Integer Divisor Results/Integer Divides its Absolute Value

Theorem

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

Then:

\(\ds n\)

\(\divides\)

\(\ds \size n\)

\(\ds \size n\)

\(\divides\)

\(\ds n\)

where:

$\size n$ is the absolute value of $n$

$\divides$ denotes divisibility.

Proof

Let $n > 0$.

Then $\size n = n$ and Integer Divides Itself applies.

Let $n = 0$.

Then Integer Divides Itself holds again.

Let $n < 0$.

Then $\size n = -n$ and Integer Divides its Negative applies.

$\blacksquare$