# Natural Number Addition is Commutative/Proof 2

## Contents

## Theorem

The operation of addition on the set of natural numbers $\N$ is commutative:

- $\forall m, n \in \N: m + n = n + m$

## Proof

Proof by induction.

Consider the natural numbers $\N$ defined as the elements of the minimal infinite successor set $\omega$.

From the definition of addition in $\omega$, we have that:

\(\displaystyle \forall m, n \in \N: \ \ \) | \(\displaystyle m + 0\) | \(=\) | \(\displaystyle m\) | ||||||||||

\(\displaystyle m + n^+\) | \(=\) | \(\displaystyle \left({m + n}\right)^+\) |

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

- $\forall m \in \N: m + n = n + m$

### Basis for the Induction

From Natural Number Addition Commutes with Zero, we have:

- $\forall m \in \N: m + 0 = m = 0 + m$

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 $P \left({k}\right)$:

- $\forall m \in \N: m + k = k + m$

Then we need to show that $P \left({k^+}\right)$ follows directly from $P \left({k}\right)$:

- $\forall m \in \N: m + k^+ = k^+ + m$

### Induction Step

This is our induction step:

\(\displaystyle k^+ + m\) | \(=\) | \(\displaystyle \left({k + m}\right)^+\) | Natural Number Addition Commutativity with Successor | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({m + k}\right)^+\) | from the induction hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle m + k^+\) | by definition |

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

Therefore:

- $\forall m, n \in \N: m + n = n + m$

$\blacksquare$

## Sources

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