Ask your own question, for FREE!
Mathematics 7 Online
ganeshie8 (ganeshie8):

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

OpenStudy (anonymous):

that ruins the fun lol

OpenStudy (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...

ganeshie8 (ganeshie8):

It doesn't, you still can have your own proof :) it seems some proofs in that link are circular..

OpenStudy (zzr0ck3r):

I only looked at the first one

OpenStudy (zzr0ck3r):

I buy everything in the one with 49 votes.

OpenStudy (zzr0ck3r):

the exponent thing is weird, but great frigging idea

OpenStudy (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

OpenStudy (zzr0ck3r):

this only shows it for one case, not all :)

OpenStudy (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 \(x\).

OpenStudy (sohailiftikhar):

we have to consider those numbers whose square root will be integer not in decimal form dude

OpenStudy (zzr0ck3r):

1.0

OpenStudy (sohailiftikhar):

means by dividing the numbers we get a number with complete square root

OpenStudy (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.

OpenStudy (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 \(\sqrt s\in N \)

OpenStudy (sohailiftikhar):

you can say that XD ..

OpenStudy (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} \(\Large a^2 =\prod_{i=1}^{l}p_i^{2f_i} ~~such~ that ~~ f_i<=k_i ~\forall i \) \(\Large a =\prod_{i=1}^{l}p_i^{ f_i} = \prod_{i=1}^{n}p_i^{k_i} \times something \)

OpenStudy (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

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!