If \(d^2\) divides \(a^2\), show that \(d\) divides \(a\) \(d,a\in \mathbb{Z}\)
Slick way with FTA http://math.stackexchange.com/questions/182988/if-a2-divides-b2-then-a-divides-b
that ruins the fun lol
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...
It doesn't, you still can have your own proof :) it seems some proofs in that link are circular..
I only looked at the first one
I buy everything in the one with 49 votes.
the exponent thing is weird, but great frigging idea
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
this only shows it for one case, not all :)
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 \(x\).
we have to consider those numbers whose square root will be integer not in decimal form dude
1.0
means by dividing the numbers we get a number with complete square root
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.
\( \Large \begin{matrix} \\ d = \prod_{i=1}^{n}p_i^{k_i} \\ d^2 = \prod_{i=1}^{n}p_i^{2k_i} \\ d^2|a^2\Rightarrow a^2=sd^2=S\prod_{i=1}^{n}p_i^{2k_i} \end{matrix} \) since a^2 is perfect square then sd^2 is also perfect square, now we need to prove \(\sqrt s\in N \)
you can say that XD ..
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} \(\Large a^2 =\prod_{i=1}^{l}p_i^{2f_i} ~~such~ that ~~ f_i<=k_i ~\forall i \) \(\Large a =\prod_{i=1}^{l}p_i^{ f_i} = \prod_{i=1}^{n}p_i^{k_i} \times something \)
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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