# Natural Numbers have No Proper Zero Divisors

## Theorem

Let $\N$ be the natural numbers.

Then for all $m, n \in \N$:

$m \times n = 0 \iff m = 0 \lor n = 0$

That is, $\N$ has no proper zero divisors.

## Proof

### Necessary Condition

Suppose that $n = 0$ or $m = 0$.

$m \times n = 0$

$\Box$

### Sufficient Condition

Let $m \times n = 0$.

Suppose WLOG that $n \ne 0$.

 $\displaystyle n$ $\ne$ $\displaystyle 0$ $\displaystyle \implies \ \$ $\displaystyle 1$ $\le$ $\displaystyle n$ Definition of One $\displaystyle \implies \ \$ $\displaystyle m \times n$ $=$ $\displaystyle m \times \left({\left({n - 1}\right) + 1}\right)$ Definition of Difference $\displaystyle$ $=$ $\displaystyle m \times \left({n - 1}\right) + m$ Natural Number Multiplication Distributes over Addition $\displaystyle \implies \ \$ $\displaystyle 0 \le m$ $\le$ $\displaystyle m \times \left({n - 1}\right) + m$

But as:

$m \times \left({n - 1}\right) \circ m = m \times n = 0$

it follows that:

$0 \le m \le 0$

and so as $\le$ is antisymmetric, it follows that $m = 0$.

$\blacksquare$