Natural Number Multiplication is Commutative

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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 \paren {n + 1} & = 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}

\map f 0 = 0 \\ \map f {n + 1} = \map f n + m \end{cases}$

Consider now $f'$ defined as $\map {f'} n = n \times m$.

Then by Zero is Zero Element for Natural Number Multiplication:

$\map {f'} 0 = 0 \times m = 0$

Furthermore:

\(\ds \map {f'} {n + 1}\) \(=\) \(\ds \paren {n + 1} \times m\)
\(\ds \) \(=\) \(\ds n \times m + m\) Natural Number Multiplication Distributes over Addition
\(\ds \) \(=\) \(\ds \map {f'} n + m\)

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:

\(\ds \forall m, n \in \N: \, \) \(\ds m \times 0\) \(=\) \(\ds 0\)
\(\ds m \times n^+\) \(=\) \(\ds \paren {m \times n} + m\)


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

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

Thus $\map P 0$ is seen to be true.


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 $\map P k$:

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


Then we need to show that $\map P {k^+}$ follows from $\map P k$:

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


Induction Step

This is our induction step:


\(\ds m \times k^+\) \(=\) \(\ds \paren {m \times k} + m\) Definition of Natural Number Multiplication
\(\ds \) \(=\) \(\ds m + \paren {m \times k}\) Natural Number Addition is Commutative
\(\ds \) \(=\) \(\ds m + \paren {k \times m}\) Induction Hypothesis
\(\ds \) \(=\) \(\ds k^+ \times m\) Natural Number Multiplication Distributes over Addition

So $\map P k \implies \map P {k^+}$ 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:

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


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

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


Basis for the Induction

$\map P 1$ is the case:

\(\ds a \times 1\) \(=\) \(\ds a\) Axiom $\text A$
\(\ds \) \(=\) \(\ds 1 \times a\) 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 $\map P {k + 1}$ is true.


So this is our induction hypothesis:

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


Then we need to show:

$\forall a \in \N: a \times \paren {k + 1} = \paren {k + 1} \times a$


Induction Step

This is our induction step:

\(\ds a \times \paren {k + 1}\) \(=\) \(\ds \paren {a \times k} + \paren {a \times 1}\) Left Distributive Law for Natural Numbers
\(\ds \) \(=\) \(\ds \paren {k \times a} + \paren {a \times 1}\) Induction hypothesis
\(\ds \) \(=\) \(\ds \paren {k \times a} + \paren {1 \times a}\) Basis for the Induction
\(\ds \) \(=\) \(\ds \paren {k + 1} \times a\) 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 $C = D$.

Euclid-VII-16.png

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 $B$ since $B \times A = D$.

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

Therefore $E$ 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

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