Using proof by contraposition to show that if $n\in\mathbb Z$ and $3n+2$ is even, then $n$ is even I have my answer below but there is one step that I am not understanding...and maybe my brain is just not trained to understand it.

Prove that if $n$ is an integer and $3n+2$ is even, then $n$ is even using a proof by contraposition.
The contraposition would read: if $n$ is an integer and $3n+2$ is odd, then $n$ is odd.
  $$
\begin{align*}
n
&= 2k+1 \\
&= 3\left(2k+1\right) + 2 \\
&= 6k+5 \\
&= 2\left(3k+2\right) + 1 \text{(this is the step I don't understand)}
\end{align*}
$$

Ok, so I understand everything up until the last step. I understand how I got to $6k + 5$, but what I don't understand is how I can prove that last statement? I know that $2\left(2k + 2\right) + 1$ means that the number is odd and the contraposition has now be proofed. And I understand that we got to $2\left(3k+2\right)+1$, then dividing by $2$, but $5/2$ isn’t $2$. 
 A: Notice that$$6k + 5 = 6k + 4 + 1 = 2(3k + 2) + 1$$
A: Well, the contrapositive of '$A$ implies $B$' is 'not $B$ implies not $A$'.
In this case, now -staying withing the realm of integers- it would read

If $n$ is odd, then $3n+2$ is odd.

And this is being proved. The equality sign on the second line is not correct, as $n=2k+1\ne 3n+2$, so it should be

Suppose $n$ is odd, i.e. $n=2k+1$. 
  Then $3n+2=2(3k+2)+1$, i.e. can be written of the form $2x+1$ with $x\in\Bbb Z$, so it is odd.

Well, it would be also enough to note that $3n+2=6k+5$ and $6k$ is even, $5$ is odd, their sum is odd.
Even simpler: If $n$ is odd, then $3n$ is odd, so $3n+2$ is odd.
A: The contrapositive of $P \to Q$ is $\lnot Q \to \lnot P$ so in your case the contrapositive would be : If $n$ is odd, then $3n+2$ is odd. (It seems you have this reversed in your statement of the contrapositive.)
That's why the start of the proof of the contrapositive is $n=2k+1$ (i.e. $n$ is odd), and then the computation of $3n+2=3(2k+1)+2=6k+5=6k+4+1=2(3k+2)+1$ shows the desired thing, that $3n+2$ is odd.
A: It suffices to use the language of congruences: being odd means being congruent to 1 modulo $2$, being even is being congruent to $0$.
Anyway, since $3$ is odd, $3n+2\equiv n \mod 2$, so the result is trivial.
A: Q5) Prove that   " if $n$ is an  integer and $3n + 2$ is odd,  then  $n$ is odd "
(Hint  using  proof by contraposition ; i.e. $\neg q\to\neg p\equiv p\to q$ )
