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:
- $z x < z y$
$\blacksquare$
Sources
- 1951: Nathan Jacobson: Lectures in Abstract Algebra: 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): $\S 2.1$: The integers: $\mathbf Z. \, 8$