Improving the proof by contraposition / why it works

Question

This is the problem

 Prove that if n is an integer and 3n+2 is odd, then n is odd

So for this I should take $3n+2$ to be true and assume $\lnot q$, therefore I can say $n = 2k$. Then $$3n+2=$$$$3(2k)+2=$$ $$6k+2 =$$ $$2(3k+1)$$

Now because at the outset I said $3n+2$ is odd and then showed that if $n$ is even then $3n+2 = 2(3k+1)$ which is even. Why does this prove $p \to q$ ?

Also, my professor is a bit strict and if we don't write it all out "formally" he takes a significant amount of points off even though the proof might be correct. Could someone rewrite this in a "formal" way?

Answer 1

Here's a simple contraposition proof without any fluff:

Suppose $n$ is even. Then $n=2\ell$, where $\ell\in\mathbb{Z}$. Thus, $$ 3n+2=3(2\ell)+2=2(3\ell+1)=2m,\quad m\in\mathbb{Z} $$ Thus, if $n$ is even, then $3n+2$ is even; that is, if $3n+2$ is odd, then $n$ is odd.