(related to Proposition: Product of an Even and an Odd Number)

- Let the \(n\) be even and let $m$ be odd.
- By definition, $2$ is a divisor of $n$, but not of $m.$
- Therefore $n=2k$ and $m=2l+1$ for some integers $k,l.$
- For the product, it follows \(nm=2k\cdot (2l+1)=2(k(2l+1))\), because the multiplication of integers is associative.
- Therefore, $2\mid nm.$
- Thus, the product $nm$ is even.∎