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ganeshie8
 one year ago
If \(d^2\) divides \(a^2\), show that \(d\) divides \(a\)
\(d,a\in \mathbb{Z}\)
ganeshie8
 one year ago
If \(d^2\) divides \(a^2\), show that \(d\) divides \(a\) \(d,a\in \mathbb{Z}\)

This Question is Closed

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Slick way with FTA http://math.stackexchange.com/questions/182988/ifa2dividesb2thenadividesb

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that ruins the fun lol

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2then don't look:) That is why I did not regurgitate the proof. I know I have seen it in NT but if someone has another way...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1It doesn't, you still can have your own proof :) it seems some proofs in that link are circular..

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2I only looked at the first one

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2I buy everything in the one with 49 votes.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2the exponent thing is weird, but great frigging idea

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0simple let take a=1 b=2 now b^2/a^2=2^2/1^2 now take square root on both sides b/a=2/1

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2this only shows it for one case, not all :)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2that would be like saying take \(4\), well \(\sqrt{4}\in \mathbb{Z}\) so it must be that \(\sqrt{x}\in \mathbb{Z}\) for all real numbers \(x\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we have to consider those numbers whose square root will be integer not in decimal form dude

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0means by dividing the numbers we get a number with complete square root

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2you are correct in saying the theorem is true, we are saying that you have not sufficiently proved it. You should look at one of the proofs on the page I posted to see what the formal proof would look like.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\( \Large \begin{matrix} \\ d = \prod_{i=1}^{n}p_i^{k_i} \\ d^2 = \prod_{i=1}^{n}p_i^{2k_i} \\ d^2a^2\Rightarrow a^2=sd^2=S\prod_{i=1}^{n}p_i^{2k_i} \end{matrix} \) since a^2 is perfect square then sd^2 is also perfect square, now we need to prove \(\sqrt s\in N \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can say that XD ..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok we just say \(\Large a^2=sd^2=\prod_{i=1}^{m}p_i^{g_i}.\prod_{i=1}^{n}p_i^{2k_i} \) W.L.G wither m>n or else, lets interpret in different variable l=max{n,m} \(\Large a^2 =\prod_{i=1}^{l}p_i^{2f_i} ~~such~ that ~~ f_i<=k_i ~\forall i \) \(\Large a =\prod_{i=1}^{l}p_i^{ f_i} = \prod_{i=1}^{n}p_i^{k_i} \times something \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0suppose d has value 2 and a has value 4. After d2 divides a2, we get value 1/4. http://www.acalculator.com/mathcalculators.html
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