Integer Divisor Results/Integer Divides its Negative

Theorem
Let $n \in \Z$, i.e. let $n$ be an integer.

Then:

That is, $n$ divides its negative, and $n$ is a multiple of its negative.

Proof
Follows directly from Negative Product:


 * $\forall n \in \Z: \exists -1 \in \Z: n = \left({-1}\right) \times \left({-n}\right)$

and:


 * $\forall n \in \Z: \exists -1 \in \Z: -n = \left({-1}\right) \times \left({n}\right)$