Show that if n is an integer and 3n+ 2 is even, then n is even using contradiction Show that if $n$ is an integer and $3n+ 2$ is even, then $n$ is even,  using a proof by contradiction.

That's the question. So since we're using contradiction, I need to show that N is odd and prove a contradiction exists, right?

$N = 2k + 1$

$3(2k + 1) + 1=6k + 5\quad$ ?

I think I'm off base, or else I don't know how to proceed.
Use these facts:

$\mbox{odd}\times\mbox{odd}=\mbox{odd}$
$\mbox{odd}+\mbox{even}=\mbox{odd}$

Thus, if $n$ was odd, then:

$3n+2=\mbox{odd}\!\times\!\mbox{odd}+\mbox{even}=\mbox{odd}+\mbox{even}=\mbox{odd}$

But we know that $3n+2$ is even - thus, $n$ can't be odd.



Another way of doing it: Say $n$ is odd. Thus, there exists a $k$ such that $n=2k+1$. Then:

$$3n+2\overset{n=2k+1}=3(2k+1)+2=6k+3+2=6k+5=6k+4+1=2(3k+2)+1$$

Anything of the form $2j+1$ is odd. (In this case, $j=3k+2$.) Thus, we have that $3n+2$ is odd.

But we know that $3n+2$ is even - thus, $n$ can't be odd.
$3n+2=2k$

$n=2k-2n-2$

$n=2(k-n-1)$

$n=2b$

Hence $3n+2$ is even when $n$ is even therefore also odd when $n$ is odd.