# Natural Number Multiplication is Commutative/Proof 2

## 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

Proof by induction:

From the definition of natural number multiplication, we have that:

 $\displaystyle \forall m, n \in \N: \ \$ $\displaystyle m \times 0$ $=$ $\displaystyle 0$ $\displaystyle m \times n^+$ $=$ $\displaystyle \left({m \times n}\right) + m$

For all $n \in \N$, let $P \left({n}\right)$ be the proposition:

$\forall m \in \N: m \times n = n \times m$

### Basis for the Induction

From Zero is Zero Element for Natural Number Multiplication, we have:

$\forall m \in \N: m \times 0 = 0 = 0 \times m$

Thus $P \left({0}\right)$ is seen to be true.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k^+}\right)$ is true.

So this is our induction hypothesis $P \left({k}\right)$:

$\forall m \in \N: m \times k = k \times m$

Then we need to show that $P \left({k^+}\right)$ follows from $P \left({k}\right)$:

$\forall m \in \N: m \times k^+ = k^+ \times m$

### Induction Step

This is our induction step:

 $\displaystyle m \times k^+$ $=$ $\displaystyle \left({m \times k}\right) + m$ Definition of Natural Number Multiplication $\displaystyle$ $=$ $\displaystyle m + \left({m \times k}\right)$ Natural Number Addition is Commutative $\displaystyle$ $=$ $\displaystyle m + \left({k \times m}\right)$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle k^+ \times m$ Natural Number Multiplication Distributes over Addition

So $P \left({k}\right) \implies P \left({k^+}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\forall m, n \in \N: m \times n = n \times m$

$\blacksquare$