# Natural Number Multiplication is Associative

## Theorem

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

$\forall x, y, z \in \N: \paren {x \times y} \times z = x \times \paren {y \times z}$

## Proof 1

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

$+^{z \times y} x = \map {+^z} {+^y x}$

By definition of multiplication, this amounts to:

$x \times \paren {z \times y} = \paren {x \times y} \times z$

From Natural Number Multiplication is Commutative, we have:

$x \times \paren {z \times y} = x \times \paren {y \times z}$

$\blacksquare$

## Proof 2

We are to show that:

$\paren {x \times y} \times n = x \times \paren {y \times n}$

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

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

 $\ds \forall m, n \in \N: \,$ $\ds m \times 0$ $=$ $\ds 0$ $\ds m \times \paren {n + 1}$ $=$ $\ds \paren {m \times n} + m$

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

For all $n \in \N$, let $\map P n$ be the proposition:

$\paren {x \times y} \times n = x \times \paren {y \times n}$

### Basis for the Induction

$\map P 0$ is the case:

 $\ds \paren {x \times y} \times 0$ $=$ $\ds 0$ $\ds$ $=$ $\ds x \times 0$ $\ds$ $=$ $\ds x \times \paren {y \times 0}$

and so $\map P 0$ holds.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

$\paren {x \times y} \times k = x \times \paren {y \times k}$

Then we need to show:

$\paren {x \times y} \times \paren {k + 1} = x \times \paren {y \times \paren {k + 1} }$

### Induction Step

This is our induction step:

 $\ds \paren {x \times y} \times \paren {k + 1}$ $=$ $\ds \paren {\paren {x \times y} \times k} + \paren {x \times y}$ Definition of Natural Number Multiplication $\ds$ $=$ $\ds \paren {x \times \paren {y \times k} } + \paren {x \times y}$ Induction Hypothesis $\ds$ $=$ $\ds \paren {x \times y} + \paren {x \times \paren {y \times k} }$ Natural Number Addition is Commutative $\ds$ $=$ $\ds x \times \paren {y + \paren {y \times k} }$ Natural Number Multiplication Distributes over Addition $\ds$ $=$ $\ds x \times \paren {\paren {y \times k} + y}$ Natural Number Addition is Commutative $\ds$ $=$ $\ds x \times \paren {y \times \paren {k + 1} }$ Definition of Natural Number Multiplication

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

$\blacksquare$

## Proof 3

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

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

Using the following axioms:

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

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

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$\paren {x \times y} \times n = x \times \paren {y \times n}$

### Basis for the Induction

$\map P 1$ is the case:

 $\ds \paren {x \times y} \times 1$ $=$ $\ds x \times y$ Axiom $\text A$ $\ds$ $=$ $\ds x \times \paren {y \times 1}$ Axiom $\text A$

and so $\map P 1$ holds.

This is our basis for the induction.

### Induction Hypothesis

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

So this is our induction hypothesis:

$\paren {x \times y} \times k = x \times \paren {y \times k}$

Then we need to show:

$\paren {x \times y} \times \paren {k + 1} = x \times \paren {y \times \paren {k + 1} }$

### Induction Step

This is our induction step:

 $\ds \paren {x \times y} \times \paren {k + 1}$ $=$ $\ds \paren {\paren {x \times y} \times k} + \paren {x \times y}$ Axiom $\text B$ $\ds$ $=$ $\ds \paren {x \times \paren {y \times k} } + \paren {x \times y}$ Induction Hypothesis $\ds$ $=$ $\ds \paren {x \times y} + \paren {x \times \paren {y \times k} }$ Natural Number Addition is Commutative $\ds$ $=$ $\ds x \times \paren {y + \paren {y \times k} }$ Natural Number Multiplication Distributes over Addition $\ds$ $=$ $\ds x \times \paren {\paren {y \times k} + y}$ Natural Number Addition is Commutative $\ds$ $=$ $\ds x \times \paren {y \times \paren {k + 1} }$ Definition of Natural Number Multiplication

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

$\blacksquare$