Solution
= Solution
Suppose, for a <proof by contradiction>, that an <integer> $n$ is both an <even number> and an <odd number>. Then $n=2u$ and $n=2v+1$ for some <integers> $u,v$. Subtracting yields $u-v=1/2$. The left side is an <integer>, while $1/2$ is not. This contradiction proves that \b[no <integer> is both even and odd].