# Natural Number Addition is Commutative/Proof 2

## 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:

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

\(\ds m + n^+\) | \(=\) | \(\ds \paren {m + n}^+\) |

For all $n \in \N$, let $\map P n$ 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 $\map P 0$ 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 $\map P k$:

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

Then we need to show that $\map P {k^+}$ follows directly from $\map P k$:

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

### Induction Step

This is our induction step:

\(\ds k^+ + m\) | \(=\) | \(\ds \paren {k + m}^+\) | Natural Number Addition Commutativity with Successor | |||||||||||

\(\ds \) | \(=\) | \(\ds \paren {m + k}^+\) | from the induction hypothesis | |||||||||||

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

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

Therefore:

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

$\blacksquare$

## Sources

- 1960: Paul R. Halmos:
*Naive Set Theory*... (previous) ... (next): $\S 13$: Arithmetic - 1982: Alan G. Hamilton:
*Numbers, Sets and Axioms*... (previous) ... (next): $\S 1$: Numbers: $1.1$ Natural Numbers and Integers: Theorem $1.3$