Natural Number Multiplication is Commutative/Proof 1

Theorem

The operation of multiplication on the set of natural numbers $\N$ is commutative:

$\forall x, y \in \N: x \times y = y \times x$

Proof

Natural number multiplication is recursively defined as:

$\forall m, n \in \N: \begin{cases}

m \times 0 & = 0 \\ m \times \paren {n + 1} & = m \times n + m \end{cases}$

From the Principle of Recursive Definition, there is only one mapping $f$ satisfying this definition; that is, such that:

$\forall n \in \N: \begin{cases}

\map f 0 = 0 \\ \map f {n + 1} = \map f n + m \end{cases}$

Consider now $f'$ defined as $\map {f'} n = n \times m$.

Then by Zero is Zero Element for Natural Number Multiplication:

$\map {f'} 0 = 0 \times m = 0$

Furthermore:

\(\ds \map {f'} {n + 1}\)

\(=\)

\(\ds \paren {n + 1} \times m\)

\(\ds \)

\(=\)

\(\ds n \times m + m\)

Natural Number Multiplication Distributes over Addition

\(\ds \)

\(=\)

\(\ds \map {f'} n + m\)

showing that $f'$ also satisfies the definition of $m \times n$.

By the Principle of Recursive Definition it follows that:

$m \times n = n \times m$

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.10$