# 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:

$\map g n = m + n$

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

$\map f n = \begin{cases} 0 & : n = 0 \\ m + \map f k & : n = k + 1 \end{cases}$

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

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

$\blacksquare$