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 embedding 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 Natural Numbers is Compatible with Multiplication:
 * $z \left({y - x}\right) \in \N$

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

That is:
 * $z x < z y$