# Multiplicative Ordering on Integers

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## 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 $z \in \N$.

$y - x \in \N$
$z \paren {y - x} \in \N$

Therefore:

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

That is:

$z x < z y$

$\blacksquare$