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.
Best Answer
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.