyde, 2.5% glutaraldehyde in 0.1M cacodylate buffer (pH 7.2) for 1h. Cells were post-fixed for 30 min (in the dark) in 1 % osmium tetroxide (OsO4), in 0.1 M cacodylate buffer (pH 7.2), washed and incubated in 2% uranyl acetate for 1h at room temperature. Samples were washed, dehydrated in a series of acetone solutions of ascending concentrations, and embedded in Polybed 812 resin. Cytoskeletons were extracted by treating cells with Nonidet 1% with protease inhibitor cocktail (Sigma, P8340) diluted in PBS for 20 min, washed in PHEM 0.1M pH7.2 for 10min. Fixation was performed in 2.5% glutaraldehyde, 4% paraformaldehyde, and 0.5% tannic acid in 0.1M cacodylate buffer (pH 7.2). Cells were post-fixed, incubated in uranyl acetate, dehydrated and infiltrated as cited above. Ultrathin sections were stained with uranyl acetate and lead citrate. 15 RESULTS Clinical characterization of families Individual IV:1, Family 1 was born at term by normal vaginal delivery, there was no family history of note. After 36 hours, she developed tachypnoea, and was nursed for 48 hours in 40% humidified Oxygen environment . She was discharged home on day 5 of life. At age 3 months, she developed rhinorrheoa and experienced several episodes of documented respiratory tract infection, the first at age 5 months. She subsequently developed a chronic cough and recurrent middle ear infections from age 8 months. At age 18 months, she developed pneumonia. Chest X-ray and CT scan revealed bronchiectasis and dextrocardia (Figure 1B), Nasal nitric oxide levels measured under conditions of tidal breathing averaged 6nl/Min. The clinical presentation and investigations suggested a diagnosis of PCD as outlined in the European Respiratory Society guidelines for the diagnosis of PCD9 However, TEM analysis of nasal biopsy samples along different segments of the cilia (tip, middle and base) showed apparently intact dynein arms (Figure 1C). Direct high-speed videomicroscopy analysis of biopsy material revealed a ciliary beat frequency of 10.87 Hz (10.4 Hz-11.34 Hz) and particulate clearance was observed (Supplementary Video 1 and 2). Although cell culture at an air-liquid interface produced a healthy ciliated epithelium, the ciliary beat pattern was in fact dyskinetic (19% twitching, 33% stiff, 48% static), with a beat frequency of 5.39 Hz (95% CI 4.29 Hz-6.49 Hz, normal range > 11Hz36 ) (Supplementary Video 3 and 4, and control Supplementary video 5). TEM revealed no obvious structural defect along the shaft of the cilia. The findings were not compatible with the ERS guidelines on the diagnosis of PCD9. However the combination of chronic lung infections, middle ear infection, and cardiac laterality defect suggested an underlying disorder of motile cilia. 16 Family 2 were from Kuwait and included two affected fetuses. Both pregnancies were terminated because the fetuses were found to have lethal congenital heart disease. Autopsy revealed similar findings in both individuals, which included situs inversus of the thoracic and abdominal organs, with a complex congenital heart malformation characterized by double outlet right ventricle (data not shown). Fetus 1 showed atrial situs inversus. There was a persistent left sided superior vena cava, that along with the inferior vena cava, drained to the left sided (morphologically right) atrium. There was also a right sided superior vena cava that drained to the right sided (morphologically left) atrium. There was a large dominant right sided ventricle showing right morphology; no definite, even rudimentary, left ventricle was identified. The main outflow tract of the large dominant ventricle was to the aorta. There was pulmonic hypoplasia. The aortic arch was right sided. Fetus two showed almost the exact same cardiac phenotype except that a hypoplastic left ventricle was identified, likely because there was a high muscular ventricular septal defect (VSD) through which it could decompress. Thus, atrial situs was inversus but ventricular situs was solitus, with atrioventricular discordance The hypoplastic ventricle had no outflow other than the VSD. In addition fetus two did not have a right sided superior vena cava. The third case consisted of a single individual clinically investigated by NGS. Individual II:1 family 3 is 27 years of age, he was the product of a normal pregnancy born to unrelated UK parents. Only brief clinical details are available. He was found to have situs inversus soon after birth and suffered neonatal respiratory disease. During the first year of life he developed a chronic cough and recurrent lower respiratory tract infections (LRTI), he also suffered recurrent middle ear disease. These symptoms have persisted into adult life. The combination of laterality defects, recurrent LRTI and 17 middle ear disease led to further investigation, including nasal ciliary biopsies. The first biopsy at age 11 months reported normal ciliary structure, although at that time (1992) no motility studies were performed. Subsequent investigation at age 26 revealed a normal NO estimation (384ppb), a further ciliary biopsy was obtained at this age. Investigation revealed normal ciliary beat pattern with good particulate clearance. CBF was in the normal range at 13.4Hz. EM revealed normal dynein arms and microtubules with no evidence if ciliary disorientation. Cilia length was normal at 5.6microns. The clinical report concluded that this was not consistent with a diagnosis of PCD. However the ciliary phenotype was very similar to that seen in family 1. This individual was subsequently investigated by targeted next generation sequencing. Genetic analyses revealed mutations in LRRC56 Autozygosity mapping identified 46 and 33 regions of homozygosity in each proband from Family 1 and 2 respectively (Table S1). In neither family did we identify disease-causing variants in known PCD and mucociliary clearance disorder genes (Table S2). Assessment of variant pathogenicity was evaluated according to ACMG best practice guidelines37. Variant filtering, using the autozygous intervals, and public/in-house databases (to eliminate variants with minor allele frequency > 1%) identified a single homozygous variant in each family, located in the same gene, LRRC56 (NM_198075.3: Family 1 c.423+1G>A; Family 2 c.419T>C, p.(Leu140Pro)). Neither variant is recorded in ExAC. A search of an ethnically matched control exome cohort revealed a single heterozygous carrier of the c.423+1G>A mutation among 1,541 normal subjects. Sanger sequencing confirmed the variants as well as segregation in both families. c.423+1G>A is predicted 18 to abrogate the intron 7 donor site; a high-throughput sequencing analysis of RT-PCR products was designed to quantify the proportion of correctly or incorrectly spliced transcripts as determined by the presence of the ĞǆŽŶ ϴ ͞GA͟ ŶƵĐůĞŽƚŝĚĞƐ and consequently confirmed that the LRRC56 variant is aberrantly spliced and not predicted to encode a functional protein (Figure S1). The missense variant c.419T>C, p.(Leu140Pro) affects a highly conserved residue (Figure S2), predicted to be deleterious by SIFT (0)38, Polyphen (1.0)39, and CADD (24.7) scores40. NGS investigation using a targeted reagent consisting of 32 genes revealed that indvidual II:1, family 3 was heterozygous for c.760G>T p.(Glu254Ter), and the splicing variant c.326+1G>A. c.760G>T is found at a frequency of 5/140890 in the GnomAD database, c.326+1G>A is not recorded on GnomAD. LRRC56 protein is conserved in eukaryotes with motile cilia constructed by intraflagellar transport (IFT) Human LRRC56 encodes a 542 amino acid protein determined from transcriptomic studies and immunohistochemistry to be expressed in many organs, mostly in testis and pituitary gland41 although significant number of tags were also detected in other organs including lungs and respiratory epithelial cells (The Human Protein Atlas)41. Interestingly, data from single-cell RNA sequencing indicated that LRRC56 transcription in lungs is restricted to ciliated cells42. The protein contains 5 leucine-rich repeat domains conserved across species whose motile cilia are assembled by intraflagellar transport (IFT)43. LRRC56 is the human ortholog of the Chlamydomonas reinhardtii gene oda8, which is thought to play a role in the maturation and/or transport of outer dynein arm (ODA )complexes during flagellum assembly, and has a biochemical distribution similar to IFT proteins43. ODA8 is proposed to function together with two 19 other proteins termed ODA5 and ODA10 to form an accessory complex involved in assembly and transport of ODA during axoneme assembly43. However, evidence for a physical interaction is lacking, and the exact role of ODA8 remains to be defined. To evaluate a possible association of LRRC56 with the multi-subunit IFT machinery, HEK293 cells were co-transfected with plasmids encoding human LRRC56 tagged with V5 and the reference IFT protein IFT88 fused to eYFP14. Immunoprecipitation using an anti-GFP antibody co-precipitated both IFT88-eYFP and LRRC56-V5, supporting an association of LRRC56 with the IFT machinery (Figure 1D) The functional role of LRRC56 was further investigated in T. brucei, an organism that possesses a 9+2 axoneme and is amenable to genetic manipulation24,44. The T. brucei LRRC56 ortholog (Tb427.10.15260) comprises 751 amino acids and shares 42% sequence identity in the conserved leucine-rich repeat (LRR) region (Figure S2). Since T. brucei maintains its mature flagellum during formation of the new one, it is possible to monitor both maintenance and assembly in the same cell 45. YFP-tagged LRRC56 (YFP::LRCC56) was expressed in cells upon endogenous tagging and the protein was detected in the distal portion of the new flagellum, whereas the old flagellum lacked the protein (Figure 2A). Following division, cells with one flagellum displayed two different profiles: either a strong signal towards the distal end of the flagellum (Figure 2B) or no signal (not shown). The first scenario reflects daughter cells that inherited a new flagellum, while the latter reflects those daughter cells that inherited an old flagellum. This observation is consistent with LRRC56 being recruited during flagellar assembly before its removal during flagellum maturation, after cell division. In cells with a single flagellum, the LRRC56 protein is present on 2 distinct 20 parallel tracks (green signal on merged image) only in the distal flagellar portion of the axoneme (in blue) and colocalize partially with IFT proteins also found on these 2 tracks but all along the flagellum (red signal on merged image). IFT proteins (red) are clearly detected where no green signal is present46. Co-localisation with the IFT protein IFT172, but not with the axoneme marker mAb25 suggests that LRRC56 associates with IFT and not dynein arms. Many, but not all, IFT trains contained LRRC56 (Figure 2), suggesting it is a cargo rather than a core component. The LRRC56 signal is lost upon detergent extraction of the cytoskeleton, which strips the membrane and IFT particles but not the dynein arms (data not shown), further supporting association with IFT. To further investigate the link with IFT, the distribution of YFP::LRRC56 was studied in trypanosome mutant cell lines in which tetracycline-inducible RNAi knockdown of an IFT-B (IFT88RNAi) or an IFT-A (IFT140RNAi) member results in either defective anterograde or retrograde transport, respectively21,22. In IFT88RNAi cells, the flagellar YFP::LRRC56 signal disappeared when anterograde transport was disrupted and the protein was found predominantly in the cytoplasm (Figure S3A and S3B). In the retrograde mutant IFT140RNAi, trains travel into the new flagellum but fail to be recycled to the base22. The YFP::LRRC56 distribution pattern turned out to be more complex (Figure S3C and S3D) and required more detailed investigation. Cultures are not synchronised and RNAi knockdown can impact cells at different stages of flagellum construction. If this happened when flagella started to grow, IFT proteins accumulated in very short flagella, but LRRC56 was rarely detected (Figure 2C). However, when IFT arrest took place at later stages of elongation, LRRC56 was frequently found in high concentration always associated with stalled IFT trains (Figure 2C). The presence of such an amount of IFT material in a spatially constrained area is never seen in control cell lines. 21 Formal demonstration of stalled IFT material in the IFT140RNAi cell line has been published using live cell analysis47 These results reflect the association of LRRC56 as cargo of IFT trains, with a progressive increase during flagellar construction. After 2 days of tetracycline-inducible RNAi knockdown, the LRRC56 is no longer detected in the flagellum (Figure S3C and S3D). LRRC56 localisation was unchanged in DNAI1RNAi and ODA7/DNAAF1 RNAi mutants that are defective in their dynein arm constitution or cytoplasmic preassembly, respectively (Figure S4). This shows that LRRC56 does not require the presence of dynein arms to be associated with flagella. Together, these findings reveal that LRRC56 likely performs an IFT-associated function in motile cilia. Absence or mutation of T. brucei LRRC56 is responsible for motility defects explained by absence of outer dynein arms in the distal segment of the axoneme We next chose to assess any impact caused by the absence, or mutation of LRRC56. An LRCC56 null mutant was generated by double gene knockout (Figure S5A). Given the small size of the T.brucei genome (35Mb), whole-genome sequencing was performed and unambiguously demonstrated that LRRC56 had been replaced by the drug resistance cassettes (Figure S5B). Not a single LRRC56 read was detected whereas hundreds were found in control cells. Visual examination showed a significant reduction in flagellar beating and cell swimming, with ~10% of cells struggling to complete cell division and remaining attached by their posterior extremities (Figure S5C). These phenotypes indicate slow cytokinesis and increased generation times which, in trypanosomes, is typical of motility defects (Figure S5D).23,48,49 Microscopy revealed that lrrc56-/- cell motility is characterised by an erratic swimming pattern 22 with altered propagation of the tip-to-base wave, increased frequency of base-to-tip waves, and frequent tumbling typical of outer dynein arm mutants (Figure 3) 23,50. Examination of the axoneme structure by TEM using cytoskeletons extracted with detergent, a routine procedure to analyse the fine structure of the axoneme51 revealed that more than one third of sections analysed from lrrc56-/- cells showed absence of 6 to 9 ODAs (Figure 4). However, information on the positioning of ODAs along the length of the flagellum is not available from this assay. Hence a second analysis was performed using whole cell samples directly fixed in suspension52. This gives lower quality images for the fine structural details of the axoneme, but the position of the flagellum along the cell body can be determined as described53 . Briefly, sections close to the proximal part of the flagellum are found at the posterior end where the cell body is larger and those close to the distal end are towards the anterior portion of the cell body that is much thinner. The distal tip of the flagellum is not attached and so sections there are easy to identify. A total of 33 sections were obtained in these conditions. When they were in the proximal region, only 20% of the images indicated a defect in outer dynein arms. By contrast, this proportion increased to 75% and 88% when sections were in the distal region or at the tip of the flagellum, respectively. This was found in the distal portions of both old and new flagella. This suggests a role for LRRC56 in the assembly of dynein arms in the distal portion of the axoneme. Immunostaining with an anti-DNAI1 (dynein arm intermediate chain 1 also known as IC78) antibody23,24 that recognises an essential structural component of the ODA, confirmed that only the proximal axonemal region stained positively in lrrc56 -/- cells (Figure 4H). Control cells stained from the base to the tip of all axonemes (Figure 4G), confirming that LRRC56 is essential for distal assembly of ODA. Compared to control cells (Figure 4G), LRRC56 null mutant cells also appear to have shorter cilia (Figure 4H). Indeed, a small but significant flagellum length 23 size reduction of about 3.8 µm as determined by IFA using mAb25 antibody was observed (data not shown). Although we donǯt have clear explanation for this size reduction, this phenotype is unlikely to impinge upon our conclusion about distal loss of ODA, since a mutant defective for IFT kinesin motors assemble short flagella that still harbour unaltered ODA all along the flagellum46. We next evaluated in trypanosomes the impact of the p.Leu259Pro substitution which corresponds to the p.Leu140Pro observed in Family 2. One LRRC56 allele was deleted and the other allele was replaced endogenously with a L259P modification in YFP-tagged LRRC56 (YFP::LRRC56L259P cells). The YFP::LRRC56L259P protein localises normally to the distal portion of the new flagellum, showing that this residue is not required for proper expression and localisation of LRRC56 (Figure S6). However, cells showed reduced motility (Figure 3B), albeit not to the same extent as the double knockout (Figure 3C,D). Ultra-structural analysis of YFP::LRRC56L259P cells by TEM revealed a reduction in the number of ODAs, but that was less frequent compared to lrrc56-/- flagella (Figure 4B,C,F). We conclude that the protein carrying the substitution can associate with IFT trains, but functions less efficiently that its normal counterpart in supporting dynein arm assembly. 24 DISCUSSION Whole exome sequencing in two unrelated consanguineous families enabled us to identify homozygous variants in LRRC56. The common clinical phenotype in the two families was cardiac laterality defects, while in Family 1 the affected individual also presented with recurrent pulmonary infections, and on investigation was found to have bronchiectasis. This combination of clinical features was suggestive of PCD, but a uniformly dyskinetic beat pattern was not observed. Investigation of cultured nasal epithelial cells from the affected individual in family 1 revealed a dyskinetic ciliary beat pattern, but no structural ciliary defects. We have observed this phenomenon before in our analysis of samples sent for the investigation of PCD (R.Hirst, unpublished data). In Family 1, individual IV:1 with a homozygous splicing variant in LRRC56, further analysis confirmed that the mutated transcript is aberrantly spliced and is not predicted to encode a functional protein. In Family 2, a homozygous missense variant was identified. Functional investigation demonstrated a role for LRRC56 in the assembly of ODA in the protist T. brucei. The affected individual in family 3 displayed a combination of laterality defects, recurrent LRTI and middle ear disease. The respiratory cellular phenotype observed on his nasal ciliary biopsies recalled that seen in individual IV:1, Family 1. This individual was found to be a compound heterozygote for variants predicted to result in absence of functional LRRC56. We then modelled the effect of a null variant and the homozygous missense variant in this organism. We demonstrate that absence of LRRC56 or the presence of the homozygous missense variant seen in Family 2, both result in motility defects, caused by loss of ODAs in the distal segment of the axoneme. This observation is consistent with our interrogation of the role of LRRC56 in this organism. 25 Previous studies have shown that LRRC56 is found only in species with motile cilia that rely on IFT for axonemal assembly43. LRRC56 shows considerable species variability, the human protein is only 61% identical with its mouse counterpart, compared with 86% for DNAI1 for instance. We have also noted that LRRC56 is nore divergent between close species of trypanosomes than IFT proteins or dynein arm components. The reported biochemical distribution of LRRC56 is similar to that of IFT subunits, with about 50% of the protein associated with the membrane and matrix fraction43. Our pull-down experiments carried out in human cells expressing tagged LRRC56 and IFT88 revealed that LRRC56 interacts with the IFT machinery. In T. brucei, LRRC56 is recruited to the flagellar compartment at advanced stages of construction and co-localises with many (but not all) IFT trains. This flagellar localisation is dependent on IFT, as confirmed by analysis of cell lines defective in either anterograde or retrograde IFT. These results support a model in which LRRC56 associates with IFT trains and may function as a cargo adaptor to transport dynein arms towards the distal end of cilia and flagella. The absence of LRRC56 has an impact on ciliary motility, as observed here in cultured cells from one affected individual and in T. brucei. This is consistent with the oda8 (LRRC56 homologue) null mutant in Chlamydomonas, although the impact on the presence of dynein arms was remarkably variable. ODAs are absent throughout the length of the flagellum in Chlamydomonas 54 whereas they are missing at only the distal part of the T. brucei flagellum. In both cases, this loss of ODA interferes with proper initiation of flagellum beating resulting in reduced motility and dyskinetic flagellar beating. In samples from the only person that was available for analysis, sections were analysed in different segments of the cilia (proximal, intermediate, distal), but dynein 26 arms were not visibly affected. Although the central core of the protein containing the leucine-rich repeat domains is well conserved, other segments are variable across species, with large and often unique N- and C-terminal extensions. This may translate into a common central function, such as association of LRRC56 with the IFT machinery for dynein arm transport, but could also result in considerable structural variation, potentially supporting species-specific interaction partners. Indeed, the composition of dynein arms (number and type of heavy, intermediate and light dynein chains) varies between species, as demonstrated by both biochemical and genetic analysis48. The outer dynein arm composition differs between the Chlamydomonas ODAs that contain 3 heavy dynein chains chains (,  and ), and the human and trypanosome ODA which each contain only 2 heavy chains ( and )48. In Chlamydomonas, early flagellum growth is very rapid (up to 10 µm per hour)55, and might necessitate a greater contribution from ODA8 to ODA transport compared to the slower growth rate observed in trypanosomes56, and respiratory epithelial cells 57. This could explain the absence of ODA throughout the axoneme compared to only the distal part in trypanosomes. We cannot exclude the possibility that discrete structural defects of dynein arms occur in the individual whose cilia were analysed. For example, DNAH11 nonsense mutations are associated with subtle dynein arm modifications that can only be detected by advanced TEM imaging and tomography 11. In this regard, approximately 30% of PCD individuals showing ciliary dysfunction have no or very subtle ciliary ultrastructure abnormalities, when investigated with standard transmission electron microscopy58 27 Overall, our findings underline the evolutionary complexity of outer dynein arms assembly and trafficking. We have shown that bi-allelic mutations in LRRC56 are responsible for laterality defect in three unrelated families. We provide evidence of a disorder of mucociliary clearance with accompanying cardiac laterality defects caused by mutations in a gene encoding a protein required for ODA transport, rather than composition or assembly. Our work expands the list of ciliary genes involved in human disorders59, while also providing insight into the role of LRRC56 and cilia biology in human development. 28 SUPPLEMENTAL DATA Supplementary data consists of six figures, 5 videos and three tables CONFLICT OF INTEREST The authors declare no competing interests. ACKNOWLEDGEMENTS We wish to thank the families reported here for their willingness to participate in these research efforts. We thank the Ultrastructural BioImaging Plateforme for providing access to the TEM equipment. We thank Bruno Louis and Jean-François Papon, Institut National de la Santé et de la Recherche Médicale, U955, Créteil, France, for help in video-microscopy. Research at the Institut Pasteur is funded by an ANR grant (ANR-14-CE35-0009-ͲͳȌǡ by a French Government )nvestissement dǯAvenir programmeǡ Laboratoire dǯExcellence ǲ)ntegrative Biology of Emerging )nfectious Diseasesǳ ȋANR-10-LABX-62-IBEID) and by La Fondation pour la Recherche Médicale (Equipe FRM DEQ20150734356). This work was supported, in part, by the Care4Rare Canada Consortium funded by Genome Canada, the Canadian Institutes of Health Research (CIHR), the Ontario Genomics Institute, Ontario Research Fund, Genome Quebec, and Childrenǯs (ospital of Eastern Ontario Foundation and the Canadian Rare Diseasesǣ Models and Mechanisms Network funded by CIHR and Genome Canada. The authors wish to acknowledge the contribution of the high-throughput sequencing platform of the McGill University and Génome Québec Innovation Centre, Montréal, Canada. Research at the University of Leeds was supported by grants MR/M009084/1 and MR/L01629X/1 awarded by the UK Medical Research Council. 29 WEB RESOURCES AgileMultiIdeogram, Burrows-Wheeler Aligner, CASA Tracking V.5.5, CLUSTAL Omega, dbSNP, Exome Aggregation Consortium (ExAC) Browser, broadinstitute.org/ Genome Analysis Toolkit (GATK), org/gatk/ The Human Protein Atlas (LRRC56), ImageJ, NHLBI Exome Variant server, OMIM, Picard, Protein BLAST, SAMtools, TriTrypDB: The Kinetoplastid Genomics Resource, SIFT, POLYPHEN2, CADD, 30 REFERENCES 1. Ralston, K.S., Lerner, A.G., Diener, D.R., and Hill, K.L. (2006). Flagellar motility contributes to cytokinesis in Trypanosoma brucei and is modulated by an evolutionarily conserved dynein regulatory system. Eukaryotic Cell 5, 696–711. 2. Mitchison, H.M., and Valente, E.M. (2017). Motile and non-motile cilia in human pathology: from function to phenotypes. J. Pathol. 241, 294–309. 3. Ahmed, N.T., Gao, C., Lucker, B.F., Cole, D.G., and Mitchell, D.R. 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Loss-of-function mutations in the human ortholog of Chlamydomonas reinhardtii ODA7 disrupt dynein arm assembly and cause primary ciliary dyskinesia. Am. J. Hum. Genet. 85, 890–896. 25. Wang, Z., Morris, J.C., Drew, M.E., and Englund, P.T. (2000). Inhibition of Trypanosoma brucei gene expression by RNA interference using an integratable vector with opposing T7 promoters. J. Biol. Chem. 275, 40174–40179. 32 26. Wirtz, E., Leal, S., Ochatt, C., and Cross, G.A. (1999). A tightly regulated inducible expression system for conditional gene knock-outs and dominant-negative genetics in Trypanosoma brucei. Mol. Biochem. Parasitol. 99, 89–101. 27. Kelly, S., Reed, J., Kramer, S., Ellis, L., Webb, H., Sunter, J., Salje, J., Marinsek, N., Gull, K., Wickstead, B., et al. (2007). Functional genomics in Trypanosoma brucei: a collection of vectors for the expression of tagged proteins from endogenous and ectopic gene loci. Mol. Biochem. Parasitol. 154, 103–109. 28. 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Value of transmission electron microscopy for primary ciliary dyskinesia diagnosis in the era of molecular medicine: Genetic defects with normal and non-diagnostic ciliary ultrastructure. Ultrastruct Pathol 41, 373–385. 59. Reiter, J.F., and Leroux, M.R. (2017). Genes and molecular pathways underpinning ciliopathies. Nat. Rev. Mol. Cell Biol. 18, 533–547. 60. Ishikawa, T. (2017). Axoneme Structure from Motile Cilia. Cold Spring Harb Perspect Biol 9,. 35 FIGURES Figure 1. LRRC56 mutations cause chronic infective lung disease and laterality defects (A) The homozygous splice-site mutation (c.423+1G>A, NM_198075.3) disrupts an invariant splice site in Family 1 individual IV:1. The unaffected sibling IV:2 is heterozygous for the mutation. The homozygous missense mutation (c.419T>C, NM_198075.3) was identified in the two affected siblings V:3 and V:4 from Family 2. Consistent with autosomal-recessive inheritance, the mutations described were detected in a heterozygous state in the unaffected parents (data not shown). The Third individual (Family 3 II:1) was a compound heterozygote for the variants c.760G>T and c.326+1G>A (NM_198075.3) (B) The Family 1 proband (IV:1) had dextrocardia, documented by chest X-ray (left panel). High-resolution axial computed tomography of the thorax in the same individual demonstrates mild bronchiectasis (red arrows) with adjacent inflammatory consolidation in the right lower lobe. Dextrocardia is also visible (CT-scan; right panel). (C) Cross section through the axoneme from cultured respiratory cells from Family 1 individual IV:1. The position of the section is indicated, bar = 100 nm. Normal axonemal structure is visible, with intact dynein arms. (D) Recombinant human LRRC56 and intraflagellar transport protein IFT88 interact in vitro. HEK293 cells were co-transfected with plasmids encoding human LRRC56 and IFT88 tagged with V5 and YFP respectively (1.5µg of each). After 48 hours, immunoprecipitation (IP) was performed with transfected and untransfected cells using Cell-TRAP magnetic beads bound to an anti-GFP antibody fragment. Protein from input whole cell extracts (WCE, left) and immunoprecipitated proteins (IP, right) were blotted using anti-V5 or anti-GFP. The IFT88-GFP fusion is a fairly large protein and in our 36 conditions, slight differences in migration are frequently observed. This does not impact on the underlying hypothesis being tested. A 廐-actin control is also shown. Figure 2. LRRC56 associates with IFT trains and not with the axoneme (A) An YFP::LRRC56 expressing cell that assembles its new flagellum (yellow arrow) shows staining (anti-GFP, green on merged images) in the new flagellum (blue arrowheads) that co-localises with the anti-IFT172 (red on merged images) but not with the anti-axoneme marker (mAb25, blue on merged images). No YFP staining is visible in the old flagellum (white arrow) whereas IFT172 positive trains are clearly present. IFT trains are predominantly found on microtubules doublets 4 and 7,22 hence the visual aspect of two separate tracks around the axoneme. DNA is stained with DAPI (cyan) showing the presence of two kinetoplasts (mitochondrial DNA) and two nuclei typical of cells at late stage of their cycle45. (B) Cytokinesis results in two daughter cells each containing a unique flagellum45. The one inheriting the new flagellum remains positive for YFP::LRRC56 that still shows association with IFT trains (same staining as in A). (C) The same immunofluorescence assay was performed on IFT140RNAi cells expressing YFP::LRRC56 after 24h in RNAi conditions. DNA staining shows that the top cell is mitotic and assembles a new flagellum. The IFT172 staining reveals the presence of a stalled IFT train that contains a considerable quantity of YFP::LRRC56 (arrowhead). The bottom cell is at a further stage of its cell cycle, yet its new flagellum is much shorter, indicating that RNAi knockdown occurred at a very early phase of construction (star). In these conditions, IFT172 staining shows that the very short flagellum contains a large amount of accumulated IFT material, yet no signal is visible for YFP::LRRC56 (star). 37 Figure 3. Absence of LRRC56 or expression of the L259P mutation reduces flagellum beating and cell motility (A-C) Tracking analysis35 showing the movement of individual trypanosomes in wild-type control (A), in cells expressing only YFP::LRRC56L259P (B) and in lrrc56-/- cells (C). Sustained motility is only observed in control conditions. (D) Quantification of the straight-line movement confirms the visual impression that motility was reduced in a statistically significant manner in YFP::LRRC56L259P cells and almost abolished in lrrc56-/-. Total number of cells, mean and standard deviation are indicated on the figure. Statistical analysis was performed using t test. Figure 4. LRRC56 is required for assembly of ODA in the distal portion of the axoneme (A-E) Sections are shown through the flagella of detergent-extracted cytoskeletons from various cell lines. Stripping the flagellum membrane and matrix facilitates the analysis of structures23,51,60. Sections through control YFP::LRRC56-expressing cells (A) possess all 9 ODA, whereas a mixture of sections with normal profiles or with several missing ODA (orange arrowheads) is encountered in YFP::LRRC56L259P-expressing cells (B-C) or lrrc56-/- cells (D-E). (F) Sections were grouped in three categories: defects in 2 ODA or less (blue), in 5 ODA or less (green) and in 6 ODA or more (red). The total counted number of sections is 50 for each sample. Full details are given in Table S3. (G-H) IFA with the anti-DNAI1 antibody stains the whole axoneme of wild-type cells (G) as expected24. However, the staining was limited to the proximal portion in lrrc56-/- cells in both growing (missing portion shown by yellow arrowheads) and mature flagella (white arrowheads) (H). 38 Supplementary Video 1: Nasal biopsy of ciliated epithelium taken from subject X shown in real time (recorded at 30 fps under 1000x magnification). Cilia can be seen clearly moving rapidly, which also moved the edge and overlying particles are cleared. This indicated normal ciliated activity. Supplementary Video 2: Nasal biopsy of ciliated epithelium taken from subject X shown in slow motion (recorded at 500 fps under 1000x magnification). Cilia can be seen clearly moving particles on the left hand side of the edge. This indicates that the cilia show normal function. Supplementary Video 3: Cultured cilia from patient X shown in real time (recorded at 30 fps under x1000 magnification). Here the cilia seem uncoordinated and no particulate clearance can be seen. This is indicated abnormal cilia movement. Supplementary Video 4: Cultured cilia from patient X shown in slow motion (recorded at 500fps under x1000 magnification). The cilia can be seen clearly and their movement showed uncoordinated, slow and stiff cilia. This indicated abnormal cilia movement. Supplementary Video 5: Nasal biopsy taken from a control subject that shows cilia (recorded at 500 fps under x1000 magnification) with a normal beat pattern, note the particulate clearance.
16363	https://www.youtube.com/watch?v=RuHPKQtNBz0	Solve 2a^2=-6+8a by Completing the Square Minute Math 85700 subscribers 40 likes Description 2526 views Posted: 9 Jun 2020 Solve 2a^2=-6+8a by Completing the Square. We learn how to solve 2a^2=-6+8a by completing the square. We learn how to solve a quadratic equation by completing the square. This is a great introduction into quadratic equations where we learn how to complete the square to solve a quadratic equation. This is under the topic of Solving Quadratic Equations by Completing the Square with this free online math video lesson. #solve #quadraticequations #completingthesquare #completethesquare #onlinemathlessons #minutemath Every Month we have a new GIVEAWAY that is FREE to Enter. See link below for details. You can enter Every Month! Subscribe to our YouTube Channel: Visit our website for links to all of our videos: Need a Math Tutor? Visit our STEM Store for all your Math and Science Gear: Great Educational Resources Here: Nominate a Great Teacher Here: Have a Math Lesson you would like to add to our community of videos? Click here: Subscribe to our other Life and Finance/Business YouTube Channel: Consider supporting us on Patreon... Follow us for... Tweets: Instagram: Facebook: Instagram Shop: Facebook Shop: Business and Personal Instagram: Transcript: Intro hi I'm Shaun Gannon and this is minute math and today we're learning about solving quadratic equations by completing the square minute math when you need healthy newsman head back we're giving this Solving the problem problem right here to a square equals a negative six plus ETA and we need to solve this by completing the square first thing I want to do is I'm gonna bring the 8a over to the left so you have to a squared minus 1/8 a and that's equal to just a negative 6 now luckily my coefficient here of two right I can see goes into each part so I'm gonna divide all parts now by two because I want the first coefficient to be x over to be one so we have an a squared here - hey divided by 4 or 2 is 4 4 minus 4a I'm gonna leave a space here equals and negative 6 divided by 2 is negative 3 and now look at my B value which is negative 4 I use my formula 1/2 times a negative 4 law squared well 1/2 times the negative Answer 4 is a negative 2 we square that then we get a positive 4 here so I'll take that 4 and I'm gonna add it to both sides of the equation the left-hand side now is a perfect square a minus 2 squared and then right-hand side well negative 3 plus 4 is a positive 1 I now can square root both sides and I have an a minus 2 here is equal to well plus or minus and we take a square root over an equal sign and square root of 1 is just 1 add a 2 to both sides and I equals a positive 2 plus or minus 1 and so we have a solved out each part individually 2 plus 1 is 8 3 & 2 minus 1 is a 1 and so our final answer here is just a equals 3 and 1 hope this video is helpful for you if it was please subscribe to this YouTube channel and like this video this helps us make more free math lessons for you and for everyone else so as always thanks for watching minute left when you need help he hears men in math [Music]
16364	https://fiveable.me/fluid-dynamics/unit-4/irrotational-flow/study-guide/iUcrAXIyWRzRHcYC	Irrotational flow \| Fluid Dynamics Class Notes \| Fiveable \| Fiveable new!Printable guides for educators Printable guides for educators. Bring Fiveable to your classroom ap study content toolsprintablespricing my subjectsupgrade 💨Fluid Dynamics Unit 4 Review 4.1 Irrotational flow All Study Guides Fluid Dynamics Unit 4 – Incompressible inviscid flows Topic: 4.1 💨Fluid Dynamics Unit 4 Review 4.1 Irrotational flow Written by the Fiveable Content Team • Last updated September 2025 Written by the Fiveable Content Team • Last updated September 2025 print study guide copy citation APA 💨Fluid Dynamics Unit & Topic Study Guides Fluid properties and statics Kinematics of fluids Conservation Laws in Fluid Dynamics Incompressible inviscid flows 4.1 Irrotational flow 4.2 Velocity potential 4.3 Stream function 4.4 Circulation and vorticity 4.5 Kelvin's circulation theorem Viscous flows and boundary layers Compressible flows Turbulence Computational fluid dynamics Aerodynamics Hydrodynamics Multiphase & Non-Newtonian Fluid Flows Environmental & Geophysical Fluid Dynamics print guide report error Irrotational flow is a fascinating concept in fluid dynamics where fluid particles don't rotate as they move. It's characterized by zero vorticity and can be described using a velocity potential function, simplifying complex flow analyses. This topic connects to broader fluid dynamics principles by exploring how irrotational flow behaves around objects. It introduces key concepts like the velocity potential, Laplace's equation, and the superposition principle, which are crucial for understanding fluid behavior in various applications. Definition of irrotational flow Irrotational flow is a type of fluid flow where the fluid particles do not rotate about their own axes as they move along streamlines Characterized by the absence of vorticity, meaning that the curl of the velocity field is zero ($\nabla \times \vec{V} = 0$) In irrotational flow, the fluid elements may translate and deform, but they do not undergo net rotation Mathematical representation Velocity potential function Irrotational flows can be described using a scalar function called the velocity potential ($\phi$) The velocity field is the gradient of the velocity potential: $\vec{V} = \nabla \phi$ This relationship ensures that the curl of the velocity field is zero, satisfying the irrotational condition The existence of a velocity potential simplifies the analysis of irrotational flows Laplace's equation for irrotational flow For incompressible, irrotational flows, the velocity potential satisfies Laplace's equation: $\nabla^2 \phi = 0$ Laplace's equation is a second-order partial differential equation that governs the behavior of the velocity potential Solving Laplace's equation with appropriate boundary conditions allows for the determination of the velocity potential and, consequently, the velocity field in irrotational flows Properties of irrotational flow Zero vorticity Irrotational flows have zero vorticity at every point in the fluid domain Vorticity is a measure of the local rotation of fluid particles and is defined as the curl of the velocity field ($\vec{\omega} = \nabla \times \vec{V}$) In irrotational flows, the vorticity is identically zero, indicating that fluid particles do not experience net rotation Path independence of velocity potential In irrotational flows, the change in velocity potential between two points is independent of the path taken between those points This path independence property allows for the definition of a unique velocity potential at each point in the fluid domain The path independence of the velocity potential is a consequence of the irrotational nature of the flow Circulation in irrotational flow Circulation is defined as the line integral of the velocity field along a closed curve In irrotational flows, the circulation around any closed curve is always zero This is a direct consequence of the path independence of the velocity potential The zero circulation property of irrotational flows has important implications for lift generation in aerodynamics Irrotational vs rotational flow Irrotational flow is characterized by zero vorticity, while rotational flow has non-zero vorticity In rotational flows, fluid particles can experience net rotation as they move along streamlines Rotational flows are more complex to analyze compared to irrotational flows due to the presence of vorticity Many real-world flows exhibit a combination of irrotational and rotational regions, with irrotational flow being an idealization that simplifies the analysis Potential flow theory Applicability to irrotational flow Potential flow theory is a mathematical framework that describes the behavior of irrotational flows It is based on the assumption that the flow is inviscid (no viscosity), incompressible, and irrotational Potential flow theory allows for the calculation of velocity fields, pressure distributions, and forces acting on bodies immersed in irrotational flows The theory provides valuable insights into the characteristics of irrotational flows and is widely used in aerodynamics and hydrodynamics Limitations of potential flow theory Potential flow theory has some limitations due to its idealized assumptions It does not account for viscous effects, which can be significant in real flows, especially near solid boundaries The theory assumes irrotational flow throughout the domain, which may not hold true in regions with flow separation or vortex shedding Potential flow theory cannot predict the onset of flow separation or the formation of wakes behind bodies Despite its limitations, potential flow theory remains a powerful tool for understanding and analyzing irrotational flows in many practical applications Elementary flows in irrotational flow Uniform flow Uniform flow is the simplest type of irrotational flow, where the velocity field is constant in both magnitude and direction The velocity potential for uniform flow in the x-direction is given by $\phi = U_\infty x$, where $U_\infty$ is the freestream velocity Uniform flow is often used as a building block for more complex irrotational flows Source/sink flow A source flow represents fluid emanating from a single point, while a sink flow represents fluid converging to a single point The velocity potential for a source/sink flow is given by $\phi = \pm \frac{Q}{4\pi r}$, where $Q$ is the strength of the source/sink and $r$ is the distance from the source/sink The velocity field for a source/sink flow is radially outward/inward and decays with distance from the source/sink Doublet flow A doublet flow is formed by placing a source and a sink of equal strength infinitesimally close to each other The velocity potential for a doublet flow is given by $\phi = \frac{\mu \cos \theta}{2\pi r^2}$, where $\mu$ is the doublet strength, $\theta$ is the angle measured from the doublet axis, and $r$ is the distance from the doublet Doublet flows are used to model the flow around solid bodies, such as cylinders and spheres Vortex flow A vortex flow represents the flow field induced by a concentrated vortex The velocity potential for a vortex flow is given by $\phi = \frac{\Gamma \theta}{2\pi}$, where $\Gamma$ is the circulation strength and $\theta$ is the angle measured from a reference direction The velocity field for a vortex flow is tangential to concentric circles and decays with distance from the vortex center Superposition principle for irrotational flows The superposition principle states that the velocity potential of a combination of irrotational flows is the sum of the individual velocity potentials This principle allows for the construction of complex irrotational flow fields by superimposing elementary flows (uniform, source/sink, doublet, vortex) The resulting velocity field is obtained by taking the gradient of the superposed velocity potential The superposition principle greatly simplifies the analysis of irrotational flows around complex geometries Irrotational flow around simple geometries Flow past a cylinder The flow past a cylinder can be modeled using a combination of a uniform flow and a doublet flow The velocity potential for the flow past a cylinder is given by $\phi = U_\infty (r + \frac{a^2}{r}) \cos \theta$, where $U_\infty$ is the freestream velocity, $a$ is the cylinder radius, $r$ is the distance from the cylinder center, and $\theta$ is the angle measured from the freestream direction The resulting flow field exhibits streamlines that divide and reconnect downstream of the cylinder, forming a symmetrical pattern Flow past a sphere The flow past a sphere can be modeled using a combination of a uniform flow and a doublet flow The velocity potential for the flow past a sphere is given by $\phi = U_\infty (r + \frac{a^3}{2r^2}) \cos \theta$, where $U_\infty$ is the freestream velocity, $a$ is the sphere radius, $r$ is the distance from the sphere center, and $\theta$ is the angle measured from the freestream direction The flow field around a sphere is similar to that of a cylinder, with streamlines dividing and reconnecting downstream of the sphere Kutta-Joukowski theorem Lift generation in irrotational flow The Kutta-Joukowski theorem relates the lift generated by a body in an irrotational flow to the circulation around the body According to the theorem, the lift per unit span is given by $L' = \rho_\infty U_\infty \Gamma$, where $\rho_\infty$ is the freestream density, $U_\infty$ is the freestream velocity, and $\Gamma$ is the circulation around the body The circulation is a measure of the net rotation of the fluid around the body and is responsible for the generation of lift Circulation and lift relationship The Kutta-Joukowski theorem establishes a direct relationship between circulation and lift A positive circulation (counterclockwise) results in a positive lift force, while a negative circulation (clockwise) results in a negative lift force The magnitude of the lift force is proportional to the circulation, freestream velocity, and fluid density The circulation around a body can be controlled by the shape of the body and the angle of attack, allowing for the manipulation of lift generation in aerodynamic applications Kelvin's circulation theorem Conservation of circulation in irrotational flow Kelvin's circulation theorem states that the circulation around a closed curve moving with the fluid remains constant in an inviscid, barotropic flow In irrotational flows, the circulation around any closed curve is always zero, and this property is conserved as the fluid moves and deforms The conservation of circulation has important implications for the generation and maintenance of lift in aerodynamic applications Implications for lift generation Kelvin's circulation theorem implies that the circulation around a body cannot be generated or destroyed within the fluid itself The circulation necessary for lift generation must be introduced by the motion of the body or by the presence of a sharp trailing edge (Kutta condition) Once the circulation is established, it is conserved and continues to provide lift as long as the flow remains irrotational and inviscid The conservation of circulation also explains the persistence of lift-generating vortices shed from the trailing edges of wings and other lifting bodies Bernoulli's equation in irrotational flow Pressure-velocity relationship Bernoulli's equation relates the pressure, velocity, and elevation along a streamline in an inviscid, steady, and incompressible flow For irrotational flows, Bernoulli's equation takes the form: $\frac{p}{\rho} + \frac{1}{2}V^2 + gz = constant$, where $p$ is the pressure, $\rho$ is the fluid density, $V$ is the velocity magnitude, $g$ is the acceleration due to gravity, and $z$ is the elevation The equation states that the sum of the pressure term, kinetic energy term, and potential energy term remains constant along a streamline Applications of Bernoulli's equation Bernoulli's equation is a powerful tool for analyzing the pressure distribution in irrotational flows It can be used to calculate the pressure difference between two points along a streamline, such as the pressure difference between the upper and lower surfaces of an airfoil The equation also explains the relationship between velocity and pressure in irrotational flows: an increase in velocity is accompanied by a decrease in pressure, and vice versa Bernoulli's equation finds applications in various fields, including aerodynamics (lift and drag calculations), hydrodynamics (flow through pipes and channels), and wind engineering (wind loads on structures) Streamlines and equipotential lines Orthogonality of streamlines and equipotential lines In irrotational flows, streamlines and equipotential lines form an orthogonal network Streamlines are lines tangent to the velocity vector at every point, representing the path followed by fluid particles Equipotential lines are lines along which the velocity potential is constant, representing lines of constant velocity magnitude The orthogonality property means that streamlines and equipotential lines intersect at right angles at every point in the flow field Visualization of irrotational flow patterns The orthogonal network of streamlines and equipotential lines provides a useful tool for visualizing irrotational flow patterns Streamlines help to understand the direction and path of fluid motion, while equipotential lines provide information about the velocity magnitude distribution The density of streamlines and equipotential lines can indicate regions of high or low velocity, as well as the presence of sources, sinks, or other flow singularities Visualization of the streamline-equipotential line network aids in the analysis and interpretation of irrotational flow fields around various geometries Conformal mapping techniques Transformation of irrotational flows Conformal mapping is a mathematical technique that transforms a complex irrotational flow in one plane (z-plane) into a simpler flow in another plane (w-plane) The transformation preserves the orthogonality of streamlines and equipotential lines, as well as the local angles between them Conformal mapping allows for the simplification of complex flow geometries into more manageable shapes, such as circles or straight lines The velocity potential and stream function in the transformed plane can be obtained using the Cauchy-Riemann equations, which relate the real and imaginary parts of the complex potential Examples of conformal mapping applications Joukowski transformation: Maps the flow around a cylinder to the flow around an airfoil-like shape, enabling the analysis of lift generation Schwarz-Christoffel transformation: Maps the flow in a polygonal domain to the flow in a half-plane or a strip, simplifying the analysis of flows around corners and edges Karman-Trefftz transformation: Maps the flow around a flat plate with a flap to the flow around a circle, facilitating the study of high-lift devices Conformal mapping techniques have been extensively used in aerodynamics, hydrodynamics, and other fields to analyze and design flow geometries with desired characteristics BackNext 4.2 Study Content & Tools Study GuidesPractice QuestionsGlossaryScore Calculators Company Get $$ for referralsPricingTestimonialsFAQsEmail us Resources AP ClassesAP Classroom every AP exam is fiveable history 🌎 ap world history🇺🇸 ap us history🇪🇺 ap european history social science ✊🏿 ap african american studies🗳️ ap comparative government🚜 ap human geography💶 ap 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16365	https://www.quora.com/What-is-the-M1V1-dilution-formula-in-chemistry	Something went wrong. Wait a moment and try again. Dilution Calculations Scientific Formula Chemistry Calculations Molarity Dilution Science Chemistry Chemical Formulas Chemical Dilutions 5 What is the M1V1 dilution formula in chemistry? Trevor Hodgson Knows English · Author has 11.8K answers and 12.3M answer views · 2y This is the dilution formula: The complete formula is : M1V1 = M2V2 M1 = molarity of solution 1 V1 = volume of solution 2 M2 = Molarity of solution 2 V2 = volume of solution 2 How would you use this equation ? Typically to solve a question such as : A 250. cm³ sample of 0.45 M NaCl solution is diluted to 875 cm³ . What is the molarity of the final solution : I always keep the unknown on the left of the equation- this makes calculation easier M1 875 cm³ = 0.45 M 250 cm³ M1 = 0.45 M 250. cm³ / 875 cm³ M1 = 0.12857 M Answer can have 2 significant digits - Molarity = 0.13 M Now see if you can solve : A 25 This is the dilution formula: The complete formula is : M1V1 = M2V2 M1 = molarity of solution 1 V1 = volume of solution 2 M2 = Molarity of solution 2 V2 = volume of solution 2 How would you use this equation ? Typically to solve a question such as : A 250. cm³ sample of 0.45 M NaCl solution is diluted to 875 cm³ . What is the molarity of the final solution : I always keep the unknown on the left of the equation- this makes calculation easier M1 875 cm³ = 0.45 M 250 cm³ M1 = 0.45 M 250. cm³ / 875 cm³ M1 = 0.12857 M Answer can have 2 significant digits - Molarity = 0.13 M Now see if you can solve : A 250. cm³ sample of 0.45 M NaCl solution is diluted to a final concentration of 0.175 M . What is the volume of the final solution? Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) · Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. It always sounded like a hassle. Dozens of tabs, endless forms, phone calls I didn’t want to take. But recently I decided to check so I used this quote tool, which compares everything in one place. It took maybe 2 minutes, tops. I just answered a few questions and it pulled up offers from multiple big-name providers, side by side. Prices, coverage details, even customer reviews—all laid out in a way that made the choice pretty obvious. They claimed I could save over $1,000 per year. I ended up exceeding that number and I cut my monthly premium by over $100. That’s over $1200 a year. For the exact same coverage. No phone tag. No junk emails. Just a better deal in less time than it takes to make coffee. Here’s the link to two comparison sites - the one I used and an alternative that I also tested. If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Related questions What is the difference between the formulas M1V1=M2V2 and M= (M1V1+M2V2) / (V1+V2)? How can you tell if your methadone has been diluted? What is the best trick to remember a chemistry formula? Does vinegar dilute heroin? What is the lidocaine dilution formula? Daniel Iyamuremye Former Senior Lecturer (Retired) (2000–2018) · Author has 12.1K answers and 2M answer views · 2y The formula is that when you dilute for example V1 volume of an M1 conc. solution to volume V2 with conc. M2, the following relation applies: M1V1 = M2V2 This equality relation results from the fact that the amount of the solute in both volumes is the same. For example, you have a solution of 0. 10M and you want 100ml of a dilution solution of 0.025M. How many ml of the stock solution (0.10M) are you going to take? M1V1 = M2V2 0.10M x V1 = 0.025M x 100 ml V1 = (0.025M x 100 ml)/0.10M = 25 ml. This means that you take 25ml of 0.10M solution and dilute it with distilled water up to 100 ml. Another meth The formula is that when you dilute for example V1 volume of an M1 conc. solution to volume V2 with conc. M2, the following relation applies: M1V1 = M2V2 This equality relation results from the fact that the amount of the solute in both volumes is the same. For example, you have a solution of 0. 10M and you want 100ml of a dilution solution of 0.025M. How many ml of the stock solution (0.10M) are you going to take? M1V1 = M2V2 0.10M x V1 = 0.025M x 100 ml V1 = (0.025M x 100 ml)/0.10M = 25 ml. This means that you take 25ml of 0.10M solution and dilute it with distilled water up to 100 ml. Another method consists of using the dilution factor. In the example above, the dilution factor is: M1/M2 = 0.10M/0.025M = 4 This means that you take a volume of 100 ml/4 = 25 ml of the solution M1 and dilute it with distilled water up to 100ml Ravi Shankar BSc.chemistry from Degree College KGM (Graduated 2022) · 2y In chemistry sometimes we are given specific concentrated solution whose molarity is known but in practical chemistry lab we use such chemical(particularly acid or base ) by varying it's concentration . For example if solution of HCl has molarity =2M but for its reaction we need 4M solution then what we should do ? Should we go back and take another sample of it from supplier?Certainly not ,we have a process in chemistry to solve this problem.Actually we add some water in it (Keeping amount of HCL constant ) In this way degree of concentration decreases or say solution now becomes diluted and t In chemistry sometimes we are given specific concentrated solution whose molarity is known but in practical chemistry lab we use such chemical(particularly acid or base ) by varying it's concentration . For example if solution of HCl has molarity =2M but for its reaction we need 4M solution then what we should do ? Should we go back and take another sample of it from supplier?Certainly not ,we have a process in chemistry to solve this problem.Actually we add some water in it (Keeping amount of HCL constant ) In this way degree of concentration decreases or say solution now becomes diluted and the process is referred as dilution . Solution before the process of dilution is known as stock solution and after this process is known bas diluted solution .This process is governed by formula which relates molarity (concentration ) and volume of stock and diluted solution . This formula is known as dilution formulation or dilution equation . Mathematically it is written as M1V2=M2V2 Where M1. and V1 are molarity and volume of stock solution and M2 and V2 are molarity and volume of diluted solution. Armando Viray B.S Biochemistry from University of Santo Tomas (Graduated 1983) · Author has 104 answers and 64.3K answer views · 2y M1V1 = M2V2 M1V1 is the molar concentration and volume of the stock solution that you will use for preparing the desired concentration of solution which is M2V2. Sponsored by Grubhub For Merchants Ready to reach Amazon Prime customers? Reach more customers through our Amazon partnership. Prime members receive free Grubhub+. Related questions In chemistry, what are dilution problems? What is the dilution formula in accounting? What is the formula of dilute? How can we do simple dilution in chemistry? What is the formula m1u1+m2u2=m1v1+m2v2 used for in physics? Ludeman Eng Former Chair, Department of Basic Science at Virginia Tech Carilion School of Medicine and Research Institute (2011–2015) · Author has 1.2K answers and 765.2K answer views · 3y Related What does 1:10 mean in solution dilution lingo? A 1:10 dilution ( 1 to 10 dilution) means that a stock solution is diluted to 1/10th (one tenth) its concentration. This is accomplished by taking 1 volume of stock and adding it to 9 volumes of diluent - or vice versa (adding 9 volumes of diluent to 1 volume of stock). So if you want to make a 1:10 dilution of 10 ml of stock, mix it with 90 ml diluent. If you want to make a 1:10 dilution of 5 ml of stock, mix it with 45 ml diluent. Ronald Andrews Former Former Engineer at Eastman Kodak and Bausch + Lomb · Author has 5.4K answers and 5.6M answer views · 3y Related What does 1:10 mean in solution dilution lingo? Dilutions need to be carefully defined. 1:10 refers to a 1 to 10 ratio. The “1” is most likely an active ingredient. The “10” could be solvent or solution. If it is one part active ingredient to 10 parts solution, that is a 10% concentration. If it is 1 part active ingredient to 10 parts solvent, that is a 9.09% concentration. Different operations have different conventions. It is also necessary to define whether the “parts” are by volume, by mass, or by some other metric. Sponsored by Amazon Business Solutions and supplies to support learning. Save on essentials and reinvest in students and staff. Aarush Samer B.SC in Science & General Knowledge, Banaras Hindu University · 6y Related How does molar conductivity vary with dilution? First of all, it's not about molecules. You need ions to conduct electricity, not molecules. Conductivity changes with the concentration of the electrolyte. The number of ions per unit volume carrying the current decreases on dilution, so conductivity always decreases with decrease in concentration. However, molar conductivity increases with dilution. This is because conductivity of a solution is the conductance of one unit volume of the solution, but molar conductivity is the conductance of that volume solution containing one mole of electrolyte. As the total volume of solution containing one mo First of all, it's not about molecules. You need ions to conduct electricity, not molecules. Conductivity changes with the concentration of the electrolyte. The number of ions per unit volume carrying the current decreases on dilution, so conductivity always decreases with decrease in concentration. However, molar conductivity increases with dilution. This is because conductivity of a solution is the conductance of one unit volume of the solution, but molar conductivity is the conductance of that volume solution containing one mole of electrolyte. As the total volume of solution containing one mole of electrolyte increases, the decrease in conductivity due to dilution is more than compensated by increase in its volume of solution that contains one mole of the electrolyte. Myriam Kobylkevich Analytical Chemist (1978–present) · Author has 2.8K answers and 4.1M answer views · 3y Related Is the first volume (V1) in the molality law (M1V1) the same as the volume from which the first molarity (M1=mol/V1) was extracted, as is the case with the second party after dilution (M2 V2)? First of all, M1V1=M2V2 refers to molarity (concentration expressed in moles of solute per liter of solution. A 1M (1 molar) solution of some compound contains 1 mole of that something per 1 liter. If you wanted to be very precise you would state the temperature at which you measured the volume. If you do not, I would assume that it is 20 degrees Celsius, and that a samll variation in lab temperature will not matter for the purpose of the solution. Molality (with L, not R) is moles of solute per kilogram of solvent used to dissolve it (not per kg or liter of solution obtained). A 1m (1 molal First of all, M1V1=M2V2 refers to molarity (concentration expressed in moles of solute per liter of solution. A 1M (1 molar) solution of some compound contains 1 mole of that something per 1 liter. If you wanted to be very precise you would state the temperature at which you measured the volume. If you do not, I would assume that it is 20 degrees Celsius, and that a samll variation in lab temperature will not matter for the purpose of the solution. Molality (with L, not R) is moles of solute per kilogram of solvent used to dissolve it (not per kg or liter of solution obtained). A 1m (1 molal with lowercase m) solution contains 1 mol of solute dissolved in 1 kilogram of solvent. Second of all, I would not call M1V1=M2V2 a law. It is just the result of applying common sense to a word problem that every fifth grader should understand and know how to solve. It is just a mass balance calculation where you equate mass of solute before adding solvent to mass after adding solvent. Third of all, the same reasoning would give you C1V1=C2V2, with C1 and C2 both meaning concentration in the same units (mg/L or mg/mL or lb/gallon, or whatever mass/volume you choose), and of course V1 and V2 being . When you multiply concentration times volume you are going to dilute, of course you get mass of solute in that volume of solution. As you add pure solvent, that mass of solute does not change. after dilution, the new concentration times volume will give you the new mass, which is the same as the mass you started with. 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Use the dilution formula: C1·V1 = C2·V2, where: C1 = initial concentration V1 = initial volume C2 = final concentration V2 = final volume If the concentration is molarity, the formula is: M1·V1 = M2·V2 Michael Flynn BSc Hons Newcastle University · Upvoted by Toby Block , Ph.D. Chemistry, University of Wisconsin - Madison (1976) · Author has 901 answers and 2.7M answer views · 4y Related Why doesn’t ammonium nitrate obey the dilution formula m1v1=m2v2? This so called formula has nothing to do with ammonium nitrate. It just means that if a solution is diluted the total moles of solute remains the same. Hcbiochem Former Professor of Biology and Chemistry (1982–2019) · Author has 4.6K answers and 1.9M answer views · 4y Related Using the 1.8M stock KClO3 solution and the dilutions equation (M1V1=M2V2), how much stock solution and water you would need to make 50. mL of the below dilutions of KClO3? (please show how you got the answer) Since you didn’t provide any of the concentrations you want, I can’t give you an answer. However, you are told to use this equation: M1V1 = M2V2 1.8 M (V1) = M2 (50 mL) Plug in the molarity desired and solve for V1. Your answer will be in mL of the stock solution. Subtract that volume from 50 to get the volume of water needed. Jose Castellanos Knows French · Author has 704 answers and 320.7K answer views · 3y Related Is the first volume (V1) in the molality law (M1V1) the same as the volume from which the first molarity (M1=mol/V1) was extracted, as is the case with the second party after dilution (M2 V2)? First volume: V1 There is a solute concentration of n1 that is diluted in a solution with a volume V1 M1 = n1/V1 So, to modify M1 it is required to modify C1 New n2 = n1 + dn1 CASE 1: if dn1 does not increase V1, i.e. V1 = constant M2 = (n1) / V1 . CASE 2: if dn1 increases V1, V2 is different from V1, i.e not the same volume. M2 = (n1 + dn1) / V2 John Britto Current Professor of Chemistry at Delaware County Community College · Author has 125 answers and 17.4K answer views · 1y Related What is the formula for calculating concentration in a dilution if only volume and initial concentration are known? You would first need to know the volume of the final diluted sample. And then you can use: M1 x V1 = M2 x V2 M1 and V1 are the initial concentration and volume M2 is the diluted concentration - this is what you are solving for V2 is the volume after dilution. Related questions What is the difference between the formulas M1V1=M2V2 and M= (M1V1+M2V2) / (V1+V2)? How can you tell if your methadone has been diluted? What is the best trick to remember a chemistry formula? Does vinegar dilute heroin? What is the lidocaine dilution formula? In chemistry, what are dilution problems? What is the dilution formula in accounting? What is the formula of dilute? How can we do simple dilution in chemistry? What is the formula m1u1+m2u2=m1v1+m2v2 used for in physics? What are the chemistry formula sheets for the NEET? How many ml of 1.10 M H2SO4 should be diluted in water to prepare 0.500 L of 0.025 M H2SO4.? How do I prepare a 5% 50 ml NaOH solution from a 98% stock solution? What are some examples of a dilute solution? How much dilution pre-series A is too much dilution? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
16366	https://scicomp.stackexchange.com/questions/10919/is-there-any-numerical-reason-for-not-using-repeated-multiplication-instead-of-i	matlab - Is there any numerical reason for not using repeated multiplication instead of integer powers? - Computational Science Stack Exchange Join Computational Science By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Is there any numerical reason for not using repeated multiplication instead of integer powers? Ask Question Asked 11 years, 7 months ago Modified11 years, 7 months ago Viewed 393 times This question shows research effort; it is useful and clear 6 Save this question. Show activity on this post. I recently discovered that, in MATLAB 2013b at least, it is significantly faster to do repeated multiplication rather than integer powers. That is, matlab tic;test4p = a.^4;toc is much slower than matlab tic;test4m = a.a.a.a;toc See for full details. Is there any numerical reason (accuracy etc) why repeated multiplication might be a bad idea? matlab Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Improve this question Follow Follow this question to receive notifications edited Feb 27, 2014 at 0:18 WalkingRandomlyWalkingRandomly asked Feb 26, 2014 at 23:49 WalkingRandomlyWalkingRandomly 263 1 1 silver badge 7 7 bronze badges 2 Are you sure that matlab is using integer powers? If you try with a biger number (20) test4p is taking amost the same time but test4m is taking much more than before.sebas –sebas 2014-02-27 00:36:36 +00:00 Commented Feb 27, 2014 at 0:36 What happens if you try a.^3.14 ? If MATLAB is using a general purpose exponentiation routine (e.g. by using log() and exp()), then it should take the same time as a.^4.Brian Borchers –Brian Borchers 2014-02-27 04:58:47 +00:00 Commented Feb 27, 2014 at 4:58 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. I'm surprised that MATLAB does not do exponentiation-by-squaring for positive integer powers, since that algorithm should require fewer multiplications. I think a consequence of that should be that numerical errors accumulate less quickly. Optimal addition-chain multiplication of exponents is an NP-complete problem, so it's not worth it to figure out the sequence of multiplies and exponentiations that minimizes the total number of multiplicative operations. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Feb 27, 2014 at 2:18 Geoff OxberryGeoff Oxberry 30.7k 9 9 gold badges 65 65 silver badges 132 132 bronze badges Add a comment\| This answer is useful 6 Save this answer. Show activity on this post. the numeric reason would be the loss precision. at each multiplication you lose a little bit of precision. If you want to implement power by multiplication, then it's better to look up the algorithm. I once coded the algorithm from Stepanov's "Elements of Programming" book, it's the fastest possible way. UPDATE: I found the earlier version of Stepanov's book as Notes Search for fast_power algorithm in section 10.3. It's so beautiful, one of the most elegant pieces of code ever Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Feb 27, 2014 at 3:25 answered Feb 27, 2014 at 3:19 Aksakal almost surely binaryAksakal almost surely binary 336 1 1 silver badge 6 6 bronze badges 2 His algorithm is essentially exponentiation by squaring with some additional implementation-level optimizations.Geoff Oxberry –Geoff Oxberry 2014-02-27 04:42:01 +00:00 Commented Feb 27, 2014 at 4:42 Yes. It's much faster than the stock algorithm. I had an accounting application where the powers were always integer numbers, so it worked fine for me.Aksakal almost surely binary –Aksakal almost surely binary 2014-02-27 12:36:46 +00:00 Commented Feb 27, 2014 at 12:36 Add a comment\| Your Answer Thanks for contributing an answer to Computational Science Stack Exchange! Please be sure to answer the question. 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16367	https://repository.uwl.ac.uk/id/eprint/8919/1/Lizarondo_et_al._2022_j.jclinepi._Five_common_pitfalls_in_mixed_methods_systematic_reviews_%E2%80%93_lessons_learned.pdf	UWL REPOSITORY repository.uwl.ac.uk Five common pitfalls in mixed methods systematic reviews – lessons learned Lizarondo, Lucylynn, Stern, Cindy, Apostolo, Joao, Carrier, Judith, de Borges, Kelli, Godfrey, Christina, Kirkpatrick, Pamela, Pollock, Danielle, Rieger, Kendra, Salmond, Susan, Vandyk, Amanda and Loveday, Heather ORCID: (2022) Five common pitfalls in mixed methods systematic reviews – lessons learned. Journal of Clinical Epidemiology. ISSN 0895-4356 This is the Accepted Version of the final output. UWL repository link: Alternative formats: If you require this document in an alternative format, please contact: open.research@uwl.ac.uk Copyright: Creative Commons: Attribution-Noncommercial-No Derivative Works 4.0 Copyright and moral rights for the publications made accessible in the public portal are retained by the authors and/or other copyright owners and it is a condition of accessing publications that users recognise and abide by the legal requirements associated with these rights. Take down policy: If you believe that this document breaches copyright, please contact us at open.research@uwl.ac.uk providing details, and we will remove access to the work immediately and investigate your claim. 1 Five common pitfalls in mixed methods systematic reviews – lessons learned Authors Lucylynn Lizarondo (lucylynn.lizarondo@adelaide.edu.au) – FIRST AUTHOR a. JBI, Faculty of Health and Medical Sciences, University of Adelaide, 55 Norwich House, King William Road, Adelaide, Australia Cindy Stern (cindy.stern@adelaide.edu.au) – FIRST AUTHOR a. JBI, Faculty of Health and Medical Sciences, University of Adelaide, 55 Norwich House, King William Road, Adelaide, Australia Joao Apostolo (apostolo@esenfc.pt) a. Health Sciences Research Unit: Nursing, Escola Superior de Enfermagem de Coimbra, Coimbra, Portugal Centre for Evidence Based Practice: A JBI Centre of Excellence, Portugal. Judith Carrier (CarrierJA@cardiff.ac.uk) a. School of Healthcare Sciences, Cardiff University, The Wales Centre For Evidence Based Care: A JBI Centre of Excellence, Cardiff CF10 3AT, United Kingdom Kelli de Borges (kelli.bsantos@gmail.com) a. Universidade Federal de Juiz de For a, Campus Universitário, Rua José Lourenço Kelmer, s/n - São Pedro, Juiz de Fora - MG, 36036-900, Brazil Christina Godfrey (godfreyc@queensu.ca) a. School of Nursing, Queen's University, Queen's Collaboration for Health Care Quality: A JBI Centre of Excellence Kingston, 99 University Ave, Kingston, Ontario K7L 3N6, Canada Pamela Kirkpatrick (p.kirkpatrick@rgu.ac.uk) a. School of Nursing and Midwifery, Robert Gordon University, The Scottish Centre for Evidence-based, Multi-professional Practice: A JBI Centre of Excellence, Garthdee House, Garthdee Rd, Garthdee, Aberdeen AB10 7AQ, United Kingdom 2 Danielle Pollock (danielle.pollock@adelaide.edu.au) a. JBI, Faculty of Health and Medical Sciences, University of Adelaide, 55 Norwich House, King William Road, Adelaide, Australia Kendra Rieger (Kendra.Rieger@twu.ca) a. School of Nursing, Trinity Western University, 22500 University Dr, Langley, British Columbia, V2Y 1Y1, Canada Susan Salmond (salmonsu@sn.rutgers.edu) a. School of Nursing, Rutgers, The State University of New Jersey, The Northeast Institute for Evidence Synthesis and Translation (NEST): A JBI Centre of Excellence, New Brunswick, New Jersey, USA Amanda Vandyk (Amanda.Vandyk@uottawa.ca) a. School of Nursing, University of Ottawa, 75 Laurier Ave. E, Ottawa, Ontario K1N 6N5, Canada Heather Loveday (Heather.Loveday@uwl.ac.uk) a. College of Nursing, Midwifery and Healthcare, University of West London, The University of West London Centre for Evidence-Based Healthcare: A JBI Centre of Excellence, St Mary's Rd, London W5 5RF, United Kingdom Corresponding author: Lucylynn Lizarondo Email: Lucylynn.lizarondo@adelaide.edu.au Telephone: +61415431825 Postal Address: JBI, Faculty of Health and Medical Sciences, University of Adelaide, 55 Norwich House, King William Road, Adelaide, Australia 3 Abstract Objective: Mixed methods systematic reviews (MMSR) combine quantitative and qualitative evidence within a single review. Since the revision of the JBI Methodology for MMSRs in 2020, there has been an increasing number of reviews published that claim to follow this approach. A preliminary examination of these indicated that authors frequently deviated from the methodology. This paper outlines five common ‘pitfalls’ associated with undertaking MMSR and provides direction for future reviewers attempting MMSR. Methods: Forward citation tracking identified 17 reviews published since the revision of the JBI mixed methods methodological guidance. Methods used in these reviews were then examined against the JBI methodology to identify deviations. Results: The issues identified related to the rationale for choosing the methodological approach; incorrect synthesis and integration approach chosen to answer the review question/s posed; the exclusion of primary mixed methods studies in the review; the lack of detail regarding the process of data transformation and a lack of ‘mixing’ of the quantitative and qualitative components. Conclusion: This exercise was undertaken to assist systematic reviewers considering conducting a MMSR as well as MMSR users to identify potential areas where authors tend to deviate from the methodological approach. Based on these findings a series of recommendations are provided. Keywords: mixed methods systematic review, evidence synthesis, mixed methods review methodology, systematic review, mixed methods research, research methodology Running title: Common pitfalls in mixed methods review 4 Background Systematic reviews are considered the gold standard in supporting evidence-based practice. By combining quantitative and qualitative evidence together within a single review, the value of a mixed method systematic review (MMSR) is undeniable. In this context, a MMSR integrates the findings of effectiveness (quantitative evidence) and patient, family, staff or other’s experiences (qualitative evidence) to enhance the meaningfulness of evidence for decision-makers.(1) While their usefulness is acknowledged, (2) the process of undertaking a MMSR remains complex and fraught with difficulties. MMSRs follow the same internationally recognised steps of all other types of systematic review, the main difference being the need to integrate the quantitative and qualitative data/evidence. The JBI Mixed Methods Methodology Group was formed in 2014 and is comprised of an international panel of researchers, clinicians, and academics in the fields of primary evidence generation, evidence synthesis and evidence implementation. The responsibility of this group is to provide methodological guidance for the JBI Collaboration (JBIC) (an international group of self-governing entities that advocate the synthesis, transfer and utilisation of evidence to improve healthcare outcomes) on how to conduct MMSR. In 2020, the JBI Mixed Methods Methodology Group revised its guidance on the conduct of MMSR,(3, 4) leading to an increase in the number of reviews recently published, which claim to follow the JBI approach (i.e. convergent integrated and convergent segregated, Figure 1). A preliminary examination of these reviews indicated that authors frequently deviated from the methodology, highlighting a need to explore the application of our approach. In this paper, we outline common ‘pitfalls’ associated with undertaking MMSRs, and provide direction for future reviewers attempting MMSRs. Methods Forward citation tracking of the two seminal papers(3, 4) published by the JBI Mixed Methods Methodology Group was undertaken in Google Scholar in June 2021 by one of the authors (LL) to locate all published MMSRs that had cited the JBI methodology and reported following its guidance. All published reviews were included, but protocols were excluded. Two of the authors (LL and CS, co-conveners of the JBI Mixed Methods Methodology Group) independently reviewed each of the reviews in full to examine and extract information regarding a) the review question/s, b) the review methods including the framework used (e.g. PICO) for the inclusion/exclusion criteria, type of studies included in the review, approach taken (i.e. whether segregated or integrated), process for data transformation and steps involved in the integration of the quantitative and qualitative evidence, c) if the review questions and methods reported aligned to the JBI approach and d) where methods deviated from the JBI approach. A meeting was held to compare and discuss the results of the two authors which demonstrated 100% agreement on whether the published reviews did or did not follow the JBI methodology. Reviewing the reasons why a review did not follow the JBI guidance allowed key problems to be identified, which were then presented to the rest of the JBI Mixed Methods Methodology Group for further discussion and refinement where a consensus was reached. 5 Findings Using a targeted literature searching strategy (described above), 63 references that cited the JBI methodology were found. Of these, 39 were review protocols, three were methodology papers, two were editorials, one was a book chapter and another one was a case study; only 17 were MMSRs. Eight of the 17 published MMSRs that indicated using the JBI methodology did not follow the methods of the JBI approach. Five pitfalls are outlined next, including a description of the issue as well as examples to narratively illustrate the problem identified and how it should be addressed. Is a MMSR the best methodological approach to answer the review question/s? Regardless of the type of systematic review undertaken, once a topic is identified, a specific, answerable question/s needs to be developed.(5) The review question/s dictate which methodological approach should be followed and the rationale for selecting a MMSR approach should be clearly described in the introduction section of the review, with both the qualitative and quantitative evidence bases adequately outlined. The essence of MMSR is dependent on the nature of the review question/s and how the investigation can allow for examination of the level of agreement between quantitative results and qualitative findings (triangulation), identification of discrepancies within the available evidence, determination of whether the quantitative and qualitative data address different aspects of a phenomenon of interest, or one type of data can explore, contextualize, or explain the findings of the other type of data.(3) If none of these elements apply, a MMSR may not be the best approach to follow. Young et al 2021 sought to evaluate and summarise the findings of relevant qualitative and quantitative studies on genetic counselling and genetic testing in Asian Americans.(6) While it is clear from the aim of the review that the authors wanted to summarise both the quantitative and qualitative evidence bases, it was unclear if their intention was to synthesise and integrate the two perspectives. If a summary of the evidence was the intent, a scoping review would have been a more appropriate approach.(7) In contrast, Clari et al 2020 (8) investigated mindfulness-based programs for people with chronic obstructive pulmonary disease and while their primary aim was to identify and summarise the qualitative and quantitative evidence (Like Young et al 2021(6)), their additional aim was to aggregate both types of evidence to understand why these programs could be effective and appropriate in this population and under which conditions and modes,(8) thereby justifying the need for a MMSR. Was the right synthesis and integration approach followed to answer the review question/s? While the aim of a MMSR is to combine qualitative and quantitative data/evidence together, integration can occur either following the process of data transformation or following independent qualitative and quantitative syntheses. The approach that should be followed depends on the nature of the review question/s posed with the aim/s or questions serving as an anchor.(3, 9-11) (See Figure 1 below). One of the most frequently identified methodological issues in MMSR is the mismatch 6 between the review question/s and the methodology used to combine quantitative and qualitative data/evidence. PICO (Population, Intervention, Comparison, Outcome); PICo (Population, phenomenon of Interest, Context) Figure 1: The JBI Framework for MMSR (adapted from (3)) Clari et al 2020(8) stated they followed the convergent segregated approach to synthesis and integration which involves separate qualitative and quantitative synthesis followed by integration of the qualitative and quantitative evidence. This approach should be used when the review question/s examine different aspects of a particular phenomenon (e.g., effectiveness and experiences). However, in this review the focus was on the application of mindfulness-based programs in order to describe favorable and unfavorable implementation factors, indicating the question could be addressed by both quantitative and qualitative studies, thus a convergent integrated approach should have been followed. On the other hand, Rose et al 2021(12) conducted a MMSR to (1) examine the impact of European school food interventions on nutrition, weight status and wellbeing outcomes and (2) explore the experiences and perceptions of adolescents in Europe who have been subject to a school food intervention or national school food policy. Unlike the review by Clari et al 2020(8), this MMSR addressed two different aspects associated with school food interventions, i.e. the effects these interventions have on adolescents, and how adolescents experience or perceive them.(12) Given that the review aimed to address different dimensions of a particular phenomenon of interest (effectiveness and experiences of school food interventions), a configurative method of synthesis – i.e., segregated approach, as opposed to aggregation only or the integrated approach, was chosen, which is the appropriate approach for the MMSR.(3) 7 Where is the mixed methods primary research in the MMSR? A MMSR combines findings from qualitative, quantitative, and mixed methods primary studies.(13) Although many published MMSRs include mixed methods studies in their review, there are a number of published reviews that do not explicitly state this in their inclusion criteria.(14) This inadvertent or deliberate decision to exclude mixed methods primary studies in a MMSR may be due to the challenges that this type of study design pose to the methods of the systematic review. For example, in some instances, the author’s rationale for using a mixed methods design may not match how they combined the methods, which can lead to unnecessary or redundant data that do not address the research question.(14) There are also intrinsic issues with reporting of findings from mixed methods research, specifically the sequence of data analysis and integration of findings.(14) However, given the development of guidelines for reporting of,(15) and critical appraisal tools for mixed methods studies,(16, 17) the inclusion of this type of study in MMSRs is made more feasible and less challenging. Mixed method primary studies are usually ‘disaggregated’ into quantitative and qualitative data for the purpose of synthesis in a MMSR.(3) This approach of categorising data into quantitative and qualitative is particularly simple and uncomplicated when the mixed methods research is presented as two separate publications, as in the case of a sequential explanatory design where the aim is to use one method to explain the findings of the other.(18) Where mixed methods research findings are presented in a single publication (as in the case of a combined survey and interview data), as long as data from the quantitative or qualitative components could be clearly extracted, ‘disaggregation’ can occur. A good example of how mixed methods studies have been included in a MMSR is Clari et al.’s 2020 (8) review on mindfulness-based programs for people with chronic obstructive pulmonary disease. This review included not only primary quantitative and qualitative studies but also mixed methods studies, where each component (quantitative and qualitative) of the mixed methods study was extracted so that they had two streams of data ready to be combined to the data of the primary quantitative and primary qualitative studies.(8) In contrast, the review by McKenzie et al 2021 looking at barriers and facilitators of physical activity participation for young people and adults with childhood-onset physical disability did not report that primary mixed methods studies were eligible for inclusion, simply stating ‘original quantitative or qualitative studies’ were of interest.(19) This implies the authors may have potentially missed evidence that could have potentially complemented or strengthened the overall evidence for this systematic review. Transforming the data – where is it and how was it done? MMSRs that follow an integrated approach to synthesis require a process known as data transformation which enables integration of the two different types of data. Transformation occurs when one type of data is transformed into another format. This can involve transforming qualitative data into a quantitative format (known as quantitising) or converting quantitative data into a qualitative format (qualitising).(3) JBI guidance advocates data transformation involve ‘qualitising’, where 8 quantitative data are converted into ‘textual descriptions’ such as themes, categories, or narratives, to allow integration with the qualitative data.(3) This approach is recommended as codifying quantitative data is generally considered less error-prone than attributing numerical values to qualitative data.(20) Gonzalez-Gonzalez et al 2021(21) reported using the JBI approach in their review looking at end of life care preferences in older patients with multi-morbidities. The JBI’s approach to data transformation involves ‘qualitisation.’ However, this review quantitised the qualitative data by using ‘meta-analysis to convert qualitative data into a numerical format for quantitative synthesis by transforming verbal counts into numbers.’ It remains unclear how this process was undertaken as transformed data are simply presented as a range of those who stated their preferences with percentages calculated e.g. Yes (n (%)) = 12-18(60-100). When transformation is necessary, the specific approach used as well as the process taken should be detailed in the review. Additionally, the original data as well as the transformed data should be provided within the review to allow readers to see the ‘before and after’ of the process. A good example of this is found in Gray et al’s 2021 review on factors influencing physical activity engagement following coronary artery bypass graft surgery, where the authors explicitly detail the qualitisation process including how each member contributed.(22) The authors also provide a table which includes both the qualitative data and the qualitised data, demonstrating a transparent approach to data transformation. Lack of integration – where is the mixing of the quantitative and qualitative components? A core element of MMSRs is the integration of quantitative evidence and qualitative evidence to create a breadth and depth of understanding and corroboration that answers the systematic review question/s.(2, 10) Despite being a key feature of MMSRs, this process is either lacking or poorly described. Depending on the systematic review question, the integration of these two sets of data or evidence allows reviewers to triangulate or determine discrepancies within the available evidence, and/or use one type of evidence to explore, contextualise, or explain the findings of the other type of evidence.(3) In a review investigating women’s knowledge about prostate cancer presentation and screening practices, Wiafe et al 2021 (23) chose to follow a convergent segregated approach to synthesis and integration, where qualitative and quantitative evidence were integrated together following independent syntheses. This should be done by attempting to answer the five ‘trigger’ questions indicated in the JBI methodology which include: (1) Are the results/findings from individual synthesis supportive or contradictory? (2) Does the qualitative evidence explain why the intervention is or is not effective? (3) Does the qualitative evidence help explain differences in the direction and size of effect across the included quantitative studies? (4) Which aspects of the quantitative evidence are/are not explored in the qualitative studies? And 9 (5) Which aspects of the qualitative evidence are/are not tested in the quantitative evidence? While Wiafe et al 2021 present a narrative synthesis of the qualitative and quantitative results, they do not attempt to answer questions 2-5.(23) Davis et al evaluated the efficacy of telehealth and mobile health interventions in adults with inflammatory bowel disease and explored their benefits and challenges as they are experienced by this population group.(24) While the authors also followed a convergent segregated approach to synthesis and integration like Wiafe et al 2021,(23) these reviewers used the five ‘trigger’ questions to formally integrate the findings. Addressing these questions allowed the reviewers to make the quantitative and qualitative evidence ‘speak’ to each other while maintaining their epistemological foundation. Conclusion The information above outlines five issues the authors have encountered in reviews that claim to align themselves to the JBI MMSR methodological approach. This is by no means a comprehensive critique of the MMSR literature published since the revised guidance, but simply our collective experiences when reading a sample of MMSR, and the items described here are relevant to all types of MMSRs regardless of methodological approach taken (refer to references 9, 10, 11 for other approaches). A more formal evaluation of adherence to the JBI methodological approach will be undertaken in the future. This paper is intended to assist systematic reviewers considering conducting a MMSR as well as MMSR users to identify potential areas where authors may deviate from the methodological approach. Based on these findings we would recommend the following: ➢ Reviewers should ensure the review question/s can be answered by a MMSR, and that they are aligned with the appropriate approach to synthesis and integration. ➢ The inclusion of primary mixed methods studies should be considered essential and explicitly detailed in the review’s inclusion criteria. ➢ If transformation of data is necessary, the process for doing so is sufficiently described including the specific approach used, the process undertaken and a description of both the original data and the transformed data. ➢ Regardless of the approach taken, the quantitative and qualitative components should be integrated appropriately. In the case of those reviews following a convergent segregated approach this should entail the five trigger questions being answered following independent synthesis of the qualitative and quantitative data. For the convergent integrated approach, this will involve data transformation allowing the quantitative and qualitative data to be combined. References 1. Bressan V, Bagnasco A, Aleo G, Timmins F, Barisone M, Bianchi M, et al. Mixed-methods research in nursing - a critical review. J Clin Nurs. 2017;26(19-20):2878-90. 10 2. Dixon-Woods M, Agarwal S, Jones D, Young B, Sutton A. Synthesising qualitative and quantitative evidence: a review of possible methods. J Health Serv Res Policy. 2005;10(1):45-53. 3. Lizarondo L, Stern C, Carrier J, Godfrey C, Rieger K, Salmond S, et al. Chapter 8: Mixed methods systematic reviews. In: Aromataris E, Munn Z, editors. JBI Manual for Evidence Synthesis: JBI; 2020. 4. Stern C, Lizarondo L, Carrier J, Godfrey C, Rieger K, Salmond S, et al. Methodological guidance for the conduct of mixed methods systematic reviews. JBI Evid Synth. 2020;18(10):2108-18. 5. Stern C. Q is for Question…. JBI Database System Rev Implement Rep. 2015;13(9):1-2. 6. Young JL, Mak J, Stanley T, Bass M, Cho MK, Tabor HK. Genetic counseling and testing for Asian Americans: a systematic review. Genet Med. 2021;23(8):1424-37. 7. Peters MDJ, Marnie C, Tricco AC, Pollock D, Munn Z, Alexander L, et al. Updated methodological guidance for the conduct of scoping reviews. JBI Evid Synth. 2020;18(10):2119-26. 8. Clari M, Conti A, Fontanella R, Rossi A, Matarese M. Mindfulness-Based Programs for People with Chronic Obstructive Pulmonary Disease: a Mixed Methods Systematic Review. Mindfulness 2020;11:1848–67. 9. Hong QN, Pluye P, Bujold M, Wassef M. Convergent and sequential synthesis designs: implications for conducting and reporting systematic reviews of qualitative and quantitative evidence. Syst Rev. 2017;6(1):61. 10. Sandelowski M, Voils CI, Barroso J. Defining and Designing Mixed Research Synthesis Studies. Res Sch. 2006;13(1):29. 11. Noyes J, Booth A, Moore G, Flemming K, Tunçalp Ö, Shakibazadeh E. Synthesising quantitative and qualitative evidence to inform guidelines on complex interventions: clarifying the purposes, designs and outlining some methods. BMJ Glob Health. 2019;4(Suppl 1):e000893. 12. Rose K, O'Malley C, Eskandari F, Lake AA, Brown L, Ells LJ. The impact of, and views on, school food intervention and policy in young people aged 11-18 years in Europe: A mixed methods systematic review. Obes Rev. 2021;22(5):e13186. 13. Leeman L, Voils CI, Sandelowski M. Conducting Mixed Methods Literature Reviews: Synthesizing the Evidence Needed to Develop and Implement Complex Social and Health Interventions. Nagy Hesse-Biber S, Johnson RB, editors: Oxford Handbooks Online; 2015. 14. Atkins S, Launiala A, Kagaha A, Smith H. Including mixed methods research in systematic reviews: Examples from qualitative syntheses in TB and malaria control. BMC Medical Research Methodology. 2012;12(1):62. 15. O'Cathain A, Murphy E, Nicholl J. The quality of mixed methods studies in health services research. J Health Serv Res Policy. 2008;13(2):92-8. 16. Pluye P, Gagnon MP, Griffiths F, Johnson-Lafleur J. A scoring system for appraising mixed methods research, and concomitantly appraising qualitative, quantitative and mixed methods primary studies in Mixed Studies Reviews. Int J Nurs Stud. 2009;46(4):529-46. 17. Heyvaert M, Maes B, Onghena P. Mixed methods research synthesis: definition, framework, and potential. Quality and Quantity. 2013;47(2):659-76. 11 18. Ivankova NV, Creswell JW, Stick SL. Using Mixed-Methods Sequential Explanatory Design: From Theory to Practice. Field Methods. 2006;18(1):3-20. 19. McKenzie G, Willis C, Shields N. Barriers and facilitators of physical activity participation for young people and adults with childhood-onset physical disability: a mixed methods systematic review. Dev Med Child Neurol. 2021;63(8):914-24. 20. The Joanna Briggs Institute. Joanna Briggs Institute Reviewers’ Manual: 2014 edition / Supplement Methodology for JBI Mixed Methods Systematic Reviews. Adelaide, Australia.2014. 21. González-González AI, Schmucker C, Nothacker J, Nury E, Dinh TS, Brueckle MS, et al. End-of-Life Care Preferences of Older Patients with Multimorbidity: A Mixed Methods Systematic Review. J Clin Med. 2020;10(1). 22. Gray E, Dasanayake S, Sangelaji B, Hale L, Skinner M. Factors influencing physical activity engagement following coronary artery bypass graft surgery: A mixed methods systematic review. Heart Lung. 2021;50(5):589-98. 23. Wiafe E, Mensah KB, Mensah ABB, Bangalee V, Oosthuizen F. Knowledge of prostate cancer presentation, etiology, and screening practices among women: a mixed-methods systematic review. Systematic Reviews. 2021;10(1):138. 24. Davis SP, Ross MSH, Adatorwovor R, Wei H. Telehealth and mobile health interventions in adults with inflammatory bowel disease: A mixed-methods systematic review. Res Nurs Health. 2021;44(1):155-72.
16368	https://www.effortlessmath.com/math-topics/multi-step-word-problems-with-remainders/?srsltid=AfmBOorXcQOD9AiqAXqohs69s7akIQr5sgGdni2Bqw2mgho2awuUq7pm	How to Navigating Complex Scenarios: Multi-step Word Problems with Remainders - Effortless Math: We Help Students Learn to LOVE Mathematics Effortless Math X +eBooks +ACCUPLACER Mathematics +ACT Mathematics +AFOQT Mathematics +ALEKS Tests +ASVAB Mathematics +ATI TEAS Math Tests +Common Core Math +CLEP +DAT Math Tests +FSA Tests +FTCE Math +GED Mathematics +Georgia Milestones Assessment +GRE Quantitative Reasoning +HiSET Math Exam +HSPT Math +ISEE Mathematics +PARCC Tests +Praxis Math +PSAT Math Tests +PSSA Tests +SAT Math Tests +SBAC Tests +SIFT Math +SSAT Math Tests +STAAR Tests +TABE Tests +TASC Math +TSI Mathematics +Worksheets +ACT Math Worksheets +Accuplacer Math Worksheets +AFOQT Math Worksheets +ALEKS Math Worksheets +ASVAB Math Worksheets +ATI TEAS 6 Math Worksheets +FTCE General Math Worksheets +GED Math Worksheets +3rd Grade Mathematics Worksheets +4th Grade Mathematics Worksheets +5th Grade Mathematics Worksheets +6th Grade Math Worksheets +7th Grade Mathematics Worksheets +8th Grade Mathematics Worksheets +9th Grade Math Worksheets +HiSET Math Worksheets +HSPT Math Worksheets +ISEE Middle-Level Math Worksheets +PERT Math Worksheets +Praxis Math Worksheets +PSAT Math Worksheets +SAT Math Worksheets +SIFT Math Worksheets +SSAT Middle Level Math Worksheets +7th Grade STAAR Math Worksheets +8th Grade STAAR Math Worksheets +THEA Math Worksheets +TABE Math Worksheets +TASC Math Worksheets +TSI Math Worksheets +Courses +AFOQT Math Course +ALEKS Math Course +ASVAB Math Course +ATI TEAS 6 Math Course +CHSPE Math Course +FTCE General Knowledge Course +GED Math Course +HiSET Math Course +HSPT Math Course +ISEE Upper Level Math Course +SHSAT Math Course +SSAT Upper-Level Math Course +PERT Math Course +Praxis Core Math Course +SIFT Math Course +8th Grade STAAR Math Course +TABE Math Course +TASC Math Course +TSI Math Course +Puzzles +Number Properties Puzzles +Algebra Puzzles +Geometry Puzzles +Intelligent Math Puzzles +Ratio, Proportion & Percentages Puzzles +Other Math Puzzles +Math Tips +Articles +Blog How to Navigating Complex Scenarios: Multi-step Word Problems with Remainders Word problems often mirror real-life situations, requiring multiple steps to reach a solution. When these problems involve division, remainders can add an extra layer of complexity. In this guide, we’ll tackle multi-step word problems that result in remainders, offering strategies to interpret and solve them effectively. Step-by-step Guide to Solve Multi-step Word Problems with Remainders: 1. Understanding the Problem: Begin by reading the problem carefully. Identify the information given and determine what you’re being asked to find. 2. Breaking Down the Steps: Determine the sequence of operations required. This might involve addition, subtraction, multiplication, and division. 3. Handling Remainders: When you encounter a division step that results in a remainder: – Understand its significance in the context of the problem. – Decide whether to round up, round down, or use the remainder as is. 4. Solving the Problem: Execute the operations in the correct sequence, keeping track of any remainder. 5. Interpreting the Remainder: In the context of the problem, decide how to represent the remainder. For instance, if you divide candies among children, the remainder represents the leftover candies. Example 1: Anna has 53 apples. She wants to pack them in bags with 8 apples in each bag. How many full bags will she have, and how many apples will be left over? Solution: Dividing 53 by 8 gives 6 with a remainder of 5. Anna will have 6 full bags, with 5 apples left over. The Absolute Best Book for 5th Grade Students ### Mastering Grade 5 Math The Ultimate Step by Step Guide to Acing 5th Grade Math Download ~~$29.99~~Original price was: $29.99.$14.99 Current price is: $14.99. Example 2: A school is organizing a trip for 89 students. Each bus can carry 30 students. How many buses are needed, and how many seats will be empty on the last bus? Solution: Dividing 89 by 30 gives 2 with a remainder of 29. The school needs 3 buses (2 full buses and 1 more for the remaining 29 students). In the last bus, 1 seat will be empty. Practice Questions: A bakery has 125 muffins and wants to pack them in boxes of 10. How many full boxes will they have, and how many muffins will be left unpacked? There are 74 students in a competition. They are to be grouped into teams of 9. How many teams will there be, and how many students will not be in a team? A Perfect Book for Grade 5 Math Word Problems! ### Mastering Grade 5 Math Word Problems The Ultimate Guide to Tackling 5th Grade Math Word Problems Download ~~$29.99~~Original price was: $29.99.$14.99 Current price is: $14.99. Answers: 12 full boxes, 5 muffins left unpacked. 8 teams, 2 students will not be in a team. 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Other Topics Puzzle – Challenge 95 The Ultimate PERT Math Formula Cheat Sheet How to Unravel the Essential Properties of Rectangles How to Find Domain and Range of Relations How to Write the Equation of a Cosine Graph? FREE 4th Grade PSSA Math Practice Test How to Use Memory Tricks to Memorize Math Formulas? What people say about "How to Navigating Complex Scenarios: Multi-step Word Problems with Remainders - Effortless Math: We Help Students Learn to LOVE Mathematics"? No one replied yet. Leave a Reply Cancel reply You must be logged in to post a comment. ### Mastering Grade 6 Math The Ultimate Step by Step Guide to Acing 6th Grade Math ~~$29.99~~Original price was: $29.99.$14.99 Current price is: $14.99. Download ### Mastering Grade 5 Math The Ultimate Step by Step Guide to Acing 5th Grade Math ~~$29.99~~Original price was: $29.99.$14.99 Current price is: $14.99. 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16369	https://en.wikipedia.org/wiki/Analytic_geometry	Jump to content Analytic geometry Afrikaans Alemannisch العربية অসমীয়া Asturianu Azərbaycanca تۆرکجه বাংলা Башҡортса Беларуская Беларуская (тарашкевіца) Български Català Чӑвашла Čeština Cymraeg Dansk Deutsch Eesti Ελληνικά Español Esperanto Euskara فارسی Français Gaeilge Galego 한국어 Հայերեն हिन्दी Hrvatski Ido Bahasa Indonesia Italiano עברית Қазақша Кыргызча Latina Latviešu Lietuvių Magyar Македонски മലയാളം Монгол Nederlands 日本語 Norsk bokmål Norsk nynorsk Oʻzbekcha / ўзбекча Piemontèis Polski Português Qaraqalpaqsha Romnă Русский Scots Simple English Slovenčina کوردی Српски / srpski Srpskohrvatski / српскохрватски Suomi Svenska Tagalog தமிழ் ไทย Тоҷикӣ Türkçe Українська اردو Tiếng Việt 文言吴语粵語中文 Edit links From Wikipedia, the free encyclopedia Study of geometry using a coordinate system This article is about coordinate geometry. For the geometry of analytic varieties, see Algebraic geometry § Analytic geometry. \| Geometry \| \| Projecting a sphere to a plane \| \| Branches Euclidean Non-Euclidean + Elliptic - Spherical + Hyperbolic Non-Archimedean geometry Projective Affine Synthetic Analytic Algebraic + Arithmetic + Diophantine Differential + Riemannian + Symplectic + Discrete differential Complex Finite Discrete/Combinatorial + Digital Convex Computational Fractal Incidence Noncommutative geometry + Noncommutative algebraic geometry \| \| Concepts Features Dimension Straightedge and compass constructions Angle Curve Diagonal Orthogonality (Perpendicular) Parallel Vertex Congruence Similarity Symmetry \| \| Zero-dimensional Point \| \| One-dimensional Line + segment + ray + Curve Length \| \| Two-dimensional \| \| \| Surface + Plane Area Polygon \| \| Triangle \| \| Altitude Hypotenuse Pythagorean theorem \| \| Quadrilateral \| \| Parallelogram + Square + Rectangle + Rhombus + Rhomboid Trapezoid Kite \| \| Circle \| \| Diameter Circumference Area \| \| \| Three-dimensional \| \| \| Volume \| \| Polyhedron \| \| Platonic Solid Tetrahedron cuboid + Cube Octahedron Dodecahedron Icosahedron Pyramid \| \| Solid of revolution \| \| Sphere Cylinder Cone \| \| \| Four-/other-dimensional Tesseract Hypersphere \| \| Geometers \| \| by name Aida Aryabhata Ahmes Alhazen Apollonius Archimedes Atiyah Baudhayana Bolyai Brahmagupta Cartan Chern Coxeter Descartes Euclid Euler Gauss Gromov Hilbert Huygens Jyeṣṭhadeva Kātyāyana Khayyám Klein Lobachevsky Manava Minkowski Minggatu Pascal Pythagoras Parameshvara Poincaré Riemann Sakabe Sijzi al-Tusi Veblen Virasena Yang Hui al-Yasamin Zhang List of geometers \| \| by period \| BCE \| \| Ahmes Baudhayana Manava Pythagoras Euclid Archimedes Apollonius \| \| 1–1400s \| \| Zhang Kātyāyana Aryabhata Brahmagupta Virasena Alhazen Sijzi Khayyám al-Yasamin al-Tusi Yang Hui Parameshvara \| \| 1400s–1700s \| \| Jyeṣṭhadeva Descartes Pascal Huygens Minggatu Euler Sakabe Aida \| \| 1700s–1900s \| \| Gauss Lobachevsky Bolyai Riemann Klein Poincaré Hilbert Minkowski Cartan Veblen Coxeter Chern \| \| Present day \| \| Atiyah Gromov \| \| \| v t e \| In mathematics, analytic geometry, also known as coordinate geometry or Cartesian geometry, is the study of geometry using a coordinate system. This contrasts with synthetic geometry. Analytic geometry is used in physics and engineering, and also in aviation, rocketry, space science, and spaceflight. It is the foundation of most modern fields of geometry, including algebraic, differential, discrete and computational geometry. Usually the Cartesian coordinate system is applied to manipulate equations for planes, straight lines, and circles, often in two and sometimes three dimensions. Geometrically, one studies the Euclidean plane (two dimensions) and Euclidean space. As taught in school books, analytic geometry can be explained more simply: it is concerned with defining and representing geometric shapes in a numerical way and extracting numerical information from shapes' numerical definitions and representations. That the algebra of the real numbers can be employed to yield results about the linear continuum of geometry relies on the Cantor–Dedekind axiom. History [edit] Ancient Greece [edit] The Greek mathematician Menaechmus solved problems and proved theorems by using a method that had a strong resemblance to the use of coordinates and it has sometimes been maintained that he had introduced analytic geometry. Apollonius of Perga, in On Determinate Section, dealt with problems in a manner that may be called an analytic geometry of one dimension; with the question of finding points on a line that were in a ratio to the others. Apollonius in the Conics further developed a method that is so similar to analytic geometry that his work is sometimes thought to have anticipated the work of Descartes by some 1800 years. His application of reference lines, a diameter and a tangent is essentially no different from our modern use of a coordinate frame, where the distances measured along the diameter from the point of tangency are the abscissas, and the segments parallel to the tangent and intercepted between the axis and the curve are the ordinates. He further developed relations between the abscissas and the corresponding ordinates that are equivalent to rhetorical equations (expressed in words) of curves. However, although Apollonius came close to developing analytic geometry, he did not manage to do so since he did not take into account negative magnitudes and in every case the coordinate system was superimposed upon a given curve a posteriori instead of a priori. That is, equations were determined by curves, but curves were not determined by equations. Coordinates, variables, and equations were subsidiary notions applied to a specific geometric situation. Persia [edit] The 11th-century Persian mathematician Omar Khayyam saw a strong relationship between geometry and algebra and was moving in the right direction when he helped close the gap between numerical and geometric algebra with his geometric solution of the general cubic equations, but the decisive step came later with Descartes. Omar Khayyam is credited with identifying the foundations of algebraic geometry, and his book Treatise on Demonstrations of Problems of Algebra (1070), which laid down the principles of analytic geometry, is part of the body of Persian mathematics that was eventually transmitted to Europe. Because of his thoroughgoing geometrical approach to algebraic equations, Khayyam can be considered a precursor to Descartes in the invention of analytic geometry.: 248 Western Europe [edit] \| \| \| Part of a series on \| \| René Descartes \| \| \| \| Philosophy Cartesianism Rationalism Foundationalism Mechanism Doubt and certainty Existence of God Dream argument Cogito, ergo sum Evil demon Trademark argument Causal adequacy principle Mind–body dichotomy Analytic geometry Coordinate system Cartesian circle Folium Rule of signs Cartesian diver Balloonist theory Wax argument Res cogitans Res extensa \| \| Works Rules for the Direction of the Mind The Search for Truth by Natural Light The World Discourse on the Method La Géométrie Meditations on First Philosophy Principles of Philosophy Passions of the Soul \| \| People Christina, Queen of Sweden Francine Descartes \| \| v t e \| See also: René Descartes § Analytic geometry Analytic geometry was independently invented by René Descartes and Pierre de Fermat, although Descartes is sometimes given sole credit. Cartesian geometry, the alternative term used for analytic geometry, is named after Descartes. Descartes made significant progress with the methods in an essay titled La Géométrie (Geometry), one of the three accompanying essays (appendices) published in 1637 together with his Discourse on the Method for Rightly Directing One's Reason and Searching for Truth in the Sciences, commonly referred to as Discourse on Method. La Geometrie, written in his native French tongue, and its philosophical principles, provided a foundation for calculus in Europe. Initially the work was not well received, due, in part, to the many gaps in arguments and complicated equations. Only after the translation into Latin and the addition of commentary by van Schooten in 1649 (and further work thereafter) did Descartes's masterpiece receive due recognition. Pierre de Fermat also pioneered the development of analytic geometry. Although not published in his lifetime, a manuscript form of Ad locos planos et solidos isagoge (Introduction to Plane and Solid Loci) was circulating in Paris in 1637, just prior to the publication of Descartes' Discourse. Clearly written and well received, the Introduction also laid the groundwork for analytical geometry. The key difference between Fermat's and Descartes' treatments is a matter of viewpoint: Fermat always started with an algebraic equation and then described the geometric curve that satisfied it, whereas Descartes started with geometric curves and produced their equations as one of several properties of the curves. As a consequence of this approach, Descartes had to deal with more complicated equations and he had to develop the methods to work with polynomial equations of higher degree. It was Leonhard Euler who first applied the coordinate method in a systematic study of space curves and surfaces. Coordinates [edit] Main article: Coordinate system In analytic geometry, the plane is given a coordinate system, by which every point has a pair of real number coordinates. Similarly, Euclidean space is given coordinates where every point has three coordinates. The value of the coordinates depends on the choice of the initial point of origin. There are a variety of coordinate systems used, but the most common are the following: Cartesian coordinates (in a plane or space) [edit] Main article: Cartesian coordinate system The most common coordinate system to use is the Cartesian coordinate system, where each point has an x-coordinate representing its horizontal position, and a y-coordinate representing its vertical position. These are typically written as an ordered pair (x, y). This system can also be used for three-dimensional geometry, where every point in Euclidean space is represented by an ordered triple of coordinates (x, y, z). Polar coordinates (in a plane) [edit] Main article: Polar coordinate system In polar coordinates, every point of the plane is represented by its distance r from the origin and its angle θ, with θ normally measured counterclockwise from the positive x-axis. Using this notation, points are typically written as an ordered pair (r, θ). One may transform back and forth between two-dimensional Cartesian and polar coordinates by using these formulae: This system may be generalized to three-dimensional space through the use of cylindrical or spherical coordinates. Cylindrical coordinates (in a space) [edit] Main article: Cylindrical coordinate system In cylindrical coordinates, every point of space is represented by its height z, its radius r from the z-axis and the angle θ its projection on the xy-plane makes with respect to the horizontal axis. Spherical coordinates (in a space) [edit] Main article: Spherical coordinate system In spherical coordinates, every point in space is represented by its distance ρ from the origin, the angle θ its projection on the xy-plane makes with respect to the horizontal axis, and the angle φ that it makes with respect to the z-axis. The names of the angles are often reversed in physics. Equations and curves [edit] Main articles: Solution set and Locus (mathematics) In analytic geometry, any equation involving the coordinates specifies a subset of the plane, namely the solution set for the equation, or locus. For example, the equation y = x corresponds to the set of all the points on the plane whose x-coordinate and y-coordinate are equal. These points form a line, and y = x is said to be the equation for this line. In general, linear equations involving x and y specify lines, quadratic equations specify conic sections, and more complicated equations describe more complicated figures. Usually, a single equation corresponds to a curve on the plane. This is not always the case: the trivial equation x = x specifies the entire plane, and the equation x2 + y2 = 0 specifies only the single point (0, 0). In three dimensions, a single equation usually gives a surface, and a curve must be specified as the intersection of two surfaces (see below), or as a system of parametric equations. The equation x2 + y2 = r2 is the equation for any circle centered at the origin (0, 0) with a radius of r. Lines and planes [edit] Main articles: Line (geometry) and Plane (geometry) Lines in a Cartesian plane, or more generally, in affine coordinates, can be described algebraically by linear equations. In two dimensions, the equation for non-vertical lines is often given in the slope-intercept form: where: m is the slope or gradient of the line. b is the y-intercept of the line. x is the independent variable of the function y = f(x). In a manner analogous to the way lines in a two-dimensional space are described using a point-slope form for their equations, planes in a three dimensional space have a natural description using a point in the plane and a vector orthogonal to it (the normal vector) to indicate its "inclination". Specifically, let be the position vector of some point , and let be a nonzero vector. The plane determined by this point and vector consists of those points , with position vector , such that the vector drawn from to is perpendicular to . Recalling that two vectors are perpendicular if and only if their dot product is zero, it follows that the desired plane can be described as the set of all points such that (The dot here means a dot product, not scalar multiplication.) Expanded this becomes which is the point-normal form of the equation of a plane.[citation needed] This is just a linear equation: Conversely, it is easily shown that if a, b, c and d are constants and a, b, and c are not all zero, then the graph of the equation is a plane having the vector as a normal.[citation needed] This familiar equation for a plane is called the general form of the equation of the plane. In three dimensions, lines can not be described by a single linear equation, so they are frequently described by parametric equations: where: x, y, and z are all functions of the independent variable t which ranges over the real numbers. (x0, y0, z0) is any point on the line. a, b, and c are related to the slope of the line, such that the vector (a, b, c) is parallel to the line. Conic sections [edit] Main article: Conic section In the Cartesian coordinate system, the graph of a quadratic equation in two variables is always a conic section – though it may be degenerate, and all conic sections arise in this way. The equation will be of the form As scaling all six constants yields the same locus of zeros, one can consider conics as points in the five-dimensional projective space The conic sections described by this equation can be classified using the discriminant If the conic is non-degenerate, then: if , the equation represents an ellipse; if and , the equation represents a circle, which is a special case of an ellipse; if , the equation represents a parabola; if , the equation represents a hyperbola; if we also have , the equation represents a rectangular hyperbola. Quadric surfaces [edit] Main article: Quadric surface A quadric, or quadric surface, is a 2-dimensional surface in 3-dimensional space defined as the locus of zeros of a quadratic polynomial. In coordinates x1, x2,x3, the general quadric is defined by the algebraic equation Quadric surfaces include ellipsoids (including the sphere), paraboloids, hyperboloids, cylinders, cones, and planes. Distance and angle [edit] Main articles: Distance and Angle In analytic geometry, geometric notions such as distance and angle measure are defined using formulas. These definitions are designed to be consistent with the underlying Euclidean geometry. For example, using Cartesian coordinates on the plane, the distance between two points (x1, y1) and (x2, y2) is defined by the formula which can be viewed as a version of the Pythagorean theorem. Similarly, the angle that a line makes with the horizontal can be defined by the formula where m is the slope of the line. In three dimensions, distance is given by the generalization of the Pythagorean theorem: while the angle between two vectors is given by the dot product. The dot product of two Euclidean vectors A and B is defined by where θ is the angle between A and B. Transformations [edit] Transformations are applied to a parent function to turn it into a new function with similar characteristics. The graph of is changed by standard transformations as follows: Changing to moves the graph to the right units. Changing to moves the graph up units. Changing to stretches the graph horizontally by a factor of . (think of the as being dilated) Changing to stretches the graph vertically. Changing to and changing to rotates the graph by an angle . There are other standard transformation not typically studied in elementary analytic geometry because the transformations change the shape of objects in ways not usually considered. Skewing is an example of a transformation not usually considered. For more information, consult the Wikipedia article on affine transformations. For example, the parent function has a horizontal and a vertical asymptote, and occupies the first and third quadrant, and all of its transformed forms have one horizontal and vertical asymptote, and occupies either the 1st and 3rd or 2nd and 4th quadrant. In general, if , then it can be transformed into . In the new transformed function, is the factor that vertically stretches the function if it is greater than 1 or vertically compresses the function if it is less than 1, and for negative values, the function is reflected in the -axis. The value compresses the graph of the function horizontally if greater than 1 and stretches the function horizontally if less than 1, and like , reflects the function in the -axis when it is negative. The and values introduce translations, , vertical, and horizontal. Positive and values mean the function is translated to the positive end of its axis and negative meaning translation towards the negative end. Transformations can be applied to any geometric equation whether or not the equation represents a function. Transformations can be considered as individual transactions or in combinations. Suppose that is a relation in the plane. For example, is the relation that describes the unit circle. Finding intersections of geometric objects [edit] Main article: Intersection (geometry) For two geometric objects P and Q represented by the relations and the intersection is the collection of all points which are in both relations. For example, might be the circle with radius 1 and center : and might be the circle with radius 1 and center . The intersection of these two circles is the collection of points which make both equations true. Does the point make both equations true? Using for , the equation for becomes or which is true, so is in the relation . On the other hand, still using for the equation for becomes or which is false. is not in so it is not in the intersection. The intersection of and can be found by solving the simultaneous equations: Traditional methods for finding intersections include substitution and elimination. Substitution: Solve the first equation for in terms of and then substitute the expression for into the second equation: We then substitute this value for into the other equation and proceed to solve for : Next, we place this value of in either of the original equations and solve for : So our intersection has two points: Elimination: Add (or subtract) a multiple of one equation to the other equation so that one of the variables is eliminated. For our current example, if we subtract the first equation from the second we get . The in the first equation is subtracted from the in the second equation leaving no term. The variable has been eliminated. We then solve the remaining equation for , in the same way as in the substitution method: We then place this value of in either of the original equations and solve for : So our intersection has two points: For conic sections, as many as 4 points might be in the intersection. Finding intercepts [edit] Main articles: x-intercept and y-intercept One type of intersection which is widely studied is the intersection of a geometric object with the and coordinate axes. The intersection of a geometric object and the -axis is called the -intercept of the object. The intersection of a geometric object and the -axis is called the -intercept of the object. For the line , the parameter specifies the point where the line crosses the axis. Depending on the context, either or the point is called the -intercept. Geometric axis [edit] Axis in geometry is the perpendicular line to any line, object or a surface. Also for this may be used the common language use as a: normal (perpendicular) line, otherwise in engineering as axial line. In geometry, a normal is an object such as a line or vector that is perpendicular to a given object. For example, in the two-dimensional case, the normal line to a curve at a given point is the line perpendicular to the tangent line to the curve at the point. In the three-dimensional case a surface normal, or simply normal, to a surface at a point P is a vector that is perpendicular to the tangent plane to that surface at P. The word "normal" is also used as an adjective: a line normal to a plane, the normal component of a force, the normal vector, etc. The concept of normality generalizes to orthogonality. Spherical and nonlinear planes and their tangents [edit] Tangent is the linear approximation of a spherical or other curved or twisted line of a function. Tangent lines and planes [edit] Main article: Tangent In geometry, the tangent line (or simply tangent) to a plane curve at a given point is the straight line that "just touches" the curve at that point. Informally, it is a line through a pair of infinitely close points on the curve. More precisely, a straight line is said to be a tangent of a curve y = f(x) at a point x = c on the curve if the line passes through the point (c, f(c)) on the curve and has slope f'(c) where f' is the derivative of f. A similar definition applies to space curves and curves in n-dimensional Euclidean space. As it passes through the point where the tangent line and the curve meet, called the point of tangency, the tangent line is "going in the same direction" as the curve, and is thus the best straight-line approximation to the curve at that point. Similarly, the tangent plane to a surface at a given point is the plane that "just touches" the surface at that point. The concept of a tangent is one of the most fundamental notions in differential geometry and has been extensively generalized; see Tangent space. See also [edit] Applied geometry Cross product Rotation of axes Translation of axes Vector space Notes [edit] ^ Boyer, Carl B. (1991). "The Age of Plato and Aristotle". A History of Mathematics (Second ed.). John Wiley & Sons, Inc. pp. 94–95. ISBN 0-471-54397-7. Menaechmus apparently derived these properties of the conic sections and others as well. Since this material has a strong resemblance to the use of coordinates, as illustrated above, it has sometimes been maintained that Menaechmus had analytic geometry. Such a judgment is warranted only in part, for certainly Menaechmus was unaware that any equation in two unknown quantities determines a curve. In fact, the general concept of an equation in unknown quantities was alien to Greek thought. It was shortcomings in algebraic notations that, more than anything else, operated against the Greek achievement of a full-fledged coordinate geometry. ^ Boyer, Carl B. (1991). "Apollonius of Perga". A History of Mathematics (Second ed.). John Wiley & Sons, Inc. pp. 142. ISBN 0-471-54397-7. The Apollonian treatise On Determinate Section dealt with what might be called an analytic geometry of one dimension. It considered the following general problem, using the typical Greek algebraic analysis in geometric form: Given four points A, B, C, D on a straight line, determine a fifth point P on it such that the rectangle on AP and CP is in a given ratio to the rectangle on BP and DP. Here, too, the problem reduces easily to the solution of a quadratic; and, as in other cases, Apollonius treated the question exhaustively, including the limits of possibility and the number of solutions. ^ Boyer, Carl B. (1991). "Apollonius of Perga". A History of Mathematics (Second ed.). John Wiley & Sons, Inc. pp. 156. ISBN 0-471-54397-7. The method of Apollonius in the Conics in many respects are so similar to the modern approach that his work sometimes is judged to be an analytic geometry anticipating that of Descartes by 1800 years. The application of references lines in general, and of a diameter and a tangent at its extremity in particular, is, of course, not essentially different from the use of a coordinate frame, whether rectangular or, more generally, oblique. Distances measured along the diameter from the point of tangency are the abscissas, and segments parallel to the tangent and intercepted between the axis and the curve are the ordinates. The Apollonian relationship between these abscissas and the corresponding ordinates are nothing more nor less than rhetorical forms of the equations of the curves. However, Greek geometric algebra did not provide for negative magnitudes; moreover, the coordinate system was in every case superimposed a posteriori upon a given curve in order to study its properties. There appear to be no cases in ancient geometry in which a coordinate frame of reference was laid down a priori for purposes of graphical representation of an equation or relationship, whether symbolically or rhetorically expressed. Of Greek geometry we may say that equations are determined by curves, but not that curves are determined by equations. Coordinates, variables, and equations were subsidiary notions derived from a specific geometric situation; [...] That Apollonius, the greatest geometer of antiquity, failed to develop analytic geometry, was probably the result of a poverty of curves rather than of thought. General methods are not necessary when problems concern always one of a limited number of particular cases. ^ Jump up to: a b Boyer (1991). "The Arabic Hegemony". A History of Mathematics. pp. 241–242. ISBN 9780471543978. Omar Khayyam (ca. 1050–1123), the "tent-maker," wrote an Algebra that went beyond that of al-Khwarizmi to include equations of third degree. Like his Arab predecessors, Omar Khayyam provided for quadratic equations both arithmetic and geometric solutions; for general cubic equations, he believed (mistakenly, as the sixteenth century later showed), arithmetic solutions were impossible; hence he gave only geometric solutions. The scheme of using intersecting conics to solve cubics had been used earlier by Menaechmus, Archimedes, and Alhazan, but Omar Khayyam took the praiseworthy step of generalizing the method to cover all third-degree equations (having positive roots). For equations of higher degree than three, Omar Khayyam evidently did not envision similar geometric methods, for space does not contain more than three dimensions, ... One of the most fruitful contributions of Arabic eclecticism was the tendency to close the gap between numerical and geometric algebra. The decisive step in this direction came much later with Descartes, but Omar Khayyam was moving in this direction when he wrote, "Whoever thinks algebra is a trick in obtaining unknowns has thought it in vain. No attention should be paid to the fact that algebra and geometry are different in appearance. Algebras are geometric facts which are proved." ^ Cooper, Glen M. (2003). "Review: Omar Khayyam, the Mathmetician by R. Rashed, B. Vahabzadeh". The Journal of the American Oriental Society. 123 (1): 248–249. doi:10.2307/3217882. JSTOR 3217882. ^ Mathematical Masterpieces: Further Chronicles by the Explorers, p. 92 ^ Cooper, G. (2003). Journal of the American Oriental Society,123(1), 248-249. ^ Stillwell, John (2004). "Analytic Geometry". Mathematics and its History (Second ed.). Springer Science + Business Media Inc. p. 105. ISBN 0-387-95336-1. the two founders of analytic geometry, Fermat and Descartes, were both strongly influenced by these developments. ^ Boyer 2004, p. 74 ^ Cooke, Roger (1997). "The Calculus". The History of Mathematics: A Brief Course. Wiley-Interscience. pp. 326. ISBN 0-471-18082-3. The person who is popularly credited with being the discoverer of analytic geometry was the philosopher René Descartes (1596–1650), one of the most influential thinkers of the modern era. ^ Boyer 2004, p. 82 ^ Jump up to: a b Katz 1998, pg. 442 ^ Katz 1998, pg. 436 ^ Pierre de Fermat, Varia Opera Mathematica d. Petri de Fermat, Senatoris Tolosani (Toulouse, France: Jean Pech, 1679), "Ad locos planos et solidos isagoge," pp. 91–103. Archived 2015-08-04 at the Wayback Machine ^ "Eloge de Monsieur de Fermat" Archived 2015-08-04 at the Wayback Machine (Eulogy of Mr. de Fermat), Le Journal des Scavans, 9 February 1665, pp. 69–72. From p. 70: "Une introduction aux lieux, plans & solides; qui est un traité analytique concernant la solution des problemes plans & solides, qui avoit esté veu devant que M. des Cartes eut rien publié sur ce sujet." (An introduction to loci, plane and solid; which is an analytical treatise concerning the solution of plane and solid problems, which was seen before Mr. des Cartes had published anything on this subject.) ^ Jump up to: a b Stewart, James (2008). Calculus: Early Transcendentals, 6th ed., Brooks Cole Cengage Learning. ISBN 978-0-495-01166-8 ^ Percey Franklyn Smith, Arthur Sullivan Gale (1905)Introduction to Analytic Geometry, Athaeneum Press ^ William H. McCrea, Analytic Geometry of Three Dimensions Courier Dover Publications, Jan 27, 2012 ^ Vujičić, Milan; Sanderson, Jeffrey (2008), Vujičić, Milan; Sanderson, Jeffrey (eds.), Linear Algebra Thoroughly Explained, Springer, p. 27, doi:10.1007/978-3-540-74639-3, ISBN 978-3-540-74637-9 ^ Fanchi, John R. (2006), Math refresher for scientists and engineers, John Wiley and Sons, pp. 44–45, ISBN 0-471-75715-2, Section 3.2, page 45 ^ Silvio Levy Quadrics Archived 2018-07-18 at the Wayback Machine in "Geometry Formulas and Facts", excerpted from 30th Edition of CRC Standard Mathematical Tables and Formulas, CRC Press, from The Geometry Center at University of Minnesota ^ M.R. Spiegel; S. Lipschutz; D. Spellman (2009). Vector Analysis (Schaum's Outlines) (2nd ed.). McGraw Hill. ISBN 978-0-07-161545-7. ^ While this discussion is limited to the xy-plane, it can easily be extended to higher dimensions. References [edit] Books [edit] Boyer, Carl B. (2004) , History of Analytic Geometry, Dover Publications, ISBN 978-0486438320 Cajori, Florian (1999), A History of Mathematics, AMS, ISBN 978-0821821022 John Casey (1885) Analytic Geometry of the Point, Line, Circle, and Conic Sections, link from Internet Archive. Katz, Victor J. (1998), A History of Mathematics: An Introduction (2nd Ed.), Reading: Addison Wesley Longman, ISBN 0-321-01618-1 Mikhail Postnikov (1982) Lectures in Geometry Semester I Analytic Geometry via Internet Archive Struik, D. J. (1969), A Source Book in Mathematics, 1200-1800, Harvard University Press, ISBN 978-0674823556 Articles [edit] Bissell, Christopher C. (1987), "Cartesian geometry: The Dutch contribution", The Mathematical Intelligencer, 9 (4): 38–44, doi:10.1007/BF03023730 Boyer, Carl B. (1944), "Analytic Geometry: The Discovery of Fermat and Descartes", Mathematics Teacher, 37 (3): 99–105, doi:10.5951/MT.37.3.0099 Boyer, Carl B. (1965), "Johann Hudde and space coordinates", Mathematics Teacher, 58 (1): 33–36, doi:10.5951/MT.58.1.0033 Coolidge, J. L. (1948), "The Beginnings of Analytic Geometry in Three Dimensions", American Mathematical Monthly, 55 (2): 76–86, doi:10.2307/2305740, JSTOR 2305740 Pecl, J., Newton and analytic geometry External links [edit] Coordinate Geometry topics with interactive animations \| v t e Major mathematics areas \| \| --- \| \| History + Timeline + Future Lists Glossary \| \| \| Foundations \| Category theory Information theory Mathematical logic Philosophy of mathematics Set theory Type theory \| \| Algebra \| Abstract Commutative Elementary Group theory Linear Multilinear Universal Homological \| \| Analysis \| Calculus Real analysis Complex analysis Hypercomplex analysis Differential equations Functional analysis Harmonic analysis Measure theory \| \| Discrete \| Combinatorics Graph theory Order theory \| \| Geometry \| Algebraic Analytic Arithmetic Differential Discrete Euclidean Finite \| \| Number theory \| Arithmetic Algebraic number theory Analytic number theory Diophantine geometry \| \| Topology \| General Algebraic Differential Geometric Homotopy theory \| \| Applied \| Engineering mathematics Mathematical biology Mathematical chemistry Mathematical economics Mathematical finance Mathematical physics Mathematical psychology Mathematical sociology Mathematical statistics Probability Statistics Systems science + Control theory + Game theory + Operations research \| \| Computational \| Computer science Theory of computation Computational complexity theory Numerical analysis Optimization Computer algebra \| \| Related topics \| Mathematicians + lists Informal mathematics Films about mathematicians Recreational mathematics Mathematics and art Mathematics education \| \| Mathematics portal Category Commons WikiProject \| \| \| Authority control databases \| \| --- \| \| International \| \| \| National \| United States France BnF data Japan Czech Republic Latvia Israel \| \| Other \| Encyclopedia of Modern Ukraine Yale LUX \| Retrieved from " Category: Analytic geometry Hidden categories: Webarchive template wayback links Articles with short description Short description matches Wikidata Pages using sidebar with the child parameter All articles with unsourced statements Articles with unsourced statements from April 2022
16370	https://en.wikipedia.org/wiki/ISO/IEC_80000	Jump to content ISO/IEC 80000 Български Català Čeština Deutsch Eesti Español Français 한국어 हिन्दी Bahasa Indonesia Italiano Magyar Македонски മലയാളം 日本語 Polski Português Українська Yorùbá 中文 Edit links From Wikipedia, the free encyclopedia International standard on physical quantities and units of measurement ISO/IEC 80000, Quantities and units, is an international standard describing the International System of Quantities (ISQ). It was developed and promulgated jointly by the International Organization for Standardization (ISO) and the International Electrotechnical Commission (IEC). It serves as a style guide for using physical quantities and units of measurement, formulas involving them, and their corresponding units, in scientific and educational documents for worldwide use. The ISO/IEC 80000 family of standards was completed with the publication of the first edition of Part 1 in November 2009. Overview [edit] By 2021, ISO/IEC 80000 comprised 13 parts, two of which (parts 6 and 13) were developed by IEC and the remaining 11 were developed by ISO, with a further three parts (15, 16 and, 17) under development. Part 14 was withdrawn. \| Part \| Year \| Name \| Replaces \| Status \| --- --- \| ISO 80000-1 \| 2022 \| General \| ISO 31-0, IEC 60027-1, and IEC 60027-3 \| published \| \| ISO 80000-2 \| 2019 \| Mathematics \| ISO 31-11 and IEC 60027-1 \| published \| \| ISO 80000-3 \| 2019 \| Space and time \| ISO 31-1 and ISO 31-2 \| published \| \| ISO 80000-4 \| 2019 \| Mechanics \| ISO 31-3 \| published \| \| ISO 80000-5 \| 2019 \| Thermodynamics \| ISO 31-4 \| published \| \| IEC 80000-6 \| 2022 \| Electromagnetism \| ISO 31-5 \| published \| \| ISO 80000-7 \| 2019 \| Light and radiation \| ISO 31-6 \| published \| \| ISO 80000-8 \| 2020 \| Acoustics \| ISO 31-7 \| published \| \| ISO 80000-9 \| 2019 \| Physical chemistry and molecular physics \| ISO 31-8 \| published \| \| ISO 80000-10 \| 2019 \| Atomic and nuclear physics \| ISO 31-9 and ISO 31-10 \| published \| \| ISO 80000-11 \| 2019 \| Characteristic numbers \| ISO 31-12 \| published \| \| ISO 80000-12 \| 2019 \| Condensed matter physics \| ISO 31-13 \| published \| \| IEC 80000-13 \| 2025 \| Information science and technology \| subclauses 3.8 and 3.9 of IEC 60027-2:2005 \| published \| \| IEC 80000-14 \| 2008 \| Telebiometrics related to human physiology \| IEC 60027-7 \| withdrawn \| \| IEC 80000-15 \| \| Logarithmic and related quantities \| \| under development \| \| IEC 80000-16 \| \| Printing and writing rules \| \| under development \| \| IEC 80000-17 \| \| Time dependency \| \| under development \| Subject areas [edit] By 2021 the 80000 standard had 13 published parts. A description of each part is available online, with the complete parts for sale. Part 1: General [edit] ISO 80000-1:2022 revised ISO 80000-1:2009, which replaced ISO 31-0:1992 and ISO 1000:1992. This document gives general information and definitions concerning quantities, systems of quantities, units, quantity and unit symbols, and coherent unit systems, especially the International System of Quantities (ISQ). The descriptive text of this part is available online. According to the standard, symbols for quantities are "generally single letters from the Latin or Greek alphabet" and are "written in italic (sloping) type". Examples include density of heat flow rate: q = Φ / A electric current density: J = I / A magnetic flux density: B = Φ / A Part 2: Mathematics [edit] ISO 80000-2:2019 revised ISO 80000-2:2009, which superseded ISO 31-11. It specifies mathematical symbols, explains their meanings, and gives verbal equivalents and applications. The descriptive text of this part is available online. Part 3: Space and time [edit] ISO 80000-3:2019 revised ISO 80000-3:2006, which supersedes ISO 31-1 and ISO 31-2. It gives names, symbols, definitions and units for quantities of space and time. The descriptive text of this part is available online. A definition of the decibel, included in the original 2006 publication, was omitted in the 2019 revision, leaving ISO/IEC 80000 without a definition of this unit; a new part of the standard, IEC 80000-15 (Logarithmic and related quantities), is under development. Part 4: Mechanics [edit] ISO 80000-4:2019 revised ISO 80000-4:2006, which superseded ISO 31-3. It gives names, symbols, definitions and units for quantities of mechanics. The descriptive text of this part is available online. Part 5: Thermodynamics [edit] ISO 80000-5:2019 revised ISO 80000-5:2007, which superseded ISO 31-4. It gives names, symbols, definitions and units for quantities of thermodynamics. The descriptive text of this part is available online. Part 6: Electromagnetism [edit] IEC 80000-6:2022 revised IEC 80000-6:2008, which superseded ISO 31-5 as well as IEC 60027-1. It gives names, symbols, and definitions for quantities and units of electromagnetism. The descriptive text of this part is available online. Part 7: Light and radiation [edit] ISO 80000-7:2019 revised ISO 80000-7:2008, which superseded ISO 31-6. It gives names, symbols, definitions and units for quantities used for light and optical radiation in the wavelength range of approximately 1 nm to 1 mm. The descriptive text of this part is available online. Part 8: Acoustics [edit] ISO 80000-8:2020 revised ISO 80000-8:2007, which revised ISO 31-7:1992. It gives names, symbols, definitions, and units for quantities of acoustics. The descriptive text of this part is available online. It has a foreword, scope introduction, scope, normative references (of which there are none), as well as terms, and definitions. It includes definitions of sound pressure, sound power, and sound exposure, and their corresponding levels: sound pressure level, sound power level, and sound exposure level. It includes definitions of the following quantities: logarithmic frequency range static pressure sound pressure sound particle displacement sound particle velocity sound particle acceleration volume flow rate, volume velocity sound energy density sound energy sound power sound intensity sound exposure characteristic impedance for longitudinal waves acoustic impedance sound pressure level sound power level sound exposure level reverberation time Part 13: Information science and technology [edit] IEC 80000-13:2025 revised IEC 80000-13:2008, which replaced subclauses 3.8 and 3.9 of IEC 60027-2:2005 and IEC 60027-3. It defines quantities and units used in information science and information technology, and specifies names and symbols for these quantities and units. It has a scope; normative references; names, definitions, and symbols; and prefixes for binary multiples. Quantities defined in this standard are: traffic intensity [A]: number of simultaneously busy resources in a particular pool of resources traffic offered intensity [A0]: traffic intensity ... of the traffic that would have been generated by the users of a pool of resources if their use had not been limited by the size of the pool traffic carried intensity [Y]: traffic intensity ... of the traffic served by a particular pool of resources mean queue length [L, (Ω)]: time average of queue length loss probability [B]: probability for losing a call attempt waiting probability [W]: probability for waiting for a resource call intensity, calling rate [λ]: number of call attempts over a specified time interval divided by the duration of this interval completed call intensity [μ]: call intensity ... for the call attempts that result in the transmission of an answer signal storage capacity, storage size [M] equivalent binary storage capacity [Me] transfer rate [r, (ν)] period of data elements [T] binary digit rate, bit rate [rb, rbit (νb, νbit)] period of binary digits, bit period [Tb, Tbit] equivalent binary digit rate, equivalent bit rate [re, (νe)] modulation rate, line digit rate [rm, u] quantizing distortion power [TQ] carrier power [Pc, C] signal energy per binary digit [Eb, Ebit] error probability [P] Hamming distance [dn] clock frequency, clock rate [fcl] decision content [Da] information content [I(x)] entropy [H] maximum entropy [H0, (Hmax)] relative entropy [Hr] redundancy [R] relative redundancy [r] joint information content [I(x, y)] conditional information content [I(x\|y)] conditional entropy, mean conditional information content, average conditional information content [H(X\|Y)] equivocation [H(X\|Y)] irrelevance [C] transinformation content [T(x, y)] mean transinformation content [T] character mean entropy [H′] average information rate [H] character mean transinformation content [T′] average transinformation rate [T] channel capacity per character; channel capacity [C′] channel time capacity; channel capacity [C] The standard also includes definitions for units relating to information technology, such as the erlang (E), bit (bit), octet (o), byte (B), baud (Bd), shannon (Sh), hartley (Hart), and the natural unit of information (nat). Clause 4 of the standard defines standard binary prefixes used to denote powers of 1024 as 10241 (kibi-), 10242 (mebi-), 10243 (gibi-), 10244 (tebi-), 10245 (pebi-), 10246 (exbi-), 10247 (zebi-), 10248 (yobi-), 10249 (robi-), and 102410 (quebi-). International System of Quantities [edit] Main article: International System of Quantities Part 1 of ISO 80000 introduces the International System of Quantities and describes its relationship with the International System of Units (SI). Specifically, its introduction states "The system of quantities, including the relations among the quantities used as the basis of the units of the SI, is named the International System of Quantities, denoted 'ISQ', in all languages." It further clarifies that "ISQ is simply a convenient notation to assign to the essentially infinite and continually evolving and expanding system of quantities and equations on which all of modern science and technology rests. ISQ is a shorthand notation for the 'system of quantities on which the SI is based'." Units of the ISO and IEC 80000 series [edit] The standard includes all SI units but is not limited to only SI units. Units that form part of the standard but not the SI include the units of information storage (bit and byte), units of entropy (shannon, natural unit of information and hartley), and the erlang (a unit of traffic intensity). See also [edit] International Vocabulary of Metrology International System of Units BIPM – publishes freely available information on SI units NIST – official U.S. representative for SI; publishes freely available guide to use of SI References [edit] ^ Standards Catalogue TC/12 Quantities and Units ^ Schirn, Alexandra (9 January 2023). "ISO 80000-1:2022- Quantities And Units – ANSI Blog". The ANSI Blog. Retrieved 16 May 2023. ^ Jump up to: a b "ISO 80000-1:2022 Quantities and units — Part 1: General". International Organization for Standardization. Retrieved 2 November 2023. ^ Jump up to: a b "ISO 80000-2:2019 Quantities and units — Part 2: Mathematics". International Organization for Standardization. Retrieved 15 September 2019. ^ Jump up to: a b "ISO 80000-3:2019 Quantities and units — Part 3: Space and time". International Organization for Standardization. Retrieved 23 October 2019. ^ Jump up to: a b "ISO 80000-4:2019 Quantities and units — Part 4: Mechanics". International Organization for Standardization. Retrieved 15 September 2019. ^ Jump up to: a b "ISO 80000-5:2019 Quantities and units — Part 5: Thermodynamics". International Organization for Standardization. Retrieved 15 September 2019. ^ Jump up to: a b "IEC 80000-6:2022 Quantities and units — Part 6: Electromagnetism". International Organization for Standardization. Retrieved 20 November 2022. ^ Jump up to: a b "ISO 80000-7:2019 Quantities and units — Part 4: Light and radiation". International Organization for Standardization. Retrieved 23 October 2019. ^ "ISO 80000-8:2020 Quantities and units — Part 8: Acoustics". International Organization for Standardization. Retrieved 20 March 2020. ^ "ISO 80000-9:2019 Quantities and units — Part 9: Physical chemistry and molecular physics". International Organization for Standardization. Retrieved 15 September 2019. ^ "ISO 80000-10:2019 Quantities and units — Part 10: Atomic and nuclear physics". International Organization for Standardization. Retrieved 15 September 2019. ^ "ISO 80000-11:2019 Quantities and units — Part 11: Characteristic numbers". International Organization for Standardization. Retrieved 11 February 2020. ^ "ISO 80000-12:2019 Quantities and units — Part 12: Condensed matter physics". International Organization for Standardization. Retrieved 15 September 2019. ^ Jump up to: a b "IEC 80000-13:2025 Quantities and units — Part 13: Information science and technology". International Organization for Standardization. Retrieved 21 February 2025. ^ "IEC 80000-14:2008 Quantities and units — Part 14: Telebiometrics related to human physiology". International Organization for Standardization. Retrieved 15 September 2019. ^ "IEC/CD 80000-15 Quantities and units — Part 15: Logarithmic and related quantities". International Organization for Standardization. Retrieved 24 April 2022. ^ "IEC/CD 80000-16: Quantities and units — Part 16: Printing and writing rules". International Organization for Standardization. Retrieved 1 May 2021. ^ "IEC/CD 80000-17 Quantities and units — Part 17: Time dependency". International Organization for Standardization. Retrieved 1 May 2021. ^ "International standards for quantities and units are under revision". EE Publishers. 31 January 2017. Archived from the original on 4 December 2019. Retrieved 4 December 2019. ^ "Standards by ISO/TC 12 – Quantities and units". International Organization for Standardization. Retrieved 24 April 2020. ^ "ISO 80000-1:2009 Quantities and units – Part 1: General". International Organization for Standardization. Retrieved 2 November 2023. ^ "ISO 80000-1:2022(en) Quantities and units – Part 1: General". International Organization for Standardization. Retrieved 2 November 2023. ^ "ISO 80000-1:2009(en) Quantities and units – Part 1: General". International Organization for Standardization. Retrieved 2 November 2023. ^ "ISO 80000-2:2009". International Organization for Standardization. Retrieved 15 September 2019. ^ "ISO 80000-2:2019(en) Quantities and units – Part 2: Mathematics". International Organization for Standardization. Retrieved 23 October 2019. ^ "ISO 80000-3:2006". International Organization for Standardization. Retrieved 20 July 2013. ^ "ISO 80000-3:2019(en) Quantities and units – Part 3: Space and time". International Organization for Standardization. Retrieved 23 October 2019. ^ "ISO 80000-4:2006". International Organization for Standardization. Retrieved 15 September 2019. ^ "ISO 80000-4:2019 Quantities and units – Part 4: Mechanics". International Organization for Standardization. Retrieved 23 October 2019. ^ "ISO 80000-5:2007". International Organization for Standardization. Retrieved 15 September 2019. ^ "ISO 80000-5:2019(en) Quantities and units – Part 5: Thermodynamics". International Organization for Standardization. Retrieved 23 October 2019. ^ "IEC 80000-6:2008". International Organization for Standardization. Retrieved 20 November 2022. ^ "ISO 80000-6:2022(en,fr) Quantities and units – Part 6: Electromagnetism". International Organization for Standardization. Retrieved 21 November 2022. ^ "ISO 80000-7:2008". International Organization for Standardization. Retrieved 15 September 2019. ^ "ISO 80000-7:2019(en) Quantities and units – Quantities and units – Part 7: Light and radiation". International Organization for Standardization. Retrieved 25 March 2021. ^ "ISO 80000-8:2007". International Organization for Standardization. Retrieved 23 April 2020. ^ "ISO 31-7:1992". International Organization for Standardization. Retrieved 23 April 2020. ^ "ISO 80000-8:2020(en) Quantities and units – Part 8: Acoustics". International Organization for Standardization. Retrieved 23 April 2020. External links [edit] BIPM SI Brochure ISO TC12 standards – Quantities, units, symbols, conversion factors NIST Special Publication 330 – The International System of Units NIST Special Publication 811 – Guide for the Use of the International System of Units \| International Organization for Standardization (ISO) standards \| \| --- \| \| List of ISO standards – ISO romanizations – IEC standards \| \| \| 1–9999 \| 1 2 3 4 6 7 9 16 17 31 + -0 + -1 + -3 + -4 + -5 + -6 + -7 + -8 + -9 + -10 + -11 + -12 + -13 68-1 128 216 217 226 228 233 259 261 262 302 306 361 500 518 519 639 + -1 + -2 + -3 + -5 + -6 646 657 668 690 704 732 764 838 843 860 898 965 999 1000 1004 1007 1073-1 1073-2 1155 1413 1538 1629 1745 1989 2014 2015 2022 2033 2047 2108 2145 2146 2240 2281 2533 2709 2711 2720 2788 2848 2852 2921 3029 3103 3166 + -1 + -2 + -3 3297 3307 3601 3602 3864 3901 3950 3977 4031 4157 4165 4217 4909 5218 5426 5427 5428 5725 5775 5776 5800 5807 5964 6166 6344 6346 6373 6385 6425 6429 6438 6523 6709 6943 7001 7002 7010 7027 7064 7098 7185 7200 7498 + -1 7637 7736 7810 7811 7812 7813 7816 7942 8000 8093 8178 8217 8373 8501-1 8571 8583 8601 8613 8632 8651 8652 8691 8805/8806 8807 8820-5 8859 + -1 + -2 + -3 + -4 + -5 + -6 + -7 + -8 + -8-I + -9 + -10 + -11 + -12 + -13 + -14 + -15 + -16 8879 9000/9001 9036 9075 9126 9141 9227 9241 9293 9314 9362 9407 9496 9506 9529 9564 9592/9593 9594 9660 9797-1 9897 9899 9945 9984 9985 9995 \| \| 10000–19999 \| 10006 10007 10116 10118-3 10160 10161 10165 10179 10206 10218 10279 10303 + -11 + -21 + -22 + -28 + -238 10383 10585 10589 10628 10646 10664 10746 10861 10957 10962 10967 11073 11170 11172 11179 11404 11544 11783 11784 11785 11801 11889 11898 11940 (-2) 11941 11941 (TR) 11992 12006 12052 12182 12207 12234-2 12620 13211 + -1 + -2 13216 13250 13399 13406-2 13450 13485 13490 13567 13568 13584 13616 13816 13818 14000 14031 14224 14289 14396 14443 14496 + -2 + -3 + -6 + -10 + -11 + -12 + -14 + -17 + -20 14617 14644 14649 14651 14698 14764 14882 14971 15022 15189 15288 15291 15398 15408 15444 + -3 + -9 15445 15438 15504 15511 15686 15693 15706 + -2 15707 15897 15919 15924 15926 15926 WIP 15930 15938 16023 16262 16355-1 16485 16612-2 16750 16949 (TS) 17024 17025 17100 17203 17369 17442 17506 17799 18004 18014 18181 18245 18629 18760 18916 19005 19011 19092 + -1 + -2 19114 19115 19125 19136 19407 19439 19500 19501 19502 19503 19505 19506 19507 19508 19509 19510 19600 19752 19757 19770 19775-1 19794-5 19831 \| \| 20000–29999 \| 20000 20022 20121 20400 20802 20830 21000 21001 21047 21122 21500 21778 21827 22000 22275 22300 22301 22395 22537 23000 23003 23008 23009 23090-3 23092 23094-1 23094-2 23270 23271 23360 23941 24517 24613 24617 24707 24728 25178 25964 26000 26262 26300 26324 27000 series 27000 27001 27002 27005 27006 27729 28000 29110 29148 29199-2 29500 \| \| 30000+ \| 30170 31000 32000 37001 38500 39075 40314 40500 42010 45001 50001 55000 56000 80000 \| \| \| \| \| v t e IEC standards \| \| --- \| \| IEC \| 60027 60034 60038 60062 60063 60068 60112 60228 60269 60297 60309 60320 60364 60446 60559 60601 60870 + 60870-5 + 60870-6 60906-1 60908 60929 60958 60980-344 61030 61131 + 61131-3 + 61131-9 61158 61162 61334 61355 61360 61400 61499 61508 61511 61784 61850 61851 61883 61960 61968 61970 62014-4 62026 62056 62061 62196 62262 62264 62304 62325 62351 62365 62366 62379 62386 62455 62680 62682 62700 63110 63119 63382 \| \| ISO/IEC \| 646 1989 2022 4909 5218 6429 6523 7810 7811 7812 7813 7816 7942 8613 8632 8652 8859 9126 9293 9496 9529 9592 9593 9899 9945 9995 10021 10116 10165 10179 10279 10646 10967 11172 11179 11404 11544 11801 12207 13250 13346 13522-5 13568 13816 13818 14443 14496 14651 14882 15288 15291 15408 15444 15445 15504 15511 15693 15897 15938 16262 16485 17024 17025 18004 18014 18181 19752 19757 19770 19788 20000 20802 21000 21827 22275 22537 23000 23003 23008 23270 23360 24707 24727 24744 24752 26300 27000 27000 family 27002 27040 29110 29119 33001 38500 39075 42010 80000 81346 \| \| Related \| International Electrotechnical Commission \| Retrieved from " Categories: ISO/IEC 80000 Physical quantities Measurement ISO/IEC standards Hidden categories: Articles with short description Short description is different from Wikidata Use dmy dates from March 2021
16371	https://ocw.mit.edu/courses/16-30-feedback-control-systems-fall-2010/a2e9eccea56da4ba879a5a3e15cf8cde_MIT16_30F10_lec09.pdf	Topic #9 16.30/31 Feedback Control Systems State-Space Systems • State-space model features • Observability • Controllability Minimal Realizations • " # " # Fall 2010 16.30/31 9–2 State-Space Model Features • There are some key characteristics of a state-space model that we need to identify. • Will see that these are very closely associated with the concepts of pole/zero cancelation in transfer functions. x x • Example: Consider a simple system 6 G(s) = s + 2 for which we develop the state-space model Model # 1 x ˙ = −2x + 2u y = 3x But now consider the new state space model ¯ T • ¯ = [x x2] −2 0 2 ˙ ¯ x y = 3 0 Model # 2 = + u 1 ¯ x 0 −1 which is clearly diﬀerent than the ﬁrst model, and larger. But let’s looks at the transfer function of the new model: • G ¯(s) = C(sI − A)−1B " + D #!−1 = 3 0 sI − −2 0 −1 0 1 2 2 s+2 6 = 3 0 = !! 1 s + 2 s+1 September 30, 2010 " # Fall 2010 16.30/31 9–3 • This is a bit strange, because previously our ﬁgure of merit when comparing one state-space model to another (page 6–??) was whether they reproduced the same same transfer function • Now we have two very diﬀerent models that result in the same transfer function • Note that I showed the second model as having 1 extra state, but I could easily have done it with 99 extra states!! • So what is going on? • A clue is that the dynamics associated with the second state of the model x2 were eliminated when we formed the product ¯ G(s) = 3 0 2 s+2 1 s+1 because the A is decoupled and there is a zero in the C matrix • Which is exactly the same as saying that there is a pole-zero cancelation in the transfer function G ˜(s) 6 6(s + 1) = , G ˜(s) s + 2 (s + 2)(s + 1) Note that model #2 is one possible state-space model of G ˜(s) • (has 2 poles) • For this system we say that the dynamics associated with the second state are unobservable using this sensor (deﬁnes C matrix). • There could be a lot “motion” associated with x2, but we would be unaware of it using this sensor. September 30, 2010 " # Fall 2010 16.30/31 9–4 • There is an analogous problem on the input side as well. Consider: Model # 1 x ˙ = −2x + 2u y = 3x with ¯ T x = [ x x2] ¯ x −2 0 2 ¯ ˙ y = 3 2 x Model # 3 = + u 0 0 −1 ¯ x which is also clearly diﬀerent than model #1, and has a diﬀerent form from the second model. ˆ " −2 0 #!−1 2 G(s) = 3 2 sI − 0 −1 0 2 6 = 3 2 = !! s+2 s+1 0 s + 2 • Once again the dynamics associated with the pole at s = −1 are canceled out of the transfer function. But in this case it occurred because there is a 0 in the B matrix • • So in this case we can “see” the state x2 in the output C = 3 2 , but we cannot “inﬂuence” that state with the input since 2 B = 0 • So we say that the dynamics associated with the second state are uncontrollable using this actuator (deﬁnes the B matrix). September 30, 2010 " # Fall 2010 16.30/31 9–5 • Of course it can get even worse because we could have ¯ ˙ ¯ x x −2 0 2 = + u 0 0 −1 ¯ x y = 3 0 So now we have • " #!−1 ] = 0 sI − −2 0 −1 0 0 2 G(s) 3 2 6 = 3 0 = !! s+2 s+1 0 s + 2 • Get same result for the transfer function, but now the dynamics as sociated with x2 are both unobservable and uncontrollable. • Summary: Dynamics in the state-space model that are uncon trollable, unobservable, or both do not show up in the transfer function. • Would like to develop models that only have dynamics that are both controllable and observable V called a minimal realization • A state space model that has the lowest possible order for the given transfer function. • But ﬁrst need to develop tests to determine if the models are observ able and/or controllable September 30, 2010 Fall 2010 16.30/31 9–6 Observability • Deﬁnition: An LTI system is observable if the initial state x(0) can be uniquely deduced from the knowledge of the input u(t) and output y(t) for all t between 0 and any ﬁnite T > 0. • If x(0) can be deduced, then we can reconstruct x(t) exactly because we know u(t) V we can ﬁnd x(t) ∀ t. • Thus we need only consider the zero-input (homogeneous) solution to study observability. y(t) = CeAt x(0) • This deﬁnition of observability is consistent with the notion we used before of being able to “see” all the states in the output of the de-coupled examples • ROT: For those decoupled examples, if part of the state cannot be “seen” in y(t), then it would be impossible to deduce that part of x(0) from the outputs y(t). September 30, 2010 " # Fall 2010 16.30/31 9–7 • Deﬁnition: A state x? 6= 0 is said to be unobservable if the zero-input solution y(t), with x(0) = x?, is zero for all t ≥ 0 • Equivalent to saying that x? is an unobservable state if CeAt x ? = 0 ∀ t ≥ 0 • For the problem we were just looking at, consider Model #2 with x? = [ 0 1 ]T = 0 6 , then x ¯ ˙ = −2 0 x ¯ + 2 u 0 −1 1 y = 3 0 x ¯ so e−2t 0 0 CeAt x ? = 3 0 0 e−t 1 0 = 3e−2t 0 = 0 ∀ t 1 So, x? = [ 0 1 ]T is an unobservable state for this system. • But that is as expected, because we knew there was a problem with the state x2 from the previous analysis September 30, 2010 • 6 Fall 2010 16.30/31 9–8 • Theorem: An LTI system is observable iﬀ it has no unob servable states. • We normally just say that the pair (A,C) is observable. Pseudo-Proof: Let x? = 0 be an unobservable state and compute the outputs from the initial conditions x1(0) and x2(0) = x1(0) + x? Then • y1(t) = CeAt x1(0) and y2(t) = CeAt x2(0) but y2(t) = CeAt(x1(0) + x ?) = CeAt x1(0) + CeAt x ? = CeAt x1(0) = y1(t) • Thus 2 diﬀerent initial conditions give the same output y(t), so it would be impossible for us to deduce the actual initial condition of the system x1(t) or x2(t) given y1(t) • Testing system observability by searching for a vector x(0) such that CeAtx(0) = 0 ∀ t is feasible, but very hard in general. Better tests are available. • September 30, 2010 Fall 2010 16.30/31 9–9 Theorem: The vector x? is an unobservable state iﬀ • ⎤ ⎡ C CA ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ CA2 . . . CAn−1 ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ x ? = 0 • Pseudo-Proof: If x? is an unobservable state, then by deﬁnition, CeAt x ? = 0 ∀ t ≥ 0 But all the derivatives of CeAt exist and for this condition to hold, all derivatives must be zero at t = 0. Then =0 t CeAt x ? d CeAt x ? dt = 0 Cx ? = 0 ⇒ CAeAt x ? = CAx ? = 0 = 0 ⇒ t=0 t=0 d2 CeAt x ? dt2 CA2 eAt x ? = CA2 x ? = 0 = 0 ⇒ t=0 t=0 . . . dk CeAt x ? dtk CAkeAt x ? = CAk x ? = 0 = 0 ⇒ t=0 t=0 • We only need retain up to the n − 1th derivative because of the Cayley-Hamilton theorem. September 30, 2010 • 6 Fall 2010 16.30/31 9–10 • Simple test: Necessary and suﬃcient condition for observability is that ⎤ ⎡ rank Mo , rank ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ C CA CA2 . . . CAn−1 ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = n • Why does this make sense? The requirement for an unobservable state is that for x? = 0 Mox ? = 0 • Which is equivalent to saying that x? is orthogonal to each row of Mo. • But if the rows of Mo are considered to be vectors and these span the full n-dimensional space, then it is not possible to ﬁnd an n-vector x? that is orthogonal to each of these. • To determine if the n rows of Mo span the full n-dimensional space, we need to test their linear independence, which is equiv alent to the rank test1 1Let M be a m × p matrix, then the rank of M satisﬁes: 1. rank M ≡ number of linearly independent columns of M 2. rank M ≡ number of linearly independent rows of M 3. rank M ≤ min{m, p} September 30, 2010 MIT OpenCourseWare 16.30 / 16.31 Feedback Control Systems Fall 2010 For information about citing these materials or our Terms of Use, visit:
16372	https://www.researchgate.net/post/What_are_the_factors_which_affect_the_self_purification_process_in_river2	Published Time: Thu, 23 May 13 19:56:02 +0000 What are the factors which affect the self purification process in river? \| ResearchGate Question Answers 10 Similar questions Research that mentions Aquatic Ecosystems Question Asked 16 April 2013 Mophin Kani Kani UKF College of Engineering & Technology What are the factors which affect the self purification process in river? In developing countries most of the developmental activities are along the banks of the river, which increase the pollution threat to the ecosystem and its functioning too. Nature provides self healing system to all ecosystem but it doesn't work in most of the times? Freshwater Ecology Ecology Aquatic Ecosystems Ecosystem Ecology Share FacebookTwitterLinkedInReddit Most recent answer Azubuike Victor Chukwuka National Environmental Standards and Regulations Enforcement Agency (NESREA) @Mark its unlikely for self purification to occur more in the temperate than in the tropics because for instance the breakdown of organic matter via microbes is more optimal and occurs all year round in the tropics. Cite Top contributors to discussions in this field Ljubomir Jacić Technical College Požarevac Yogesh Chandra Tripathi Forest Research Institute Arvind Singh Banaras Hindu University Dhyaa Allban Krishnan Umachandran WRW Services LLP Powered By 10 Another potential side effect to the pill: making it harder for women to read certain emotions Share Next Stay Get help with your research Join ResearchGate to ask questions, get input, and advance your work. Join for free Log in All Answers (10) Dola Bhattacharjee Ministry of Environment, Forests and Climate Change, Govt. of India Dear Dr. Kani There are 3 major factors influencing the self purification process in river: Biochemical processes lead to the degradation and decomposition of organic wastes by organisms (including microbes and plants) in the water. Hydrochemical processes like oxidation, mineralization etc. may purify the water. Physical processes like adsorption, dilution etc. also aid in the same. Well, there is a limit to the natural healing process. Not all man made pollutants (e.g. plastics) are degradable. Also natural healing process is a function that is time dependent. Therefore if the rate of introduction of pollutants into the ecosystem is faster than the rate of its natural decomposition, the toxicants get piled up and result is detrimental. Cite 1 Recommendation Mophin Kani Kani UKF College of Engineering & Technology Thank you so much for your interesting notes dear dola, is there any other external factors ? Cite Praman Kumar Jawaharlal Nehru Architecture and Fine Arts University Hi i would like to add few more points the self purification process is some times delayed and and some times fastened in different seasons the factors like water temperature - leads to change in oxygen exchange rate, amount of precipitation, intensity of precipitation in an area, characteristics of the precipitation like pH, physical form - fog, rain, hailstones also impact water temperatures, change in water contents like increase in silt brought by runoff decrease the sunlight infiltration thus delay in process, sunlight penetration also helps bacteria for fastening the decomposition, increase in nutrient content brought by runoff also increases the algae etc thus decrease in oxygen and delay or stagnation of self purification etc.. These won't happen in isolation - the velocity of water in a channel, the depth of water, the turbulence created by the particles are carried from upstream or eroded loose rock at the bottom, the change in velocity w.r.t width and the curvature of the stream, river basin topography etc and human impact factors like the effluent load per unit length and comparison of load with above all factors. Cite Dola Bhattacharjee Ministry of Environment, Forests and Climate Change, Govt. of India Please refer to: 5. Lernlec, L.: Self purification of freshwater streams as affected by temperature and oxygen, nitrogen and other substances. Pp. 91-96. In: Advances in Water Pollution Research. B. A. southgate (Ed.). Proceedings of International Conference, held in London /Sept 1962). Macmillan, London (1964). Cite 1 Recommendation Sharma Dhruv Assam Down Town University Safeguarding of river catchment area, regulation of industrial effluents and maintenance of free flow of water in the river will ensure self purification to a greater extent. Cite Mophin Kani Kani UKF College of Engineering & Technology Nice to hear from you all about "self purification" , thank you so much for your nice contributions to my query, Cite Azubuike Victor Chukwuka National Environmental Standards and Regulations Enforcement Agency (NESREA) ...other factors affecting the self-purification process in a river include the geochemical profile of the substratum, ion exchange properties that determine mobilization or retention of metal pollutants for instance. the existing pollutant load of the sediment before another pollution event can determine the self purification process the incidence of other sources of pollution or anthropogenic impact down stream will actually will actually determine the reduced or increased recovery ability and recovery process of the river. the depth of the river and the steepness of the gradient along which it flows will determine the influence of oxygen in oxidizing volatile pollutants which is one of the recovery mechanisms or river. Fast flowing rivers due to steep gradients will have more interactions with atmospheric oxygen. Cite Jörg Schaller Leibniz Centre for Agricultural Landscape Research I agree with Dola Bhattacharjee, but the input of organic matter to fix the pollutant during orgaic matter decay is the most important Cite Mark Martinez Philippine Orthopedic center This is an interesting question. :) Based from an article about wastewater management written by M.M. Ghangrekar and IIT Kharagpur, the self purification of natural water systems is a complex process that often involved physical, chemical, and biological processes working simultaneously. The factors that affect self purification would be dilution, wherein a sufficient amount of water can suffice and be available in a receiving amount of water. The dissolved oxygen level will be available since there is a sufficient amount of water for dilution. Also, temperature and sunlight can be factors that affect self purification in rivers since both acts and correlates with each other very well. It is a given fact that nutrients including dissolved oxygen is more present in cold temperatures. Self purification can occur more in cold temperatures rather than hot temperatures. For sunlight on the other hand, more sunlight produces more photosysthesis by algae. Sunlight helps add oxygen in the stream, thus leading to self purification. For more information, this site could help :) www.bio.auth.gr/river/mod/pp/Chapter%206.doc Cite Azubuike Victor Chukwuka National Environmental Standards and Regulations Enforcement Agency (NESREA) @Mark its unlikely for self purification to occur more in the temperate than in the tropics because for instance the breakdown of organic matter via microbes is more optimal and occurs all year round in the tropics. Cite Similar questions and discussions Can I trust Sciencedomain International publishing Group? I please inform me details about.Science domain? Question 89 answers Asked 28 February 2017 Md. Nekmahmud Argon can I publish my paper in Science domain publishing Group? View Is there a special name for the aquatic environment of a pond or lake which is under the ice? Question 25 answers Asked 26 July 2013 Randolph Lauff I'm looking for the term analogous to "subnivean." Now I know if I read that question, I'd answer it by saying, "really cold water", but I'm looking for something a bit more scientific than that! View What is the best method for germinating propagules trapped using AstroTurf mats along river banks? Question 5 answers Asked 24 March 2014 Zarah Pattison I have used AstroTurf mats to trap propagules (seeds and vegetative fragments) deposited over winter along river banks. The propagules deposited on these mats will be used in a germination trial to assess abundance and species richness of viable propagules. I have reviewed the methods used in similar studies, such as those by Gurnell, Goodson, and Cockel. The authors in these various studies germinated propagules directly on the mat in a seedling tray, which in some instances was supplemented with sterile soil when needed. Other studies assessing sediment deposition and propagules wash the mats and germinate in seed trays. As I am not assessing sediment is there any benefit to washing the mats prior to germination, so that they germinate in sterile compost rather than directly on the turf mats? I.e. Will it aid the germination of propagules? View Is there (scientific) proof that water fowl can transport fish eggs from one water body to an other? Question 60 answers Asked 28 February 2014 Harald Ahnelt I am interested in the dispersal of freshwater fishes in artificial water bodies like pit gravels. There is some anecdotic rumor but I could not find proof for the transport of fish eggs by water fowl. 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16373	https://en.wikipedia.org/wiki/Chemical_synapse	Jump to content Chemical synapse Afrikaans العربية Català Español فارسی Galego 한국어 Kurdî Latina Romnă Русский Simple English Slovenščina کوردی Српски / srpski Українська Tiếng Việt Edit links From Wikipedia, the free encyclopedia Biological junctions through which neurons' signals can be sent This article is about chemical synapses of the nervous system. For general information, see synapse. For other uses, see synapse (disambiguation). Chemical synapses are biological junctions through which neurons' signals can be sent to each other and to non-neuronal cells such as those in muscles or glands. Chemical synapses allow neurons to form circuits within the central nervous system. They are crucial to the biological computations that underlie perception and thought. They allow the nervous system to connect to and control other systems of the body. At a chemical synapse, one neuron releases neurotransmitter molecules into a small space (the synaptic cleft) that is adjacent to another neuron. The neurotransmitters are contained within small sacs called synaptic vesicles, and are released into the synaptic cleft by exocytosis. These molecules then bind to neurotransmitter receptors on the postsynaptic cell. Finally, the neurotransmitters are cleared from the synapse through one of several potential mechanisms including enzymatic degradation or re-uptake by specific transporters either on the presynaptic cell or on some other neuroglia to terminate the action of the neurotransmitter. The adult human brain is estimated to contain from 1014 to 5 × 1014 (100–500 trillion) synapses. Every cubic millimeter of cerebral cortex contains roughly a billion (short scale, i.e. 109) of them. The number of synapses in the human cerebral cortex has separately been estimated at 0.15 quadrillion (150 trillion) The word "synapse" was introduced by Sir Charles Scott Sherrington in 1897. Chemical synapses are not the only type of biological synapse: electrical and immunological synapses also exist. Without a qualifier, however, "synapse" commonly refers to chemical synapses. Structure [edit] Further information on formation of synapses: Synaptogenesis Synapses are functional connections between neurons, or between neurons and other types of cells. A typical neuron gives rise to several thousand synapses, although there are some types that make far fewer. Most synapses connect axons to dendrites, but there are also other types of connections, including axon-to-cell-body, axon-to-axon, and dendrite-to-dendrite. Synapses are generally too small to be recognizable using a light microscope except as points where the membranes of two cells appear to touch, but their cellular elements can be visualized clearly using an electron microscope. Chemical synapses pass information directionally from a presynaptic cell to a postsynaptic cell and are therefore asymmetric in structure and function. The presynaptic axon terminal, or synaptic bouton, is a specialized area within the axon of the presynaptic cell that contains neurotransmitters enclosed in small membrane-bound spheres called synaptic vesicles (as well as a number of other supporting structures and organelles, such as mitochondria and endoplasmic reticulum). Synaptic vesicles are docked at the presynaptic plasma membrane at regions called active zones. Immediately opposite is a region of the postsynaptic cell containing neurotransmitter receptors; for synapses between two neurons the postsynaptic region may be found on the dendrites or cell body. Immediately behind the postsynaptic membrane is an elaborate complex of interlinked proteins called the postsynaptic density (PSD). Proteins in the PSD are involved in anchoring and trafficking neurotransmitter receptors and modulating the activity of these receptors. The receptors and PSDs are often found in specialized protrusions from the main dendritic shaft called dendritic spines. Synapses may be described as symmetric or asymmetric. When examined under an electron microscope, asymmetric synapses are characterized by rounded vesicles in the presynaptic cell, and a prominent postsynaptic density. Asymmetric synapses are typically excitatory. Symmetric synapses in contrast have flattened or elongated vesicles, and do not contain a prominent postsynaptic density. Symmetric synapses are typically inhibitory. The synaptic cleft—also called synaptic gap—is a gap between the pre- and postsynaptic cells that is about 20 nm (0.02 μ) wide. The small volume of the cleft allows neurotransmitter concentration to be raised and lowered rapidly. An autapse is a chemical (or electrical) synapse formed when the axon of one neuron synapses with its own dendrites. Signaling in chemical synapses [edit] Overview [edit] Here is a summary of the sequence of events that take place in synaptic transmission from a presynaptic neuron to a postsynaptic cell. Each step is explained in more detail below. Note that with the exception of the final step, the entire process may run only a few hundred microseconds, in the fastest synapses. The process begins with a wave of electrochemical excitation called an action potential traveling along the membrane of the presynaptic cell, until it reaches the synapse. The electrical depolarization of the membrane at the synapse causes channels to open that are permeable to calcium ions. Calcium ions flow through the presynaptic membrane, rapidly increasing the calcium concentration in the interior. The high calcium concentration activates a set of calcium-sensitive proteins attached to vesicles that contain a neurotransmitter chemical. These proteins change shape, causing the membranes of some "docked" vesicles to fuse with the membrane of the presynaptic cell, thereby opening the vesicles and dumping their neurotransmitter contents into the synaptic cleft, the narrow space between the membranes of the pre- and postsynaptic cells. The neurotransmitter diffuses within the cleft. Some of it escapes, but some of it binds to chemical receptor molecules located on the membrane of the postsynaptic cell. The binding of neurotransmitter causes the receptor molecule to be activated in some way. Several types of activation are possible, as described in more detail below. In any case, this is the key step by which the synaptic process affects the behavior of the postsynaptic cell. Due to thermal vibration, the motion of atoms, vibrating about their equilibrium positions in a crystalline solid, neurotransmitter molecules eventually break loose from the receptors and drift away. The neurotransmitter is either reabsorbed by the presynaptic cell, and then repackaged for future release, or else it is broken down metabolically. Neurotransmitter release [edit] The release of a neurotransmitter is triggered by the arrival of a nerve impulse (or action potential) and occurs through an unusually rapid process of cellular secretion (exocytosis). Within the presynaptic nerve terminal, vesicles containing neurotransmitter are localized near the synaptic membrane. The arriving action potential produces an influx of calcium ions through voltage-dependent, calcium-selective ion channels at the down stroke of the action potential (tail current). Calcium ions then bind to synaptotagmin proteins found within the membranes of the synaptic vesicles, allowing the vesicles to fuse with the presynaptic membrane. The fusion of a vesicle is a stochastic process, leading to frequent failure of synaptic transmission at the very small synapses that are typical for the central nervous system. Large chemical synapses (e.g. the neuromuscular junction), on the other hand, have a synaptic release probability, in effect, of 1. Vesicle fusion is driven by the action of a set of proteins in the presynaptic terminal known as SNAREs. As a whole, the protein complex or structure that mediates the docking and fusion of presynaptic vesicles is called the active zone. The membrane added by the fusion process is later retrieved by endocytosis and recycled for the formation of fresh neurotransmitter-filled vesicles. An exception to the general trend of neurotransmitter release by vesicular fusion is found in the type II receptor cells of mammalian taste buds. Here the neurotransmitter ATP is released directly from the cytoplasm into the synaptic cleft via voltage gated channels. Receptor binding [edit] Receptors on the opposite side of the synaptic gap bind neurotransmitter molecules. Receptors can respond in either of two general ways. First, the receptors may directly open ligand-gated ion channels in the postsynaptic cell membrane, causing ions to enter or exit the cell and changing the local transmembrane potential. The resulting change in voltage is called a postsynaptic potential. In general, the result is excitatory in the case of depolarizing currents, and inhibitory in the case of hyperpolarizing currents. Whether a synapse is excitatory or inhibitory depends on what type(s) of ion channel conduct the postsynaptic current(s), which in turn is a function of the type of receptors and neurotransmitter employed at the synapse. The second way a receptor can affect membrane potential is by modulating the production of chemical messengers inside the postsynaptic neuron. These second messengers can then amplify the inhibitory or excitatory response to neurotransmitters. Termination [edit] After a neurotransmitter molecule binds to a receptor molecule, it must be removed to allow for the postsynaptic membrane to continue to relay subsequent EPSPs and/or IPSPs. This removal can happen through one or more processes: The neurotransmitter may diffuse away due to thermally-induced oscillations of both it and the receptor, making it available to be broken down metabolically outside the neuron or to be reabsorbed. Enzymes within the subsynaptic membrane may inactivate/metabolize the neurotransmitter. Reuptake pumps may actively pump the neurotransmitter back into the presynaptic axon terminal for reprocessing and re-release following a later action potential. Synaptic strength [edit] The strength of a synapse has been defined by Bernard Katz as the product of (presynaptic) release probability pr, quantal size q (the postsynaptic response to the release of a single neurotransmitter vesicle, a 'quantum'), and n, the number of release sites. "Unitary connection" usually refers to an unknown number of individual synapses connecting a presynaptic neuron to a postsynaptic neuron. The amplitude of postsynaptic potentials (PSPs) can be as low as 0.4 mV to as high as 20 mV. The amplitude of a PSP can be modulated by neuromodulators or can change as a result of previous activity. Changes in the synaptic strength can be short-term, lasting seconds to minutes, or long-term (long-term potentiation, or LTP), lasting hours. Learning and memory are believed to result from long-term changes in synaptic strength, via a mechanism known as synaptic plasticity. Receptor desensitization [edit] Desensitization of the postsynaptic receptors is a decrease in response to the same neurotransmitter stimulus. It means that the strength of a synapse may in effect diminish as a train of action potentials arrive in rapid succession – a phenomenon that gives rise to the so-called frequency dependence of synapses. The nervous system exploits this property for computational purposes, and can tune its synapses through such means as phosphorylation of the proteins involved. Synaptic plasticity [edit] Main article: Synaptic plasticity Synaptic transmission can be changed by previous activity. These changes are called synaptic plasticity and may result in either a decrease in the efficacy of the synapse, called depression, or an increase in efficacy, called potentiation. These changes can either be long-term or short-term. Forms of short-term plasticity include synaptic fatigue or depression and synaptic augmentation. Forms of long-term plasticity include long-term depression and long-term potentiation. Synaptic plasticity can be either homosynaptic (occurring at a single synapse) or heterosynaptic (occurring at multiple synapses). Homosynaptic plasticity [edit] Main article: Homosynaptic plasticity Homosynaptic plasticity (or also homotropic modulation) is a change in the synaptic strength that results from the history of activity at a particular synapse. This can result from changes in presynaptic calcium as well as feedback onto presynaptic receptors, i.e. a form of autocrine signaling. Homosynaptic plasticity can affect the number and replenishment rate of vesicles or it can affect the relationship between calcium and vesicle release. Homosynaptic plasticity can also be postsynaptic in nature. It can result in either an increase or decrease in synaptic strength. One example is neurons of the sympathetic nervous system (SNS), which release noradrenaline, which, besides affecting postsynaptic receptors, also affects presynaptic α2-adrenergic receptors, inhibiting further release of noradrenaline. This effect is utilized with clonidine to perform inhibitory effects on the SNS. Heterosynaptic plasticity [edit] Main article: Heterosynaptic plasticity Heterosynaptic plasticity (or also heterotropic modulation) is a change in synaptic strength that results from the activity of other neurons. Again, the plasticity can alter the number of vesicles or their replenishment rate or the relationship between calcium and vesicle release. Additionally, it could directly affect calcium influx. Heterosynaptic plasticity can also be postsynaptic in nature, affecting receptor sensitivity. One example is again neurons of the sympathetic nervous system, which release noradrenaline, which, in addition, generates an inhibitory effect on presynaptic terminals of neurons of the parasympathetic nervous system. Integration of synaptic inputs [edit] Main article: Summation (neurophysiology) In general, if an excitatory synapse is strong enough, an action potential in the presynaptic neuron will trigger an action potential in the postsynaptic cell. In many cases the excitatory postsynaptic potential (EPSP) will not reach the threshold for eliciting an action potential. When action potentials from multiple presynaptic neurons fire simultaneously, or if a single presynaptic neuron fires at a high enough frequency, the EPSPs can overlap and summate. If enough EPSPs overlap, the summated EPSP can reach the threshold for initiating an action potential. This process is known as summation, and can serve as a high pass filter for neurons. On the other hand, a presynaptic neuron releasing an inhibitory neurotransmitter, such as GABA, can cause an inhibitory postsynaptic potential (IPSP) in the postsynaptic neuron, bringing the membrane potential farther away from the threshold, decreasing its excitability and making it more difficult for the neuron to initiate an action potential. If an IPSP overlaps with an EPSP, the IPSP can in many cases prevent the neuron from firing an action potential. In this way, the output of a neuron may depend on the input of many different neurons, each of which may have a different degree of influence, depending on the strength and type of synapse with that neuron. John Carew Eccles performed some of the important early experiments on synaptic integration, for which he received the Nobel Prize for Physiology or Medicine in 1963. Volume transmission [edit] When a neurotransmitter is released at a synapse, it reaches its highest concentration inside the narrow space of the synaptic cleft, but some of it is certain to diffuse away before being reabsorbed or broken down. If it diffuses away, it has the potential to activate receptors that are located either at other synapses or on the membrane away from any synapse. The extrasynaptic activity of a neurotransmitter is known as volume transmission. It is well established that such effects occur to some degree, but their functional importance has long been a matter of controversy. Recent work indicates that volume transmission may be the predominant mode of interaction for some special types of neurons. In the mammalian cerebral cortex, a class of neurons called neurogliaform cells can inhibit other nearby cortical neurons by releasing the neurotransmitter GABA into the extracellular space. Along the same vein, GABA released from neurogliaform cells into the extracellular space also acts on surrounding astrocytes, assigning a role for volume transmission in the control of ionic and neurotransmitter homeostasis. Approximately 78% of neurogliaform cell boutons do not form classical synapses. This may be the first definitive example of neurons communicating chemically where classical synapses are not present. Relationship to electrical synapses [edit] An electrical synapse is an electrically conductive link between two abutting neurons that is formed at a narrow gap between the pre- and postsynaptic cells, known as a gap junction. At gap junctions, cells approach within about 3.5 nm of each other, rather than the 20 to 40 nm distance that separates cells at chemical synapses. As opposed to chemical synapses, the postsynaptic potential in electrical synapses is not caused by the opening of ion channels by chemical transmitters, but rather by direct electrical coupling between both neurons. Electrical synapses are faster than chemical synapses. Electrical synapses are found throughout the nervous system, including in the retina, the reticular nucleus of the thalamus, the neocortex, and in the hippocampus. While chemical synapses are found between both excitatory and inhibitory neurons, electrical synapses are most commonly found between smaller local inhibitory neurons. Electrical synapses can exist between two axons, two dendrites, or between an axon and a dendrite. In some fish and amphibians, electrical synapses can be found within the same terminal of a chemical synapse, as in Mauthner cells. Effects of drugs [edit] Main article: Neuropharmacology One of the most important features of chemical synapses is that they are the site of action for the majority of psychoactive drugs. Synapses are affected by drugs, such as curare, strychnine, cocaine, morphine, alcohol, LSD, risperidone, and countless others. These drugs have different effects on synaptic function, and often are restricted to synapses that use a specific neurotransmitter. For example, curare is a poison that stops acetylcholine from depolarizing the postsynaptic membrane, causing paralysis. Strychnine blocks the inhibitory effects of the neurotransmitter glycine, which causes the body to pick up and react to weaker and previously ignored stimuli, resulting in uncontrollable muscle spasms. Morphine acts on synapses that use endorphin neurotransmitters, and alcohol increases the inhibitory effects of the neurotransmitter GABA. LSD interferes with synapses that use the neurotransmitter serotonin. Risperidone is a blocker of various receptors including several dopamine and serotonin receptors, and it can bind with high affinity to some types of serotonin receptors. Cocaine blocks reuptake of dopamine and therefore increases its effects. History and etymology [edit] During the 1950s, Bernard Katz and Paul Fatt observed spontaneous miniature synaptic currents at the frog neuromuscular junction. Based on these observations, they developed the 'quantal hypothesis' that is the basis for our current understanding of neurotransmitter release as exocytosis and for which Katz received the Nobel Prize in Physiology or Medicine in 1970. In the late 1960s, Ricardo Miledi and Katz advanced the hypothesis that depolarization-induced influx of calcium ions triggers exocytosis. Sir Charles Scott Sherringtonin coined the word 'synapse' and the history of the word was given by Sherrington in a letter he wrote to John Fulton: 'I felt the need of some name to call the junction between nerve-cell and nerve-cell... I suggested using "syndesm"... He [ Sir Michael Foster ] consulted his Trinity friend Verrall, the Euripidean scholar, about it, and Verrall suggested "synapse" (from the Greek "clasp").'–Charles Scott Sherrington See also [edit] Acclimatisation (neurons) Neuroscience Neurexin Ribbon synapse Notes [edit] ^ Drachman D (2005). "Do we have brain to spare?". Neurology. 64 (12): 2004–5. doi:10.1212/01.WNL.0000166914.38327.BB. PMID 15985565. S2CID 38482114. ^ Alonso-Nanclares L, Gonzalez-Soriano J, Rodriguez JR, DeFelipe J (September 2008). "Gender differences in human cortical synaptic density". Proc. Natl. Acad. Sci. U.S.A. 105 (38): 14615–9. Bibcode:2008PNAS..10514615A. doi:10.1073/pnas.0803652105. PMC 2567215. 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PMID 4320287. References [edit] Carlson, Neil R. (2007). Physiology of Behavior (9th ed.). Boston, MA: Pearson Education. ISBN 978-0-205-59389-7. Kandel, Eric R.; Schwartz, James H.; Jessell, Thomas M. (2000). Principles of Neural Science (4th ed.). New York: McGraw-Hill. ISBN 978-0-8385-7701-1. Llinás R, Sugimori M, Simon SM (April 1982). "Transmission by presynaptic spike-like depolarization in the squid giant synapse". Proc. Natl. Acad. Sci. U.S.A. 79 (7): 2415–9. Bibcode:1982PNAS...79.2415L. doi:10.1073/pnas.79.7.2415. PMC 346205. PMID 6954549. Llinás R, Steinberg IZ, Walton K (1981). "Relationship between presynaptic calcium current and postsynaptic potential in squid giant synapse". Biophysical Journal. 33 (3): 323–352. Bibcode:1981BpJ....33..323L. doi:10.1016/S0006-3495(81)84899-0. PMC 1327434. PMID 6261850. Bear, Mark F.; Connors, Barry W.; Paradiso, Michael A. (2001). Neuroscience: Exploring the Brain. Hagerstown, MD: Lippincott Williams & Wilkins. ISBN 978-0-7817-3944-3. Hormuzdi, SG; Filippov, MA; Mitropoulou, G; Monyer, H; Bruzzone, R (March 2004). "Electrical synapses: a dynamic signaling system that shapes the activity of neuronal networks". Biochim Biophys Acta. 1662 (1–2): 113–137. doi:10.1016/j.bbamem.2003.10.023. PMID 15033583. Karp, Gerald (2005). Cell and Molecular Biology: concepts and experiments (4th ed.). Hoboken, NJ: John Wiley & Sons. ISBN 978-0-471-46580-5. Nicholls, J.G.; Martin, A.R.; Wallace, B.G.; Fuchs, P.A. (2001). From Neuron to Brain (4th ed.). Sunderland, MA: Sinauer Associates. ISBN 978-0-87893-439-3. External links [edit] Listen to this article (7 minutes) This audio file was created from a revision of this article dated 19 June 2005 (2005-06-19), and does not reflect subsequent edits. Synapse Review for Kids Synapse – Cell Centered Database Atlas of Ultrastructure Neurocytology An electron microscope picture gallery assembled by Kristen Harris' lab of synapses and other neuronal structures. \| Nervous tissue \| \| --- \| \| CNS \| \| \| \| --- \| \| Tissue Types \| Grey matter White matter + Projection fibers + Association fiber + Commissural fiber + Lemniscus + Nerve tract + Decussation Neuropil Meninges \| \| Cell Types \| \| \| \| --- \| \| Neuronal \| Pyramidal Purkinje Granule Von Economo Medium spiny Interneuron \| \| Glial \| Astrocyte Ependymal cells + Tanycyte Oligodendrocyte progenitor cell Oligodendrocyte Microglia \| \| \| \| PNS \| \| \| \| --- \| \| General \| Dorsal + Root + Ganglion + Ramus Ventral + Root + Ramus Ramus communicans + Gray + White Autonomic ganglion (Preganglionic nerve fibers Postganglionic nerve fibers) Nerve fascicle Funiculus \| \| Connective tissues \| Epineurium Perineurium Endoneurium \| \| Neuroglia \| Myelination: Schwann cell + Neurilemma + Myelin incisure + Node of Ranvier + Internodal segment Satellite glial cell \| \| \| Neurons/ nerve fibers \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| \| Parts \| \| \| \| --- \| \| Soma \| Axon hillock \| \| Axon \| Telodendron Axon terminals Axoplasm Axolemma Neurofibril/neurofilament \| \| Dendrite \| + Nissl body + Dendritic spine + Apical dendrite/Basal dendrite \| \| \| Types \| Bipolar Unipolar Pseudounipolar Multipolar Interneuron + Renshaw \| \| Afferent nerve fiber/ Sensory neuron \| GSA GVA SSA SVA fibers + Ia or Aα + Ib or Golgi or Aα + II or Aβ and Aγ + III or Aδ or fast pain + IV or C or slow pain \| \| Efferent nerve fiber/ Motor neuron \| GSE GVE SVE Upper motor neuron Lower motor neuron + α motorneuron + β motorneuron + γ motorneuron \| \| \| Termination \| \| \| \| --- \| \| Synapse \| Electrical synapse/Gap junction Chemical synapse + Synaptic vesicle + Active zone + Postsynaptic density Autapse Ribbon synapse Neuromuscular junction \| \| Sensory receptors \| Meissner's corpuscle Merkel nerve ending Pacinian corpuscle Ruffini ending Muscle spindle Free nerve ending Nociceptor Olfactory receptor neuron Photoreceptor cell Hair cell Taste receptor \| \| Retrieved from " Categories: Cell signaling Signal transduction Neural synapse Hidden categories: Articles with short description Short description matches Wikidata Articles with hAudio microformats Spoken articles Articles containing video clips
16374	https://www.examples.com/maths/multiples-of-3.html	Home Multiples of 3 Multiples of 3 – 100+ List, Examples Created by: Team Maths - Examples.com, Last Updated: July 29, 2024 Notes Multiples of 3 – 100+ List, Examples Multiples of 3 are integers that can be obtained by multiplying the number 3 by any whole number. These numbers, such as 3, 6, 9, 12, and so on, are the result of the multiplication process involving 3. The concept of multiples is closely related to factors and divisors; specifically, a multiple of 3 is any number that 3 can evenly divide without leaving a remainder. Understanding multiples helps in various mathematical applications, including finding common multiples and solving problems involving divisibility. What are Multiples of 3? Multiples of 3 are numbers that result from multiplying 3 by any integer. They are the sequence 3, 6, 9, 12, 15, and so on, where each number is a product of 3 and another whole number. These numbers are divisible by 3 without leaving a remainder. Prime factorization of 3: 3 = 3¹ First 10 multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 First 50 Multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48. For example, 70, 80, 20 and 60 are all multiples of 3, 22 is not a multiple of 3 for the following reasons: \| Number \| Calculation \| Remainder \| --- \| 15 \| 15 ÷ 3 = 5 \| 0 \| \| 27 \| 27 ÷ 3 = 9 \| 0 \| \| 33 \| 33 ÷ 3 = 11 \| 0 \| \| 60 \| 60 ÷ 3 = 20 \| 0 \| \| 22 \| 22 ÷ 3 = 7.33 \| 1 \| , List of First 100 Multiples of 3 with Remainders \| Number \| Calculation \| Remainder \| --- \| 3 \| 3÷3 = 1, so 3 is divisible by 3. \| 0 \| \| 6 \| 6÷3 = 2, so 6 is divisible by 3. \| 0 \| \| 9 \| 9÷3 = 3, so 9 is divisible by 3. \| 0 \| \| 12 \| 12÷3 = 4, so 12 is divisible by 3. \| 0 \| \| 15 \| 15÷3 = 5, so 15 is divisible by 3. \| 0 \| \| 18 \| 18÷3 = 6, so 18 is divisible by 3. \| 0 \| \| 21 \| 21÷3 = 7, so 21 is divisible by 3. \| 0 \| \| 24 \| 24÷3 = 8, so 24 is divisible by 3. \| 0 \| \| 27 \| 27÷3 = 9, so 27 is divisible by 3. \| 0 \| \| 30 \| 30÷3 = 10, so 30 is divisible by 3. \| 0 \| \| 33 \| 33÷3 = 11, so 33 is divisible by 3. \| 0 \| \| 36 \| 36÷3 = 12, so 36 is divisible by 3. \| 0 \| \| 39 \| 39÷3 = 13, so 39 is divisible by 3. \| 0 \| \| 42 \| 42÷3 = 14, so 42 is divisible by 3. \| 0 \| \| 45 \| 45÷3 = 15, so 45 is divisible by 3. \| 0 \| \| 48 \| 48÷3 =16, so 48 is divisible by 3. \| 0 \| \| 51 \| 51÷3 = 17, so 51 is divisible by 3. \| 0 \| \| 54 \| 54÷3 = 18, so 54 is divisible by 3. \| 0 \| \| 57 \| 57÷3 = 19, so 57 is divisible by 3. \| 0 \| \| 60 \| 60÷3 = 20, so 60 is divisible by 3. \| 0 \| \| 63 \| 63÷3 = 21, so 63 is divisible by 3. \| 0 \| \| 66 \| 66÷3 = 22, so 66 is divisible by 3. \| 0 \| \| 69 \| 69÷3 = 23, so 69 is divisible by 3. \| 0 \| \| 72 \| 72÷3 = 24, so 72 is divisible by 3. \| 0 \| \| 75 \| 75÷3 = 25, so 75 is divisible by 3. \| 0 \| \| 78 \| 78÷3=26, so 78 is divisible by 3. \| 0 \| \| 81 \| 81÷3 = 27, so 81 is divisible by 3. \| 0 \| \| 84 \| 84÷3 = 28, so 84 is divisible by 3. \| 0 \| \| 87 \| 87÷3 = 29, so 87 is divisible by 3. \| 0 \| \| 90 \| 90÷3 = 30, so 90 is divisible by 3. \| 0 \| \| 93 \| 93÷3 = 31, so 93 is divisible by 3. \| 0 \| \| 96 \| 96÷3 = 32, so 96 is divisible by 3. \| 0 \| \| 99 \| 99÷3 = 33, so 99 is divisible by 3. \| 0 \| \| 102 \| 102÷3 = 34, so 102 is divisible by 3. \| 0 \| \| 105 \| 105÷3 = 35, so 105 is divisible by 3. \| 0 \| \| 108 \| 108÷3 = 36, so 108 is divisible by 3. \| 0 \| \| 111 \| 111÷3 = 37, so 111 is divisible by 3. \| 0 \| \| 114 \| 114÷3 = 38, so 114 is divisible by 3. \| 0 \| \| 117 \| 117÷3 = 39, so 117 is divisible by 3. \| 0 \| \| 120 \| 120÷3 = 40, so 120 is divisible by 3. \| 0 \| \| 123 \| 123÷3 = 41, so 123 is divisible by 3. \| 0 \| \| 126 \| 126÷3 = 42, so 126 is divisible by 3. \| 0 \| \| 129 \| 129÷3 = 43, so 129 is divisible by 3. \| 0 \| \| 132 \| 132÷3 = 44, so 132 is divisible by 3. \| 0 \| \| 135 \| 135÷3 = 45, so 135 is divisible by 3. \| 0 \| \| 138 \| 138÷3 = 46, so 138 is divisible by 3. \| 0 \| \| 141 \| 141÷3 = 47, so 141 is divisible by 3. \| 0 \| \| 144 \| 144÷3 = 48, so 144 is divisible by 3. \| 0 \| \| 147 \| 147÷3 = 49, so 147 is divisible by 3. \| 0 \| \| 150 \| 150÷3 = 50, so 150 is divisible by 3. \| 0 \| \| 153 \| 153÷3 = 51, so 153 is divisible by 3. \| 0 \| \| 156 \| 156÷3 = 52, so 156 is divisible by 3. \| 0 \| \| 159 \| 159÷3 = 53, so 159 is divisible by 3. \| 0 \| \| 162 \| 162÷3 = 54, so 162 is divisible by 3. \| 0 \| \| 165 \| 165÷3 = 55, so165 is divisible by 3. \| 0 \| \| 168 \| 168÷3 = 56, so 168 is divisible by 3. \| 0 \| \| 171 \| 171÷3 = 57, so 171 is divisible by 3. \| 0 \| \| 174 \| 174÷3 = 58, so 174 is divisible by 3. \| 0 \| \| 177 \| 177÷3 = 59, so 177 is divisible by 3. \| 0 \| \| 180 \| 180÷3 = 60, so 180 is divisible by 3. \| 0 \| \| 183 \| 183÷3 = 61, so 183 is divisible by 3. \| 0 \| \| 186 \| 186÷3 = 62, so 186 is divisible by 3. \| 0 \| \| 189 \| 189÷3 = 63, so 189 is divisible by 3. \| 0 \| \| 192 \| 192÷3 = 64, so 192 is divisible by 3. \| 0 \| \| 195 \| 195÷3 = 65, so 1195 is divisible by 3. \| 0 \| \| 198 \| 198÷3 = 66, so 198 is divisible by 3. \| 0 \| \| 201 \| 201÷3 = 67, so 201 is divisible by 3. \| 0 \| \| 204 \| 204÷3 = 68, so 204 is divisible by 3. \| 0 \| \| 207 \| 207÷3 = 69, so 207 is divisible by 3. \| 0 \| \| 210 \| 210÷3 = 70, so 210 is divisible by 3. \| 0 \| \| 213 \| 213÷3 = 71, so 213 is divisible by 3. \| 0 \| \| 216 \| 216÷3 = 72, so 216 is divisible by 3. \| 0 \| \| 219 \| 219÷3 = 73, so219 is divisible by 3. \| 0 \| \| 222 \| 222÷3 = 74, so 222 is divisible by 3. \| 0 \| \| 225 \| 225÷3 = 75, so 225 is divisible by 3. \| 0 \| \| 228 \| 228÷3 = 76, so 228 is divisible by 3. \| 0 \| \| 231 \| 231÷3 = 77, so 231 is divisible by 3. \| 0 \| \| 234 \| 234÷3 = 78, so 234 is divisible by 3. \| 0 \| \| 237 \| 237÷3 = 79, so 237 is divisible by 3. \| 0 \| \| 240 \| 240÷3 = 80, so 240 is divisible by 3. \| 0 \| \| 243 \| 243÷3 = 81, so 243 is divisible by 3. \| 0 \| \| 246 \| 246÷3 = 82, so 246 is divisible by 3. \| 0 \| \| 249 \| 249÷3=83, so 249 is divisible by 3. \| 0 \| \| 252 \| 252÷3 = 84, so 252 is divisible by 3. \| 0 \| \| 255 \| 255÷3 = 85, so 255 is divisible by 3. \| 0 \| \| 258 \| 258÷3 = 86, so 258 is divisible by 3. \| 0 \| \| 261 \| 261÷3 = 87, so 261 is divisible by 3. \| 0 \| \| 264 \| 264÷3 = 88, so 264 is divisible by 3. \| 0 \| \| 267 \| 267÷3 = 89, so 267 is divisible by 3. \| 0 \| \| 270 \| 270÷3 = 90, so 270 is divisible by 3. \| 0 \| \| 273 \| 273÷3 = 91, so 273 is divisible by 3. \| 0 \| \| 276 \| 276÷3 = 92, so 276 is divisible by 3. \| 0 \| \| 279 \| 279÷3 = 93, so 279 is divisible by 3. \| 0 \| \| 282 \| 282÷3 = 94, so 282 is divisible by 3. \| 0 \| \| 285 \| 285÷3 = 95, so 285 is divisible by 3. \| 0 \| \| 288 \| 288÷3 = 96, so 288 is divisible by 3. \| 0 \| \| 291 \| 291÷3 = 97, so 291 is divisible by 3. \| 0 \| \| 294 \| 294÷3 = 98, so 94 is divisible by 3. \| 0 \| \| 297 \| 297÷3 = 99, so 297 is divisible by 3. \| 0 \| \| 300 \| 300÷3 = 100, so 300 is divisible by 3. \| 0 \| Read More About Multiples of 3 Table of 3 Important Notes Definition: Multiples of a number are obtained by multiplying the number by integers (e.g., multiples of 2 are 2, 4, 6, 8, …). Divisibility: A number is a multiple of another if it can be divided by that number without leaving a remainder. Sequence: The sequence of multiples is infinite. For instance, multiples of 2 are 2, 4, 6, 8, 10, and so on, continuing indefinitely. Applications: Understanding multiples is crucial for solving problems in arithmetic, algebra, and number theory, such as finding the least common multiple (LCM) and greatest common divisor (GCD). Pattern Recognition: Recognizing patterns in multiples helps in simplifying calculations and solving mathematical puzzles and problems effectively. Examples on Multiples of 3 Example 1: Checking if a Number is a Multiple of 3 To determine if 45 is a multiple of 3, divide 45 by 3: 45÷3 = 15 Since the result is an integer with no remainder, 45 is a multiple of 3. Example 2: Finding Multiples of 3 To find the first five multiples of 3, multiply 3 by the first five integers: 3×1 = 3 3×2 = 6 3×3 = 9 3×4 = 12 3×5 = 15 Thus, the first five multiples of 3 are 3, 6, 9, 12, and 15. Example 3: Real-Life Application Consider you have 27 apples, and you want to distribute them equally into groups of 3: 27÷3 = 9 You will have 9 groups with 3 apples each. This shows that 27 is a multiple of 3. Example 4: Summing Multiples of 3 Find the sum of the multiples of 3 up to 15: Multiples: 3, 6, 9, 12, 15 Sum: 3+6+9+12+15 = 45 Therefore, the sum of the multiples of 3 up to 15 is 45. Example 5: Multiples of 3 within a Range Identify multiples of 3 between 10 and 30: Starting from 12 (the first multiple of 3 greater than 10) 12, 15, 18, 21, 24, 27, 30 Practical Examples of Multiples of 3 Example 1: Sharing Candies You have 18 candies and want to divide them equally among 3 friends. Each friend gets: 18÷3 = 6 So, 18 is a multiple of 3. Example 2: Counting Steps If you take 3 steps at a time, after 5 times, you will have taken: 3×5 = 15 Therefore, 15 steps is a multiple of 3. Example 3: Packing Items You have 27 books to pack into boxes, with 3 books per box. You will need: 27÷3 = 9 This means 27 is a multiple of 3, requiring 9 boxes. Example 4: Buying Tickets Concert tickets are sold in packs of 3. If you need 12 tickets, you will buy: 3×4 = 12 So, 12 is a multiple of 3, and you buy 4 packs. Example 5: Arranging Chairs You want to arrange 30 chairs in rows of 3. The number of rows will be: 30÷3 = 10 Thus, 30 is a multiple of 3, and you will have 10 rows. FAQs What is a multiple of 3? A multiple of 3 is any number that can be evenly divided by 3 without leaving a remainder. Examples include 3, 6, 9, 12, and so on. How can I determine if a number is a multiple of 3? To determine if a number is a multiple of 3, divide the number by 3. If the result is a whole number with no remainder, then it is a multiple of 3. What are the first five multiples of 3? The first five multiples of 3 are 3, 6, 9, 12, and 15. Are all multiples of 3 also multiples of other numbers? Not necessarily. While some multiples of 3 might also be multiples of other numbers (e.g., 6 is a multiple of both 3 and 2), others may not be (e.g., 9 is only a multiple of 3). Why are multiples of 3 important in mathematics? Multiples of 3 are important in mathematics for understanding patterns, solving problems involving divisibility, and finding the least common multiple (LCM) and greatest common divisor (GCD). Can negative numbers be multiples of 3? Yes, negative numbers can also be multiples of 3. Examples include -3, -6, -9, and so on. How are multiples of 3 used in real life? Multiples of 3 are used in various real-life applications such as event planning, packaging, cooking, and financial planning, where grouping or dividing items into sets of 3 is practical. Is zero considered a multiple of 3? Yes, zero is considered a multiple of 3 because 0÷3=0 with no remainder. What is the relationship between multiples of 3 and the number 9? Multiples of 9 are also multiples of 3. This is because 9 itself is a multiple of 3 ( 9 = 3×3). Thus, any multiple of 9 can be evenly divided by 3. What is the rule for finding multiples of 3 using digit sum? A number is a multiple of 3 if the sum of its digits is a multiple of 3. For example, for 123, the sum of digits is 1+2+3 = 6, which is a multiple of 3, so 123 is a multiple of 3. Share : AI Generator Text prompt Add Tone Select a Tone Friendly Formal Casual Instructive Professional Empathetic Humorous Serious Optimistic Neutral 10 Examples of Public speaking 20 Examples of Gas lighting Practice Test Which of the following numbers is a multiple of 3? Choose the correct answer 11 14 18 22 of 10 What is the smallest multiple of 3 that is greater than 10? Choose the correct answer 12 13 14 15 of 10 Which number is not a multiple of 3? Choose the correct answer 27 30 34 38 of 10 What is the result of multiplying 3 by 7? Choose the correct answer 18 21 24 27 of 10 Which of these numbers is the result of adding two multiples of 3? Choose the correct answer 25 30 35 40 of 10 Which number below is the largest multiple of 3 less than 50? Choose the correct answer 48 50 55 60 of 10 Which of the following is a multiple of 3 and also a prime number? Choose the correct answer 3 6 9 12 of 10 If you subtract 6 from 15, which of the following is true? Choose the correct answer The result is a multiple of 3 The result is not a multiple of 3 The result is 9 The result is 12 of 10 What is the next multiple of 3 after 57? Choose the correct answer 60 61 of 10 Which number is a multiple of 3 and also a perfect square? Choose the correct answer 9 11 13 of 10 school Ready to Test Your Knowledge? close Before you leave, take our quick quiz to enhance your learning! assessment Assess Your Mastery emoji_events Boost Your Confidence speed Instant Results memory Enhance Retention event_available Prepare for Exams repeat Reinforce Learning 👉 Start the Quiz Now
16375	https://www.quora.com/Can-a-linear-combination-have-a-constant-equal-to-zero	Can a linear combination have a constant equal to zero? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Linear Relationship Permutations and Combinat... Vectors (mathematics) Linear Algebra 1 Basic Algebra Linear Operations Mathematical Sciences Linear Equations 5 Can a linear combination have a constant equal to zero? All related (32) Sort Recommended Jordan Berg Professor of Mechanical Engineering · Author has 72 answers and 367.8K answer views ·7y Can a linear combination have a constant equal to zero? Let’s say we have a bunch of vectors, N N of them in all, call them x i x i, where i=1,⋯,N i=1,⋯,N. A linear combination of those vectors looks like ∑N i=1 c i x i∑i=1 N c i x i, where all the c i c i are scalars. If you are asking whether any or all of the c i c i can be zero, the answer is yes. In fact, this plays an important role in a commonly used definition of linear independence, which says that the only way that the linear combination of a linearly independent set of vectors can be equal to zero is for all the coefficients to be zero. Upvote · 9 3 Pablo Emanuel MSc in Math and IMO medalist · Upvoted by Nathan Hannon , Ph. D. Mathematics, University of California, Davis (2021) · Author has 847 answers and 5.5M answer views ·3y Related Why's a constant function a linear function if a linear function is a function that the power of x is equal to 1 (in constant functions it's zero)? A linear function is a function for which we have f(a x)=a f(x)f(a x)=a f(x) for any scalar a a, and f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y). With that definition, the only constant function that is linear is f(x)=0 f(x)=0. Having said that, many people call affine functions linear, and, based on your question, this is what you have in mind. An affine function is, by definition, any function for which there is a constant C such that g(x)=f(x)−C g(x)=f(x)−C is linear. Since f(x)=0 f(x)=0 is linear, any constant function is indeed affine. If we’re talking about functions from the real line to itself, the linear functions are exactly the functions of the form Continue Reading A linear function is a function for which we have f(a x)=a f(x)f(a x)=a f(x) for any scalar a a, and f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y). With that definition, the only constant function that is linear is f(x)=0 f(x)=0. Having said that, many people call affine functions linear, and, based on your question, this is what you have in mind. An affine function is, by definition, any function for which there is a constant C such that g(x)=f(x)−C g(x)=f(x)−C is linear. Since f(x)=0 f(x)=0 is linear, any constant function is indeed affine. If we’re talking about functions from the real line to itself, the linear functions are exactly the functions of the form f(x)=a x f(x)=a x where a is any real number including 0, and the affine functions are the functions of the form f(x)=a x+b f(x)=a x+b, where both a and b are real numbers that can be 0 or not. Upvote · 99 12 Danya Rose I'm an Aussie feller, I've been one all my life. It may make me seem funny... · Author has 2K answers and 7M answer views ·8y Related What does linear combination mean in mathematics? If you have a collection {x i}{x i} of things that you can add together and multiply by scalars a i a i, then the sum ∑i a i x i=a 1 x 1+a 2 x 2+…∑i a i x i=a 1 x 1+a 2 x 2+… is a linear combination of whatever the x i x i s are (they could represent anything from numbers of one sort or another, to vectors, to matrices, to tensors, to functions, or whatever). For example, if x 1∈R 2 x 1∈R 2 is the vector (1 0)(1 0), and x 2∈R 2 x 2∈R 2 is the vector (−1 2 3)(−1 2 3), then \displaystyle{\qquad a_1 x_1+a_2 x_2=a_1\begin{pmatrix}1\0\end{pmatrix}\displaystyle{\qquad a_1 x_1+a_2 x_2=a_1\begin{pmatrix}1\0\end{pmatrix} Continue Reading If you have a collection {x i}{x i} of things that you can add together and multiply by scalars a i a i, then the sum ∑i a i x i=a 1 x 1+a 2 x 2+…∑i a i x i=a 1 x 1+a 2 x 2+… is a linear combination of whatever the x i x i s are (they could represent anything from numbers of one sort or another, to vectors, to matrices, to tensors, to functions, or whatever). For example, if x 1∈R 2 x 1∈R 2 is the vector (1 0)(1 0), and x 2∈R 2 x 2∈R 2 is the vector (−1 2 3)(−1 2 3), then a 1 x 1+a 2 x 2=a 1(1 0)+a 2(−1 2 3)a 1 x 1+a 2 x 2=a 1(1 0)+a 2(−1 2 3), where a 1 a 1 and a 2 a 2 are real numbers, then you can think of the set of linear combinations of x 1 x 1 and x 2 x 2 as the set of points in R 2 R 2 that you can “reach” by any choice of a 1 a 1 and a 2 a 2. In this case, because x 1 x 1 and x 2 x 2 are not scalar multiples of one another, this set is indeed all of R 2 R 2. For example, the vector (0 1)(0 1) can be obtained as a linear combination of x 1 x 1 and x 2 x 2 by choosing a 1=1 6 a 1=1 6 and a 2=1 3 a 2=1 3, as 1 6×1+1 3×−1 2=0 1 6×1+1 3×−1 2=0, and 1 6×0+1 3×3=1 1 6×0+1 3×3=1. In short, a linear combination of some things is a sum of scalar multiples of those things. Upvote · 9 8 Related questions More answers below What precisely is a linear combination? What is the linear combination of zero (0) vectors? Is the zero vector a linear combination? Why is a zero vector linearly dependent? Can you provide an example of two vectors whose sum is zero and whose difference is equal to the sum of any other two vectors? Assistant Bot · 1y Yes, a linear combination can have a constant equal to zero. In general, a linear combination of a set of vectors (or functions) involves multiplying each vector by a corresponding scalar (coefficient) and then summing the results. The expression for a linear combination can be written as: c 1 v 1+c 2 v 2+…+c n v n c 1 v 1+c 2 v 2+…+c n v n where c 1,c 2,…,c n c 1,c 2,…,c n are the scalars and v 1,v 2,…,v n v 1,v 2,…,v n are the vectors. If any of the scalars c i c i is equal to zero, that means the corresponding vector v i v i does not contribute to the sum. Therefore Continue Reading Yes, a linear combination can have a constant equal to zero. In general, a linear combination of a set of vectors (or functions) involves multiplying each vector by a corresponding scalar (coefficient) and then summing the results. The expression for a linear combination can be written as: c 1 v 1+c 2 v 2+…+c n v n c 1 v 1+c 2 v 2+…+c n v n where c 1,c 2,…,c n c 1,c 2,…,c n are the scalars and v 1,v 2,…,v n v 1,v 2,…,v n are the vectors. If any of the scalars c i c i is equal to zero, that means the corresponding vector v i v i does not contribute to the sum. Therefore, it is entirely possible for the sum of the linear combination to equal zero, depending on the values of the other coefficients and vectors. For example, consider the linear combination of two vectors v 1 v 1 and v 2 v 2: c 1 v 1+c 2 v 2=0 c 1 v 1+c 2 v 2=0 If we set c 1=1 c 1=1 and c 2=−1 c 2=−1, then: 1⋅v 1+(−1)⋅v 2=v 1−v 2=0 1⋅v 1+(−1)⋅v 2=v 1−v 2=0 This indicates that v 1=v 2 v 1=v 2. Thus, a linear combination can indeed equal zero, and the coefficients can include zeros as well. Upvote · Jared Ronning Student at University of Maryland · Upvoted by BowTangey , PhD Mathematics, Iowa State University (1988) · Author has 81 answers and 251.6K answer views ·Updated 8y Related How would you show that a determinant can be zero if and only if its rows are linearly dependent? Suppose the rows of an n×n n×n matrix A A are given by →r 1,…,→r n r 1→,…,r n→. First we want to show that if the rows are linearly dependent, then the determinant of A A is 0 0. Assume the rows are linearly dependent. Then there exists scalars, c 1,…,c n c 1,…,c n, not all zero, such that c 1→r 1+⋯+c n→r n=→0 c 1 r 1→+⋯+c n r n→=0→. Take c k c k to be a nonzero weight, for k∈[1,n]k∈[1,n]. Then →r k r k→ is a linear combination of the other vectors. It follows that we can do a sequence of row operations to get a 0 0 row in the k k th row (adding multiples of one row to another does not change the determina Continue Reading Suppose the rows of an n×n n×n matrix A A are given by →r 1,…,→r n r 1→,…,r n→. First we want to show that if the rows are linearly dependent, then the determinant of A A is 0 0. Assume the rows are linearly dependent. Then there exists scalars, c 1,…,c n c 1,…,c n, not all zero, such that c 1→r 1+⋯+c n→r n=→0 c 1 r 1→+⋯+c n r n→=0→. Take c k c k to be a nonzero weight, for k∈[1,n]k∈[1,n]. Then →r k r k→ is a linear combination of the other vectors. It follows that we can do a sequence of row operations to get a 0 0 row in the k k th row (adding multiples of one row to another does not change the determinant). Thus, doing a cofactor expansion along the k k th row in the appropriate row equivalent matrix will leave us with a determinant of 0 0. Next we want to show that if the determinant is zero, then the rows are linearly dependent. That’s equivalent to showing that if the rows are linearly independent, then the determinant is not zero (Proof by contrapositive). Suppose that the rows of A A are linearly independent and that U U is an echelon form of A A. Then the determinant of A A is some scalar multiple of the product of the diagonal entries in U. None of these entries can be zero, because otherwise one of the rows could be reduced to a row of zeros and by a similar argument from before, that would entail a relation of linear dependence on the rows. Thus, det(A)≠0 det(A)≠0. A shorter argument would be to note that A A is invertable if and only if det(A)≠0 det(A)≠0 if and only if det(A T)≠0 det(A T)≠0 if and only if the columns of A A are linearly independent if and only if the columns of A T A T are linearly independent. Which works because the columns of A T A T are the rows of A A. Upvote · 99 10 Related questions More answers below What is 0 0 0 0 (the zeroth power of zero)? Is it possible to add three vector of equal magnitude and get zero? What is the difference between a linear transformation and a linear combination? What does linear combination mean in mathematics? Why is linear algebra called "linear"? Pranesh Muddebihal Studied at Indian Institute of Science, Bangalore (IISc) · Author has 199 answers and 498.3K answer views ·8y Related How would you show that a determinant can be zero if and only if its rows are linearly dependent? This is NOT a proof, but hope it helps you get the concept right. A 2×2 2×2 determinant basically represents area of a parallelogram formed by 2 vectors, →a=a 1^i+a 2^j a→=a 1 i^+a 2 j^ and →b=b 1^i+b 2^j b→=b 1 i^+b 2 j^ The area is ∣∣∣a 1 a 2 b 1 b 2∣∣∣\|a 1 a 2 b 1 b 2\| Now, if the vectors are linearly dependent, they will lie on the same line, which makes the area of the parallelogram zero. Similarly, a 3×3 3×3 determinant represents volume of a parellelopiped , whose adjacent sides are →a=a 1^i+a 2^j+a 3^k a→=a 1 i^+a 2 j^+a 3 k^ , →b=b 1^i+b 2^j+b 3 b→=b 1 i^+b 2 j^+b 3 Continue Reading This is NOT a proof, but hope it helps you get the concept right. A 2×2 2×2 determinant basically represents area of a parallelogram formed by 2 vectors, →a=a 1^i+a 2^j a→=a 1 i^+a 2 j^ and →b=b 1^i+b 2^j b→=b 1 i^+b 2 j^ The area is ∣∣∣a 1 a 2 b 1 b 2∣∣∣\|a 1 a 2 b 1 b 2\| Now, if the vectors are linearly dependent, they will lie on the same line, which makes the area of the parallelogram zero. Similarly, a 3×3 3×3 determinant represents volume of a parellelopiped , whose adjacent sides are →a=a 1^i+a 2^j+a 3^k a→=a 1 i^+a 2 j^+a 3 k^ , →b=b 1^i+b 2^j+b 3^k b→=b 1 i^+b 2 j^+b 3 k^and →c=c 1^i+c 2^j+c 3^k c→=c 1 i^+c 2 j^+c 3 k^ The volume is ∣∣ ∣∣a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3∣∣ ∣∣\|a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3\| Now, if the vectors are linearly dependent, they will lie on on plane, which makes the parallelopiped a 2D figure. Thus, its volume will be zero and hence the determinant. NOTE : If you want to use the properties of determinants, keep this in mind : For 2 vectors →A,→B,A→,B→,→A+t→B A→+t B→ is always co-planar with →A,→B.A→,B→. So, effectively, you are checking for the linear dependence of vectors. Upvote · 99 11 9 1 Charles S. Former mathematician, current patent lawyer · Upvoted by Alexey Godin , Ph.D. Mathematics & Economics, Moscow State University (1998) · Author has 7.6K answers and 60.4M answer views ·8y Related What is the difference between “linear equation” and “linear combination”? An equation is like a sentence. It says s omething: this stuff over here is equal to that stuff over there. A linear combination is just a word or a phrase. On its own, it doesn’t say anything. And I guess carrying the analogy a little further, a linear combination is a certain type of word. In more mathematical detail, a linear combination of some objects — say, objects x,y,x,y, and z z — is an expression of the form a x+b y+c z a x+b y+c z, where the coefficients a,b,a,b, and c c come from some field. That field is typically understood from the context. If there’s no specific context (like in this question), it’s Continue Reading An equation is like a sentence. It says s omething: this stuff over here is equal to that stuff over there. A linear combination is just a word or a phrase. On its own, it doesn’t say anything. And I guess carrying the analogy a little further, a linear combination is a certain type of word. In more mathematical detail, a linear combination of some objects — say, objects x,y,x,y, and z z — is an expression of the form a x+b y+c z a x+b y+c z, where the coefficients a,b,a,b, and c c come from some field. That field is typically understood from the context. If there’s no specific context (like in this question), it’s just an abstract field F F. If you hate that, just think of the coefficients as real numbers. The point is, a x+b y+c z a x+b y+c z is a linear combination, but a b c x y z a b c x y z or x 2+y 2+z 2 x 2+y 2+z 2 isn’t. All the expressions in the previous sentence are perfectly valid. But only one of them is a linear combination. On the other hand, a linear equation is… well… an equation. It’s something that has an equals sign. And it’s of the form (or can be re-written in the form) a x=b a x=b, with the footnote that a a, x x, and b b (as well as the concept of what multiplication is — that is, how to make sense of a x a x even if you know what a a and x x are) can all be kind of fancy. In other words, 7 x=14 7 x=14 is a linear equation. 7 x−15=1 7 x−15=1 is also a linear equation, because it can be re-written in the form a x=b a x=b. But Δ f=1 ρ 2∂∂ρ(ρ 2∂f∂ρ)+1 ρ 2 sin θ∂∂θ(sin θ∂f∂θ)+1 ρ 2 sin 2 θ∂2 f∂φ 2=0 Δ f=1 ρ 2∂∂ρ(ρ 2∂f∂ρ)+1 ρ 2 sin⁡θ∂∂θ(sin⁡θ∂f∂θ)+1 ρ 2 sin 2⁡θ∂2 f∂φ 2=0 is also a linear equation. (It’s Laplace’s equation in spherical coordinates, for anyone interested in that kind of thing.) It’s hard to figure out what the a,x,a,x, and b b are, but I promise they’re there. :) Upvote · 99 10 9 1 Kurt Behnke PhD in Mathematics&Theoretical Physics, University of Hamburg (Graduated 1981) · Author has 8.3K answers and 14.2M answer views ·May 27 Related When will linear equations with rational coefficients and constant terms equal to zero have no solution? Linear equations with rational coefficients and constant term zero are called homogeneous. And the rational numbers are not essential. Any field will do. Think of the matrix of coefficients as a linear map A from a vector space V to a vector space W and apply the standard dimension formula from linear algebra: the dim V=dim dim ker A+dim im A.dim⁡V=dim⁡dim⁡ker⁡A+dim⁡im A. Since the kernel of A A consists exactly of the solutions of your equation this tells you: the number of independent solutions equals the difference between the dimension of V (the number of variables) and the number of linearly independent ro Continue Reading Linear equations with rational coefficients and constant term zero are called homogeneous. And the rational numbers are not essential. Any field will do. Think of the matrix of coefficients as a linear map A from a vector space V to a vector space W and apply the standard dimension formula from linear algebra: the dim V=dim dim ker A+dim im A.dim⁡V=dim⁡dim⁡ker⁡A+dim⁡im A. Since the kernel of A A consists exactly of the solutions of your equation this tells you: the number of independent solutions equals the difference between the dimension of V (the number of variables) and the number of linearly independent rows or columns of the matrix. So if you have got as many linearly independent matrix columns as you have got variables, there is only the trivial solution (all variables zero, which always exists). Upvote · Alon Amit 30 years of Linear Algebra. · Upvoted by Sam Sinai , Ph.D student in Mathematical Biology · Author has 8.8K answers and 173.8M answer views ·11y Related What is a linear combination? A linear combination of things is the sum of these things, each multiplied by some number (or "scalar"). Upvote · 99 13 Dmitriy Genzel PhD in CS · Author has 2.9K answers and 35.7M answer views ·9y Related What precisely is a linear combination? If you have some set of mathematical objects {x 1…x n}{x 1…x n}that support multiplication by a scalar and addition (members of some ring or a vector space, or something), then any y=a 1 x 1+…+a n x n y=a 1 x 1+…+a n x n is their linear combination (where a i a i are some sort of "scalars" - whatever that means in your context - likely real numbers). The most common illustration is to use normal Euclidean-space 3D vectors. A linear combination of any two such vectors is a vector that lies in the same plane through the origin as the original two vectors placed at the origin. Upvote · 9 5 Siddhartha Bhattacharjee passionate about equations and inequations · Author has 141 answers and 126.7K answer views ·5y Related Why are linear equations equal to zero? An equation itself can’t equal anything. If there’s two quantities a a and b b, and we have it that a=b a=b, we can write the same statement as a−b=0 a−b=0 In the case of linear equations, this is done only for ease. Using this transposition of all terms to the left hand side of an equation, we get a systematic structure like a x+b y+c=0 a x+b y+c=0, or any higher or lower order structure. Upvote · 9 2 9 1 Keyura Sree Velavarthipati M.P.C from Sri Chaitanya Educational Institutions (Graduated 2016) ·4y Related Must the initial condition be zero for linear systems? Yes, for a system to be linear its initial conditions must be zero. If not, the system is said to be incrementally linear but not linear. The basic tool to analyse a linear system is Transfer function and it is defined only for those systems with zero initial conditions. Initial conditions are due to the system elements like capacitors and inductors and zero initial conditions means the capacitor has zero initial voltage and inductor has zero initial current, indicating that no energy is stored in the circuit. Note that the resistor is a memoryless element i.e., it doesn't store energy but diss Continue Reading Yes, for a system to be linear its initial conditions must be zero. If not, the system is said to be incrementally linear but not linear. The basic tool to analyse a linear system is Transfer function and it is defined only for those systems with zero initial conditions. Initial conditions are due to the system elements like capacitors and inductors and zero initial conditions means the capacitor has zero initial voltage and inductor has zero initial current, indicating that no energy is stored in the circuit. Note that the resistor is a memoryless element i.e., it doesn't store energy but dissipiates. The problem with a non-linear system is that the output signal will contain a term that is independent of the input signal. So, it generates a non-zero output even when zero input signal is applied. But a linear system must generate a zero output signal for a zero input signal. Even the properties of linearity like additivity and homogeneity are violated making the system non-linear. To analyse a system with initial conditions(non-linear, also time variant systems), we have tools like state space analysis. Upvote · 9 2 Scott Brickner I regret not majoring in math, so I do it as a hobby · Author has 19.1K answers and 6.5M answer views ·6y Related Does a zero determinant of a set of vectors in matrix form always indicate linear dependence? If not, why, and what are the exceptions? Yes, a zero determinant means the columns of the matrix are linearly dependent as vectors. Think of it this way: the matrix represents a linear transformation. It maps the standard basis vectors to the vector in the corresponding column of the matrix. The standard basis vectors represent a unit polytope—a square in 2d, a cube in 3d, and so on. If the column vectors are linearly independent, then that unit polytope must be mapped to a polytope of the same dimension. When they’re linearly dependent, it’s mapped to a polytope of strictly smaller dimension. So, if you’re in 3d, your unit cube is map Continue Reading Yes, a zero determinant means the columns of the matrix are linearly dependent as vectors. Think of it this way: the matrix represents a linear transformation. It maps the standard basis vectors to the vector in the corresponding column of the matrix. The standard basis vectors represent a unit polytope—a square in 2d, a cube in 3d, and so on. If the column vectors are linearly independent, then that unit polytope must be mapped to a polytope of the same dimension. When they’re linearly dependent, it’s mapped to a polytope of strictly smaller dimension. So, if you’re in 3d, your unit cube is mapped to either a parallelepiped, a square, a line, or a point. All of those except the parallelepiped has a zero volume. Sure, some of them may have non-zero area or length, but only the parallelepiped has a non-zero volume. That happens in higher dimensions, too—in N dimensions, there’s an N-dimensional analog of volume, and only N-dimensional objects have N-volumes—objects with fewer dimensions have zero N-volume. If the matrix has a non-zero determinant, then it maps the N-dimensional unit polytope to an N-dimensional object. If it has a zero determinant, it maps it to a less-than-N-dimensional object—but that means the images of the standard basis are linearly dependent. Upvote · 9 9 9 3 Related questions What precisely is a linear combination? What is the linear combination of zero (0) vectors? Is the zero vector a linear combination? Why is a zero vector linearly dependent? Can you provide an example of two vectors whose sum is zero and whose difference is equal to the sum of any other two vectors? What is 0 0 0 0 (the zeroth power of zero)? Is it possible to add three vector of equal magnitude and get zero? What is the difference between a linear transformation and a linear combination? What does linear combination mean in mathematics? Why is linear algebra called "linear"? What is the relationship between mutually perpendicular non-zero pairs of vectors and linear independence? How can I learn linear algebra from zero? Why is 0/1 equal to zero? How does zero times zero equal to zero make any sense? Why is every equation always equal to zero? Related questions What precisely is a linear combination? What is the linear combination of zero (0) vectors? Is the zero vector a linear combination? Why is a zero vector linearly dependent? Can you provide an example of two vectors whose sum is zero and whose difference is equal to the sum of any other two vectors? What is 0 0 0 0 (the zeroth power of zero)? Is it possible to add three vector of equal magnitude and get zero? What is the difference between a linear transformation and a linear combination? What does linear combination mean in mathematics? Why is linear algebra called "linear"? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
16376	https://study.com/learn/lesson/stop-codon-mutations-concept-function.html	Stop Codon Overview, Diagram & Amino Acid - Lesson \| Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up Copyright Science Courses / Biology: High School Course Stop Codon Overview, Diagram & Amino Acid Lesson Transcript Vernell Leavings, Angela Lynn Swafford Author Vernell Leavings Vernell attended the University of Central Florida, where he worked as a lab and lecture teaching assistant and personal tutor. He graduated with a bachelor's degree in Biology before becoming a middle school science teacher. Vernell taught 7th and 8th grade science for several years before leaving to pursue other options. View bio Instructor Angela Lynn Swafford Lynn has a BS and MS in biology and has taught many college biology courses. View bio Learn what the three stop codons are. Identify the functions of the start and stop codons, study codon diagrams, and discover mutations of stop codons in DNA. Updated: 11/21/2023 Table of Contents What Are the Stop Codons In DNA? How Many Codons Are There? Functions and Mutations of Termination Codes Types of Stop Codon Mutations Lesson Summary Show Frequently Asked Questions What are the 3 stopping codons? There are three (3) stop codons that immediately halt the protein synthesis process. These 3 codons are: UAA, UAG, and UGA. What are start and stop codons? Start and stop codons are unique triplets that being and end the process of protein synthesis. The universal start codon is AUG, easy to remember because school starts in AUGust. The three stop codons are UAA, UAG, and UGA. These are also called termination codons. Create an account Table of Contents What Are the Stop Codons In DNA? How Many Codons Are There? Functions and Mutations of Termination Codes Types of Stop Codon Mutations Lesson Summary Show What Are the Stop Codons In DNA? -------------------------------- What Is DNA? Deoxyribonucleic acid (DNA) is the form of genetic material in eukaryotes, or organisms whose cells contain a nucleus. DNA is a rather long molecule that is made of smaller pieces bonded together. These smaller pieces are called nucleotides and are made of three (3) individual parts: a nitrogenous base, phosphate group, and a sugar. A string of nucleotides that code for a trait (characteristic) is called a gene, each gene is unique and requires a specific sequence of bases. There are four nitrogenous bases: adenine, thymine, guanine, and cytosine; the order of these bases on DNA strand is what determines the specific gene and protein. What Is a Codon? A triplet of nitrogenous bases (for example, GAG) is called a codon and is specific to an amino acid. Amino acids are the "building blocks" of proteins, and synthesizing the correct protein depends strictly on having the correct order of amino acids. In turn, the sequencing of amino acids depends on the order of bases that make the codons. If the order of bases is altered, it will change the codons, which can lead to a change in protein and a possible mutation. The image below shows each codon and the amino acids they code for. Making Proteins Protein synthesis is the process of translating the information contained in a gene into a product the cell needs for proper function. In order for this to happen, the gene on the DNA strand must first be converted to an older form of genetic material, ribonucleic acid or RNA. The transcribed (copied) messenger RNA (mRNA) is sent to the cytoplasm to bind with free-floating transfer RNA (tRNA) molecules that carry amino acids. The base codon on the mRNA will complement the anticodon (complementary base triplet) on the tRNA and will signal for the tRNA to bind to that location only. This lock-and-key fit ensures that the correct tRNA and amino acid joins where it should and builds the correct protein. Remember: adenine will pair with thymine (uracil in RNA), and guanine will pair with cytosine. Start and Stop Codons In order for the process of protein synthesis to begin, it needs to have a starting point, or a "start" codon. The universal start codon, AUG (a triplet of adenine, uracil, and guanine,) codes for the amino acid methionine and is therefore found in most proteins. AUG should be easy to remember because these are the the first three letters of August when the school year typically starts. While start codons begin the process of building proteins, stop codons of course will end it; these are also known as termination codons. There are three (3) stop codons: UAA, UAG, and UGA. Each of these will be read during protein synthesis and bind the tRNA automatically end the process. These three codons may be a bit more difficult to remember but try to relate them to college football teams (Alabama, Auburn, and Georgia) that typically make it to bowl games at the end of the season. Notice that one of the bases in the start and stop codons, uracil, is unusual for DNA sequences. Uracil is found in RNA and not in DNA and takes thymine's place of pairing with adenine. Remember that DNA is a mutation of RNA and so the transcription, translation and even replication processes require RNA bases. To unlock this lesson you must be a Study.com Member. Create your account Click for sound 4:01 You must c C reate an account to continue watching Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Create Your Account To Continue Watching As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. Try it now Try it now. Already registered? Log in here for access Back Resources created by teachers for teachers Over 30,000 video lessons& teaching resources‐all in one place. Video lessons Quizzes & Worksheets Classroom Integration Lesson Plans I would definitely recommend Study.com to my colleagues. It’s like a teacher waved a magic wand and did the work for me. I feel like it’s a lifeline. Jennifer B. Teacher Try it now Back Coming up next: Beneficial Mutation \| Overview & Examples You're on a roll. Keep up the good work! Take QuizWatch Next Lesson Replay Just checking in. Are you still watching? Yes! Keep playing. Your next lesson will play in 10 seconds 0:00 What are Mutations? 0:25 What are Stop Codons? 2:00 Nonsense Mutations 2:40 Nonstop Mutations 3:16 Lesson Summary View Video Only Save Timeline 68K views Video Quiz Course Video Only 68K views How Many Codons Are There? -------------------------- There are 64 unique codons, each based on the different combinations of the four nitrogenous bases. There are 61 triplets that code for amino acids and the remaining 3 are stop codons. The amino acid coded for can be checked on a codon chart, like the one below. To properly read the chart below, start in the center with your first base, move outward to the second, and then your third base will be the outer ring; be careful to select the correct letter for the final base. The amino acids coded for are written on the outermost portion of the codon wheel. As shown on the codon chart, each triplet codes for a single amino acid, but each amino acid can be signaled by multiple codons. This property, known as redundancy, helps to guard against errors when replicating or transcribing DNA. Having multiple options for each amino acid increases the chances that the intended one will bind when signaled. However, when insertions or deletions create a stop codon, protein synthesis ends prematurely, and the intended protein is not created. You can use a codon chart to figure out what amino acid is being coded for on a strand of mRNA. To unlock this lesson you must be a Study.com Member. Create your account Functions and Mutations of Termination Codes -------------------------------------------- When mRNA is translated, tRNA molecules that carry amino acids look for the codon that signals they should bind there to initiate and continue the chain. Remember, mRNA is read three bases at a time, and the order of these bases determines a codon. Each codon corresponds to only one amino acid, so changing the codon will change the amino acid that joins the protein chain. The tRNA molecules carry an anticodon that is complementary to the triplet on the mRNA strand. The strand is continuously read, and amino acids are added to the protein chain until a stop codon is reached. A single strand of DNA is an incredibly long molecule, and the process of copying DNA is not perfect. Sometimes there are mutations, or changes, that occur that can alter the reading frame of bases. Insertions add an extra base, while deletions remove existing bases; both changes result in a shift in how the bases are read as triplets. The image below shows how these changes appear on DNA strand. To unlock this lesson you must be a Study.com Member. Create your account Types of Stop Codon Mutations ----------------------------- The term mutation often has a negative connotation meaning something bad or harmful. Sometimes these changes have no impact whatsoever, these are called silent mutations. Silent mutations are due to a change in the order of nitrogenous bases; however, the property of redundancy makes the effect neutral. Other times the change in bases can have a potentially minor impact, while other times they can lead to a stop codon and an unfinished protein product. The pink letters show the changes made from the original strand. The results of these changes are also visible. Missense Mutations One of the potential results of a frameshift change is a missense mutation. When missense mutations happen, a different amino acid is added to a protein chain instead of the intended one. For example, instead of having the codon AUA for the isoleucine amino acid, it would end up with something like GUA for the amino acid valine. Missense mutations may not cause any real damage but can also result in an inactive protein, which will cause problems if a protein product is required and cannot be synthesized. Nonsense Mutations Another result of changes in the codon reading frame is a nonsense mutation, where a premature stop codon is reached. Because stop codons immediately end the protein synthesis process, the result is almost always an incomplete and inactive protein. These mutations are the most dangerous because protein products are necessary for proper health and function. If they the products cannot be synthesized, cells will lack what they require and may die as a result. To unlock this lesson you must be a Study.com Member. Create your account Lesson Summary -------------- Codons consist of three nucleotide bases together, and these triplets code for specific amino acids. There are four nitrogenous bases: adenine, guanine, thymine, and cytosine. These four bases together create 64 unique codons, with 61 that code for amino acids and 3 stop (or termination) codons that end the process of protein synthesis. During protein synthesis, the different amino acids bond to a strand of messenger RNA (mRNA) that has the instructions for a gene product. The process begins with the near universal start codon AUG which codes for the amino acid methionine and is found in almost all proteins. Although each codon only signals for one amino acid, there can be multiple codons for an amino acid; this property is known as redundancy and increases the likelihood of getting the correct sequence of amino acids in a protein chain. When reading the codons and translating the gene instructions, sometimes there are errors due to a change in the reading frame of bases (frameshift.) This shift can happen due to changes, such as insertions or deletions, of bases from the strand of DNA or mRNA. Remember: if the order of the base changes, the codon and the amino acids they code for will change. Sometimes these changes are less serious, like with silent mutations, where base change doesn't affect the amino acid binding order, or missense mutations that do change the order of the amino acids that bind in the chain. Other times, changes in bases lead to a nonsense mutation where a stop codon is reached too early. Both missense and nonsense mutations can result in an inactive and nonfunctional protein. To unlock this lesson you must be a Study.com Member. Create your account Video Transcript What Are Mutations? DNA is the unit of heredity of all organisms and is arranged into genes. Genes, including any mutations in them, can be passed on from parents to children. A mutation is any permanent change in a DNA sequence. Mutations don't have to be bad; some are beneficial while others have no effect. However, in this lesson, we will focus on potentially detrimental mutations involving stop codons. What Are Stop Codons? We'll get to what a stop codon is in a little bit. First, we need to learn more about DNA and genes. Genes are sequences of DNA that code for proteins, and proteins are made up of many amino acids. Nucleotides are the repeating units of a DNA sequence. There are four nucleotides, each with a different nitrogenous base: thymine (T), adenine (A), guanine (G), and cytosine (C). The order of these nitrogenous bases is what determines the sequences of amino acids in a protein. The first step in making a protein, called transcription, is to use the DNA sequence of a gene to make an RNA molecule. RNA is made up of nucleotides and nitrogenous bases just like DNA, except that thymine is replaced with uracil (U). Then, this RNA molecule is used to assemble a chain of amino acids, or a protein, in a process called translation. Amino acids are specified by a 3-base sequence called a codon. Amino acids continue to be added to the chain until a stop codon is reached. A stop codon does not specify an amino acid but instead stops translation, just like a stoplight halts traffic. Transcription makes a single-stranded RNA molecule from a double-stranded DNA molecule. Translation uses each codon to determine which amino acid belongs in the protein. (CAG codes for glutamine and CCC codes for proline. UAA is a stop codon.) There are three different stop codons in RNA and DNA: \| RNA \| DNA \| --- \| \| UAG \| TAG \| \| UAA \| TAA \| \| UGA \| TGA \| There are two types of mutations involving stop codons: nonsense mutations and nonstop mutations. Nonsense mutations Nonsense mutations occur when a codon that is supposed to specify an amino acid is changed to a stop codon instead. This results in a protein that is shorter than normal because it cannot finish adding all the necessary amino acids. These truncated proteins are often not functional. A nonsense mutation would be like if you were sitting at a stoplight and it never turned green. You would never get to finish your trip, just like translation would not be able to finish making the whole protein. Many genetic disorders can be caused by having short, nonfunctioning proteins. For example, nonsense mutations occur in some cases of cystic fibrosis. An example of a nonsense mutation. Nonstop Mutations Nonstop mutations are kind of the opposite of nonsense mutations. They occur when the sequence of a stop codon is changed to specify an amino acid instead. When this happens, translation will continue until another stop codon is found. This results in a long protein that, again, is not usually able to function. Nonstop mutations would be like a traffic light that is always green. This sounds great for driving, but for making proteins that need to be a certain length to function, this would not be a good thing. While nonstop mutations do cause genetic disorders, these disorders tend to be very rare. An example of a nonstop mutation. Lesson Summary Stop codons are sequences of DNA and RNA that are needed to stop translation or the making of proteins by stringing amino acids together. There are three RNA stop codons: UAG, UAA, and UGA. In DNA, the uracil (U) is replaced by thymine (T). If a change in the DNA sequence, or mutation, of a gene occurs that creates a stop codon, this is called a nonsense mutation. A nonstop mutation is a change in the sequence of a stop codon that turns it into a codon that specifies an amino acid. Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Unlock Your Education See for yourself why 30 million people use Study.com Become a Study.com member and start learning now. Become a Member Already a member? Log In Back Resources created by teachers for teachers Over 30,000 video lessons& teaching resources‐all in one place. 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16377	https://users.wpi.edu/~bservat/book.html	"Combinatorial Rigidity", AMS, 1993. Title: Combinatorial Rigidity Authors: Jack E. Graver, Brigitte Servatiusand Herman Servatius Publisher: Graduate Studies in Math., AMS, 1993. Card Catalog: QA 166.6.G73 Assorted figures from the text. The Preface Click here to order Some Figures from the Text: Preface to Text: A framework in m-space is a triple (V,E,p), where (V,E) is a finite graph and p is an embedding of V into real m-space. A framework is a mathematical model for a physical structure in which each vertex v corresponds to an idealized ball joint located at p(v), and each edge corresponds to a rigid rod connecting the joints corresponding to its endpoints. Obviously this concrete realization is meaningful only for $m < 4$, and may be used to describe a very general class of physical structures, including rigid ones such as pedestals or bridges, as well as moving structures such machines or organic molecules. Making the distinction between frameworks whose realization is rigid and those which can move is the fundamental problem of rigidity theory, which can also be considered for frameworks in higher dimensions. In low dimensions one could construct an appropriate realization of a given framework and test the model for rigidity. Of course, the mathematical task is to develop a method for predicting rigidity without building a model. One would expect that whether a framework is rigid or not depends on both the graph (V,E) and the embedding p; or, in more general terms, that the question of rigidity has both combinatorial and geometric aspects. Our primary interest is in the combinatorial part of rigidity theory, which we call combinatorial rigidity. However, the two parts of rigidity theory are not so easily separated. In fact, only in dimensions one and two has total separation of the two parts been achieved. In the first chapter we will give an overview of the subject, developing both aspects of the theory of rigidity informally in an historical context. This chapter stands apart from the rest of the book in that it contains no formal proofs. Most of the concepts introduced here will be reintroduced in a more formal setting later on. The second chapter is devoted to a study of infinitesimal rigidity, a linear approximation which stands at the boundary of the combinatorial and geometric nature of rigidity. The infinitesimal approach offers at least a partial separation of the combinatorial and geometric aspects by regarding the matrix of the derivative of a framework motion as a matroid on the edges of the framework. In general, depending on the dimension and the embedding, the edges of a graph are the underlying set of several such matroids, all of which belong to the class of abstract rigidity matroids, which are defined at end of chapter 2. The fundamental combinatorial structures used to study rigidity are the various rigidity matroids. The second chapter consists of a development of matroid theory, the theoretical foundation for much of modern combinatorics. There will of course be a special emphasis on those parts of the subject most applicable to rigidity matroids. Chapter 3 is devoted to an extensive study of combinatorial rigidity dimension 2, which has a nice analogy, via the 1-dimensional case, with ``traditional'' graph theory from a slightly different point of view. A thorough knowledge of planar rigidity is essential to developing a good intuition for rigidity as a whole, and provides an extensive collection of tractable examples. Algorithmic and computational aspects are also treated. In the last chapter, we discuss combinatorial rigidity in higher dimensions. Special attention is paid to dimension~3, in which there is the most practical interest, but where the characterization problem is still unsolved. Many of the results in this chapter have not yet appeared elsewhere. The book concludes with an extensive annotated bibliography. This text is suitable for a second graduate course in combinatorics and was already used as such at Syracuse University and at Worcester Polytechnic Institute by the authors. Each chapter contains a variety of exercises, some letting the reader fill in the details of the theory, some working through examples, as well as many which point the way to aspects of rigidity theory not covered in the text. Exercises are placed so that the reader can check his understanding of each concept before going on to the next one. The annotations in the bibliography are not only a valuable research tool, but also meant to stimulate a project oriented course of study. Each of the chapters is mathematically self-contained, and the reader may safely peruse them in the order best suited to his background and interest. Other articles on Structural Rigidity (Back tobooks)
16378	https://math.stackexchange.com/questions/1246643/intersection-of-integer-sets	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Intersection of integer sets Ask Question Asked Modified 10 years, 5 months ago Viewed 833 times 2 $\begingroup$ This is probably a trivial question for mathematicians but I am not seeing how to approach the following problem: Imagine two sets defined by: set1 = (x: x Where: all variables are integers step != 0 Example of what I mean by this: for start = 3, stop = 20, step = 5: set = [3, 8, 13, 18] I am looking for: whether the two sets intersect what start, stop, step are for the intersection set algorithms diophantine-equations linear-diophantine-equations Share edited Apr 22, 2015 at 22:50 mvw 35.2k22 gold badges3434 silver badges6565 bronze badges asked Apr 22, 2015 at 14:00 MaxMax 2122 bronze badges $\endgroup$ 3 1 $\begingroup$ What are $m$ and $n$? $\endgroup$ Todd Wilcox – Todd Wilcox 2015-04-22 14:04:34 +00:00 Commented Apr 22, 2015 at 14:04 $\begingroup$ @ToddWilcox m and n are integer. I added an example of what I mean with this notation to the question. $\endgroup$ Max – Max 2015-04-22 14:12:45 +00:00 Commented Apr 22, 2015 at 14:12 $\begingroup$ @Max I added a Ruby implementation to my answer. Have a look at the excerpt from the log to better understand how it works. $\endgroup$ mvw – mvw 2015-04-22 22:41:57 +00:00 Commented Apr 22, 2015 at 22:41 Add a comment \| 2 Answers 2 Reset to default 2 $\begingroup$ Mathematical Model: You could model the sets $S_1$ and $S_2$ like this \begin{align} I_1 &= [a_1, b_1) \quad (a_1, b_1 \in \mathbb{N})\ I_2 &= [a_2, b_2) \quad (a_2, b_2 \in \mathbb{N}) \ Z_1 &= { x = a_1 + n \Delta x_1 \mid n \in \mathbb{Z} } \quad (\Delta x_1 \in \mathbb{N}) \ Z_2 &= { x = a_2 + m \Delta x_2 \mid m \in \mathbb{Z} } \quad (\Delta x_2 \in \mathbb{N}) \ S_1 &= Z_1 \cap I_1 \subseteq I_1 \ S_2 &= Z_2 \cap I_2 \subseteq I_2 \end{align} Then for an element of the intersection $S_1 \cap S_2$ we have the condition $$ a_1 + n \Delta x_1 = a_2 + m \Delta x_2 \iff \ (\Delta x_1) n - (\Delta x_2) m = a_2 - a_1 \quad () $$ This is a linear Diophantine equation $$ a x + b y = c \quad () $$ plus the limiting condition $$ a_1 + n \Delta x_1 \in I_1 \cap I_2 $$ A linear Diophantine equation $()$ has either no or infinite many solutions. It has solutions if $\gcd(a,b) \mid c$. Applying the Solution of the Diophantine Equation: Translating this to the case of equation $()$ we have the criterion $$ \gcd(\Delta x_1, \Delta x_2) \mid (a_2 - a_1) \quad (#) $$ In the case of solutions, we can find a particular solution by using the extended Euclidean algorithm to find integers $s$, $t$ with the property $$ s \Delta x_1 + t \Delta x_2 = \gcd(\Delta x_1, \Delta x_2) \quad (##) $$ Then $$ (n_0, m_0) = \left( \frac{a_2-a_1}{\gcd(\Delta x_1, \Delta x_2)} s, -\frac{a_2-a_1}{\gcd(\Delta x_1, \Delta x_2)} t \right) $$ is a solution of equation $()$. Then all solutions are $$ (n, m) = \left( n_0 + k \frac{\Delta x_2}{\gcd(\Delta x_1, \Delta x_2)}, m_0 + k \frac{\Delta x_1}{\gcd(\Delta x_1, \Delta x_2)}, \right) \quad (k \in \mathbb{Z}) $$ Determining the Unbounded Intersection: So we can use equation $(#)$ to find out, if the intersection can have solutions. If it can, then we use the $s$ obtained by $(##)$ to calculate $n_0$ and use the first components of the above equation to calculate all $n$ and all elements $$ x = a_1 + n \Delta x_1 = a_1 + \left( \frac{a_2-a_1}{\gcd(\Delta x_1, \Delta x_2)} s + k \frac{\Delta x_2}{\gcd(\Delta x_1, \Delta x_2)} \right) \Delta x_1 \quad (k \in \mathbb{Z}) \quad (+) $$ of the unbounded intersection $Z_1 \cap Z_2$. These then have to be tested against the boundaries of $I_1$ and $I_2$ to determine $S_1 \cap S_2$. $$ a_1 \le x < b_1 \wedge a_2 \le x < b_2 $$ So it might a good idea to look even before $(#)$ at $I_1 \cap I_2$ first, where we need $$ a_1 < b_2 \wedge a_2 < b_1 \quad (++) $$ for a non-empty intersection $I_1\cap I_2$. Start, Stop and Step of the Intersection: From the above we take that the step of the intersection set is $$ \frac{\Delta x_1 \Delta x_2}{\gcd(\Delta x_1, \Delta x_2)} \quad (\$) $$ Start is found via the smallest feasible $k$, let us call it $k_1$ such that $$ \left( a_i - a_1 - \frac{a_2-a_1}{\gcd(\Delta x_1, \Delta x_2)} s \Delta x_1 \right)\frac{\gcd(\Delta x_1, \Delta x_2)}{\Delta x_1\Delta x_2} \le k $$ thus $$ k_1 = \max_{i \in {1,2}} \left\lceil \left( a_i - a_1 - \frac{a_2-a_1}{\gcd(\Delta x_1, \Delta x_2)} s \Delta x_1 \right)\frac{\gcd(\Delta x_1, \Delta x_2)}{\Delta x_1\Delta x_2} \right\rceil $$ The element $x$ just before stop is found via the largest feasible $k$, let us call it $k_2$ such that $$ k < \left( b_i - a_1 - \frac{a_2-a_1}{\gcd(\Delta x_1, \Delta x_2)} s \Delta x_1 \right)\frac{\gcd(\Delta x_1, \Delta x_2)}{\Delta x_1\Delta x_2} $$ thus $$ k_2 = \min_{i \in {1,2}} \left\lfloor \left( b_i - a_1 - \frac{a_2-a_1}{\gcd(\Delta x_1, \Delta x_2)} s \Delta x_1 \right)\frac{\gcd(\Delta x_1, \Delta x_2)}{\Delta x_1\Delta x_2} \right\rfloor $$ or one less if the argument of floor was already an integer. Plugging the found $k_1$ and $k_2$ into equation $(+)$ then gives start and stop values of the intersection. Summary: We gave two criteria, equations $(#)$ and $(++)$ to determine if there is a non-empty intersection at all. We further listed the solution process which allows to determine all solutions enumerated by an integer parameter $k$ and gave conditions how to find the parameters $k_1$ for the smallest and $k_2$ for the largest feasible solution, which can be used via equation $(+)$ to get start and stop values. We also listed the step value of the intersection, see equation $(\$)$. The method looks complicated, but has the big advantage of being fast, the sizes of the sets do not really matter, so I would consider it $O(1)$. Implementation: I implemented a sample program in Ruby, to find out if this works as planned. You can find the source code here. I am happy to not have to correct the above formulas. The biggest bug I was running into during implementation was not using floating point arithmetics for the calculation of $k_1$ and $k_2$. The rest only needs integer arithmetics. Testing went against a brute force implementation of the set intersection. Over one million random test cases passed, so I think it mostly works. See an excerpt from a log here. The output should help to understand this approach. Share edited Apr 22, 2015 at 22:36 answered Apr 22, 2015 at 14:21 mvwmvw 35.2k22 gold badges3434 silver badges6565 bronze badges $\endgroup$ 3 $\begingroup$ Thank you so much! It seems the problem is not quite as trivial as I first thought. It is going to take me a bit of time to digest your answer... $\endgroup$ Max – Max 2015-04-22 14:57:24 +00:00 Commented Apr 22, 2015 at 14:57 $\begingroup$ You write "solutions, we can find a particular solution by using the extended Euclidean algorithm to find integers s, t with the property ..." Why do the solutions satisfy this property? I am surprised since the offsets a1 and a2 do not seem to come into play. $\endgroup$ Max – Max 2015-04-22 15:46:23 +00:00 Commented Apr 22, 2015 at 15:46 $\begingroup$ The solutions of finding working $n$ and $m$ show up after that, and involve the difference of $a_1$ and $a_2$. $\endgroup$ mvw – mvw 2015-04-22 15:51:37 +00:00 Commented Apr 22, 2015 at 15:51 Add a comment \| 0 $\begingroup$ If I understand correctly you are considering the sets $$A= {x\| x= a+ns_A, \space n\in\mathbb{Z}}$$ and $$B= {x\| x= b+ns_B, \space m\in\mathbb{Z}}$$ with the given boundaries (in other words intersected with $[start, stop]$). To solve $A\cap B$, you need to solve the diophantine equation $$s_An - s_Bm = b-a$$ See for example here how this can be done: Linear Diophantine Equation Then take every $a+ns_A$, for the $n$'s that are in a solution of the diophantine equation. Lastly intersect with the interval. Of course the way to actually do this is to start with the first non-negative $n$ and go on until $a+ns_A$ becomes larger than the smaller of the boundaries. Share answered Apr 22, 2015 at 14:21 ploosu2ploosu2 12.5k22 gold badges2222 silver badges4242 bronze badges $\endgroup$ Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algorithms diophantine-equations linear-diophantine-equations See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Related Finding sets that do not intersect. 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16379	https://www.vocabulary.com/lists/259673	corpulent - Vocabulary List \| Vocabulary.com Compete in the Vocabulary Bowl: 192 countries, 50 states, countless prizes. Register now!X SKIP TO CONTENT Log inSign up Dictionary Vocabulary Lists VocabTrainer™ Lists by Grade Literature Non-Fiction Textbooks & Curricula Test Prep Current Events Roots & Affixes Just For Fun Speeches Historical Documents OthersNew list Speeches Historical Documents corpulent Aldo D. (Belgium) Share Copy link 4 words 0 learners Learn words with Flashcards and other activities Definition first Print Flashcards corpulent /ˈkɔrpjələnt/ adj.excessively large obese /oʊˈbis/ /əʊˈbis/ adj.excessively large weighty /ˈweɪdi/ adj.having relatively great weight; heavy rotund /roʊˈtʌnd/ adj.spherical in shape Continue learning with flashcards! Log inSign up Shuffle Previous 1/4 Next View fullscreen Other learning activities Practice Answer a few questions about each word. Use this to prep for your next quiz!Vocabulary Jam Compete with other teams in real time to see who answers the most questions correctly!Spelling Bee Test your spelling acumen. Read the definition, listen to the word and try spelling it! Teaching tools Quiz Create and assign quizzes to your students to test their vocabulary.Assign activities Assign learning activities including Practice, Vocabulary Jams and Spelling Bees to your students, and monitor their progress in real-time. Full list of words from this list: words onlydefinitions & notes corpulentexcessively large obeseexcessively large weightyhaving relatively great weight; heavy rotundspherical in shape Created on May 1, 2013 Sign up now (it’s free!) Whether you’re a teacher or a learner, Vocabulary.com can put you or your class on the path to systematic vocabulary improvement. Get started Learn with us Learner subscriptions Vocabulary lists Dictionary Test Prep Join a Vocabulary Jam Commonly confused words Word of the day Teach with us For educators For schools and districts How it works Success stories Research Professional development Contact sales Resources Help articles/FAQ Teaching resources Learner resources ESL/ELL resources Grade level resources IPA Pronunciation Contact support Leaderboards Vocabulary Bowl Today's leaders Weekly leaders Monthly leaders About Our Mission Blog Tell us what you think Privacy Policy Terms of Use Comprehensive K-12 personalized learning Immersive learning for 25 languages Trusted tutors ready to help in 300+ subjects 35,000 worksheets, games, and lesson plans Marketplace for millions of educator-created resources Fun educational games for kids Spanish-English dictionary, translator, and learning Diccionario inglés-español, traductor y sitio de aprendizaje Fast and accurate language certification French-English dictionary, translator, and learning Copyright © 2025 Vocabulary.com, Inc. A division of IXL Learning • All Rights Reserved. Learn with us Learner subscriptions Vocabulary lists Dictionary Test Prep Join a Vocabulary Jam Commonly confused words Word of the day Teach with us For educators For schools and districts How it works Success stories Research Professional development Contact sales Resources Help articles/FAQ Teaching resources Learner resources ESL/ELL resources Grade level resources IPA Pronunciation Contact support Leaderboards Vocabulary Bowl Today's leaders Weekly leaders Monthly leaders About Our Mission Blog Tell us what you think Privacy Policy Terms of Use My Account Log inSign up My Account Log Out My Learning My Proficiency Report My Profile Schools & Teachers My Classes Assignments & Activities My Lists Find a List to Learn... Create a New List... My Progress Words I'm Learning My Trouble Words Words I've Mastered My Achievements User Administration User Authentication My Account Comprehensive K-12 personalized learning Immersive learning for 25 languages Trusted tutors ready to help in 300+ subjects 35,000 worksheets, games, and lesson plans Marketplace for millions of educator-created resources Fun educational games for kids Spanish-English dictionary, translator, and learning Diccionario inglés-español, traductor y sitio de aprendizaje Fast and accurate language certification French-English dictionary, translator, and learning Copyright © 2025 Vocabulary.com, Inc. A division of IXL Learning • All Rights Reserved.
16380	https://fiveable.me/key-terms/college-algebra/radicals	printables 📈college algebra review key term - Like Radicals Citation: MLA Definition Like radicals are square roots, cube roots, or other radical expressions that have the same index and are raised to the same power. They can be combined through addition, subtraction, or multiplication to simplify radical expressions. 5 Must Know Facts For Your Next Test Like radicals can be added or subtracted by combining the coefficients and keeping the same radical expression. Multiplying like radicals involves multiplying the coefficients and adding the exponents of the radical expression. Dividing like radicals involves dividing the coefficients and subtracting the exponents of the radical expression. Raising a like radical to a power involves raising the coefficient to that power and multiplying the exponent by that power. Simplifying a radical expression often involves identifying and combining like radicals. Review Questions Explain how to add or subtract like radicals and provide an example. To add or subtract like radicals, you combine the coefficients while keeping the same radical expression. For example, $\sqrt{5} + 3\sqrt{5} = 4\sqrt{5}$, and $7\sqrt{3} - 2\sqrt{3} = 5\sqrt{3}$. The key is that the radicals must have the same index and be raised to the same power in order to be considered like radicals. Describe the process of multiplying like radicals and explain how it differs from adding or subtracting them. When multiplying like radicals, you multiply the coefficients and add the exponents of the radical expressions. For instance, $\sqrt{2} \cdot 3\sqrt{2} = 3\sqrt{2^2} = 3\sqrt{4} = 6\sqrt{2}$. This is different from adding or subtracting like radicals, where you only combine the coefficients. Multiplying like radicals involves a more complex operation that takes into account the exponents of the radical expressions. Analyze how simplifying radical expressions often requires identifying and combining like radicals. Provide an example to illustrate this concept. Simplifying radical expressions frequently involves recognizing and combining like radicals. For example, to simplify the expression $\sqrt{8} + 2\sqrt{2} - \sqrt{18} + 3\sqrt{2}$, we first identify the like radicals: $\sqrt{8}$ and $\sqrt{2}$. We can then combine them by adding the coefficients: $\sqrt{8} + 2\sqrt{2} + 3\sqrt{2} = \sqrt{8} + 5\sqrt{2}$. Recognizing and combining like radicals is a crucial step in simplifying more complex radical expressions. Related terms Radical Expression: A mathematical expression containing one or more radicals, such as $\sqrt{x}$ or $\sqrt{y}$. Radical Index: The number that indicates the root being taken, such as 2 for a square root or 3 for a cube root. Rational Exponent: An exponent that can be expressed as a fraction, such as $x^{2/3}$, which is equivalent to $\sqrt{x^2}$. "Like Radicals" also found in: Subjects (3) Elementary Algebra Inorganic Chemistry II Intermediate Algebra
16381	https://www.nagwa.com/en/explainers/236182094534/	Lesson Explainer: Similarity of Polygons \| Nagwa Lesson Explainer: Similarity of Polygons \| Nagwa Sign Up Sign In English English العربية English English العربية My Wallet Sign Up Sign In My Classes My Messages My Reports My Wallet My Classes My Messages My Reports Lesson Explainer: Similarity of Polygons Mathematics • Second Year of Preparatory School Join Nagwa Classes Attend live Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher! Check Available Classes Next Session: Sunday 14 September 2025 • 4:00pm Remaining Seats: 12 Try This In this explainer, we will learn how to identify and prove the similarity of polygons, write the order of the corresponding vertices, and use the similarity to solve problems. We can begin by recalling that polygons are two-dimensional shapes with straight sides. For example, squares, rectangles, triangles, hexagons, and octagons are all polygons. Polygons that have exactly the same shape and size are congruent, whereas similar polygons have the same shape and may have a different size. We can define similar polygons more formally below. Definition: Similar Polygons Two polygons are similar if their corresponding angles are congruent and their corresponding sides are in proportion. Let’s consider the two quadrilaterals and below. If we are given that ( is similar to ), we have We can also observe the corresponding sides. These are and , and , and , and and . Since corresponding sides are in the same proportion, we can write The proportional relationship can also be given with all the numerators and denominators swapped in the entire statement; that is, We should use the similarity statement to identify corresponding vertices, rather than solely using any given diagrams. For example, if we have two triangles such that , then , , and . We could also note that side would be corresponding to . In the first example, we will use corresponding sides and angles to identify whether two polygons are similar. Example 1: Verifying Whether Two Given Polygons Are Similar Are the two polygons similar? Answer We recall that two polygons are similar if their corresponding angles are congruent and their corresponding sides are in proportion. Inspecting the angles in the figure, we have two pairs of congruent angles: We can calculate in quadrilateral using the property that the internal angles in a quadrilateral sum to . Hence, we have We can use the same property of the angle measures in quadrilaterals to calculate . We have Therefore, we have 4 pairs of corresponding angles that are congruent. We now check whether we have a proportional relationship between the lengths of corresponding sides; that is, we check whether We could also write the proportionality as Substituting the given measurements, we have Although we have two pairs of side lengths in the same proportion, we do not have all four pairs of sides in the same proportion. Hence, we can give the answer: no, the two polygons are not similar. Similar polygons can also be considered as a dilation of each other. If the scale factor is 1, then the polygons are congruent. We can use the scale factor of this dilation to work out the measure of unknown sides. This scale factor may also be referred to as the ratio of enlargement. This may be particularly useful when the ratio of sides is clearer, or more intuitive. Look at the figure below, where . To find the length of , we could observe that the lengths of must be double the lengths of . This is because we can write that The scale factor from to is 2. The scale factor in the opposite direction is found by dividing by 2. But, we must express scale factors in terms of a multiplier, and dividing by 2 is equivalent to multiplying by . Therefore, to find the length of , we multiply the corresponding length in by . This gives us This is an equivalent approach to finding the length of by writing a single equation involving the 2 pairs of sides: We will now check whether another pair of polygons are similar. Example 2: Finding the Scale Factor between Two Similar Polygons Are these two polygons similar? If yes, find the scale factor from to . Answer Two polygons are similar if their corresponding angles are congruent and their corresponding sides are in proportion. We can see from the diagram that there are pairs of angles that are marked as having equal measures: We recall that the angles in a quadrilateral sum to ; hence, we can write the measures of and as Using the congruent angles above, we can also write as Thus, Therefore, we have found that there are four congruent angles. This alone is insufficient to prove similarity, so, we must also determine whether the corresponding sides of the polygon are in proportion. The sides that are corresponding are and , and , and , and and . The sides are in proportion if Considering each ratio in turn, we have As each of these proportions can be simplified to , the proportions are equal. Since the angles are also congruent, we have proved that the polygons are similar. Although it is not required here to write a similarity statement between the polygons, we do need to take into account the letter ordering if doing so. If we write polygon with the letters in that order, then we must place the congruent vertices in the similar polygon in the same order. Hence, . Alternatively, would be another valid statement. In order to find the scale factor from to , we can take any pair of corresponding sides and divide the length of the side in by the corresponding side length in . We have Alternatively, we have already calculated that the proportion of the corresponding sides is . We must note, however, that this proportion was calculated by dividing a side in by a corresponding side in . This would give us the scale factor from to , and we need the scale factor in the reverse direction, from to . To do this, we instead multiply by the reciprocal of , which is . This is equivalent to the decimal 0.8. We can give the answer: yes, the polygons are similar, and the scale factor from to is 0.8. It is worth noting some facts about similarity in regular polygons. Recall that a regular polygon is a polygon where all angles are congruent and all sides are congruent. Regular polygons include equilateral triangles, squares, regular pentagons, regular hexagons, and so on. Considering two squares of different side lengths, since within each square all angles are congruent, then each of these angles are also congruent to their corresponding angles in the other square. Furthermore, the proportion of corresponding side lengths will be the same for each side length. Therefore, we can say that any regular -sided polygon is similar to another regular -sided polygon, where the values of are the same. That is, all equilateral triangles are similar, all squares are similar, and so on. Of course, since in similar polygons corresponding angles must be congruent and corresponding sides must be in the same proportion, we cannot say that squares are similar to rectangles (as the corresponding side lengths will not be in the same proportion), nor can we say that rhombuses are similar to squares (as corresponding angles are not congruent). In the next example, we will see how we can use the information that polygons are similar to determine an unknown side length. Example 3: Finding the Length of a Side in a Quadrilateral given the Corresponding Sides in a Similar Quadrilateral and Their Lengths Given that , determine the length of . Answer We are given that the two polygons and are similar. This means that their corresponding angles are congruent, and their corresponding sides are in proportion. We can use the proportionality of the sides to help us find the unknown side length, . In the figure, we are given the lengths of the two sides, and . We can identify from the similarity statement that these two sides are corresponding, as . We can then use the side to help us work out the corresponding length of . We can write that Substituting the given lengths, we have We have determined that the length of is 105 in. It is a common error to confuse similarity and congruence. Congruent polygons have equal corresponding pairs of angles and equal corresponding sides. Two errors commonly seen when dealing with similarity are either mistakenly writing that corresponding sides are equal, rather than in proportion, or writing that corresponding sides are in proportion and corresponding angles are in proportion. If we have two similar polygons, for example, triangles, the angle measures in both triangles must still sum to , regardless of the difference in their sizes. In the next example, we will see how we use a similarity relationship to determine an unknown side length and an unknown angle measure. We will consider how to find the side length both by using the proportionality relationship between two corresponding pairs of sides and by calculating the scale factor. Example 4: Finding the Side Length and Angle Measure in Similar Quadrilaterals Given that , find and the length of . Answer We are given the information that the two polygons are similar. Their corresponding angles are congruent and their corresponding sides are in proportion. To find , we note that we do not have enough information about the angles in polygon to work out . However, because , we know that the given angle measure of is corresponding to . It must also be . Using the property that the sum of the internal angle measures in a quadrilateral is , we have Next, the length of can be determined by using the corresponding side, , in . The proportion of these sides will be the same proportion as that between all other pairs of corresponding sides in the polygons. We are given the lengths of another pair of corresponding sides, and . Therefore, Substituting the lengths, we have Alternatively, we could have calculated the length of by finding the scale factor from to . In order to find the scale factor, we use a known pair of side lengths. Hence, Using this scale factor, the length of must be multiplied by to give the length of . This is given by Thus, using either method to find the side length, we have determined that and the length of is 123.1 cm. We will now see how we can solve a problem involving similar polygons and a perimeter. Example 5: Finding the Side Lengths of a Polygon given Its Perimeter and the Side Lengths of a Similar Polygon A polygon has sides of lengths 2 cm, 4 cm, 3 cm, 8 cm, and 4 cm. A second similar polygon has a perimeter of 31.5 cm. What are the lengths of its sides? Answer We recall that similar polygons have corresponding angles that are congruent and corresponding sides in proportion. We are given that the side lengths of one polygon, a pentagon, are 2 cm, 4 cm, 3 cm, 8 cm, and 4 cm. We are required to determine the side lengths of a similar polygon using only the information about its perimeter, which is the distance around the edge of the polygon. As the sides of similar polygons are in proportion, then the perimeter, which is also a measure of length, will be in the same proportion. We calculate the perimeter of the first polygon as follows: The scale factor from the first polygon to the second polygon can be found by In order to find the sides in the second polygon, we multiply each corresponding side length in the first polygon by the scale factor of . Hence, we have The sides in the second polygon can be given as 3 cm, 6 cm, 4.5 cm, 12 cm, and 6 cm. We can now summarize the key points below. Key Points Two polygons are similar if their corresponding angles are congruent and their corresponding sides are in proportion. We can use the similarity statement to identify corresponding sides and angles, and we must ensure that the letter ordering is correct when writing a similarity relationship between polygons. We can calculate an unknown side by writing the proportional relationship between the side and its corresponding side, along with the proportion between another pair of corresponding sides, or by first calculating the dilation scale factor. The scale factor between the perimeters of two similar polygons is the same as that between corresponding side lengths. All regular polygons are similar to other regular polygons with the same number of sides. Lesson Menu Lesson Lesson Plan Lesson Video Lesson Explainer Lesson Playlist Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! 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16382	https://biosignaling.biomedcentral.com/articles/10.1186/s12964-025-02216-w	Cell Communication and Signaling Glycosylation as an intricate post-translational modification process takes part in glycoproteins related immunity Download PDF Download PDF Review Open access Published: Glycosylation as an intricate post-translational modification process takes part in glycoproteins related immunity Meng Tian1, Xiaoyu Li1, Liuchunyang Yu1, JinXiu Qian1, XiuYun Bai1, Jue Yang1, RongJun Deng1, Cheng Lu2, Hongyan Zhao3 & … Yuanyan Liu1 Cell Communication and Signaling volume 23, Article number: 214 (2025) Cite this article 2200 Accesses 1 Citations Metrics details Abstract Protein glycosylation, the most ubiquitous and diverse type of post-translational modification in eukaryotic cells, proteins are input into endoplasmic reticulum and Golgi apparatus for sorting and modification with intricate quality control, are then output for diverse functional glycoproteins that are utilized by cells to precisely regulate various biological processes. In order to maintain the precise spatial structure of glycoprotein, misfolded and unfolded glycoproteins are recognized, segregated and degraded to ensure the fidelity of protein folding and maturation. This review enumerates the role of five immune-related glycoproteins and reveals the relevance of glycosylation to their antigen presentation, immune effector function, immune recognition, receptor binding and activation, and cell adhesion and migration. With the knowledgement of glycoproteins in immune responses and etiologies, we propose several relevant therapeutic strategies on targeting glycosylation process for immunotherapy. Introduction Glycosylation as an enzymatic process is likely as ancient as life itself that may produce glycosidic linkages between saccharides and other saccharides, proteins or lipids, which work out abundant repertoire of glycans to regulate critical cellular processes . In the present review, protein glycosylation as intricate protein post-translational modification (PPTM) pathway was further discussed. Endoplasmic reticulum (ER) and Golgi apparatus as predominant location that intricate glycosylation modification processes are occurred, in where native glycosyltransferases and glycosidases attach different glycans to proteins using active nucleotide sugars as substrates, then transport from ER to Golgi apparatus and cell membrane to yield various functional glycoproteins [2, 3, 4]. As a conserved PPTM, protein glycosylation makes indispensable contributions to abundant repertoire of glycoproteins while ensuring their stability through rigorous quality control for folding and degradation. Glycoproteins, formed by the combination of glycans and proteins, represent crucial biomacromolecules with distinct structural properties and a diverse array of biological functions that are involved in various physiological processes as well as disease mechanisms . Nearly all essential molecules that participate in innate and adaptive immune responses may rely on the involvement of glycoproteins . The process of protein glycosylation allows glycoproteins to fold correctly and mature to maintain their precise spatial structure and quality control, forming a wide variety of glycoproteins with different immunological roles, which provide immune-related substrates for subsequent immune responses such as antigen presentation, immune effector functions, immune recognition, receptor binding and activation, as well as cell adhesion and migration. Given the variety and intricacy of glycoprotein structures and corresponding functions, we cannot summarize all immune-related glycoproteins here, but instead focus on a few illustrative examples, which include major histocompatibility complex (MHC) proteins, immunoglobulin (Ig), complement, immune cell receptors, and cell adhesion molecules (CAMs). This review discusses the occurrence of protein glycosylation process and its biological significance, focusing on five types of glycoproteins that are vital to the immune system. Moreover, since the significant roles of glycoproteins in medicine which are associated with various human diseases, as cancer , several relevant therapeutic strategies to target glycosylation are also introduced. Background of protein glycosylation While the amino acid sequences of most proteins are determined by the genetic codes encrypted in DNA sequences, many undergo various PPTM, including phosphorylation, sulphation, lipidation, acylation, alkylation and glycosylation . Among them, glycosylation stands out as one of the most common and intricate forms that significantly influences protein spatial structure, and is crucial for the biosynthesis and biological activity of glycoproteins involved in numerous biological recognition events [5, 9]. Of note, the synthesis of glycans occurs without a template, and a massive of factors involved in this cellular glycosylation machinery, collectively of which make the complex repertoire of glycogens found on glycoproteins. As the most intricate PPTM, glycosylation brings diversity and heterogeneity of glycoprotein spatial structure to suit the desired function within the cell . Generally speaking, protein glycosylation reflects both the repertoire of glycosyltransferases and the capacity for glycosylation within producing cells. However, some individual proteins may not undergo efficient glycosylation, leading to cause diseases, and specific characteristics of glycosylation are directed towards particular proteins rather than being universally found. Furthermore, various cellular and environmental factors that affect glycosylation efficiency influence both the secretory pathway and glycosylation machinery. Requirements for glycosylation Protein glycosylation is the most ubiquitous and intricate PPTM that occurs primarily in ER and Golgi apparatus [10, 11]. The inherent structural variations of glycans makes this process an effective way to generate protein diversity and regulate its properties . This process not only requires a highly efficient and well-coordinated system to ensure accuracy and quality control during the processing of sugar chains, but also necessitates precise allocation of various mechanisms including glycosyltransferases (that add sugars), glycosidases (that remove sugars), nucleotide sugars transporters (that provide sugar substrates) and nascent polypeptide (that connect with glycans) (Fig. 1A) . These activated nucleotide sugars mainly include uridine diphosphate N-acetylglucosamine (UDP-GlcNAc), uridine diphosphate N-acetylgalactosamine (UDP-GalNAc), uridine diphosphate galactose (UDP-gal), uridine diphosphate glucose (UDP-glc), uridine diphosphate xylose (UDP-xyl), uridine diphosphate glucuronic acid (UDP-GlcA), guanosine diphosphate mannose (GDP-man), guanosine diphosphate fucose (GDP-fuc), and cytidine monophosphate N-acetyl neuraminic acid (CMP-Neu5Ac) (Fig. 1B and C) [13, 14, 15]. They are primarily produced in cytoplasm and nucleus, then transported into ER and Golgi apparatus by nucleotide sugar transporters (NSTs) located within their membranes since nucleotide sugars are unable to cross the organellular membranes [16, 17]. Primary sites where glycosylation occurs The glycosylation of proteins mainly accomplished by the collaboration between ER and Golgi apparatus [18, 19]. There are many covalent modifications of transporters between these two organelles, including glycosylation, selective proteolysis, sulfation, phosphorylation, and fatty acid addition, which can be biochemical markers of this process. Among them, the most in-depth research is glycosylation . About one-third of the cellular proteins are directed to ER lumen for proper folding and initial modification before being transported to Golgi apparatus for further PPTM like glycosylation, since they contains a series of native components assisting glycoprotein synthesis and maturation, such as chaperones, folding enzymes, glucosidases, and carbohydrate transferases [21, 22, 23, 24]. Between ER and Golgi apparatus, proteins also must be sorted into coated vesicles in order to flow . The COPII transports properly folded glycoproteins from ER to Golgi apparatus via a complex involving ERGIC53 and oxidoreductase ERp44, and then being sorted for delivery to their final destinations . However, If native components that assist in glycoprotein production in the ER is accidentally reached to Golgi apparatus, the COPI recognize and transport it back to the ER (Fig. 1A) . On the whole, except that mucin-type O-glycosylation starts from the Golgi apparatus, most of the rest takes place primarily within ER and ends in Golgi apparatus [28, 29]. Main types of glycosylation Glycosylation refers to the template-free and continuous process of attaching glycans to proteins, and the variety of glycans attached immensely increases the complexity of the resulting protein structure. Within ER and Golgi apparatus, protein glycosylation primarily includes N-glycosylation, O-glycosylation, C-mannosylation, and the formation of glycosylphosphatidylinositol (GPI) -anchored proteins, of which N- and O-glycosylation are two most prevalent types [29, 30]. Newly synthesized proteins undergo proper folding, as well as initial N- and O-glycosylation within the ER before being transported to the Golgi apparatus for further processing, maturation of glycans, trafficking, and sorting . Many proteins undergo N-glycosylation initiating in ER and maturing in Golgi apparatus, in which GlcNAc forms a covalent bond with nitrogen atom of Asn side chain through an N-glucosidic linkage within consensus sequence Asn-X-Ser/Thr, where X represents any amino acid except proline [19, 32, 33]. Based on their side chain branches, N-glycans can be categorized into three major subtypes like high mannose N-glycans (characterized by elongated by mannose residues), complex N-glycans (which undergo further chain elongation with the addition of GlcNAc to Golgi apparatus), and hybrid N-glycans (incorporating galactose or fucose residues alongside mannose in Golgi apparatus) . The entire process of protein N-glycosylation occurs in the ER and Golgi apparatus through four stages. First, the precursor molecule containing 14 sugar molecules (Glc3Man9GlcNAc2) is synthesized within the ER membrane, then the glycan is attached to its substrate. Subsequently, the glycoprotein undergoes initial processing within the ER lumen, followed by further maturation of the glycan in the Golgi apparatus. . O-glycosylation refers to adding glycans to serine or threonine in Golgi apparatus and involves initially linking several monosaccharides including Gal, Man, Fuc, and GalNAc, which occurs during the post-translational stage of the Golgi apparatus [36, 37]. Unlike N-glycosylation, O-glycosylation lacks specific consensus sequence. Its synthesis involves the progressive addition of individual monosaccharides along the exocytic pathway . In higher eukaryotic cells, two primary types of O-glycans synthesized in Golgi apparatus: shorter mucin-type glycans and longer glycosaminoglycan chains found on proteoglycans . Reversible O-glycosylation is prevalent modification observed in various proteins including transcription factors, cytoskeletal proteins, oncogenes and kinases . Compared with N/O-glycosylation, the extent of PPTM in C-mannosylation is significantly lower, which is referred to as C-glycosylation due to the attachment of α-mannopyranose monosaccharide to a tryptophan residue within the polypeptide sequence Trp-X-X-Trp, where X represents any amino acid, through carbon-carbon bonds . And formation of GPI-anchored protein is the use of glycans as linker to connect proteins to phosphatidylinositol anchors in the membrane, also known as glypiation . The biological significance of protein glycosylation Protein glycosylation plays a crucial role in connecting intricate metabolic pathways to different proteoforms, refining protein structures, and performing biological functions . The modification involves the addition of complex oligosaccharides, known as glycans, which have diverse effects on overall structure and function . Moreover, glycans play a direct role in numerous biological processes, including intracellular trafficking, interactions between cells and extracellular matrix, signalling pathways, development, host-pathogen interactions, and immune responses . One significant role of glycosylation is promoting protein folding and subsequent trafficking, in addition to modulate interactions with receptors and ligands, and influencing innate and adaptive immune responses . Generally, accurate protein glycosylation affects almost every aspect relevant to cellular processes that control protein stability , assist in protein folding , promote protein secretion and trafficking , and protect protein from degradation , facilitate recognition between cells and extracellular matrix . All of these participates in immune responses . Studies have confirmed that abnormal glycosylation is either cause or result of various diseases such as autoimmune diseases, diabetes, cancer, cardiovascular, and cystic fibrosis [19, 31]. For the process of protein glycosylation, inputs the form of given post-translational protein enter ER and Golgi apparatus for sorting and modification with intricate glycosylation, are then output diverse functional glycoproteins, which is utilized by cells to precisely regulate critical cellular immune responses. Role of glycosylation in the quality control of glycoprotein folding and degradation Glycoproteins are essential in regulating various immune responses. If not correctly folded and matured within ER and Golgi apparatus, misfolded and unfolded dysfunctional glycoproteins are expressed on cell surface or extracellular spaces, resulting in disruptions in signaling pathways that respond to the function of immune cells . In mammalian systems, the toxic accumulation of abnormal proteins necessitates sophisticated quality control mechanisms to identify, isolate and degrade these misfolded proteins, thereby ensuring proper protein folding and maturation [46, 47]. Among them, a key part of ER protein quality control is glycosylation, which facilitates the delivery of properly folded glycoproteins. Glycan structures present on newly synthesized glycoproteins play a crucial role in their secretion by affecting protein folding, assisting in quality control monitoring within ER and enable transport along the secretion pathway with selective targeting [48, 49]. The glycoprotein quality control system efficiently also employs many chaperone enzymes and lectins, including UDP-glucose: glycoprotein glucosyltransferase (UGGT), calnexin (CNX), calreticulin (CRT), protein disulfide bond isomerase (ERp57 or PDIs), and glucosidases to convert nascent glycopolypeptides into properly folded native forms . These intricate glycosylation processes of proteins may exert a repertoire of biological function in glycoprotein maturation and accurate folding to maintain their precise spatial structure on surveillance of quality control for subsequent immunological recognition, antigen presentation, immune effector functions, immune checkpoint, cell adhesion and migration, and receptor binding and activation. Glycosylation influences glycoprotein spatial structure through protein folding processes It is well established that glycans can profoundly influence the conformation of short glycopeptides . The majority of protein molecules need to adopt specific three-dimensional structures in order to acquire function properly, therefore it is necessary to ensure protein folding properly . When proteins are in the translocon complex, glycans are added to unfolded proteins to aid protein folding [12, 49, 51]. In addition, newly synthesized proteins encounter several chaperones or folding enzymes that work together to help correct protein fold before being released from the ER . Chaperones identify immature, abnormal, or easily aggregated proteins through exposed hydrophobic fragments, assisting in the maturation of non-glycosylated proteins or unmodified domains on glycosylated proteins . In the process of N-glycosylation (Fig. 2), after the eukaryotes transfer Glc3Man9GlcNAc2 glycans to the new polypeptide chains, the two Glu residues are pruned to form the GlcMan9GlcNAc2 structure, which serves as a ligand for CNX and CRT [14, 48]. When glucosidases I and II trim these two glucoses from N-linked core glycans to form GlcMan9GlcNAc2 structure, nascent glycoproteins associate with CNX and/or CRT. In this way, this interaction introduces another folding factor, ERp57 that thiol oxidoreductase that associates with CNX and CRT. If cysteine is present in the glycoprotein, it forms disulfide bonds with ERp57, thereby stabilizing the spatial structure of the peptide chain [5, 49]. Once glucosidases II remove third glucose residue to yield Man9GlcNAc2, these complexes separate. If glycoprotein remains improperly folded, UGGT will re-glycosylate oligosaccharides by recognizing exposed hydrophobic regions on its surface through its unique multiple domains, which allows misfolded glycoproteins to be identified and rebound to lectins. This cycle continues until either proper folding occurs or quality control mechanisms break it [46, 54]. Once folded correctly, glycoproteins are no longer recognized by glucosyltransferase, meanwhile, they can exit ER successfully, and most of glycoproteins will enter the Golgi apparatus by various membrane-bound lectins, namely VIPL, VIP36 or ERGIC53 [8, 51, 55]. In this organelle, enzymes like glycosyltransferase and glycosidase facilitate O-glycosylation while converting high mannose-type glycans into mature acidic sialyloligosaccharides along with additional O-glycosylation . Generally speaking, interactions among calnexin, calreticulin, ERp57, and UGGT will slow down protein folding rates but increase overall efficiency . Glycoprotein degradation The function of chaperones, folding catalysts and protein-modifying enzymes in ER lumen are responsible for identifying and binding misfolded proteins, facilitate refolding until they form the correct conformation for exiting ER . After protein reaches its native structure, modified glycan will act as a signal that directs its transport to various cellular compartments such as endosomes, lysosomes, or defaults to secretion at plasma membrane . However, due to environmental factors and inherent complexities, protein folding is one of error prone processes during gene expression . If protein fails to fold correctly, it will be degraded by the ER-associated degradation (ERAD) pathway (Fig. 2) . ERAD is a rather complex and ordered process in which misfolded proteins are recognized by resident factors within ER in protein secretion pathway and directed by transport mechanisms to reverse shuttle into the cytoplasm, where they undergo ubiquitination before being degraded by proteasomes via the ubiquitin-proteasome system [56, 59]. Glycans also have a significant role in ERAD of proteins . If protein folded incorrectly, ER degradation-enhancing α-mannosidase-like proteins (EDEM) begins removing mannose residues from their core glycans, resulting in substrate demannosidation, which will be dislocated into the cytoplasm and degraded by ERAD [60, 61, 62]. Once in the cytoplasm, these substrates are immediately ubiquitinated and then degraded by 26 S proteasomes . In addition, EDEM is upregulated upon unfolded protein responses (UPR) activation, which allows directly targeting of substrate glycoproteins to ERAD without needing prior mannose trimming, thereby expediting reduction of excess protein load within the ER . Soluble ER-resident proteins specifically identify trimmed oligosaccharide generated by ERManI/EDEM by mannose-6-phosphate receptor homologous domain . In certain organisms, cytoplasmic peptide-N-glycanase can remove N-glycans from ubiquitinated misfolded proteins destined for proteasomal degradation . When glycosylation processes are inhibited, one commonly observed outcome is an accumulation of aggregated misfolded proteins that do not reach functional states [49, 51]. Misfolding or mistargeting proteins in early secretory pathways pose significant risk to cells . Disruption of ER-associated functions, such as dysregulation of glycoprotein quality control, leads to occurrence of UPR by activating intricate cytoplasmic and nuclear signaling pathways . The UPR initiates an adaptive response aimed at restoring homeostasis within the ER by reducing protein expression levels while increasing molecular chaperone production to manage excess misfolded proteins. It also promotes the degradation of ER-associated proteins to eliminate misfolded entities, but if stress conditions persist or worsen, the UPR may ultimately lead to apoptosis as a means to resolve ongoing issues . Glycosylation modification maintain glycoprotein stability The presence of bulky oligosaccharide groups hinders interaction formation in unfolded state, forcing polypeptide chain to adopt more extended conformations that enhance protein stability [12, 46]. Research on soybean agglutinin, a glycoprotein, indicates that its non-glycosylated monomeric is less stable than glycosylated counterpart at both normal and higher temperatures. This increased stability in glycosylated proteins is due to non-covalent interactions between proteins and carbohydrate components . CNX-CRT cycle facilitates proper folding, prevents the aggregation of intermediates and premature oligomerization. Meanwhile, the ERAD pathway removes aberrant proteins from ER through degradation that relies on ubiquitin-proteasome system. Together, they maintain glycoprotein stability and provide quality control via blocking incomplete folded glycoprotein from entering Golgi apparatus . In general, glycosylation processes within this quality control system monitor maturation fidelity, through regulating the accurate temporal and spatial folding, maintaining the stability of glycoprotein to respond to internal and external cues and retaining misfolded proteins within the ER for efficient targeting towards degradation . The role of glycoproteins in immune responses Cell-surface glycoproteins are essential for immune responses. These immune related glycoproteins are very dynamic and diversity, which allow the cell to adapt to over-changing environment . The diversity of protein glycosylation generates the complex spatial structures of glycoproteins that have profound effects on immune responses. As the most fundamental modification conjugate of macromolecular components, glycans located on the cell membrane can delicately shield or exposure large areas of protein surfaces, which may interfere the lateral protein-protein interactions and influence the orientation of binding sites on attached proteins. Moreover, the specific spatial configuration of individual glycoprotein may anchor one typical molecular pathway to govern some extent immune processes. Based on the diversified glycogen pool as they conjugate to specific substrates and receptors, a diverse and abundant repertoire of glycoprotein are produced. This intricate glycosylation modification precisely manipulate the processes of immunological recognition, antigen presentation, immune effector functions, cell adhesion and migration, along with receptor binding and activation. Major histocompatibility complex (MHC) in antigen presentation MHC class I and II are glycoproteins that display endogenous and exogenous antigens on cell surface, facilitating the recognition and activation of circulating T lymphocytes . The oligosaccharides linked to glycoproteins at the junction of T cell and antigen-presenting cells (APCs) assist in orienting the binding surface, offering protection from proteases, and limiting nonspecific lateral protein-protein interactions . In adaptive immunity, glycans play a role in organizing immunological synapses, and are involved in producing and loading of MHC I antigenic peptides, as well as presenting of MHC II antigens [69, 70]. Mature MHC I consists of three subunits: a transmembrane heavy chain glycoprotein; the small soluble non-glycosylated β2-microglobulin (β2M) protein that forms the heterodimer essential for presenting antigenic peptides; and antigenic peptides that are necessary for transporting MHC I to the cell membrane [68, 71]. For MHC I, after the core Glc3Man9GlcNAc2 structure is attached to Asn86, it undergoes rapid processing by glucosidases I and II, leading to the formation of monoglucosylated Glc1Man9GlcNAc2. This modification enables interaction with CNX and CRT. Together with additional accessory molecules, these chaperones play a role in the ER to ensure proper folding of the heavy chain, association with β2M, and loading of antigenic peptides into the binding groove (Fig. 3) [72, 73]. Blocking MHC I N-glycosylation may cause the failure of β2M and antigenic peptides to bind, leading to an increase in intracellular misfolded proteins and a reduction in cell surface expression . Unlike MHC I, which utilizes different peptide pools derived from endogenous sources, MHC II processes exogenous proteins internalized through endocytosis. These proteins are subsequently degraded by resident proteases within endosomes or lysosomes into specific peptides ranging from 10 to 25 residues in length before being loaded onto MHC II complexes [68, 75]. MHC II has two highly conserved N-glycosylation acceptor sites at approximately Asn78 on the alpha chain and at Asn19 on the beta chain, but these are variable between different species, which may be related to separate functional roles of MHC II allotypes . N-glycans are thought to guide the MHC II molecules to the Golgi and then to endocytic compartments, protecting them from premature proteolysis in non-acidic vesicles to ensure that the MHC II groove is accessible for antigenic peptide binding [76, 77]. In addition, the elimination of N-glycans on MHC II reduces glycoantigen presentation in live APCs, virtually eliminates their binding to recombinant MHC II in vitro, and significantly limits glycoantigen-mediated T cell recognition and activation in vitro and in vivo, resulting in dysregulation of intestinal immune homeostasis [78, 79]. However, the basic binding properties of MHC to peptides and TCR to MHC-peptide complexes are largely unaffected by the presence or absence of complex N-glycan . In brief, MHC molecules are glycoproteins that display antigenic peptides on the cell surface to recognize and activate circulating T lymphocytes . The presence of glycans is imperative for the functionality of the immune system. For example, tumor cells exhibit epitopes derived from the Mucin 1 core domain with truncated glycosylation, in conjunction with MHC-I molecules, resulting in natural MHC-restricted recognition of ‘hypoglycation’ epitopes . Alterations in N-glycan branching may influence the capacity of MHCII to present glycoantigens and stimulate anti-inflammatory T cells . Also, the evasion of pathogens through MHC glycan modification has been shown to effectively prevent antigen presentation to CD8 + T cells. MHC glycan modification makes tumor cells less efficient in presenting antigenic peptides, thus escaping cytotoxic T-lymphocytes lysis and promoting metastasis . Immune effector functions of Immunoglobulins Immunoglobulins are an important type of glycoproteins whose glycans are associated particularly with the Fc domain influencing their immune effector functions. The Fc fragments links various sugar moieties, including fuc, gal, and sialic acid [82, 83]. Human IgG is categorized into four subclasses with 36 potential glycoforms, resulting in up to 144 functional states, making the regulation of antibody effects more complex and precise . Among them, the bi-antennary complex N-glycan is typical sugar linked to asparagine 297 in the Fc of IgG antibodies (Fig. 4). The glycans attached to IgG are crucial for preserving structural integrity, enabling interactions with Fc receptors, and triggering subsequent immune responses . Afucosylation of IgG enhances the affinity of Fc for type I FcγRIIIa through interaction between the carbohydrate moieties on FcγRIIIa and IgG1 Fc and increases antibody-dependent cellular cytotoxicity (ADCC) by NK cells [82, 86, 87, 88]. Terminal galactosylation increases C1q binding, enhancing the classical complement pathway [89, 90]. Hypo-galactosylation (G0 glycans) is associated with inflammatory conditions like rheumatoid arthritis [91, 92]. Fucosylation and sialylation are two major Fc modifications that affect the recruitment of inflammatory effector cell responses via FcγRIIb and DC-SIGN [82, 93]. Sialylation of IgG (α2,6-linked sialic acid) promotes binding to DC-SIGN on dendritic cells, leading to immunosuppressive effects . IVIG (intravenous immunoglobulin) relies on sialylated IgG for its anti-inflammatory function [95, 96]. Furthermore, altered IgG glycosylation patterns are linked to autoimmune diseases [97, 98, 99]. IgE glycans play critical roles in modulating immune responses, particularly in allergic reactions and parasite defensewith multiple N-glycosylation sites . It binds to FcεRI on mast cells and basophils, triggering degranulation and histamine release. Glycosylation of IgE Fc enhances its affinity for FcεRI, prolonging mast cell sensitization . Removing or altering glycans reduces FcεRI binding and weakens IgE-mediated allergic responses . IgE also interacts with CD23 (FcεRII) on B cells and dendritic cells. Data show that non-N-glycosylated CD23 has higher affinity for IgE than glycosylated CD23 . Glycans influence IgE homeostasis, antigen presentation, and B cell regulation. Certain glycan modifications affect FcεRI signaling and the extent of mast cell activation. Glycosylation may influence IgE’s ability to recruit eosinophils and macrophages for parasite clearance [104, 105]. In addition, IgA glycans also play essential roles in shaping immune responses, like modulating FcαR affinity, thus affecting inflammation or immune suppression [106, 107]. IgM glycans induce internalisation of IgM by T cells, which in turn cause severe inhibition of T cell responses . IgD glycans are related to the stage of immune response and cell maturity . The glycoproteins of complement system may participate in pathogens phagocytosis The complement system composed of a repertoire of glycoproteins acts major roles in innate immune system and serves as one of the initial defenses encountered by pathogens during infection . The glycoproteins involved in this system can trigger immune responses that result in cell lysis and mark pathogens by forming the membrane attack complex for phagocytosis. Almost all complement components are primarily produced in the liver with varying degrees of glycosylation . C1q, which initiates the classical pathway, consists of six trimers each containing one N-glycan located within its globular head domain. These N-glycans are oriented to maximize binding between globular heads and targets while minimizing nonspecific interaction (Fig. 5) [9, 110]. C3, a complement component necessary to initiate the terminal pathway, contains two N-glycan sites on both its α/β-chain, and both chains are modified with high-mannose glycans . CD59, a GPI-anchored glycoprotein found on cell surfaces that contributes to protecting host cells from complement-induced lysis by preventing membrane attack complex formation [111, 113]. Factor B, a component of the alternative pathway of complement, whose N-glycans and Asp254 are typically involved in shielding C3b binding sites, thereby regulating complement activation through alternative pathway . Changes in the glycosylation of factor H, a serum glycoprotein, can improve the affinity for the factor H receptor, thereby increasing the efficacy of factor H in regulating pathogen-related complement activation [115, 116]. In addition, C6, C7, C8 and C9 have been reported to be characterized by C-mannosylation. It is hypothesized that these mannoses can bind mannose receptors, with the result that properdin and MAC proteins can recruit macrophages or locate mannose-binding pathogens [117, 118, 119]. Although almost all proteins in this system undergo glycosylation processes, only a limited number have been thoroughly characterized functionally . Glycosylation regulates the process of immune cell receptors binding and activation A wide range of membrane receptors derived from glycoproteins are involved in modulating immune responses. Numerous immune cell receptors are either positively or negatively regulated by N-glycosylation processes for their binding and activation. After translation, the antigen receptors on T/B cell undergo modifications involving N- and O-glycan chains, thus T cell receptor (TCR) and B cell receptor (BCR) are highly glycosylated multisubunit complex . The selective deletion of conserved N-glycosylation sites in constant regions of both α and β-chains of TCR resulted in enhanced multimerization and reduced TCR-MHC dissociation, which ultimately improving recognition of tumour cell carrying target antigens (Fig. 6) [120, 121]. During the deficiency of Mgat5, which is responsible for the initiation of GlcNAc-β-(1,6)-branching on N-glycans, and therefore the reduction of N-acetyllactosamine, lowers the threshold of T cell activation in vitro by enhancing TCR clustering . Alterations in the distribution of deglycosylated BCR in the plasma membrane may affect its binding to other membrane proteins, potentially hindering signalling and involvement in downstream oncogenic pathways . For example, inhibition of BCR glycosylation reduces BCR clustering and internalization, while promoting its binding to CD22, thereby weakening the activation of PI3 kinase and NF-κB . CD28, a glycoprotein located on T cell surface, functions as an immune receptor that modulates immune responses by interacting with CD80/CD86 (B7-1/B7-2) present on APCs to transmit secondary signals of T cell activation. Removing of N-glycosylation on CD28 significantly increases its binding affinity to CD80 and amplifies downstream signaling, suggesting that N-glycosylation negatively regulates the function of CD28 (Fig. 6) . NKp30, a natural cytotoxicity receptor found on NK cell, is responsible for clearing cancerous cells is activated upon stimulation by tumour-expressed B7-H6. It has been shown that B7-H6 binds more effectively to glycosylated NKp30 compared to de-glycosylated mutants , moreover, N-glycosylation is required for the oligomerization of NKp30 which triggers receptor activation . In addition, mammals have evolved phyloglycomic recognition system to identify glycans from lower organisms as part of non-self recognition mechanisms leading to immune activation . Cell adhesion molecules are involved in immune cells adhesion and migration The glycosylation modification governed protein folding processes make glycoproteins with delicate spatial structures, which enlarge the chances of protein-glycan interactions, and furtherly conduct the immune response of cells to communicate with each other . Most cell adhesion molecules (CAMs) are glycoproteins, including families such as selectins, integrins, cadherins and immunoglobulin superfamily, which can mediate binds to other cells and extracellular matrix within microenvironment, contributing for the migration and adhesion of immune cells . E-cadherin, a key transmembrane glycoprotein involved in epithelial cells adhesion, is influenced by branched N-glycans, which can disrupt cell-cell adhesion and downstream signaling pathways, thereby promoting invasion and metastasis (Fig. 7) . The terminal sialylation of integrins on tumor cell regulates intracellular signalling and cell adhesion, whilst, core fucosylation is crucial in integrin-mediated cell proliferation and migration [130, 131, 132]. The combination of selectins with their glycan ligands creates a system for cell-cell adhesion primarily between leukocyte and endothelial cell, which can mediate the rolling and migration of leukocyte at the inflammatory site . Research has indicated that E-selectin within bone marrow vascular niche encourages breast cancer metastasis by triggering mesenchymal to epithelial transition through signaling activation (Fig. 7) . ICAM-1, a member of the immunoglobulin superfamily, has typical chain of polysialic acid glycan that plays a role in leukocyte adhesion. Understanding that different N-glycoforms of ICAM-1 can mediate recruitment of different monocyte subsets is also of therapeutic interest . Glycoproteins serve a variety of functions, including roles as enzymes, hormones, antibodies and lectins. The diversity in their functions arises from their structural features, especially the diversified glycogen pool . The intricate glycosylation processes indispensably participate in immune responses, pathogen antigen recognition and ligand-receptor interactions resulting in cellular activities [45, 85]. The microbial molecule is usually identified in glycoconjugate patterns at the beginning of innate immune response. In adaptive immunity, glycans play a role in organizing immunological synapses, and are involved in producing and loading of MHC I antigenic peptides and processing of MHC II antigens [69, 70]. On cell surface, protein glycosylation is believed to function as molecular spacer that positions signaling and adhesion molecules for effective intercellular communication, meanwhile, the extended glycan chains as spaced apart of glycosylated immune molecules to restrict lateral association of glycoproteins . Glycoproteins on immune cell surfaces can detect pathogens and function either as receptors or transporters to facilitate communication between cells while recruiting various immune cells to infection sites [67, 135]. Therapeutic strategies targeting glycosylation Protein glycosylation has become a promising focus for immune-related diseases therapy as cancer. In the field of therapeutics and biomarkers, glycosylation has proven to be an crucial element that enhances the available tools and strategies for precision medicine . Most best-selling biotherapeutics are glycoproteins and numerous clinically and therapeutically important proteins are glycosylated, including monoclonal antibodies (mAbs), hormones, growth factors and vaccines [5, 136]. Consequently, approaches that utilize abnormal glycosylation patterns in cancer cells may offer new therapeutic avenues and synergize with existing targeted therapies to improve their specificity and effectiveness . Immunotherapy for intervention in protein glycosylation process of tumour and immune cell Glycosylation is significantly related to tumour development and abnormal glycosylation of proteins usually signals tumorigenesis . For one thing, the glycosylation of tumour cells contributes to evade host immune detection, a hallmark feature of cancer . Enveloped viruses like HIV may exploit host glycosylation processes to shield potential protein antigenic epitopes from immune recognition, thereby employing glycosylation inhibitors could serve as a viable antiviral strategy by disrupting proper folding of viral envelope proteins such as those found in hepatitis B virus and HIV [6, 75]. Additional, one significant alteration that increased expression of sialylated glycans linked to cancer progression has been observed in tumour glycosylation. Abnormal sialylation contributes to tumour growth, metastasis and evasion from immune responses. Thereby, blocking sialic acid biosynthesis can enhance interactions between tumour cells and T cells while improving cytotoxic T cell-mediated destruction of these tumour cells, which is of high therapeutic value in cancer [138, 139]. For another thing, abnormal glycosylation on cancer cell can be detected by immune cell, leading to immunosuppressive outcomes . For example, in T cell, tumour cells may downregulate surface MHCI levels to alter TCR-mediated activation signals ; simultaneously, they can also have the capacity to upregulate PD-L1 levels that modulate PD-1 inhibitory signaling pathways . CTLA-4 serves as immune checkpoint receptor on T cell that competes against CD28, a glycoprotein located on T lymphocytes, for binding sites on CD80/CD86 on APCs, resulting in diminished immune responses. The quantity and branching structure of N-glycans associated with CTLA-4 affect its expression at cell surface and TCR signaling can be mediated through the hexosamine and N-glycan branching pathways, which upregulate the level of CTLA-4 on cell surface impacting overall T-cell functionality . Thus, enhancing the stimulatory immune receptors or blocking the activation of inhibitory immune receptors by altering glycosylation could revive the antitumour function of immune cells. Immunotherapy targeting glycosylation modification of immune checkpoint molecules Given that the majority of immune checkpoints are glycoproteins located in membrane, their glycosylation is necessary for proper ligand-receptor interactions and antitumour immune function across various cancers . PD-1 acts inhibitory receptor located on activated T cell, which suppresses TCR signaling through interacting with the ligand PD-L1 present on tumour cells . N-glycans is essential for maintaining the levels and localization of PD-1 at cell surface. When PD-1 lacks glycosylation, it becomes more susceptible to ubiquitination leading to rapid degradation within the cytoplasm before reaching cell surface. Studies have shown that genetic deletion or pharmacologic inhibitioncan reduce PD-1 levels by disrupting its glycosylation, which subsequently impedes T-cell-mediated immunity . Furthermore, the glycosylation status of PD-1 influences its binds to PD-L1, a highly glycosylated member of the B7 family found on malignant and nucleated cells within tumour microenvironment . Most of the PD-L1 present in human tumour tissue and cancer cell line undergoes glycosylation modifications, which stabilize this protein and enhances its ability to suppress immune responses [145, 146]. Cancer cells exploit immune checkpoints to evade and inhibit antitumour immune response, thus immunotherapies that target these checkpoints, especially PD-1 and PD-L1, have marked a significant advancement in cancer treatment [142, 146]. Glycans can indirectly promote immune evasion by enhancing immune checkpoints, thereby therapeutic interventions using mAbs to block PD-1/PD-L1 interaction can rejuvenate T cell functionality and have shown effectiveness in fostering long-lasting antitumour immune responses . Intervention of glycosylation in therapeutic monoclonal antibodies Therapeutic monoclonal antibodies (mAbs) are a type of glycoproteins generated by living cell systems . Recombinant monoclonal antibodies based therapeutics, which can capable of recognizing and binding to specific epitopes on identical or different antigenic surfaces, have been developed for treating tumour, infections and inflammation . Optimisation and management of N-glycan profiles is a crucial aspect of bioprocess development for these antibodies . Detailed glycan structure and function analysis help to confirm the presence of particular glycans within antibody and their influence on drug safety, effectiveness, and clearance, thereby understanding their role as critical quality attributes. For the past few years, impressive advancements have been achieved in therapeutic monoclonal antibodies whose therapeutic effects are largely mediated by interactions between Fc domain of these antibodies and their receptors on immune cells, and which can initiate various immunomodulatory responses. The nature of these effector functions is heavily influenced by the glycosylation patterns present in the Fc fragments which affect both antibody binding affinity and complement system activation. For example, afucosylated or asialylated of Fc region has a high binding affinity with Fc receptors to promote ADCC , terminal galactosylated IgG1 has a high binding affinity with C1q and promotes CDC , and sialylated antibodies are strongly associated with enhanced ADCP activity by increasing the binding affinity with Fc receptors . Consequently, research into Fc glycosylation profiles alongside associated effector functions has sparked interest in engineering therapeutic antibodies. With advancements in glycosylation engineering targeting specific proteins or glycans themselves becoming increasingly popular strategies within therapeutic antibody development [41, 154]. Conclusions and perspectives Protein glycosylation that is mainly occurred in ER and Golgi apparatus is an intricate PPTM that involves dynamic and non-templating processes. Proteins serve as fundamental components of life, and diversified forms of glycosylation grossly expand the spatial structure and diverse function of newly produced glycoproteins . Intriguingly, a diverse repertoire of glycogen will be modified in protein, which may direct the glycoprotein folding to maintain specific exposure epitopes for recognition by the immune system. Meanwhile, these diverse and abundant glycoproteins precisely manipulate the processes of including antigen presentation, immune effector functions, immune recognition, receptor binding and activation, as well as cell adhesion and migration. Therefore, protein glycosylation it has become a promising target for immunological therapy of cancers. In fact, glycosylation on proteins, lipids and carbohydrates are as integral to the immune pathway and function. Research that connects immunology with glycobiology will persist in yielding fresh insights into immunity and uncover novel therapeutic strategies for various disease. Based on the diversity of glycan, their regulation of glycoprotein interactions may be involved in mechanisms that control multiple immune responses to a variety of extracellular stimuli in ways that are currently unknown . Moreover, due to the complex and diverse structure of glycans, it is difficult to achieve accurate analysis and functional study through conventional techniques, resulting in the field of glycoimmunology still faces many challenges, but also brings exciting opportunities. With the increasing availability of new technologies, applying glycobiology to explore fundamental aspects of immune function offers significant potential for acknowledging novel etiologies and providing precise therapeutic strategies on immune-related diseases. Data availability No datasets were generated or analysed during the current study. Abbreviations PPTM: : Protein post-translational modification ER: : Endoplasmic reticulum MHC: : Major histocompatibility complex Ig: : Immunoglobulin GlcNAc: : N-acetylglucosamine GalNAc: : N-acetylgalactosamine gal: : galactose glc: : glucose xyl: : xylose man: : mannose fuc: : fucose COP I: : Coat protein I COP II: : Coat protein II CNX: : Calnexin CRT: : Calreticulin ERAD: : ER-associated degradation UPR: : Unfolded protein responses TCR: : T cell receptor BCR: : B cell receptor mAbs: : monoclonal antibodies References Pinho SS, Alves I, Gaifem J, Rabinovich GA. 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Article CAS PubMed Google Scholar Download references Acknowledgements A special thanks for the long-term subsidy mechanism from the Ministry of Finance and the Ministry of Education of PRC (People’s Republic of China) for BUCM (Beijing University of Chinese Medicine), and thanks for National Administration of Traditional Chinese Medicine High level Key Discipline Construction Project of Traditional Chinese Medicine. Funding This work was supported by the Beijing Science and Technology New Star Program Cross-cooperation Project (No. 20240484711), the Beijing Natural Science Foundation (No. 7252275). Author information Authors and Affiliations School of Chinese Materia Medica, Beijing University of Chinese Medicine, Beijing, 100029, China Meng Tian, Xiaoyu Li, Liuchunyang Yu, JinXiu Qian, XiuYun Bai, Jue Yang, RongJun Deng & Yuanyan Liu 2. Institute of Basic Research in Clinical Medicine, China Academy of Chinese Medical Sciences, Beijing, 100700, China Cheng Lu 3. Beijing Key Laboratory of Research of Chinese Medicine on Prevention and Treatment for Major Diseases, Experimental Research Center, China Academy of Chinese Medical Sciences, Beijing, China Hongyan Zhao Authors Meng Tian View author publications Search author on:PubMed Google Scholar 2. Xiaoyu Li View author publications Search author on:PubMed Google Scholar 3. Liuchunyang Yu View author publications Search author on:PubMed Google Scholar 4. JinXiu Qian View author publications Search author on:PubMed Google Scholar 5. XiuYun Bai View author publications Search author on:PubMed Google Scholar 6. Jue Yang View author publications Search author on:PubMed Google Scholar 7. RongJun Deng View author publications Search author on:PubMed Google Scholar 8. Cheng Lu View author publications Search author on:PubMed Google Scholar 9. Hongyan Zhao View author publications Search author on:PubMed Google Scholar 10. Yuanyan Liu View author publications Search author on:PubMed Google Scholar Contributions MT and YYL wrote and conceived the manuscript. LCYY, JXQ, XYB and JY collected literature. RJD and XYL designed the figures. MT, YYL, HYZ and CL supervised and revised the manuscript. All authors read and approved the final manuscript. Corresponding authors Correspondence to Cheng Lu, Hongyan Zhao or Yuanyan Liu. Ethics declarations Ethics approval and consent to participate Not applicable. Consent for publication Not applicable. Competing interests The authors declare no competing interests. Clinical trial number Not applicable. Additional information Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. 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To view a copy of this licence, visit Reprints and permissions About this article Cite this article Tian, M., Li, X., Yu, L. et al. Glycosylation as an intricate post-translational modification process takes part in glycoproteins related immunity. Cell Commun Signal 23, 214 (2025). Download citation Received: Accepted: Published: DOI: Share this article Anyone you share the following link with will be able to read this content: Sorry, a shareable link is not currently available for this article. Provided by the Springer Nature SharedIt content-sharing initiative Keywords Glycosylation Glycans Quality control Glycoprotein Immune response Immunotherapy Cell Communication and Signaling ISSN: 1478-811X Contact us General enquiries: journalsubmissions@springernature.com
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Subjects MathAlgebraAlgebra 2CalculusGeometryPre-CalculusAP CalculusStatisticsELA Exam Prep SAT TutorsSAT MathSAT EnglishACT TutorsACT Math Grades High SchoolCollegeElementaryHomeschool8th Grade7th Grade6th Grade5th GradeAll Grades Company AboutCareersContact Learn BlogAI in Education Log in 888.405.2553FIND YOUR TUTOR 888.405.2553FIND YOUR TUTORLog In Fact checked A complex conjugate is a pair of two-component numbers that are called complex numbers. Each complex conjugate possesses a real... Mathematics 2 minutes Fact checked What Is a Complex Conjugate In Mathematics? A complex conjugate is a pair of two-component numbers that are called complex numbers. Each complex conjugate possesses a real... Mathematics 2 minutes Written by: Michael Brown Updated on: May 16, 2025 In this article Real Number ComponentImaginary Number ComponentWhere Did Complex Numbers Come From?Multiplication of Two Complex ConjugatesEquations in Quantum Mechanics‍”Born Probability”Which Professions Use Complex Numbers?Do High School Students Use Complex Conjugation?Are Complex Numbers on the SAT Test?How Learner Math Tutors Can HelpFrequently Asked Questions This is also a headingThis is a heading Get started with a hand-selected tutor for your child: Find your tutor In mathematics, every complex number (a two-component number involving a real number added to an imaginary number) has a complex conjugate. This complex conjugate will have the same real part, while the imaginary part will have the same magnitude but the opposite sign. A complex number can be represented by the expression a + bi, and its complex conjugate can be represented by a - bi, where a and b are real numbers and i is an imaginary number (specifically, the square root of negative 1). Despite having imaginary components, complex conjugates are used to describe physical realities. The use of complex conjugates works despite the presence of imaginary components, because when the two components are multiplied together, the result is a real number. In terms of their mathematical applications, complex conjugates are used in the rationalization of complex numbers. Math tutoring is an excellent way to help students understand complex conjugates as they’re also necessary for finding the amplitude of the polar form of a complex number. Find your tutor Real Number Component Every complex number possesses a real number component added to an imaginary component. In the expression a + bi, a is the real number component . Imaginary Number Component An imaginary number is defined as any number that when squared results in a real negative number. This may be restated in other terms for simplification. An imaginary number is any real number multiplied by the square root of negative one (-1). In this form, a complex conjugate is a pair of numbers that can be written y = a+bi and ӯ = a–bi, where “i” is the square root of -1. To distinguish the two y-values, one is generally written with a bar over the letter, ӯ, although occasionally an asterisk is used. Though their magnitude is equal, the sign of one of the imaginary components in the pair of complex conjugate numbers is opposite to the sign of the other. Where Did Complex Numbers Come From? The basic idea that a complex number exists in a number system containing real numbers and imaginary numbers originated with the mathematician Gerolamo Cardano in 1545, though many other mathematicians have also contributed to the concept of complex numbers, such as Casper Wessel in 1799. The earliest reference to the square root of a negative number (i.e., an imaginary number) is thought to appear in the work of the ancient Greek mathematician Hero of Alexandria. Regardless of the lengthy history of complex numbers, there are many ways in which they’re applied today, including in multiplication. Multiplication of Two Complex Conjugates To demonstrate that the multiplication of two complex conjugate numbers produces a real result, consider the example y = 7+2i and ӯ = 7–2i. Multiplying them together gives yӯ = 49+14i–14i–4i2 = 49+4 = 53. Such a real result from complex conjugate multiplication is important, particularly when considering systems at the atomic and subatomic levels. Frequently, the mathematical expressions for minuscule physical systems include an imaginary component. This is especially important in quantum mechanics, which is the non-classical physics of the very small. Equations in Quantum Mechanics In quantum mechanics, the characteristics of a physical system consisting of a particle are described by a wave equation. These equations reveal everything that can be learned about the particle in its system. Frequently, wave equations feature an imaginary component. Multiplying the equation by its complex conjugate results in a physically interpretable “probability density.” The characteristics of the particle may be determined by mathematically manipulating this probability density. ‍”Born Probability” As an example, the use of probability density is important in the discrete spectral emission of radiation from atoms. This application of probability density is called “Born probability,” after the German physicist Max Born. The closely-related statistical interpretation that the measurement of a quantum system will give certain specific results is called the Born rule. Max Born received the 1954 Nobel Prize in Physics for his work in this area. Unfortunately, attempts to derive the Born rule from other mathematical derivations have been met with mixed results. Which Professions Use Complex Numbers? There are a few obvious professionals who use complex numbers, including scientists, engineers, and, of course, mathematicians. Physicists, quantum physicists, and electrical engineers also use complex numbers on a regular basis. And there are other, less obvious professions that rely on complex numbers. Interestingly, sales analysts and economists regularly use complex numbers in their line of work. Complex numbers can also be used in signal processing to find certain notes with audio-editing software. They are involved in AC circuit analysis, as well. Complex numbers are frequently utilized in calculations of current, voltage, and resistance in AC circuits. This data can then be used to calculate the differences between two AC power supplies over time. And as mentioned above, complex numbers play an important role in quantum mechanics, the branch of physics that studies the motions and interactions between subatomic particles. These are just a few examples of how complex numbers (and complex conjugates) are used for practical, real-world purposes. It’s important for students to have a firm grasp of complex numbers and related topics because they are essential to a wide variety of professions. Math problems involving complex numbers may also appear on certain college entrance exams. Do High School Students Use Complex Conjugation? Whether a high school student learns complex conjugation depends on which mathematics classes are taught in a particular school. Classes often vary from state to state, so it’s hard to say exactly what will be taught in high school classrooms. Often, complex conjugates are taught in Algebra 1 or Algebra 2. However, complex conjugates are not just restricted to Algebra courses, as instructors could discuss real and imaginary numbers in varying contexts. Typically, Algebra I is taught to ninth grade students or eighth grade honor students. Some students may take Algebra 2 during their junior year of high school. Are Complex Numbers on the SAT Test? Most of the SAT Mathsection is focused on numbers, operations, algebra, functions, geometry, measurement, data analysis, statistics, and probability. That being said, there is a chance that the concept of complex numbers could appear on the SAT in one or two questions. In order to be well prepared, it’s a good idea to have a firm grasp of complex numbers and complex conjugates before taking the SAT. There are many components that work together to yield a correct answer in mathematics, so it’s important for students to know how to do certain procedures and how to use certain techniques across various mathematical subjects. It can be difficult to determine what an SAT question is asking if a student does not have an in-depth knowledge of several branches of mathematics.One of the best ways to prepare for college entrance exams and ensure success is to work with a tutor. A tutor can tailor their instruction to a student’s specific needs. Contrary to what some might think, tutors can greatly benefit all students, regardless of whether they’re struggling with certain concepts or whether they excel in class and need something more challenging. How Learner Math Tutors Can Help While math may come easily to some, there are many mathematical topics that can be confusing for high school and college students. Mathematics is a complex, multifaceted subject. Certain concepts may seem overwhelming, but an experienced math tutor can help. A tutor can assist students by providing individualized help and instruction – depending on which tutoring service the parent or student chooses. Some companies only offer generic lessons based on the state curriculum and standardized tests, while others provide a far more individualized experience. The best tutoring services will first ascertain the strengths and weaknesses of the student and then base tutoring sessions on their learning needs and current academic performance. Learner’s tutoring model is based on personalized lessons that are intended to help students grasp the concepts they find most difficult, yet need to master for their classes or exams. The process of connecting with a Learner tutor is very simple. First, parents answer questions about their child’s academic background and unique needs. Then, they schedule a call with an Academic Coach to identify their child’s learning goals and objectives. Lastly, Learner will custom match each child with the right tutor. The first session with Learner is always free, so it’s a great opportunity to try out tutoring and see if it’s a good fit. Learner understands that every student has their own learning style, unique way of thinking, and academic aspirations. Working with a highly experienced Learner tutor is a great way to give specific, direct feedback to students on their performance. Tutoring is also an excellent way to encourage students and build their confidence as they tackle new and unfamiliar topics, such as complex conjugates. Keep in mind that tutors aren’t just for students who need extra academic help. Tutoring is a great resource for students who are excelling academically, too. Tutors can encourage these students to think creatively, give them additional practice, and stretch and challenge them beyond the basics of the school curriculum. Looking for private one-on-one math tutoring? Speak with our academic advisor to get custom matched with youronline math tutor today! Frequently Asked Questions Complex conjugates can be a confusing topic. Here are some common questions and answers about complex conjugates and related topics. What Is the Square Root of a Complex Number? The square root of a complex number can be determined by using a formula. The square root of a natural number comes in pairs, and the square root of a complex number uses x and y to represent real numbers. What Is a Complex Plane in Math? A complex plane is the plane formed by complex numbers. The complex plane has an x-axis, which is called the real axis and is formed by real numbers. The complex plane also has a y-axis, which is called the imaginary axis because it is formed by imaginary numbers. What Are Real Coefficients? A coefficient is any modifying value associated with a variable by multiplication. A real number is any non-imaginary number. What Is a Notation in Math? In math, notations are symbols used to represent mathematical objects and ideas. Mathematical notations are essential to the physical sciences, engineering, and economics. Common notations include symbols representing approximation, greater than, less than, inequality, addition, subtraction, multiplication, and division. What Is the Product of Two Complex Numbers? When two imaginary numbers are multiplied together, the product is a real number. Can Complex Numbers Be Used in Fractions? If you have a fraction with a complex number in the denominator, you can multiply both the numerator and the denominator by the complex conjugate of the denominator. This simplifies the fraction rather than changing the value of the fraction. Can Complex Numbers Be Negative? A complex number can be the negative of another complex number, but it cannot itself be negative. Get started with a custom-matched tutor for your child. Find your tutor About the author: Michael Brown Mike developed his passion for education as a math instructor at Penn State University. He expanded his educational experience launching and running an Executive Education business - training over 100,000 students per year. As the CEO of Learner, Mike focuses on accelerating learning and unleashing the potential of students. Related Articles All Articles ### Are Your District's Math Scores Keeping Pace? Comparing PSSA Results Across Pennsylvania We analyzed PSSA math data from across Pennsylvania in order to understand how the current state of math education compares to previous years, and what can be expected moving forward. Evan Bender ### What Can You Do with a Math Degree? Learn about the career paths that a math degree can lead to, from finance to data analysis. This guide will teach you how to make the most of your education. Michael Brown ### What is Inequality in Math? Explore the concept of inequality in math and its real-world applications in finance, physics, and economics. 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16384	https://ocw.mit.edu/courses/18-022-calculus-of-several-variables-fall-2010/pages/lecture-notes/	Browse Course Material Course Info Instructor Prof. James McKernan Departments Mathematics As Taught In Fall 2010 Level Undergraduate Topics Mathematics Calculus Differential Equations Linear Algebra Learning Resource Types assignment_turned_in Problem Sets with Solutions grading Exams with Solutions notes Lecture Notes co_present Instructor Insights Download Course search GIVE NOW about ocw help & faqs contact us 18.022 \| Fall 2010 \| Undergraduate Calculus of Several Variables Lecture Notes \| LEC# \| TOPICS \| LECTURE NOTES \| --- \| 1 \| Vectors in R2 and R3 \| (PDF) \| \| 2 \| Dot product \| (PDF) \| \| 3 \| Cross product \| (PDF) \| \| 4 \| Planes and distances \| (PDF) \| \| 5 \| n-dimensional space \| (PDF) \| \| 6 \| Cylindrical and spherical coordinates \| (PDF) \| \| 7 \| Functions \| (PDF) \| \| 8 \| Limits \| (PDF) \| \| 9 \| The Derivative \| (PDF) \| \| 10 \| More about derivatives \| (PDF) \| \| 11 \| Higher derivatives \| (PDF) \| \| 12 \| Chain rule \| (PDF) \| \| 13 \| Implicit functions \| (PDF) \| \| 14 \| Parametrised curves \| (PDF) \| \| 15 \| Arclength \| (PDF) \| \| 16 \| Moving frames \| (PDF) \| \| 17 \| Vector fields \| (PDF) \| \| 18 \| Div grad curl and all that \| (PDF) \| \| 19 \| Taylor polynomials \| (PDF) \| \| 20 \| Maxima and minima: I \| (PDF) \| \| 21 \| Maxima and minima: II \| (PDF) \| \| 22 \| Double integrals \| (PDF) \| \| 23 \| Inclusion-exclusion \| (PDF) \| \| 24 \| Triple integrals \| (PDF) \| \| 25 \| Change of coordinates: I \| (PDF) \| \| 26 \| Change of coordinates: II \| (PDF) \| \| 27 \| Line integrals \| (PDF) \| \| 28 \| Manifolds with boundary \| (PDF) \| \| 29 \| Conservative vector fields revisited \| (PDF) \| \| 30 \| Surface integrals \| (PDF) \| \| 31 \| Flux \| (PDF) \| \| 32 \| Stokes theorem \| (PDF) \| \| 33 \| Gauss theorem \| (PDF) \| \| 34 \| Forms on Rn \| (PDF) \| Course Info Instructor Prof. James McKernan Departments Mathematics As Taught In Fall 2010 Level Undergraduate Topics Mathematics Calculus Differential Equations Linear Algebra Learning Resource Types assignment_turned_in Problem Sets with Solutions grading Exams with Solutions notes Lecture Notes co_present Instructor Insights Download Course Over 2,500 courses & materials Freely sharing knowledge with learners and educators around the world. Learn more © 2001–2025 Massachusetts Institute of Technology Creative Commons License Terms and Conditions Proud member of: © 2001–2025 Massachusetts Institute of Technology You are leaving MIT OpenCourseWare Please be advised that external sites may have terms and conditions, including license rights, that differ from ours. MIT OCW is not responsible for any content on third party sites, nor does a link suggest an endorsement of those sites and/or their content. Continue
16385	http://mathcentral.uregina.ca/qq/database/qq.09.07/s/marija1.html	\| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| \| \| --- \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- \| \| \| ← BACK \| PRINT \| + TEXT SIZE – \| SEARCH \| HOME \| \| \| \| \| \| \| \| \| --- \| \| Math Central \| Quandaries & Queries \| \| \| \| \| \| \| \| Question from Marija, a student: Is there a formula for finding the diameter of the base of the cone? \| \| \| \| \| Lots of them. It depends on what you know about the cone. Here is a cone with a few things identified on it. I'm assuming you have what we call a right circular single-napped cone, which is a cone that has a circle for a base and the tip of it is centered over the circle. The calculations mostly rely on the fact that their is a right angle at the bottom, so you can make a right triangle with the radius r, the height h and the slant s. The angle X is also in the triangle. If you know any two things on this diagram, you can find any of the others. For example, if you know the height h and the angle X, you can use the trig function TAN (tangent) to find the radius: r = h tan(X). And of course since the diameter is just twice the radius, diameter = 2 h tan(X). If you know want the area of a circle, it is calculated using A = π r2, so we can put the two equations together and we get this: A = π ( h tan(X) )2. The volume V of a cone is V = (1/3) h A. So if you knew the height h and the volume V and wanted the area, you would re-arrange this algebraically into: A = 3V / h. If you know the slant length s and the angle X, you can use the trig function SIN (sine) to find the radius r: r = s sin(X). And we can use this to find the area A again by combining it with A = π r2 : A = π ( s sin(X) )2 Let's say you wanted to know the volume and all you have is the slant s and the angle X. This takes a few steps: First find the Area A in terms of s and X, which we just did: A = π ( s sin(X) )2. Now we need h because V = (1/3) h A. But we can use the COS (cosine) function to find h if we know X and s: h = s cos(X). If we combine these three equations, we get: V = (1/3) ( s cos(X) ) π ( s sin(X) )2 which will work, but we can reduce it using trig identities to: V = [π s2 sin (2X) ] / 6. See what else you can figure out. Cheers, Stephen La Rocque. \| \| \| \| \| \| --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. \|
16386	https://unacademy.com/content/jee/study-material/mathematics/middle-term-of-an-arithmetic-progression/	Access free live classes and tests on the app Login Join for Free Profile Settings Refer your friends Sign out Terms & conditions • Privacy policy About • Careers • Blog © 2023 Sorting Hat Technologies Pvt Ltd JEE Exam » JEE Study Material » Mathematics » Middle Term of an Arithmetic Progression Middle Term of an Arithmetic Progression In this study material, we will study how to find the middle term of an arithmetic progression. If we talk about the meaning and introduction of an arithmetic progression, ‘Arithmetic’ means mathematical numbers. On the other hand, ‘Progression’ is referred to as moving from one number to another. Let us take an example to clear this: 3, 5, 7, 9, 11, 13 are said to be mathematical numbers. If we are moving from one number to another (progression), we have to add ‘2’ to get to the other number. For instance: 9+2=11 11+2=13 2 (the number that we are adding) is a constant number. We cannot add 1 or 3 to get this series. Adding a constant number is compulsory. So, this series in which we add a particular number and move from one number to another is called Arithmetic Progression. Similarly, suppose we have a series of 3, 5, 7, 10, 14. Then, it is not an arithmetic progression. Because 3+2=5, 5+2=7, 7+3=10, 10+4=14 We are adding different numbers to get a series. Moreover, we should only add a constant number to get a series. So, it will not be called an Arithmetic Progression. Middle term of an Arithmetic Progression Above, we have discussed the meaning and introduction of Arithmetic Progression. Now, before discussing how to find the middle term of an Arithmetic Progression, let us see some of the basic formulas: Formula to Find the nth Term of an A.P. an = a + (n-1) × d an is the nᵗʰ term in the sequence. Moreover, the first term in the sequence is ‘a. In addition to this, ‘d’ is the common difference between terms. Sum of n terms in an A.P. Sn = n/2 [2a + (n-1) × d] Sum of all terms of an A.P. S all terms = n/2 ( a+l) Middle term of an Arithmetic Progression with ‘n’ terms If ‘n’ is odd, then the middle term= (n+1)/2 th term For instance, 15=odd=n (15+1)/2 th term 16/2 th term 8th term So, the answer is the 8th term will be the middle term of an Arithmetic Progression. If ‘n’ is even, then the two middle terms= n/2 th term and n/2 + 1 th term. For instance, 16=even=n 16/2 th term 8th term and, 16/2 + 1 th term 8+1 th term 9th term So, the 8th and 9th terms will be a middle term of an Arithmetic Progression. Find the Middle Term of an Arithmetic Progression Let us discuss some questions about finding the middle term of an Arithmetic Progression. Q1) Find the middle term of an Arithmetic Progression 6, 13, 20,……, 216. Here, a = 6 (a is the first term of the sequence) d = (common difference between terms)= a² – a¹ = 13-6 = 7 a(n) = 216 a + (n-1)d = 216 So, according to the above-given details, 6+(n-1)7=216 6+7n-7=216 7n-1=216 7n=216+1 7n=217 n=217/7 n=31=odd So, the value of ‘n’ will be 31= odd Therefore, Middle term = (n+1)/2th (31+1)/2th 32/2th =16th term a16= a+15d 6+157 6+105 111 So, 111 will be the answer to the above question. Q2) Find the middle term of an Arithmetic Progression 3,8,13,18,….., 73? According to the question, a= 3 (first term of the question) l= 73 (last term of the question) d= a² – a¹ (difference between the both) =8-3 =5 l= a+(n-1)d 73= 3+(n-1).5 73-3= (n-1).5 70= 5n-5 70+5=5n 75=5n n=75/5 n=15 Therefore, the total no. of terms is 15 (odd). According to the formula: If ‘n’ is odd, then the middle term= (n+1)/2 th term (15+1)/2 th term 16/2 th term 8th term a+(8-1)d a+7d 3+75 3+35 38 So, after following the question, we’ll get 38 as our answer. Conclusion In this article, we have the introduction and meaning of A.P., the middle term of A.P., and how to find the middle term of A.P. Moreover, these notes also discuss some examples to clarify the topic. Arithmetic Progression is not a difficult topic. To solve the questions, one has to be clear with the concept and formulas. Once you understand the concept, you will be able to score well in the exams. Frequently asked questions Get answers to the most common queries related to the JEE Examination Preparation. What is Arithmetic Progression? Ans. As discussed above, arithmetic progression is a series of numbers, which differ from each other by a constant n...Read full How to solve the middle term of Arithmetic Progression? Ans. The easy and simple way to solve A.P. is first to check the difference between the numbers. Then, determine whe...Read full Discuss the important formulas in solving the middle term of Arithmetic Progression. Ans. Some of the important formulas of an arithmetic progression are as follows: ...Read full State the rules and important points in finding the middle term of A.P. Ans. Some of the rules and important points while solving the middle term of A.P. are as follows:- ...Read full Discuss the types of Arithmetic Progression. Ans. There are two types of Arithmetic Progression. It is as follows: ...Read full Ans. As discussed above, arithmetic progression is a series of numbers, which differ from each other by a constant number. It means that the difference between any term and the number next to it should be constant. For instance, consider the following series: 7, 9, 11, 13, 15 The difference between these numbers is 2, which becomes the constant number in this series. Ans. The easy and simple way to solve A.P. is first to check the difference between the numbers. Then, determine whether this difference between each term in the series is the same/constant. After this, follow the question as said. Apply the appropriate formulas as mentioned above to solve the question further. Ans. Some of the important formulas of an arithmetic progression are as follows: an=a+(n–1)×d S = n/2 [2a+(n −1)×d]. Finite = n/2(a+l) Tn=Sn– Sn-1 These are some of the basic formulas that will help solve the Arithmetic Progression questions. As all A.P. questions are dependent on formulas, it is very important to learn and apply appropriate formulas accordingly. Ans. Some of the rules and important points while solving the middle term of A.P. are as follows:- Check the difference between the terms is constant. If the difference between the terms is not the same or varies, then it is not said to be A.P. Apply appropriate formulas, seeing to the requirements of the questions. Find the term like a, d, l, n. Most importantly, learn to analyse the question properly and be good at mathematical calculations to avoid any error. Ans. There are two types of Arithmetic Progression. It is as follows: Finite Arithmetic Progression: It is said to be a limit series. Moreover, in this last term of the series is provided. Infinite Arithmetic Progression: The last term is not provided in this kind of series. These series continue to infinity, as there is no last term. That is why these series are called infinite arithmetic progression series. Share via
16387	https://www.youtube.com/watch?v=_EXi_75z7og	Parity Puzzles Zach Star 1400000 subscribers 6541 likes Description 190311 views Posted: 9 May 2020 Sign up with brilliant and get 20% off your annual subscription: STEMerch Store: Support the Channel: PayPal(one time donation): ►Follow me Instagram: Twitter: Animations: Brainup Studios ( ) ►My Setup: Space Pictures: Magnetic Floating Globe: Camera: Mic: Tripod: Equilibrium Tube: ►Check out the my Amazon Store: 326 comments Transcript: this video was sponsored by brilliant here we have 11 gears looped together the question is can all of these gears rotate at the same time like if I go and rotate one will everything rotate together well let's look at any two of the gears here if we rotate one of them let's say clockwise the other one will move counterclockwise and vice versa same if we put another one down here so we can see that any two neighboring gears must have opposite rotations thus if we go back given an attempted clockwise rotation on one gear the next would be counterclockwise causing the next to be clockwise and so on until we get to the end and reach a problem as these two would have the same rotation which isn't possible so the answer is no these gears cannot all rotate together if you had a loop like this there'd have to be an even number of gears or else would be jammed this was an example of a parity puzzle parody just being whether a number is even or odd because that's really what the solution came down to not the exact quantity of gears or their size but the parity and the same thing applies to all the problems to come so there's your hint now let's see the next question let's say you have three marbles labeled a B and C arranged on the table as Shem you're going to move one of the marbles through the other two or the line connecting them and you do this all the way through so the marbles are never collinear then you repeat this process however you like moving one through the other two over and over the question is can you get back the original configuration after 15 moves before I just say the answer a hint would be this is similar to the last problem in that it comes down to clockwise versus counterclockwise orientation for example right now going from A to B to C involves clockwise rotation after I move one marble through the other two going still from A to B to C involves counterclockwise rotation it's switched after another move it switches again so regardless of what we do the same walk in the same order will involve the opposite rotation that means if we start here with this configuration after an odd number of moves you'll have that backwards orientation meaning it would be impossible to get back in 15 moves or any odd number it would require an even number okay for this next question let's say we have a rook starting at the top-left corner of a chessboard the question is can the rook move around the board such that it touches each square once and only once and end up on the bottom right tile so it can't go over any tiles that already traced over and you have to use legal rook moves so no going diagonally now you don't need to start trying different paths to figure this out there are 64 tiles here in total meaning 63 moves must take place for this to be accomplished basically 63 tiles need to be touched ending on the bottom right now each move of the rook switches the color it's on if it's on a black square the next single move must transition it to a white square and vice versa since the rook started on a white square and must make an odd number of moves 63 then it must end on a black square and since that final Square is white then this puzzle is impossible so that wasn't too bad but I found another slightly more difficult version we're given a five by five board and a rook that is allowed to start an end on any square can you touch each tile once and only once using again legal rook moves well here since there are an odd number of tiles in total 25 then the white and black squares won't be equal in number in this case there's one more white square than black also since there are 25 squares then it will take 24 moves to complete the task if it is possible and since 24 is an even number then the rook will end on the same color it starts on it starts on white then it will go white black white black until you end on white and if you start on black and same idea in each case though the color you start on shows up one extra time like on the left there would be 13 white and 12 black squares that show up whereas on the right it'd be 12 white and 13 black meaning to match the colors of the actual board you need to start on white if you start on black then it's definitely impossible but as you can see we start on white and is possible in this case to go around touching every square once and only once then end on white as expected no this does not prove it's always possible but it definitely is impossible if we were to start on a black square now this next puzzle I showed in a video years ago plus numberphile recently did an even more advanced version but it fits really well here so if you've seen it skip to this time stamp otherwise here we go here we have an eight puzzle a game where you mix up the tiles and you have to put it back to the original configuration of one through eight now let's say from the solved configuration we just switch the last two tiles the question is is this solvable using only legal moves this is definitely the toughest puzzle in this video if you've never seen it before but it still has to do with parity first thing I'm gonna do is put the puzzle back to how it was and list the numbers horizontally the same way they're read left to right going down one row at a time now notice that if I move any number side-to-side that does not change the horizontal ordering of the numbers since we don't care about the blank spot the puzzle order is preserved however if I do a vertical move then the orders changed and for any vertical move that number will move two spots over on the horizontal line and if we do another vertical move like with the seven then you'll see the same thing happens it moves two spots over now going back currently there are zero numbers out of order of course after the vertical move there are now two numbers out of order what I mean by that is out of all the comparisons you can make two are out of order you compare 1 & 2 these are in order one comes before 2 1 & 3 are in order and same with 1 & 4 simply because still one is before 4 as it should be we don't care out by how much but compare 4 & 5 this is an out of order pair 4 should come first but it doesn't and therefore 4 & 6 are also out of order these are the only two pairs out of order though pick any other 2 numbers and you'll find the smaller number does show up first as it should now realize that any vertical move the AKS shift over by 2 causes the out of order pairs to change by 2 or 0 like if I move the 5 over that does nothing to the horizontal list but moving the 7 up we again shift it over by 2 to the spot here before the move these two numbers are in order thus afterwards they're out of order these 2 are in order so we see the same thing so the out of order pairs have gone up by 2 see we're only changing the order of two comparisons after a vertical move takes place that's why the out of order pairs can change by a maximum of two now if I move the 8 over and before the 6 goes down we have an out of order pair and an inorder pair after the move they both switch leaving still an out of order and an inorder pair meaning this value has changed by 0 so that's all that can happen a change of 2 or 0 and or to win the game this value must be 0 by the end all numbers in order so when we started with this scenario the out of order pairs was 1 just the 7 & 8 all other comparisons are in order and since that number can only change by 2 or 0 then this will never reach a value of 0 or any even number the out of order pairs will always be odd meaning this puzzle is impossible now moving on this next puzzle is more of a geometric one the goal here is to create a polygon any polygon you want such that you can draw a single line segment that crosses every single side but does not touch any of the corners so what I did here actually failed the line segment touches a few sides but not all of them in fact no matter what line segment I draw here I can't touch every side so now the question is would this even be possible and if so what criteria would have to be met for it to be possible ok spoiler I'm just going to immediately answer the is it possible question and that answer is yes it is here's the polygon for example a strange-looking one where I can draw a single line segment and touch every side to see what makes this possible I found it more intuitive to first just draw that line segment this here is the line segment that will intersect all sides so now let's construct that polygon since the line can intersect corners we have to start the polygon on one side of the line as there are two regions this line segment divided our plane into now every side of the polygon must intersect the line so it doesn't matter what I draw but it has to cross that line if it didn't then that edge would not be intersected when we complete the polygon so at this point we've drawn one side the next side also has to cross the line so I'll draw this and now we've constructed two sides in order to have a closed polygon we need at least three sides which would form a triangle but if I were to make one right now we'd have a polygon that is not intersected on all sides that last edge fails so we have to keep going and trace another third edge over the line segment and here I can't even connect back without overlap so I'll draw a fourth one and a fifth one notice that when we end on the opposite side we started on we at that moment have an odd number of sides when we land on the same side that value is even but since we have to connect back to the starting point then that last connection has to come from the right region leaving us with an even number of sides in total in this case six so it won't always be possible but a necessary condition is that the polygon has an even number of sides you need an odd number to get to that opposite region then one more to connect back to the beginning having an odd number of sides would make this impossible so these are just some examples of puzzles I really like and I'm gonna do another similar video soon but for now if you want to keep challenging yourself with similar questions and problems I definitely recommend checking out brilliant the sponsor of this video if you enjoyed what you saw here their contest math courses would be exactly what you're looking for these cover a variety of topics and specific problems which show up in advanced math competitions concepts include inequalities modular arithmetic combinatorics probability and much more plus they also include problem-solving strategies such as using symmetry or finding invariants which can reveal some hidden properties about the question at hand it really is true that just sitting down and challenging yourself with new types of questions is the best way to improve your problem-solving abilities and with all their intuitive animations along with constant practice problems brilliant a great resource for doing just that then on top of all this they have dozens of other courses in math science and engineering for you to choose from also the first 200 people to go sign up with the link below or by going to brilliant org slash Zack star will get 20% off their annual premium subscription and with that I'm going to end that video there thanks as always to my supporters on patreon social media links to follow me or down below and I'll see you guys in the next video
16388	https://www.onlinemathlearning.com/completing-the-square-worksheet.html	Completing the Square Worksheets (printable, online, answers, examples) OnlineMathLearning.com - Do Not Process My Personal Information If you wish to opt-out of the sale, sharing to third parties, or processing of your personal or sensitive information for targeted advertising by us, please use the below opt-out section to confirm your selection. Please note that after your opt-out request is processed you may continue seeing interest-based ads based on personal information utilized by us or personal information disclosed to third parties prior to your opt-out. You may separately opt-out of the further disclosure of your personal information by third parties on the IAB’s list of downstream participants. This information may also be disclosed by us to third parties on the IAB’s List of Downstream Participants that may further disclose it to other third parties. Personal Data Processing Opt Outs CONFIRM × Completing the Square Worksheets Related Topics: More Math Worksheets Printable Math Worksheets Share this page to Google Classroom Printable “Quadratic Equations” worksheets: Solve Quadratic Equation (use factoring) Solve Word Problems using Quadratic Equations Sketching Quadratic Graphs Printable Math Worksheets Rewrite Expressions in Completed-Squared Form Solve Quadratic Equation (use completing the square) Solve Quadratic Equation (use quadratic formula) Discriminant in the Quadratic Formula Examples, solutions, videos, and worksheets to help Algebra 1 students learn how to rewrite quadratic expressions given in standard form, ax 2 + bx + c in the equivalent completed-square form, [a(x–h)]2 + k. How to rewrite expressions in completed-square form? Completing the square is a method used to rewrite a quadratic expression in a form that reveals its minimum or maximum point, which is helpful for solving equations and graphing. Printable Math Worksheets Learning Software Steps to Complete the Square When 𝑎 = 1 Start with the quadratic expression in the form: 𝑥 2 + 𝑏𝑥 + 𝑐 Take half of the coefficient of 𝑥, square it to get (𝑏/2)2. Add and subtract (𝑏/2)2 in the expression. Rewrite the expression as a perfect square. Steps to Complete the Square When 𝑎 ≠ 1 When the coefficient of 𝑥 2 is not equal to 1, you’ll need to factor 𝑎 out before completing the square. Start with the quadratic expression in the form: a𝑥 2 + 𝑏𝑥 + 𝑐 Factor out 𝑎 a(x 2 + b/a x) + c Take half of the coefficient of 𝑥, square it to get (𝑏/(2a))2. Add and subtract (𝑏/(2a))2 in the expression. Complete the square inside the parentheses: Take half of the coefficient of 𝑥, square it, and add and subtract this squared term inside the parentheses. This gives: 𝑎(𝑥 2 + 𝑏/𝑎 𝑥 + (𝑏/(2𝑎))2 − (𝑏/(2𝑎))2 ) + c Rewrite as a perfect square and simplify: Rewrite the expression in completed square form, adjusting any constants outside the square. The following example shows how to rewrite an expression in completed-square form. Discover more Math Mathematics College application assistance Online learning platforms Calculus learning materials math College prep math General Science Educational software games Reference books dictionaries Have a look at this video if you need to review how to complete the square. Completing the Square Click on the following worksheet to get a printable pdf document. Scroll down the page for more Completing the Square Worksheets. Printable Math Worksheets More Completing the Square Worksheets Printable (Answers on the second page.) 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16389	https://www.youtube.com/watch?v=SqFj06Y3tl4	Point Slope Form of a Line - How to Write Mario's Math Tutoring 458000 subscribers 392 likes Description 46076 views Posted: 17 Mar 2016 Learn how to write equations of lines in Point Slope Form and also how to graph them in this free math video tutorial by Mario's Math Tutoring. We discuss the formula as well as how to rewrite an equation in point slope form into slope intercept form. Timestamps: 00:00 Intro 0:22 Formula for the Point Slope of the Equation of a Line 0:46 Showing Where the Point Slope Formula Comes From 1:16 Find the Point Slope Form of the Line through (-1,4) with Slope 2 2:06 Equation of the Line through (3,1),(6,10) 2:56 How to Rearrange Into Slope Intercept Form Related Videos: Writing the Equation of a Line in Slope Intercept Form Writing the Equation of a Line in Standard Form Organized List of My Video Lessons to Help You Raise Your Scores & Pass Your Class. Videos Arranged by Math Subject as well as by Chapter/Topic. (Bookmark the Link Below) ➡️JOIN the channel as a CHANNEL MEMBER at the "ADDITIONAL VIDEOS" level to get access to my math video courses(Algebra 1, Algebra 2/College Algebra, Geometry, and PreCalculus), midterm & final exam reviews, ACT and SAT prep videos and more! (Over 390+ videos) 28 comments Transcript: Intro hi this is Mario with Mario's math tutoring coming to you with another math video to help you boost your score in your math class improve your understanding and hopefully make math a lot less stressful so let's get right into this video this is a second video of a three part series talking about lines in this video we're going to talk about the point-slope form of a line okay and so let's go to the equation Formula for the Point Slope of the Equation of a Line here it's y minus y1 equals M times X minus x1 so what the M represents is the slope that's the angle of the line and what the x1 and y1 represent okay these are the coordinates of the point that the line goes through okay so if you know that there's a point on the line you know they according to that point and you know the angle of the line you can quickly find the equation using this point-slope form just a little Showing Where the Point Slope Formula Comes From background on where this formula comes from we know that the slope formula equals y2 minus y1 over x2 minus x1 okay anything divided by 1 is itself if we cross multiply across the equal sign we get y2 minus y1 equals M times x2 minus x1 and you can see the formulas are the same so it comes from that formula for slope let's look at some example problems and talk about how to write the equations of lines so in this first Find the Point Slope Form of the Line through (-1,4) with Slope 2 example we're given the slope is 2 and the point that the line goes through is negative 1/4 so this one set up pretty nice for us so what we're gonna do is we're going to do y minus the y coordinate which is 4 ok the slope which is 2x minus the x coordinate which is negative 1 but when you subtract a negative it's really just like adding a positive so I'm just gonna write that as plus 1 okay now if we wanted to graph this line all we would have to do is say alright it goes through the point negative 1 4 so negative 1/4 okay right there and the slope of the line is 2 2 of course is like 2 over 1 so the rise is - the run is 1 and you can repeat that process right to run 1 and you can see there's your there's your lines ok let's take a look at another example say Equation of the Line through (3,1),(6,10) we want to find the equation of the line through these two points 3-1 and 610 so first thing we need to do is we to find the slope of the line by using our slope formula so we have y2 minus y1 so 10 minus 1 over x2 minus x1 which is 6 minus 3 so that's 9 over 3 which equals 3 so slope of our line is 3 and then what we're going to do now this is the part that's an H student see a little bit confused by they say do we use this point or do we use this point and it doesn't actually matter which point that you use so let's just pick this point over here okay since the numbers a little bit smaller it'd be a little bit easier to work with so we've got Y minus the y-coordinate which is 1 equals the slope which was 3 times X minus the x-coordinate which is 3 okay now this is point-slope form if we want to rearrange it into the slope-intercept How to Rearrange Into Slope Intercept Form form what we can do is we can distribute the 3 so that's 3x minus 9 we can add the 1 to both sides so that's going to give you 3x minus 8 and now we have the equation in the slope-intercept form okay so you can rearrange these equations in two different forms but sometimes it's easier to start with one form and in this case we want with the point-slope form because we knew one of the points was on the line and we were able to find the slope very easily but then if we want to go ahead and rearrange it we can do so so I hope you're finding these videos helpful if you are go ahead and subscribe to the channel go ahead and check out the the other videos in this series on graphing lines and writing equations of lines and I look forward to seeing you in the next video I'll talk to you soon
16390	https://www.storyofmathematics.com/3d-vector/	JUMP TO TOPIC [show] What Is A 3-D Vector? 3-D Coordinate System How To Find The Magnitude Of A 3-D Vector? What Is A Displacement Vector? Direction Of A Vector Determined By The Unit Vector Angle Between Two 3-D Vectors How To Graph A 3-D Vector? Practice Problems Answers 3D Vectors – Explanation and Examples Vectors are very useful in daily life. However, in the real world, things happen in three-dimension. Generally, we learn to solve vectors in two-dimensional space. Still, to expand and develop the use of vectors in more realistic applications, it is essential to explain the vectors in terms of three-dimensional planes. A 3-D vector is defined as: “A three-dimensional vector is a line segment drawn in a 3-D plane having an initial point referred to as tail, and final point referred to as the head. Like a normal vector in the 2-D plane, a 3-D vector also has some magnitude and direction”. In this topic, we shall discuss the following points in detail: What is a 3-D vector? How to find the magnitude of a 3-D vector? How to calculate the angle between two 3-D vectors? How to draw a 3-D vector? Examples Problems What Is A 3-D Vector? A 3-D vector is a vector represented in a 3-D plane having three coordinates; x, y, and z. As in the previous sections, we have learned and discussed the vectors in 2-dimensional space. To avoid the computational complexity and simplify the idea so that we can understand the concept easily, it’s time to learn about 3-D vectors. For example, if we need to specify the direction of any rigid object or body such as cars, airplanes, robots, etc., one would normally think that he needs three coordinates to define the position of the objects x, y, and z-axis and that is completely correct. So, to describe the impact of all the features, we need to use three-dimensional space. Similarly, if we consider a map in 2-D, it’s only useful for navigating from one point to another. Still, if we need to specify various landscapes and environments, only a 2-D description of a map is not enough. That’s why it is necessary to understand the concept of 3-D vectors in a 3-D coordinate system and their properties. A 3-D vector is like a 2-D vector in all aspects, but in the case of a 3-D vector, we need to keep track of one more direction. 3-D vector operations are analogous to 2-D operations with just an added computational step. We can do various computations like finding the angle between two vectors, scalar multiplications, etc. 3-D Coordinate System Now, the first question is, “What is a 3-D coordinate system?” A 3-D coordinate system has 3 dimensions or can be regarded as having 3 perpendicular axes: x, y, and z-axes. Such a system is called a 3-dimensional rectangular coordinate system. A vector drawn in a 3-D plane and has three coordinate points is stated as a 3-D vector. There are three axes now, so this means that there are three intersecting pairs of axes. Each pair forms a plane, xy-plane, yz-plane, and xz-plane. A 3-D vector can be represented as u (ux, uy, uz) or or uxi + uyj + uzk. How To Find The Magnitude Of A 3-D Vector? The magnitude of 3-D vectors is calculated in a similar way with the addition of one more coordinate. \|u\| = √((ux)^2 + (uy)^2 + (uz)^2) Where ux, uy, and uz are the magnitudes of coordinate axes. As we have already discussed, the concept of a 3-D vector is not different from that of a 2-D vector, except now there is one more dimension in the 3-D vector. The magnitude of a vector is always positive, as the common mistake in computing the magnitude of a vector is that we forget the absolute sign. Only the magnitude of the null vector is zero. Let us have a better understanding of the concept with the help of an example. Example 1 Calculate the magnitude of the following 3-D vectors. u = (3,4,5) v = <2,5,6,> s = 3i + 8k Solution Let’s first consider equation 1: u = (3,4,5) \|u\| = √ ((3)2 + (4)2 + (5)2) \|u\| = √ (9 + 16 + 25) \|u\| = 7.07 Now, consider the equation 2: v = <2,5,6,> \|v\| = √ ((2)2 + (5)2 + (6)2) \|v\| = √ (4 + 25 + 36) \|v\| = 8.06 Let’s evaluate for the equation 3: \|s\| = √ ((3)2 + (0)2 + (8)2) \|s\| = √ (9 + 0 + 64) \|s\| = 9.05 So, in the above examples we have calculated magnitudes of 3-D vectors. What Is A Displacement Vector? The displacement vector is defined as: “A vector that explains about change in the position of the object is called a displacement vector.” Let us consider a vector AB whose starting point is A (x1, y1, z1), and the ending point is B (x2, y2, z2). It has some magnitude and direction, and in this case, the direction is defined to be from A to B. The coordinates of the displacement vector are AB = (x2 – x1 , y2 – y1 , z2 – z1) Therefore, the magnitude is given as: \|AB\| = √ ((x2 – x1)^2 + (y2 – y1)^2 + (z2 – z1)^2) Let’s conduct some examples. Example 2 Given that the coordinates of two points are A (4,6,8) and B (7,8,4). Find out the distance between two points. Solution To find the distance between two points in a 3-dimensional plane, we will use the following formula: \|AB\| = √ ((x2 – x1)^2 + (y2 – y1)^2 + (z2 – z1)^2) \|AB\| = √ ((7 – 4)^2 + (8 – 6)^2 + (4 – 8)^2) \|AB\| = √ ((3)^2 + (2)^2 + (-4)^2) \|AB\| = √ (9 + 4 + 16) \|AB\| = √ (29) \|AB\| = 5.38 The distance between the two points is 5.38 m. Direction Of A Vector Determined By The Unit Vector A unit vector is defined as a type of vector whose magnitude is always equal to 1. So, the unit vector describes the direction of a vector v given that the magnitude of the vector is \|v\|. Then, the direction vector is given as, Û = U / \|U\| Let’s solve some examples to imply this concept on 3-D vectors. Example 3 Find out the direction and magnitude of the given 3-D vector PQ (3,5,6). Solution The magnitude of the given vector is given as: \|PQ\| = √ ((3)2 + (5)2 + (6)2) \|PQ\| = √ (9 + 25 + 36) \|PQ\| = 8.366 The direction of the 3-D vector is given by unit vector as follow: UPQ = PQ / \|PQ\| UPQ = [3, 5, 6] / 8.366 Example 4 Find out the direction and magnitude of the given vector AB = 5i + 3j + 2k Solution The magnitude of the given vector is given as: \|AB\| = √ ((5)^2 + (3)^2 + (2)^2) \|AB\| = √ (25 + 9 + 4) \|AB\| = 6.166 The direction of the vector is given by unit vector as follow: UAB = AB / \| AB \| UAB = (5i + 3j + 2k) / 6.166 Angle Between Two 3-D Vectors Let us consider two 3-D vectors u and v. The scalar product of two vectors in 3-D space is given as: u.v = \|u\| \|v\|.cosθ where \|u\| and \|v\| are the magnitudes of the two vectors u and v and θ is the angle between the two vectors. To understand the concept of the angle between two 3-D vectors, let’s revise the concept of a scalar product or dot product. The scalar product is defined as the product of two 3-D vectors, which gives a scalar quantity in return. So, the angle between two 3-D vectors is given as the dot product of the two vectors divided by the product of the magnitudes of two vectors. The following steps must be followed to calculate the angle between two 3-D vectors: Firstly, calculate the magnitude of the two vectors. Now, start with considering the generalized formula of dot product and make angle θ as the main subject of the equation and model it accordingly, u.v = \|u\| \|v\|.cosθ cosθ = u.v / \|u\| \|v\| θ = arccos (u.v / \|u\| \|v\|) Use the standard algebraic formula to calculate the dot product of two vectors. Similarly, the angle between two 3-D vectors can also be calculated by using a cross product by following the same steps as discussed above, and the only difference is that it will have sin instead of cos and generalized formula of cross-product in order two find out the result. Let us understand the concept with the help of an example. Example 5 Given that there are two vectors u = 2i + 2j + 3k and v = 6i + 3j + 1k. using the formula of dot product calculate the angle between the two vectors. Solution Follow the following steps to calculate the angle between two vectors. Start with the formula of the dot product. Find out the magnitude of the two vectors. Calculate the dot product of two vectors. Divide the product of two vectors by the product of the magnitude of two vectors. Calculate the value of θ by putting into the equation given below θ = arccos (u.v / \|u\| \|v\|) Magnitude of u is given as, \|u\| = √ ((2)^2 + (2)^2 + (3)^2) \|u\| = √ (4 + 4 + 9) \|u\| = √ (17) Magnitude of v is given as, \|v\| = √ ((6)^2 + (3)^2 + (1)^2) \|v\| = √ (36 + 9 + 1) \|v\| = √ (46) Now, calculating the dot product of two vectors, u.v = (2i + 2j + 3k) . (6i + 3j + 1k) u.v = ((2.6)(1) + (2.3)(1) + (3.1)(1)) u.v = 12 + 6 +3 u.v = 21 Now, as a final step put all the values into the formula in order to calculate the value of θ. θ = arccos (u.v / \|u\| \|v\|) θ = arccos (21 /√ (17).√ (46) ) θ = arccos (21 / (4.12). (6.78) ) θ = arccos (0.75) θ = 0.7227 rad So, converting the angle into degrees, θ = 41.36º How To Graph A 3-D Vector? To graph a 3-D vector, we will consider the following analogy. Let us consider a 3-D coordinate system with 3 axes x, y, and x-axes, which can also be denoted in standard unit vectors such as i, j, and k. As shown in the figure, the labeled sides are positive x-axes, positive y-axes, and positive z-axis, and the unlabeled sides are regarded as negative axes. The intersection of three perpendicular axes is called origin O. So, with these axes, any point A in space can be assigned three coordinates A = (A1, A2, A3). Let’s consider a person standing near the corner of a room and looking down at the point where walls meet the floor. So, that intersection can be visualized as a 3-D axis. The floor and the wall to the left of the person intersecting each other in a line can be regarded as positive x-axes. The floor and the wall intersecting towards the right side of the person are y-axes. The walls intersecting in a vertical line are positive z-axis. The opposite part of each is regarded as a negative part of each axis. A vector is drawn as blue with its tail fixed at the origin and arrowhead pointing in the direction in the figure below. Now, draw the vector’s projection on three axes, which are shown in red, which are the coordinates of the given vector. Just as in two-dimension, we can also denote a three-dimensional vector in terms of a unit vector i, j, and k. These are the unit vectors in the above positive axes. A 3-D vector can be dented as A = A1i + A2j + A3k where A1, A2, and A3 are the coordinates of a 3-D vector. There are various 3-D vectors plotting and graphing software that can be used to visualize and draw 3-D vectors and understand their specifications properly. Practice Problems Calculate the magnitude of the following 3-D vectors: u = 5i + 10j + 8k AB = 1i + 2j + 5k <3,5,8> Given that the coordinates of two points are A (5,0,8) and B (9,5,4). Find out the distance between two points. Find out the angle between the given vectors u <-8,7,2> and v <-6,4,3>. Find out the direction vector of u <2,6,5> Find out the direction and magnitude of the given vector AB = -8i + 5j + 9k Given that there are two vectors u = 8i + 6j + 9k and v = 3i + 3j + 5k. using the formula of dot product calculates the angle between the two vectors. A book is lying on the table such that a force F1 = 1i + 1j + 1k acting in an upward direction and a force F2 = -(1i + 1j + 1k) acting in the downward direction so that two forces are equal in magnitude and opposite in direction. Calculate the angle between the two forces. Answers 13.8 5.5 9.9 7.54 55.6° (<2, 6, 5>)/ (√65) \|AB\| = 13 , UAB =(-8i + 5j + 9k)/ (13) 17.2° 180° All the vector diagrams are constructed using GeoGebra. Previous Lesson \| Main Page \| Next Lesson
16391	https://www.jmp.com/content/dam/jmp/documents/en/technical-reports/powerAndSampleSize.pdf	Power and Sample Size Calculations in JMP R ⃝ Clay Barker, Senior Research Statistician Developer Summary. JMP provides tools for performing prospective power and sample size calculations for a variety of settings. These capabilities range from aiding the user in planning a basic test on the mean or variance, to more complicated situations such as designing a reliability study. This document provides details on the theory behind the power and sample size features in JMP . 1. Introduction JMP provides tools for conducting prospective power and sample size calculations for a variety of set-tings. This document provides details on the strategies and formulas that are used in the JMP Power and Sample Size Calculator. Sample JSL functions are also provided to mimic what is happening behind the scenes in JMP. Section 2 provides details on calculations for planning tests on normal means and vari-ances. Section 3 provides details on calculations for planning tests on a binomial proportion. Section 4 covers some basic notation for reliability data that will be useful in Section 5 and Section 6, which cover reliability demonstration plans and reliability test plans respectively. Closing comments are in Section 7, and a convenient summary of formulas appears in Appendix A. 2. Calculations for normal means and variances The Power and Sample Size Calculator in JMP provides functionality for planning tests on normal means and variances. This section summarizes the strategies employed by JMP for performing such calculations. Topics covered in this section include: test of k sample means, test of a standard deviation, and a test on the number of counts per unit. For each topic, we discuss the motivation behind the methods used by JMP and provide sample JSL code for replicating JMP results. 2.1. One Sample Mean Test Suppose we have a random sample of n normal variates, Xi ∼N(µ, σ2). We want to test the hypothesis H0 : µ = µ0 vs Ha : µ ̸= µ0. One way to conduct this test is to use the traditional full-vs-reduced F test F ⋆= SSE(Reduced)-SSE(Full) df(Reduced)-df(Full) × 1 MSE(Full) (1) which has an F(1, n−1) distribution under the null hypothesis. Here the full model uses the sample mean for µ and the reduced model uses µ0 for µ, leading to SSE(Reduced) = n X i=1 (xi −µ0)2 SSE(Full) = n X i=1 (xi −¯ x)2. 2 JMP Development Under the alternative hypothesis, we know that F ⋆has an F(1, n−1, nδ2/σ2) distribution where nδ2/σ2 is a noncentrality parameter. Here δ is an assumed difference between the true mean and the hypothesized mean µ0. How do we use F ⋆to do power and sample size calculations? To calculate power, ﬁrst we need to get a critical value for testing H0. Given a testing level α, fcrit = Φ−1 (1 −α, 1, n −1) where Φ−1 (1 −α, 1, n −1) is the inverse CDF of the F(1, n −1) distribution evaluated at 1 −α. So to calculate power, we use power = Pr (reject H0 given µ = µ0 + δ) = Pr f > fcrit f ∼F 1, n −1, nδ2 σ2 (2) given sample size n, testing level α, error variance σ2, and difference to detect δ. We can use built-in JSL functions to calculate power directly. For n = 20, δ = .75, and σ = 1.25, we use the following example JSL function to calculate power. f_power = function({n, delta, sigma, alpha=.05},{fcrit, dsr, pow}, fcrit = fquantile(1-alpha, 1, n-1); dsr = delta/sigma; pow = 1 - f distribution(fcrit, 1, n-1, ndsrdsr); pow; ); f_power(20, .75, 1.25); Given (2), obtaining a sample size that yields power of 1 −β is straightforward. We could start at n = 10, for example, and increase n until the power is greater than or equal to 1 −β. The following JSL function uses the power function that we just deﬁned to calculate sample size. f_ss = function({delta, sigma, its, dpower, alpha}, {n, i, pow}, n = 10; // place to start for(i=1, i<=its, i++, pow = f_power(n, delta, sigma, alpha); if(pow>=dpower, break()); n++; ); n; ); f_ss(.75, 1.25, 1000, .95, .05); Fortunately JMP uses a more efﬁcient strategy based on (2) than this brute force method for determining sample size. JMP does not provide the option to plan a one-sided test of the mean. But we can obtain calculations for the one-sided case simply by doubling the desired α level and using the Power and Sample Size Calculator in JMP. So if we wanted to do a one-sided sample size calculation with α = .05, we would use a two-sided calculation with α = .1. Power and Sample Size 3 2.1.1. Distributional Results How do we arrive at the distributional results for F ⋆stated earlier? First we realize that under the null and alternative hypotheses, SSE(Full)/σ2 has a χ2 n−1 distribution. Then we rewrite the difference in sums of squares: SSE(Reduced) −SSE(Full) = n X i=1 (xi −µ0)2 − n X i=1 (xi −¯ x)2 = n X i=1 (xi −¯ x)2 + n(¯ x −µ0)2 − n X i=1 (xi −¯ x)2 = n(¯ x −µ0)2. Under the null hypothesis, µ = µ0 and ¯ x −µ0 σ/√n ∼ N(0, 1) ⇒n ¯ x −µ0 σ 2 = ¯ x −µ0 σ/√n 2 ∼ χ2 1. So under the null hypothesis, F ⋆is the ratio of χ2 variates divided by their degrees of freedom. The resulting distribution is F(1,n −1). Similarly under the alternative hypothesis, µtrue = µ0 + δ and ¯ x −µ0 σ/√n ∼ N(δ, 1) ⇒n ¯ x −µ0 σ 2 = ¯ x −µ0 σ/√n 2 ∼ χ2 1 nδ2/σ2 where χ2 1 nδ2/σ2 is a non-central χ2 distribution with non-centrality parameter nδ2/σ2. So under the alternative hypothesis, F ⋆is the ratio of a non-central χ2 variate (divided by the appropriate degrees of freedom) and a central χ2 variate (divided by the appropriate degrees of freedom). The resulting distribution is the non-central F distribution: F(1,n −1,nδ2/σ2). 2.2. Two Samples: Comparing Means Suppose we have two independent samples (each of size n) with common variance σ2 Xi1 ∼N(µ1, σ2) and Xi2 ∼N(µ2, σ2) and we are interested in testing that the means are equal: H0 : µ1 −µ2 = 0 vs Ha : µ1 −µ2 ̸= 0. One way to test this hypothesis is to use (1) again. In this case we have SSE(Full) = n X i=1 (xi1 −¯ x1)2 + n X i=1 (xi2 −¯ x2)2 where ¯ xj = 1 n n X i=1 xij and SSE(Reduced) = n X i=1 (xi1 −¯ x0)2 + n X i=1 (xi2 −¯ x0)2 where ¯ x0 = ¯ x1 2 + ¯ x2 2 . 4 JMP Development The difference in sums of squares can be written SSE(R) - SSE(F) = n X i=1 (xi1 −¯ x0)2 + n X i=1 (xi2 −¯ x0)2 − n X i=1 (xi1 −¯ x1)2 − n X i=1 (xi2 −¯ x2)2 = n(¯ x1 −¯ x0)2 + n(¯ x2 −¯ x0)2 = n 4 (¯ x1 −¯ x2)2 + n 4 (¯ x2 −¯ x1)2 = n 2 (¯ x1 −¯ x2)2. Because of independence, we know that ¯ x1 −¯ x2 ∼N(µ1 −µ2, 2σ2/n). Then under the null hypothesis µ1 = µ2 and we have ¯ x1 −¯ x2 σ p 2/n ∼ N(0, 1) ⇒ " ¯ x1 −¯ x2 σ p 2/n #2 ∼ χ2 1. Under the alternative hypothesis, µ1 −µ2 = δ and ¯ x1 −¯ x2 ∼ N(δ, 2σ2/n) ⇒ " ¯ x1 −¯ x2 σ p 2/n #2 ∼ χ2 1 nδ2 2σ2 where nδ2/(2σ2) is a non-centrality parameter. We know that under the null and alternative hypotheses SSE(Full) σ2 ∼χ2 2n−2, so F ⋆has a central F-distribution under the null hypothesis and a non-central F-distribution under the alternative hypothesis: F ⋆∼F(1, 2n −2) under H0 and F ⋆∼F 1, 2n −2, nδ2 2σ2 under Ha. We can use this information to perform power calculations. Given testing level α, fcrit = Φ−1(1 −α, 1, 2n −2) where Φ−1(1 −α, 1, 2n −2) is the inverse CDF of the F(1,2n-2) distribution evaluated at 1 −α. Then the power is expressed power = Pr(reject H0 given µ1 −µ2 = δ ) = Pr f > fcrit f ∼F 1, 2n −2, nδ2 2σ2 Power and Sample Size 5 for constant sample size n, testing level α, error variance σ2, and assumed difference in means δ. We can write our own JSL function to calculate power to detect a difference in means. For two independent samples of size n = 25 each, with error variance σ2 = 1.252, and assumed difference in means δ = .8, the following JSL function tells us that we have about a 60% chance of rejecting the null hypothesis. f_power = function({n, delta, sigma, alpha=.05},{fcrit, dsr, pow}, fcrit = f quantile(1-alpha, 1, 2n-2); dsr = delta/sigma; pow = 1 - f distribution(fcrit, 1, 2n-2, (n/2)dsrdsr); pow; ); f_power(25, .8, 1.25); This power function can then be used to solve for the sample size required to achieve a desired power level. 2.3. k Samples: Testing for a Common Mean If we have k independent samples (each of size n) of random normals Xij ∼N(µj, σ2) for i = 1, . . . , n; j = 1, . . . , k we may be interested in testing whether or not the means are all equal H0 : µ1 = µ2 = . . . = µk vs Ha : not all means equal. Once again, we can test this hypothesis using (1). Using techniques similar to the previous two subsec-tions, we can show that F ⋆ ∼ F (k −1, kn −k) under H0 and F ⋆ ∼ F k −1, kn −k, n Pk j=1(µj −¯ µ)2 σ2 ! under Ha, where ¯ µ = Pk j=1 µj/k. For example, suppose that we have four samples (n = 20 each) of random normals and want to test to see if all of the means are equal. We will assume that σ2 = 42 = 16 and we are interested in the power to detect a difference in means when µ1 = 10, µ2 = 11, µ3 = 13, and µ4 = 14. We can use the following JSL function to ﬁnd that we will have about an 84% chance of concluding that the means are not all equal. 6 JMP Development k_power = function({nper, muvec, sigma, alpha=.05}, {k, fcrit, mubar, sigma2, del}, k = nrow(muvec); fcrit = f quantile(1-alpha, k-1, nperk-k); mubar = mean(muvec); sigma2 = sigmasigma; del = sum( (muvec-mubar)‘(muvec-mubar) )nper/sigma2; 1-f distribution(fcrit, k-1, nperk-k, del); ); k_power(20, [10, 11, 13, 14], 4); This power function could then be used to determine the sample size needed to obtain a desired power level. 2.4. One Sample Test of Standard Deviation Suppose we have a sample of n random normal variates with mean µ and variance σ2: Xi ∼N(µ, σ2). We are interested in testing the hypothesis H0 : σ = σ0 vs Ha : σ > σ0 (3) using the test statistic T = Pn i=1(xi −¯ x)2 σ2 0 . Large values of T provide evidence against the null hypothesis. Under the null hypothesis, σ = σ0 and T ∼χ2 n−1. Under the alternative hypothesis, the true standard deviation is greater than σ0 and T = Pn i=1(xi −¯ x)2 σ2 0 = Pn i=1(xi −¯ x)2 σ2 \| {z } ∼χ2 n−1 ×σ2 σ2 0 ⇒σ2 0 σ2 T ∼ χ2 n−1 ⇒ σ2 0 (σ0 + δ)2 T ∼ χ2 n−1 where δ = σ −σ0. So under the alternative hypothesis, T does not have a χ2 distribution but is instead a constant multiple of a χ2 distribution. Then the power to detect this difference in standard deviation is written power = Pr(reject H0 given σ = σ0 + δ) = Pr X > σ2 0 (σ0 + δ)2 × χ1−α where χ1−α is the 1 −α quantile of the χ2 n−1 distribution and X ∼χ2 n−1. For example, suppose we want to test (3) at the α = .05 level with n = 20 observations when σ0 = 1.2 and we want to detect a difference of at least .1 (when the true standard deviation is at least 1.3). The following JSL function tells us that we only have about a 14% chance of rejecting the null hypothesis. Power and Sample Size 7 sd_power_larger = function({sig0, delta, n, alpha=.05}, {crit, spd}, crit = chi square quantile(1-alpha, n-1); spd = sig0+delta; 1-chi square distribution(critsig0sig0/(spdspd), n-1); ); sd_power_larger(1.2, .1, 20); So we should consider increasing the sample size in this example in order to have greater power to reject the null hypothesis. Calling this JSL function repeatedly, we could increase n until we reach a more acceptable power level. We can also rearrange the expression for power to solve for δ. Solving for δ tells us the difference that we would be able to detect given a speciﬁed sample size and power level. Using 1 −β to denote power, 1 −β = Pr X > σ2 0 (σ0 + δ)2 × χ1−α where X ∼χ2 n−1 ⇒χβ = σ2 0 (σ0 + δ)2 χ1−α ⇒δ = σ0 rχ1−α χβ −σ0 (4) where χp is the pth quantile of the χ2 n−1 distribution. For example, we may be interested in the magnitude of the change from σ0 = 2 that we can detect with 90% power when we have n = 50 observations. The following JSL function tells us that we could detect an increase of about δ = .68. delta_larger = function({sigma0, pow, n, alpha=.05},{chi1, chi2}, chi1 = chi square quantile(1-alpha, n-1); chi2 = chi square quantile(1-pow, n-1); sigma0sqrt(chi1/chi2) - sigma0; ); delta_larger(2, .9, 50); Instead of (3), we may be interested in testing H0 : σ ≥σ0 vs Ha : σ < σ0 (5) using the same test statistic T that we used when testing (3). Now small values of T provide evidence against the null hypothesis. The power to reject the null hypothesis is power = Pr(reject H0 given σ = σ0 −δ) = Pr X < σ2 0 (σ0 −δ)2 × χα where χα is the α quantile of the χ2 n−1 distribution, σ + δ = σ0, and X ∼χ2 n−1. For example, suppose we want to test (5) at the α = .05 level with n = 20 observations when σ0 = 2 and we want to detect a difference of δ = .5 (so true standard deviation of 1.5). The following JSL function tells us that the power to reject the null hypothesis is about 48%. sd_power_smaller = function({sig0, delta, n, alpha=.05},{crit, smd}, crit = chi square quantile(alpha, n-1); 8 JMP Development smd = sig0-delta; chi square distribution(critsig0sig0/(smdsmd), n-1 ); ); sd_power_smaller(2, .5, 20); Again, we could increment n and use this JSL function to help us ﬁnd the sample size needed to have a desired power to reject the null hypothesis. Similar to (4), we are able to use δ = σ0 r χα χ1−β −σ0 to solve for δ when given n, σ0, and a desired power. 2.5. Counts per Unit Suppose we are interested in the number of defects (or any other meaningful count) per unit of some product. For example, we may be interested in the number of leather imperfections on each basketball that we produce. Since we are working with counts, we can assume that the number of defects for unit i, di, has a Poisson distribution with parameter λ di ∼Poisson(λ) ⇒E(di) = Var(di) = λ. We are interested in detecting a change in the average number of defects per unit. For convenience we will focus on detecting an increase in the average, which means testing the hypothesis H0 : λ ≤λ0 vs Ha : λ > λ0. (6) One way to test (6) would be to use the test statistic T = ¯ d −λ0 p λ0/n where n is our sample size and ¯ d is the mean number of defects per unit in our sample. Because of the central limit theorem, we assume that T is approximately standard normal under the null hypothesis. To calculate power to detect an increase in the mean counts per unit, we need to ﬁnd the distribution of T under the alternative hypothesis. Noting that λ = λ0 + δ under the alternative, we can rearrange the test statistic as T = ¯ d p λ0/n − p nλ0 = ¯ d p λ/n × p λ/n p (λ −δ)/n − p nλ0 = ¯ d −λ p λ/n \| {z } ∼N(0,1) × r λ λ −δ + δ r n λ −δ . Rewriting T in this manner reveals that T ∼N δ r n λ0 , λ0 + δ λ0 Power and Sample Size 9 under the alternative hypothesis of the true mean being λ0 + δ. Now we can use this result to perform power and sample size calculations. If we are performing an α level test, the power to reject the null hypothesis is written power = Pr(reject H0 given λ = λ0 + δ) = Pr T > Z1−α λ = λ0 + δ = 1 −Φ Z1−α −δ p n/λ0 p (λ0 + δ)/λ0 ! (7) where Z1−α is the 1−α quantile of the standard normal distribution and Φ() is the cumulative distribution function of the standard normal distribution. For example, say that we are interested in testing that the mean number of defects per unit is λ0 = 4 and we want to know the power to detect an increase of one defect per unit. We are limited to a sample of n = 50 units. We can use the following JSL function to determine that we have about a 55% chance of detecting this increase. cpu_power = function({lambda0, del, n, alpha=.05},{z, pow}, z = normal quantile(1-alpha); pow = 1 - normal distribution( (z -delsqrt(n/lambda0))/sqrt((lambda0+del)/lambda0) ); pow; ); l_0 = 4; delta = .5; n=50; cpu_power(l_0, delta, n); We can also use (7) to solve for the sample size necessary to obtain a speciﬁed power to reject the null hypothesis. Using 1 −β to denote the desired power to reject the null hypothesis, the sample size is n = λ0 δ2 Z1−α −Zβ r λ0 + δ λ0 !2 where Z1−p is the 1 −p quantile of the standard normal distribution. For example, suppose we want to test that the average number of defects per unit is λ0 = 4 and we want to ﬁnd the sample size needed to provide 90% power to detect an increase of one defect per unit. Using the following JSL function, we ﬁnd that we need n = 38 units to have the speciﬁed power to detect the increase. 10 JMP Development cpu_n = function({lambda0, del, pow, alpha=.05},{za, zb, const, n}, za = normal quantile(1-alpha); zb = normal quantile(1-pow); const = (za-zbsqrt((lambda0+del)/lambda0))/del; n = lambda0constconst; ceiling(n); ); l_0 = 4; delta = 1; pow = .9; cpu_n(l_0, delta, pow); 3. Calculations for a binomial proportion When conducting an experiment where the response is binomially distributed, researchers are typically interested in testing the hypothesis H0 : p = p0 vs Ha : p ̸= p0 (8) or a one-sided version of the alternative. When planning such a study, the normal approximation to the binomial distribution provides a simple way to calculate the sample size, n, necessary to have power 1 −β to reject the null hypothesis. For small sample sizes, or more speciﬁcally when np or n(1 −p) is small, the normal distribution cannot adequately approximate the binomial distribution. This results in unwarranted optimism. To avoid these mistaken conclusions, we advocate the use of exact power calculations. While exact power calculations do not have as simple a form as the normal approximation, they can be calculated efﬁciently with current computing resources. 3.1. An adjusted test of the binomial proportion The usual Wald test statistic for the hypothesis stated in (8) is T(y) = (ˆ p −p0)2 ˆ p(1−ˆ p) n where ˆ p = y/n. (9) Agresti and Coull (1998) discuss some of the weaknesses of Wald-based tests and conﬁdence intervals for p. Wald tests of p are often liberal, meaning that the Type I error rate is higher than the nominal level. Because of the poor performance of the Wald-based tests and intervals, Agresti and Coull (1998) suggest using a slight adjustment to the Wald test statistic T(y) = (ˆ p −p0)2 ˆ p(1−ˆ p) n+4 where ˆ p = y + 2 n + 4. (10) This simple modiﬁcation of (9) is motivated by the score test for p, but is more simple to use than the score statistic. The adjusted Wald test statistic performs better than the Wald test statistic here in terms of maintaining the desired Type I error rate for tests and coverage rates for conﬁdence intervals. Figure 1 shows how Type I error rate varies as a function of p for the Wald and adjusted Wald tests when n = 25. The tests were conducted at the 0.05 level and we can see that the Wald test is liberal for nearly all values of p. For p close to zero or one, the Type I error rate can be higher than 50%. The adjusted Wald test comes much closer to achieving the nominal testing level. Power and Sample Size 11 0.0 0.2 0.4 0.6 0.8 1.0 0.00 0.05 0.10 0.15 p Type I Error Rate Wald Adjusted Wald Fig. 1. Plot of Type I error rate as a function of p for n = 25. We consider a simple example to investigate the difference between (9) and (10). Suppose that Y ∼ Binomial(n = 5, p = 0.005) and we want to test the hypothesis H0 : p = 0.1 vs Ha : p ̸= 0.1. If we use (9) to test this hypothesis, then the test statistic goes to inﬁnity when y = 0 or y = n = 5. So when y = 0 is observed, should the null hypothesis be rejected? Intuitively it makes sense to reject since the test statistic is approaching inﬁnity. But under H0 (p = 0.1), there is a 59% chance of observing y = 0. We certainly would not want to use a test with a Type I error rate that high. Now consider using (10) to test the very same hypothesis. The adjusted Wald test statistic has the favorable quality of taking ﬁnite values for y between zero and n. In our simple example, (10) gives us T(0) = 0.78 which would not reject at any reasonable level. This example shows that the adjusted test resolves any ambiguity when observations fall at the boundary (y = 0 or y = n). 3.2. Review of the normal approximation Here we show how to use the normal approximation to the binomial distribution to determine the appro-priate sample size for a test to have a desired power. 3.2.1. Obtaining the approximate sample size Suppose we want to test the hypothesis stated in (8) at level α. In the previous subsection we discussed the advantages of using the adjusted Wald statistic for testing this hypothesis, but for the sake of simplicity we are going to use the usual Wald statistic for calculating sample size. If we want power 1 −β to reject the null hypothesis, we can calculate the corresponding sample size as follows Pr ˆ p −p0 p p0(1 −p0)/n > Φ−1(1 −α/2) ! = 1 −β (11) Pr ˆ p −p p p0(1 −p0)/n + p −p0 p p0(1 −p0)/n > Φ−1(1 −α/2) ! = 1 −β Pr ˆ p −p p p(1 −p)/n > Φ−1(1 −α/2) s p0(1 −p0) p(1 −p) − p −p0 p p(1 −p)/n ! = 1 −β 12 JMP Development Pr z > Φ−1(1 −α/2) s p0(1 −p0) p(1 −p) − p −p0 p p(1 −p)/n ! = 1 −β p −p0 p p(1 −p)/n −Φ−1(1 −α/2) s p0(1 −p0) p(1 −p) = Φ−1(1 −β). where Φ() is the standard normal CDF. Solving for n yields n = " Φ−1(1 −β) p p(1 −p) + Φ−1(1 −α/2) p p0(1 −p0) p −p0 #2 . (12) The National Institute of Standards and Technology (2009) provides the formula in (12). Given the true proportion p, the hypothesized proportion p0, the testing level α, and desired power, the approximate sample size is a simple function of the normal cumulative distribution function. This formula is appealing because it is simple enough to calculate using a spreadsheet application like Excel. The following JSL function calculates the approximate sample size as a function of p, p0, power (denoted ‘pow’ in the JSL function), and testing level α. n_approx = function({p,p0,pow,alpha=.05}, {n}, n = (sqrt(p(1-p))normal quantile(pow) + sqrt(p0(1-p0))normal quantile(1-alpha/2) )/abs(p-p0); nn; ); A continuity corrected version of (12) also exists for calculating approximate sample size. The deriva-tion of the continuity corrected sample size is nearly identical to that of (12), except the sample proportion is estimated using ˆ p = y + 1 2 n and we proceed just as in (11). This modiﬁcation leads to the approximate sample size n = " Φ−1(1 −β) p p(1 −p) + Φ−1(1 −α/2) p p0(1 −p0) p −p0 #2 + 1 \|p −p0\|. (13) Here (13) is just as simple to use as (12), but less prone to underestimating the sample size. The following JSL function uses the function for the raw normal approximation to calculate the continuity corrected sample size. ncc = function({p,p0,pow,alpha=.05}, {}, n_approx(p,p0,alpha, pow) + 1/abs(p-p0); ); Using the two JSL functions we just deﬁned, we see that the continuity corrected approximation can be quite a bit different from the raw normal approximation. For example, try p = .995, p0 = .95, and request 40% power to reject the null hypothesis. show(n_approx(.995, .95, .4)); show(ncc(.995,.95,.4)); Power and Sample Size 13 0 1 2 3 4 0 2 4 6 8 y T(y) Fig. 2. Plot of adjusted Wald test statistic as a function of y. 3.2.2. A cautionary example We have seen that both (12) and (13) provide a simple solution for approximate binomial sample sizes. Unfortunately the normal approximation to the binomial breaks down for small n or p. When this hap-pens, the result can be misleading tests (rejecting when you should not) or insufﬁcient sample sizes. The remainder of this section will look at examples of the latter. Suppose that we want to test H0 : p = 0.0001 vs Ha : p ̸= 0.0001, assuming that the truth is p = 0.1. Plugging these values into (12), we ﬁnd that n = 27 is the approximate sample size that gives us 95% power to reject the null hypothesis. The adjusted Wald statistic for testing this hypothesis would look like T(y) = ( y+2 31 −0.0001)2 y+2 31 (1−y+2 31 ) 31 which has a χ2 1 distribution under the null hypothesis. Using χ to denote the 1-α quantile of the χ2 1 distribution, we can solve for the values of Y where the test statistic equals the critical value. T(y) = χ y + 2 31 −0.0001 2 = χy + 2 31 1 −y + 2 31 1 31 y + 2 31 2 −0.0002 y + 2 31 + 0.00012 −χ y + 2 31 + χ y + 2 31 2 = 0. Continuing in this line and using the quadratic formula, we ﬁnd that T(y) = χ at y1 = −2.00 and y2 = 1.42. These calculations are reﬂected in Figure 2. This result tells us that the test will reject when Y ≤y1 and when Y ≥y2. Since we are dealing with a binomially distributed random variable, obviously Y ≤−2.00 is not possible. So the test only rejects when Y ≥1.42. Since we assumed that the true distribution is binomial(n=27, p = 0.1), we can calculate the probability that the test will reject. Pr{Y ≥1.42\|Y ∼binomial(27, 0.1)} = Pr{Y ≥2\|Y ∼binomial(27, 0.1)} = 1 −Pr{Y ≤1\|Y ∼binomial(27, 0.1)} ≈ 0.767. 14 JMP Development Even though we used a nominal 95% power to obtain an approximate sample size, we ended up with a sample size that only provides 76.7% power to reject the null hypothesis. Despite the more precise adjusted Wald test used here, n = 27 does not come close to providing the desired power to reject the null hypothesis. Using the continuity corrected estimate instead results in n = 37, yielding 89.6% power to reject the null hypothesis. So for this example, (12) is extremely misleading while (13) yields a sample size that comes closer to the nominal power level. Now suppose that we would like to test H0 : p = 0.1 vs Ha : p ̸= 0.1, with 95% power and assuming that the truth is p = 0.001. Formula (12) yields n = 42 and formula (13) yields n = 52, resulting in approximately 0.0% power and 94.9% power respectively. Here the continuity corrected approximation is able to obtain the desired power level, but (12) yields practically zero power to reject the null hypothesis. The examples in this section have shown us the potential danger of using the normal approximation to calculate sample sizes needed for testing the binomial proportion. Neither approximation is able to guarantee that the sample size is sufﬁcient to obtain the desired power level. Underestimating the sample size can result in a signiﬁcant decrease in power. In practice, this could result in erroneous decisions as well as wasted money and resources. 3.3. Exact power calculations We have seen that the normal approximation does not always perform well when used to calculate sample sizes needed for testing the binomial proportion. Instead, exact power calculations should be used to calculate sample size. Exact calculations guarantee that the desired power level is obtained. Here we will look at how to calculate exact power and how to use that information to calculate sample size. 3.3.1. How to calculate power exactly Because the binomial is a discrete distribution, we have the luxury of being able to calculate power exactly using a simple summation. The basic idea is to ﬁnd all of the values of y that lead to a rejected test. Then add up the probabilities of observing those values of y. More explicitly, let’s say that we want to calculate the power to reject the null hypothesis H0 : p = p0 when testing at the α level. We will use the adjusted Wald statistic (10) and let χ denote the 1 −α quantile of the χ2 1 distribution. To calculate power, use the summation Power = n X y=0 Pr{Y = y \| Y ∼Binomial(n, p)}I{T(y) ≥χ} (14) where I{T(y) ≥χ} equals one when T(y) ≥χ and zero otherwise. Let’s go through a simple example. Suppose we have a sample of n = 5 observations and want to test the hypothesis H0 : p = 0.8 vs Ha : p ̸= 0.8. Testing at the α = 0.05 level, we will calculate the power to reject the null hypothesis when the true proportion is p = .1. Plugging these values into (14) gives us Power = 5 X y=0 Pr{Y = y \| Y ∼Binomial(5, 0.1)}I{T(y) ≥3.84}. The calculations are summarized in Table 1 and result in power approximately equal to 0.99. The follow-ing JSL functions can be used to perform exact power calculations and double-check the calculations in Table 1. Tstat = function({y,n,p_0}, {phat, T}, phat = (y+2)/(n+4); T = (phat - p_0)/sqrt(phat(1-phat)/(n+4)); T; Power and Sample Size 15 Table 1. Example power calculation. y T(y) I{T(y) ≥χ} Pr(Y = y) 0 17.4 1 0.59 1 8.8 1 0.33 2 4.6 1 0.07 3 2.2 0 0.008 4 .72 0 4.5e-4 5 .03 0 1e-5 ); power_2sided = function({n, p, p_0, level=.05}, {pow, chi, T, Tsqr}, pow = 0; chi = chi square quantile(1-level, 1); for(y=0, y<=n, y++, T = Tstat(y,n,p_0); Tsqr = TT; if( Tsqr > chi, pow += binomial probability(p,n,Y)); ); pow; ); power_2sided(5,.1,.8); 3.3.2. Why did the power drop? When working with the binomial distribution, power calculations for testing the proportion behave in a somewhat counter-intuitive manner. We are accustomed to seeing the power to reject the null hypothesis increase monotonically as sample size increases, but this is not the case when working with the binomial distribution. For a particular choice of n, increasing the sample size to n + 1 may actually result in a decrease in power. For example, consider testing the hypothesis H0 : p = 0.8 vs Ha : p ̸= 0.8 when the truth is p = 0.5. A sample size of eight yields about 64% power to reject the null hypothesis, while a sample size of nine only has 50% power to reject the null hypothesis (tests conducted at the 0.05 level). Why does this happen? This strange behavior is due to the discreteness of the binomial distribution. For any sample size, we can solve for the values of Y that will lead to rejecting the null hypothesis. In the problem described above, the test rejects for Y ≤4.21 when n = 8 and Y ≤4.87 for n = 9. Because the binomial is a discrete distribution, this means the test would reject for Y ≤4 for both n = 8 and n = 9. However, Pr{Y ≤4\|Y ∼binomial(8, 0.5)} > Pr{Y ≤4\|Y ∼binomial(9, 0.5)} which means the power drops when going from n = 8 to n = 9. Although the behavior may seem quite counter-intuitive, it is not a cause for concern. It does mean that after planning an experiment, researchers should be careful when adding extra runs to the experiment. One option is to add enough runs to the experiment to ensure that the power does not go down; this could mean that a set of two, three, or more runs would be added. Another option would be to not worry about 16 JMP Development 10 20 30 40 50 60 0.2 0.4 0.6 0.8 Sample Size Power Fig. 3. Plot of power as a function of sample size. adding a single extra run, since the drop in power is rarely severe. As long as an experiment is carefully planned, the non-monotonic nature of power should not be a cause for concern. 3.3.3. Sample size calculations In Section 3.2, we looked at examples where the normal approximation did a poor job of calculating sample size. Since it is not hard to calculate power exactly, using exact power calculations to obtain sample size is not a problem. Consider the following situation. Suppose we have a coin and want to test that the probability of landing on heads is 0.6. That is, we want to test H0 : p = 0.6 vs Ha : p ̸= 0.6. If the true probability is p = 0.4, we want 75% power to reject the null hypothesis. So what sample size is needed to obtain this power level? A simple approach would be to start at a particular sample size (say, n = 10) and calculate power for a range of values. Once the power gets above 75%, we have an adequate sample size. A more efﬁcient way to do this is to use a bisection search. Figure 3 shows the power as a function of sample size for our example. Judging by the plot, we see that n = 43 is adequate to provide 75% power to reject the null hypothesis. So if the true probability of landing on heads is 0.4, we need 43 trials to have at least 75% power to reject the hypothesis H0 : p = 0.6. Similarly, given a sample size and desired power, we can calculate the null proportion (p0) that yields the desired power level. The approach is very similar to calculating sample size. 3.4. Actual testing level In Section 3.3.3, we saw that in general when solving for sample size, we will not be able to ﬁnd an n that will yield the desired power precisely. Instead, we ﬁnd a sample size that yields at least the desired power. Similarly, the actual testing level will be different from the nominal testing level. For example, in Section 3.3.1, we went through an example of how to calculate power for a test of H0 : p = 0.8 at the α = .05 level when n = 5. We can go through similar steps to calculate the actual level of this test, this time using 0.8 as the true proportion: Actual α = 5 X y=0 Pr{Y = y \| Y ∼Binomial(5, 0.8)}I{T(y) ≥3.84}. Power and Sample Size 17 0.0 0.2 0.4 0.6 0.8 1.0 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 Hypothesized p Actual size of test Fig. 4. Plot of actual test size as a function of p0 for n = 20 and nominal testing level α = .05. For this example the actual testing level is 0.0579, slightly higher than the nominal level of 0.05. Figure 4 again shows that the actual Type I error rate ﬂuctuates around the nominal test size. The actual test size is an important value to consider. If it is too far from the nominal α, one may consider rethinking their experimental set-up. For example if we perform a sample size calculation and ﬁnd that the actual test size is quite a bit larger than the nominal α, we may consider increasing the sample size until we ﬁnd an n that achieves the desired power level and yields a type I error rate below the desired level. 3.5. Summary and extension to two-sample proportions We have looked into issues related to performing sample size calculations for tests of the binomial propor-tion. Calculations based on the normal distribution are attractive because of their simple form. However, we saw that these approximate values can be misleading, especially when np or n(1 −p) is small. A more reliable alternative is to use exact power calculations, which guarantee that the observed power will be greater than or equal to the desired power. Although the approximate sample sizes may perform well in some situations (especially the continuity corrected version), exact calculations are the only way to be sure that the desired power level is obtained. Exact calculations are not quite as convenient as the normal approximation, but they are by no means complicated. In this paper we have only considered the one sample test of proportion, but the same strategies can be easily extended to the two-sample situation testing H0 : p1 −p2 = δ vs Ha : p1 −p2 ̸= δ. Agresti and Caffo (2000) discuss an adjusted Wald test of the two-sample hypothesis test that is very similar in nature to the test statistic in (10). 4. Reliability Calculations JMP 9 introduced two new tools for designing reliability studies. Reliability demonstrations allow the re-liability of a new product to be compared to a standard. Reliability tests allow us to make inference about either failure probabilities or quantiles of the assumed failure time distribution. This section introduces some basic reliability concepts that will be used in the following sections that provide details about how these features are implemented in the JMP power and sample size platform. 18 JMP Development Table 2. Relationships between distributions. Distribution of x Cumulative Distribution Function Φ(x) Distribution of log(x) Frechet exp h −exp −log(x)−µ σ i Largest Extreme Value Loglogistic 1 2 + 1 2tanh hlog(x)−µ 2σ i Logistic Lognormal ΦN hlog(x)−µ σ i Normal Weibull 1 −exp n −exp hlog(x)−µ σ io Smallest Extreme Value 4.1. Location scale vs. log-location scale JMP allows the user to assume that failure times take several different forms, either from a location scale distribution or a log-location scale distribution. For convenience, we are going to assume that we are working with log-location scale distributions throughout the remainder of this section. The log-location scale distributions are a natural choice for survival/reliability data because they do not allow for negative failure times. Although we are not going to cover location scale distributions here, going from a log-location scale distribution to a location scale distribution is trivial: Just carefully place some exp() and log() arguments. Let x denote failure time. If x has a log-location scale distribution, then log(x) has a location scale distribution. Table 2 provides a summary of the distributions currently available in JMP, using ΦN(·) to denote the cumulative distribution function of the standard normal distribution. 4.2. Constant censoring times To make things easier, we assume a constant censor time for all observations. This assumption is appro-priate in reliability studies because the reliability study is conducted for a ﬁxed amount of time. Any units that have not failed by the end of the study are censored at the same time (c). The constant censoring time simpliﬁes the likelihood function considerably L(µ, σ\|X) = Φ log(c) −µ σ nc Y xi<c φ log(xi) −µ σ , (15) where xi are failure times, µ is the location parameter, σ is the scale parameter, nc is the number of censored observations, and Φ(·) and φ(·) are the assumed standard cumulative distribution and probability density functions respectively. The assumption of constant censoring times may not be appropriate outside of engineering studies. For example, in clinical trials a participant may be lost to follow-up at any point during the study. 4.3. What’s up with the Weibull? The Weibull distribution has two different parameterizations. Internally, we treat the Weibull distribution as a typical log-location scale distribution with probability density function f(x\|µ, σ) = 1 xσ exp log(x) −µ σ exp −exp log(x) −µ σ because that makes calculations much more convenient. Unfortunately, the ‘α, β’ parameterization of the Weibull distribution is more familiar to many engineers f(x\|µ, σ) = β αβ xβ−1exp − x α β . (16) Power and Sample Size 19 So we convert from the ‘α, β’ parameterization to the location scale parameterization using the following relationship µ = log(α) and σ = 1 β . When we report the covariance matrix of the Weibull parameters, we use the location scale parameters since we use those parameters in further calculations. 5. Reliability Demonstration Plans The goal of a reliability demonstration is to show that a new product meets or exceeds some reliability standard. For example, suppose that we are producing cars with a GPS system in the dash. We currently use GPS system A, which we have found to have a 90% probability of surviving one year of use. But GPS system B is less expensive, so we are considering putting it in our cars but are not willing to sacriﬁce reliability. We would use a reliability demonstration to ﬁnd out if system B is at least as reliable as system A. The ﬁrst step in planning the demonstration is assuming a failure time distribution (Weibull, log-normal, ...) and a scale parameter σ. Often in these situations, engineers have a good idea about the distribution and scale either from prior experience or expert opinion. After specifying these quantities, the GPS reliability standard can be written .9 = 1 −Φ log(1) −µ σ⋆ where Φ(·) is the CDF of the assumed failure time distribution and σ⋆is the assumed scale parameter. More generally, the reliability standard is stated p⋆= 1 −Φ log t⋆−µ⋆ σ⋆ (17) where p⋆is the standard probability of survival at time t⋆and µ⋆is the location parameter associated with the reliability standard. Typically µ⋆is not explicitly stated; instead it is solved for using µ⋆= log(t⋆) −σ⋆Φ−1 (1 −p⋆) . (18) By stating the reliability standard as in (17), we are able to solve for the location parameter of the assumed reliability standard using (18). 5.1. Reliability standard as a hypothesis test In order to be able to formulate a demonstration plan, we pose the reliability demonstration as a hypothesis test. We do that by testing H0 : p < p⋆vs Ha : p ≥p⋆ (19) where p is the probability of survival at time t⋆for the new product. If p ≥p⋆, then the new product is at least as reliable as the standard and we would like to reject the null hypothesis with high probability. When the test rejects, we say that the new product passed the reliability demonstration. So how do we decide when to reject the null hypothesis? A natural choice is to test n units for t units of time. If one or more units fail before time t, we do not reject the null hypothesis. If all n units survive to time t, then we reject the null hypothesis and conclude that the product is at least as reliable as the standard. 20 JMP Development More generally, we can say that we are willing to tolerate k failures during the demonstration. If more than k units fail before time t, then we do not reject the null hypothesis. If k or fewer units fail, then we reject the null hypothesis and conclude that the new product is at least as reliable as the standard. Choosing k = 0 will give us the quickest/cheapest experiment. Choosing larger k will result in a more expensive study in terms of time spent and units tested, but also more power to detect a small increase in reliability. We will see an example of this trade-off in Section 5.4. The two quantities that deﬁne a demonstration are the number of units tested (n) and the length of time for testing them (t). If we deﬁne one of these quantities, we can solve for the other using information from the reliability standard and the assumed failure time distribution. In general n and t are inversely related, meaning that increasing the number of units tested will lower the amount of time needed to test them and vice versa. 5.2. Solving for the study length and the number of units tested In the previous subsection we showed how to view a reliability demonstration as a hypothesis test. Now all that remains is to ﬁnd the combination of sample size (n) and length of study (t) that will give the reliability study the desired false-positive rate (α). Since we intend to test n independent units in the demonstration, the number of failures has a binomial(n, p) distribution where p is equal to the probability of a unit failing before time t. We want to test (19) at the α level, so we know that Pr( reject H0 \| H0 true ) = Pr( k or fewer failures \| H0 true ) α = Pr x ≤k \| x ∼Bin n, Φ log(t) −µ⋆ σ⋆ (20) where µ⋆and σ⋆come from the reliability standard. Plugging (18) into (20), we ﬁnd that α = Pr x ≤k \| x ∼Bin n, Φ log(t) −log(t⋆) + σ⋆Φ−1(1 −p⋆) σ⋆ , (21) which includes the two unknown values t and n. Given the length of the demonstration, we use (21) to solve for the number of units tested. Similarly, given the number of units tested, (21) will lead us to the required length of the demonstration. Before going any further with (21), we must take advantage of some properties of the binomial and beta distributions. Recall that if x has a binomial(n,p) distribution, Pr(x ≤k) = I1−p(n −k, k + 1) where Ix(a, b) is the regularized incomplete beta function Ix(a, b) = R x 0 ta−1(1 −t)b−1dt R 1 0 ta−1(1 −t)b−1dt which can also be rewritten in terms of the gamma function. Writing the cumulative distribution function in this manner is useful because it provides a connection to the Beta distribution. Recall that Pr(Y ≤y \| Y ∼Beta(α, β)) = Iy(α, β) = B(y; α, β). This relationship will allow us to use C++ routines written for the Beta distribution, which will make calculations much more convenient. Power and Sample Size 21 Now using the properties of the binomial and beta distributions, we can solve for t given that there are n units to be tested. Continuing on from (21), α = I1−δ(n −k, k + 1) where δ = Φ log(t) −log(t⋆) + σ⋆Φ−1(1 −p⋆) σ⋆ B−1(α; n −k, k + 1) = 1 −Φ log(t) −log(t⋆) + σ⋆Φ−1(1 −p⋆) σ⋆ log(t) −log(t⋆) + σ⋆Φ(1 −p⋆) = σ⋆Φ−1 1 −B−1(α; n −k, k + 1) t = t⋆exp σ⋆Φ−1 1 −B−1(α; n −k, k + 1) exp [σ⋆Φ−1(1 −p⋆)] (22) where B−1(α; n −k, k + 1) is the α quantile of the Beta(n −k, k + 1) distribution. Given a testing level, number of failures tolerated, number of units tested, and reliability standard, formula (22) provides a convenient solution for the necessary length of the demonstration. Unfortunately, solving for n when given t is not quite as convenient. Now we are interested in ﬁnding the sample size n such that α = I1−δ(n −k, k + 1) where δ = Φ log(t) −log(t⋆) + σ⋆Φ−1(1 −p⋆) σ⋆ where t is known. This is equivalent to ﬁnding n that satisﬁes B−1(α; n −k, k + 1) −1 + Φ log(t) −log(t⋆) + σ⋆Φ−1(1 −p⋆) σ⋆ = 0. (23) Given α, n, k, and the reliability standard, (23) is a function of only n. So we can ﬁnd the n that satisﬁes (23) using standard root-ﬁnding methods. In JMP we use Brent’s method to ﬁnd the root of (23), giving us the required sample size given that the length of the demonstration is t. 5.3. Example Demonstration Plan We will go back to the GPS example at the beginning of this section. We want to show that the new GPS system has at least a 90% chance of surviving after 12 months. We assume that the failure times have a lognormal distribution with scale parameter σ = 3 and that we only have n = 50 GPS units to test. The assumption on the scale parameter may come from prior experience or from an expert opinion. If a passing demonstration is deﬁned by zero failures during the study, how long do we need to test the GPS units to assure that there is only a 5% chance of passing the demonstration when the new unit is not as reliable as the standard? Given the requirements, we know that we are working with the lognormal distribution and that α = .05, t⋆= 12, p⋆= .9, σ⋆= 3, k = 0, and n = 50. Plugging these values into (22), we get t = 12exp 3Φ−1 1 −B−1(.05; 50, 1) exp[3Φ−1(1 −.9)] = 5.044, where Φ(·) is the cumulative distribution function of the standard normal distribution. So the GPS units need to be tested for just over ﬁve months. 22 JMP Development How would the demonstration plan change if we decided to deﬁne success as observing two or fewer failures during the study? Then k = 2 and the rest of the parameters stay the same, leading to t = 12exp 3Φ−1 1 −B−1(.05; 50 −2, 1 + 2) exp[3Φ−1(1 −.9)] = 16.673. So tolerating more failures means testing the 50 units for more than three times as long. The longer testing time means greater power to detect an increase in reliability, but may not be feasible due to cost or time constraints. 5.4. Determining power to reject the null hypothesis After planning a reliability demonstration as described in the previous subsection, it is important to con-sider the power to detect an increase in reliability. That is, what is the probability of passing the demon-stration given that the new product is at least as reliable as the standard? If the true reliability of the new product is greater than the standard, then we can express the true reliability at time t⋆as rp⋆= 1 −Φ log(t⋆) −µt σ⋆ where 0 ≤r ≤1/p⋆is the true reliability ratio and µt is the true location parameter of the failure time distribution of the new product. When r ≥1, then the true reliability is at least that of the standard. Then the true location parameter is µt = log(t⋆) −σ⋆Φ−1(1 −rp⋆). We can use the true location parameter to determine the probability of passing the demonstration given a change in reliability. Then power = Pr(pass demonstration \| true reliability at time t⋆is rp⋆) = Pr x ≤k \| x ∼Bin n, 1 −Φ log(t) −log(t⋆) + σ⋆Φ−1(1 −rp⋆) σ⋆ (24) is the formula for determining the probability of passing the demonstration given that the true reliability at time t⋆is rp⋆. For the GPS example in the previous section, we decided to test 50 units for 5.044 months (allowing for zero failures) to determine whether or not the reliability at one year is at least .9. What if the true reliability at one year is .99 for the new product? We would plug r = 1.1 into (24) and ﬁnd that the probability of passing the demonstration is .7998. We also considered a demonstration plan that tested 50 units for 16.673 months and allowed for two failures during the demonstration. If the true reliability at one year is .99, then the probability of passing the demonstration is .9708. So extending the study and allowing for more failures provides a much greater probability of passing the demonstration. For the two demonstration plans considered for the GPS example, Figure 5 plots the probability of passing the demonstration as a function of the improvement in reliability. 5.5. Demo plan summary The reliability demonstration is a straightforward way to compare the reliability of a product to a standard. The basic idea is not complicated. We want to test n units for t units of time and if k or fewer units fail, then the product passes the demonstration, and we conclude that it is at least as reliable as the standard. After making a few distributional assumptions and specifying the reliability standard, we can solve for the necessary length of the study analytically as stated in (22). We cannot solve for the necessary sample size analytically, but a root-ﬁnding algorithm (like Brent’s method) will get the job done. Power and Sample Size 23 1.00 1.02 1.04 1.06 1.08 1.10 1.12 0.0 0.2 0.4 0.6 0.8 1.0 Ratio of improvement Probability of passing demonstration k=0, n=50, t=5.044 k=2, n=50, t=16.673 Fig. 5. Plot of power as a function of the reliability ratio of improvement 6. Reliability Test Plans Suppose that we are interested in estimating the probability of a GPS unit failing before it has been in use for one year. Similarly, we may be interested in the time at which 50% of the units will have failed. In either situation, we would want to achieve some level of precision (based on conﬁdence limits) about the estimate. The reliability test plan is designed to solve this problem. Given some distributional assumptions, it provides the sample size or length of study necessary to be able to estimate either a quantile or failure probability with the desired level of precision. 6.1. What is the objective? In reliability tests, the objective is to either estimate the pth quantile or the cumulative distribution func-tion evaluated at time t. More explicitly, we are interested in q(p) = exp σΦ−1(p) + µ or p(t) = Φ log(t) −µ σ where µ and σ are the location and scale parameters, respectively, and Φ(·) is the standard cumulative distribution function of the assumed failure time distribution. In order to deﬁne precision about ˆ q(p) and ˆ p(t) (which we will do in the following subsection), we need to have conﬁdence intervals about the two quantities. Either way, we ﬁrst have to ﬁnd the expected information matrix for (µ, σ). As usual, we obtain the information matrix by taking the ﬁrst two derivatives of (15). To get the expected information matrix, we use numerical integration tools that exist internally in JMP. From there we can invert the expected information matrix to get the expected covariance matrix Σ. Since we do not necessarily know the sample size, we calculate the expected information using a single observation and can later scale it by the deﬁned sample size. 24 JMP Development Once we have the expected information matrix, we can ﬁnd the expected variance and Wald conﬁ-dence interval (denoted [cl, cu]) for Φ−1(p)σ + µ using the delta method. Then we use [ecl, ecu] as the expected conﬁdence interval for q(p). Note that this interval will not be symmetric because of the expo-nentiation. But the upper and lower endpoints of the interval are guaranteed to be non-negative, which is attractive because we know that log-location scale distributions are only deﬁned for non-negative values. For p(t), ﬁrst we use the delta method to ﬁnd the expected variance of z = log(t)−µ σ . From there we get the Wald conﬁdence interval for z, denoted [zl, zu]. Then we use [Φ(zl), Φ(zu)] as the expected conﬁdence limits for p(t). Once again, the conﬁdence interval around p(t) will not be symmetric be-cause of the Φ(·) transformation, but it is guaranteed to be contained in [0,1] (a favorable property for probabilities). 6.2. Measures of precision In Section 6.1 we found ways to obtain expected conﬁdence intervals. We will use [ql, qu] to denote the conﬁdence interval about q(p) and [pl, pu] to denote the conﬁdence interval about p(t). Now that the conﬁdence intervals are deﬁned, we deﬁne three different measures of precision based on them. The Interval Ratio is simply the square root of the upper conﬁdence limit divided by the square root of the lower conﬁdence limit. For the quantile case, that results in p qu/ql. This precision deﬁnition allows us to say that the upper limit is only, say, 10% greater than the lower limit. Although this precision measure may not seem intuitive, it leads to convenient sample size formulas for quantiles that appear in Meeker and Escobar (1998). We know that the interval ratio will always be greater than one because the upper conﬁdence limit has to be greater than or equal to the lower limit. Larger values of the interval ratio correspond to less precise measurements. The Two-sided Interval Absolute Width is the width of the conﬁdence interval. For example, the precision is qu −ql in the case of the quantile. This precision measure allows us to say that we want the conﬁdence interval around q(p) to be only two, for example. Similarly, the Lower One-sided Interval Absolute Width is the width of the lower conﬁdence limit. More speciﬁcally, if we are interested in a quantile, then the precision is deﬁned as q(p)−ql. The lower interval width is of interest because in some situations we may want to be able to say that “the .9 quantile is at least 100,” for example. For these two measures, values closer to zero correspond to more precise measurements. The last two precision measures can also be stated in a relative way. Instead of stating that you want the conﬁdence interval around q(.9) to be 1, we could say that we want the conﬁdence interval around q(.9) to be .1 ∗q(.9). In other words, we want the width of the conﬁdence interval to be some percentage of the point estimate. JMP will allow for the precision to be stated using these relative measures as well. To keep the internal calculations relatively simple, we simply convert these relative precisions back into absolute precisions and continue as usual. 6.3. Solving for sample size or length of study In the two previous subsections, we looked at objectives of the reliability test plan and ways to deﬁne the desired precision. Given the failure time distributional assumptions, desired precision, and objectives, we are interested in determining the sample size or the length of the test. Given the length of the test, we can solve for the required sample size. Given the sample size, we can solve for required length of the test. Given the length of the study is t, we can ﬁnd the expected information matrix for (µ, σ) using numer-ical integration techniques that already exist in JMP. Then a naive way to ﬁnd the sample size would be to start at say n = 3 and try out different sample sizes until we reach the desired precision level. Calculating the precision for each n is straightforward, using n in conjunction with the expected information matrix gives us the expected covariance matrix for (µ, σ). Then we use the expected covariance matrix to get an expected conﬁdence interval, which leads to the precision calculation. Naturally, as n increases, we Power and Sample Size 25 get more precise estimates. Rather than using this brute force to ﬁnd n, we use Brent’s method, which is much more efﬁcient for ﬁnding roots. Similarly, if we are given n, we could start at say t = .1 and increase t until we reach the desired precision. Once we get to the desired precision level, we stop and declare that t⋆is the necessary study length. Rather than using this brute force approach, we use Brent’s method to go about ﬁnding t⋆much more efﬁciently. But there is a bump in the road when solving for t. For certain combinations of dis-tributional assumptions, sample size, and desired precision, a sufﬁcient study length may not exist. For example, if we are only testing n = 5 units but want an extremely precise measurement of the .5 quantile, even if we never stopped the study (no censoring, so t = ∞) we would not be able to achieve the desired precision. When this happens, the only option is to reconsider the distributional assumptions, increase the sample size, or choose a more realistic precision. So for most of the situations that we will encounter with reliability test plans, we have to use numerical techniques to do the calculations. However, there are several situations where we can ﬁnd an analytical solution for either n or t. The following subsections provide the details. 6.3.1. Sample size calculations for estimated quantiles Luckily when we want to calculate the n that gives us precision δ for estimating qp, we have analytical expressions for each deﬁnition of precision. For the sample size that yields interval ratio δ, we solve for n using δ2 = exp h σΦ−1(p) + µ + p v/nz1−α/2 i exp h σΦ−1(p) + µ − p v/nz1−α/2 i δ2 = exp 2z1−α/2 p v/n 2log(δ) = 2z1−α/2 p v/n √n = z1−α/2 √v log(δ) n = z2 1−α/2v [log(δ)]2 (25) where v is the expected variance of σΦ−1(p) + µ when there is a single observation, α is the conﬁdence level, and z1−α/2 is the 1 −α/2 quantile of the standard normal distribution. For the sample size that yields lower tail precision δ, we solve for n using δ = exp σΦ−1(p) + µ −exp h σΦ−1(p) + µ − p v/nz1−α/2 i δ = γ −γexp −z1−α/2 p v/n γ γ −δ = exp(z1−α/2 p v/n) n = v z1−α/2 log(γ) −log(γ −δ) 2 where z1−α/2, α, and v are deﬁned as in (25) and γ = exp(σΦ−1(p) + µ). For the sample size that provides interval width equal to δ, the simplest strategy is to convert it back into an interval ratio problem. First recall that for interval ratio R, R2 = exp(2z1−α/2 p v/n) 26 JMP Development and therefore log(R) = z1−α/2 p v/n. Then we can convert from the interval width precision measure to the interval ratio precision measure using δ = exp [log(γ) + log(R)] −exp [log(γ) −log(R)] δ = γR −γ R δ γ = R −1 R 0 = R2 −R δ γ −1 which is a prime candidate for being solved by the quadratic formula. Using the quadratic formula yields R = δ/γ + p (δ/γ)2 + 4 2 . (26) We know that R = (δ/γ− p (δ/γ)2 + 4)/2 is not a useful solution because we know that the ratio cannot be negative. So if we want to ﬁnd the sample size that yields expected interval width δ, we can use (26) to convert the problem into an interval ratio problem and use (25) to solve for sample size. 6.3.2. Sample size for estimated failure probabilities When our goal is to estimate the failure probability p(t), we are only able to ﬁnd an analytical expression for n when we are using lower interval tail width as the precision measure. The expected Wald conﬁdence interval for p(t) is h Φ γ −z1−α/2 p v/n , Φ γ + z1−α/2 p v/n i where γ = (log(t) −µ)/σ and v is the expected variance of γ obtained using the delta method on the expected information matrix. Then to ﬁnd the sample size needed for lower tail length δ, we solve δ = Φ(γ) −Φ γ −z1−α/2 p v/n γ −z1−α/2 p v/n = Φ−1 [Φ(γ) −δ] z1−α/2 p v/n = γ −Φ−1 [Φ(γ) −δ] n = vz2 1−α/2 [δ −Φ−1(γ −δ]2 . This formula provides the sample size needed to have expected lower tail width δ. 6.4. Example reliability test plan Suppose we work for a battery company and are interested in estimating the median failure time of our AA batteries. We are going to assume that the failure time (in months) of these batteries has a lognormal distribution with location parameter µ = 2 and scale parameter σ = .75. Figure 6 shows the assumed cumulative distribution function and median value at approximately t = 7.389. We can only wait ﬁve months for the test to run and we want the 95% conﬁdence interval around the median to be approximately δ = 2. So we need to determine the sample size needed to ﬁt these requirements. Power and Sample Size 27 0 5 10 15 20 25 0.0 0.2 0.4 0.6 0.8 1.0 Time Failure probability Fig. 6. Plot of the assumed failure time probability for the battery example. Median failure time is highlighted. First we must convert the problem into an interval ratio problem using (26): R = 2/7.389 + p (2/7.389)2 + 4 2 ≈1.144. Then we can use δ = 1.144 in (25): n = 1.962v [log(1.144)]2 = 356.13 where v = 1.6877 is the approximate variance of the median obtained by using the delta method with the expected covariance matrix. Then we round up and say that testing n = 357 batteries for ﬁve months will yield the desired precision about the median failure time. 6.5. Some important diagnostics The goal of the reliability test is to make inferences about a certain quantile or failure probability. We make these inferences based on the ﬁtted failure time distribution. But in order to ﬁt a distribution to our data, we must have enough information to be able to estimate all of the parameters. Since we use location scale distributions in JMP, that means that we have two parameters to estimate. And that means that we need to observe at least three failures during the reliability test to be able to safely estimate the model parameters. So when planning the reliability test, it is crucial to consider the number of failures that are expected during the test. If there are no failures during the test, the test will have been a waste of our resources. In JMP we use two diagnostics to help us decide whether or not the test is going to be worthwhile. First we report the expected number of failures during the study E(number of failures) = nPr(unit fails before time ts) = nΦ log(ts) −µ σ , 28 JMP Development where ts is the length of the study, Φ(·) is the assumed CDF, n is the number of units being tested, and µ and σ are the assumed location and scale parameters, respectively. We also provide the user with the probability of observing fewer than three failures during the test: Pr(fewer than three failures) = Pr x ≤2 \| x ∼Bin n, Φ log(ts) −µ σ . These two diagnostics are very useful in determining whether or not the test is worth pursuing. For example, suppose again that we are making batteries and that we are interested in the median failure time. We assume (from past experience) that the failure time in months is lognormally distributed with µ = 1.25 and σ = 1.5. Unfortunately because of cost constraints, we only want to test 20 batteries for one month. Then we compute our diagnostics E(number of failures) = 20Φ log(1) −1.25 1.5 ≈ 4.05 where Φ(·) is the CDF of the standard normal CDF. And the probability of observing fewer than three failures is Pr(fewer than three failures) = Pr x ≤2 \| x ∼Bin 20, Φ log(1) −1.25 1.5 ≈ .20. So with only testing 20 batteries for one month, we expect to observe four failures, and there is a 20% chance that the test will not yield enough information to make any inferences. Given this information, we would probably decide not to risk wasting our time on this test and instead lobby for more resources. 6.6. Test plan summary The reliability test involves testing a sample of units for a set period of time and recording failure times. Once the data are collected, we ﬁt a distribution to the data and make inferences about failure probabilities or population quantiles. The test plan feature in JMP allows us to determine either the sample size or length of study necessary to achieve a desired level of precision about those inferences (based on the expected conﬁdence interval). We must keep in mind, however, that for a given sample size, we can only reach a certain level of precision, even if we let all of the units run until failure. When planning a reliability test, it is also important to make sure that we expect to observe enough failures to be able to ﬁt the assumed failure distribution. 7. Summary The Power and Sample Size Calculator in JMP provides tools for planning experiments in a variety of settings. This document has covered some of the theory and resulting formulas that are used in the Power and Sample Size Calculator. The example JSL code and functions are useful for gaining better insight into these features. References Agresti, A. and Caffo, B. (2000), “Simple and effective conﬁdence intervals for proportions and differences of proportions result from adding two successes and two failures.” The American Statis-tician, 54, 280-288. Power and Sample Size 29 Agresti, A. and Coull, B. (1998), “Approximate is better than ‘exact’ for interval estimation of binomial proportions.” The American Statistician, 52, 119-126. Meeker, W.Q. and Escobar, L. A. (1998), Statistical Methods for Reliability Data. Wiley: New York. National Institute of Standards and Technology (2009), “Engineering Statis-tics Handbook”. Gaithersburg: National Institute of Standards and Technology, (accessed 10 June 2009). 30 JMP Development A. Power and sample size formula cheat-sheet This appendix provides a convenient summary of some of the formulas presented in the body of the paper. Throughout the appendix, we use α to denote the testing level ‘Alpha’. One Sample Mean σ = Std Dev p = Extra Parameters δ = Difference to Detect n = Sample Size Power = 1 −F f1−α, 1, n −p −1, nδ2 σ2 • f1−α is the 1 −α quantile of the F(1,n −p −1) distribution. • F(x, df1, df2, nc) is the cumulative distribution function of the non-central F distribution with degrees of freedom df1 and df2 and non-centrality parameter nc evaluated at x. Analytical solutions for δ and n do not exist, so we use numerical techniques in conjunction with the power formula to solve for them. Power and Sample Size 31 Two Sample Means σ = Std Dev p = Extra Parameters δ = Difference to Detect n = Sample Size per group Power = 1 −F f1−α, 1, 2n −p −2, nδ2 2σ2 • f1−α is the 1 −α quantile of the F(1,2n −p −2) distribution. • F(x, df1, df2, nc) is the cumulative distribution function of the non-central F distribution with degrees of freedom df1 and df2 and non-centrality parameter nc evaluated at x. Analytical solutions for δ and n do no exist, so we use numerical techniques in conjunction with the power formula to solve for them. k Sample Means σ = Std Dev p = Extra Parameters µ1, µ2, . . . , µk : assumed mean for each of the k groups n = Sample Size per group Power = 1 −F f1−α, k −1, kn −p −k, n Pk j=1(µj −¯ µ)2 σ2 ! • ¯ µ = Pk j=1 µj/k • f1−α is the 1 −α quantile of the F(k-1,kn −p −k) distribution. • F(x, df1, df2, nc) is the cumulative distribution function of the non-central F distribution with degrees of freedom df1 and df2 and non-centrality parameter nc evaluated at x. 32 JMP Development One Sample Standard Deviation σ0 = Hypothesized Standard Deviation δ = Difference to Detect n = Sample Size 1 −β = Power Case 1: Alternative Standard Deviation Is Larger 1 −β = 1 −F σ2 0χ1−α (σ0 + δ)2 , n −1 δ = σ0 rχ1−α χβ −σ0 Case 2: Alternative Standard Deviation Is Smaller 1 −β = 1 −F σ2 0χα (σ0 + δ)2 , n −1 δ = σ0 r χα χ1−β −σ0 • χp is the pth quantile of the χ2 n−1 distribution. • F(x, n −1) is the CDF of the χ2 n−1 distribution evaluated at x. An analytical solution for n does not exist for either case, so numerical methods are used to solve for sample size. Power and Sample Size 33 One Sample Proportion p = Proportion p0 = Null Proportion n = Sample Size Here we will show how to calculate power for the two-sided alternative. Calculations for the one-sided alternative are carried out in a similar manner. Power = n X y=0 Pr n Y = y Y ∼Binomial(n, p) o I {T(y) ≥χ1−α} where T(y) = (ˆ p −p0)2 ˆ p(1−ˆ p) n+4 and ˆ p = y + 2 n + 4. • χ1−α is the 1 −α quantile of the χ2 1 distribution. • I {T(y) ≥χ1−α} equals one when T(y) ≥χ1−α and zero otherwise. Here we do not have closed-form expressions to solve for n or p0, so numerical techniques are used to solve for them. 34 JMP Development Two Sample Proportions p1 = Proportion 1 p2 = Proportion 2 δ0 = Null Difference in Proportion n1 = Sample Size 1 n2 = Sample Size 2 Here we will show how to calculate power for the two-sided alternative. Calculations for the one-sided alternative are carried out in a similar manner. Power = n1 X y1=0 n2 X y2=0 Pr (Y1 = y1) Pr (Y2 = y2) I {T(y1, y2) ≥χ1−α} where Y1 ∼Binomial(n1, p1) , Y2 ∼Binomial(n2, p2), T(y1, y2) = (ˆ p1 −ˆ p2 −δ0)2 ˆ p1(1−ˆ p1) n1+2 + ˆ p2(1−ˆ p2) n2+2 , and ˆ pj = yj + 1 nj + 2 for j = 1, 2. • χ1−α is the 1 −α quantile of the χ2 1 distribution. • I {T(y1, y2) ≥χ1−α} equals one when T(y1, y2) ≥χ1−α and zero otherwise. Here we do not have closed-form expressions to solve for n or p0, so numerical techniques are used to solve for them. Power and Sample Size 35 Counts per Unit λ0 = Baseline Count per Unit δ = Difference to Detect n = Sample Size 1 −β = Power Power = 1 −Φ Z1−α −δ p n/λ0 p (λ0 + δ)/λ0 ! n = λ0 δ2 Z1−α −Zβ r λ0 + δ λ0 !2 • Φ() is the standard normal CDF. • Z1−p is the 1 −p quantile of the standard normal distribution. An analytical solution for δ does not exist, so numerical methods must be used to solve for δ given power and n. Sigma Quality Level σq = Sigma Quality Level n = number of opportunities d = number of defects σq = Z1−d/n + 1.5 d = n 1 −Φ σq −1.5 n = d 1 −Φ σq −1.5 −1 • Φ() is the standard normal CDF. • Z1−d/n is the 1 −d/n quantile of the standard normal distribution.
16392	https://helpingwithmath.com/graphing-linear-inequalities/	Skip to content Helping with Math Home » Math Theory » Algebra » Graphing Linear Inequalities: A Comprehensive Guide Graphing Linear Inequalities: A Comprehensive Guide Introduction Graphing linear inequalities is a fundamental topic in algebra that teaches students to represent and analyze relationships between variables. These inequalities are essential for solving real-world problems, enabling us to express various possibilities rather than one exact value. This article shall tackle graphing linear inequalities, their key concepts, illustrative examples, and real-life applications. Grade Appropriateness Graphing linear inequalities is typically taught in Pre-algebra and Algebra I courses, usually taken by middle school students (grades 6-8) and early high school (grades 9-10). Math Domain The study of graphing linear inequalities falls within the broader mathematical domain of algebra. Applicable Common Core Standards The following Common Core Standards for Mathematics apply to the topic of graphing linear inequalities: 7.EE.B.4: Use variables to represent quantities in a real-world or mathematical problem and construct simple equations and inequalities to solve problems by reasoning about the quantities. 6.EE.B.8: Solve real-world and mathematical problems by writing and solving equations of the form x + p = q and px = q for cases in which p, q and x are all nonnegative rational numbers 8.EE.C.7: Solve linear equations in one variable. Definition A linear inequality is an inequality that involves linear expressions. It can be written in the form: Ax + By < C,Ax + By ≤ C,Ax + By > C, orAx + By ≥ C,where, x and y are variables and,A, B, and C are constants. Graphing linear inequalities is the process of plotting the solution set of the inequality on a coordinate plane. Key Concepts Linear inequality: A mathematical statement that involves linear expressions connected by inequality symbols (<, >, ≤, or ≥). Graph: An organized visual representation of data and values using lines, shapes, and colors. Graphs may also be called charts, usually with two or more points showing the relationship between values. Coordinate plane: A two-dimensional plane with x and y axes to represent and analyze relationships between variables. Boundary line: The line that divides the coordinate plane into two half-planes, representing the equation obtained by replacing the inequality symbol with an equal sign. Shading: The process of highlighting the region in the coordinate plane containing the inequality’s solution set. Key Steps Graph the boundary line. Change the inequality to the equality symbol and isolate the variable y. Write the equation in slope intercept form (y=mx+b). The boundary line that you will graph will depend on the inequality (>,≥, <, ≤ ). A less than < or greater than (>) is a strict inequality. Therefore, you need to draw a dashed line. A less than or equal to (≤ ) or greater than or equal to ( ≥) sign is a non-strict inequality. Therefore, you need to draw a solid line. Select a test point, not on the boundary line. Select a point not on the line and test its coordinates in the original inequality to determine whether to shade above or below it. Substitute the test point for the inequality. The region above or below the line containing the point will be shaded if the coordinates of the point satisfy the inequality. Shade one side of the boundary line. The solution set for the inequality is the entire shaded region. Discussion with Illustrative Examples Example 1: Graph the linear inequality y> 2x – 1 Solution: Graph the boundary line: y = 2x – 1. Since the equation is written in the slope intercept form y=mx+b, therefore, m (slope)=2 while b (y-intercept)=-1 Since the given is a strict inequality greater than (>), then a dashed line must be used. Select a test point not on the boundary line, like (0,0). Substitute the test point into the inequality: 0 > 2(0) – 1, which is true. Since the test point satisfies the inequality, shade the region containing the test point. Example 2: Graph the linear inequality 3x + 4y ≤ 12 Solution: Graph the boundary line: 3x + 4y = 12. Isolate the variable y and write the equation in the slope-intercept form (y=mx+b). 3x+4y=12 4y=-3x+12 y=-¾x+3 Select a test point not on the boundary line, like (0,0). Substitute the test point into the inequality: 3(0) + 4(0) ≤ 12, which is true. Shade the side of the boundary line that includes the test point since the test point satisfies the given inequality. Since the given is a non-strict inequality greater than or equal to (≤), then a solid line must be used. Example 3: Graph the linear inequality x – 2y > 4. Solution: Graph the boundary line: x – 2y = 4. Isolate the variable y and write the equation in the slope-intercept form (y=mx+b). x-2y=4 -2y=-x+4 y=½x-2 Select a test point not on the boundary line, like (0,0). Substitute the test point into the inequality: 0 – 2(0) > 4, which is false. Shade the region of the boundary line that does not include the test point because the test point does not satisfy the inequality. Since the given is a strict inequality greater than (>), then a dashed line must be used. Real-life Application with Solution During his break, Joseph has a maximum meal budget of $9. He must pay $2 for drinks and $3 for food. How many different drinks and food combinations are available for purchase? Solution Let x be the number of drinks, and y be the number of food. Since Joseph has a maximum meal budget of $9, we will use the inequality 2x+3y≤9. Below is the graph of 2x+3y=9, which in slope-intercept form is y=-2⁄3x+3. Substituting the test point 0,0 into the inequality, 2x+3y ≤ 9 20+30 ≤ 9 0 ≤ 9 True Shade the side of the boundary line that includes the test point because the test point (0,0) satisfies the inequality. Since the given is a non-strict inequality, less than or equal to (≤), then a solid line must be used. Since we want to identify the number of possible drinks and food combinations, we must locate points in the shaded region that are positive integers. The table below shows what these points represent. \| \| \| Combinations \| \| Number of drinks($2) \| Number of food($3) \| Total \| \| 1 drink = $2 \| 1 food = $3 \| $5 \| \| 1 drink = $2 \| 2 food = $6 \| $8 \| \| 2 drinks = $4 \| 1 food = $3 \| $7 \| \| 3 drinks = $6 \| 1 food = $3 \| $9 \| Notice that when we substitute the x and y values to the inequality, the total results to less than or equal to $9. Therefore, there are four different combinations formed for drinks and food. Practice Test Graph the linear inequality y < x + 3. Graph the linear inequality 2x – y ≥ 6. A company produces two products, A and B. They need at least three units of A and four units of B to meet demand. If x is the number of units (A) and y represents the number of units (B), graph the inequality representing the minimum production requirements. Graph the system of linear inequalities: x + y ≤ 5, x ≥ 1, y ≥ 2. The sum of two numbers is less than 10, and the first is at least 3. Graph the inequality representing this situation. Frequently Asked Questions (FAQs) Can linear inequalities have infinitely many solutions? Yes, the solution set for a linear inequality typically consists of infinitely many points in the coordinate plane. How do I determine whether to use a solid or dashed line for the boundary line? Use a solid line when the inequality includes “≤” or “≥” and a dashed line when it has “<” or “>.” How do I graph a linear inequality with two variables? First, graph the boundary line, the equation you get by replacing the inequality symbol with an equal sign. Then, choose a test point not on the boundary line and substitute it into the inequality. Shade the side of the boundary line that satisfies the inequality. What is the difference between a linear equation and a linear inequality? A linear equation is a statement that two expressions are equal, while a linear inequality is a statement that one expression is >, <, ≥, or ≤ to another expression. How to solve a given system of linear inequalities? In solving a system of linear inequalities, graph each inequality individually, and find the region where the shaded areas of all inequalities overlap. This overlapping region represents the solution set for the system. Recommended Worksheets Graphing Linear Inequalities (Civil Rights Movement Themed) Math WorksheetsGraphing and Solving Systems of Linear Equations in Two Variables 8th Grade Math WorksheetsSolving Word Problems Involving Linear Equations and Linear Inequalities 7th Grade Math Worksheets Browse All Worksheets Link/Reference Us We spend a lot of time researching and compiling the information on this site. If you find this useful in your research, please use the tool below to properly link to or reference Helping with Math as the source. We appreciate your support! Graphing Linear Inequalities: A Comprehensive Guide "Graphing Linear Inequalities: A Comprehensive Guide". Helping with Math. Accessed on September 28, 2025. "Graphing Linear Inequalities: A Comprehensive Guide". Helping with Math, Accessed 28 September, 2025. Graphing Linear Inequalities: A Comprehensive Guide. Helping with Math. Retrieved from Additional Algebra Theory: Equation of a Straight Line Absolute Value Point-Slope Form Rate of Change & Slope of a Line Addition & Subtraction Of Square Roots Quadratic Equation Equation Identifying Arithmetic Patterns of Numbers Symbols in Algebra Union of Sets Latest Worksheets The worksheets below are the mostly recently added to the site. Permutation Math Activities Hypothesis Testing Math Activities Estimating Population Mean Math Activities Triangle Congruence Math Activities Area of the Sector Math Activities Arc Length Math Activities Rational Functions Math Activities Piecewise Functions Math Activities Evaluating Functions Math Activities Relative Frequencies Math Activities
16393	http://mathcentral.uregina.ca/QQ/database/QQ.09.06/richard1.html	\| \| \| \| \| --- --- \| \| \| \| \| \| \| \| \| --- \| \| SEARCH \| HOME \| \| \| \| \| \| \| --- \| \| Math Central \| Quandaries & Queries \| \| \| \| \| \| \| Subject: Math/AreaName: Richard Who are you: Other An octagon is inscribed in a square so that the vertices of the octagon trisect the sides of the square. The perimeter of the square is 108 centimeters. What is the number of square centimeters in the area of the octagon? This information is for my granddaughter. I am trying to help with her school work. If you could show me how to solve this problem, it would be greatly appreciated. Thanking you in advance for your cooperation. \| \| \| This is best done by three simple steps:- the answer you want is the area of the square minus the area of four 'corner triangles' (see the diagram): Now for the square: the four sides are each 1/4 of the perimeter (the sum of the sides): 27 centimeters.From this you can calculate the area of the whole square. For the triangles: the short sides are each 1/3 of the sides of the square: 9 cm.From this, and the right angle in the corner, you can calculate the area of each of the triangles.Alternatively, each pair of the triangles forms a square of side 9 cm. So you now have the pieces. You can now complete the pieces and subtract. There is a second way to check this answer: With this you can see that the whole square is divided into 9 smaller squares. Four triangles (or two of the squares) are left out. So you want the area of seven of the squares, each of which is 9 cm by 9 cm. Again, you can complete the calculations! Walter Whiteley \| \| \| \| \| \| \| Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. \|
16394	https://www.wcupa.edu/sciences-mathematics/mathematics/sParsell/documents/mat161/MAT161Rogawski.pdf	MAT 161—West Chester University—Fall 2010 Notes on Rogawski’s Calculus: Early Transcendentals Scott Parsell §2.1—Limits, Rates of Change, and Tangent Lines Example 1. The distance in feet that an object falls in t seconds under the inﬂuence of gravity is given by s(t) = 16t2. If a pumpkin is dropped from the top of a tall building, calculate its average velocity (a) between t = 3 and t = 4 (b) between t = 3 and t = 3.5 (c) between t = 3 and t = 3.1 (d) between t = 3 and t = 3.01 (e) between t = 3 and t = 3.001 What appears to be the instantaneous velocity of the pumpkin at t = 3? In general, if an object’s position is given by s(t), then the average velocity over the interval [t0, t1] is ∆s ∆t = s(t1) −s(t0) t1 −t0 . By allowing the point t1 to move closer and closer to t0, we are able to estimate the instan-taneous velocity at t0. This process of computing what happens as t1 gets very close to (but never equal to) t0 is an example of taking a limit. Even more generally, the average rate of change of the function y = f(x) over the interval [x0, x1] is ∆y ∆x = f(x1) −f(x0) x1 −x0 . Geometrically, it is the slope of the secant line connecting the points (x0, f(x0)) and (x1, f(x1)). By allowing the point x1 to move closer and closer to x0, we obtain the tangent line to the graph of y = f(x) at the point x = x0. The slope of the tangent line at x0 is the instantaneous rate of change of y with respect to x at the point x0. Example 2. Find the slope of the secant line on the graph of f(x) = ex for each of the following intervals. (a) [0, 0.5] (b) [0, 0.1] (c) [0, 0.01] (d) [0, 0.001] What appears to be the slope of the tangent line to the graph at x = 0? What is the equation of this tangent line? 2 §2.2—Limits: A Numerical and Graphical Approach As we saw in the previous section, in order to make sense of instantaneous rates of change, we need to understand the concept of limits. Deﬁnition. If we can make f(x) as close as we like to L by taking x suﬃciently close to c, then we say that the limit of f(x) as x approaches c is equal to L, and we write lim x→c f(x) = L. In other words, this statement means that the quantity \|f(x) −L\| becomes arbitrarily small (but not necessarily zero) whenever x is suﬃciently close to (but not equal to) c. Note: In the previous section, we used numerical data to analyze lim t→3 16t2 −144 t −3 and lim x→0 ex −1 x . Important points: 1. We must consider values on both sides of c. 2. The limit may or may not exist. 3. The value of f at x = c is irrelevant. Example 1. Use numerical and graphical data to guess the values of the following limits. (a) lim x→1 x2 −1 x −1 (b) lim x→0 sin x x 3 One-sided limits lim x→c+ f(x) = L means f(x) approaches L as x approaches c from the right lim x→c−f(x) = L means f(x) approaches L as x approaches c from the left We have lim x→c f(x) = L if and only if lim x→c+ f(x) = L and lim x→c−f(x) = L. Example 2. Compute each of the following limits for the function graphed below. (a) lim x→2+ f(x) (b) lim x→2−f(x) (c) lim x→2 f(x) (d) lim x→5+ f(x) (e) lim x→5−f(x) (f) lim x→5 f(x) Inﬁnite Limits We write lim x→c f(x) = ∞if the values of f(x) become arbitrarily large and positive as x approaches c. Similarly, we write lim x→c f(x) = −∞if the values of f(x) become arbitrarily large and negative as x approaches c. Similar deﬁnitions apply to one-sided inﬁnite limits. Notice that if lim x→c+ f(x) = ±∞or lim x→c−f(x) = ±∞then the line x = c is a vertical asymptote for the graph of y = f(x). Example 3. Evaluate each of the following limits. (a) lim x→0− 1 x (b) lim x→2 1 (x −2)4 (c) lim x→5+ x + 3 x −5 (d) lim x→0+ ln x 4 §2.3—Basic Limit Laws While numerical data and graphs often provide useful intuition about limits, relying exclusively on this type of information can give misleading results if a function has subtle behavior that is not captured by our data. Therefore, we need to develop explicit methods for computing limits. The following example illustrates some of the basic principles. Example 1. Evaluate each of the following limits. (a) lim x→4 7 (c) lim x→4 (3x + 7) (b) lim x→4 x From the reasoning in parts (a) and (b) we see that for any constants k and c lim x→c k = k and lim x→c x = c. We can also generalize part (c). Assuming lim x→c f(x) and lim x→c g(x) both exist, we have (1) Sums and Diﬀerences: lim x→c (f(x) ± g(x)) = lim x→c f(x) ± lim x→c g(x) (2) Constant Multiples: For any constant k, we have lim x→c kf(x) = k lim x→c f(x) (3) Products: lim x→c f(x)g(x) = ( lim x→c f(x) ) ( lim x→c g(x) ) (4) Quotients: lim x→c f(x) g(x) = lim x→c f(x) lim x→c g(x) provided that lim x→c g(x) ̸= 0. Example 2. Use the above limit laws to evaluate lim x→2 x4 + 3x2 4x + 5 . 5 §2.4—Limits and Continuity The reasoning from Example 2 of §2.3 shows that the limit of any polynomial or rational function can be found by direct substitution, provided the limit of the denominator is not zero. This property is known as continuity and is shared by many familiar types of functions. Deﬁnition. We say that f is continuous at x = c if lim x→c f(x) = f(c). This implicitly requires checking three things: (i) f(c) exists (ii) lim x→c f(x) exists (iii) the numbers in (i) and (ii) are equal. If c is an endpoint of the domain, we use the appropriate one-sided limit instead. A point where f is not continuous is called a discontinuity. Example 1. At what points does the function graphed below fail to be continuous? Notice that there are various ways in which a function can fail to be continuous–for example, a hole in the graph, a ﬁnite jump, or a vertical asymptote. Example 2. Sketch the graphs of the following functions near x = 2. (a) f(x) = x2 −4 x −2 (c) f(x) = x2 + 4 x −2 (b) f(x) = x2 −4 \|x −2\| 6 Useful facts: 1. Sums, diﬀerences, products, quotients, powers, roots, and compositions of continuous functions are continuous at all points of their domains. 2. Polynomials, rational functions, root functions, trigonometric functions, exponentials, and logarithms are continuous at all points of their domains. Example 3. For what values of x is the function f(x) = 2 + sin(x2) √ x2 −4 −1 continuous? Example 4. For what values of x is the function f(x) = { x + 1 if x ≤3 x2 −1 if x > 3 continuous? Note that the deﬁnition of f(x) in Example 4 automatically ensures that the limit as x approaches 3 from the left is equal to f(3), so we say that f is left-continuous at x = 3. Example 5. For what values of a is f(x) = { ax + 1 if x ≤3 ax2 −1 if x > 3 continuous at x = 3? 7 §2.5—Evaluating Limits Algebraically We saw in Example 2 of §2.4 that the behavior of a function near a point where both the numerator and denominator approach zero can sometimes be analyzed by cancelling a common factor. These “0/0” limits occur frequently when dealing with instantaneous rates of change, so we illustrate here some of the algebraic manipulations that can be useful. Example 1. Evaluate lim x→3 x2 −9 x2 −2x −3. Example 2. Evaluate lim x→25 25 −x 5 −√x. Example 3. Evaluate lim x→1 x3 −1 x −1 . 8 Example 4. Evaluate lim x→4 √2x + 1 −3 x −4 . Example 5. Evaluate lim h→0 f(2 + h) −f(2) h , where f(x) = 1 x. Note: This is the instantaneous rate of change of f at x = 2. 9 §2.6—Trigonometric Limits Example 1. Calculate lim x→0 x2 sin(1/x). The reasoning used in Example 1 is a special case of the following theorem. The Squeeze Theorem. Suppose that l(x) ≤f(x) ≤u(x) for all x ̸= c in some open interval containing c. If lim x→c l(x) = lim x→c u(x) = L then lim x→c f(x) = L. An important application of the Squeeze Theorem is the following result, which we pre-dicted numerically in Example 1(b) of §2.2. Theorem. If θ is measured in radians, then lim θ→0 sin θ θ = 1. 10 Example 2. Evaluate the following limits. (a) lim x→0 sin 3x x (b) lim x→0 tan x 2x Example 3. Evaluate lim θ→0 1 −cos θ θ Review of the Unit Circle. If your trigonometry is rusty, now would be a good time to check out §1.4 in the text. In particular, recall that the x and y coordinates of a point on the unit circle at an angle θ (measured counter-clockwise from the positive x-axis) are given by x = cos θ and y = sin θ. Hence the Pythagorean Theorem immediately gives the identity cos2 θ + sin2 θ = 1. Moreover, by generalizing to a circle of radius r, we obtain the familiar right-triangle rela-tionships given by SOHCAHTOA. You should also remember the values of sin θ and cos θ at the special angles in the ﬁrst quadrant: θ 0 π/6 π/4 π/3 π/2 sin θ 0 1/2 √ 2/2 √ 3/2 1 cos θ 1 √ 3/2 √ 2/2 1/2 0 It’s then easy to move to other quadrants using a reference angle, remembering that sine is positive in Quadrants I and II and cosine is positive in Quadrants I and IV. Everything about the other four trig functions follows from what we know about sine and cosine via tan θ = sin θ cos θ, cot θ = cos θ sin θ , sec θ = 1 cos θ, csc θ = 1 sin θ. In particular, it’s easy to show that 1 + tan2 θ = sec2 θ and cot2 θ + 1 = csc2 θ. 11 §2.7—The Intermediate Value Theorem An important property of continuous functions is that they do not “skip over” any y-values. The precise statement is as follows: The Intermediate Value Theorem. Suppose that f is continuous on [a, b] and f(a) ̸= f(b). Then for every value M between f(a) and f(b), there exists at least one value c in the interval (a, b) for which f(c) = M. Example 1. Show that the function f(x) = x4 + x2 + 1 takes on the value 10 for some x in the interval (1, 2). Root-ﬁnding. An important corollary of the IVT is that if f is continuous on [a, b] and f(a) and f(b) have opposite signs, then the equation f(x) = 0 has a solution in (a, b). By applying this repeatedly, one can ﬁnd roots of equations to arbitrary accuracy. The algorithm, known as the Bisection Method, is illustrated in the following example. Example 2. Find an interval of length 1/4 in which the equation x3 + x + 1 = 0 has a solution. 12 §3.1—Deﬁnition of the Derivative The slope of the secant line connecting the points P(a, f(a)) and Q(a + h, f(a + h)) on the graph of f is ∆y ∆x = f(a + h) −f(a) h . This is the average rate of change of f over the interval [a, a + h]. The slope of the tangent line to the curve y = f(x) at the point P(a, f(a)) is f ′(a) = lim h→0 f(a + h) −f(a) h , provided the limit exists. This is the instantaneous rate of change of f at x = a and is also called the derivative of f at x = a. By setting x = a + h, we can alternatively write f ′(a) = lim x→a f(x) −f(a) x −a , which is sometimes easier to work with. Example 1. Find the derivative of the function f(x) = 16x2 at x = 3. 13 Example 2. Find the equation of the tangent line to f(x) = √x + 1 at the point (3, 2). Example 3. Find the equation of the tangent line to f(x) = 1/x2 at the point (−1, 1). 14 §3.2—The Derivative as a Function The derivative of the function f(x) with respect to the variable x is the function f ′ whose value at x is f ′(x) = lim h→0 f(x + h) −f(x) h . The process of calculating a derivative is called diﬀerentiation. We sometimes write dy dx or d dx[f(x)] instead of f ′(x). Example 1. Use the above deﬁnition to ﬁnd the derivative of the following functions. (a) f(x) = x3 (b) f(x) = 1 √x 15 Some Basic Rules: 1. Derivative of a Constant Function: d dx(c) = 0 2. Power Rule: d dx(xn) = nxn−1 when n is a constant. 3. Derivative of the Natural Exponential Function: d dx(ex) = ex 4. Constant Multiples: d dx[cf(x)] = cf ′(x) 5. Sums and Diﬀerences: d dx[f(x) ± g(x)] = f ′(x) ± g′(x) Example 2. Find the derivative of each of the following functions. (a) f(x) = 3x5 −4x3 + 5 x2 + 6 (b) f(x) = 4ex + 10 3 √x + e x3 A function f(x) is diﬀerentiable at x = c if f ′(c) exists. There are several ways a function can fail to be diﬀerentiable: 1. Corner 2. Cusp 3. Vertical tangent 4. Discontinuity Theorem. Diﬀerentiability implies continuity. In other words, if f has a derivative at x = c then f is continuous at x = c. The converse of this theorem is false! Continuity does NOT imply diﬀerentiability—see the corner, cusp, and vertical tangent examples above. Example 3. At what points does the function graphed below fail to be diﬀerentiable? 16 §3.3—The Product and Quotient Rules Diﬀerentiating products and quotients is not quite as simple as diﬀerentiating sums and diﬀerences. For example, consider writing x5 as the product x3 · x2. The product of the derivatives of the two factors in the second expression is 3x2 · 2x = 6x3, but we know that the derivative of this product is really 5x4. This shows that the derivative of a product is NOT equal to the product of the derivatives. Instead we have: The Product Rule: d dx[f(x)g(x)] = f(x)g′(x) + g(x)f ′(x) Example 1. Find the derivative of each of the following functions. (a) h(x) = (x2 + 3x + 1)ex (b) P(x) = (3x2/3 + 2ex)(4 −x−5) Example 2. A company’s revenue from t-shirt sales is given by R(x) = xq(x), where q(x) is the number of shirts it can sell at a price of $x apiece. If q(10) = 200 and q′(10) = −13, what is R′(10)? 17 Likewise, simple examples show that the derivative of a quotient is NOT equal to the quotient of the derivatives. The correct result is as follows: The Quotient Rule: d dx [f(x) g(x) ] = g(x)f ′(x) −f(x)g′(x) [g(x)]2 Example 3. Find the derivative of each of the following functions. (a) h(x) = x x5 + 3 (b) F(x) = xex 7 −√x Example 4. Find the equation of the tangent line to the curve y = ex x + 2 at the point (0, 1/2). 18 §3.4—Rates of Change Recall that the instantaneous rate of change of y = f(x) with respect to x at x = a is dy dx x=a = f ′(a) = lim h→0 f(a + h) −f(a) h , provided the limit exists. This is the limit of the average rates of change of f over smaller and smaller intervals of the form [a, a + h]. Some examples: s(t) = position s′(t) = velocity v(t) = velocity v′(t) = acceleration Q(t) = charge Q′(t) = current W(t) = work/energy W ′(t) = power P(t) = population P ′(t) = population growth rate R(x) = revenue from producing x units R′(x) = marginal revenue C(x) = cost of producing x units C′(x) = marginal cost Example 1. The position (in meters) of a particle moving along the s-axis after t seconds is given by s(t) = 1 3t3 −2t2 + 3t for t ≥0. (a) When is the particle moving forward? Backward? (b) When is the particle’s velocity increasing? Decreasing? 19 Example 2. A rock thrown vertically upward from the surface of the moon at a velocity of 24 m/s reaches a height of s = 24t −0.8t2 meters in t seconds. (a) Find the rock’s velocity and acceleration at time t. (b) How long does it take the rock to reach its highest point? What is its maximum height? Example 3. Suppose that the cost of producing x washing machines is C(x) = 2000 + 100x −0.1x2. (a) Find the marginal cost when 100 washing machines are produced. (b) Compare the answer to (a) with the cost of producing the 101st machine. 20 §3.5—Higher Derivatives When we compute the derivative of a function f(x), we get a new function f ′(x). If we take the derivative of the function f ′(x), we get another new function, which is called the second derivative of f(x) and denoted f ′′(x). For example, the derivative of position with respect to time is velocity, and the derivative of velocity with respect to time is acceleration; therefore we say that acceleration is the second derivative of position: a(t) = v′(t) = s′′(t). We can continue this process to get higher derivatives: f ′(x) = d dx[f(x)] = dy dx (1st derivative) f ′′(x) = d dx[f ′(x)] = d2y dx2 (2nd derivative) f ′′′(x) = d dx[f ′′(x)] = d3y dx3 (3rd derivative) f (4)(x) = d dx[f ′′′(x)] = d4y dx4 (4th derivative) and so on. Example 1. Find the ﬁrst, second, third, and fourth derivatives of the function f(x) = x10 −5x4 + 3x + 2. Example 2. Find f ′′(1) for the function f(x) = x3ex. 21 §3.6—Derivatives of Trigonometric Functions Our goal in this section is to ﬁnd formulas for the derivatives of the 6 basic trig functions: d dx(sin x) = d dx(cot x) = d dx(cos x) = d dx(sec x) = d dx(tan x) = d dx(csc x) = To ﬁnd a formula for the derivative of the function f(x) = sin x we must return to the deﬁnition of the derivative in terms of a limit, which we studied in §3.2: f ′(x) = lim h→0 f(x + h) −f(x) h . Here the trigonometric identity sin(x + h) = sin x cos h + cos x sin h will help us get started, and we will need to recall two special trigonometric limits that we calculated back in §2.6: lim h→0 sin h h = 1 and lim h→0 cos h −1 h = 0. 22 Example 1. Find the derivatives of the following functions. (a) f(x) = x2 sin x + 2 cos x (b) g(t) = et t −sin t Example 2. Use the quotient rule to ﬁnd the derivatives of tan x and sec x. 23 Example 3. Find the derivatives of the following functions. (a) f(x) = ex sec x + 2 tan x (b) r(θ) = 1 + cot θ θ3 −4 csc θ §3.7—The Chain Rule How do we diﬀerentiate compositions of functions like e2x, cos(x2), √ x3 + 1, or sin4 x? Suppose that y = f(u) and u = g(x), so that y = f(g(x)). It is helpful to think of f as the “outer” function and g as the “inner” function. If we have du dx = g′(x) = 2 and dy du = f ′(u) = 3, then a 1 unit change in x gives approximately 2 units change in u, which then gives approximately 6 units change in y. This heuristic argument suggests that dy dx = dy du · du dx = f ′(u)g′(x) = f ′(g(x))g′(x). This is in fact true whenever f and g are diﬀerentiable and is known as the Chain Rule. The Chain Rule: d dx[f(g(x))] = f ′(g(x))g′(x) In words, this says that the derivative of a composition is the derivative of the outer function, evaluated at the inner function, times the derivative of the inner function. 24 Example 1. Find the derivatives of the following functions. (a) h(x) = e2x (b) h(x) = cos(x2) (c) h(x) = √ x3 + 1 (d) h(x) = sin4 x Example 2. Find formulas for the velocity and acceleration of a particle whose position is given by s(t) = 5 cos(2t). 25 In many problems, the Chain Rule must be applied in combination with other rules such as the product and quotient rules. It is also possible to have a composition within a composition, f(g(h(x)), which requires more than one application of the Chain Rule. The following examples illustrate these more challenging situations. Example 3. Find dy dx for the following functions. (a) y = ex2 cos 3x (b) y = sin( √ x4 + 1) (c) y = tan(e2x) (x2 + 1)6 (d) y = √ 1 + √ 1 + √x (e) y = cos5(sin3 x) 26 §3.8—Implicit Diﬀerentiation Example 1. Find the equation of the tangent line to the circle x2 + y2 = 4 at (1, √ 3). Solution #1 (Solving for y): Solution #2 (Diﬀerentiating implicitly): In many of our examples it will not be possible to solve for y, so we’ll be forced to use the second method. Basic procedure for implicit diﬀerentiation: 1. Take the derivative of both sides with respect to x. In doing this, we think of y as a function of x, so derivatives of expressions involving y require the Chain Rule. 2. Solve algebraically for dy dx by collecting all the terms containing dy dx on one side of the equation and then factoring and dividing. 27 Remark. Implicit diﬀerentiation can be used to prove the power rule for rational expo-nents once it has been proved for integer exponents. For instance, if y = x2/3, then we can write y3 = x2 and hence 3y2 dy dx = 2x = ⇒dy dx = 2x 3y2 = 2x 3x4/3 = 2 3x−1/3. Example 2. Find the slope of the tangent line to the curve 3x4y2 −7xy3 = 4 −8y at the point (0, 1/2). Example 3. Find dy dx for the curve x cos y + y cos x = 1. 28 §3.9—Derivatives of Inverse Functions Review of Inverses. A function is one-to-one if no y value occurs for two diﬀerent values of x. For example, f(x) = x3 is one-to-one, but f(x) = x2 is not. This deﬁnition is captured graphically by the Horizontal Line Test: a function is one-to-one if and only if no horizontal line intersects its graph more than once. If f is one-to-one, then there is an inverse function f −1 deﬁned by f −1(y) = x ⇐ ⇒f(x) = y. The domain of f −1 is the range of f, and the range of f −1 is the domain of f. The graph of f −1 is the reﬂection of the graph of f across the line y = x. Note that f and f −1 “undo” each other, meaning that f(f −1(x)) = x and f −1(f(x)) = x. Warning: f −1(x) is NOT the same as 1 f(x). Example 1. Find the inverse of the function f(x) = 2x + 1. The Inverse Trig Functions. Even though the trigonometric functions are not one-to-one, we can deﬁne inverses for them by restricting their domains to intervals on which the functions are one-to-one. For example, sin x is one-to-one on the interval −π/2 ≤x ≤π/2 and cos x is one-to-one on the interval 0 ≤x ≤π. Moreover, these functions cover the full range of y values between −1 and 1 as x runs over these restricted intervals. It is often helpful to think of the values of inverse trig functions as angles. • y = sin−1 x is the number in [−π/2, π/2] for which sin y = x • y = cos−1 x is the number in [0, π] for which cos y = x • y = tan−1 x is the number in (−π/2, π/2) for which tan y = x • y = cot−1 x is the number in (0, π) for which cot y = x • y = sec−1 x is the number in [0, π/2) ∪(π/2, π] for which sec y = x • y = csc−1 x is the number in [−π/2, 0) ∪(0, π/2] for which csc y = x The inverse trig functions are sometimes denoted by arcsin x, arccos x, arctan x, etc. Example 2. Evaluate each of the following. (a) sin−1( 1 2) (b) cos−1(−1 2) (c) tan−1(1) 29 Example 3. Convert cos ( tan−1( x 3) ) to an algebraic expression in x. The Derivative Formulas: d dx(sin−1 x) = 1 √ 1 −x2 d dx(tan−1 x) = 1 x2 + 1 d dx(sec−1 x) = 1 \|x\| √ x2 −1 The derivatives of the inverse “co” functions are just the negatives of these. For instance, sin−1 x + cos−1 x = π 2 = ⇒cos−1 x = π 2 −sin−1 x = ⇒(cos−1 x)′ = −(sin−1 x)′. Example 4. Find the derivatives of the following functions. (a) y = (sin−1 x)3 + cos−1(x3) (b) y = x2 tan−1 x + √ sec−1 x Derivatives of inverse functions in general. If f is one-to-one and we write g = f −1, then we have f(g(x)) = x, so diﬀerentiating both sides using the Chain Rule gives f ′(g(x))g′(x) = 1 = ⇒g′(x) = 1 f ′(g(x)). For instance, in Example 1 we have f(x) = 2x + 1 and g(x) = 1 2x −1 2, so f ′(x) = 2 implies that g′(x) = 1/f ′(g(x)) = 1/2. This is exactly the technique we used for sin−1 x above, and we will use it again in the next section to ﬁnd derivatives of logarithmic functions. 30 §3.10—Derivatives of General Exponential and Logarithmic Functions Review of logarithms. Suppose that b > 0 and b ̸= 1. The function y = bx is one-to-one, so it has an inverse, namely f −1(x) = logb x. The domain of logb x is (0, ∞), and the range is (−∞, ∞). Thus we have y = logb x ⇐ ⇒by = x. In other words, logb x is the power that we must raise b to in order to get x. In particular, we have blogb x = x for all x > 0 and logb bx = x for all x. There are 3 main algebraic properties of logs to remember: (1) logb(xy) = logb x + logb y (2) logb(x/y) = logb x −logb y (3) logb xr = r logb x The case where the base b is e ≈2.71828 occurs so frequently that we use the special notation ln x to stand for loge x, so that y = ln x ⇐ ⇒ey = x. Derivatives of Logarithmic Functions. We already know how to diﬀerentiate ex, and we can use this to ﬁnd the derivative of ln x via implicit diﬀerentiation: Example 1. Compute the derivatives of the following functions. (a) y = x3 ln x + ln(cos x) (b) y = (ln x)7 + ln(ln(ln x)) Remark. By writing xn = eln(xn) = en ln x we can use the Chain Rule and the formula for the derivative of ln x to prove the power rule for any real exponent n: d dx(xn) = d dx(en ln x) = en ln x · n x = xn · n x = nxn−1. 31 Example 2. Find the derivatives of the following functions. (a) y = 2x (b) y = log2 x The calculations in Example 2 generalize to show that d dx(bx) = (ln b)bx and d dx(logb x) = 1 (ln b)x whenever b ̸= 1 is a positive constant. Note that the formulas for the derivative of ex and ln x are special cases of this, since ln e = 1. Example 3. Find the derivatives of the following functions. (a) y = sec(3x log10 x) (b) y = 5sin x + log5(tan−1(x5)) Example 4 (Logarithmic diﬀerentiation). Find the derivative of the function y = xx by ﬁrst taking the natural log of both sides and then diﬀerentiating implicitly. 32 §4.1—Linear Approximation and Applications The tangent line to the curve y = f(x) at x = a is given by y −f(a) = f ′(a)(x −a), or y = f(a) + f ′(a)(x −a). Thus when x is close to a we have the approximation f(x) ≈f(a) + f ′(a)(x −a). This is called the linear approximation or tangent line approximation to f at a. The function L(x) = f(a) + f ′(a)(x −a) is called the linearization of f at x = a. Example 1. Let f(x) = √x and a = 25. (a) Find the equation of the tangent line to f at x = 25, and write down the linearization. (b) Use the linearization from part (a) to estimate √ 26, √ 23, and √ 28. (c) Analyze the quality of the approximations from (b) by completing the following table. x Linear approx L(x) f(x) = √x via calculator Error = \|f(x) −L(x)\| 26 23 28 33 Example 2. Use a linear approximation to estimate each of the following: (a) sin(0.02) (b) 3 √ 8.06 (c) e0.03 (d) ln(0.95) When we are primarily interested in estimating the change in a given quantity, it is sometimes more convenient to rewrite the linear approximation in the form f(x) −f(a) ≈f ′(a)(x −a) or ∆f ≈f ′(a)∆x, where ∆x = x−a and ∆f = f(x)−f(a). A typical application involves the analysis of error propagation, as in the following example. Example 3. The edge of a cube is measured at 30 cm, with a possible error of ±0.1 cm. Estimate the maximum possible error in computing the cube’s volume. 34 §3.11—Related Rates Suppose that two or more quantities are related by some equation. For instance, if C is the circumference of a circle and r is the radius, then C = 2πr. As another example, if a and b are the legs of a right triangle with hypotenuse c, then a2 + b2 = c2. If the quantities involved change with time, then we can diﬀerentiate both sides of the equation with respect to t to derive a relationship between the rates of change: e.g. dC dt = 2πdr dt or 2ada dt + 2bdb dt = 2cdc dt. If some of these rates of change are known, then we may be able to use these equations to solve for the unknown rates of change. Example 1. The radius of a circular oil spill is increasing at a constant rate of 1.5 meters per second. How fast is the area of the spill increasing when the radius is 30 meters? Example 2. Boyle’s Law states that when a sample of gas is compressed at constant temperature the product of the pressure and the volume remains constant. At a certain instant, the volume of a gas is 600 cubic centimeters, the pressure is 150 kPa, and the pressure is increasing at a rate of 20 kPa per minute. How fast is the volume decreasing at this instant? 35 Example 3. A ladder 25 feet long is leaning against a vertical wall. The bottom of the ladder is being pulled horizontally away from the wall at a constant rate of 3 feet per second. At the instant when the bottom of the ladder is 15 feet from the wall, determine (a) how fast the top of the ladder is sliding down the wall (b) how fast the angle between the top of the ladder and the wall is changing Example 4. A tank has the shape of an inverted cone with height 16 meters and base radius 4 meters. Water is being pumped into the tank at a constant rate of 2 cubic meters per minute. How fast is the water level rising when the water is 5 meters deep? 36 §4.2—Extreme Values Let f be a function with deﬁned on some interval I. We say that f(a) is the absolute maximum of f on I if f(a) ≥f(x) for all x in I. We say that f(a) is the absolute minimum of f on I if f(a) ≤f(x) for all x in I. Note: The absolute maximum and minimum values refer to the largest and smallest y values on the graph, not the x values at which they occur. We say that f has a local maximum at x = c if f(c) ≥f(x) for all x in some open interval containing c. We say that f has a local minimum at c if f(c) ≤f(x) for all x in some open interval containing c. Maxima and minima are sometimes called extrema. Absolute extrema are sometimes called global extrema, and local extrema are sometimes called relative extrema. Example 1. Identify the coordinates of all absolute and local extrema for the function graphed below on the interval [0, 10]. Example 2. Determine the absolute extrema of each function on the given intervals. (a) y = x2 (b) y = 1/x (i) [0, 2] (i) (0, 3] (ii) [0, 2) (ii) [3, ∞) (iii) (0, 2] (iii) [−3, 3] 37 Extreme Value Theorem. If f is continuous on the closed interval [a, b], then f attains both an absolute maximum and an absolute minimum value in [a, b]. A number c in the domain of f for which f ′(c) = 0 or f ′(c) does not exist is called a critical point of f. The only possible places where local and absolute extrema occur are at critical points or at endpoints of the domain. Example 3. Find the absolute maximum and minimum values of the function f(x) = x3 −12x + 1 on the interval [−3, 5]. Example 4. Find the absolute maximum and minimum values of the function f(x) = x5/3 −10x2/3 on the interval [−8, 8]. Example 5. Find the absolute maximum and minimum values of the function f(x) = x2 ln x on the interval [ 1 2, ∞). 38 §4.3—The Mean Value Theorem and Monotonicity The Mean Value Theorem. Suppose that y = f(x) is continuous on the closed interval [a, b] and diﬀerentiable on the open interval (a, b). Then there is at least one point c in (a, b) for which f(b) −f(a) b −a = f ′(c). The picture that makes this “obvious”: Consequences of the MVT. One useful interpretation of the theorem is that there is some point in the interval at which the instantaneous rate of change is equal to the average rate of change. Another important consequence is the following: Increasing/Decreasing Test: If f ′(x) > 0 for all x in some interval, then f is increasing on that interval (i.e., the y values are getting larger). If f ′(x) < 0 for all x in some interval, then f is decreasing on that interval (i.e., the y values are getting smaller). Example 1. Show that the function f(x) = x3 + 2x + 4 is always increasing. Example 2. For what values of x is the function f(x) = xe−x decreasing? 39 First Derivative Test for Local Extrema: Suppose that c is a critical point of a diﬀerentiable function f. 1. If f ′ changes from negative to positive at c, then f(c) is a local minimum. 2. If f ′ changes from positive to negative at c, then f(c) is a local maximum. 3. If f ′ does not change sign at c, then f(c) is neither a local maximum nor a local minimum. Example 3. Find the intervals on which the function f(x) = 3x4 −4x3 −12x2 + 5 is increasing and decreasing, and identify all local extrema of the function. Example 4. Find the intervals on which the function f(x) = x5 −15x3 + 4 is increasing and decreasing, and identify all local extrema of the function. 40 §4.4—The Shape of a Graph Concavity f is concave up ⇐ ⇒f ′ is increasing ⇐ ⇒f ′′ > 0 f is concave down ⇐ ⇒f ′ is decreasing ⇐ ⇒f ′′ < 0 Four basic shapes of graphs: (a) f ′ > 0, f ′′ > 0 (b) f ′ > 0, f ′′ < 0 (c) f ′ < 0, f ′′ > 0 (d) f ′ < 0, f ′′ < 0 A point on the graph of f where the concavity changes is called an inﬂection point of f. These can only occur where f ′′ = 0 or f ′′ is undeﬁned. Note that these are the local maxima and minima of f ′. Example 1. Find the points of inﬂection of f(x) = x3 −6x2 + 1, and determine the intervals on which the curve is concave up and concave down. Second Derivative Test for Local Extrema: Suppose f ′′ is continuous and f ′(c) = 0. 1. If f ′′(c) > 0, then f has a local minimum at x = c. 2. If f ′′(c) < 0, then f has a local maximum at x = c. 3. If f ′′(c) = 0, then the test gives no information. In this case, we must go back to the ﬁrst derivative test. e.g. f(x) = x3 versus f(x) = x4 Example 2. Use the second derivative test to determine the location of all local maxima and local minima of f(x) = 3x4 −4x3 −12x2 + 5. [Compare with Example 3 in §4.3.] 41 §4.5—Graph Sketching and Asymptotes Example 1. Sketch the graph of each of the following functions. (a) f(x) = x4 −4x3 (b) f(x) = x2/3(x −5) 42 Asymptotic behavior. We say that lim x→∞f(x) = L if f(x) can be made as close as we like to L by taking x suﬃciently large. Similarly, we say that lim x→−∞f(x) = L if f(x) can be made as close as we like to L by taking −x suﬃciently large (that is, \|x\| suﬃciently large and x < 0). If lim x→∞f(x) = L or lim x→−∞f(x) = L, then the line y = L is called a horizontal asymptote for the graph of y = f(x). Example 2. Evaluate each of the following limits. (a) lim x→∞ 8x3 + 5x + 1 2x3 + 4 (b) lim x→−∞ 3x2 −10 x5 + 3x + 1 Example 3. Sketch the graph of the rational function f(x) = x + 2 x + 1. 43 §4.6—Applied Optimization Example 1. A farmer with 600 feet of fencing wants to construct a rectangular pen and then divide it in half with a fence parallel to one of the sides. What dimensions maximize the area of the pen? Example 2. You are asked to design a cylindrical can (with top and bottom) of volume 500 cubic centimeters. What dimensions should the can have in order to minimize the amount of metal used? 44 When arguing that a critical number actually yields the optimal result, we frequently make use of the following principle: First Derivative Test for Absolute Extrema. Suppose that f is continuous and that c is the only critical number of f. If f(c) is a local maximum (resp. minimum), then it is also the absolute maximum (resp. minimum). Example 3. You are asked to design an athletic complex in the shape of a rectangle with semi-circular ends. A running track 400 meters long is to go around the perimeter. What dimensions will give the rectangular playing ﬁeld in the center the largest area? Example 4. A box with no top is to have volume 4 cubic meters, and its base is to be a rectangle twice as long as it is wide. If the material for the bottom costs $3 per square meter and the material for the sides costs $1.50 per square meter, ﬁnd the dimensions that minimize the total cost of constructing the box. 45 Example 5. You are standing on a sidewalk at the corner of a muddy rectangular ﬁeld of length 1 mile and width 0.2 miles. You can run along the sidewalk bordering the long side of the ﬁeld at 8 mph, and you can run through the mud at 5 mph. Assuming there is no sidewalk along the short side of the ﬁeld, ﬁnd the quickest route to the opposite corner. Example 6. Find the volume of the largest cylinder that can be inscribed in a sphere of radius R. What percentage of the sphere’s volume is occupied by such a cylinder? 46 §4.7—L’Hˆ opital’s Rule A general method for evaluating “0/0” or “∞/∞” type limits: L’Hˆ opital’s Rule. Suppose that either (i) lim x→a f(x) = 0 and lim x→a g(x) = 0 or (ii) lim x→a f(x) = ±∞and lim x→a g(x) = ±∞. Then we have lim x→a f(x) g(x) = lim x→a f ′(x) g′(x). Here a can be a real number or ±∞. Example 1. Evaluate the following limits. (a) lim x→0 ex −1 x (b) lim x→∞ ln x √x Warning: L’Hˆ opital’s Rule does not apply unless (i) or (ii) holds. For example, 0 = lim x→0 sin x x + 1 ̸= lim x→0 cos x 1 = 1. Sometimes it’s necessary to apply l’Hˆ opital’s Rule more than once: Example 2. Evaluate the following limits. (a) lim x→0 1 −cos x x2 (b) lim x→∞ ex x4 47 We can sometimes deal with other indeterminate forms like 0 · ∞, 00, ∞−∞, and 1∞ by converting them to 0/0 or ∞/∞and then applying l’Hˆ opital’s Rule. Example 3. Evaluate the following limits. (a) lim x→0+ x ln x (b) lim x→1 ( 1 ln x − 1 x −1 ) (c) lim x→∞ ( 1 + 1 x ) x 48 §4.9—Antiderivatives We say that F is an antiderivative of f if F ′(x) = f(x) for all x. For example, x2 and x2 + 1 are antiderivatives of 2x sin x and sin x −17 are antiderivatives of cos x If F is any antiderivative of f, then it follows from the Mean Value Theorem that the most general antiderivative of f is F(x) + C, where C is an arbitrary constant. The set of all antiderivatives of f is denoted ∫ f(x) dx and is called the indeﬁnite integral of f with respect to x. For example, ∫ 2x dx = x2 + C and ∫ cos x dx = sin x + C. Example 1. Evaluate the following indeﬁnite integrals. (a) ∫ (x2 + 2 cos x) dx (b) ∫ (sin x + x−6) dx (c) ∫ (e3x + 3 sec2 x) dx Some Useful Indeﬁnite Integrals ∫ xn dx = xn+1 n + 1 + C (n ̸= −1) ∫1 x dx = ln \|x\| + C ∫ sin kx dx = −1 k cos kx + C ∫ cos kx dx = 1 k sin kx + C ∫ ekx dx = 1 kekx + C 49 Example 2. Evaluate ∫( cos 2x −5√x + 7 x + 1 √ 1 −x2 ) dx. Note that we can split up sums and diﬀerences of indeﬁnite integrals: ∫ (f(x) ± g(x)) dx = ∫ f(x) dx ± ∫ g(x) dx However, there is no such law for products: ∫ f(x)g(x) dx ̸= ∫ f(x) dx ∫ g(x) dx. For example, ∫ x cos x dx ̸= x2 2 sin x + C. Initial Conditions. If we know a function’s derivative and the value of the function at one point, we can determine the function by ﬁrst ﬁnding the general antiderivative and then using the known value to solve for C. Example 3. Suppose that f ′(x) = 3x2 and f(1) = 5. Find a formula for f(x). Example 4. A particle’s acceleration is given by a(t) = 5 + 4t −2t2, and its initial velocity and position are v(0) = 3 and s(0) = 10. Find formulas for v(t) and s(t). 50 §5.1—Approximating and Computing Area Example 1. Estimate the area of the region bounded by the curve y = x2 and the x-axis between x = 0 and x = 2 by approximating the region with 4 rectangles of equal width whose heights are determined using (a) left endpoints (b) right endpoints (c) midpoints Using a larger number of rectangles gives a better estimate of the area, and we deﬁne the exact area to be the limit of these approximations as the number of rectangles tends to inﬁnity. In order to add up a large number of terms, it is convenient to use sigma notation: N ∑ j=1 aj = a1 + a2 + a3 + · · · + aN−1 + aN Example 2. Evaluate the following: (a) 7 ∑ j=1 j (b) 5 ∑ j=1 j2 51 Example 3. Use sigma notation to write the right-endpoint approximation RN for the area of the region bounded by the curve y = x2 and the x-axis between x = 0 and x = 2. To ﬁnd the exact area under the curve, we need to ﬁnd a way to express RN (or LN or MN) in closed form so that we can compute the limit as N →∞. In general, it is quite diﬃcult to do this, but there are many special cases that can be handled; for instance: N ∑ j=1 j = N(N + 1) 2 and N ∑ j=1 j2 = N(N + 1)(2N + 1) 6 . Example 4. Calculate the exact area of the region bounded by the curve y = x2 and the x-axis between x = 0 and x = 2. Finding distance traveled. By applying the same reasoning as above and using the fact that distance = velocity × time when velocity is constant, we see that the net change in position of an object over an interval is the area under its velocity curve. Example 5. A car’s velocity during a 1-hour period is measured at 12-minute intervals: time (hours) 0 0.2 0.4 0.6 0.8 1.0 velocity (miles per hour) 66 75 78 82 79 74 Estimate the total distance traveled by the car during the hour using (a) left endpoints (b) right endpoints 52 §5.2—The Deﬁnite Integral To approximate the area bounded by a continuous function y = f(x) and the x-axis on the interval [a, b], we divide into N subintervals of width ∆x = b −a N . The jth subinterval is the interval [xj−1, xj], where xj = a + j∆x. For each j, we use a rectangle of height f(xj) and width ∆x to approximate the area under that portion of the curve. The Riemann sum RN = N ∑ j=1 f(xj)∆x = f(x1)∆x + f(x2)∆x + · · · + f(xN)∆x approximates the total area under the curve on the interval [a, b]. We get the exact area by letting N →∞, which gives the deﬁnite integral of f from a to b: ∫b a f(x) dx = lim N→∞ N ∑ j=1 f(xj)∆x. Here the function f is called the integrand and the numbers a and b are the limits of integration. Note that the choice to use right-endpoints here is simply a convenience; for continuous functions, we could choose points randomly in each subinterval and still get the same result as ∆x →0. Example 1. Calculate the following deﬁnite integrals directly from the deﬁnition. (a) ∫3 0 x dx 53 (b) ∫b 0 x2 dx If f(x) takes both positive and negative values on [a, b], then the deﬁnite integral gives the “signed area” under the curve. That is, areas above the x-axis are counted positively, and areas below the x-axis are counted negatively. Example 2. Evaluate the following integrals. (a) ∫2π 0 sin x dx (b) ∫3 −2 x dx Properties of the deﬁnite integral • Conventions: ∫a b f(x) dx = − ∫b a f(x) dx and ∫a a f(x) dx = 0 • Linearity: ∫b a (kf(x) ± mg(x)) dx = k ∫b a f(x) dx ± m ∫b a g(x) dx (k, m constant) • Additivity: ∫b a f(x) dx + ∫c b f(x) dx = ∫c a f(x) dx • Comparison: If f(x) ≤g(x) for all x in [a, b], then ∫b a f(x) dx ≤ ∫b a g(x) dx. Example 3. Suppose that ∫1 0 f(x) dx = 2 and that f(x) ≤4 for all x in [1, 3]. What is the largest possible value that the integral ∫3 0 f(x) dx could have? 54 §5.3—The Fundamental Theorem of Calculus, Part I It turns out that the key to evaluating deﬁnite integrals eﬃciently is ﬁnding an antideriva-tive for the integrand. We actually observed a special case of this in Example 5 of Section 5.1 when we saw that the area under a velocity graph gives the net change in position. More generally, if f has an antiderivative F, then we can view f as a rate of change of F and apply the same reasoning to establish the following: Fundamental Theorem of Calculus, Part I. If f is continuous on [a, b] and F is any antiderivative of f, then ∫b a f(x) dx = F(b) −F(a). This theorem (often called the FTC for short) may be interpreted as saying that the deﬁnite integral of a rate of change gives the total change. For example, if f represents velocity and F represents position, then the deﬁnite integral of velocity is change in position. Example 1. Use the FTC to calculate each of the following. (a) ∫2 0 x2 dx (b) ∫π/2 0 cos x dx (c) ∫1 0 dx 1 + x2 55 Example 2. Evaluate each of the following deﬁnite integrals. (a) ∫3 0 e5x dx (b) ∫1 0 √x(x2 + 3) dx Example 3. Find the area bounded by the curve y = 1/x and the x-axis between x = 2 and x = 6. Example 4. What is wrong with the following calculation? ∫2 −1 1 x2 dx = −1 x 2 −1 = −1 2 −1 = −3 2. Example 5. Evaluate ∫5 0 √ 25 −x2 dx. 56 §5.4—The Fundamental Theorem of Calculus, Part II Example 1. Consider the function f graphed below, and let A(x) denote the signed area under the curve on the interval [0, x]. Calculate each of the following. (a) A(0) (b) A(2) (c) A(5) Fundamental Theorem of Calculus, Part II. Suppose that f is continuous on [a, b], and let A(x) = ∫x a f(t) dt. Then A′(x) = d dx ∫x a f(t) dt = f(x). That is, A(x) is the antiderivative of f(x) satisfying the initial condition A(a) = 0. Why is this true? Interpretation: Diﬀerentiation and integration are “inverse” operations, i.e., ∫x a f(t) dt is an antiderivative of f(x). Example 2. Calculate each of the following derivatives. (a) d dx ∫x 1 sin t t dt (b) d dx ∫x −5 t3et dt 57 Example 3. Find the derivative of each function. (a) F(x) = ∫x3 4 √ 1 + t2 dt (b) G(x) = ∫10 e2x cos3 t dt Example 4. For what values of x is the function F(x) = ∫x 0 1 1 + t + t2 dt concave up? Example 5. Find a function F(x) such that F ′(x) = ln x and F(1) = 3. 58 §5.5—Net or Total Change as the Integral of a Rate A useful interpretation of the FTC (Part I) is that the deﬁnite integral of a rate of change gives total change. For instance, if s(t) represents position, then s′(t) is velocity (or rate of change of position), and we have ∫b a s′(t) dt = s(b) −s(a). That is, the deﬁnite integral of velocity gives the net change in position. The same principle applies when integrating any function that can be viewed as a rate of change. Example 1. A particle’s velocity is given by v(t) = t2 −4t + 3. (a) Find the object’s net change in position over the interval 0 ≤t ≤3. (b) Find the total distance traveled by the object over the interval 0 ≤t ≤3. Example 2. The rate of energy consumption in a certain home (in kilowatts) is modeled by the function R(t) = 2 + 0.5 cos(πt/3), where t is measured in months since January 1. According to this model, how many kilowatt-hours of energy will be used in a typical year? 59 §5.6—The Substitution Method In earlier sections, we obtained formulas like ∫ cos 2x dx = 1 2 sin 2x + C and ∫ e5x dx = 1 5e5x + C by mentally attempting to reverse the eﬀect of the chain rule. A more systematic approach is to substitute a new variable for the inner function. For instance, if we let u = 2x in the ﬁrst integral above, then du = 2dx, and thus dx = 1 2du, so we get ∫ cos 2x dx = ∫ (cos u)1 2du = 1 2 ∫ cos u du = 1 2 sin u + C = 1 2 sin 2x + C. In general, we can evaluate ∫ f(g(x))g′(x) dx by substituting u = g(x) and du = g′(x) dx. Example 1. Evaluate the following indeﬁnite integrals. (a) ∫ 2xex2 dx (b) ∫√ 3x + 4 dx (c) ∫ x4 cos(x5) dx 60 Example 2. What is wrong with the following calculation of ∫ cos(x5) dx? Let u = x5, so that du = 5x4 dx. Then dx = du 5x4, so ∫ cos(x5) dx = ∫ (cos u) du 5x4 = 1 5x4 ∫ cos u du = 1 5x4 sin u + C = sin(x5) 5x4 + C. Example 3. Evaluate the following indeﬁnite integrals. (a) ∫(1 + ln x)10 x dx (b) ∫sin √x √x dx (c) ∫ sec2 x 1 + tan x dx 61 Substitution in Deﬁnite Integrals: ∫b a f(g(x))g′(x) dx = ∫g(b) g(a) f(u) du Example 4. Use substitution to evaluate the following deﬁnite integrals. (a) ∫1 0 x3 √ x4 + 9 dx (b) ∫π/2 0 (1 + sin3 x) cos x dx (c) ∫4 2 dx x ln x (d) ∫π/2 π/4 3cot θ csc2 θ dθ 62 §5.7—Further Transcendental Functions We record here for reference two important integrals involving the inverse trig functions: ∫ dx √ a2 −x2 dx = sin−1 (x a ) + C and ∫ dx a2 + x2 = 1 a tan−1 (x a ) + C Example. Evaluate each of the following integrals. (a) ∫ dx 25 + x2 (b) ∫ dx √ 4 −9x2 (c) ∫1/2 0 x dx 16x4 + 1 (d) ∫√e 1 dx x √ 1 −(ln x)2 63
16395	https://www.sciencedirect.com/science/article/abs/pii/S1574013716300077	Skip to article My account Sign in Access through your organization Purchase PDF Article preview Abstract Introduction Section snippets References (102) Cited by (16) Computer Science Review Volume 22, November 2016, Pages 89-105 Review Article Sorting on graphs by adjacent swaps using permutation groups Author links open overlay panel rights and content Abstract This paper is a review of sorting on several well-known graphs by adjacent swaps using permutation groups. Given a graph with a line, star, complete, or ring topology having vertices (numbered from 1 to ), we place objects (numbered from 1 to ) arbitrarily in such a way that exactly one object is placed on each vertex. A sorting procedure is intended to reach the target object placement in which each object is placed on vertex for . Now, the problem is to find a minimum-length sequence of adjacent swaps needed to sort the initial arrangement of objects on vertices in the graph, which employs a known permutation sorting or permutation decomposition procedure using a generating set of a symmetric group . This paper reviews the existing permutation sorting and permutation decomposition procedures for sorting on graphs with line, star, complete, and ring topologies by adjacent swaps. We also provide concrete examples and our own implementation for this review paper in order to describe how abstract group-theoretical methods are used to find a minimum-length sequence of adjacent swaps needed to sort objects on those graphs. Introduction Given a graph with a line, star, complete, or ring topology having vertices, consider that objects (numbered from 1 to ) are placed on vertices (numbered from 1 to ) in the graph in such a way that exactly one object is placed on each vertex. This is represented by a bijective assignment between objects and vertices, which in turn can be represented by a permutation (i.e. a permutation of objects). A permutation of objects can be considered as an element of symmetric group on letters, denoted by , , . Using swaps of objects between two vertices, the goal of a sorting procedure is to sort those objects so that the permutation corresponding to the target assignment is the identity permutation. Sorting plays an important role in algorithm design , , , robotics , , , graph theory , , , genomic rearrangements , , , , and so on. In particular, sorting objects (e.g. tokens, boxes, markers, tasks, etc.) on graphs has been researched extensively in recent years , , , , , . This review paper is concerned with swaps of objects on adjacent vertices in several well-known graphs, where a swap depends on a graph with a given topology. The basic technique used here is that if a permutation denotes an initial bijective assignment between objects and vertices, the right multiplication of by transposition represents a swap of the object on vertex and vertex . Now, the problem of finding a minimum-length sequence of swaps needed to sort objects is reduced to find a minimum-length permutation decomposition of the inverse of permutation using transpositions from the set . However, if only swaps on adjacent vertices in a graph are allowed, not all transpositions can be used, but transpositions in a chosen generating set of can be used. This generating set of relies on how vertices are interconnected by a graph with a given topology. In Akers and Krishnamurthy proposed a puzzle consisting of markers on vertices in an arbitrary transposition tree with vertices, which is basically the problem of sorting an initial arrangement of markers on vertices in a tree with vertices by adjacent swaps. They also discussed some tight upper bounds of the minimum number of adjacent swaps needed to sort markers on vertices in a certain type of transposition trees with vertices. In Kim discussed the problem of finding a minimum-length sequence of adjacent task swaps needed from a (bijective) source task assignment to reach a (bijective) target task assignment on a task swapping graph, which is reduced to find a minimum-length sequence of adjacent swaps needed to sort tasks on a task swapping graph. In Yamanaka et al. showed that sorting an initial arrangement tokens on vertices in a simple and connected graph of order by adjacent swaps is solvable in token swaps. They also showed a polynomial-time 2-approximation algorithm for the problem of finding the minimum number of adjacent swaps needed to sort tokens on vertices in a tree with vertices. The problem of sorting objects on vertices by using the minimum number of adjacent swaps in a graph of order with a line, star, complete, or a ring topology can also be viewed as the problem of finding a minimum-length generator sequence of with respect to the corresponding generating set of , where a minimum-length generator sequence of a permutation is well-studied in permutation group theory. The minimum-length generator sequence problem is NP-hard in general, but the problem is polynomial for some generating sets and their generated permutations groups. In this review paper we focus on polynomial-time exact algorithms for sorting by swaps of objects on adjacent vertices in graphs with line, star, complete, and ring topologies. In the remainder of this paper by sorting objects on a graph we mean sorting an arrangement of objects on vertices in a simple and connected graph of order . We also assume that both objects and vertices in a graph are numbered from 1 to in such a way that each object is assigned to exactly one vertex, allowing an arrangement of objects on vertices to be represented by a permutation, i.e., a permutation of objects. The remainder of this review paper is organized as follows. Section 2 gives a brief overview of the necessary background used in this review paper. Sorting objects on graphs with line, star, complete, and ring topologies by means of a minimum-length sequence of adjacent swaps is discussed in Section 3. The object sorting procedures described in this section are direct consequences of the known permutation sorting and permutation decomposition (or permutation factorization) methods used in permutation group theory, in which we simply apply them to find a minimum-length sequence of adjacent swaps needed to sort objects on graphs with some well-known topologies. We also discuss upper bounds of the minimum number of adjacent swaps needed to sort objects on those graphs in this section. Section 4 contains a discussion of Cayley graphs for describing the possible arrangements of objects at the vertices of graphs with line, star, complete, and ring topologies. We also discuss the word metric on a Cayley graph in this section, in order to describe the distance between two arrangements of objects at the vertices of the associated graph. Section 5 illustrates some applications of the object sorting procedures discussed in Section 3 to genomic rearrangements. In particular, we discuss sorting permutations (or circular permutations) by reversals of length 2, sorting by prefix-exchanges, and sorting by block-interchanges of length 1 in genomic rearrangements as applications of the object sorting procedures described in Section 3. Discussion and a survey of related works are given in Section 6. In Section 7 we give a brief description of our implementation. We provide concluding remarks in Section 8. Section snippets Preliminaries This section describes the necessary background used in this review paper. The definitions and results used in this section are found in , , , , , , , , , , , , , , . A group is a nonempty set , closed under a binary operation , such that the following axioms are satisfied: (i) for all ; (ii) there is an element , called an identity of , such that for all ; (iii) for each element , Adjacent swap-based sorting methods for graphs with line, star, complete, and ring topologies We first consider object sorting on a graph with a line topology using adjacent swaps (see Fig. 1). In this review paper we assume that the label of each vertex in a graph denotes an object with that label, while the label next to each vertex in a graph denotes a vertex label. We also assume that each graph with a line topology is labeled in ascending order from left to right (see Fig. 1(a)). Therefore, we see that Fig. 1(a) illustrates the bijective assignment between eight objects (numbered Cayley graphs as state diagrams for sorting objects on graphs with line, star, complete, and ring topologies Cayley graphs , , play an important role in both mathematics (e.g. combinatorics , , group theory , , etc.) and computer science (e.g. interconnection network design , , , data center design , , distributed systems , , multi-agent systems , , etc.). In Section 3 we used the diameters of several kinds of Cayley graphs to find upper bounds of the minimum number of adjacent swaps needed to sort objects on graphs with line, Applications to genomic rearrangements In biology a genome is defined as the entire collection of genes encoded by a particular organism . A series of genomic rearrangement events may alter the genomic architecture of a species, where each genomic rearrangement event can change the ordering of genes . The main genomic rearrangement operations include transpositions , , , reversals (also called inversions) , , , block-interchanges , , , etc. Genomes are often represented by Discussion and related works In previous sections we focus on optimal sorting procedures by adjacent swaps. We discussed sorting objects on vertices in a graph of order with a line, star, complete, or ring topology using the minimum number of adjacent swaps. When sorting objects at the vertices of graphs with some other topologies, we may consider a -approximation algorithm which guarantees that the output solution is no greater than times optimal solution, where for minimization problems. We first Implementation Permutation sorting and permutation decomposition procedures discussed in Section 3 run in (deterministic) polynomial time , , , , , which can be implemented as software modules. Although there are other utilities for permutation groups, such as GAP and MAGMA , our implementation is specifically designed for procedures described in Section 3. The purpose of our implementation is to provide a tool for solving the adjacent swap-based object sorting problems Concluding remarks This paper reviewed adjacent swap-based sorting methods for graphs with several well-known topologies. For a given permutation representing an initial arrangement of objects on vertices in a graph of order with a line, star, complete, or ring topology, the problem of finding a minimum-length sequence of adjacent swaps needed to sort objects is reduced to find a minimum-length permutation decomposition of using the generating set of corresponding to a given topology. The References (102) K. 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BjÃ¶rner et al. ### Combinatorics of Coxeter Groups (2005) Cited by (16) Improving Quantum Computation by Optimized Qubit Routing 2023, Journal of Optimization Theory and Applications ### Token Swapping on Trees 2023, Discrete Mathematics and Theoretical Computer Science ### Hardness of Token Swapping on Trees 2022, Leibniz International Proceedings in Informatics Lipics ### Sorting permutations with a transposition tree 2019, 2019 8th International Conference on Modeling Simulation and Applied Optimization Icmsao 2019 ### Analysis of algorithm Î´ ast on full binary trees 2019, 2019 2nd International Conference on Advanced Computational and Communication Paradigms Icaccp 2019 ### Bounding the diameter of cayley graphs generated by specific transposition trees 2017, 2017 International Conference on Advances in Computing Communications and Informatics Icacci 2017 View all citing articles on Scopus View full text © 2016 Elsevier Inc. All rights reserved.
16396	https://medium.com/@amit25173/probability-density-function-of-binomial-distribution-dc0e4f1ade8a	Sitemap Open in app Sign in Sign in Probability Density Function of Binomial Distribution Amit Yadav 9 min readJul 16, 2024 Hey, is this you? You want to learn Data Science but have no idea where to start? I understand there’s an overwhelming amount of information out there, making it hard to even find a starting point. Resources like courses and coaching programs that promise well-structured information often cost thousands of dollars. But it doesn’t have to be this way. I’ve curated 101+ free resources for you to learn Data Science in 90 days. It covers: Programming Mathematics Data Analytics Machine Learning And much more… Why don’t you give it a try? Download Free Resources. Now let’s get back to the blog: When learning about statistics, one of the first concepts you’ll come across is the binomial distribution. Let’s explore what it is and why it’s so crucial for your data analysis toolkit. What is a Binomial Distribution? A binomial distribution is a probability distribution that summarizes the likelihood that a value will take one of two independent states and is most commonly associated with processes that involve a fixed number of trials, where each trial has a constant probability of success. In simpler terms, imagine you’re flipping a coin. Each flip can either result in heads (success) or tails (failure), and you’re flipping the coin a set number of times. The binomial distribution helps you determine the probability of getting a certain number of heads (or tails) in those flips. The key parameters of a binomial distribution are: n: The number of trials or experiments. p: The probability of success in each trial. The mathematical formula for the probability of getting exactly k successes in nnn trials is given by: Where (n/k) is the binomial coefficient, which you might know as “n choose k,” representing the number of ways to choose k successes from n trials. Real-World Examples of Binomial Distribution To make this more tangible, let’s look at a few real-world examples where you’d apply the binomial distribution: Quality Control in Manufacturing: Imagine you work in a factory that produces light bulbs, and you know from historical data that the probability of a bulb being defective is 0.05. If you sample 100 bulbs, you can use the binomial distribution to calculate the probability of finding exactly 5 defective bulbs. Clinical Trials: Suppose you’re analyzing the effectiveness of a new drug. If the probability of a patient recovering after taking the drug is 0.8, and you test the drug on 50 patients, the binomial distribution can help you determine the likelihood of exactly 40 patients recovering. Marketing Campaigns: You’re running an email marketing campaign and you know that, on average, 10% of recipients open your emails. If you send emails to 1000 people, the binomial distribution allows you to estimate the probability that exactly 100 people will open the email. These examples illustrate how versatile and practical the binomial distribution is for various fields, from manufacturing and healthcare to marketing. Importance in Statistical Analysis and Probability Theory Understanding the binomial distribution is fundamental for several reasons: Foundation for Other Distributions: Many other probability distributions build upon the principles of the binomial distribution. For instance, the normal distribution, which is pivotal in statistics, can be approximated using the binomial distribution under certain conditions (large n and p not too close to 0 or 1). Hypothesis Testing: The binomial distribution plays a critical role in hypothesis testing, particularly in tests of proportions. When you want to determine if there’s a significant difference between observed and expected outcomes, the binomial distribution is often your go-to model. Decision Making: In business and research, decisions are frequently based on probabilities derived from binomial distributions. Whether it’s forecasting sales, predicting outcomes of clinical trials, or quality control, understanding the binomial distribution empowers you to make informed decisions. As you delve deeper into statistics and probability, you’ll find that the binomial distribution serves as a building block for more complex analyses. By mastering this concept, you’ll enhance your ability to interpret data, conduct rigorous research, and make data-driven decisions in your professional field. Understanding Probability Density Function (PDF) First, let’s clarify the difference between a Probability Density Function (PDF) and a Probability Mass Function (PMF): Probability Mass Function (PMF): The PMF applies to discrete random variables. It gives you the probability that a discrete random variable is exactly equal to some value. For example, in the context of the binomial distribution, which is a discrete distribution, the PMF tells you the probability of observing exactly kkk successes in nnn trials. Probability Density Function (PDF): The PDF, on the other hand, is used for continuous random variables. It describes the likelihood of a random variable to take on a particular value, but it doesn’t give probabilities directly. Instead, it gives a density that you would integrate over an interval to get the probability. For instance, in a normal distribution, the PDF helps you understand the density of the data points along the distribution curve. So, when we talk about the binomial distribution, we’re really referring to its PMF, but sometimes you’ll see the term PDF used interchangeably. It’s essential to understand this distinction, especially as you move into more advanced statistical analyses. Calculation Examples To truly grasp the PMF of the binomial distribution, let’s walk through a detailed calculation example. This hands-on approach will make the concept clearer. Step-by-Step Calculation Example Imagine you’re conducting a survey, and you want to determine the probability of a certain number of respondents answering “yes” to a question. Suppose you survey 10 people, and the probability of any one person answering “yes” is 0.3. Get Amit Yadav’s stories in your inbox Join Medium for free to get updates from this writer. Here’s how you calculate the probability of exactly 4 people answering “yes”: Identify the parameters: Number of trials (n): 10 Probability of success (p): 0.3 Number of successes (k): 4 2. Use the binomial PMF formula: 3. Calculate the binomial coefficient (n/k): 4. Calculate pkp^kpk and (1−p)^{n-k}: Put it all together: So, the probability of exactly 4 out of 10 people answering “yes” is approximately 0.2001, or 20.01%. Visualizing with Different Values of nnn and ppp To better understand how the binomial distribution behaves, let’s visualize it with different values of nnn (number of trials) and ppp (probability of success). Let’s say you want to compare two scenarios: Scenario 1: n=10n = 10n=10 and p=0.5p = 0.5p=0.5 Scenario 2: n=20n = 20n=20 and p=0.7p = 0.7p=0.7 Using Python, you can plot the PMF for these scenarios to see how the distribution changes: ``` import matplotlib.pyplot as pltfrom scipy.stats import binom# Define parameters for both scenariosn1, p1 = 10, 0.5n2, p2 = 20, 0.7# Create arrays of possible outcomesx1 = range(n1 + 1)x2 = range(n2 + 1)# Calculate PMF for each outcomepmf1 = binom.pmf(x1, n1, p1)pmf2 = binom.pmf(x2, n2, p2)# Plot the PMF for both scenariosplt.figure(figsize=(12, 6))plt.subplot(1, 2, 1)plt.bar(x1, pmf1, color='blue')plt.title(f'Binomial Distribution PMF (n={n1}, p={p1})')plt.xlabel('Number of Successes')plt.ylabel('Probability')plt.subplot(1, 2, 2)plt.bar(x2, pmf2, color='green')plt.title(f'Binomial Distribution PMF (n={n2}, p={p2})')plt.xlabel('Number of Successes')plt.ylabel('Probability')plt.tight_layout()plt.show() import as from import Define parameters for both scenarios 100.5 200.7 Create arrays of possible outcomes range 1 range 1 Calculate PMF for each outcome Plot the PMF for both scenarios 12 6 1 2 1 'blue'f'Binomial Distribution PMF (n={n1}, p={p1})'{n1}{p1} 'Number of Successes' 'Probability' 1 2 2 'green'f'Binomial Distribution PMF (n={n2}, p={p2})'{n2}{p2} 'Number of Successes' 'Probability' ``` In the first scenario with n=10n = 10n=10 and p=0.5p = 0.5p=0.5, you’ll see a symmetric distribution centered around 5 successes, reflecting the equal likelihood of success and failure. In the second scenario with n=20n = 20n=20 and p=0.7p = 0.7p=0.7, the distribution skews towards the higher number of successes due to the higher probability of success. Visualizing these distributions helps you understand how changing the number of trials and the probability of success impacts the shape and spread of the binomial distribution. It’s a powerful tool for interpreting data and making predictions based on your binomial model. Graphical Representation Visualizing the binomial distribution is an excellent way to grasp its characteristics and behavior. Let’s dive into how you can plot the binomial distribution using Python and interpret the resulting graphs. Plotting the Binomial Distribution To plot the binomial distribution, you can use Python with libraries like matplotlib and scipy. Here’s a step-by-step guide to help you create a plot: Install the necessary libraries: Ensure you have matplotlib and scipy installed. You can install them using pip: pip install matplotlib scipy Import the libraries and define your parameters: ``` import matplotlib.pyplot as pltfrom scipy.stats import binom# Parameters for the binomial distributionn = 10 # Number of trialsp = 0.5 # Probability of successx = range(n + 1) # Possible outcomes import as from import Parameters for the binomial distribution 10 Number of trials0.5 Probability of success range 1 Possible outcomes ``` Calculate the PMF: pmf = binom.pmf(x, n, p) pmf 4. Plot the distribution: plt.bar(x, pmf, color='blue')plt.xlabel('Number of Successes')plt.ylabel('Probability')plt.title(f'Binomial Distribution PMF (n={n}, p={p})')plt.show() 'blue' 'Number of Successes' 'Probability'f'Binomial Distribution PMF (n={n}, p={p})'{n}{p} This code will generate a bar plot of the binomial distribution, showing the probability of each possible number of successes. Interpretation of the Graph When you look at the graph, you’ll see how the probabilities are distributed across different numbers of successes: Symmetry: For p = 0.5, the distribution is symmetric around n/. This means that the probabilities of getting k successes are the same as getting n − k successes. Skewness: For p≠0.5p, the distribution will be skewed. If p > 0.5, the distribution will be skewed to the right, indicating higher probabilities for more successes. Conversely, if p < 0.5, it will be skewed to the left. Spread: The spread of the distribution depends on n. A larger n results in a wider spread, showing a broader range of possible outcomes. Cumulative Distribution Function (CDF) CDF of Binomial Distribution The Cumulative Distribution Function (CDF) gives you the probability that a random variable takes on a value less than or equal to a certain value. In the context of the binomial distribution, the CDF helps you understand the probability of having up to a certain number of successes. Formula for the CDF: The CDF is the sum of the PMF values up to a given point k: Example Calculation and Interpretation Let’s continue with our previous example where n=10n = 10n=10 and p=0.5p = 0.5p=0.5. Suppose you want to find the probability of having up to 4 successes. Calculate the CDF: cdf = binom.cdf(4, n, p)print(cdf) print This will give you the probability that you get 4 or fewer successes in 10 trials. Interpret the CDF: If the CDF value is 0.376, this means there is a 37.6% chance of having 4 or fewer successes out of 10 trials. Applications and Use Cases The binomial distribution is not just a theoretical concept; it has numerous practical applications across various fields. Here are a few real-world scenarios: Quality Control: In manufacturing, you might use the binomial distribution to determine the probability of a certain number of defective items in a batch. For example, if a factory produces light bulbs with a 2% defect rate, you can calculate the probability of finding 3 defective bulbs in a sample of 100. Finance: In finance, the binomial distribution can model the probability of a certain number of profitable trades out of a series of trades. If each trade has a 60% chance of being profitable, you can use the binomial distribution to assess the risk and return of a trading strategy. Medicine: In clinical trials, you can use the binomial distribution to model the probability of a certain number of patients recovering from a treatment. If a drug has a 70% success rate, you can calculate the likelihood of different numbers of successful treatments in a group of 50 patients. Case Studies or Hypothetical Examples Case Study: Manufacturing: Suppose a car manufacturer wants to ensure that no more than 5% of the cars produced are defective. They test 200 cars and find 8 defects. Using the binomial distribution, they can calculate the probability of this outcome and determine if the defect rate is within acceptable limits. Hypothetical Example: Marketing Campaign: Imagine you’re running an email campaign where the open rate is 15%. If you send out 1,000 emails, you can use the binomial distribution to predict the probability of different numbers of opens. This helps you set realistic expectations and evaluate the campaign’s performance. Conclusion In this exploration of the binomial distribution, you’ve learned about its definition, real-world applications, and how to calculate and visualize it. You’ve also discovered the differences between the PMF and PDF, and how the CDF adds another layer of understanding to the distribution. Understanding the binomial distribution and its PMF is essential for your statistical analysis toolkit. It helps you model binary outcomes in various fields, from quality control to finance and medicine. By mastering these concepts, you’re better equipped to make data-driven decisions and conduct rigorous statistical analyses. Pmf Probability Density Machine Learning Ai Statistics ## Written by Amit Yadav 1.5K followers ·141 following Get Data Science Roadmap for Your First Data Science Job : No responses yet Write a response What are your thoughts? More from Amit Yadav Amit Yadav ## Best Laptops For Data Science in 2025 I’ve found that laptops with Intel i7 or i9 processors, as well as AMD Ryzen 7 or 9, offer the best performance. But are they good overall? Jun 23, 2024 232 4 Amit Yadav ## What is pandas.pipe() and Why Should You Use It? You know how messy things can get when you’re trying to juggle multiple tasks at once? 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16397	https://math.answers.com/statistics/For_an_odd_number_of_data_how_is_the_median_computed	For an odd number of data how is the median computed? - Answers Create 0 Log in Subjects>Math>Statistics For an odd number of data how is the median computed? Anonymous ∙ 8 y ago Updated: 4/28/2022 For an odd number of data values, the median is the middle number, the [(n+1)/2]th number i.e. for 7 data values, (7 +1)/ 2 = 4, and the 4th is the middle value, or median. For an even number of values, the median is the mean of the two middle numbers, i.e. one-half the sum of the two middle values (add n/2nd value and n/2+1st values and divide by 2). Examples: Median of 1, 3, 2 reordered as 1, 2, 3 = median 2 Median of 6, 5, 3, 1 reordered as 1, 3, 5, 6 = median 4 (3+5 divided by 2) Wiki User ∙ 8 y ago Copy Add Your Answer What else can I help you with? Search Continue Learning about Statistics ### What is the middle number in set of data? The median. If there are an odd number of elements in the set, there is a middle number which is the median. If there are an even number of elements in the set, the median is the mean of the middle two numbers. ### Is median always sometimes or never in a data set? If the sample has an odd number of items in it then the median will definitely be in the sample at least once because the median is value of the set of data items whose value(s) are in the middle of the sample when the sample is sorted from smallest to largest. If the sample has an even number of items in it then if the middle items are different the median will be their average, and it will differ from all of the items in the data set. I could continue in this vein but already you can see that the median sometimes occurs in a data set but not always. ### How is Median determined both in data sets with an odd and even number of values? Odd: Line numbers up in numerical order, middle number. Even: Same as above, find two middle numbers, find mean of two middle numbers. ### What is the Mathematical meaning of the median? The median is the middle value in a set of data. It is the value that separates the higher half of the data from the lower half. If there is an even number of data points, the median is the average or mean of the two middle values. If there is an odd number, just arrange the data in order from smallest values to largest and the middle one is the median. ### Step by step in finding the median of ungrouped data? Steps:1.) Arrange the data either ascending or descending order of their values.2.) Determine the total number of observations, say n.3.) If the set of number is odd then the middle number will be the median. And if the set of number is even then mean of middle two numbers will be the median.a.) After that we'll be starting to discuss and give examples on how to find the median of ungrouped data.Example1:1.) Find the median of 12, 15, 10, 18, 8.2.) Arrange the data either ascending or descending order of their values.8, 10, 12, 15, 18(ascending)18, 15, 12, 10, 8(descending)3.) If the set of numbers is an odd integer, find the middle number in the set of numbers.From the above middle number 12 then 12 is the median.Example2:1.) Find the median of 23, 46, 18, 32, 65, 20.2.) Arrange the data either ascending or descending order of their values.18, 20, 23, 32, 46, 65(ascending)65, 46, 32, 23, 20, 18(descending)3.) If the set of numbers is an even integer, find the two middle numbers in the set of numbers.From the above we can say that there are two middle numbers23 and 32. So,Md = 23+32/2 = 27.5By: HuebosFb: Jupete02@yahoo.comTwitter: @kimjupetehuebos Related Questions Trending Questions What is the pattern of a variability within a data set called?Is 300 pounds 6.2 male consider to fat?What is the probability of 6 any numbers in lottery of 40 numbers?Is 4000 reasonable estimate for the difference of 9215-5022?Use of statistics in information technology?How many 3 letter combinations can be made from 10 letters?What is the answer to a pie chart of main goods that minnesota produces?What increases the chances of rejecting null hypothesis?Why would you end up with a double bar graph if you would plot the frequencies of two phenotypes of a single-gene trait?What is the definition of nonlinear plot?What is the theoretical probability of drawing a joker out of a deck of cards?What part of speech is the word estimated?What is the difference between static white box testing and dynamic white box testing?What percentage of wealth is held by women?What is 3 percent of 325000?Value received and value parted with about Transaction?How can government officials use the crime statistics data from the 1970's to predict crime twenty years later?Is it bad for a guy to have unprotected sex when a girl has her period?How does a graph show me that i have anomalous result?How do you conduct the process of requirement analysis? 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16398	https://statproofbook.github.io/P/lognorm-qf.html	Quantile function of the log-normal distribution \| The Book of Statistical Proofs The Book of Statistical Proofs- [x] AboutCitationsContributeCredits Proof: Quantile function of the log-normal distribution Index:The Book of Statistical Proofs ▷ Probability Distributions ▷ Univariate continuous distributions ▷ Log-normal distribution ▷ Quantile function Theorem: Let X X be a random variable following a log-normal distribution: X∼ln N(μ,σ 2).(1)(1)X∼ln⁡N(μ,σ 2). Then, the quantile function of X X is Q X(p)=exp(μ+√2 σ⋅e r f−1(2 p−1))(2)(2)Q X(p)=exp⁡(μ+2 σ⋅e r f−1(2 p−1)) where e r f−1(x)e r f−1(x) is the inverse error function. Proof: The cumulative distribution function of the log-normal distribution is: F X(x)=1 2[1+e r f(ln x−μ√2 σ)].(3)(3)F X(x)=1 2[1+e r f(ln⁡x−μ 2 σ)]. From this point forward, the proof is similar to the derivation of the quantile function for the normal distribution. Because the cumulative distribution function (CDF) is strictly monotonically increasing, the quantile function is equal to the inverse of the CDF: Q X(p)=F−1 X(x).(4)(4)Q X(p)=F X−1(x). This can be derived by rearranging equation (3)(3): p=1 2[1+e r f(ln x−μ√2 σ)]2 p−1=e r f(ln x−μ√2 σ)e r f−1(2 p−1)=ln x−μ√2 σ x=exp(μ+√2 σ⋅e r f−1(2 p−1)).(5)(5)p=1 2[1+e r f(ln⁡x−μ 2 σ)]2 p−1=e r f(ln⁡x−μ 2 σ)e r f−1(2 p−1)=ln⁡x−μ 2 σ x=exp⁡(μ+2 σ⋅e r f−1(2 p−1)). ∎ Sources: Wikipedia (2022): "Log-normal distribution"; in: Wikipedia, the free encyclopedia, retrieved on 2022-07-08; URL: Metadata: ID: P326 \| shortcut: lognorm-qf \| author: majapavlo \| date: 2022-07-09, 11:05. view/edit this proof cite this proof The Book of Statistical Proofs StatProofBook@gmail.com StatProofBook StatProofBook License The Book of Statistical Proofs – a centralized, open and collaboratively edited archive of statistical theorems for the computational sciences; available under CC-BY-SA 4.0.
16399	https://mathoverflow.net/questions/136479/distribution-of-moduli-of-quadratic-residues	Skip to main content Distribution of moduli of quadratic residues Ask Question Asked Modified 11 years, 7 months ago Viewed 891 times This question shows research effort; it is useful and clear 9 Save this question. Show activity on this post. Let D be a fixed positive squarefree integer. For a positive integer x, define S(D,x)={q<x:Dis a quadratic residue(modq)}. Here q can be any integer, not necessarily a prime. Are elements of S(D,x) evenly distributed? In other words, let 0<a<b<1 be constants and consider an interval I=[ax,bx]. Ideally I would like to see some result that says that the number of elements of S(D,x) in I is proportional to the length of I on the average as x goes to infinity (is this true?). I am aware of classical 1917-1918 results of Vinogradov and Polya (and some later developments) about distribution of quadratic residues, which in particular imply that quadratic residues modulo a fixed prime p are evenly distributed in the interval [0,p] in the same sense as I described above. What I need, however, is a result on the distribution of moduli with respect to which a fixed integer is a quadratic residue, and I cannot find anything like this in the literature. In other words, I am wondering how the divisors of x2−D are distributed on the average as x goes to infinity. It is a well-known fact that for an arbitrary integer y, there are unproportionally many (on the average) small and large divisors of y as y goes to infinity, i.e., divisors are not uniformly distributed. But what if we take y in the special form x2−D? Any thoughts on the subject are much appreciated! nt.number-theory Share CC BY-SA 3.0 Improve this question Follow this question to receive notifications edited Jan 12, 2014 at 21:03 tc1729 14577 bronze badges asked Jul 12, 2013 at 7:08 Lenny FukshanskyLenny Fukshansky 35511 silver badge55 bronze badges 4 1 If you were to restrict to q prime then for D>1, quadratic reciprocity, the Chinese remainder theorem, and the equidistribution of primes in residue classes (Vallée-Poussin) give you equidistribution. For example, suppose D=6. 2 is a quadratic residue mod q>2 iff q=±1mod8, and 3 is a quadratic residue iff q=±1mod12, so 6 is a quadratic residue when q=1,5,19,23mod24, and it's a quadratic nonresidue when q=7,11,13,17mod24. – Douglas Zare Commented Jul 12, 2013 at 8:13 Here’s a guess: D is a square modulo a random prime (or prime power) with probability 1/2, hence it is a square modulo a random number with k prime factors with probability 2−k. Since a random integer below n has k≈loglogn prime factors on average, which is in particular nonconstant, I’d expect that limx→∞\|S(D,x)∩[ax,bx]\|\|[ax,bx]\|=0. – Emil Jeřábek Commented Jul 12, 2013 at 11:27 @EmilJeřábek Your initial argument suggests that \|S(D,x)\| is asymptotic to x/(logx)log2, no? If that holds with a reasonable error term (which is not clear at all) then \|S(D,bx)\|−\|S(D,ax)\| will be asymptotic to (a−b)\|S(D,x)\|. – Felipe Voloch Commented Jul 12, 2013 at 11:43 1 @FelipeVoloch: The OP asked for it to be proportional to the length of I, not to \|S(D,x)\|. Anyway, I’d be very cautious with such calculations. For one thing, that the average order of k is loglogx does not mean that the average order of 2−k is (logx)−log2. In fact, for constant k, there are ∼x(loglog)k−1(k−1)!logx numbers in [1,x] with k prime factors, which would rather suggest that \|S(D,x)\| is of order x/logx−−−−√. It’s still unclear to me whether this is the right exponent, and whether \|S(D,x)\|logx−−−−√/x actually converges. – Emil Jeřábek Commented Jul 12, 2013 at 14:13 Add a comment \| 1 Answer 1 Reset to default This answer is useful 8 Save this answer. Show activity on this post. For D=−1, Landau proved that S(−1,x)∼Kxlogx−−−−√ where K=12√∏p≡3mod411−p−2√. This shows that, for fixed 0<a<b, S(−1,x)∩[ax,bx]∼K(bxlog(bx)−−−−−−√−axlog(ax)−−−−−−√) and K(bxlog(bx)−−−−−−√−axlog(ax)−−−−−−√)=K(bxlogx+logb−−−−−−−−−−√−axlogx+loga−−−−−−−−−−√) =K(bxlogx−−−−√+O(x(logx)3/2)−axlogx−−−−√−O(x(logx)3/2))∼K(b−a)xlogx−−−−√ So they are equidistributed in that #(S(−1,x)∩[ax,bx])#S(−1,x) approaches b−a but, if you ask for the denominator to be x, then the limit just goes to zero. I believe the same should be true for any nonsquare D. Let ZD(s)=∏(Dp)=111−p−s=∑D is a square modulo n1ns. The proof of Landau's theorem in Leveque's "Topics in number theory" comes down to analyzing the behavior of ZD(s) on Re(s)=1. The key facts are that Z−1(s)=C(s−1)1/2+⋯ and Z−1(s) is otherwise bounded on Re(s)=1. These results hold for any nonsquare D. Let ζD(s) be the zeta function of Q(D−−√). Then ZD(s)2ζD(s)=(factors for primes dividing 2D)×∏(Dp)=−1(1−1p2s). The right hand side is clearly convergent to the right of Re(s)=1/2, and ζD(s) is well known to have a simple pole at s=1 and no other poles on the line Re(s)=1. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Jul 15, 2013 at 14:19 David E SpeyerDavid E Speyer 159k1414 gold badges432432 silver badges779779 bronze badges 1 David, great -- thank you very much! In fact, Gang Yu pointed out to me that this result can be obtained by an argument entirely analogous to the proof of the main theorem in: James, R. D.; The Distribution of Integers Represented by Quadratic Forms. Amer. J. Math. 60 (1938), no. 3, 737–744 jstor.org/discover/10.2307/… Here your function ZD(s) is the first product in the formula for f(s) in Lemma 3 of James' paper. – Lenny Fukshansky Commented Jul 15, 2013 at 21:35 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions nt.number-theory See similar questions with these tags. Related 67 Perron number distribution 7 Are there Carlitz analogues of quadratic residues and reciprocity? 1 A distribution related to Fermat's two squares theorem 4 Distance between primes that are quadratic residues modulo an other prime Question feed By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.

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