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:

$\forall x \in \Z: M_z \left({x}\right) = z x$


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