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ganeshie8

  • one year ago

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

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  1. zzr0ck3r
    • one year ago
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    Slick way with FTA http://math.stackexchange.com/questions/182988/if-a2-divides-b2-then-a-divides-b

  2. anonymous
    • one year ago
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    that ruins the fun lol

  3. zzr0ck3r
    • one year ago
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    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...

  4. ganeshie8
    • one year ago
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    It doesn't, you still can have your own proof :) it seems some proofs in that link are circular..

  5. zzr0ck3r
    • one year ago
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    I only looked at the first one

  6. zzr0ck3r
    • one year ago
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    I buy everything in the one with 49 votes.

  7. zzr0ck3r
    • one year ago
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    the exponent thing is weird, but great frigging idea

  8. sohailiftikhar
    • one year ago
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    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

  9. zzr0ck3r
    • one year ago
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    this only shows it for one case, not all :)

  10. zzr0ck3r
    • one year ago
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    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 \(x\).

  11. sohailiftikhar
    • one year ago
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    we have to consider those numbers whose square root will be integer not in decimal form dude

  12. zzr0ck3r
    • one year ago
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    1.0

  13. sohailiftikhar
    • one year ago
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    means by dividing the numbers we get a number with complete square root

  14. zzr0ck3r
    • one year ago
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    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.

  15. anonymous
    • one year ago
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    \( \Large \begin{matrix} \\ d = \prod_{i=1}^{n}p_i^{k_i} \\ d^2 = \prod_{i=1}^{n}p_i^{2k_i} \\ d^2|a^2\Rightarrow a^2=sd^2=S\prod_{i=1}^{n}p_i^{2k_i} \end{matrix} \) since a^2 is perfect square then sd^2 is also perfect square, now we need to prove \(\sqrt s\in N \)

  16. sohailiftikhar
    • one year ago
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    you can say that XD ..

  17. anonymous
    • one year ago
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    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} \(\Large a^2 =\prod_{i=1}^{l}p_i^{2f_i} ~~such~ that ~~ f_i<=k_i ~\forall i \) \(\Large a =\prod_{i=1}^{l}p_i^{ f_i} = \prod_{i=1}^{n}p_i^{k_i} \times something \)

  18. anonymous
    • one year ago
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    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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