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:

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

All we need to do is show that $M_z$ is an order embedding from $\struct {\Z, +, \le}$ 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:

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

Therefore

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

That is:

$z x < z y$

$\blacksquare$

Sources

• 1951: Nathan Jacobson: Lectures in Abstract Algebra: Volume $\text { I }$: Basic Concepts ... (previous) ... (next): Introduction $\S 5$: The system of integers
• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): $\S 20$: Theorem $20.10$
• 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: $\mathbf Z. \, 8$