Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

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.

share|improve this answer
    
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
add comment

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\ $

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.