# Natural Number Multiplication is Commutative

## Contents

## Theorem

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

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

In the words of Euclid:

*If two numbers by multiplying one another make certain numbers, the numbers so produced will be equal to one another.*

(*The Elements*: Book $\text{VII}$: Proposition $16$)

## Proof 1

Natural number multiplication is recursively defined as:

- $\forall m, n \in \N: \begin{cases} m \times 0 & = 0 \\ m \times \left({n + 1}\right) & = m \times n + m \end{cases}$

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

- $\forall n \in \N: \begin{cases} f \left({0}\right) = 0 \\ f \left({n + 1}\right) = f \left({n}\right) + m \end{cases}$

Consider now $f'$ defined as $f' \left({n}\right) = n \times m$.

Then by Zero is Zero Element for Natural Number Multiplication:

- $f' \left({0}\right) = 0 \times m = 0$

Furthermore:

\(\displaystyle f' \left({n + 1}\right)\) | \(=\) | \(\displaystyle \left({n + 1}\right) \times m\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle n \times m + m\) | Natural Number Multiplication Distributes over Addition | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle f' \left({n}\right) + m\) |

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

By the Principle of Recursive Definition it follows that:

- $m \times n = n \times m$

$\blacksquare$

## Proof 2

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:

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

$\blacksquare$

## Proof 3

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

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

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.

$\blacksquare$

## Euclid's Proof

Let $A, B$ be two (natural) numbers, and let $A$ by multiplying $B$ make $C$, and $B$ by multiplying $A$ make $D$.

We need to show that $C = D$.

We have that $A \times B = C$.

So $B$ measures $C$ according to the units of $A$.

But the unit $E$ also measures $A$ according to the units in it.

So $E$ measures $A$ the same number of times that $B$ measures $C$.

Therefore from Proposition $15$ of Book $\text{VII} $: Alternate Ratios of Multiples $E$ measures $B$ the same number of times that $A$ measures $C$.

We also have that $A$ measures $D$ according to the units of $B$.

But the unit $E$ also measures $B$ according to the units in it.

Therefore from Proposition $15$ of Book $\text{VII} $: Alternate Ratios of Multiples $E$ measures $B$ the same number of times that $A$ measures $D$.

But we also have that $E$ measures $B$ the same number of times that $A$ measures $C$.

So $A$ measures $C$ and $D$ the same number of times.

Therefore $C = D$.

$\blacksquare$

## Sources

- 1951: Nathan Jacobson:
*Lectures in Abstract Algebra: I. Basic Concepts*... (previous) ... (next): Introduction $\S 4$: The natural numbers: $\text{M} 2$ - 1972: A.G. Howson:
*A Handbook of Terms used in Algebra and Analysis*... (previous) ... (next): $\S 4$: Number systems $\text{I}$: Peano's Axioms