As the title says, prove:
$m \in Z$, if $m$ is even then then $mn$ is even
My attempt:
By definition of an even number, $2i$, and definition of an odd number, $2j+1$ I did the following by cases. I thought it was logical to create cases for if $n$ was even or if it was odd.
Case 1: Even
$$mn=(2i)(2j)$$
$$=4(ij)$$
Case 2: Odd
$$mn=(2i)(2j+1)$$ $$= (2i)(2j)+2i$$ $$=4(ij)+2i$$
However, after my attempt the books answer was the following:
"Suppose $x,y \in Z$ and $x$ is even. Then $x=2a$ for some integer a, by definition of an even number. Thus $xy = (2a)(y) = 2(ay).$ Therefore, $xy=2b$ where $b$ is the integer $ay$, so $xy$ is even. "
Are both methods acceptable? Any suggestions to the way I did this are more then welcome!