Invertible Elements under Natural Number Multiplication

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Theorem

Let $\N$ be the natural numbers.

Let $\times$ denote multiplication.


Then the only invertible element of $\N$ for $\times$ is $1$.


Proof

Suppose $m \in \N$ is invertible for $\times$.

Let $n \in \N: m \times n = 1$.

Then from Natural Numbers have No Proper Zero Divisors:

$m \ne 0$ and $n \ne 0$

Thus, $1 \le m$ and $1 \le n$.

If $1 \le m$ then from Ordering on Natural Numbers is Compatible with Multiplication:

$1 \le n < m \times n$

This contradicts $m \times n = 1$.

The result follows.

$\blacksquare$


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