Multiplicative Ordering on Integers

Theorem
Let $$x, y, z \in \mathbb{Z}$$ 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 $$M_z: \mathbb{Z} \to \mathbb{Z}$$ be the mapping defined as: $$\forall x \in \mathbb{Z}: M_z \left({x}\right) = z x$$.

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

By Monomorphism from Total Ordering, we just need to show that $$x < y \Longrightarrow z x < z y$$.

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

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

Therefore $$0 < z \left({y - x}\right) = z y - z x$$, that is, $$z x < z y$$.