# 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 minimally inductive st $\omega$.

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:

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

$\blacksquare$