Integer Divisor Results/Integer Divides its Negative

Theorem

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

Then:

\(\ds n\)

\(\divides\)

\(\ds -n\)

\(\ds -n\)

\(\divides\)

\(\ds n\)

where $\divides$ denotes divisibility.

Proof

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

Hence we can apply Product of Ring Negatives:

$\forall n \in \Z: \exists -1 \in \Z: n = \paren {-1} \times \paren {-n}$

and Product with Ring Negative:

$\forall n \in \Z: \exists -1 \in \Z: -n = \paren {-1} \times \paren n$

$\blacksquare$

Sources

• 1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {2-2}$ Divisibility: Example $\text {2-2}$
• 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor