ganeshie8
  • ganeshie8
If \(d^2\) divides \(a^2\), show that \(d\) divides \(a\) \(d,a\in \mathbb{Z}\)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
zzr0ck3r
  • zzr0ck3r
Slick way with FTA http://math.stackexchange.com/questions/182988/if-a2-divides-b2-then-a-divides-b
anonymous
  • anonymous
that ruins the fun lol
zzr0ck3r
  • zzr0ck3r
then 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...

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ganeshie8
  • ganeshie8
It doesn't, you still can have your own proof :) it seems some proofs in that link are circular..
zzr0ck3r
  • zzr0ck3r
I only looked at the first one
zzr0ck3r
  • zzr0ck3r
I buy everything in the one with 49 votes.
zzr0ck3r
  • zzr0ck3r
the exponent thing is weird, but great frigging idea
sohailiftikhar
  • sohailiftikhar
simple 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
  • zzr0ck3r
this only shows it for one case, not all :)
zzr0ck3r
  • zzr0ck3r
that 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 \(x\).
sohailiftikhar
  • sohailiftikhar
we have to consider those numbers whose square root will be integer not in decimal form dude
zzr0ck3r
  • zzr0ck3r
1.0
sohailiftikhar
  • sohailiftikhar
means by dividing the numbers we get a number with complete square root
zzr0ck3r
  • zzr0ck3r
you 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
  • anonymous
\( \Large \begin{matrix} \\ d = \prod_{i=1}^{n}p_i^{k_i} \\ d^2 = \prod_{i=1}^{n}p_i^{2k_i} \\ d^2|a^2\Rightarrow a^2=sd^2=S\prod_{i=1}^{n}p_i^{2k_i} \end{matrix} \) since a^2 is perfect square then sd^2 is also perfect square, now we need to prove \(\sqrt s\in N \)
sohailiftikhar
  • sohailiftikhar
you can say that XD ..
anonymous
  • anonymous
ok 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
  • anonymous
suppose d has value 2 and a has value 4. After d2 divides a2, we get value 1/4. http://www.acalculator.com/math-calculators.html

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