Ordering on Natural Numbers is Compatible with Multiplication

Theorem
Let $m, n, k \in \N$ where $\N$ is the set of natural numbers.

Let $k \ne 0$.

Then:
 * $m < n \iff m \times k < n \times k$

Proof
Proof by induction:

First we note that if $k = 0$ then $m \times k = 0 = n \times k$ whatever $m$ and $n$ are, and the proposition clearly does not hold.

So, for all $k \in \N \setminus \left\{{0}\right\}$, let $P \left({k}\right)$ be the proposition:
 * $m < n \iff m \times k < n \times k$

$P \left({1}\right)$ is true, as this just says $m \times 1 = m < n = n \times 1$ from Natural Number Multiplication Identity is One.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({j}\right)$ is true, where $j \ge 1$, then it logically follows that $P \left({j^+}\right)$ is true.

So this is our induction hypothesis:
 * $m < n \iff m \times j < n \times j$

Then we need to show:
 * $m < n \iff m \times j^+ < n \times j^+$

Induction Step
This is our induction step:

Let $m < n$.

Then:

Similarly:
 * $m < n \iff \left({m \times j}\right) + n < \left({n \times j}\right) + n$

But then from Ordering on Natural Numbers is Compatible with Addition, we also have:
 * $m < n \iff \left({m \times j}\right) + m < \left({m \times j}\right) + n$

So from Natural Number Ordering is Transitive, we have:
 * $m < n \iff \left({m \times j}\right) + m < \left({n \times j}\right) + n$

This gives:

So $P \left({j}\right) \implies P \left({j^+}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall m, n, k \in \N, k \ne 0: m \times n \iff m \times k < n \times k$