Integer Divisor Results/Integer Divides its Negative

Theorem
Let $n \in \Z$ be an integer.

Then:

where $\backslash$ denotes divisibility.

Proof
From Integers form Integral Domain, the integers are an integral domain.

Hence we can apply Product of Ring Negatives:
 * $\forall n \in \Z: \exists -1 \in \Z: n = \left({-1}\right) \times \left({-n}\right)$

and Product with Ring Negative:
 * $\forall n \in \Z: \exists -1 \in \Z: -n = \left({-1}\right) \times \left({n}\right)$