Natural Number Multiplication is Commutative/Proof 3

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Theorem

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

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


Proof

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.

$\blacksquare$


Sources

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