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:
 * $\forall x \in \Z: \map {M_z} x = z x$

It is sufficient to show that $M_z$ is an order embedding from $\struct {\Z, +, \le}$ to itself.

By Monomorphism from Total Ordering, it is sufficient to show that:
 * $x < y \implies z x < z y$

Let $x < y$.

Then:
 * $0 < y - x$

So $z \in \N$.

Hence by Natural Numbers are Non-Negative Integers:
 * $y - x \in \N$

Thus by Ordering on Natural Numbers is Compatible with Multiplication:
 * $z \paren {y - x} \in \N$

Therefore:
 * $0 < z \paren {y - x} = z y - z x$

That is:
 * $z x < z y$