Natural Number Multiplication is Closed

Theorem
Let $m$ and $n$ be natural numbers.

Then:


 * $m \times n \in \N$

where $\times$ denotes natural number multiplication.

Proof
Let $g: \N \to \N$ be defined by:


 * $g \left({n}\right) = m + n$

Applying the Principle of Recursive Definition to $0$ and $g$, we get the following function $f: \N \to \N$:


 * $f \left({n}\right) = \begin{cases}

0 &: n = 0\\ m + f \left({k}\right) &: n = k + 1 \end{cases}$

which is seen to be equivalent to $m \times n$ for all $m,n \in \N$.

Hence $m \times n \in \N$.