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Prove that for every integer $n,$ if $n$ is odd then $n^2$ is odd.

I wonder whether my answer to the question above is correct. Hope that someone can help me with this.

Using contrapositive, suppose $n^2$ is not odd, hence even. Then $n^2 = 2a$ for some integer $a$, and $$n = 2(\frac{a}{n})$$ where $\frac{a}{n}$ is an integer. Hence $n$ is even.

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    How do you know a/n is an integer?2012-03-12
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    You have to consider the two cases of n being even or odd, and show that only the even case is possible.2012-03-12
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    Is this homework? If not, then it might be simpler not to use a contrapositive approach. Let $n = 2k+1$ and work from there.2012-03-12
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    Ops! I made a mistake in the title edit: I wrote "If my.." instead of "Is my..".2012-03-12
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    A number is even if it has 2 as a factor and odd otherwise, elevating a number to an integer power doesn't add factors to a number...2012-03-12
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    `(5+5+5+5)+5 = even+odd = odd`.2012-03-12
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    Since @Patrick mentioned it in a comment to an answer below, I'll link to my one-line direct proof of the recent "$n^2$ is even iff $n$ is even" question: http://math.stackexchange.com/a/119358/4092012-03-13

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