Chapter 4 - Elementary Number Theory and Methods of Proof - Exercise Set 4.1 - Page 162: 46

$x = 2k$ $\begin{split} x.y & = (2k)y \\ & = 2(k.y) \\ \end{split}$ $x.y$ is even.

Let $x$ be a even integer, and $y$ just a integer, $x$ can be rewriten: $x = 2k$ (with $k$ being a integer) $k \in \mathbb{Z}$ Now we can multiply $x.y$: $\begin{split} x.y & = (2k)y \\ & = 2(k.y) \\ \end{split}$ Since $k.y$ is integer, then x.y is even, by the definition of even numbers.