Integer Divisor Results

Theorem
Let $$m, n \in \mathbb{Z}$$, i.e. let $$m, n$$ be an integer.

The following results all hold:


 * $$1 \backslash n$$, i.e. $$1$$ divides $$n$$;
 * $$n \backslash n$$, i.e. $$n$$ divides itself;
 * $$n \backslash -n$$, i.e. $$n$$ divides its negative;
 * $$n \backslash \left|{n}\right|$$;
 * $$\left|{n}\right| \backslash n$$;
 * $$n \backslash 0$$, i.e. $$0$$ divides zero;
 * $$m \backslash n \iff -m \backslash n \iff m \backslash -n$$.

Proof
As the set of integers form an integral domain, the concept "divides" is fully applicable to the integers.

Therefore many of these results can be derived as direct applications of the theorems proved for integral domains.


 * $$1 \backslash n$$: Follows directly from Unity Divides All Elements.


 * $$n \backslash n$$: Follows directly from Every Element Divisor of Itself.


 * $$n \backslash -n$$:

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


 * $$n \backslash \left|{n}\right|$$ and $$\left|{n}\right| \backslash n$$:

Let $$n > 0$$. Then $$\left|{n}\right| = n$$ and $$n \backslash n$$ applies.

Let $$n = 0$$. Then $$n \backslash n$$ holds again.

Let $$n < 0$$. Then $$\left|{n}\right| = -n$$ and $$n \backslash -n$$ applies.


 * $$n \backslash 0$$: Follows directly from Every Element Divides Zero.


 * $$m \backslash n \iff -m \backslash n \iff m \backslash -n$$:

Let $$m \backslash n$$.

From the above, we have $$-m \backslash m$$, and from Divides is Ordering on Positive Integers it follows that $$-m \backslash n$$.

From the above, we have $$n \backslash -n$$, and from Divides is Ordering on Positive Integers it follows that $$m \backslash -n$$.

The rest of the proof follows similarly.