# Understanding Proof by Contrapositive

This is a question from my textbook, and I'm pretty sure I have the answer, but I am having trouble writing out what I am thinking.

Prove: If $n$ is an integer and $3n+2$ is odd, then $n$ is odd.

Contrapositive: If $n$ is even, then $3n+2$ is even.

$n$ is even, by definition $n = 2k$, where $k$ is an integer.

Plugging in $2k$ into $3n+2$, we have $3(2k)+2$.

$3(2k)+2 = 6k+2 = 2(3k+1)$, which means $3n+2$ is even.

Since if $n$ is even, then $3n+2$ is even. Negating the conclusion of the original conditional statement implies that it's hypothesis "$3n+2$ is odd" is false, therefor if $3n+2$ is odd, $n$ is odd.

What am I missing, or what can I do to make this less awkward? Or am I wrong about the proof on the most basic levels?

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It can be proved directly: $\rm\ 3n\!+\!2\,=\, 2k\!+\!1\,\ is\ odd\:\Rightarrow\: n\,=\,2(k\!−\!n)-1\,\ is\ odd.$ –  Math Gems Feb 26 '13 at 6:03

Your proof is correct, though you could improve on it by writing it in essay style, cutting down on some verbosity, and by mentioning that $3k+1$ is also an integer. For example, you could write something like this:
We try to prove the contrapositive. If $n$ is even, then $n=2k$ for some integer $k$, so $3n+2=3(2k)+2=6k+2=2(3k+1)$ where $3k+1$ is an integer, making $3n+2$ even.