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 $m \times n = 0$.

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 $m = 0$.

$\blacksquare$


Sources

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