Natural Number Multiplication is Commutative/Proof 3

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Theorem

The operation of multiplication on the set of natural numbers $\N_{> 0}$ is commutative:

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


Proof

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.

$\blacksquare$


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