Multiplicative Ordering on Integers

Theorem
Let $$x, y, z \in \Z$$ such that $$z > 0$$.

Then:


 * $$x < y \iff z x < z y$$
 * $$x \le y \iff z x \le z y$$

Proof
Let $$z > 0$$.

Let $$M_z: \Z \to \Z$$ be the mapping defined as:
 * $$\forall x \in \Z: M_z \left({x}\right) = z x$$

All we need to do is show that $$M_z$$ is an order monomorphism from $$\left({\Z, +, \le}\right)$$ to itself.

By Monomorphism from Total Ordering, we just need to show that:
 * $$x < y \implies z x < z y$$

If $$x < y$$, then $$0 < y - x$$, so $$z \in \N$$ and $$y - x \in \N$$ by Natural Numbers are Non-Negative Integers.

Thus by Ordering on Naturally Ordered Semigroup Product:
 * $$z \left({y - x}\right) \in \N$$

Therefore
 * $$0 < z \left({y - x}\right) = z y - z x$$

That is:
 * $$z x < z y$$