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
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): $\S 16$: Theorem $16.13: \ 1^\circ$