# Natural Number Multiplication is Commutative/Proof 3

## Theorem

The operation of multiplication on the set of natural numbers $\N_{> 0}$ is commutative:

$\forall x, y \in \N_{> 0}: x \times y = y \times x$

## Proof

Using the following axioms:

 $(A)$ $:$ $\displaystyle \exists_1 1 \in \N_{> 0}:$ $\displaystyle a \times 1 = a = 1 \times a$ $(B)$ $:$ $\displaystyle \forall a, b \in \N_{> 0}:$ $\displaystyle a \times \paren {b + 1} = \paren {a \times b} + a$ $(C)$ $:$ $\displaystyle \forall a, b \in \N_{> 0}:$ $\displaystyle a + \paren {b + 1} = \paren {a + b} + 1$ $(D)$ $:$ $\displaystyle \forall a \in \N_{> 0}, a \ne 1:$ $\displaystyle \exists_1 b \in \N_{> 0}: a = b + 1$ $(E)$ $:$ $\displaystyle \forall a, b \in \N_{> 0}:$ $\displaystyle$Exactly one of these three holds: $\displaystyle a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y}$ $(F)$ $:$ $\displaystyle \forall A \subseteq \N_{> 0}:$ $\displaystyle \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0}$

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:

$\forall a \in \N_{> 0}: a \times n = n \times a$

### Basis for the Induction

$P \left({1}\right)$ is the case:

 $\displaystyle a \times 1$ $=$ $\displaystyle a$ Axiom $A$ $\displaystyle$ $=$ $\displaystyle 1 \times a$ Axiom $A$

and so $P \left({1}\right)$ holds.

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 $P \left({k + 1}\right)$ is true.

So this is our induction hypothesis:

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

Then we need to show:

$\forall a \in \N: a \times \left({k + 1}\right) = \left({k + 1}\right) \times a$

### Induction Step

This is our induction step:

 $\displaystyle a \times \left({k + 1}\right)$ $=$ $\displaystyle \left({a \times k}\right) + \left({a \times 1}\right)$ Left Distributive Law for Natural Numbers $\displaystyle$ $=$ $\displaystyle \left({k \times a}\right) + \left({a \times 1}\right)$ Induction hypothesis $\displaystyle$ $=$ $\displaystyle \left({k \times a}\right) + \left({1 \times a}\right)$ Basis for the Induction $\displaystyle$ $=$ $\displaystyle \left({k + 1}\right) \times a$ Right Distributive Law for Natural Numbers

The result follows by the Principle of Mathematical Induction.

$\blacksquare$