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n’t help in finding eigenvalues in the infinite dimensional case, it is still true, by the same argument, that the eigenvalues of a self adjiont operator are real. Our inner product space will be a certain subspace of function space C[0, 2π], the continuous functions f : [0, 2π] → R with the usual inner product (f, g) = ... |
a local minimum at r. The contribution of linear algebra is to provide the theorem which classifies when a general quadratic form q satisfies q(x) > 0 if x 6= 0 (see Proposition 12.6). In critical point theory, this result is called the second derivative test. The following example is a preview of the general result. 359... |
s positive definite. Similarly, one can also show Proposition 12.8. A symmetric A ∈ Rn×n is negative definite if and only if (−1)mdet(Am) > 0 for all m ≤ n. Proof. The proof is yet another an exercise. Here is another example. Example 12.5. Consider the matrix 1 0 1 1 1 0 2 −1 0 0 2 2 0 . One finds that det(A1) = ... |
ular. The point of the QR-algorithm, as we will see, is that in the eigenvalues of A are sometimes the diagonal entries of R. 12.3.2 The QR-Convergence Theorem To understand the QR-algorithm, we need to define what is meant by a convergent sequence of matrices. Definition 12.2. A sequence Am = (a(m) be convergent if and ... |
and moreover, it’s also crucial in proving many of the deep results about the space L(V ) of all linear transformations T : V → V . Our proof of the Jordan Decomposition Theorem illustrates the importance of the Cayley-Hamilton Theorem, which we proved in Chapter 9, but only for diagonalizable matrices. This Theorem i... |
) = C(2, −1, 1)T . Now (T − I3)2 = −2 0 1 −1 0 6 0 −3 3 , which clearly has rank one. Its kernel, which is spanned by 0 1 and 0 , 3 0 1 is therefore the other invariant subspace of T . Hence the semi-simple linear transformation S is determined by ) = S( 3 0 1 ) = 3 0 1 ... |
6 are . Thus there are 10 partition of 6, so π(6) = 10. The partition function grows very rapidly. The upshot of the Jordan Canonical Form is that to each partition (n1, n2, . . . , ns) of n, there is a nilpotent matrix of the form (13.10) (with only zeros on the diagonal, of course), and every n × n nilpotent matrix ... |
s a family of planes a · x = d completely filling up R3 such that two planes either coincide or don’t have any points in common. Hence the family of planes ax + by + cz = d (a, b, c fixed and d arbitrary) are all parallel. By drawing a picture, one can see from the Parallelogram Law that every vector on ax + by + cz = ... |
In this case we call (a, b, n) a right handed triple. (Otherwise, it’s a left handed triple.) Let θ be the angle between a and b, so 0 < θ < π. Then we put a × b = |a||b| sin θn. (14.11) If a and b are collinear, we set a × b = 0. While this definition is very elegant, and is useful for revealing the geometric properti... |
tion of the planes 3x+2y−z = 2 and 4x+5y+z = 1. Exercise 14.31. Is x × y orthogonal to 2x − 3y? Generalize this. Exercise 14.32. Find the distance from (1, 2, 1)T to the plane containing 1, 3, 4)T , (2, −2, −2)T , and (7, 0, 1)T using the cross product. Exercise 14.33. Formulate a definition for the angle between two pl... |
sin(1/x) over the interval (0, 1] extended by the single point (0,0). This set is then equipped with the topology induced from the Euclidean plane. It is connected but not path-connected. It is the continuous image of a locally compact space (namely, let V be the space -1 ∪ (0, 1], and use the map f from V to T defined ... |
ntify the entire horizontal fiber (x, 1) down to on point. (x, 1 − ) is in a neighborhood of this point for all x, so we have the space, as below. And from this transformation we can see why it is called as cone, the product space is compressed to one point as we approach (x, 1). Example 2.4.8. The suspension SX of X is... |
ean topology. Then an example of a compact set in R1, for example, would be [0,1]. This can be seen from the definition of a compact set and the example above where it is illustrated that this interval has a finite subcover. 40 Mathematics 490 – Introduction to Topology Winter 2007 In fact, when Rn has the Euclidian topo... |
perty of a space being Hausdorff provides a nice, easy test to see if a space is not metrizable. If a topological space X is not Hausdorff, then it is not metrizable. Note that being Hausdorff is a necessary but not sufficient condition. All of this will be proved in a moment, but for the time being let’s get the formal defi... |
ntioned before, this ensures that ˜X is not ”too large”, that is, not ”too much larger” than X. Proof. First let’s show that X is homeomorphic to the set {X} ⊂ ˜X. But that’s easy: construct a function that sends each point of X to the corresponding point in {X}. This function is obviously one-to-one and onto, and it i... |
point and let f : X → Y be an onto continuous function. Define the nonempty open sets U , V such that Y = U ∪ V . Since f is continuous, X = f −1(U ) ∪ f −1(V ), but this contradicts our assumption (c). Therefore, Y can’t be a discrete space with more than one point, i.e., there is no onto continuous function from X to... |
plane To see this, place a plane above the sphere as shown in figure 4.2. Then, note that we can define a function from a subset of RP2 to R2 by sending each line to its intersection with R2. However, by doing this, we actually miss the lines that lie in the xy-plane. To remedy this, we add another line that lies in the... |
us, and the projective plane. We are working towards the proof of this fact, but first, we must find the identification polygons of our two other building blocks: the sphere and the torus, just as we did for the projective plane. Once we have these polygons, we will be able to take any arbitrary identification polygon and ... |
e it. Proof: There are three actions which can modify any good graph. These three actions are: 1. Add a vertex 2. Add an edge 3. Add a vertex and an edge Given an arbitrary surface, in this case, a sphere with the good graph in figure 4.32 on it. Counting the number of vertices, edges and faces gives us 6 vertices, 7 ed... |
d each base of the cylinder, connecting the vertex on the border to itself. Finally we add an edge from the vertex on one edge of the cylinder to the vertex on the other edge of the cylinder. By adding these edges we divide the cylinder so that the complement of the graph is a disc. The final graph is shown in figure 4.5... |
p based at p. Sliding from the bottom of the square to the top will give us a continuous family of loops starting at α and ending at β. Figure 5.3 illustrates this situation for two loops on a torus. Figure 5.3: Two loops on the torus Theorem 5.1. Homotopy is an equivalence relation. Proof: We need to show that a homot... |
) is some point in D, say the center for simplicity. So, f ◦ g = P , and therefore f ◦ g ’ IY . For g ◦ f , we have that (g ◦ f )(α) = C, where C is the center of the disk. Radial retraction shows that the map g ◦ f is homotopic to IX . Informally, spaces are homotopy equivalent when one can be continuously “shrunk” i... |
en the fundamental group of the original space can be easily obtained. The following is the precise statement of the Seifert-Van Kampen Theorem: Theorem 5.5.1 (Seifert-Van Kampen). Given a space X, let X1 and X2 be subspaces of X such that X1 ∪X2 = X and is open, X1 ∩X2 6= {φ} and is open, and X1 ∩X2 is path-connected.... |
actually isn’t. Consider the map from the plane to the projective plane, π : R2 −→ RP 2 which is defined as shown in Figure 5.26, which also shows the preimage of a disc. Note that adjacent projective planes1 are flipped so that the correct sides match. This is needed so that π is continuous. Figure 5.26: Map from the p... |
ors of n = 8 are … 1; 2; 4; 8. (Don’t forget negative divi- sors.) • Very often we can write this generically, so for example n j x + 1 means that x + 1 can be written as the product of n and some other integer m. We occasionally use the term proper divisor to denote a positive divisor of n which is not n. When n = 8, ... |
are several things to note about this early computation. First, note that the answer to divmod came in parentheses, a so-called tuple data type. Second, there is another way to approach this computation, more programmatically so that it’s easier to reuse. What do you think the [0] and [1] mean? divmod (281376 ,29) [0] ... |
addition to being the main geometry textbook in the Western and Islamic worlds for two millennia (as late a teacher as Charles Dodgson a.k.a. Lewis Carroll extolled its virtues in print in Euclid and His Modern Rivals7), there are substantial number-theoretic portions as well. No one really knows how much of the Elemen... |
̸= 1, a j c and b j c, but ab does not divide c? (See Exercise 2.5.20.) 2.5 Exercises 1. Try stating and proving the division algorithm (Theorem 2.1.1) but for b < 0. 2. Can you find an n such that the possible remainders of a perfect square when divided by n are all numbers between zero and n 1? If you can, how many ... |
f the answers are right.) • It’s easy to check that any particular solution works. ( a x0 + ) ( n + b b d y0 a d ) n = ax0 + abn d + by0 abn d and ax0 + by0 = c by hypothesis. • Why does this give all solutions? First note that since the only common d must be relatively divisors of a and b are divisors of d, the intege... |
ncrete, let’s assume that the equation is ax+by = c, and gcd(a; b) = 1. Then, using our technique from last time, from the solution (x0; y0) we get a new solution (x0 + b; y0 a), so the distance between any two solutions is, by the Pythagorean Theorem, √ √ [(x0 + b) x0]2 + [(y0 a) + y0]2 = a2 + b2. Our strategy is to a... |
even, (which means z is odd). 3.4.2.2 An intricate argument We have now reduced our investigation to the following case: we assume that gcd(x; y; z) = 1, that x; z are odd, and that y is even. Now we will do a somewhat intricate, but familiar, type of argument about factorization and divisibility. Let’s rewrite our si... |
CHAPTER 3. FROM LINEAR EQUATIONS TO GEOMETRY 36 that unless p = 0, c is strictly less than z. But p = 0 doesn’t give a triangle at all! So we have our strictly smaller triangle satisfying the same properties. ■ Corollary 3.4.12 No Pythagorean triangles can have area a perfect square. Proof. If so, we can use the previ... |
ch satisfy x2 xy + y2 = z2. If Pythagorean triples correspond to right triangles, what sort of triangles do these triples correspond to? CHAPTER 3. FROM LINEAR EQUATIONS TO GEOMETRY 40 21. Find a (fairly) obvious solution to the equation mn = nm for m ̸= n. Are there other such solutions? 22. Show that gcd(x; y)2 = gcd... |
l of brains, like the Scarecrow in L. Frank Baum’s Oz books? Of course, the reason is not that I am clever, but that congruence can be turned into arithmetic! Unlike the Wizard, I will give away my secret. I just used the following useful property. Fact 4.2.3 If a b (mod n), then am bm (mod n) no matter how huge m is. ... |
ules for using modular arithmetic. • First off, always first reduce modulo n, then do your arithmetic (add, multiply, exponentiate). We have seen lots of examples of this. • Secondly, always use the most convenient residue (recall Definition 4.4.6) of a number modulo n. Example 4.5.1 For example, to add [22] + [21] mod... |
0; 1 and n ̸= 0. 17. Try to find some a and n such that an (mod 5) equals an+5 (mod 5), where a ̸ 0; 1 and n ̸= 0. 18. Find some a and n such that an (mod 6) equals an+6 (mod 6), where a ̸ 0; 1 and n ̸= 0. Then try to find an example where they are not equal. 19. Explore, using the interact after Question 4.6.7 or ‘by ... |
3). □ CHAPTER 5. LINEAR CONGRUENCES 59 Example 5.2.4 Try completely solving one of the following two congruences (Exercise 5.6.3) on your own now, before moving on. The rest of the Exercises provide other interesting practice. • 7x 8 (mod 15) • 6x 8 (mod 14) □ Example 5.2.5 Finally, let’s see examples of using the stra... |
basically this proof in any text; I use the notation in [E.2.1], while that in [E.2.4] basically uses the letter m instead of n. Algorithm 5.4.1 The following steps not only yield the solution, but mostly indicate the proof as well. 1. First, let’s call the product of the moduli n1n2 nk = N . 2. Take the quotient N /n... |
olutions in the form x ai (mod ni). Then you can use the CRT to get a solution of that system, which is also a solution of the ‘big’ system. 4www-history.mcs.st-andrews.ac.uk/Biographies/Lebesgue_Victor.html CHAPTER 5. LINEAR CONGRUENCES Example 5.5.1 For instance, try now to solve this system: • 2x 2 (mod 5) • 5x 4 (m... |
ime (f (40) ) , factor (1681) ( False , 41^2) This example is due to Euler2. For this form of polynomial it is the best known3, but you may have thought (based on the scanty evidence of this one example) that one could eventually find a polynomial which just gives primes. Quite the opposite is true! Fact 6.1.4 There is... |
same sense as the others; see Joyce’s commentary16. □ Example 6.3.4 Usually we will implicitly assume the primes are in nondecreasing order, and write 32 instead of 3 3 (with the primes now necessarily in increasing order), so the following notation is common to express a prime power factorization: n∏ N = i=1 pei i . S... |
N ), and N = i . Then we have an equivalence between this congruence and a related system of congruences. ∏ • Certainly if N divides X Y , so does every factor of N , so X Y (mod pei i ) for each of the prime power factors of N . (Once again, solutions to the “big” congruence are also solutions to a system of many lit... |
g the relatively simple (and computable) cases of quadratic congruences. Much later, we will return to a full investigation of this. You may recall that we looked at one particular quadratic congruence in Question 4.6.7 and Exercise 4.7.19, and saw that the solution depended at least partly on the modulus in Exercises ... |
od practice to try the same process with x1 = 18 instead in Exercise 7.7.3. □ 1One way or another one of these series will have to enter in, unfortunately; [E.2.1, Section 4.3] has more of a binomial theorem-esque treatment, while [E.2.13, Theorem 4.7] and [E.5.1, Theorem 6.2] more explicitly invoke Taylor series. CHAP... |
n the future. 7.5.1 Wilson’s Theorem Theorem 7.5.1 Wilson’s Theorem. If p is a prime, then (p 1)! 1 (mod p), where the exclamation point here indicates the factorial. Proof. If p = 2 this is very, very easy to check. So assume p ̸= 2, hence p 1 is even. Now we will think of all the numbers from 1 to p 1, which will be ... |
at the modulus n then has factors m other than 1 or n. 12. Prove that it is impossible for p j x2 + 1 if a prime p has p 3 (mod 4) – that is, if p is of the form 4n + 3. (Hard7.) 13. Prove that x2 + y2 = p has no (integer) solutions for prime p with that same form 4n + 3. 14. Show that y2 = x3 + 999 has no (integer) so... |
osely by our previous experience. 15mathworld.wolfram.com/Cauchy-FrobeniusLemma.html CHAPTER 7. FIRST STEPS WITH GENERAL CONGRUENCES 105 5. Two towering theorems giving theoretical tools to harness more complex congruences are Wilson’s Theorem and Fermat’s Little Theorem. 6. Finally, we explore Mordell curves again in ... |
is.org/ 02468101357913579 CHAPTER 8. THE GROUP OF INTEGERS MODULO N 112 Blues BrBG BuGn ... tab20b_r tab20c_r What color patterns can you see here? To say it another way, what potential theorems do you see? (Again, do you see any that we already have discussed?) In a classroom or self-study situation, I strongly recomm... |
e pigeonhole principle, among other names.) To be concrete, let’s say xs = xt, with s < t. Now we can do a very curious thing. Take the inverse of x, written x1. If we multiply it together s times, we get (x1)s which we can write xs. Then multiply xs = xt by xs; xsxs = xsxt, or e = xts. We are almost there! This means ... |
ll have solutions to ax 1 according to Proposition 5.1.1; see also Question 5.3.6 and Exercise 5.6.5. Finally, we do need to check whether the multiplication is closed on this set. After all, it’s not obvious that if ax 1 and bx 1 have solutions, then so does (ab)x 1! But if gcd(a; n) and gcd(b; n) are both 1, then ab ... |
, m_3 )) print ( b_1 ^( euler_phi ( m_1 ) -1) * c_1 * d_1 ^( euler_phi ( m_1 )) + b_2 ^( euler_phi ( m_2 ) -1) * c_2 * d_2 ^( euler_phi ( m_2 )) + b_3 ^( euler_phi ( m_3 ) -1) * c_3 * d_3 ^( euler_phi ( m_3 ))) It’s worth trying to time this – recall that we can use %time for this in notebooks, see Sage note 4.2.1. Th... |
: : : ; [p 2]; [p 1]g have gcd(a; p) = 1. 6. Verify Euler’s Theorem by hand for n = 15 for all relevant a (note that ϕ(15) = 8, and remember that a8 = ((a2)2)2 so we can use modulo reduction at each squaring). 7. Get the inverse of 29 modulo 31, 33, and 34 using Euler’s Theorem. CHAPTER 9. THE GROUP OF UNITS AND EULER... |
lib . pyplot as plt from matplotlib . ticker import IndexLocator , FuncFormatter @interact def power_table_plot (p =(13 , prime_range (5 ,50) ) ): mycmap = plt . get_cmap ( ' gist_earth ' ,p -1) myloc = IndexLocator ( floor (p /5) ,.5) myform = FuncFormatter ( lambda x ,y: int (x +1) ) cbaropts = { ' ticks ' : myloc , ... |
one of n a) come to the same thing. Since ϕ(n) is already assumed to be even, the only possible odd ϕ(n)/q comes from q = 2, but ϕ(n) is assumed to be divisible by four, so ϕ(n)/q will be even. ■ Example 10.2.6 If you did the table at the beginning of this subsection properly, you will note that 3 and 14 are a pair of ... |
inally, for those with more experience with groups, a good exercise would be to see whether Claim 10.4.4 converts into a statement about the number of elements of each order of any cyclic group. Now let’s prove our claim. Claim 10.4.4 If p is prime, the number of elements of Up of order d is ϕ(d) (where of necessity d ... |
). 6. Prove Lemma 10.3.4. Suppose p is prime and the order of a modulo p is d. Prove that if b and d are coprime, then ab also has order d modulo p. Hint: actually write down the powers of ab, and figure out which ones could actually be 1. Lagrange’s (group) Theorem 8.3.12 could also be useful. 7. Prove Lemma 10.3.5. S... |
blocks of letters, and then turn those into numbers. After all, presumably there are a lot more three-letter (or longer) possible blocks of letters in English than would make it too easy to decrypt them. (Can you think of exceptions, though?) For pairs, we will represent the first letter as a number from 1 to 26, and ... |
. Recall that we have methods to solve modular exponential congruences (such as using primitive roots). That gives us tools sufficient to implement these subtle techniques. Sage note 11.3.1 Another reminder to evaluate definitions. Don’t forget to evaluate the commands below so we can use words as messages instead of j... |
ist You should have gotten an error (in fact, a ZeroDivisionError, which It turns out not even to be possible to go backshould sound relevant). wards. Be warned that you must know the mathematics to use cryptography CHAPTER 11. AN INTRODUCTION TO CRYPTOGRAPHY 167 wisely. 11.4 An Interesting Application: Key Exchange Th... |
␣ product ␣ of ␣ primes ␣ bigger ␣ than ␣ that ␣ is ␣ $pq =% s \ cdot %s =% s$ " %(p ,q ,n))) pretty_print ( html (r"( which ␣ means ␣ my ␣ secret ␣ $\ phi (n) =\ phi (% s\ cdot ␣%s) =(% s -1) (%s -1) $ ␣ is ␣ $% s$ )" %(p ,q ,p ,q , phi ))) pretty_print ( html (" And ␣I␣ chose ␣ exponent ␣$ % s$ " %e )) pretty_print ... |
olleague, Russ Tuck, for this observation. CHAPTER 11. AN INTRODUCTION TO CRYPTOGRAPHY 176 My encoded message is 197108322 A big product of primes bigger than that is pq =(197108347) (591324977) =116555088756283019 ( which means my secret phi (n)= phi ((197108347) (591324977) ) =(197108347 -1) (591324977 -1) is 1165550... |
OGRAPHY 180 Proof. What good do these do us? Well, the Chinese Remainder Theorem allows us to reconstruct K0 modulo mimj with any two keys ki and kj. That may not seem like a lot; that just gives us things to within multiples of mimj. But by our choice of M = m1m2 > pm3, we know that M /p > m3 (and hence M /p > mi as w... |
mma. Lemma 12.1.2 Suppose ℓ = jk is even, and k is an even factor. Then 2ℓ 1 factors as 2ℓ 1 = 2jk 1 = )k1 Proof. Multiply and/or apply a little induction. (See Exercise 12.7.1.) )k3 + 2j + 1 )k2 2j 2j 2j 2j + ( ( ( ( ) (( ) ) 1 ■ Example 12.1.3 For instance, 26 1 = 63 factors as 232 1 = (23 + 1)(23 1) which correspond... |
e ancient Chinese to Leibniz used this test for the base a = 2 to assert numbers are prime. And it doesn’t do a bad job. As some former students pointed out, it’s sort of like internet date matching for primes; it doesn’t always work but can succeed reasonably often. 02468101357913579 CHAPTER 12. SOME THEORY BEHIND CRY... |
tself by two and then taking a to that new power. Once again, if the result is 1 we say n passes the test, and if it is not 1, we say it fails. • If the result is +1 and we can continue dividing the power by two, do so and check the result, as often as need be. If we arrive at the point where we have divided n 1 by all... |
) 2(1) = 30 p 3600 3596 2 = 30 1 = 29; 31. □ 12.5.2 Trial division The first, and oldest, method of factoring is one you already know, and maybe used a few minutes ago – trial factorization, or trial division. It is the method we used with the Sieve of Eratosthenes; you just try each prime number, one by one. In Algori... |
ain significance is that if a sizable quantum computer implementing this algorithm could be built, it could perform the specific tasks of factoring large RSA moduli in polynomial (rather than exponential) time. As of this writing, we’re still a long way from reliable physical implementations of any size, but they are a... |
ation algorithms use several different methods to attack different types of factors. We can try to simulate this in a basic way by creating a Sage interact. Evaluate the first cell to define things (don’t forget to keep the rho method defined); then you can evaluate the second cell, which is the interact. def TrialDivF... |
a sum of squares, how might you prove this? A separate question to at least keep track of is this. Question 13.0.5 Assuming you can indeed write it in this way, how many □ ways you can write a number as a sum of squares? This chapter is completely devoted to continuing to address questions about writing numbers as a s... |
licit_plot (g -n , ( -1 , viewsize ) , ( -1 , viewsize ) , plot_points = 100) lattice_pts = [[i ,j] for i in [ -1.. viewsize ] for j in [ -1.. viewsize ]] plot_lattice_pts = points ( lattice_pts , rgbcolor =(0 ,0 ,0) , pointsize =2) curve_pts = [ coords for coords in lattice_pts if g( coords [0] , coords [1]) == n] if ... |
e , xmax = viewsize , ymin = - viewsize , ymax = viewsize , aspect_ratio =1) For one final preliminary, let’s define one more thing for any old point (x; y) in the integer lattice (and especially for our blue dots). Definition 13.4.4 We call the norm of a point (x; y) the sum of squares, ♢ N (x; y) = x2 + y2. 13.4.2 Pr... |
e large interact in Example 13.4.13 which ends the section. p Claim 13.4.11 Consider the circle of radius 2p centered at the origin. The interior of the disk bounded by this circle has two points “repeated” by shifting the parallelogram L. Proof. Recall from Fact 13.4.9 that the disk is composed of all its intersection... |
b to get M = c2 + d2 CHAPTER 13. SUMS OF SQUARES 233 The power of p we divided by (so that N = M pℓ) must be an even power, since each term on the right-hand side is a perfect square and can only contribute even powers of primes by the Fundamental Theorem of Arithmetic. Since N had an odd power of p, we know M still h... |
p 2 Z of the form p = 4n + 3, p 2 Z[i] is prime. • Given a prime p 2 Z of the form p = 4n + 3, p i 2 Z[i] is also prime. • Given a prime p 2 Z not of the form p = 4n + 3, any factors a + bi and a bi in Z[i] corresponding to writing p = a2 + b2 are prime (recall Theorem 13.5.5). The last point can be confusing. Since a ... |
of). There is a huge field which developed from these observations, but we will not digress much further upon it. If you recall the discussion in Subsection 11.6.4, it turns out Germain originally investigated n in the case where it is one of the numbers now known as Germain primes (recall Subsection 11.6.4); see [E.5.... |
will change. It is not a hard exercise to see that the line through two rational points on a curve will have rational slope, nor what its formula is, so that every rational point on the circle is gotten by intersecting (1; 0) with a line with rational slope. This is not necessarily visible in Figure 15.1.1! Figure 15.... |
8671682660 ) which does seem a bit excessive but sure is fun3. There are endless variations on such questions. If we consider Dudeney’s problem as an example of summing two perfect squares to make a perfect cube, we have a more general question that Diophantus and al-Karaji explored for their rational rational solution... |
re tractable. This is because we understand conic sections so well, after having worked with them for two thousand years! Figure 15.4.1 Integer points on x2 + 2y2 = 9 In Figure 15.4.1 we see our second prototype, x2+2y2 = 9. You can see that, in addition to the obvious solution where y = 0, there is the (nearly as obvi... |
, xmin = x_0 -4 , xmax = x_1 +4 , ymin = y_0 -4 , ymax = y_1 +4 , aspect_ratio =1) pretty_print ( html (" The ␣ new ␣ points ␣ are ␣ $x_1 =% s$ ␣ and ␣ $y_1 =% s$ " %( x_1 , y_1 ))) As it turns out, this is quite an old idea. Finding integer solutions to this hyperbola is called solving Pell’s equation, and has been s... |
+ ny0y1; y = x0y1 + y0x1 solves x2 ny2 = kℓ. If x2 0 1 = ℓ ny2 ny2 Proof. See Exercise 15.7.17. ■ This is particularly nice if k = ℓ = 1, because getting a solution for that would then give a solution to the Pell equation! Brahmagupta used analogous techniques for his time (and more sophisticated things) to solve very... |
(i ,l , sols ))) Let’s see a few examples of how to be more systematic about this. Example 16.1.3 Prime power modulus. For instance, let’s go from p to p2 by trying a bit of Example 7.2.5 from earlier. Here, f (x) = x2 + 1 is what we want a solution for. If we are looking (mod 25), then we already know that (mod 5) we ... |
aper on the linguistics of higher mathematics waiting to be written. CHAPTER 16. SOLVING QUADRATIC CONGRUENCES 280 to even know whether there is a solution to a given congruence without such machinery! 16.3.3 Some history In fact, it is even hard to conjecture patterns for harder cases unless you are quite clever. Eule... |
s symmetric about the middle of the rows, with two of each of the QR colors appearing. Further, these are the same colors as the ones appearing in every other column in rows with a primitive root (the rows with every color represented); naturally, the order may be quite different. Finally, the second column’s color in ... |
ter writing (p1)! and factoring out 2(p1)/2, the “repeated” numbers will be the odd numbers between 1 and (p 1)/2. Clearly the only “missing” numbers are even ones between (p1)/2 and p, which are just congruent to the negatives of the “repeated” odd numbers, so the same argument as above with (p 1)! will work. It remai... |
ly enough to deal with his chronic illness. His work in several areas of algebra and function theory is still considered forward-looking. Of particular interest for this text is the Eisenstein integers2, a generalization of the Gaussian integers (see Exercise 14.4.2). I say “yet another” because this is similar to the ... |
1 if p 5 (mod 12) Proof. Combine the facts in Claim 17.3.3. We see that • If p = 12ℓ + 1 we add two even numbers, so 3 is a QR. • If p = 12ℓ + 5, we add an even number and 1, so 3 is not a QR. • If p = 12ℓ + 6 + 1 = 12ℓ + 7, we add an odd and zero, so 3 is not a QR. • If p = 12ℓ + 6 + 5 = 12ℓ + 11, we add an odd and 1,... |
ted with them, so n must obey all of the congruence conditions for for all the p which divide it. This might be a lot of conditions, which narrows the field considerably. a p ( ) Then we can use a variant on the Fermat factoring method to check for possible a for which a prime divisor p of n definitely is or definitely... |
s 2618163402417 21290000 1 which is a number with close to four hundred thousand digits. (The previous record had about half as many, so this is a huge advance.) @interact def _(q =(11 ,[ r for r in prime_range (3 ,100) if is_prime (2* r +1) ]) ): p = 2* q +1 a= mod (p -4 , p) if a. multiplicative_order () ==p -1: pret... |
set of green points has the same parity as that of the set of blue points outside the green box. Again, refer to the interactive graphic and try it with different primes for best understanding. Claim 17.6.5 Either set of green points has the same parity as the set of blue points outside the green box. Proof. There are ... |
ems okay nonetheless. We can test this property in the following Sage cell. @interact def _(a =25 , b =11) : pretty_print ( html (r"$\ phi (% s) =% s\ text {␣ and ␣ }\ phi (% s) =% s$ " %(a , euler_phi (a) , b , euler_phi (b)))) if gcd (a ,b) ==1: CHAPTER 18. AN INTRODUCTION TO FUNCTIONS 321 pretty_print ( html (r" And... |
’s true. And there’s more to come. 18.3 Exercises Exercise Group. We see in Subsection 18.2.2 that r is not multiplicative. But could some related properties still be true? 1. 2. Look at the cases where zero is involved. State the broadest possible multiplicativity result you can for this case. Look at the second two e... |
trying larger and larger input numbers, we might use a little theory to get a higher ratio. To wit, if a number has lots of small prime divisors, we might think it has lots of factors. So taking big powers of these would have even more small prime divisors and might get us big ratios. @interact def _(n =[1..15]) : pre... |
VISORS 336 • If 1(n) = a b in lowest terms, then b j n. • If r is “caught” between (n) and n (such that n < r < (n)) and is relatively prime to n, then r/n is not an abundancy index. Proof. We skip the proof, but proving the first two facts is left as Exer■ cise 19.6.22. Holdener and Stanton picturesquely call rational... |
me factor at least 108. • Have a second largest prime exceeding 10000. • Have the sum of the reciprocals of the prime divisors of the number between about 0:6 and 0:7. • Have the sum of the reciprocals of odd perfect numbers be finite (since the sum of the reciprocals of all perfect numbers is finite!). In fact, the 12... |
4)8,16j31,64,N,0,P,Q,R,S.3,S»9>l7>33>6$.-15>»,2-3,47-z,4,8,16.6,11,14,48.71,187,iifi.jLorsqueIvndesnombresdudernierordíPauecsonoppo-sé,&leprécédentdupremierordrescrótnombrespremiers,l'ontrouueradesnombressemblables;àçéuxdontilestque-stion.Parexemple,lenpmbréduderniérrangji,Scndupre-mierordre,Se$quileprecede,sontnombres... |
20.2.4 For n = 24 this idea is actually true. We can line these up in pairs as (1; 24), (2; 12), (3; 8), (4; 6), and that gives 2 = 8 total divisors. □ It means we can get a sense of a first That estimate is very important! ⌊p 24 ⌋ bound on the average value of . At the very least we have that 1 n n∑ k=1 (k) 1 n n∑ p ... |
Lattice points, , and symmetry Notice we have now divided the lattice points up into three parts, two of which are ‘the same’: • The ones on the line y = x. • The lattice points above the line and below the hyperbola. • The lattice points to the right of the line and below the hyperbola. Try it interactively, and perha... |
the ϕ function? You can try out various ideas in the following interact. Note that a is the coefficient and n is the power of a model axn. def L(n): ls = [] out = 0 for i in range (1 , n +1) : out += euler_phi (i) ls . append ((i , out /i)) return ls LS = L (1000) P = line ( LS ) @interact def _(a =.01 , n =2 , view =... |
nting argument to prove that p (n) = ( p n) + ϕ(n; ( n)) 1. This is the typical way to calculate in software without actually counting primes, and with some speedups it can be quite efficient. Interestingly, this is also how one finds the nth prime5. You use an approximation to the nth prime like n log(n) and then chec... |
try to understand some of the intermediate steps. This is a good place to highlight the contributions of the great Russian mathematician Chebyshev. Historical remark 21.3.3 Pafnuty Chebyshev. Chebyshev14 (Чебышёв) was a prominent Russian mathematician of the mid-19th century, but his most important legacy may be bringi... |
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