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:

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

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:

\(\displaystyle f' \left({n + 1}\right)\) \(=\) \(\displaystyle \left({n + 1}\right) \times m\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle n \times m + m\) $\quad$ Natural Number Multiplication Distributes over Addition $\quad$
\(\displaystyle \) \(=\) \(\displaystyle f' \left({n}\right) + m\) $\quad$ $\quad$

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:

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


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:


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

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:

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


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:

\(\displaystyle a \times 1\) \(=\) \(\displaystyle a\) $\quad$ Axiom $A$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 1 \times a\) $\quad$ Axiom $A$ $\quad$

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:

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

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 $B$.

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 $C = D$.

$\blacksquare$


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