Natural Number Multiplication is Commutative/Proof 2

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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

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:

$\forall m, n \in \N: m \times n = n \times m$

$\blacksquare$


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