Integer Divisor Results/Integer Divides its Absolute Value
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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.
$\blacksquare$