Natural Number Multiplication is Commutative/Proof 2

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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 by induction:

From the definition of natural number multiplication, we have that:

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

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:

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

So $P \left({k}\right) \implies P \left({k^+}\right)$ and the result follows by the Principle of Mathematical Induction.


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