Natural Number Multiplication is Associative/Proof 2

Theorem

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

$\forall x, y, n \in \N: \left({x \times y}\right) \times n = x \times \left({y \times n}\right)$

Proof

We are to show that:

$\left({x \times y}\right) \times n = x \times \left({y \times n}\right)$

for all $x, y, n \in \N$.

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

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

Let $x, y \in \N$ be arbitrary.

For all $n \in \N$, let $P \left({n}\right)$ be the proposition:

$\left({x \times y}\right) \times n = x \times \left({y \times n}\right)$

Basis for the Induction

$P \left({0}\right)$ is the case:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({x \times y}\right) \times 0\)	\(=\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x \times 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x \times \left({y \times 0}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

and so $P \left({0}\right)$ 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 $P \left({k^+}\right)$ is true.

So this is our induction hypothesis:

$\left({x \times y}\right) \times k = x \times \left({y \times k}\right)$

Then we need to show:

$\left({x \times y}\right) \times k^+ = x \times \left({y \times k^+}\right)$

Induction Step

This is our induction step:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({x \times y}\right) \times k^+\)	\(=\)	\(\displaystyle \left({\left({x \times y}\right) \times k}\right) + \left({x \times y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by definition
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({x \times \left({y \times k}\right)}\right) + \left({x \times y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by the induction hypothesis
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({x \times y}\right) + \left({x \times \left({y \times k}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Natural Number Addition is Commutative
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x \times \left({y + \left({y \times k}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Natural Number Multiplication Distributes over Addition
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x \times \left({\left({y \times k}\right) + y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Natural Number Addition is Commutative
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x \times \left({y \times k^+}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by definition

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

$\blacksquare$

Sources

• Paul R. Halmos: Naive Set Theory (1960)... (previous)... (next): $\S 13$: Arithmetic