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
$m \in \N$ 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.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.13$: Corollary