Natural Numbers have No Proper Zero Divisors

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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 $m = 0$.

Then from Zero is Zero Element for Natural Number Multiplication:

$m \times n = 0$

$\Box$


Sufficient Condition

Let $m \times n = 0$.

Without loss of generality, suppose $n \ne 0$.

\(\ds n\) \(\ne\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds 1\) \(\le\) \(\ds n\) Definition of One
\(\ds \leadsto \ \ \) \(\ds m \times n\) \(=\) \(\ds m \times \paren {\paren {n - 1} + 1}\) Definition of Difference (Natural Numbers)
\(\ds \) \(=\) \(\ds m \times \paren {n - 1} + m\) Natural Number Multiplication Distributes over Addition
\(\ds \leadsto \ \ \) \(\ds 0 \le m\) \(\le\) \(\ds m \times \paren {n - 1} + m\)

But as:

$m \times \paren {n - 1} \circ m = m \times n = 0$

it follows that:

$0 \le m \le 0$

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

$\blacksquare$


Sources

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