# Natural Number Multiplication is Associative

## Theorem

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

$\forall x, y, z \in \N: \left({x \times y}\right) \times z = x \times \left({y \times z}\right)$

## Proof 1

From Index Laws for Semigroup: Product of Indices we have:

$+^{z \times y} x = +^z \left({+^y x}\right)$

By definition of multiplication, this amounts to:

$x \times \left({z \times y}\right) = \left({x \times y}\right) \times z$

From Natural Number Multiplication is Commutative, we have:

$x \times \left({z \times y}\right) = x \times \left({y \times z}\right)$

$\blacksquare$

## Proof 2

We are to show that:

$\left({x \times y}\right) \times n = x \times \left({y \times n}\right)$

for all $x, y, n \in \N$.

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

 $\displaystyle \forall m, n \in \N: \ \$ $\displaystyle m \times 0$ $=$ $\displaystyle 0$ $\displaystyle m \times \left({n + 1}\right)$ $=$ $\displaystyle \left({m \times n}\right) + m$

Let $x, y \in \N$ be arbitrary.

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

$\left({x \times y}\right) \times n = x \times \left({y \times n}\right)$

### Basis for the Induction

$P \left({0}\right)$ is the case:

 $\displaystyle \left({x \times y}\right) \times 0$ $=$ $\displaystyle 0$ $\displaystyle$ $=$ $\displaystyle x \times 0$ $\displaystyle$ $=$ $\displaystyle x \times \left({y \times 0}\right)$

and so $P \left({0}\right)$ 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:

$\left({x \times y}\right) \times k = x \times \left({y \times k}\right)$

Then we need to show:

$\left({x \times y}\right) \times \left({k + 1 }\right) = x \times \left({y \times \left({k + 1}\right)}\right)$

### Induction Step

This is our induction step:

 $\displaystyle \left({x \times y}\right) \times \left({k + 1 }\right)$ $=$ $\displaystyle \left({\left({x \times y}\right) \times k}\right) + \left({x \times y}\right)$ Definition of Natural Number Multiplication $\displaystyle$ $=$ $\displaystyle \left({x \times \left({y \times k}\right)}\right) + \left({x \times y}\right)$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \left({x \times y}\right) + \left({x \times \left({y \times k}\right)}\right)$ Natural Number Addition is Commutative $\displaystyle$ $=$ $\displaystyle x \times \left({y + \left({y \times k}\right)}\right)$ Natural Number Multiplication Distributes over Addition $\displaystyle$ $=$ $\displaystyle x \times \left({\left({y \times k}\right) + y}\right)$ Natural Number Addition is Commutative $\displaystyle$ $=$ $\displaystyle x \times \left({y \times \left({k + 1 }\right)}\right)$ Definition of Natural Number Multiplication

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

$\blacksquare$

## Proof 3

In the Axiomatization of 1-Based Natural Numbers, this is rendered:

$\forall x, y, z \in \N_{> 0}: \left({x \times y}\right) \times z = x \times \left({y \times z}\right)$

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: $\displaystyle a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y}$ $(F)$ $:$ $\displaystyle \forall A \subseteq \N_{> 0}:$ $\displaystyle \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0}$

Let $x, y \in \N_{> 0}$ be arbitrary.

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

$\left({x \times y}\right) \times n = x \times \left({y \times n}\right)$

### Basis for the Induction

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

 $\displaystyle \left({x \times y}\right) \times 1$ $=$ $\displaystyle x \times y$ Axiom $A$ $\displaystyle$ $=$ $\displaystyle x \times \left({y \times 1}\right)$ 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 $P \left({k^+}\right)$ is true.

So this is our induction hypothesis:

$\left({x \times y}\right) \times k = x \times \left({y \times k}\right)$

Then we need to show:

$\left({x \times y}\right) \times \left({k + 1}\right) = x \times \left({y \times \left({k + 1}\right)}\right)$

### Induction Step

This is our induction step:

 $\displaystyle \left({x \times y}\right) \times \left({k + 1}\right)$ $=$ $\displaystyle \left({\left({x \times y}\right) \times k}\right) + \left({x \times y}\right)$ Axiom $B$ $\displaystyle$ $=$ $\displaystyle \left({x \times \left({y \times k}\right)}\right) + \left({x \times y}\right)$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \left({x \times y}\right) + \left({x \times \left({y \times k}\right)}\right)$ Natural Number Addition is Commutative $\displaystyle$ $=$ $\displaystyle x \times \left({y + \left({y \times k}\right)}\right)$ Natural Number Multiplication Distributes over Addition $\displaystyle$ $=$ $\displaystyle x \times \left({\left({y \times k}\right) + y}\right)$ Natural Number Addition is Commutative $\displaystyle$ $=$ $\displaystyle x \times \left({y \times \left({k + 1}\right)}\right)$ Definition of Natural Number Multiplication

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

$\blacksquare$