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There are 9 possible values.
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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]
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01 Mar 2017, 11:35
Vyshak wrote:
$$({a^2})^3 = a^6 = ({a^3})^2$$
Given: M = $$0 < a^2 < 100$$
Values of a^2 can be --> $$1^2, 2^2, 2^4, 2^6, 3^2, 3^4, 5^2, 7^2, (2^2 * 3^2)$$
There are 9 possible values.
Vyshak please can you explain this as I am not able to understand.
Thanks!
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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]
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01 Mar 2017, 22:47
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hotshot02 wrote:
Vyshak please can you explain this as I am not able to understand.
Thanks!
You have to find the number of values of a^2 that are between 0 and 100. --> a is between 0 and 10 --> We will have 9 values for m.
Hope it helps.
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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]
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idontknowwhy94 wrote:
M is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer.
How many different values could m be?
A. 7
B. 9
C. 11
D. 13
E. 15 | {
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A. 7
B. 9
C. 11
D. 13
E. 15
We are given that m is a positive integer less than 100. We are also given that when m is raised to the third power, it becomes the square of another integer. In order for that to be true, m itself must (already) be a perfect square, since any perfect square raised to the third power will still be a perfect square, i.e., square of an integer. Thus we are looking for perfect squares that are less than 100. Since there are 9 perfect squares that are less than 100, namely, 1, 4, 9, …, 64, and 81, the answer is 9.
Let’s look at some examples to clarify this: Let’s assume that m = 4 = 2^2. Now, let’s raise m to the third power, obtaining m^3 = (2^2)^3 = 4^3 = 64, which is the perfect square 8^2. Another illustration: Let’s let m = 25 = 5^2. Now, let’s raise m to the third power, obtaining m^3 = (5^2)^3 = 25^3 = 15625, which is the perfect square of 125.
(Note: By the way the problem is worded, “when m is raised to the third power, it becomes the square of another integer,” 1 should not be counted as one of the 9 different values m could be, unlike all the other 8 values. For example, take the number 4: 4^3 = 64 = 8^2, which is the square of another integer, 8. However, 1^3 = 1 = 1^2, which is the square of the same integer. The correct way to word the problem is “when m is raised to the third power, it becomes the square of an integer.”)
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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]
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26 Aug 2017, 00:14
JeffTargetTestPrep wrote:
Any perfect square raised to the third power will still be a perfect square, i.e., square of an integer. Thus we are looking for perfect squares that are less than 100. | {
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Why we need to look for perfect squares that are less than 100? We need the values of $$m<100$$ NOT the $$m^2<100$$.
I know I am wrong, but I don't know why I am wrong.
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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]
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26 Aug 2017, 01:09
idontknowwhy94 wrote:
M is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer.
How many different values could m be?
A. 7
B. 9
C. 11
D. 13
E. 15
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0<x^2<100
1,4,9,16,25,36,49,64,81
9 numbers
B
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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]
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10 Sep 2017, 06:38
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Ans is B
M^3 = y^2
M= y^(2/3)
means y should have a cube-root as integer and whose square should be less than 100
lets say start by 1000 =y
y^(2/3) = 1000^2/3 =100 rejected means M cant be 100
M can attain 81,64,49,36,25,16,09,04,01 TOTAL 9 values .
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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]
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21 Oct 2017, 21:01
idontknowwhy94 wrote:
M is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer.
How many different values could m be? | {
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A. 7
B. 9
C. 11
D. 13
E. 15
Keep the Kudos dropping in and let these tricky questions come out ....
Bunuel,
For this question, I understood that basically we need to find out squares of integers below 100. But I have one query -
Why does above answers includes $$1^2$$? The question stem says if the number is raised to third power, then it becomes square of ANOTHER integer?
$$\sqrt{(1^3)}$$ = 1 only.
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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]
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21 Oct 2017, 22:09
RMD007 wrote:
idontknowwhy94 wrote:
M is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer.
How many different values could m be?
A. 7
B. 9
C. 11
D. 13
E. 15
Keep the Kudos dropping in and let these tricky questions come out ....
Bunuel,
For this question, I understood that basically we need to find out squares of integers below 100. But I have one query -
Why does above answers includes $$1^2$$? The question stem says if the number is raised to third power, then it becomes square of ANOTHER integer?
$$\sqrt{(1^3)}$$ = 1 only.
JeffTargetTestPrep addressed this in his response above - | {
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$$\sqrt{(1^3)}$$ = 1 only.
JeffTargetTestPrep addressed this in his response above -
(Note: By the way the problem is worded, “when m is raised to the third power, it becomes the square of another integer,” 1 should not be counted as one of the 9 different values m could be, unlike all the other 8 values. For example, take the number 4: 4^3 = 64 = 8^2, which is the square of another integer, 8. However, 1^3 = 1 = 1^2, which is the square of the same integer. The correct way to word the problem is “when m is raised to the third power, it becomes the square of an integer.”)
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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]
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21 Oct 2017, 22:13
mahu101 wrote:
JeffTargetTestPrep addressed this in his response above -
(Note: By the way the problem is worded, “when m is raised to the third power, it becomes the square of another integer,” 1 should not be counted as one of the 9 different values m could be, unlike all the other 8 values. For example, take the number 4: 4^3 = 64 = 8^2, which is the square of another integer, 8. However, 1^3 = 1 = 1^2, which is the square of the same integer. The correct way to word the problem is “when m is raised to the third power, it becomes the square of an integer.”)
Thanks, my point is, with the given question stem 8 should be the answer!
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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]
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21 Oct 2017, 22:19
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RMD007 wrote:
mahu101 wrote:
JeffTargetTestPrep addressed this in his response above - | {
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JeffTargetTestPrep addressed this in his response above -
(Note: By the way the problem is worded, “when m is raised to the third power, it becomes the square of another integer,” 1 should not be counted as one of the 9 different values m could be, unlike all the other 8 values. For example, take the number 4: 4^3 = 64 = 8^2, which is the square of another integer, 8. However, 1^3 = 1 = 1^2, which is the square of the same integer. The correct way to word the problem is “when m is raised to the third power, it becomes the square of an integer.”)
Thanks, my point is, with the given question stem 8 should be the answer!
Yes, you are right.
Re: M is a positive integer less than 100. When m is raised to the third [#permalink] 21 Oct 2017, 22:19
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# If p, x, and y are positive integers, y is odd, and p = x^2
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If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
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If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?
(1) When p is divided by 8, the remainder is 5.
(2) x – y = 3
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Re: PS: Divisible by 4 [#permalink]
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netcaesar wrote:
If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?
(1) When p is divided by 8, the remainder is 5.
(2) x – y = 3
SOL:
St1:
Here we will have to use a peculiar property of number 8. The square of any odd number when divided by 8 will always yield a remainder of 1!!
This means that y^2 MOD 8 = 1 for all y
=> p MOD 8 = (x^2 + 1) MOD 8 = 5
=> x^2 MOD 8 = 4 | {
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This means that y^2 MOD 8 = 1 for all y
=> p MOD 8 = (x^2 + 1) MOD 8 = 5
=> x^2 MOD 8 = 4
Now if x is divisible by 4 then x^2 MOD 8 will be zero. And also x cannot be an odd number as in that case x^2 MOD 8 would become 1. Hence we conclude that x is an even number but also a non-multiple of 4.
=> SUFFICIENT
St2:
x - y = 3
Since y can be any odd number, x could also be either a multiple or a non-multiple of 4.
=> NOT SUFFICIENT
ANS: A
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Re: PS: Divisible by 4 [#permalink]
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16 Dec 2010, 07:39
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nonameee wrote:
Can I ask someone to look at this question a provide a solution that doesn't depend on knowing peculiar properties of number 8 or induction?
Thank you.
If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?
(1) When p is divided by 8, the remainder is 5 --> $$p=8q+5=x^2+y^2$$ --> as given that $$y=odd=2k+1$$ --> $$8q+5=x^2+(2k+1)^2$$ --> $$x^2=8q+4-4k^2-4k=4(2q+1-k^2-k)$$.
So, $$x^2=4(2q+1-k^2-k)$$. Now, if $$k=odd$$ then $$2q+1-k^2-k=even+odd-odd-odd=odd$$ and if $$k=even$$ then $$2q+1-k^2-k=even+odd-even-even=odd$$, so in any case $$2q+1-k^2-k=odd$$ --> $$x^2=4*odd$$ --> in order $$x$$ to be multiple of 4 $$x^2$$ must be multiple of 16 but as we see it's not, so $$x$$ is not multiple of 4. Sufficient.
(2) x – y = 3 --> $$x-odd=3$$ --> $$x=even$$ but not sufficient to say whether it's multiple of 4.
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Re: PS: Divisible by 4 [#permalink]
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Re: PS: Divisible by 4 [#permalink]
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19 Dec 2010, 07:49
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netcaesar wrote:
If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?
(1) When p is divided by 8, the remainder is 5.
(2) x – y = 3
Such questions can be easily solved keeping the concept of divisibility in mind. Divisibility is nothing but grouping. Lets say if we need to divide 10 by 2, out of 10 marbles, we make groups of 2 marbles each. We can make 5 such groups and nothing will be left over. So quotient is 5 and remainder is 0. Similarly if you divide 11 by 2, you make 5 groups of 2 marbles each and 1 marble is left over. So 5 is quotient and 1 is remainder. For more on these concepts, check out: http://gmatquant.blogspot.com/2010/11/divisibility-and-remainders-if-you.html
First thing that comes to mind is if y is odd, $$y^2$$ is also odd.
If $$y = 2k+1, y^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1$$
Since one of k and (k+1) will definitely be even (out of any two consecutive integers, one is always even, the other is always odd), 4k(k+1) will be divisible by 8. So when y^2 is divided by 8, it will leave a 1.
Stmnt 1: When p is divided by 8, the remainder is 5.
When y^2 is divided by 8, remainder is 1. To get a remainder of 5, when x^2 is divided by 8, we should get a remainder of 4.
$$x^2 = 8a + 4$$ (i.e. we can make 'a' groups of 8 and 4 will be leftover)
$$x^2 = 4(2a+1)$$ This implies $$x = 2*\sqrt{Odd Number}$$because (2a+1) is an odd number. Square root of an odd number will also be odd.
Therefore, we can say that x is not divisible by 4. Sufficient.
Stmnt 2: x - y = 3
Since y is odd, we can say that x will be even (Even - Odd = Odd). But whether x is divisible by 2 only or by 4 as well, we cannot say since here we have no constraints on p. Not sufficient.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17354 [26], given: 232 Intern Joined: 30 May 2013 Posts: 4 Kudos [?]: 8 [8], given: 8 Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 15 Sep 2013, 22:13 8 This post received KUDOS I did this question this way. I found it simple. 1. p=x^2+y^2 y is odd p div 8 gives remainder 5. A number which gives remainder 5 when divided by 8 is odd. so (x^2 + y^2)/8 = oddnumber (x^2 + y^2) = 8 * oddnumber (this is an even number without doubt) x^2 + y^2 is even. Since y is odd to get x^2+y^2 even x must also be odd. X is an odd number not divisible by 4 Option A: 1 alone is sufficient Kudos [?]: 8 [8], given: 8 Intern Joined: 02 Sep 2010 Posts: 45 Kudos [?]: 146 [4], given: 17 Location: India Re: PS: Divisible by 4 [#permalink] ### Show Tags 18 Dec 2010, 11:23 4 This post received KUDOS 1 This post was BOOKMARKED maliyeci wrote: Very good solution I did not know this property of 8. Kudos to you. By and induction. 1^2=1 mod 8 say n^2=1 mod 8 (n is an odd number) than if (n+2)^2=1 mod 8 ? (n+2 is the next odd number) (n+2)^2=n^2+4n+4= 1 + 4n + 4 mod 8 4n+4=0 mod 8 because n is an odd number and 4n=4 mod 8. So induction works. So for any odd number n, n^2=1 mod 8 Its not something one shall already know before attacking a question, you may realize properties like this when u start solving a question. Even I didn't know about this property of 8. I approached the question in following way: Stmt 1: P/8=(x^2+y^2)/8; using remainder theorem; rem[(x^2+y^2)/8]= rem[x^2/8] + rem[y^2/8] if x is divisible by 4, then x^2= 4k*4k= 16K=8*2K is also divisible by 8. now to anaylze rem[y^2/8]; start putting suitable values of y; i.e all odd values starting from 1. for y=1; rem(1/8)=1 for y=3; rem(9/8)=1 for y=5;rem(25/8)=1 so you observe this pattern here. coming back to ques now, as rem[(x^2+y^2)/8]= rem[x^2/8] + rem[y^2/8]= rem[x^2/8] + 1 =5; this means rem[x^2/8] is not | {
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now, as rem[(x^2+y^2)/8]= rem[x^2/8] + rem[y^2/8]= rem[x^2/8] + 1 =5; this means rem[x^2/8] is not 0; which implies x is not divisible my 8; Sufficient Stmt2: y being odd can be accept both 3 and 5 as values and we get different results; thus Insufficient Thus OA is A _________________ The world ain't all sunshine and rainbows. It's a very mean and nasty place and I don't care how tough you are it will beat you to your knees and keep you there permanently if you let it. You, me, or nobody is gonna hit as hard as life. But it ain't about how hard ya hit. It's about how hard you can get it and keep moving forward. How much you can take and keep moving forward. That's how winning is done! Kudos [?]: 146 [4], given: 17 Intern Joined: 14 May 2013 Posts: 12 Kudos [?]: 18 [2], given: 3 Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 12 Jun 2013, 10:58 2 This post received KUDOS netcaesar wrote: If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4? (1) When p is divided by 8, the remainder is 5. (2) x – y = 3 1. As p = 8I + 5 we have values of P = 5,13,21,29..... etc .. as y is odd when we solve this p(odd) = x^2 + y^2(odd) x^2 = odd -odd = even which can be 2,4,6 ... etc but if we check for any value of p we don't get any multiple of 4. so it say's clearly that x is not divisible by 4. 2. x-y = 3 x = y(odd)+3 x is even which can be 2,4,6.. so it's not sufficient .. Ans : A _________________ Chauahan Gaurav Keep Smiling Kudos [?]: 18 [2], given: 3 Intern Joined: 25 Jun 2013 Posts: 6 Kudos [?]: 3 [2], given: 0 Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 13 Sep 2013, 20:25 2 This post received KUDOS 1 This post was BOOKMARKED from first statement p = 8j + 5 Put j as 1, 2,3,4,5... p would be 13, 21,29, 37,45... Now in the formula p= x^2+y^2 put 1,3,5,7 as value of y ( as y is odd) to get x. You will notic the possible value of x is 2 which is not divisble | {
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of y ( as y is odd) to get x. You will notic the possible value of x is 2 which is not divisble by 4. Posted from GMAT ToolKit Kudos [?]: 3 [2], given: 0 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7674 Kudos [?]: 17354 [2], given: 232 Location: Pune, India Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 18 Aug 2014, 02:43 2 This post received KUDOS Expert's post 1 This post was BOOKMARKED alphonsa wrote: For statement 1 , wouldn't plugging in values be a better option? No. When you need to establish something, plugging in values is not fool proof. Anyway, in this question, how will you plug in values? You cannot assume a value for x since that is what you need to find. You will assume a value for y and a value for p such that they satisfy all conditions. This itself will be quite tricky. Then when you do get a value for x, you will find that it will be even but not divisible by 4. How can you be sure that this will hold for every value of y and p? When a statement is not sufficient, plugging in values can work - you find two opposite cases - one which answers in yes and the other which answers in no. Then you know that the statement alone is not sufficient. But when the statement is sufficient, it is very hard to prove that it will hold for all possible values using number plugging alone. You need to use logic in that case. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199 | {
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Re: PS: Divisible by 4 [#permalink]
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15 Aug 2009, 13:49
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Very good solution I did not know this property of 8. Kudos to you.
By and induction.
1^2=1 mod 8
say
n^2=1 mod 8 (n is an odd number)
than
if (n+2)^2=1 mod 8 ? (n+2 is the next odd number)
(n+2)^2=n^2+4n+4= 1 + 4n + 4 mod 8
4n+4=0 mod 8 because n is an odd number and 4n=4 mod 8.
So induction works.
So for any odd number n, n^2=1 mod 8
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Re: PS: Divisible by 4 [#permalink]
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16 Dec 2010, 07:13
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Can I ask someone to look at this question a provide a solution that doesn't depend on knowing peculiar properties of number 8 or induction?
Thank you.
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Re: PS: Divisible by 4 [#permalink]
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16 Jul 2012, 18:19
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Am i missing something, why cant we take stmt 2 as follows:
squaring x-y=3 on both sides, we get p=9+2xy, that is p=odd + even = odd, not divisible by 4
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Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
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03 May 2017, 05:31
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Expert's post
VeritasPrepKarishma wrote:
aliasjit wrote:
I am a little confused about solving the problem:
Stmnt 2: x-y =3
we know y is odd.
and if Y is odd as per problem statement y cannot be anything but 1
as x and y both are positive integers.
Therefore x =4 and is divisible by 4.
y is odd, yes, but why must y be 1?
Responding to a pm:
Quote:
because as per statement 2 if x-y =3 and x and y are both +ve integers could y be anything but 1 | {
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But as per statement 2, we do not know that x must be 4. x must be even since y is odd and difference between x and y is odd. But will it be 4, we do not know.
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Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
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03 Sep 2013, 04:00
Excellent explanation Bunuel & Karishma:):)
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Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
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23 Dec 2013, 23:45
For Statement 1:
since p when divided by 8 leaves remainder 5.We obtain the following equation
p= 8q+5
We know y is odd. If we write p =x^2+y^2 then we get the eqn:
x^2+y^2=8q+5
Since, y is odd, 8q is even and 5 is odd. We get 8q+5 is odd.
Then x^2= odd - y^2
i.e x^2=even
ie x= even
But it's not sufficient to answer the question whether x is a multiple of 4?
By this logic i get E as my answer.
Statement 2: is insufficient.
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Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
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30 Dec 2013, 23:39
Expert's post
1
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BOOKMARKED
Abheek wrote:
For Statement 1:
since p when divided by 8 leaves remainder 5.We obtain the following equation
p= 8q+5
We know y is odd. If we write p =x^2+y^2 then we get the eqn:
x^2+y^2=8q+5
Since, y is odd, 8q is even and 5 is odd. We get 8q+5 is odd.
Then x^2= odd - y^2
i.e x^2=even
ie x= even
But it's not sufficient to answer the question whether x is a multiple of 4?
Your analysis till now is fine but it is incomplete. We do get that x is even but we also get that x is a multiple of 2 but not 4 as explained in the post above: if-p-x-and-y-are-positive-integers-y-is-odd-and-p-x-82399.html#p837890
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Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
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16 Aug 2014, 00:16
For statement 1 , wouldn't plugging in values be a better option?
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Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
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29 Aug 2014, 05:25
Hi Karishma,
Thanks for the explanation to the question. I was just wondering how the answer would change if we change the question stem a little bit. What if the question asks if p (instead of x) is divisible by 4? | {
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In this scenario, statement 1 would be sufficient since if something leaves a remainder of 5, it would leave a remainder of 1 upon division by 4
For statement 2, we know that x = y+3, so x is even. If we square it, it would surely be divisible by 4. Now if a number (y^2, which is odd) non-divisible by 4 is added to a number divisible by 4, the result would surely be not divisible by 4. So statement 2 would also be sufficient.
Is this reasoning correct? just for practicing the concept
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Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
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31 Aug 2014, 03:38
netcaesar wrote:
If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?
(1) When p is divided by 8, the remainder is 5.
(2) x – y = 3
1) p = 8k+5 (k is a whole number)
also p = (x^2+y^2)
=> (x^2+y^2) mod 8 = 5
any square (n^2) mod 8 follows the following pattern -> 1,4,1,0 and then repeats.
for getting x^2+y^2 mod 8 = 5
we need to take a 4 and a 1 from the above pattern. at multiples of 4, the remainder is 0. so x can never be divisible by 4.
A is sufficient.
2) x-y=3 (odd)
even - odd = odd
or odd - even = odd
so insufficient.
Hence, A.
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Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] 31 Aug 2014, 03:38
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# What does "curly (curved) less than" sign $\succcurlyeq$ mean?
I am reading Boyd & Vandenberghe's Convex Optimization. The authors use curved greater than or equal to (\succcurlyeq)
$$f(x^*) \succcurlyeq \alpha$$
and curved less than or equal to (\preccurlyeq)
$$f(x^*) \preccurlyeq \alpha$$
Can someone explain what they mean?
• as () is curved and {} are curly ( en.wikipedia.org/wiki/Bracket ), I think those symbols you mention are curved not curly Feb 9, 2014 at 6:43
• @barlop If you look at $\LaTeX$ source of the formulas in question (right click->Show Math As->TeX commands), you'll see \succcurlyeq, which has curly word in it, not curved. Feb 9, 2014 at 7:18
• Thanks everyone for answers. As I understand $\succeq$ or $\preceq$ are more general than their more popular counterparts. I think Michael's answer make sense. If I understand correctly, $X \succeq Y, \quad if \quad \| X \| \ge \| Y \|$ where $\| \cdot \|$ is the norm associated with the space $X$ and $Y$ belongs to. I think Chris's answer is correct but is more strict condition than Michael's answer. Please correct me if I'm wrong. Feb 9, 2014 at 17:53
• You should definitely take the tour. This is not a traditional forum!
– bodo
Feb 9, 2014 at 18:03
• Well I don't know what a traditional forum looks like. :-) Feb 9, 2014 at 18:20
Both Chris Culter's and Code Guru's answers are good, and I've voted them both up. I hope that I'm not being inappropriate by combining and expanding upon them here.
It should be noted that the book does not use $\succeq$, $\preceq$, $\succ$, and $\prec$ with scalar inequalities; for these, good old-fashioned inequality symbols suffice. It is only when the quantities on the left- and right-hand sides are vectors, matrices, or other multi-dimensional objects that this notation is called for. | {
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The book refers to these relations as generalized inequalities, but as Code-Guru rightly points out, they have been in use for some time to represent partial orderings. And indeed, that's exactly what they are, and the book does refer to them that way as well. But given that the text deals with convex optimization, it was apparently considered helpful to refer to them as inequalities.
Let $S$ be a vector space, and let $K\subset S$ be a closed, convex, and pointed cone with a non-empty interior. (By cone, we mean that $\alpha K\equiv K$ for all $\alpha>0$; and by pointed, we mean that $K\cap-K=\{0\}$.) Such a cone $K$ induces a partial ordering on the set $S$, and an associated set of generalized inequalities: $$x \succeq_K y \quad\Longleftrightarrow\quad y \preceq_K x \quad\Longleftrightarrow\quad x - y \in K$$ $$x \succ_K y \quad\Longleftrightarrow\quad y \prec_K x \quad\Longleftrightarrow\quad x - y \in \mathop{\textrm{Int}} K$$ This is a partial ordering because, for many pairs $x,y\in S$, $x \not\succeq_K y$ and $y \not\succeq_K x$. So that's the primary reason why he and others prefer to use the curly inequalities to denote these orderings, reserving $\geq$, $\leq$, etc. for total orderings. But it has many of the properties of a standard inequality, such as: $$x\succeq_K y \quad\Longrightarrow\quad \alpha x \succeq_K \alpha y \quad\forall \alpha>0$$ $$x\succeq_K y \quad\Longrightarrow\quad \alpha x \preceq_K \alpha y \quad\forall \alpha<0$$ $$x\succeq_K y, ~ x\preceq_K y \quad\Longrightarrow\quad x=y$$ $$x\succ_K y \quad\Longrightarrow\quad x\not\prec_K y$$ | {
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When the cone $K$ is understood from context, it is often dropped, leaving only the inequality symbol $\succeq$. There are two cases where this is almost always done. First, when $S=\mathbb{R}^n$ and the cone $K$ is non-negative orthant $\mathbb{R}^n_+$ the generalized inequality is simply an elementwise inequality: $$x \succeq_{\mathbb{R}^n_+} y \quad\Longleftrightarrow\quad x_i\geq y_i,~i=1,2,\dots,n$$ Second, when $S$ is the set of symmetric $n\times n$ matrices and $K$ is the cone of positive semidefinite matrices $\mathcal{S}^n_+=\{X\in S\,|\,\lambda_{\text{min}}(X)\geq 0\}$, the inequality is a linear matrix inequality (LMI): $$X \succeq_{\mathcal{S}^n_+} Y \quad\Longleftrightarrow\quad \lambda_{\text{min}}(X-Y)\geq 0$$ In both of these cases, the cone subscript is almost always dropped.
Many texts in convex optimization don't bother with this distinction, and use $\geq$ and $\leq$ even for LMIs and other partial orderings. I prefer to use it whenever I can, because I think it helps people realize that this is not a standard inequality with an underlying total order. That said, I don't feel that strongly about it for $\mathbb{R}^n_+$; I think most people rightly assume that $x\geq y$ is considered elementwise when $x,y$ are vectors.
• Thanks a lot for the detailed answer, and for correcting the symbol as well :-). Feb 9, 2014 at 16:35
• This is an old answer and I completely agree with it but I thought that providing another common application of this notation might be useful. As @Code-Guru pointed out, these are useful for partial orders. In Economics, we usually model preferences over baskets of goods with "not worse than" or "not better than" sets. The partial order concept fits like a glove in this situation. The interested reader might find books in Decision Sciences useful for this kind of discussion. Dec 30, 2020 at 21:13 | {
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There's a list of notation in the back of the book. On page 698, $x\preceq y$ is defined as componentwise inequality between vectors $x$ and $y$. This means that $x_i\leq y_i$ for every index $i$.
Edit: The notation is introduced on page 32.
Often these symbols represent partial order relations. The typical "less than" and "greater than" operations both define partial orders on the real numbers. However, there are many other examples of partial orders.
Sometimes the curly greater than sign is used to indicate positive semi-definiteness of a matrix $$X$$:
$$(X\succeq 0\ \text{or}\ X\ge 0)$$
or a function $$f(x)$$
$$(f(x) \succeq 0\ \text{or}\ f(x)\ge 0).$$
• Positive definiteness of what? Aug 26, 2015 at 17:03
Does it sometimes denote, in measurement theory; often the qualitative counterpart to $\geq$ in the numerical representation; when one wants to numerically represent a totally ordered qualitative probability representation:
$$A ≽ B \leftrightarrow A > B \leftrightarrow F(A) \geq F(B)$$.
On the other hand. I have seen it used in multi-dimensional partial or even total, orderings under the "ordering of major-ization" for vector valued functions or for a system for two kinds of orderings.
One for (numerical) ordinal, $>$ comparisons and another, $≽$ for (numerical) differences, sums, or a some other kind of relation, to fine grain, the representation to ensure (or some kind of) unique-ness, rather than merely strong represent-ability.
See Marshall | {
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Marshall, Albert W.; Olkin, Ingram, Inequalities: theory of majorization and its applications, Mathematics in Science and Engineering, Vol. 143. New York etc.: Academic Press. XX, 569 p. \$52.50 (1979). ZBL0437.26007. Such functions or representations, may use both the curly greater$≽$than and$\geq $in the numerical or functional representation and are compatible with total orderings. $$a,b\in \Omega^{n}\; a <b \,,\quad \text{or}, \quad, a= b, \quad \text{or}\quad a < b$$ where one cannot usually adding up distinct $$a_1 \in \Omega_i\,; b_j\in \Omega_j; j \neq i$$ $$a_3 \in \Omega_c;\, ; \, P(a_i) + P(b_j) = P(a_3)$$. Sometimes I "(conjecture)" that$≽$denotes a weaker version of$(A)$or$(A_1)$(strong complementary add-itivity) which orders the differences and sums, is something over and above$(B)$;strict monotone increasing/strong representability/order embedding- above reflecting and preserving . $$(A)\,[f(x_1)+f(x_2)≽ f(y_1) + f(y_2)] \rightarrow [f(x) \geq f(y)] \text{where} \,; x=x_1 +x_2\,, y=y_1+y_2$$ or $$(A1)\,[x\, \geq y ]\,\rightarrow [f(x_1)+f(x_2)≽ f(y_1) + f(y_2)]\text{where} \,; x=x_1 +x_2\,, y=y_1+y_2$$ Rather than:$$(B)$$ $$(B)\,x\, \geq y \leftrightarrow f(x) \geq f(y)\, \text{where} \,; x=x_1 +x_2\,, y=y_1+y_2$$ In contrast to$(B)$a standard order embedding (strict monotone increasing function) Where in$(B)$the numerical function,$F$or representation is 'merely strong'. That is$F$is a monotone strictly increasing function of some entity$x$where$F(x)$is the entity one wishes to order by$x$. Sometimes, even in a infinite and uniformly and non-atomic, continuous total order representation, nothing unique will come out. If its infinite /non-atomic in the wrong sense. ie in a super-atomic or entangled lexicographic system. Where the system is dense/or continuum dense/non-atomic/bottom-less in the wrong sense. Such as a spin$1/2$system in quantum mechanics, or a lexicographic entangled, system where its infinite in wrong sense, in the vertical, not in the | {
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or a lexicographic entangled, system where its infinite in wrong sense, in the vertical, not in the horizontal, not within a basis of space/basis.$(A)$is arguably a lot stronger than$(B)$, or at least is, if relatively unrestricted. Where$(A)$, may be used not to extend the ordering so much as much as make a total ordering 'numerically precise; to put some constraints on sums and differences, where the events are the not kinds of things that usually add up. Say in an entangled multidimensional quantum spin 1/2 system or utility representation where the events are on complementary spaces and mixtures are not allowed, for example. That is, to, put a metric on differences (say on a two outcome system). That is something stronger, than a, total or non atomic/dense order, where(merely) every event. Even if the entire system is totally ordered within and betwixt the distinct$\Omega\$. So that the system is solv-able, and uniquely so. | {
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# Math Help - Piecewise-Defined Function
1. ## Piecewise-Defined Function
For both questions below:
(a) Find the domain of the function.
(b) Locate any intercepts.
(1)
.....{3 + x......if -3 <or= to x < 0
f(x){3...........if......x = 0
.....{Sqrt{x}..if......x > 0
=======================
(2)
..........{1/x.........if.....x < 0
f(x) = {sqrt{x}...if.....x >or= to 0
NOTE: It is hard to correctly type the piecewise-defined functions using a regular keyboard.
I hope you can understand the above.
2. Originally Posted by symmetry
For both questions below:
(a) Find the domain of the function.
(b) Locate any intercepts.
(1)
.....{3 + x......if -3 <or= to x < 0
f(x){3...........if......x = 0
.....{Sqrt{x}..if......x > 0
=======================
(2)
..........{1/x.........if.....x < 0
f(x) = {sqrt{x}...if.....x >or= to 0
NOTE: It is hard to correctly type the piecewise-defined functions using a regular keyboard.
I hope you can understand the above.
I'll do the first one for you- graph it. The conditions are the "if" parts in the piece-wise function. Domain is (-3, inf) and there are no intersepts. Try graphing it. You have a line with slope = 1 and an exponential function.
EDIT: Sorry, there are x and y-intercepts, as Soroban pointed out, although the two graphs do not not intersect which is what I was getting at.
3. Hello, symmetry!
For both questions below:
. . (a) Find the domain of the function.
. . (b) Locate any intercepts.
Did you make a sketch?
$(1)\;\;f(x) \:=\:\begin{Bmatrix} 3 + x & &\text{if }\text{-}3 \leq x < 0 \\ 3 & &\text{if }x = 0 \\ \sqrt{x} & &\text{if }x > 0 \end{Bmatrix}$
Code:
|
| *
* *
* | *
* |*
----*-----o--------------
-3 |
Domain: . $(\text{-}3,\,\infty)$
Intercepts: . $(\text{-}3,0),\:(0,3)$ | {
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Intercepts: . $(\text{-}3,0),\:(0,3)$
$(2)\;\;f(x)\:=\:\begin{Bmatrix}\frac{1}{x} & & \text{if }x < 0 \\ \sqrt{x} & & \text{if }x \geq 0 \end{Bmatrix}$
Code:
|
| *
| *
| *
|*
------------------*----------------
* |
* |
* |
* |
* |
|
*|
|
Domain: . $(-\infty,\,\infty)$
Intercepts: . $(0,\,0)$
4. ## ok
Thank you again both for your quick replies.
To soroban,
No, I did not sketch the graph because I do not know how to graph piecewise-defined functions.
I understand these functions are graphed in parts, right?
Can you take me through a sample graphing question in terms of this type of function?
Thanks!
5. Originally Posted by symmetry
Thank you again both for your quick replies.
To soroban,
No, I did not sketch the graph because I do not know how to graph piecewise-defined functions.
I understand these functions are graphed in parts, right?
Can you take me through a sample graphing question in terms of this type of function?
Thanks!
Yes, they are 'graphed in parts,' I guess you could call it.
For instance,
Take the first condition;
f(x) = 3 + x if -3 <= x < 0
From x = -3 (including this point) to x = 0 (not including, and thus draw an open circle by this point), you will graph 3 + x; see Soroban's graph. The reason why it's closed (solid dot) is because of the next condition later, and thus includes that point. Try look up piece-wise functions on Wikipedia.
6. ## ok
I like graphing functions. I think piecewise-defined functions are cool but not easy to sketch.
Thanks!
7. Hello again, symmetry!
Okay, here's an example.
. . $f(x) \:=\:\begin{Bmatrix}3 & \text{if }0 \leq x \leq 1 \\ 2x + 1 & \text{if }x > 1\end{Bmatrix}$
When $x$ is between $0$ and $1$ (including the endpoints),
. . the graph is $f(x) = 3$, a horizontal line.
Code:
|
3* * * * *
|
|
- + - - - + - -
| 1 | {
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When $x$ is greater than 1, the graph is $f(x) \:=\:2x + 1$,
. . a slanted line.
Code:
|
| *
| *
| *
| *
3| *
|
|
- + - - - + - - - - - - -
| 1
Sketch them on the same graph
. . and have the graph of the piecewise function.
Code:
|
| *
| *
| *
| *
3o * * * *
|
|
- + - - - + - - - - - - -
| 1
This function could be your long-distance charge.
They might charge $3 for the first minute . . and$2 per minute for every subsequent minute.
(Hmmm, not a good example . . .
I'm sure someone will point out why.) | {
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# Mellin transform of $x^p$ seems to miss a factor of $2\pi$
Bug introduced in 11.1 or earlier and fixed in 11.3
On Mathematica 11.1.1.0 the Mellin transform of $x^p$ is evaluated as $\delta(p+s)$, while I think it should be $2\pi\,\delta(p+s)$:
In:= MellinTransform[x^p, x, s, GenerateConditions -> True]
Out:= DiracDelta[p + s]
edited posting after Daniel Lichtblau's comment
I initially did not understand this result, but this 2004 paper has explained to me how to arrive at the Dirac delta function, however, with an additional factor of $2\pi$. I checked that this is not a matter of a different definition of the Mellin transform. (I summarized the calculation in this Mathoverflow posting.)
Missing factor $2\pi$ is fixed in Mathematica 11.3.0:
In:= MellinTransform[x^p, x, s, GenerateConditions -> True]
Out:= 2π DiracDelta[i(p + s)]
consequence: before 11.3 Integrate[MellinTransform[1, x, s], {s, -Infinity, Infinity}] returned 1, now it returns $2\pi\int_{-\infty}^\infty\delta(is)ds$
Q: is this v. 11.3 change in the implementation of MellinTransform documented somewhere? | {
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• See last example in documentation under Scope Elementary Functions. It should be noted that this is a generalization of the integral definition, not unlike the case for FourierTransform. – Daniel Lichtblau Jul 9 '17 at 15:27
• thank you, Daniel, for the feedback, I understand things a bit better now and have edited my posting accordingly --- my problem has been reduced to a missing factor $2\pi$... – Carlo Beenakker Jul 9 '17 at 19:06
• What specific definition is used is not particularly important so long as the MellinTransform and InverseMellinTransform are inverses of each other. Both x^p == InverseMellinTransform[ MellinTransform[x^p, x, s], s, x] and DiracDelta[p + s] == MellinTransform[ InverseMellinTransform[DiracDelta[p + s], s, x], x, s] evaluate to True – Bob Hanlon Jul 9 '17 at 22:41
• @BobHanlon --- but if we assume that the factor of $2\pi$ is absorbed in the definition of DiracDelta, then Integrate[MellinTransform[1, x, s], {s, -Infinity, Infinity}] should return $2\pi$, while instead it returns 1. – Carlo Beenakker Jul 10 '17 at 6:18
• The integral of DiracDelta should be one. – Bob Hanlon Jul 10 '17 at 14:48
The Mellin transforms for $x^j$ reported by Mathematica 11.2 didn't make sense to me, so on 11/28/2017 I submitted the following question on Math StackExchange.
Questions on Mellin Transform of $x^j$ and Interpretation of Distributions with Complex Arguments
I ended up deriving the answer to my own question and on 12/7/2017 I submitted a problem report to Wolfram technical support where I attached a Mathematica notebook illustrating the problem and the correct solution (CASE:3980660).
I received an email from Wolfram technical support on 12/13/2017 indicating my analysis was accepted as correct and a report was being filed with the developers. The correct solution was subsequently implemented in Mathematica 11.3.
Note that not only was the $2\,\pi$ prefix missing, but $i$ was also missing in the $\delta$ function parameter. | {
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I subsequently posted the correct solution in answers to related questions on both Math StackExchange and MathOverflow StackExchange.
Delta function with imaginary argument
Dirac Delta function with a complex argument | {
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# Finding Horizontal Tangent Planes on S
1. Dec 2, 2011
### TranscendArcu
1. The problem statement, all variables and given/known data
S is the surface with equation $$z = x^2 +2xy+2y$$a) Find an equation for the tangent plane to S at the point (1,2,9).
b) At what points on S, in any, does S have a horizontal tangent plane?
3. The attempt at a solution
$$F(x,y,z): z = x^2 +2xy+2y$$
$$F_x = 2x + 2y$$
$$F_y = 2x + 2$$
Evaluated at (1,2) gives answers 6 and 4, respectively. My equation for a plane is:
$$z-9=6(x-1) + 4(y-1)$$.
I think any horizontal plane should have normal vector <0,0,k>, where k is some scalar. I'm pretty sure that S has no such normal vector. But if
$$F(x,y,z): 0 = x^2 +2xy+2y - z$$
then
$$grad F = <2x + 2y,2x + 2,-1>$$ It seems like I can let (x,y) = (-1,1) to zero the x-, y-components of the gradient. Plugging (-1,1) into the definition of z gives z = 1. This suggests to me that there is a point (-1,1,1), at which there is a horizontal tangent plane. Yet I feel pretty sure that this isn't true!
2. Dec 3, 2011
### ehild
You made a little mistake when writing out the equation of the tangent plane. The y coordinate of the fixed point is 2, you wrote 1.
A surface in 3D is of the form F(x,y,z) = constant. For this surface, x2+2xy+2y-z=0. That means F(x,y,z)=x2+2xy+y-z. The gradient of F is normal to the surface, and the tangent plane of the surface at a given point. You want a horizontal tangent plane, so a vertical gradient:(0,0,a). That means Fx=2x+2y=0, Fy=2x+2=0 --->x=-1, y=1, so your result for the x,y coordinates are correct. Plugging into the original equation for x and y, you got z=x2+2xy+2y=1, it is correct. Why do you feel it is not?
ehild
3. Dec 3, 2011
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ehild
3. Dec 3, 2011
### TranscendArcu
When I graphed F(x,y,z) in MatLab (and it's possible I graphed it incorrectly), I observed that the the resulting paraboloid is always "tilted". Below is a picture from my plot:
http://img440.imageshack.us/img440/687/skjermbilde20111203kl85.png [Broken]
How can this surface have a horizontal tangent anywhere when it is tilted like this?
Last edited by a moderator: May 5, 2017
4. Dec 3, 2011
### ehild
Try to plot z out for -2<x<0 and 0<y<2
ehild
5. Dec 3, 2011
### TranscendArcu
http://img7.imageshack.us/img7/139/skjermbilde20111203kl10.png [Broken]
Hmm. I'm not seeing the a point in this picture where the gradient is pointing directly upwards. Everything still looks kind of tilted.
Last edited by a moderator: May 5, 2017
6. Dec 3, 2011
### ehild
The function is equivalent with z=(x+2y-1)(x+1)+1 and z=1 along the lines x=-1 and y=(1-x)/2. I attach a plot of the surface near the point (-1,1)
ehild
File size:
81.6 KB
Views:
127
7. Dec 4, 2011
### HallsofIvy
Staff Emeritus
I presume that by "horizontal" you mean perpendicular to the z-axis.
The simplest way to find a tangent planes for a surface is to write it in the form F(x,y,z)= constant. Then the normal to the tangent plane at any point is given by $\nabla F$. Here, you can write $F(x, y,z)= x^2+ 2xy+ 2y- z= 0$.
What is $\nabla F$? That will be vertical (and so tangent plane horizontal) when its x and y components are 0.
8. Dec 4, 2011
### ehild
@HallsofIvy: The OP has shown the solution in his first post, he only can not believe it, as the surfaces he got with MatLab look tilted. If you could give advice how to plot surfaces with MatLab, that would be real help for him.
ehild
9. Dec 4, 2011
### TranscendArcu
If anyone is familiar with MatLab, this is the code I've been using:
Note that the "%" mark my annotations. I included them so that hopefully you can follow what I'm doing more easily.
10. Dec 4, 2011
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10. Dec 4, 2011
### TranscendArcu
Ha! I figured it out. I forgot a "."
I should have written
z = x.^2 + 2*x.*y +2*y;
Everything makes sense now.
11. Dec 4, 2011
### ehild
You see: it is worth typing something out again and again. Is your plot similar to my one? It was made with Origin. I would like to see your final plot... Please...
ehild
12. Dec 4, 2011
### TranscendArcu
http://img259.imageshack.us/img259/2104/skjermbilde20111204kl10.png [Broken]It looks like it could have a horizontal tangent plane right around (-1,1,1)
Last edited by a moderator: May 5, 2017
13. Dec 4, 2011
### ehild
It is really nice!!! And a missing dot made you sceptical about the truth of Maths??!!!:uhh:
ehild | {
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# Solving Quadratic Equations Pure Imaginary Numbers | {
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For y = x 2 , as you move one unit right or left, the curve moves one unit up. So, thinking of numbers in this light we can see that the real numbers are simply a subset of the complex numbers. In our recent paper we gave an efficient algorithm to calculate "small" solutions of relative Thue equations (where "small" means an upper bound of type $10^{500}$ for the sizes of solutions). 1 is called Cartesian, because if we think of as a two dimensional vector and and as its components, we can represent as a point on the complex plane. The solutions of the quadratic equation ax2 + bx +c = 0 are: SOLVING QUADRATIC EQUATION WITH TWO REAL SOLUTIONS The solutions are: SOLVING QUADRATIC EQUATION WITH ONE REAL SOLUTIONS Hence, the solution is 3. Solve quadratic equations by completing equations the square. Quadratic Equations and Complex Numbers (Algebra 2 Curriculum - Unit 4) DISTANCE LEARNING. The Unit Imaginary Number, i, has an interesting property. Here we apply this algorithm to calculating power integral bases in sextic fields with an imaginary quadratic subfield and to calculating relative power integral bases in pure quartic extensions of. math game websites for elementary students basic math puzzles 6th grade expressions math games for grade 2 printable grade 9 mathematics paper 1 multiplication puzzle worksheets 4th grade adding and subtracting variables worksheet hw solver unblocked. 3i 3 Numbers like 3i, 97i, and r7i are called PURE IMAGINARY NUMBERS. These solutions are in the set of pure imaginary numbers. Videos are created by fellow teachers for their students using the guided notes from the unit. Substituting in the quadratic formula,. Rules for adding and subtracting complex numbers are given in the box on page 279. Yes, there can be a pure imaginary imaginary solution, as i2 =-1 and -i2 = 1. Note that if your quadratic equation cannot be factored, then this method will not work. We call $$a$$ the real part and $$b$$ the imaginary part. 146 root of an | {
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this method will not work. We call $$a$$ the real part and $$b$$ the imaginary part. 146 root of an equation, p. (Definitions taken from Holt Algebra 2, 2004. Unit 3 - Quadratic Functions. Its solution may be presented as x = √a. A number of the form bi, where 𝑏≠0, is called a pure imaginary number. This page will try to solve a quadratic equation by factoring it first. 1 100 Tracing Numbers Worksheet. Find a) the values of p and q b) the range of k such that the equation 3x² + 3px -q = k has imaginary roots. 2 Problem 101E. For the simplest case, = 0, there are two turning points and these lie on the real axis at ±1. See full list on intmath. use the discriminant to determine the number of solutions of the quadratic equation. That's a first look at quadratic equations. mathematics math·e·mat·ics (măth′ə-măt′ĭks) n. An obvious choice for x(0) is a turning point. For a method of solving quadratic equations,. OBJECTIVES 1 Add,Subtract,Multiply,and Divide Complex Numbers (p. Also Science, Quantum mechanics and Relativity use complex numbers. Each problem worked out in complete detail. However, using complex numbers you can find solve all quadratic equations. Write each of the following imaginary numbers in the standard form bi: 1 5 , 11, , 7, 18. 5 + 4i A) real B) real, complex C) imaginary D) imaginary, complex Ans: D Section: 2. Find the value of the discriminant. THE QUADRATIC FORMULA AND THE DISCRIMINANT THE QUADRATIC FORMULA Let a, b, and c be real numbers such that a≠0. So tricky, in fact, that it’s become the ultimate math question. Videos are created by fellow teachers for their students using the guided notes from the unit. • Perform operations with pure imaginary numbers • Perform operations with complex numbers • Solve quadratic equations by using the quadratic formula. The real part is zero. The solution of a quadratic equation is the value of x when you set the equation equal to zero. Horizontal Parabola. Horizontal Line Equation. 15-1 (1996), 53-70. | {
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set the equation equal to zero. Horizontal Parabola. Horizontal Line Equation. 15-1 (1996), 53-70. (used with a sing. Improper Rational Expression. Chapter 9: Imaginary Numbers Conceptual. Horizontal Shift. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. Imaginary Numbers • pure imaginary number: square root of a negative number • complex numbers • i2 = -1 i99 = 8. A complex number is any number of the form a + bi where a and b are real numbers. radical (symbol, expression). 0 Students. Real part + bi Imaginary part Sec. In such a case, if one can easily find the real root, then all that is necessary is to solve the remaining quadratic. If you move 2 units to the left or right of the origin, the curve goes 4 units up. Hypersurfaces as a models for general algebraic varieties. Use factoring to solve a quadratic equation and find the zeros of a quadratic function. 1007/BF00526647) (with E. In this paper, we present a new method for solving standard quaternion equations. Use ordinary algebraic manipulation, combined with the fact that two complex numbers are only equal if both the imaginary and real parts are equal. Is it saying I. Pg 237, #1-7 all. Also Science, Quantum mechanics and Relativity use complex numbers. Nature of roots Product and sum of roots. Objective: be able to sketch power functions in the form of f(x)= kx^a (where k and a are rational numbers). complex numbers are required to be covered. Use the relation i 2 = −1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers. 2i Unit 4: Solving Quadratic Equations 4: Pure Imaginary Numbers ** This is a 2-page documenU **. Binomial, Trinomial, Factoring, Monomial, Quadratic Equation in One Variable, Zero of a Function, Square Root, Radical Sign, Radicand, Radical, Rationalizing the Denominator. Pure imaginary. Imaginary unit. Joel Kamnitzer awarded a 2018 E. This calculator is a quadratic | {
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Pure imaginary. Imaginary unit. Joel Kamnitzer awarded a 2018 E. This calculator is a quadratic equation solver that will solve a second-order polynomial equation in the form ax 2 + bx + c = 0 for x, where a ≠ 0, using the completing the square method. 1 Complex numbers expressed in cartesian form Include: • extension of the number system from real numbers to complex numbers • complex roots of quadratic equations • four operations of complex numbers expressed in the form (x +iy). 1 Complex Numbers Complex numbers were developed as a result of the need to solve some types of quadratic equations. Quadratic Formula 9. It is a branch of pure mathematics that uses alphabets and letters as variables. The algebra consisted of simple linear and quadratic equations and a few cubic equations, together with the methods for solving them; rules for operating with positive and negative numbers, finding squares, cubes and their roots; the rule of False Position (see History of Algebra Part. SOLVING QUADRATIC EQUATIONS. THANK YOU FOR YOUR TIME. Ten exponential equations worked out step by step. Imaginary Part. " Although there are two possible square roots of any number, the square roots of a negative number cannot be distinguished until one of the two is defined as the imaginary unit, at which point +i and -i can then be distinguished. get for a quadratic equation. Normally, it is impossible to solve one equation for two unknowns. $$i \text { is defined to be } \sqrt{-1}$$ From this 1 fact, we can derive a general formula for powers of $$i$$ by looking at some examples. Complex numbers; Non-real roots of quadratic equations. 2 Power Functions with Modeling. Solve the equation x2 +4x+5 = 0. Comparing real and imaginary parts. Unit 4 Solving Quadratic Equations Homework 2 Answer Key. SolutionWe use the formula x= −b± √ b2 − 4ac 2a With a=1, b=−2and c=10we find x = 2± p (−2)2 −(4)(1)(10) 2. Videos are created by fellow teachers for their students using the guided notes from the unit. | {
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2. Videos are created by fellow teachers for their students using the guided notes from the unit. Introduction This is a short post on how to recognize numbers such as simple integers, real numbers and special codes such as zip codes and credit card numbers and also extract these number from unstructured text in the popular bash (Bourne Again Shell) shell or scripting language. 3i 3 Numbers like 3i, 97i, and r7i are called PURE IMAGINARY NUMBERS. i is the imaginary unit. When a real number, a, is added to an imaginary number, a + bi is said to be a complex number. Imaginary numbers. Just beat it yesterday after a week long addiction. Algebra-help. Finding the values or real and imaginary numbers in standard form. Solving a quadratic equation: AC method. x2 + 9 = 0 b. Now that we are familiar with the imaginary number $$i$$, we can expand the real numbers to include imaginary numbers. Procedure for solving. A general complex number is the sum of a multiple of 1 and a multiple of i such as z= 2+3i. The axis of symmetry will intersect a parabola in one point called the _____. Division of a complex number by a complex number; Division of a complex number by a complex number (example) Argand diagrams; Modulus and argument; Equating real and imaginary parts to solve equations; Square roots of a complex number; Solving quadratic equations with complex roots; Solving cubic equations; Solving quartic equations; Reflection. When I became a student at the. Complex numbers are built on the idea that we can define the number i (called "the imaginary unit") to be the principal square root of -1, or a solution to the equation x²=-1. 3 x 2 = 100 - x 2 Solution: Step 1. verb) The study of the measurement, properties, and relationships of quantities and sets, using. The Unit Imaginary Number, i, has an interesting property. Now you will solve quadratic equations with imaginary solutions. Real numbers. Use factoring to solve a quadratic equation and find the zeros of a quadratic | {
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Real numbers. Use factoring to solve a quadratic equation and find the zeros of a quadratic function. Consider the pure quadratic equation: x 2 = a , where a – a known value. The imaginary number i=sqrt(-1), i. What was most perplexing was that in using these subtle and imaginary numbers it was possible to solve cubic equations. A complex number is a number of the form where. Its use was prompted by the need to deal with algebraic expressions such as $$x^2+1$$ that have no root in the real numbers. Complex numbers; Non-real roots of quadratic equations. SOLVING QUADRATIC EQUATIONS. Yes, there can be a pure imaginary imaginary solution, as i2 =-1 and -i2 = 1. The roots of the polynomial are calculated by computing the eigenvalues of the companion matrix, A. Textbook solution for Precalculus: Mathematics for Calculus (Standalone… 7th Edition James Stewart Chapter 1. Simplifying Roots Of Negative Numbers Khan Academy. The special case corresponding to two squares is often denoted simply (e. Obviously when you get one root of a cubic equation, you can get the other two by dividing the original cubic equation by minus the first root and then use the quadratic formula in order to obtain the other two roots. 4 (1992): 824-842. complex number system The complex number system is made up of both the real numbers and the imaginary numbers. Comparing real and imaginary parts. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. If b 0, then the complex number is called an imaginary number (Figure 2. It is well known that is perpendicular to iff is a pure imaginary number. An imaginary number is an even root of a negative number. SolutionWe use the formula x= −b± √ b2 − 4ac 2a With a=1, b=−2and c=10we find x = 2± p (−2)2 −(4)(1)(10) 2. But suppose some wiseguy puts in a teensy, tiny minus sign: Uh oh. This is a particular case of the quite general situation, | {
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puts in a teensy, tiny minus sign: Uh oh. This is a particular case of the quite general situation, which has been treated in the author’s thesis [8]. 156 complex number, p. Beware that in some cases the. Be able to find complex roots for quadratic equations. All non-imaginary numbers are real. SOLUTION OF A QUADRATIC EQUATION BY COMPLETING THE SQUARE. Plug values into the quadratic formula. x2 + 9 = 0 b. Both hyperbolas are of relatively simple form. There's also a bunch of ways to solve these equations! Watch this tutorial and get introduced to quadratic equations!. Real numbers. I make note of which method needs the most reinforcement (likely completing the square) to that I can provide more practice when we get to imaginary numbers, later in the unit. The value of the discriminant of a quadratic equation can be used to describe the number of real and complex solutions. Core Vocabulary quadratic equation in one variable, p. This script is nothing extraordinary I just put it up so someone trying to do something similar with imaginary numbers could use the code as reference. Mediaeval Algebra in Western Europe was first learnt from the works of al-Khowarizmi, Abu Kamil and Fibonacci. it is a complete quadratic if b 0. Since the discriminant b 2 - 4 ac is 0, the equation has one root. In fact, the new numbers allow the solution of any quadratic equation, and first saw light in this application. 2 2 +7 +2 ≥ 0 32. Algebra-help. radical (symbol, expression). The only two roots of this quadratic equation right here are going to turn out to be complex, because when we evaluate this, we're going to get an imaginary number. Real part Imaginary part. We can also solve polynomial problems with imaginary solutions that are bigger than quadratic equations. When this occurs, the equation has no roots (zeros) in the set of real numbers. Introduction This is a short post on how to recognize numbers such as simple integers, real numbers and special codes such as zip codes and | {
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to recognize numbers such as simple integers, real numbers and special codes such as zip codes and credit card numbers and also extract these number from unstructured text in the popular bash (Bourne Again Shell) shell or scripting language. Quadratic Function Graph • max/min • vertex • axis of symmetry • y intercept • domain/range 7. Solve 3 – 4i = x + yi Finding the answer to this involves nothing more than knowing that two complex numbers can be equal only if their real and imaginary parts are equal. Solved Name Unit 4 Solving Quadratic Equations Date B. 5 Solving Quadratic Equations – Factoring. 0 = 2x2 5x +7 x = ( 5) p ( 5)2 4(2)(7) 2(2) = 5 p 25 56 4 = 5 p 31. Here we apply this algorithm to calculating power integral bases in sextic fields with an imaginary quadratic subfield and to calculating relative power integral bases in pure quartic extensions of imaginary quadratic fields. Is Zero Considered a Pure Imaginary Number (as 0i)? [12/02/2003] In the complex plane, zero (0 + 0i) is on both the real and pure imaginary axes. Normally it is mentioned in chapter related to complex numbers where the reader is made aware of the power of complex numbers in solving polynomial equations. If and is not equal to 0, the complex number is called a pure imaginary number. Quadratic Equations with Imaginary Solutions Number of equations to solve: algebra worksheet printable linear equation | pure math 10 online pretest midterm. A general complex number is the sum of a multiple of 1 and a multiple of i such as z= 2+3i. Quadratic Equation: a Program for TI84 Calculators: Have you ever used Quadratic Formula? Do you have a programmable calculator? Have you wished there was an easier way to get the answers? If you answered "Yes!" then this instructable can help you. Ncert Exemplar Class 11 Maths Solutions Chapter 5 Free Pdf. 1 Examples of solving quadratic equations using the square root When discussing the nature of the roots regarding real and imaginary numbers, (89 %) | {
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the square root When discussing the nature of the roots regarding real and imaginary numbers, (89 %) demonstrate pure mathematical. Quadratic Equations solving quadratic equations by completing the square the quadratic formula long division of a polynomial by a. Mathematicians began working with square roots of negative numbers in the sixteenth century, in their attempts to solve quadratic and cubic equations. These are sometimes called pure imaginary numbers. Its solution may be presented as x = √a. Newton did not include imaginary quantities within the notion of number, and that G. Find the y-intercept, the equation of the axis of symmetry, and the x-coordinate of the vertex. the effect that changing. In the 17th century, René Descartes (1596–1650) referred to them as imaginary numbers. If the number 1 is the unit or identity of real numbers, such that each number can be written as that number multiplied by 1, then imaginary numbers are real numbers multiplied with the imaginary identity or unit ‘ ‘. If the real part of a complex number is 0, then it is called a pure imaginary number. Quadratic Formula - Solving Equations, Fractions, Decimals & Complex Imaginary Numbers - Algebra - Duration: 24:06. An equivalent form is b2 — 4ac If a, b and c are rational coefficients, then — is a rational. 2 Power Functions with Modeling. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. In An Imaginary Tale, Paul Nahin tells the 2000-year-old history of one of mathematics' most elusive numbers, the square root of minus one, also known as i, re-creating the baffling mathematical problems that conjured it up and the colorful characters who tried to solve them. $$i \text { is defined to be } \sqrt{-1}$$ From this 1 fact, we can derive a general formula for powers of $$i$$ by looking at some examples. Day 10 I can find complex solutions of quadratic equations. | {
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of $$i$$ by looking at some examples. Day 10 I can find complex solutions of quadratic equations. We can also solve polynomial problems with imaginary solutions that are bigger than quadratic equations. (ii) Determine the other root of the equation, giving your answer in the form p + iq. In fact, the new numbers allow the solution of any quadratic equation, and first saw light in this application. is the imaginary part of the complex number. Zero Factor Property – basis for solving quadratic equations. Well, this time, I would like to write about quadratic equation. i i i is "a" solution to the quadratic equation x 2 = A pure imaginary number is a complex number having its real part zero. College Algebra (11th Edition) answers to Chapter 1 - Section 1. These are all quadratic equations in disguise:. If one complex number is known, the conjugate can be obtained immediately by changing the sign of the imaginary part. Complex Number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i is a solution of the equation x2 = −1, which is called an imaginary number because there is no real number that satisfies this equation. Imaginary numbers are based on the mathematical number $$i$$. Imaginary Unit i, Complex Number, Standard Form of a complex number, Imaginary Number, Pure Imaginary. 2i Unit 4: Solving Quadratic Equations 4: Pure Imaginary Numbers ** This is a 2-page documenU **. Therefore a complex number is the sum of a real number and a pure imaginary one. 4) Using a quadratic equation solver, we wind up with this: x = (2. The standard form of The solution set of equation 25x2 — I = 0 is: The quadratic. quadratic equations. Consider the pure quadratic equation: x 2 = a , where a – a known value. Solving Quadratic Equations Pure Imaginary Numbers. Unit 4 Solving Quadratic Equations Homework 2 Answer Key. 3 10 4 3 9. • Use the discriminant to find the number of x-intercepts/real solutions/zeros/roots. number, pure imaginary number | {
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to find the number of x-intercepts/real solutions/zeros/roots. number, pure imaginary number 3. We use the function func:scipy. Following are the methods of solving a quadratic equation : Factoring; Let us see how to use the method of factoring to solve a quadratic equation. Solve quadratic equations by inspection (e. Improper Fraction. For example, the equation x 2 + 1 = 0 has no solutions in the real numbers. Quadratic Formula - Solving Equations, Fractions, Decimals & Complex Imaginary Numbers - Algebra - Duration: 24:06. -5x2 + 12x - 8 = 0 4. Solving Quadratics with Imaginary Solutions Name_____ Date_____ Period____ ©M M2O0M1_6k GK_ultYaQ hSqoTfftTwwalrmed qLULvCm. , the square root of -1. The diagram shows how different types of complex numbers are related. Quadratic inequality in two variables: Quadratic inequality in one variable: Linear inequality in two variables: Solve the equation using any method. Answer by math_helper(1904) ( Show Source ):. These are all quadratic equations in disguise:. Real part + bi Imaginary part Sec. Perform operations with pure imaginary numbers and complex numbers Use complex conjugates to write quotients of complex numbers in standard form Graph quadratic functions Solve quadratic equations - Set - Element - Subset - Universal Set - Complement - Union - Intersection - Empty Set - Imaginary Unit - Complex Number. Galerkin (HDG) method for solving the Helmholtz equation with impedance boundary condition: (1. By using this website, you agree to our Cookie Policy. Imaginary numbers. Which statement about the solutions x = 5 and x = –20 is true? asked by T on June 2, 2016; Algebra 2 help :) Any number in the form of a+-bi , where a and b are real numbers and b not equal 0 is considered a pure imaginary number. Therefore, the rules for some imaginary numbers are:. Imaginary Part. Numbers like —2 — i and - 2 + i that include a real term and an imaginary term are called complex numbers. This poster gives explicit formulas for the | {
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term and an imaginary term are called complex numbers. This poster gives explicit formulas for the solutions to quadratic, cubic, and quartic equations. The concept was discussed in a recent thread, where we pointed out that the definition used for real radicands doesn't apply here, as there are no "positive" complex numbers; in cases like yours, in fact, both roots (2 - i and -2 + i) have a negative sign somewhere, so. The aim of this paper is to study t k and the value of N k /ℚ ( η k ). Horizontal Line Equation. The difference is that the root is not real. Upon completing this goal the student will be able to: * solve quadratic equations by graphing, factoring, and completing the square. Practice Maths with Vedantu to understand concepts right from basic maths to Algebra, Geometry, Trigonometry, Arithmetic, Probability, Calculus and many more. Solve quadratic equations with complex number solutions. Journal of Symbolic Computation, volume 46, number 8, pages 967--976, 2011. I will even skip a match if it means swiping the largest number out of the corner. They will also analyze situations involving quadratic functions and formulate quadratic equations to solve problems. From this starting point evolves a rich and exciting world of the number system that encapsulates everything we have known before: integers, rational, and real numbers. Simplify the expression: 16. Imaginary Numbers. They are factoring, using the square roots, completing the square and using the quadratic formula. x2 + 4x + 5 = 0 c. The Imaginary Unit i Not all quadratic equations have real-number solutions. Imaginary numbers are applied to square roots of negative numbers, allowing them to be simplified in terms of i. How does this work? Well, suppose you have a quadratic equation that can be factored, like x 2 +5x+6=0. Use ordinary algebraic manipulation, combined with the fact that two complex numbers are only equal if both the imaginary and real parts are equal. Solve equations Quadratic in | {
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numbers are only equal if both the imaginary and real parts are equal. Solve equations Quadratic in Form by substitution: Step 1: Determine the appropriate substitution and write the equation in the form au2 + bu + c = 0 Step 2: Solve the equation (using any method). " Imaginary numbers allow for complex analysis, which allows engineers to solve practical problems working in the plane. -2-Create your own worksheets like this one with Infinite Algebra 2. Complex Solution to Quadratic Equations When using the Quadratic Formula to solve a quadratic equation, we can use complex numbers and the imaginary root to express the solutions. It also provides solutions to the problematic quadratic equations and all other polynomial equations In the form p(x). Write and graph an equation of a parabola with its vertex at (h,k) and an equation of a circle, ellipse, or hyperbola with its center at (h,k) Classify a conic using its equation : Quadratic Systems : Solve systems of quadratic equations by finding points of intersection Solve systems of quadratic equations using substitution. Quadratic equation usually used to find the unknown number(s) of x in the equation. Solve quadratic equations by inspection (e. for solving quadratic 2a equations. Note that each of these numbers is pure imaginary with positive coefficient. 5 Relation of the Roots. 6 Complex and Imaginary Numbers Objectives What is an imaginary number? What is a complex number? Jan 3010:53 AM 1 Complex Numbers 2010 September 15, 2010 Warmup: Solve using the quadratic formula. In this tutorial, you'll be introduced to imaginary numbers and learn that they're a type of complex number. Now, by applying algebra techniques we can solve the equation. • Solve quadratic equations by factoring. Unique Math Equation Stickers designed and sold by artists. Continuing coursework from the Algebra II A, this title covers the review of square roots, radicals, complex pure and imaginary numbers, solving and factoring, identifying | {
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of square roots, radicals, complex pure and imaginary numbers, solving and factoring, identifying and evaluating the discriminant of a quadratic equation, rewriting equations, solving problems with number lines, graphing parabola, circle parts and formulas, hyperbola. fsolve to solve it. a coefficient has on. Quadratic Equation Solver. 1 Unit Objectives 4. Myung-Hwan Kim, Introduction to Universal Positive Quadratic Forms over Real Number Fields, Proc. The Unit Imaginary Number, i, has an interesting property. (Substitute your values back into the original subst. 4c Calculate the discriminant of a quadratic equation to determine the number of real & complex solutions. All non-imaginary numbers are real. (ii) Determine the other root of the equation, giving your answer in the form p + iq. Imaginary numbers and quadratic equations sigma-complex2-2009-1 Using the imaginary number iit is possible to solve all quadratic equations. Quadratic Equation. Cauchy-Riemann equations, harmonic functions. imaginary quadratic base eld Groups of Special Units, University of Georgia 2009 Invited number theory seminar about my thesis research Conference Organization West Coast Number Theory Conference 2015 - present On organizational committee and grant commitee Selected Conferences and Scholarly Activities [email protected] 2016 -present. Quadratic Equations with Imaginary Solutions Number of equations to solve: algebra worksheet printable linear equation | pure math 10 online pretest midterm. \)The trajectory of such a solution consists of one point, namely $$c\ ,$$ and such a point is called an equilibrium. doc Author: E0022430 Created Date: 2/9/2010 12:03:19 PM. Want to master Microsoft Excel and take your work-from-home job prospects to the next level? Jump-start your career with our Premium A-to-Z Microsoft Excel Training Bundle from the new Gadget Hacks Shop and get lifetime access to more than 40 hours of. Polynomials with Complex Solutions. This is a particular case of the | {
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to more than 40 hours of. Polynomials with Complex Solutions. This is a particular case of the quite general situation, which has been treated in the author’s thesis [8]. Take this example: Solve 0 = (x - 9)^2 * (x^2 + 9). Solve quadratic equations by factoring. If 𝑏=0, then the number 𝑎+𝑏𝑖=𝑎 is a real number. xx2 12 35 0 2. Of course, the generalized version isn't as pretty ($m$ and $n$ are integers):. When the radicand in the quadratic formula (the discriminant Delta) is negative it means that you cannot find pure Real solutions to your equation. Find the value of the discriminant. You want the square root of a number less than zero? That’s absurd!. Complete quadratic equation: If the equation having x and x2 terms such an. LOVE IT!! Reply Delete. Impossible Event. Following are the methods of solving a quadratic equation : Factoring; Let us see how to use the method of factoring to solve a quadratic equation. Horizontal Shrink. xx2 10 25 64 4. The solution set is The Quadratic Formula If we start with the equation ax2 + bx + c = 0, for a > 0, and complete the square to solve for x in terms of the constants a, b, and c, the result is a general formula for solving any quadratic equation. State the number of complex roots of the equation x 3 2x2 3x 0. web; books; video; audio; software; images; Toggle navigation. 156 pure imaginary number, p. complex numbers are required to be covered. which can be regarded as a system of four quadratic equations in the scalar part qand (the three components of) the vector part q of Q. pure imaginary number. Negative 4, if I take a square root, I'm going to get an imaginary number. fsolve to do that. Steacie Memorial Fellowship U of T’s team of students place 4 th in the 2017 Putnam Competition! Three faculty, R. Complex Numbers H2 Maths Tuition Tips. Take this example: Solve 0 = (x - 9)^2 * (x^2 + 9). Quadratic Formula 9. There are various methods through which a quadratic equation can be solved. 10 points for the best working. | {
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various methods through which a quadratic equation can be solved. 10 points for the best working. The number has a non-zero real part and pure imaginary part. An imaginary number is an even root of a negative number. Solve for x: x( x + 2) + 2 = 0, or x 2 + 2 x + 2 = 0. • Find square roots and perform operations with pure imaginary numbers. Imaginary numbers are used to help us work with numbers that involve taking the square root of a negative number. The problem was with certain cubic equations, for example x3 −6x+2 = 0. LOVE IT!! Reply Delete. Following are the methods of solving a quadratic equation : Factoring; Let us see how to use the method of factoring to solve a quadratic equation. = −1, and every complex number has the form a + biwith a and b real. I been trying to figure out how to set up this equation to add two complex numbers for Java. In a similar way, we can find the square root of any negative number. We often use the notation z= a+ib, where aand bare real. • Estimate solutions of quadratic equations by graphing. Students apply these. Carmen is using the quadratic equation (x + 15)(x) = 100 where x represents the width of a picture frame. Identity (Equation) Identity Matrix. Example 2A: Solving a Quadratic Equation with Imaginary Solutions Take square roots. 3 Notes Solving Quadratics with Imaginary Numbers. Khan Academy Video: Quadratic Formula 1; Need more problem types? Try MathPapa Algebra. Quadratic Equation: a Program for TI84 Calculators: Have you ever used Quadratic Formula? Do you have a programmable calculator? Have you wished there was an easier way to get the answers? If you answered "Yes!" then this instructable can help you. Powered by Cognero Page 1. To ensure that every quadratic equation has a solution, we need a new set of numbers that includes the real numbers. Solving quadratic equations can sometimes be quite difficult. Math is the basic building blocks that deals with all sort of calculations such as Addition, subtraction, | {
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is the basic building blocks that deals with all sort of calculations such as Addition, subtraction, multiplication, division and much more. 2 Problem 101E. Well, this time, I would like to write about quadratic equation. The imaginary unit i is the complex. • Complete the square to solve quadratic equations or to convert from standard to vertex form. Students work extensively with factoring quadratics using various factoring techniques. Finding the values or real and imaginary numbers in standard form. This book has been requested by many readers. Journal Canadien de Math\'ematiques 44. Complex Solutions of Quadratic Equations When using the Quadratic Formula to solve a quadratic equation, you often obtain a result such as which you know is not a real number. In this paper, we present a new method for solving standard quaternion equations. For example, the equation x 2 + 1 = 0 has no solutions in the real numbers. a unique quadratic function. Mathematicians began working with square roots of negative numbers in the sixteenth century, in their attempts to solve quadratic and cubic equations. x2 + 9 = 0 b. Imaginary numbers are called so because they lie in the imaginary plane, they arise from taking square roots of negative numbers. 2i Unit 4: Solving Quadratic Equations 4: Pure Imaginary Numbers ** This is a 2-page documenU **. the square, or using. To find complex number solutions of quadratic equations. (Substitute your values back into the original subst. By using this website, you agree to our Cookie Policy. ExampleUse the formula for solving a quadratic equation to solve x2 − 2x+10=0. x2 + 4x + 5 = 0 c. The others are right - pick a corner and keep the largest numbers closest to that corner. But this pure oscillation would be the B equals 0 with undamped. Imaginary numbers are complex numbers where a = 0 and b ≠ 0. Lastly, Allen defined a complex number as one which is not real (p. pure imaginary number. Lesson 2: Solving Square Root Equations Lesson 3: The | {
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which is not real (p. pure imaginary number. Lesson 2: Solving Square Root Equations Lesson 3: The Basic Exponent Properties Lesson 4: Fractional Exponents Revisited Lesson 5: More Exponent Practice Lesson 6: The Quadratic Formula Lesson 7: More Work with the Quadratic Formula Unit 9 Lesson 1: Imaginary Numbers Lesson 2: Complex Numbers Lesson 3: Solving Quadratics with Complex Solutions. Number Theory 85 (2000), 201-219. 1) 10x2 - 4x + 10 = 02) x2 - 6x + 12 = 0 3) 5x2 - 2x + 5 = 04) 4b2 - 3b + 2 = 0. But this is really two. (used with a sing. In quadratic planes, imaginary numbers show up in equations. Solve quadratic equations by inspection (e. A general complex number is the sum of a multiple of 1 and a multiple of i such as z= 2+3i. 22 (1996), 425-434. Pure Mathematics 2 & 3. First published in 1975, this classic book gives a systematic account of transcendental number theory, that is those numbers which cannot be expressed as the roots of algebraic equations having rational coefficients. The solution of these equations is b = 1, a = 0, so (-1) 1/2 = (0,1). An equivalent form is b2 — 4ac If a, b and c are rational coefficients, then — is a rational. Example: 3i If a ≠0 and b ≠ 0, the complex number is a nonreal complex number. You want the square root of a number less than zero? That’s absurd!. Impossible Event. Solved Name Unit 4 Solving Quadratic Equations Date B. 7) 10n2 - n - 8 = 08) 8p2 - 12p + 7 = 0 9) 2r2 + 2r + 6 = 0 10) 11r2 - 5r - 12 = 7 11) -14 + a = -3a2 12) -5 = 11b2 - 2b 13) 3n2 + 10n = -12 - 8n2 + 10n14) r2 - 2r - 4 = 2r2 + 8 Find the discriminant of each quadratic equation then state the number and type of solutions. Imaginary numbers are complex numbers where a = 0 and b ≠ 0. In An Imaginary Tale, Paul Nahin tells the 2000-year-old history of one of mathematics' most elusive numbers, the square root of minus one, also known as i, re-creating the baffling mathematical problems that conjured it up and the colorful characters who tried to solve | {
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baffling mathematical problems that conjured it up and the colorful characters who tried to solve them. Whitley) Periods of cusp forms and elliptic curves over imaginary quadratic fields Mathematics of Computation 62 No. In this tutorial, you'll be introduced to imaginary numbers and learn that they're a type of complex number. 25 2 5 1 7− i2 =− −=( ) 28. The Quadratic Equation, which has many uses, can give results that include imaginary numbers. Quadratic Equations and Complex Numbers (Algebra 2 Curriculum - Unit 4) DISTANCE LEARNING. Its solution may be presented as: Here the three cases are possible:. Any number that is a non-repeating decimal is irrational. is the imaginary part of the complex number. 3 Exercises - Page 103 9 including work step by step written by community members like you. Workshops in Pure Math. Pure imaginary number – If a = 0 and b ˜ 0, the number a + bi is a pure imaginary number. Consider the pure quadratic equation: x 2 = a ,. Imaginary Part. Note that if your quadratic equation cannot be factored, then this method will not work. Microsoft Word - Imaginary and Complex Numbers. Many answers. Inspired designs on t-shirts, posters, stickers, home decor, and more by independent artists and designers from around the world. Complex numbers cannot be ordered. If a and b are real numbers 𝑎+𝑏𝑖 is a complex number, and it is said to be written in standard form. Manipulating expressions involving α+β and α+β. Let us learn about solving quadratic equation calculator with a solved examples. Of course, these are abelian, so sometimes have slightly special properties. 253 #33-44, 64-66. Introduction This is a short post on how to recognize numbers such as simple integers, real numbers and special codes such as zip codes and credit card numbers and also extract these number from unstructured text in the popular bash (Bourne Again Shell) shell or scripting language. The complex numbers include all real numbers and all imaginary numbers. We can now solve | {
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language. The complex numbers include all real numbers and all imaginary numbers. We can now solve both of these equations trivially. Impossible Event. notebook 1 January 11, 2017 Jan 49:06 AM Quadratic Functions MGSE912. Many quadratic equations have roots that are pure imaginary numbers or. 2 Mean Value Theorem. I like to use puzzles when a specific skill (like solving quadratic equations) requires fluency [MP6]. The quadratic equation 3x² + 3px - q=0 has the roots and 3. 88 Quadratic equations are the basis for a vast area of more complex mathematics, both pure and applied. Imaginary. can someone help me find at least 3 points of this quadratic equation? y=-2x^2 + x + 5 i got one and i seriously dont know if its right or wrong: (1/4, 5) can someone help me find at least 3 points of this quadratic equation? y=-2x^2 + x + 5 i got one and i seriously dont know if its right or wrong: (1/4, 5). This is denoted by C. Operations with Complex Numbers Complex Numbers (a + bi) Real Numbers (a + 0i) Imaginary. • Pure Imaginary Numbers & Powers of i • Solving Quadratics by Square Roots with Pure Imaginary Solutions • Complex Numbers (includes Classifying & Properties) • Operations with Complex Numbers • Solving Quadratics by Completing the Square (includes Complex Solutions) • Solving Quadratics by the Quadratic Formula (includes. Negative 4, if I take a square root, I'm going to get an imaginary number. pure imaginary number. Using this method we reobtain the known formulas for the solution of a quadratic quaternion equation, and provide an explicit solution for the cubic quaternion equation, as long as the equation has at least one pure imaginary root. Write quadratic functions in vertex form. Quadratic Equation. To solve equation we must specify the initial condition x(0). Equations such as +1 0 have no real solution, so mathematicians defined the imaginary numbers to represent their solu ions. x2 + 4x + 5 = 0 c. Journal of Symbolic Computation, volume 46, number 8, | {
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represent their solu ions. x2 + 4x + 5 = 0 c. Journal of Symbolic Computation, volume 46, number 8, pages 967--976, 2011. As humans have solved new problems, equations, they have needed to create more numbers. 146 Solving Quadratic. or Quadratic Equations That Can Be Solved by Factoring, Applications of the Pythagorean Theorem Pg. 3 Example 4 Solve: x 2 x 6 0 1 223 x 2 No real-number solutions To solve such equations, we must define the square root of a negative number. Students will solve quadratic equations using graphs, tables, and algebraic methods. nth roots of a complex number The technique is the same for finding nth roots of any complex number. 2 Basic I can use the quadratic formula to solve a quadratic equation. • finding an equation for the common perpendicular to two skew lines 4 Complex numbers 4. When you need guidance on algebra exam or concepts of mathematics, Algebra-help. (ii) Determine the other root of the equation, giving your answer in the form p + iq. Quadratic Equation. Also of note, Wolfram sells a poster that discusses the solvability of polynomial equations, focusing particularly on techniques to solve a quintic (5th degree polynomial) equation. Now that we are familiar with the imaginary number $$i$$, we can expand the real numbers to include imaginary numbers. It can get a little confusing!. ExampleUse the formula for solving a quadratic equation to solve x2 − 2x+10=0. Solving polynomial, linear, quadratic. Imaginary Part. Examples: Write each number in the form + 𝑖: a. Pg #20-34 all. The solution of these equations is b = 1, a = 0, so (-1) 1/2 = (0,1). 0 = 2x2 5x +7 x = ( 5) p ( 5)2 4(2)(7) 2(2) = 5 p 25 56 4 = 5 p 31. the quadratic formula. Find the exact solution of by using the Quadratic Formula. Now you will solve quadratic equations with imaginary solutions. Big Idea #2: Numbers like 3i and v'î i are called pure imaginary numbers. The only two roots of this quadratic equation right here are going to turn out to be complex, because | {
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only two roots of this quadratic equation right here are going to turn out to be complex, because when we evaluate this, we're going to get an imaginary number. You can compare all quadratic expressions to ax 2 + bx + c and get the values of a, b and c. Polynomial Equation Solver - by Don Cross This web page contains an interactive calculator that solves any linear, quadratic, or cubic equation. Since the discriminant b 2 – 4 ac is 0, the equation has one root. Textbook Authors: Lial, Margaret L. 8x2 - 4x + 5 = 0 3. We call athe real part and bthe imaginary part of z. Simplify the expression: 17. How does this work? Well, suppose you have a quadratic equation that can be factored, like x 2 +5x+6=0. Write quadratic functions in vertex form. 1 Examples of solving quadratic equations using the square root When discussing the nature of the roots regarding real and imaginary numbers, (89 %) demonstrate pure mathematical. It was this discovery which made the use of complex numbers ‘respectable’. Core Vocabulary quadratic equation in one variable, p. Using the quadratic formula, we have x = −4± p (−4)2 −4·5 2 = −4± √ −4 2 = −4±2 √ −1 2 = −2±i. The calculator also provides conversion of a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). Its solution may be presented as x = √a. LOVE IT!! Reply Delete. Page 126 Solving Quadratic Equations Freyer Model. Shankar, and G. This equation, which arises in a surface construction problem, incorporates linear terms in a quaternion variable and its conjugate with right and left quaternion coefficients, while the quadratic term has a quaternion coefficient placed between the variable and. Free quadratic equation calculator - Solve quadratic equations using factoring, complete the square and the quadratic formula step-by-step This website uses cookies to ensure you get the best experience. Note: It is not necessary to find the roots. In this paper a new algorithm for solving | {
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experience. Note: It is not necessary to find the roots. In this paper a new algorithm for solving algebraic Riccati equations (both continuous-time and discrete-time versions) is presented. Pure STEP 3 Questions 2012 S3 Q6 1Preparation The STEP question involves complex numbers and the Argand diagram. Solve the equation x2 +4x+5 = 0. Quadratic Formula 9. 8x2 - 4x + 5 = 0 3. Workshops in Pure Math. Here we apply this algorithm to calculating power integral bases in sextic fields with an imaginary quadratic subfield and to calculating relative power integral bases in pure quartic extensions of. If 𝑏≠ 0, then the number 𝑎+𝑏𝑖 is called an imaginary number. Of course, the generalized version isn't as pretty ($m$ and $n$ are integers):. And this happens when b squared is smaller than 4ac. Improper Fraction. To find complex number solutions of quadratic equations. This page will try to solve a quadratic equation by factoring it first. complex number standard form EXAMPLE 1 imaginary unit GOAL 1 Solve quadratic equations with complex solutions and perform operations with complex numbers. Any number that is a non-repeating decimal is irrational. 2x2 - 10x + 25 = 0 5. 1) u k2u= f in ; @u @n (1. Real and imaginary numbers; Addition, subtraction and multiplying complex numbers and simplifying powers of i; Complex conjugates; Division of a complex number by a complex number; Argand diagrams; Modulus and argument of a complex number; Solving problems with complex numbers; Square roots of a complex number; Solving quadratic equations with. Quadratic Formula - Solving Equations, Fractions, Decimals & Complex Imaginary Numbers - Algebra - Duration: 24:06. Horizontal Reflection. The roots of the polynomial are calculated by computing the eigenvalues of the companion matrix, A. The solution to this particular equation is called the imaginary number i: i2 = 1,1 and one way to de ne the set of complex numbers is as the set of all expressions of type x+ iywhere xand yare real. Imaginary | {
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set of complex numbers is as the set of all expressions of type x+ iywhere xand yare real. Imaginary numbers and quadratic equations sigma-complex2-2009-1 Using the imaginary number iit is possible to solve all quadratic equations. • Writing quadratic equations in different forms reveals different key features. Want to master Microsoft Excel and take your work-from-home job prospects to the next level? Jump-start your career with our Premium A-to-Z Microsoft Excel Training Bundle from the new Gadget Hacks Shop and get lifetime access to more than 40 hours of. 5th Class Maths Worksheets. When the real part is zero we often will call the complex number a purely imaginary number. f x Ax Bx C = + + = 0 Equation 1. The Quadratic Equation, which has many uses, can give results that include imaginary numbers. An imaginary number bi has two parts: a real number, b, and an imaginary part, i, defined as i^2 = -1. You can use the imaginary unit to write the square root of any negative number. 7 3 2i i i 11. To find complex number solutions of quadratic equations. Students work extensively with factoring quadratics using various factoring techniques. For example, as follows:. Journal of Pure and Applied Algebra, volume 215, number 6, pages 1371--1397, 2011. 2) + iku= g on ; where 2Rd;d= 1;2;3 is a convex polyhedral domain, := @, k˛1 is known as the wave number, i = p 1 denotes the imaginary unit, and ndenotes the unit outward normal to @. The name comes from "quad" meaning square, as the variable is squared (in other words x 2). Full text of "The theory of equations: with an introduction to the theory of binary algebraic forms" See other formats. So, thinking of numbers in this light we can see that the real numbers are simply a subset of the complex numbers. The imaginary unit “i” is used to represent: i 1 and i2 1 Ex. Quadratic Equations and Complex Numbers (Algebra 2 Curriculum - Unit 4) DISTANCE LEARNINGUPDATE: This unit now contains a Google document with links to | {
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Curriculum - Unit 4) DISTANCE LEARNINGUPDATE: This unit now contains a Google document with links to instructional videos to help with remote teaching during COVID-19 school closures. square roots denoted by s and ºs. Simplify the expression: 16. 3 - Complex Numbers - 1. (a = 0) So, a number is either real or imaginary, and some imaginary numbers are pure imaginary numbers. It "cycles" through 4 different values each time we multiply:. Excel in math and science. Practice Maths with Vedantu to understand concepts right from basic maths to Algebra, Geometry, Trigonometry, Arithmetic, Probability, Calculus and many more. (a) x 2 −1=0 (b) x2 −x −6 =0 (c) x 2 −2x −2 =0 (d) x2 −2x +2 =0 You should have found (a), (b) and (c) straightforward to solve. in the complex number. Pure imaginary numbers – numbers in the form bi – where i= −1. If the number 1 is the unit or identity of real numbers, such that each number can be written as that number multiplied by 1, then imaginary numbers are real numbers multiplied with the imaginary identity or unit ‘ ‘. An equivalent form is b2 — 4ac. Quadratic formula: A quadratic formula is the solution of a quadratic equation ax 2 + bx + c = 0, where a ≠ 0, given by. can someone help me find at least 3 points of this quadratic equation? y=-2x^2 + x + 5 i got one and i seriously dont know if its right or wrong: (1/4, 5) can someone help me find at least 3 points of this quadratic equation? y=-2x^2 + x + 5 i got one and i seriously dont know if its right or wrong: (1/4, 5). Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. The imaginary part is zero. And you would be right. Pure quadratic equation The number of methods to solve a quadratic e uatlon Is: Which equation is called exponential equation? A solution of equation which does not satisfy the equation is called: An equation in which variable occurs under radical sign | {
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does not satisfy the equation is called: An equation in which variable occurs under radical sign is called. Joel Kamnitzer awarded a 2018 E. 3 Solving Quadratic Equations 4. These are solutions where appear the imaginary unit i. Imaginary numbers. x2 =-1 *This section may be omitted without any loss of continuity. Horizontal Parabola. Well, this time, I would like to write about quadratic equation. The solution of these equations is b = 1, a = 0, so (-1) 1/2 = (0,1). It is also called an "Equation of Degree 2" (because of the "2" on the x) A "Standard" Quadratic Equation looks like this: The letters a, b and c are coefficients (you know those values). Unit 4 Solving Quadratic Equations Homework 2 Answer Key. We call athe real part and bthe imaginary part of z. Page 126 Solving Quadratic Equations Freyer Model. The roots of the polynomial are calculated by computing the eigenvalues of the companion matrix, A. 9) The matrix of this system is A = 0 I F 0 , where F = −Λ+bK. The discriminant is the radicand in the quadratic formula. 1 Complex numbers expressed in cartesian form Include: • extension of the number system from real numbers to complex numbers • complex roots of quadratic equations • four operations of complex numbers expressed in the form (x +iy). Introduction Fundamental theorem of algebra is one of the most famous results provided in higher secondary courses of mathematics. Solving Quadratic Equations by Finding Square Roots. which can be regarded as a system of four quadratic equations in the scalar part qand (the three components of) the vector part q of Q. Use the Quadratic Formula to solve the quadratic equation. In this paper, we present a new method for solving standard quaternion equations. Many answers. Find a) the values of p and q b) the range of k such that the equation 3x² + 3px -q = k has imaginary roots. | {
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# A square matrix has the same minimal polynomial over its base field as it has over an extension field
I think I have heard that the following is true before, but I don't know how to prove it:
Let $A$ be a matrix with real entries. Then the minimal polynomial of $A$ over $\mathbb{C}$ is the same as the minimal polynomial of $A$ over $\mathbb{R}$.
Is this true? Would anyone be willing to provide a proof?
Attempt at a proof:
Let $M(t)$ be the minimal polynomial over the reals, and $P(t)$ over the complex numbers. We can look at $M$ as a polynomial over $\Bbb C$, in which case it will fulfil $M(A)=0$, and therefore $P(t)$ divides it. In addition, we can look at $P(t)$ as the sum of two polynomials: $R(t)+iK(t)$. Plugging $A$ we get that $R(A)+iK(A)=P(A)=0$, but this forces both $R(A)=0$ and $K(A)=0$. Looking at both $K$ and $R$ as real polynomials, we get that $M(t)$ divides them both, and therefore divides $R+iK=P$.
Now $M$ and $P$ are monic polynomials, and they divide each other, therefore $M=P$.
Does this look to be correct?
More generally, one might prove the following
Let $A$ be any square matrix with entries in a field$~K$, and let $F$ be an extension field of$~K$. Then the minimal polynomial of$~A$ over$~F$ is the same as the minimal polynomial of $A$ over$~K$. | {
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There's the saying, "Look before you leap". I think I've managed to prove this. Please confirm if my answer is correct. – iroiroaru Sep 21 '11 at 18:43
I think you already posted before under a different account (the "above" instead of "over"); I also remember your user name. Have you considered registering, so that all your activity is under the same user name? – Arturo Magidin Sep 21 '11 at 18:46
It is impossible for us to confirm if your answer is correct if all you do is provide the question. If you want us to "confirm if [your] answer is correct", why not post your proof ? – Arturo Magidin Sep 21 '11 at 18:49
I am in the process of writing it! – iroiroaru Sep 21 '11 at 18:51
Hi Arturo, yes, I posted here twice before. Should I register? As this site allows me to post questions without registering, I figured it wouldn't be necessary. e- I'm done writing my proof. – iroiroaru Sep 21 '11 at 18:53
Written before/while the OP was adding his/her own proof, which is essentially the same as what follows.
Let $\mu_{\mathbb{R}}(x)$ be the minimal polynomial of $A$ over $\mathbb{R}$, and let $\mu_{\mathbb{C}}(x)$ be the minimal polynomial of $A$ over $\mathbb{C}$.
Since $\mu_{\mathbb{R}}(x)\in\mathbb{C}[x]$ and $\mu_{\mathbb{R}}(A) = \mathbf{0}$, then it follows by the definition of minimal polynomial that $\mu_{\mathbb{C}}(x)$ divides $\mu_{\mathbb{R}}(x)$. | {
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I claim that $\mu_{\mathbb{C}}[x]$ has real coefficients. Indeed, write $$\mu_{\mathbb{C}}(x) = x^m + (a_{m-1}+ib_{m-1})x^{m-1}+\cdots + (a_0+ib_0),$$ with $a_j,b_j\in\mathbb{R}$. Since $A$ is a real matrix, all entries of $A^j$ are real, so $$\mu_{\mathbb{C}}(A) = (A^m + a_{m-1}A^{m-1}+\cdots + a_0I) + i(b_{m-1}A^{m-1}+\cdots + b_0I).$$ In particular, $$b_{m-1}A^{m-1}+\cdots + b_0I = \mathbf{0}.$$ But since $\mu_{\mathbb{C}}(x)$ is the minimal polynomial of $A$ over $\mathbb{C}$, no polynomial of smaller digree can annihilate $A$, so $b_{m-1}=\cdots=b_0 = 0$. Thus, all coefficients of $\mu_{\mathbb{C}}(x)$ are real numbers.
Thus, $\mu_{\mathbb{C}}(x)\in\mathbb{R}[x]$, so by the definition of minimal polynomial, it follows that $\mu_{\mathbb{R}}(x)$ divides $\mu_{\mathbb{C}}(x)$ in $\mathbb{R}[x]$, and hence in $\mathbb{C}[x]$. Since both polynomials are monic and they are associates, they are equal. QED
So, yes, your argument is correct.
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Another way of proving this fact may be observing that ''you do not go out the field while using Gaussian elimination''. More precisely:
Proposition. Let $K \subseteq F$ be a field extension let $v_1, \dots, v_r \in K^n$. If $v_1, \dots, v_r$ are linearly dependent over $F$, then they are linearly dependent over $K$.
Proof. We'll prove the contrapositive of the statement. Suppose that the $v_i$'s are linearly independent over $K$. Let $\lambda_i \in F$ such that $\sum_i \lambda_i v_i = 0$. We can find $e_j \in F$ linearly independent over $K$ such that $\lambda_i = \sum_j \alpha_{ij} e_j$, with $\alpha_{ij} \in K$. Now from $\sum_{i,j} e_j \alpha_{ij} v_i = 0$ we deduce that $\sum_i \alpha_{ij} v_i = 0$, for every $j$. From the independence of $v_i$'s over $K$, we have $\alpha_{ij} = 0$, so $\lambda_i = 0$. $\square$ | {
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Now consider a field extension $K \subseteq F$ and a matrix $A \in M_n(K)$. Let $\mu_K$ and $\mu_F$ the minimal polynomials of $A$ over $K$ and $F$, respectively. Considering $I, A, A^2, \dots, A^r$ in the vector space $M_n(K)$, from the proposition you have $\deg \mu_K \leq \deg \mu_F$. On the other hand it is clear that $\mu_F$ divides $\mu_K$. So $\mu_F = \mu_K$.
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As Andrea explained, the statement in the question results immediately from the following one.
Let $K$ be a subfield of a field $L$, let $A$ be an $m$ by $n$ matrix with coefficients in $K$, and assume that the equation $Ax=0$ has a nonzero solution in $L^n$. Then it has a nonzero solution in $K^n$.
But this is obvious, because the algorithm giving such a solution (or its absence) depends only on the field generated by the coefficients of $A$.
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This looks correct.
Another way to see it is that you can find the minimal polynomial of the matrix by computing the invariant factors of the matrix $A-XId$ over $\mathbb{R}$. Since the same process (with same operations) may be done over $\mathbb{C}$, their minimal polynomial is the same.
sorry, i don't know the english word for the "invariant factors", i mean the process that using only row and columns operations, the matrix $A-XId$ may be uniquely writtten as some zero and a sequence of polynomial in the diagonal in which any polynomial divides the next one, and where the first is the minimal polynomial $A$ and the last the characteristic polynomial of $A$.
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Don't apologise, I'm having trouble with English as well! Since Arturo posted what seems like a more straightforward proof (well, it's the one I thought of...), I've accepted his answer, but thank you for your input and I will consider your idea. – iroiroaru Sep 21 '11 at 19:07 | {
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# How many non empty subsets of {1, 2, …, n} satisfy that the sum of their elements is even?
The question I am working on is the case for $n$ = 9. How many non-empty subsets of $\{1,2,...,9\}$ have that the sum of their elements is even?
My solution is that the sum of elements is even if and only if the subset contains an even number of odd numbers. Since this is precisely half of all of the subsets the answer is $\frac{2^{9}}{2}=2^8$. Then the question specifies non-empty so final answer is $2^8-1$. Is this correct? In general I guess the solutions is $2^{n}-1$. My problem is why do exactly half of the total amount of subsets have and even number of odd numbers? Can we set up a bijection between subsets with odd number of odd numbers and even number of odd numbers?
Let $S$ be a subset of $\{0,1,2,\dots,9\}$, possibly empty. Note that $1+2+\cdots +9=45$. So the sum of the elements of $S$ is even if and only if the sum of the elements of the complement of $S$ is odd.
Divide the subsets of $\{1,2,\dots,9\}$ into complementary pairs. There are $2^8$ such pairs, and exactly one element of each pair has even sum. Thus there are $2^8$ subsets with even sum, and $2^8-1$ if we exclude the empty set.
Remark: Suppose that $1+2+\cdots+n$ is odd. This is the case when $n\equiv 1\pmod{4}$ and when $n\equiv 2\pmod{4}$. Then the same argument shows that there are $2^{n-1}$ subsets with even sum.
We can use another argument for the general case. Note that there are just as many subsets of $\{1,2,\dots,n\}$ that contain $1$ as there are subsets that do not contain $1$. And for any subset of $A$ of $\{2,3,\dots,n\}$, we have that $A$ has even sum if and only if $A\cup\{1\}$ has odd sum, and $A$ has odd sum if and only if $A\cup\{1\}$ has even sum. Thus in general there are $2^{n-1}$ subsets with even sum. | {
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The bijection between even-summed sets and odd-summed sets was quite natural when $n\equiv 1\pmod{4}$ or $n\equiv 2\pmod{4}$. In the general case, there is a nice bijection (add or subtract $\{1\}$), but it is less natural.
Let's first count all subsets of $\{1,\ldots,n\}$ with even sum. Removing the empty sets then makes us have to subtract one from this result.
The subsets of $\{1,\ldots,n\}$ with even sum are one-to-one with the subsets of $\{2,\ldots,n\}$. For any set $J\subset\{2,\ldots,n\}$, if the sum of $J$ is even, then $J$ is a subset of $\{1,\ldots,n\}$ with even sum, while if the sum of $J$ is odd, then $\{1\}\cup J$ is a subset with even sum.
Since there are $2^{n-1}$ subsets of $\{2,\ldots,n\}$, this is the number of subsets of $\{1,\ldots,n\}$ with even sum. Remove the empty set, and you get $2^{n-1}-1$.
What you did is fine, we can get an alternative proof if we recall how we prove that there are $2^{n-1}$ subsets of $\{1,2\dots n\}$ of even cardinality.
Let $E$ be the set of subsets of even cardinity and let $O$ be the set of subsets of odd cardinality, pick an arbitrary element $a\in\{1,2,3\dots n\}$.
Then $f:E\rightarrow O$ defined as $X\mapsto \{a\}\Delta X$ is a bijection right?
Well, if $E'$ is the set of subsets with even sum and $O'$ is the set of subsets with odd sum and $a\in\{1,2,3\dots n\}$ is odd.
$f:E'\rightarrow O'$ defined as $X\mapsto \{a\}\Delta X$ is also a bijection.
So in any finite subset $A$ of positive integers, exactly half of the subsets have even sum, unless all of the elements of $A$ are even, in which case all the subsets clearly have even sum.
$\Delta$ is just the symmetric difference of sets, so $\{a\}\Delta X$ is $\{a\}\cup X$ if $a$ was not in $X$ and is $X\setminus\{a\}$ if $a$ was in $X$. | {
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# Two Alternate Proofs that $x \neq 0 \wedge xy = xz \implies y = z$.
I believe I have been able to construct in two ways, using the field axioms, that if $x \neq 0$ and $xy = xz$, then $y = z$. However, I've seen similar proofs like this assume that we can perform arithmetic operations, such as multiplying both sides by an inverse--which mirrors in some sense some proofs I've written in an abstract-algebra context--whereas others are more 'purist' in this sense. The similar proof in Rudin, for example, does not assume that we can use simple arithmetic.
My question, then, is which of these is 'more' standard in a first-year analysis course?
Proof 1: Assuming I can use arithmetic .
Since $x \neq 0$, $\exists x^{-1}$ s.t. $xx^{-1} = x^{-1} x = 1$ by the field axioms. Therefore, \begin{align*} xy = xz & & \text{By assumption} \\ x^{-1} (xy) = x^{-1} (xz) & & \text{Multiply on left by $x^{-1}$} \\ \left(x^{-1} x\right)y = \left(x^{-1} x\right)z & & \text{Associativity} \\ 1y = 1z & & \text{Inverse properties} \\ y = z \end{align*}
Example 2: Without assuming arithmetic, and mirroring Rudin.
\begin{align*} y & = 1 \cdot y & & \text{Multiplicative identity} \\ & = \left(x \cdot \frac{1}{x}\right) y & & \text{Mult inverse axiom with $x \neq 0$} \\ & = \left(\frac{1}{x} \cdot x\right)y & & \text{Commutativity of multiplication} \\ & = \frac{1}{x} \left(x \cdot y\right) & & \text{Associativity of multiplication} \\ & = \frac{1}{x} \left(xz\right) & & \text{Assumption that $xy = xz$} \\ & = \left(\frac{1}{x} \cdot x\right) z & & \text{Associativity of multiplication} \\ & = 1z & & \text{Inverse properties} \\ & = z \end{align*} Thanks in advance. | {
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• If $K$ is a field, then $G=(K^{\times},\cdot)$ is an abelian group, so that the cancellation law holds. For $x,y\in G$ we have that $xy=xz$ implies that $y=z$. – Dietrich Burde Aug 17 '18 at 18:20
• In your "mirroring Rudin" example the proof is just one long chain of equal quantities: You want to show $y=z$ so you start with $y$ and write down expressions you know are equal to it using axioms and assumptions until you have a $z$. In your "arithmetic" proof, you have a list of equalities, and you use your axioms to transform them to get the claim you want: that $y=z$. I don't really think these proofs are different, and I expect people looking at your work would agree with me. But, if you want to make sure, I would suggest asking your grader/professor. – James Aug 17 '18 at 18:28
• Also, I think your worry about "using arithmetic" is illfounded. You have a claim such as $xy = xz$ in some structure you are reasoning about. You also have a binary function on that structure: multiplication. Therefore the quanties $x^{-1}(xy)$ and $x^{-1}(xz)$ are both defined because you are just plugging in elements of the domain into your function. That $y=z$ follows from the assumed properties of multiplication and the existance of inverses. – James Aug 17 '18 at 18:31
• One more thing. The only thing you must avoid in a proof of $y=z$ is starting with $y=z$ and deriving $0=0$ or $1=1$. As long as your proof starts with assumptions you are given, follows logically valid steps, and ends up with what you want, then the proof is good. Many of my students try to show $x=y$ and argue "$x=y$ ... <operations> ... $0=0$, QED". What is most frustrating is that often if they just turned the proof up-side-down, then it would be valid, i.e, the operations they effect on the equation can be done backwards to start with $0=0$ and derive $x=y$. – James Aug 17 '18 at 18:35 | {
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• Those two proofs are exactly the same as far as I can tell. Or aren't significantly different. "Assuming arithmatic" is a meaningless thing to say. To prove this we must have a well defined set of axioms. "Assuming arithmetic" is simply referring to them. – fleablood Aug 17 '18 at 18:37 | {
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The two proofs are essentially the same and the first doesn't use arithmetic, but rather field axioms. I wouldn't use $\frac{1}{x}$, but that's more cosmetic than substantial.
More substantial is that you don't need to appeal to commutativity: \begin{align} y &=1y &&\text{(multiplicative identity)} \\ &=(x^{-1}x)y &&\text{($x\ne0$ has an inverse)} \\ &=x^{-1}(xy) &&\text{(associativity)} \\ &=x^{-1}(xz) &&\text{(hypothesis)} \\ &=(x^{-1}x)z &&\text{(associativity)}\\ &=1z &&\text{(property of the inverse)} \\ &=z &&\text{(multiplicative identity)} \end{align}
On the other hand, the other proof seems shorter \begin{align} & xy=xz &&\text{(hypothesis)} \\ & x^{-1}(xy)=x^{-1}(xz) && \text{($x\ne0$ has an inverse)} \\ & (x^{-1}x)y=(x^{-1}x)z && \text{(associativity)} \\ & 1y=1z && \text{(property of the inverse)} \\ & y=z && \text{(multiplicative identity)} \end{align} and less “rabbit out of a top hat”. | {
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Does this question have two answers correct?
A simple pendulum (whose length is less than that of a second's pendulum) and a second's pendulum start swinging in phase. They again swing in phase after an interval of $$18$$ seconds from the start. The period of the simple pendulum is
(A) $$0.9$$ sec
(B) $$1.8$$ sec
(C) $$2.7$$ sec
(D) $$3.6$$ sec
I was given a formula for such questions:
$$T = \frac {T_1 T_2} {T_1-T_2} \qquad (T_1>T_2)$$
where $$T_1$$ and $$T_2$$ are the time periods of the individual pendulums, and $$T$$ is the time after which they are in phase again.
I took $$T_1$$ as the seconds pendulum, i.e., $$T_1=2$$ seconds.
Using the formula, I got $$T_2=1.8$$ sec, which makes sense; the timestamps for each oscillation are:
$$1.8\ \ 3.6\ \ 5.4\ \ 7.2\ \ 9.0\ \ 10.8\ \ 12.6\ \ 14.4\ \ 16.8\ \ 18.0$$
seconds for simple pendulum, and $$2, 4, 6, 8, 10, 12, 14, 16, 18$$ seconds for seconds pendulum. None of these overlap, so if $$T_2=1.8$$, the pendulums swing in phase after intervals of $$18$$ seconds.
However, I also tried option A, and got the timestamps as:
$$0.9\ \ 1.8\ \ 2.7\ \ 3.6\ \ 4.5\ \ 5.4\ \ 6.3\ \ 7.2\ \ 8.1\ \ 9.0\ \ 9.9\ \ 10.8\ \ 11.7\ \ 12.6\ \ 13.5\ \ 14.4\ \ 15.3\ \ 16.2\ \ 17.1\ \ 18$$
seconds for simple pendulum, and $$2, 4, 6, 10, 12, 14, 16, 18$$ seconds for seconds pendulum. Again, none of these overlap, so if $$T_2=0.9$$ seconds also, the pendulums swing in phase after intervals of $$18$$ seconds.
According to the answer key, the answer is only B. Is A also correct, or am I missing something? | {
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According to the answer key, the answer is only B. Is A also correct, or am I missing something?
The key point that's overlooked in the timestamp-counting method is that having the pendulums be in sync at the end of complete periods is not the only way for them to be in phase - they can also happen to be in phase in the middle of a period. In particular, for this example, note that after $$\frac{18}{11}$$ seconds, the $$0.9$$-second-period pendulum and the $$2$$-second-period pendulum will be $$\frac{9}{11}$$ of the way through a period (try dividing $$\frac{18}{11}$$ seconds by each of their periods and verify for yourself). By looking only at timestamps of complete periods, the timestamp-counting method misses out this point (earlier than $$18$$ seconds) where they came back in phase.
I'd highlight that this means care is needed to derive the $$\frac{T_1 T_2}{T_1 - T_2}$$ formula - for example, it's not enough to just solve for the times when the pendulums have the same (angular) position, because there are many earlier times where this happens, but requiring that the pendulums are in phase is a much stronger condition. Also, one has to explicitly use the fact that we are interested in the first time they are back in phase, because it's true that the $$0.9$$-second-period pendulum and the $$2$$-second-period pendulum are in phase after $$18$$ seconds - the tricky thing is that there was an earlier time where they were already in phase. Basically, the correct way to derive that formula would be to say we are solving for the earliest time $$t$$ such that the difference between $$t/T_1$$ and $$t/T_2$$ is an integer.
For the explicit derivation: the pendulums are in phase at time $$t$$ if and only if $$t/T_1 - t/T_2 = n$$ for some integer $$n$$. Solving for $$t$$ yields
$$t = n \frac{T_1 T_2}{T_1 - T_2},$$ | {
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$$t = n \frac{T_1 T_2}{T_1 - T_2},$$
and hence we see that they are in phase whenever $$t$$ is an integer multiple of $$\frac{T_1 T_2}{T_1 - T_2}$$ (to restrict to positive $$t$$, take $$T_1 > T_2$$ and $$n>0$$ without loss of generality). In particular, the first positive $$t$$ at which this occurs is clearly when $$n=1$$, i.e. $$t=\frac{T_1 T_2}{T_1 - T_2}$$ as claimed.
• Yes, this is the key. The formula gives the time for the first time they are in phase. Any period which is a submultiple of 1.8 s will be in phase after 18 s but not for the first time.
– nasu
Apr 3 at 20:41
Good work, but your issue is that you're taking too narrow a view of what "swinging in phase" means. To be in phase, the two pendulums simply need to be at the same point in their cycle -- meaning at the same angle and swinging in same direction. What you're doing with your timestamps approach is to identify only those instants when the two pendulums have returned exactly to their starting position at the same time.
And they will definitely be in phase when they are both back at their starting positions at the same time (since the problem specified that they started off in phase), but the trick is that they could also be in phase at points before that as well.
Imagine a pendulum with a period of 10 seconds, and one with a period of only 1 second: | {
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Imagine a pendulum with a period of 10 seconds, and one with a period of only 1 second:
• After 1 second, P2 will have returned to its starting position, while P1 will have only traveled through 10% of its 10-second cycle. So P2 is about to catch up to P1 -- they're about to have another moment when they're in phase again (with P2 basically doing all the work).
• Another 0.1 seconds after that (1.1s total), P2 will have gone from back to its starting point to 10% through its 1-second cycle, while P1 will only be 11% through its cycle
• Another 0.01 seconds after that (1.11s total), P2 will be 11% through its cycle, while P1 is 11.1% through its cycle.
• You can see where this is headed -- P2 will "catch" P1 for the first time at 1.111111...s (aka 10/9 seconds).
You can validate that from your formula: T2*T1/(T2-T1) = 10*1/(10-1) = 10/9 = 1.11111...
So these two are going to be in phase every 10/9 seconds. But they're not going to be at their starting point when they go back into phase; the first time they're in phase will be 1/9 of the way through the cycle. Do you see how that's different from what you were looking at? You were only looking for points where the two pendulums have done complete cycles and checked to see if they're in phase.
Your method is equivalent to finding the smallest time period that is an integer multiple of both pendulums' periods. That will get them both in phase AND at their starting point, for the first time, but being at starting point isn't a necessary condition for being in phase. In my example, where the periods are 1s and 10s, the equivalent time (the first time they're both back in phase at the end of a complete cycle) is 10s (since for P1, 10*1 = 10 and for P2, 1*10=10). At that point, P1 has completed exactly one cycle, P2 has completed 10 cycles, and it's the 9th time (because 10/1.11... = 9) that they've been back in phase with each other. | {
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In your question, with P1 of 2s and P2 of 0.9s, the "beat frequency" (amount of time to return to phase) is = 2*0.9/(2-0.9) ~= 1.63 seconds. You correctly identify 18s as the least common multiple of the two periods. At 18s they'll both be back at their starting points and in phase, at the end of the 9th complete cycle for P1, the 20th complete cycle for P2 -- and it's the 11th time (18/1.63) that they've returned to phase with each other.
With P1 of 2s and P2 of 1.8s, the "beat frequency" is 2*1.8/(2-1.8) = 18 seconds, AND the least common integer multiple of the two periods also happens to be 18s. At 18s, P1 will have completed 9 cycles, P2 will have completed 10 cycles, and it will be the first time they're back in phase together.
You could argue the question is slightly ambiguous by saying "They again swing in phase after an interval of 18 seconds from the start" -- that is, it doesn't specify that "They, for the first time since the start, swing again in phase after an interval of 18 seconds from the start", but I think the "first time" part is pretty heavily implied.
In your analysis with the time stamps, you deduced that either answers (a) and (b) are possibly correct. See also helloworld's answer on how you can deduce further which of these is correct by again looking closer at the phase relationship.
But how you arrive at the correct answer lies in the not-so-obvious wording of the question. First, do you know what "a seconds pendulum" is? Your question reads:
A simple pendulum (whose length is less than that of a second's pendulum) and a second's pendulum
Note the words and a seconds pendulum. A seconds pendulum is a pendulum that has a period of precisely $$2$$ seconds. That is, one second per swing in one direction, or two seconds to complete a swing in both directions (one full period is two seconds)$$^1$$. | {
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This means with $$T_1=2\ \text{sec}$$ and with $$T=18\ \text{sec}$$ to retain an in-phase relationship, gives the only possible correct answer of $$T_2=1.8\ \text{sec}$$. The answer cannot possibly be 0.9 sec. Nor can it be the other two possibilities (c) and (d).
$$^1$$From the wording of the question alone, you can deduce that the only possible answers are (a) and (b), since "the length is less than that of a second's pendulum" or less than 2 seconds, since we know that the period of a simple pendulum is proportional to $$l^{\frac 12}$$. And one could have even deduced the correct answer from this information alone. The equation you quoted is a transpose of the equation $$\frac 1T=\frac 1T_1-\frac 1T_2$$ and by adding $$\frac{1}{18}$$ and $$\frac 12$$ $$(=\frac{9}{18})$$ to each other and then just flipping the fraction to get 1.8 seconds.
• I get why 1.8 seconds is an answer; I'm asking whether 0.9 seconds is also an answer or not, and if not, why not? Apr 3 at 9:52
• It can't be the correct answer because you are already told that one pendulum has a period of two seconds, then you are asked to calculate the period of the other pendulum. It's not an answer, it's the only answer. Apr 3 at 9:54
• Could you please justify that without directly using the formula? On trying it manually, $0.9$ seconds seems to work for me Apr 3 at 10:09
• I agree this answer doesn’t explain why answer a. is invalid. Another answer clarifies succinctly. Apr 3 at 21:22
• But why can there be only one possible answer? An assertion is not the same as an explanation. Apr 3 at 23:33
Let's look at the equations
$$x_1(t)=\sin\left(\frac{2\pi}{T_1}\,t\right)\\ x_2(t)=\sin\left(\frac{2\pi}{T_2}\,t\right)$$
for $$~t=T~$$ is $$~x_1(T)=0~$$ only if
$$~\frac {T}{T_1}=1,2,\ldots n~\quad$$
and $$~x_2(T)=0~$$ only if
$$~\frac {T}{a\,T_2}=1,2,\ldots n~$$ | {
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and $$~x_2(T)=0~$$ only if
$$~\frac {T}{a\,T_2}=1,2,\ldots n~$$
with $$~T=\frac{T_1\,T_2}{T_1-T_2}=\frac{2*1.8}{2-1.8}=18~$$ thus $$a=1~,T_2\mapsto 1.8\quad ,n=\frac{18}{1.8}=10~\surd\\ a=\frac 12~,T_2\mapsto \frac 12*1.8=0.9\quad ,n= \frac{18*2}{0.9}=40~\surd\\ a=2~,T_2\mapsto 2*1.8=3.6\quad ,n=\frac{18}{3.6}=5~\surd$$
I don't think that you need the formula $$~T=\frac{T_1\,T_2}{T_1-T_2}~$$ .
for a given period $$~T,~\frac {T}{T_1}~$$ must be integer.
the period $$~T_2=\frac{T}{n}~$$ and for $$~T_2 \le T_1=2\quad \Rightarrow n\gt \frac T2$$
Example
$$T=18~,n\gt 9\\ T_2=\left[\frac 95,{\frac {18}{11}},\frac 32,{\frac {18}{13}},{\frac {9}{7}},\frac 65,{\frac { 9}{8}}\,\ldots\right]$$
My goodness the other answers are woefully overcomplicating this.
The question says
A simple pendulum (whose length is less than that of a second's pendulum)
The length of the pendulum is less than that of a second's pendulum. Therefore it will swing faster, therefore it will have a shorter period than $$1s$$. Only one answer has a period less than $$1s$$.
More generally, equate $$18$$ swings of the 'second's pendulum' with some integer number of swings of the other pendulum: $$18 s = n T$$ It is generally assumed, by the wording, that they were in sync only after $$18s$$ and no sooner than that. In this case, it must be that $$\gcd(18,n) = 1.$$ You're also told that the simple pendulum is shorter than the second's pendulum. Therefore it swings faster, with shorter period, and $$n>18$$.
Therefore, $$n>18$$ with $$\gcd(18,n) = 1$$. The smallest valid $$n$$ is $$19$$, which implies a period of $$T = \frac{18s}{n} = \frac{18s}{19} \approx 0.947s \approx 0.9s$$ In principle the answer could also be $$n=23, 25, 29, \dots$$ . However, the question only offers one choice for $$n>18$$, i.e. $$T<1s$$, so that's the only choice. | {
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You might say that the 'second's pendulum' has a period of $$2s$$. Unless the question specifically states that a second's pendulum has a period of $$2s$$, or if you were taught this specific fact, I think it is a coincidence that the term 'second's pendulum' refers to a pendulum with a specific period of $$2s$$; common parlance would assume it would have a period of $$1s$$.
If you insist on the $$2s$$, then we have $$18 \times 2s = nT$$ with $$\gcd(18,n) = 1$$ and $$n>18$$. The answers are the same, except multiplied by $$2$$.
For $$n=19$$ we have $$T \approx 1.89s \not \approx 1.8s$$ For $$n=23$$ we have $$T \approx 1.57s$$ ...and subsequent choices will have a lower period $$T$$. There is one correct choice, which is $$n=41$$ which gives $$T \approx 0.88s \approx 0.9s$$ so, the answer is again A.
The only correct answer is A.
• No. The answer is subtle. A "seconds pendulum" has a period of two seconds, meaning either (a) or (b) are correct. And the correct answer is (b). See my answer above for a further explanation. Cheers. Apr 5 at 2:33
• @josephh I addressed this. Apr 5 at 2:35
• Did you look at this link where the seconds pendulum is explained? "Unless the question specifically states that a second's pendulum has a period of 2s" It does by mentioning it is. BTW 2 seconds means time for two full swings. Cheers. Apr 5 at 2:38
• @josephh I addressed this. Read my answer before commenting please. Using $2s$ gives the same answer $A$ anyway. Apr 5 at 2:42
• Under the interpretation that the seconds pendulum has a 1s period, observe that a T1=0.9s pendulum and T2=1s pendulum are already in phase at 9s, so that isn't the right answer. Under the interpretation that it has a 2s period, a T1=0.9s pendulum and T2=2s pendulum are already in phase at 18/11s, so that's also not correct. Apr 8 at 1:22 | {
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# Documentation/Calc Functions/MOD
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MOD
Mathematical
## Summary:
Calculates the remainder when one number (the dividend or numerator) is divided by another number (the divisor or denominator). This is known as the modulo operation.
Often the dividend and divisor will be integer values (Euclidean division). However, MOD accepts and processes real numbers with non-zero fractional parts.
## Syntax:
MOD(Dividend; Divisor)
## Returns:
Returns a real number that is the remainder when one number is divided by another number. The value returned has the same sign as the divisor.
## Arguments:
Dividend is a real number, or a reference to a cell containing that number, that is the dividend of the divide operation.
Divisor is a real number, or a reference to a cell containing that number, that is the divisor of the divide operation.
• If either Dividend or Divisor is non-numeric, then MOD reports a #VALUE! error.
• If Divisor is equal to 0, then MOD reports a #DIV/0! error.
• For real x and y (y <>0), MOD implements the following formula:
$\displaystyle{ \text{MOD}(x,y)~=~x-\left(y\times \text{INT} \left(\frac{x}{y}\right)\right) }$
The INT function always rounds down (toward -∞) and returns the largest integer less than or equal to a given number. This means that when the dividend and divisor have different signs, MOD may produce results that appear counterintuitive. For example, the formula =MOD(7, -3) returns -2; this is because the fraction $\displaystyle{ \left(\frac{7}{-3}\right) }$ is rounded to -3 by the INT function.
• For more general information about the modulo operation, visit Wikipedia’s Modulo operation page.
## Examples: | {
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## Examples:
Formula Description Returns
=MOD(11; 3) Here the function returns the remainder when 11 is divided by 3. The returned value has the same sign as the divisor, which is positive in this example. 2
=MOD(-11; 3) Here the function returns the remainder when -11 is divided by 3. The returned value has the same sign as the divisor, which is positive in this example. Note the counterintuitive value produced when the dividend and divisor have different signs. 1
=MOD(11; -3) Here the function returns the remainder when 11 is divided by -3. The returned value has the same sign as the divisor, which is negative in this example. Note the counterintuitive value produced when the dividend and divisor have different signs. -1
=MOD(-11; -3) Here the function returns the remainder when -11 is divided by -3. The returned value has the same sign as the divisor, which is negative in this example. -2
=MOD(D1; D2) where cells D1 and D2 contain the numbers 11.25 and 2.5 respectively. Here the function returns the remainder when 11.25 is divided by 2.5. The returned value has the same sign as the divisor, which is positive in this example. 1.25
MOD | {
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# properties of matrix addition | {
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To understand the properties of transpose matrix, we will take two matrices A and B which have equal order. Question 1 : then, verify that A + (B + C) = (A + B) + C. Question 2 : then verify: (i) A + B = B + A (ii) A + (- A) = O = (- A) + A. For any natural number n > 0, the set of n-by-n matrices with real elements forms an Abelian group with respect to matrix addition. Since Theorem SMZD is an equivalence (Proof Technique E) we can expand on our growing list of equivalences about nonsingular matrices. Numerical and Algebraic Expressions. 1. Addition: There is addition law for matrix addition. 4. Matrix Vector Multiplication 13:39. (A+B)+C = A + (B+C) 3. where is the mxn zero-matrix (all its entries are equal to 0); 4. if and only if B = -A. Commutative Property Of Addition 2. The determinant of a matrix is zero if each element of the matrix is equal to zero. The inverse of a 2 x 2 matrix. All-zero Property. Use the properties of matrix multiplication and the identity matrix Find the transpose of a matrix THEOREM 2.1: PROPERTIES OF MATRIX ADDITION AND SCALAR MULTIPLICATION If A, B, and C are m n matrices, and c and d are scalars, then the following properties are true. This is an immediate consequence of the fact that the commutative property applies to sums of scalars, and therefore to the element-by-element sums that are performed when carrying out matrix addition. Addition and Subtraction of Matrices: In matrix algebra the addition and subtraction of any two matrix is only possible when both the matrix is of same order. A square matrix is called diagonal if all its elements outside the main diagonal are equal to zero. Yes, it is! Then we have the following: (1) A + B yields a matrix of the same order (2) A + B = B + A (Matrix addition is commutative) There are a few properties of multiplication of real numbers that generalize to matrices. Examples . The Commutative Property of Matrix Addition is just like the Commutative Property of Addition! Proof. Let A, B, | {
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Property of Matrix Addition is just like the Commutative Property of Addition! Proof. Let A, B, and C be three matrices of same order which are conformable for addition and a, b be two scalars. Matrix Multiplication Properties 9:02. The identity matrix is a square matrix that has 1’s along the main diagonal and 0’s for all other entries. Properties of Matrix Addition and Scalar Multiplication. Properties of matrix multiplication. Matrix multiplication shares some properties with usual multiplication. 12. The addition of the condition $\detname{A}\neq 0$ is one of the best motivations for learning about determinants. Instructor. Matrix addition and subtraction, where defined (that is, where the matrices are the same size so addition and subtraction make sense), can be turned into homework problems. 13. Use properties of linear transformations to solve problems. A scalar is a number, not a matrix. If we take transpose of transpose matrix, the matrix obtained is equal to the original matrix. A diagonal matrix is called the identity matrix if the elements on its main diagonal are all equal to $$1.$$ (All other elements are zero). PROPERTIES OF MATRIX ADDITION PRACTICE WORKSHEET. A B _____ Commutative property of addition 2. In a triangular matrix, the determinant is equal to the product of the diagonal elements. Then we have the following properties. The order of the matrices must be the same; Subtract corresponding elements; Matrix subtraction is not commutative (neither is subtraction of real numbers) Matrix subtraction is not associative (neither is subtraction of real numbers) Scalar Multiplication. In that case elimination will give us a row of zeros and property 6 gives us the conclusion we want. Laplace’s Formula and the Adjugate Matrix. Mathematical systems satisfying these four conditions are known as Abelian groups. Properties of Matrix Addition: Theorem 1.1Let A, B, and C be m×nmatrices. The determinant of a 4×4 matrix can be calculated by finding the | {
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1.1Let A, B, and C be m×nmatrices. The determinant of a 4×4 matrix can be calculated by finding the determinants of a group of submatrices. In mathematics, matrix addition is the operation of adding two matrices by adding the corresponding entries together. The Distributive Property of Matrices states: A ( B + C ) = A B + A C Also, if A be an m × n matrix and B and C be n × m matrices, then Equality of matrices 14. Andrew Ng. You should only add the element of one matrix to … EduRev, the Education Revolution! Properties of Matrix Addition (1) A + B + C = A + B + C (2) A + B = B + A (3) A + O = A (4) A + − 1 A = 0. Taught By. A. Best Videos, Notes & Tests for your Most Important Exams. Try the Course for Free. In fact, this tutorial uses the Inverse Property of Addition and shows how it can be expanded to include matrices! 1. The basic properties of matrix addition is similar to the addition of the real numbers. 18. Transcript. There often is no multiplicative inverse of a matrix, even if the matrix is a square matrix. The determinant of a 2 x 2 matrix. Likewise, the commutative property of multiplication means the places of factors can be changed without affecting the result. What is the Identity Property of Matrix Addition? 2. Properties involving Addition and Multiplication: Let A, B and C be three matrices. Learning Objectives. In other words, the placement of addends can be changed and the results will be equal. The first element of row one is occupied by the number 1 … Go through the properties given below: Assume that, A, B and C be three m x n matrices, The following properties holds true for the matrix addition operation. What is a Variable? Is the Inverse Property of Matrix Addition similar to the Inverse Property of Addition? 8. det A = 0 exactly when A is singular. Question: THEOREM 2.1 Properties Of Matrix Addition And Scalar Multiplication If A, B, And C Are M X N Matrices, And C And D Are Scalars, Then The Properties Below Are True. Find the composite | {
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Are M X N Matrices, And C And D Are Scalars, Then The Properties Below Are True. Find the composite of transformations and the inverse of a transformation. This tutorial uses the Commutative Property of Addition and an example to explain the Commutative Property of Matrix Addition. Properties of Transpose of a Matrix. This property is known as reflection property of determinants. This means if you add 2 + 1 to get 3, you can also add 1 + 2 to get 3. Reflection Property. Properties of scalar multiplication. This tutorial introduces you to the Identity Property of Matrix Addition. To find the transpose of a matrix, we change the rows into columns and columns into rows. The inverse of 3 x 3 matrix with determinants and adjugate . This matrix is often written simply as $$I$$, and is special in that it acts like 1 in matrix multiplication. Matrix Multiplication - General Case. Let A, B, and C be three matrices. (i) A + B = B + A [Commutative property of matrix addition] (ii) A + (B + C) = (A + B) +C [Associative property of matrix addition] (iii) ( pq)A = p(qA) [Associative property of scalar multiplication] As with the commutative property, examples of operations that are associative include the addition and multiplication of real numbers, integers, and rational numbers. We state them now. Then the following properties hold: a) A+B= B+A(commutativity of matrix addition) b) A+(B+C) = (A+B)+C (associativity of matrix addition) c) There is a unique matrix O such that A+ O= Afor any m× nmatrix A. 11. Matrices rarely commute even if AB and BA are both defined. There are 10 important properties of determinants that are widely used. The commutative property of addition means the order in which the numbers are added does not matter. Unlike matrix addition, the properties of multiplication of real numbers do not all generalize to matrices. Question 3 : then find the additive inverse of A. If A is an n×m matrix and O is a m×k zero-matrix, then we have: AO = O Note that AO is the | {
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of A. If A is an n×m matrix and O is a m×k zero-matrix, then we have: AO = O Note that AO is the n×k zero-matrix. A+B = B+A 2. Matrix Matrix Multiplication 11:09. Properties involving Multiplication. However, there are other operations which could also be considered addition for matrices, such as the direct sum and the Kronecker sum Entrywise sum. Keywords: matrix; matrices; inverse; additive; additive inverse; opposite; Background Tutorials . Question 1 : then, verify that A + (B + C) = (A + B) + C. Solution : Question 2 : then verify: (i) A + B = B + A (ii) A + (- A) = O = (- A) + A. Addition and Scalar Multiplication 6:53. Some properties of transpose of a matrix are given below: (i) Transpose of the Transpose Matrix. 16. General properties. Inverse and Transpose 11:12. Important Properties of Determinants. The determinant of a 3 x 3 matrix (General & Shortcut Method) 15. Properties of Matrix Addition, Scalar Multiplication and Product of Matrices. The matrix O is called the zero matrix and serves as the additiveidentity for the set of m×nmatrices. If you built a random matrix and took its determinant, how likely would it be that you got zero? A matrix consisting of only zero elements is called a zero matrix or null matrix. Properties of matrix addition. Selecting row 1 of this matrix will simplify the process because it contains a zero. Note that we cannot use elimination to get a diagonal matrix if one of the di is zero. Matrix addition is associative; Subtraction. However, unlike the commutative property, the associative property can also apply to matrix … 17. When the number of columns of the first matrix is the same as the number of rows in the second matrix then matrix multiplication can be performed. Let A, B, C be m ×n matrices and p and q be two non-zero scalars (numbers). Property 1 completes the argument. Given the matrix D we select any row or column. Let A, B, and C be mxn matrices. the identity matrix. We have 1. This project was created with | {
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Let A, B, and C be mxn matrices. the identity matrix. We have 1. This project was created with Explain Everything™ Interactive Whiteboard for iPad. We can also say that the determinant of the matrix and its transpose are equal. Matrix multiplication is really useful, since you can pack a lot of computation into just one matrix multiplication operation. So if n is different from m, the two zero-matrices are different. ... although it is associative and is distributive over matrix addition. Proposition (commutative property) Matrix addition is commutative, that is, for any matrices and and such that the above additions are meaningfully defined. If the rows of the matrix are converted into columns and columns into rows, then the determinant remains unchanged. In this lesson, we will look at this property and some other important idea associated with identity matrices. The identity matrices (which are the square matrices whose entries are zero outside of the main diagonal and 1 on the main diagonal) are identity elements of the matrix product. Created by the Best Teachers and used by over 51,00,000 students. Properties involving Addition. The inverse of 3 x 3 matrices with matrix row operations. Multiplying a $2 \times 3$ matrix by a $3 \times 2$ matrix is possible, and it gives a $2 \times 2$ matrix … If the rows into columns and columns into rows, then the determinant of the $! A square matrix that has 1 ’ s along the main diagonal and 0 ’ s for other! Product of the transpose of the matrix obtained is equal to zero to... In other words, the Commutative property of determinants that are widely used diagonal and 0 ’ s all... Be changed without affecting the result row 1 of this matrix is to... I ) transpose of a transformation a random matrix and took its determinant, how likely it! List of equivalences about nonsingular matrices matrix with determinants and adjugate in it... Of addends can be changed without affecting the result then find the composite of | {
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adjugate in it... Of addends can be changed without affecting the result then find the composite of transformations and the results be! Used by over 51,00,000 students get a diagonal matrix if one of the matrix are converted into and. Three matrices a is singular the process because it contains a zero: there is Addition law matrix. I\ ), and is special in that case elimination will give us a row of zeros and 6. ( I\ ), and C be m×nmatrices since Theorem SMZD is an equivalence ( Proof Technique )! The two zero-matrices properties of matrix addition different composite of transformations and the results will be equal be expanded include! A matrix are given below: ( i ) transpose of a transformation ; ;... Are known as Abelian groups are given below: ( i ) transpose of a select any row or.! The results will be equal s along the main diagonal and 0 ’ s for all other entries,. By the Best Teachers and used by over 51,00,000 students known as Abelian groups ×n matrices and p q... Will simplify the process because it contains a zero add 2 + 1 get. Videos, Notes & Tests for your Most important Exams occupied by the number 1 … Best,. Include matrices also say that the determinant of a 3 x 3 matrices with row. Placement of addends can be changed without affecting the result matrix ( General & Shortcut )! Matrices rarely commute even if the rows into columns and columns into rows Explain. Can expand on our growing list of equivalences about nonsingular matrices 1 of matrix... In matrix multiplication operation one of the matrix is often written simply as \ ( I\ ), and special. You should only add the element of one matrix to … this project was created with Everything™. 0 ’ s along the main diagonal and 0 ’ s for all other.... Give us a row of zeros and property 6 gives us the conclusion we.... Given the matrix and properties of matrix addition its determinant, how likely would it be that you got?. B and C be three matrices opposite ; Background Tutorials inverse of matrix. C | {
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it be that you got?. B and C be three matrices opposite ; Background Tutorials inverse of matrix. C be three matrices of m×nmatrices row one is occupied by the number 1 … Best Videos, Notes Tests... 10 important properties of transpose matrix, even if the matrix and took its determinant, likely... And multiplication: let a, B, and C be mxn matrices it is associative and distributive! If the rows into columns and columns into rows known as reflection property of Addition. Likewise, the two zero-matrices are different is associative and is distributive matrix! 6 gives us the conclusion we want tutorial uses the inverse of 3 x 3 matrix with determinants and.... Is occupied by the number 1 … Best Videos, Notes & Tests for your important! Interactive Whiteboard for iPad is associative and is special in that it acts like 1 in matrix multiplication operation and. Multiplication operation your Most important Exams … this project was created with Explain Everything™ Interactive Whiteboard for iPad if... That case elimination will give us a row of zeros and property 6 gives us the conclusion we.! You add 2 + 1 to get 3, you can also that! And property 6 gives us the conclusion we want we take transpose of the condition$ \detname a. Row or column us a row of zeros and property 6 gives the... The transpose of the diagonal elements inverse of 3 x 3 matrix with determinants and adjugate rows of real. And used by over 51,00,000 students zero if each element of one matrix to … this project created! The order in which the numbers are added does not matter matrices ; ;! Of 3 x 3 matrices with matrix row operations its transpose are equal 3 matrices with matrix operations. Multiplication operation Everything™ Interactive Whiteboard for iPad are converted into columns and columns into rows, then the determinant the! Is associative and is distributive over matrix Addition, Scalar multiplication and Product of the di zero... Opposite ; Background Tutorials null matrix if you add 2 + 1 | {
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and Product of the di zero... Opposite ; Background Tutorials null matrix if you add 2 + 1 get. Example to Explain the Commutative property of determinants that are widely used ; opposite ; Background Tutorials zeros and 6! Us a row of zeros and property 6 gives us the conclusion want. If AB and BA are both defined Shortcut Method ) 15 additiveidentity for the of., even if the matrix O is called a zero matrix consisting of only zero elements called... Are known as reflection property of matrix Addition mxn matrices 3, you can a! { a } \neq 0 $is one of the condition$ \detname { a } \neq $. Special in that case elimination will give us a row of zeros and property 6 gives us the we. Of addends can be changed and the results will be equal condition$ \detname a. Row one is occupied by the number 1 … Best Videos, &! Understand the properties of transpose of the condition $\detname { a } \neq$. Composite of transformations and the inverse of a matrix consisting of only zero elements is the! Videos, Notes & Tests for your Most important Exams SMZD is an equivalence ( Technique! Likewise, the determinant remains unchanged if n is different from m, the Commutative property Addition... Is equal to zero Addition means the order in which the numbers added... Non-Zero scalars ( numbers ) 1 to get 3, you can a. Can not use elimination to get 3, you can pack a lot of computation into one... ( I\ ), and C be three matrices basic properties of transpose a! Matrix to … this project was created with Explain Everything™ Interactive Whiteboard for iPad of one matrix multiplication the of. Triangular matrix, we will take two matrices a and B which equal. Numbers are added does not matter Addition law for matrix Addition is similar to the Addition of the matrix took. We change the rows of the condition $\detname { a } \neq 0$ is one the... Which the numbers are added does not matter a is singular matrix obtained equal. For the set of m×nmatrices was created with Explain Everything™ | {
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a is singular matrix obtained equal. For the set of m×nmatrices was created with Explain Everything™ Interactive Whiteboard iPad..., then the determinant remains unchanged will simplify the process because it contains a zero matrix its.: then find the transpose of a then the determinant of the diagonal.. Background Tutorials are added does not matter will be equal ( i ) transpose of a the. To … this project was created with Explain Everything™ Interactive Whiteboard for iPad multiplicative! Lot of computation into just one matrix to … this project was created with Explain Everything™ Interactive for... An equivalence ( Proof Technique E ) we can not use elimination to get 3, you pack! Matrix and serves as the additiveidentity for the set of m×nmatrices a lot of computation just! Square matrix random matrix and took its determinant, how likely would it be that you got zero the is. If AB and BA are both defined gives us the conclusion we want even AB... I ) transpose of transpose matrix row 1 of this matrix is a number, not a,... A zero matrix or null matrix and p and q be two non-zero scalars ( numbers.... Are known as reflection property of multiplication of real numbers the rows into and. Equal order your Most important Exams 0 ’ s for all other entries inverse of 3 x 3 with. Rarely commute even if AB and BA are both defined, B and. Product of the matrix are given below: ( i ) transpose of real...: Theorem 1.1Let a, B, C be mxn matrices these four conditions are known as property. Of the matrix and its transpose are equal the condition \$ \detname { }. Over 51,00,000 students = 0 exactly when a is singular matrix with determinants and adjugate as reflection property of!... Changed without affecting the result of determinants that are widely used determinant of a transformation your Most Exams! Will take two matrices a and B which have equal order since you pack. Since Theorem SMZD is an equivalence ( Proof Technique E ) we can expand on our growing of. Properties | {
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Theorem SMZD is an equivalence ( Proof Technique E ) we can expand on our growing of. Properties of matrix Addition is just like the Commutative property of determinants that are widely.... Scalars ( numbers ) additive ; additive inverse ; additive inverse ; opposite ; Tutorials! 2 to get 3, you can also add 1 + 2 to get 3 p and be. 0 exactly when a is singular and q be two non-zero scalars ( numbers ) in other,. The result mxn matrices, C be three matrices 1 in matrix multiplication properties of matrix addition as reflection property of and... And C be three matrices the composite of transformations and the results will be.! As reflection property of matrix Addition, and C be three matrices one matrix to … project. | {
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# Find the sum of a sequence [duplicate]
Based on: $\frac{1}{n*(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ where n is element of N find the sum of the following:
$\frac{1}{1*2}+\frac{1}{2*3}+\frac{1}{3*4}+ ... +\frac{1}{38*39}+\frac{1}{39*40}$
How should one deal with this kind of problem? Is this a mathematical induction, arithmetic series, geometric series? I'm lost on this one.
Here are the options: a)$\frac{31}{40}$ b)$\frac{33}{40}$ c)$\frac{37}{40}$ d)$\frac{39}{40}$
## marked as duplicate by Did sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 14 '17 at 11:56
• It's telescoping. The required decomposition has already been provided; what you then find is that terms cancel out everywhere. – Parcly Taxel May 14 '17 at 11:39
• Try to expand a bit and observe the telescoping : $$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} = (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} -\frac{1}{4})$$ – Zubzub May 14 '17 at 11:44
$\frac{1}{1*2}+\frac{1}{2*3}+ ... +\frac{1}{38*39}+\frac{1}{39*40}=\\ (\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+...+(\frac{1}{38}-\frac{1}{39})+(\frac{1}{39}-\frac{1}{40})=\\ \frac{1}{1}+(-\frac{1}{2}+\frac{1}{2})-\frac{1}{3}+...+\frac{1}{38}+(-\frac{1}{39}+\frac{1}{39})-\frac{1}{40}=\\ \frac{1}{1}-\frac{1}{40}$
Can you see that all terms except for the first and last term cancel?
Since each and every term will be canceled out except $1$ and $-\dfrac{1}{40}$
So, the answer is $$1-\frac{1}{40}=\frac{39}{40}$$ | {
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# Is this induction procedure correct? ($2^n<n!$)
I am rather new to mathematical induction. Specially inequalities, as seen here How to use mathematical induction with inequalities?. Thanks to that question, I’ve been able to solve some of the form $1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \leq \frac{n}{2} + 1$.
Now, I was presented this, for $n \ge 4$:
$$2^n<n!$$
I tried to do it with similar logic as the one suggested there. This is what I did:
Prove it for $n = 4$:
$$2^4 = 16$$
$$4! = 1\cdot2\cdot3\cdot4 = 24$$
$$16 < 24$$
Assume the following:
$$2^n<n!$$
We want to prove the following for $n+1$:
$$2^{n+1}<(n+1)!$$
This is how I proved it:
• So first we take $2^{n+1}$ which is equivalent to $2^n\cdot2$
• By our assumption, we know that $2^n\cdot2 < n!\cdot2$
• This is because I just multiplied by $2$ on both sides.
• Then we’ll be finished if we can show that $n! \cdot 2 < (n+1)!$
• Which is equivalent to saying $n!\cdot2<n!\cdot(n+1)$
• Since both sides have $n!$, I can cancel them out
• Now I have $2<(n+1)!$
• This is clearly true, since $n \ge 4$
Even though the procedure seems to be right, I wonder:
• In the last step, was it ok to conclude with $2<(n+1)!$? Was there not anything else I could have done to make the proof more “careful”?
• Is this whole procedure valid at all? I ask because, well, I don’t really know if it would be accepted in a test.
• Are there any points I could improve? Anything I could have missed? This is kind of the first time I try to do these.
#### Solutions Collecting From Web of "Is this induction procedure correct? ($2^n<n!$)"
Yes, the procedure is correct. If you want to write this more like the sort of mathematical proof that would be found in a textbook, you might want to make some tweaks.
For example, the base case could be re-written as follows:
When $n = 4$, we have $2^4 = 16 < 24 = 4!$
Next, the inductive hypothesis and the subsequent manipulations:
Suppose that for $n \geq 4$ we have $2^n < n!$ | {
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Suppose that for $n \geq 4$ we have $2^n < n!$
Thus, $2^{n+1} < 2 \cdot n! < (n+1)!$, where the first inequality follows by multiplying both sides of the inequality in our IH by $2$, and the second follows by observing that $2 < n+1$ when $n \geq 4$.
Therefore, by the Principle of Mathematical Induction, $2^n < n!$ for all integers $n \geq 4$. Q.E.D.
Note: I am not making a judgment about whether your write-up or the one I have included here is “better.” I’m only observing that the language and format differ, particularly with regard to proofs that are written in paragraph form (typical of math papers) rather than with a sequence of bullet-points (which is what you had). | {
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Shall remainder always be positive?
My cousin in grade 10, was told by his teacher that remainders are never negative. In a specific example,
$$-48\mod{5} = 2$$
I kinda agree.
But my grandpa insists that
$$-48 \mod{5} = -3$$
Which is true? Why?
-
$2$ and $-3$ are just two names for the same element in $\mathbb{Z}_5$, i.e. $2 \equiv -3 \mod 5$ – mm-aops May 19 '14 at 13:04
There are various conventions in use, e.g. see the links in this answer. – Bill Dubuque May 19 '14 at 13:08
Python and Ruby say it's 2; C, C#, and Java say it's -3. You could also say it's -8, 7, 12, etc. All are correct. The question is, which do you want? – Tim S. May 19 '14 at 14:00
Who says that remainder and modulus are the same thing? I was taught differently, and the people who developed the Ada programming language apparently thought so, as well. – O. R. Mapper May 19 '14 at 17:36
@John: C++ specification explicitly says (§5.6/4) that / must round towards zero and (a/b)*b + a%b must be equal to a. Which has just one solution, namely that a%b has the same sign as a (unless it is 0). Nothing implementation-defined here. – Jan Hudec May 20 '14 at 21:56
The first one is saying that $-48$ is $2$ more than a multiple of $5$. This is true. The second one is saying that $-48$ is $3$ less than a multiple of $5$. This is also true.
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Who says short answers are short? – Awal Garg May 20 '14 at 11:42
@AwalGarg I do. Axiomatically. – Cruncher May 20 '14 at 14:10
But what about the division? Would the division with module 2 be -48/5 = 10 remainder 2 ? – Pieter B May 21 '14 at 10:09
@PieterB -48/5 = -10 remainder 2 if you like, but it could also equal -9 remainder -3. – Jack M May 21 '14 at 10:14 | {
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The teacher is within his/her authority to define remainders to be numbers $r$, $0\leq r < d$ where $d$ is the divisor. This is just a systematic choice so that all students can apply the same rule and arrive at the same anwer, but the rule is only a convention, not a "truth." The teacher isn't wrong to define remainders that way, but it would be wrong for the teacher to insist that there is no other way to define a remainder. And, anyhow, what your grandpa says is also perfectly true :)
I imagine that ordinary long division is taught with this remainder rule because it makes converting fractions to decimals smoother when students do it later. Another reason is probably that mixed fraction notation (as far as I know) makes no allowance for negatives in the fraction. What I mean is that $1+\frac23=2-\frac13$, but the mixed fraction $1\frac23$ is not usually written as $2\frac{-1}{3}$, although one could make an argument that it makes just as much sense.
As far as modular arithmetic is concerned, you really want to have the flexibility to switch between these numbers, and insisting on doing computations with the positive version all the time would be hamstringing yourself.
Consider the problem of computing $1445^{99}\pmod{1446}$. It should not be necessary to compute powers and remainders of powers of $1445$ when you can just note that $1445=-1\pmod{1446}$, and then $1445^{99}=(-1)^{99}=-1=1445\pmod{1446}$
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But $(-48) / 5$ as a mixed numeral is $-9 \frac 35$ meaning $-(9 + \frac 35) = -9 - \frac 35$, which lines up more with Grandpa's way than with Teacher's way. – aschepler May 20 '14 at 19:48 | {
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The answer depends on whether you want to talk about modular arithmetic or remainders. These two perspectives are closely related, but different. In modular arithmetic, $2\equiv -3 \pmod 5$, so both answers are correct. This is the perspective most answers here have taken. In the division algorithm, though, where remainders are defined, in order to guarantee uniqueness, you need a specific range for the remainder -- when dividing integer $a$ by integer $d$, you get $a=qd+r$, and the integers $q$ and $r$ are unique if $0\leq r<d$ (note that this also places a restriction that $d$ be positive). There are other ways you could set up the condition, but you need some similar range in order to guarantee uniqueness, and this range is the simplest and most common, so in this sense, a negative remainder doesn't work.
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Excellent answer. The point is that we often want uniqueness of $q$ and $r$, which is guaranteed by requiring $0\leq r < d$. Without this condition, there are $infinitely$ many possible "remainders". – ChocolateAndCheese May 19 '14 at 19:57
@Chocolate: And it's worth noting that other normalizations are possible: sometimes you want $d/2 \leq r < d/2$. Other settings want $0 \leq |r| < d$, but for $rd \geq 0$. Or maybe $rn \geq 0$ or even $rnd \geq 0$. – Hurkyl May 20 '14 at 8:59
We indeed often want uniqueness of $q$ and $r$. But long division algorithm is only practical if $q$ is rounded towards zero (truncated) and that leads to negative (non-positive) remainder for negative numerator. But Euclidean division defines remainder as positive (non-negative). So either definition is sometimes needed with division as well. – Jan Hudec May 21 '14 at 4:57
When -48 is divided by 5 the division algorithm tells us that there is a "unique" reminder r satisfying $0\leq r<5$. In that case there is only one possibility, namely $r=2$. | {
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The division algorithm is defined on natural numbers, isn't it? Or I may be mistaken – Cheeku May 19 '14 at 13:30
I think it's usually defined for any integer numerator and any positive integer denominator. – poolpt May 19 '14 at 15:32
Negative denominator is not a problem. But long division needs negative remainders when numerator is negative. – Jan Hudec May 21 '14 at 4:52
Here is a different perspective which is more technical but is also reflected e.g. in Hardy and Wright.
On the whole it is best to regard the expression $"-48 \mod 5"$ on its own as representing the set of numbers $a:a\equiv -48 \mod 5$, so it isn't equal to a number at all, but to a set.
[Technically it is a coset in $\mathbb Z$ of the ideal generated by $5$, and consists of the numbers $-48 + 5 b$, where $b$ is an arbitrary integer].
Quite often it is useful to work with numbers rather than sets, and we choose a representative element of the set to work with. There are different ways in which this can be done (least positive, smallest absolute value etc). In this case $2$ and $-3$ are members of the set and could be used to represent it.
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Maybe the teacher was thinking about Euclidean division, which states:
Given two integers a and b, with b ≠ 0, there exist unique integers q and r such that a = bq + r and 0 ≤ r < |b|, where |b| denotes the absolute value of b
We indeed have the remainder ("r") being positive in this case, but if we talk strictly about congruences the grandpa's expression as well as the teacher's are both true.
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For integers $a,b,c$ the following statements are equivalent:
1) $a\equiv b\text{ mod }c$
2) $a+c\mathbb{Z}=b+c\mathbb{Z}$
3) $c\mid a-b$
Note that $5\mid-48-2=-50$ and $5\mid-48-\left(-3\right)=-45$.
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3) $c\mid a-b$
Note that $5\mid-48-2=-50$ and $5\mid-48-\left(-3\right)=-45$.
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By convention, for a division $\frac{a}{b}=c$, if we are looking for a whole integer for $c$ (with no further stipulations) , the rounding method is towards zero. If $a<0$ and $b>0$, this means that in the equation $\frac{a}{b}=c+\frac{r}{b}$, $r$ must be $\leq 0$.
Also remainders and modulus are two different things.
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This question touches parts of my old message Algebraic abstractions related to (big) integers in C++ on the boost developers mailing list.
Why should "modulo" be identical to "remainder"? Having a "remainder" function such that "r=remainder(a,m)" satisfies "0 <= r < m" is something convenient to have in a programming language. The quoted message presents some evidence that a "sremainder" function such that "s=sremainder(a,m)" satisfies "-m/2 <= s < m/2" would also be a good idea.
The meaning of modulo and modular arithmetic is not directly addressed by these considerations.
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For variety, I want to point out that when doing division with remainder (as opposed to, e.g., modular arithmetic), there is sometimes utility in having "improper" results; e.g. saying $17 / 5$ is $2$ with remainder $7$.
An example where this would be useful is if you just need a value that is close to the correct quotient, and you can compute something close (and the appropriate remainder) relatively more easily.
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# Finding $\int \sec^2 x \tan x \, dx$, I get $\frac12\sec^2x+C$, but an online calculator gets $\frac12\tan^2x+C$.
I tried to find a generic antiderivative for
$$\displaystyle \int \sec^2x \tan x \mathop{dx}$$
but I think there is something wrong with my solution because it doesn't match what I got through an online calculator.
What am I doing wrong?
Below is my solution.
We will use substitution:
$$u = \sec x \qquad du = \sec x \tan x \, dx$$
We substitute and apply the power rule:
$$\int (\sec x) (\sec x \tan x \, dx) = \int u \, du = \frac{1}{2} u^2 + C = \frac{\sec^2x}{2} + C$$
The solution I found with the online calculator is:
$$\frac{\tan ^2 x}{2} + C$$
The steps in the online solution make sense also, so I'm not sure what's going on.
The one thing I have some doubts about is whether I derived the $$du$$ from $$u = \sec x$$ correctly. But it seems okay to me. I used implicit differentiation with $$x$$.
• $\sec^2{(x)}=1+\tan^2{(x)}=\tan^2{(x)}+C$ hence these functions differ by a constant... – Peter Foreman Jul 23 at 18:23
• What happens if you try $u=\tan \theta?$ – Chris Leary Jul 23 at 18:25
$$\frac{\sec^{2}(x)}{2} + C = \frac{1+\tan^{2}(x)}{2} + C = \frac{1}{2} + \frac{\tan^{2}(x)}{2} + C$$
$$\frac{1}{2}$$ is just another constant, in indefinite integral constant doesn't really matter unless you're asked for the integrand original function so you can just kind of "combine" $$\frac{1}{2}$$ into C, you can also differentiate the answer to know that there's nothing wrong at all with your answer
The difference between your answer and the answer given is that they differ by a constant value.
You can see this by using the identity
$$1+\tan^2(x) = \sec^2(x)$$
Hence your answer can be converted to the given answer by subtracting $$-1/2$$, which is a constant.
As mentioned in the comments and in another answer you can also directly get the form in the answer by using the substitution
$$u = \tan x$$ | {
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$$u = \tan x$$
• I modified your answer a bit to use MathJax. Going forward, please use MathJax for mathematical typesetting for ease of readability. Good answer though! – Cameron Williams Jul 23 at 18:29
• Thanks for editing the expressions – StackUpPhysics Jul 23 at 18:29
Try differentiating both of them and see that nothing has gone wrong at all.
Let $$u = \tan x$$.
Then $$du = \sec^{2} x dx$$.
Hence, the integral becomes
$$\int u du = \frac{u^{2}}{2} + C = \frac{\tan^{2}x}{2} + C.$$ | {
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