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 $2$.

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 $n=2k+1\
e 3n+2$, 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.