[Math] Using proof by contraposition to show that if $n\in\mathbb Z$ and $3n+2$ is even, then $n$ is even

Question

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 $2$.

Answer 1

Notice that$$6k + 5 = 6k + 4 + 1 = 2(3k + 2) + 1$$