Is my proof correct: if $n$ is odd then $n^2$ is odd? 
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.


Two bits:
Firstly, to clean your proof up, you might instead go like this. If $n^2 = 2a$, then in particular $2 \mid n^2$. $2$ is a prime, and thus we have at least 1 of $2 \mid n$ or $2 \mid n$. Clearly, one happens $\implies$ both happen, and thus $2 \mid n$. So $n$ is even.
Secondly, what if we didn't use contrapositive?
$m$ is odd means $m = 2k + 1$ for some $k$. Then $m^2 = 4k^2 + 4k + 1$.
