Natural Number Multiplication is Commutative/Proof 2
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:
- $\forall m, n \in \N: m \times n = n \times m$
$\blacksquare$
Sources
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 13$: Arithmetic