Natural Number Multiplication is Commutative
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 $C = D$.
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
- 1937: Richard Courant: Differential and Integral Calculus: Volume $\text { I }$ (2nd ed.) ... (previous) ... (next): Chapter $\text I$: Introduction: $1$. The Continuum of Numbers (footnote ${}^*$)
- 1951: Nathan Jacobson: Lectures in Abstract Algebra: Volume $\text { I }$: 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
- 1982: Alan G. Hamilton: Numbers, Sets and Axioms ... (previous) ... (next): $\S 1$: Numbers: $1.1$ Natural Numbers and Integers: Examples $1.1 \ \text {(f)}$