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I just need to know that if $ a $ divides $ b $, where $ a $ and $ b $ are integers, does $ a^{2} $ divide $ b^{2} $?

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2 Answers 2

If $a$ divides $b$, then $b=ka$ for some integer $k$, so $b^2=k^2a^2$ where $k^2$ is an integer.

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How do you prove that k^2 is an integer? Is that by closure? –  amster27 Feb 22 '13 at 1:07
@amster27: If you are concerned about whether $ k^{2} $ is an integer, then you should be equally concerned about whether both $ a^{2} $ and $ b^{2} $ are integers. :) –  Haskell Curry Feb 22 '13 at 1:23
Thanks, I'm just starting to write proofs in a Mathematical Reasoning class, and I just feel so lost. But these explanations helped. –  amster27 Feb 22 '13 at 1:28
@amster: Note that $ +_{\mathbb{Z}}: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z} $ and $ \times_{\mathbb{Z}}: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z} $. Therefore, adding two integers or multiplying two integers yields an integer. –  Haskell Curry Feb 22 '13 at 1:44

Hint $\rm\ \ a\mid b\ \Rightarrow\ \dfrac{b}a\in \Bbb Z\ \Rightarrow\ \dfrac{b^2}{a^2} = \left(\dfrac{b}a\right)^2\!\in \Bbb Z^2\subseteq \Bbb Z\:\Rightarrow\: a^2\mid b^2\ $

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That's closer to a proof than a hint. –  1015 Feb 22 '13 at 1:29
The argument is elegant, but I think that the OP wants to stay within $ \mathbb{Z} $ and does not wish to jump into $ \mathbb{Q} $, as he seems to be very interested in the axioms governing the properties of $ \mathbb{Z} $. –  Haskell Curry Feb 22 '13 at 1:29
@Haskell Possibly, but there was no hint of any such constraint in the question. –  Math Gems Feb 22 '13 at 1:46

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