Ordering on Natural Numbers is Compatible with Multiplication

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Theorem

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

Let $k \ne 0$.


Then:

$m < n \iff m \times k < n \times k$


Corollary

Let $a, b, c, d \in \N$ where $\N$ is the set of natural numbers.

Then:

$a > b, c > d \implies a c > b d$


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 $k \in \N \setminus \set 0$, let $\map P k$ be the proposition:

$m < n \iff m \times k < n \times k$


From Identity Element of Natural Number Multiplication is One:

$m \times 1 = m < n = n \times 1$

and so $\map P 1$ is true.


This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P j$ is true, where $j \ge 1$, then it logically follows that $\map P {j^+}$ 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:

\(\ds m\) \(<\) \(\ds n\)
\(\ds \leadstoandfrom \ \ \) \(\ds m \times j\) \(<\) \(\ds n \times j\) Induction Hypothesis
\(\ds \leadstoandfrom \ \ \) \(\ds \paren {m \times j} + m\) \(<\) \(\ds \paren {n \times j} + m\) Ordering on Natural Numbers is Compatible with Addition


Similarly:

$m < n \iff \paren {m \times j} + n < \paren {n \times j} + n$

But then from Ordering on Natural Numbers is Compatible with Addition, we also have:

$m < n \iff \paren {m \times j} + m < \paren {m \times j} + n$

So from Natural Number Ordering is Transitive, we have:

$m < n \iff \paren {m \times j} + m < \paren {n \times j} + n$


This gives:

\(\ds m\) \(<\) \(\ds n\)
\(\ds \leadstoandfrom \ \ \) \(\ds \paren {m \times j} + m\) \(<\) \(\ds \paren {n \times j} + n\) from above
\(\ds \leadstoandfrom \ \ \) \(\ds m \times j^+\) \(<\) \(\ds n \times j^+\) Definition of Natural Number Multiplication


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


Therefore:

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

$\blacksquare$


Sources

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