Wikipedia:Reference desk/Archives/Mathematics/2010 September 29

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September 29

Proving converse of the Euclid's Lemma

The statement I'm trying to prove is that if p | ab and p | a or p | b, then p is prime. I know I'm supposed to show that if there exists a divisor which divides p, that divisor is either p or 1. I'm wondering how should I get about to proving this? —Preceding unsigned comment added by 142.244.143.233 (talk) 00:04, 29 September 2010 (UTC)

Well typically that's the definition of prime, but I guess you must be starting from some other definition. Just as a clarification, note that it should be "if for any a and b, p | ab implies p | a or p | b, then p is prime." What you wrote doesn't quite mean that. Anyway, one trick you can use if you are stuck on how to prove something is to try proving the contrapositive. Try assuming that p is not prime, and then show that there exist a and b such that p | ab but p does not divide a or b. Rckrone (talk) 01:29, 29 September 2010 (UTC)

I still seem to be getting nowhere with this proof, is there some kind of contradiction I have to use here? 142.244.143.233 (talk) 02:01, 29 September 2010 (UTC)

Actually nvm, I think I got it. 142.244.143.233 (talk) 02:09, 29 September 2010 (UTC)

What definition of prime are you using? --COVIZAPIBETEFOKY (talk) 02:08, 29 September 2010 (UTC)

Using the definition learned in elementary school, an integer that is greater than 1 that is only divisible by 1 and itself. 142.244.143.233 (talk) 02:09, 29 September 2010 (UTC)

You say you got it above, but for the benefit of anyone else for which this isn't obvious: take Rckrone's advice, and remember what it means for a number to not be a prime. Simply apply definitions, and it works out naturally. --COVIZAPIBETEFOKY (talk) 02:55, 29 September 2010 (UTC)

Also note that the theorem is false for p=1. Where does the proof hit a snag in that case? --COVIZAPIBETEFOKY (talk) 02:58, 29 September 2010 (UTC)

Euclid's lemma says that if p is prime and p|ab then p|a or p|b. A "converse" would simply reverse "if" and "then", so it would say if p|a or p|b then p is prime and p|ab. But that is clearly false. Your proposed statement, that "if p | ab and p | a or p | b, then p is prime", is not a converse, and is also clearly false. For example, 6|4×12, and 6|4 or 6|12. But 6 is not prime. Michael Hardy (talk) 03:15, 29 September 2010 (UTC)

Rckrone addressed this issue above; the op left out a universal quantifier. Euclid's lemma can be restated equivalently as "If p is prime, then (for any a, b, if p|ab then p|a or p|b)". This is the statement that we are taking the converse of. --COVIZAPIBETEFOKY (talk) 03:34, 29 September 2010 (UTC)

Linear Algebra

Question from my quantum physics class (I'm learning the math as I go, so I need some help!): An operator A is defined by A = i(I + U)(I - U)^-1, where U is unitary and I is the identity operator. Show that if U does not have the eigenvalue 1, then A is hermitian.

Well, I managed to show that the hermitian of A = i(I - U)^-1(I + U), so I guess all that's left is to show that U not having an eigenvalue of 1 implies that (I - U)^-1(I + U) = (I + U)(I - U)^-1, but I don't know how to do this. A push in the right direction would be appreciated! 74.15.136.172 (talk) 01:00, 29 September 2010 (UTC)

Also, if U doesn't have an eigenvalue 1, then doesn't that imply that det(U - I) ≠ 0? But if I - U is invertible, then don't we already know that det(U - I) ≠ 0, and hence that U doesn't have an eigenvalue 1? 74.15.136.172 (talk) 01:14, 29 September 2010 (UTC)

You wrote A = i(I + U)(I − U)⁻¹.

Are you sure it didn't say A = i(I + U)(I − U⁻¹)? Michael Hardy (talk) 03:18, 29 September 2010 (UTC)

PS: If it did say (I − U⁻¹) rather than (I − U)⁻¹, then it's easy. Michael Hardy (talk) 03:23, 29 September 2010 (UTC)

Yeah I'm sure, but it wouldn't be the first time my prof made a mistake. Is there any reason to think that the question isn't soluble if A = i(I + U)(I − U)⁻¹? 74.15.136.172 (talk) 03:25, 29 September 2010 (UTC)

PS: If my prof did make an error, then it seems that the solution doesn't require that U not have an eigenvalue equal to 1. Did you get the same thing? 74.15.136.172 (talk) 03:30, 29 September 2010 (UTC)

Okay I think I got it; that (I - U)^-1(I + U) = (I + U)(I - U)^-1 can be proved by simply distributing (I - U)^-1 into the other bracket and then combining things into a common inverse, for both sides of the equation. I'm too lazy to type the algebra, but does this seem right? 74.15.136.172 (talk) 03:47, 29 September 2010 (UTC)

I'm not sure what you mean by "distributing (I - U)^-1 into the other bracket", but what I'd do is to multiply from the left and from the right by (I - U). -- Meni Rosenfeld (talk) 06:47, 29 September 2010 (UTC)

Well I'm wondering how anyone misses the following: the expression

(I − U)(I − U)⁻¹

is the product of a matrix and its inverse matrix, and clearly that is the identity matrix:

(I − U)(I − U)⁻¹ = I.

Michael Hardy (talk) 17:32, 29 September 2010 (UTC)

Oh. One of them has a plus sign. Never mind..... Michael Hardy (talk) 17:33, 29 September 2010 (UTC)

z=cos2x+isin2x

I'm having difficulty with this question. If $z=\cos {2x}+i\sin {2x}$ show that ${\frac {z}{1-z}}={\frac {i}{2\sin {x}}}(\cos {x}+i\sin {x})$
${\frac {\cos {2x}+i\sin {2x}}{1-\cos {2x}-i\sin {2x}}}$

${\frac {(\cos {x}+i\sin {x})^{2}}{1-\cos {2x}-i\sin {2x}}}$

${\frac {\cos {x}+i\sin {x}}{1-\cos {2x}-i\sin {2x}}}(\cos {x}+i\sin {x})$

I guess now I have to show that ${\frac {\cos {x}+i\sin {x}}{1-\cos {2x}-i\sin {2x}}}={\frac {i}{2\sin {x}}}$ but I have so far not been able to do so. MrMahn (talk) 22:26, 29 September 2010 (UTC)

You might want to consider the formulae linking $\cos {x}$ and $\sin {x}$ to $e^{ix}$ and $e^{-ix}$. --81.153.109.200 (talk) 22:33, 29 September 2010 (UTC)

The key is that cos2x = cos²x - sin²x = 1 - 2 sin²x. Use the first of those on the top, the second on the bottom, replace sin2x with 2sinxcosx and simplify.--JohnBlackburne^words_deeds 22:38, 29 September 2010 (UTC)

${\frac {\cos ^{2}x-\sin ^{2}x+i\sin {2x}}{1-1+2\sin ^{2}x-i\sin {2x}}}$

${\frac {\cos ^{2}x-\sin ^{2}x+i\sin {2x}}{2\sin ^{2}x-i\sin {2x}}}$

${\frac {\cos ^{2}x-\sin ^{2}x+2i\sin {x}\cos {x}}{2\sin ^{2}x-2i\sin {x}\cos {x}}}$

${\frac {\cos ^{2}x-\sin ^{2}x+2i\sin {x}\cos {x}}{2\sin {x}(\sin {x}-i\cos {x})}}$

Hmm... what now? I can't really take a factor of i out of the expression on the numerator.MrMahn (talk) 23:00, 29 September 2010 (UTC)

You're almost there. i(sin x - i cos x) = (cos x + i sin x). Combining the last line above with your earlier expansion of the top gives

${\frac {(\cos {x}+i\sin {x})^{2}}{2\sin {x}(-i(\cos {x}+i\sin {x}))}}$

My easlier suggestion about was a bit misleading I think: you only need to use the double angle formulae on the bottom of the fraction.--JohnBlackburne^words_deeds 23:23, 29 September 2010 (UTC)

Put z = exp(2 i x) and multiply numerator and denominator by exp(-i x) Count Iblis (talk) 23:03, 29 September 2010 (UTC)

${\frac {z}{1-z}}={\frac {\exp(2ix)}{1-\exp(2ix)}}={\frac {\exp(ix)}{\exp(-ix)-\exp(ix)}}={\frac {i}{2\sin(x)}}\left[\cos(x)+i\sin(x)\right]$ Count Iblis (talk) 23:41, 29 September 2010 (UTC)

I wonder if the OP might benefit from remembering how to divide one complex number by another:

${a+bi \over c+di}={(a+bi)(c-di) \over (c+di)(c-di)}={(a+bi)(c-di) \over c^{2}+d^{2}}$

etc. Michael Hardy (talk) 23:53, 30 September 2010 (UTC)

Knowledge of Euler's formula or unwieldy trigonometric identities is not required. Let w = cis x so that w² = z (by de Moivre's theorem). Then

${\frac {iw}{2\sin x}}={\frac {iw}{-i(w-w^{-1})}}={\frac {w}{w^{-1}-w}}={\frac {w^{2}}{1-w^{2}}}.$

This is in essence what Count Iblis said, but perhaps it's easier to follow if you don't know about e^ix. —Anonymous Dissident^Talk 04:45, 1 October 2010 (UTC)

Implicit differentiation and polynomial division revisited

Hey, me from a few days ago. I have two completely unrelated questions. 1) When you differentiate both sides of a relation implicitly, for example 4x^4+y^2=3, what do you do with the y, if your only differentiating with respect to x? and 2) When you divide polynomial P(x) by f(x) there is a theorem that the remainder is P of the roots of f(x). But what if f(x) has more than 1 root, i.e., has a degree greater than 1? That wold imply that the remainder is P of the product of the roots, but there would a leftover x term that would not show up there. How can I find it? THanks. 24.92.78.167 (talk) 23:39, 29 September 2010 (UTC)

1) Differentiating y with respect to x gives you dy/dx, as you might expect. So in your example differentiating the y^2 term with respect to x gives you 2y(dy/dx) using the chain rule.

2) A polynomial P(x) divided by a linear term will always give you a remainder that is just a number, but when you take the remainder by a higher degree polynomial, that won't generally be the case. If f(x) has degree n, then the remainder of P(x) will generally be a polynomial g(x) of degree n-1 (it could be less though). If r is a root of f(x) then g(r) will still be equal to P(r), but that's definitely looks most exciting when g is just a constant.

It's easy to see why it works though. Saying g(x) is the remainder of P(x) divided by f(x) means that P(x) = q(x)f(x) + g(x) for some polynomial q(x). If we plug in r, then P(r) = q(r)f(r) + g(r), but f(r) = 0, so then P(r) = g(r). Rckrone (talk) 01:37, 30 September 2010 (UTC)