# Natural Number Addition is Commutative

## Theorem

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

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

## Proof 1

Consider the natural numbers defined as a naturally ordered semigroup.

By definition, the operation in a naturally ordered semigroup is commutative.

Hence the result.

$\blacksquare$

## Proof 2

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:

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

$\blacksquare$

## Proof 3

In the axiomatisation of $1$-based natural numbers, this is rendered:

$\forall x, y \in \N_{> 0}: x + y = y + x$

Using the following axioms:

 $(A)$ $:$ $\displaystyle \exists_1 1 \in \N_{> 0}:$ $\displaystyle a \times 1 = a = 1 \times a$ $(B)$ $:$ $\displaystyle \forall a, b \in \N_{> 0}:$ $\displaystyle a \times \paren {b + 1} = \paren {a \times b} + a$ $(C)$ $:$ $\displaystyle \forall a, b \in \N_{> 0}:$ $\displaystyle a + \paren {b + 1} = \paren {a + b} + 1$ $(D)$ $:$ $\displaystyle \forall a \in \N_{> 0}, a \ne 1:$ $\displaystyle \exists_1 b \in \N_{> 0}: a = b + 1$ $(E)$ $:$ $\displaystyle \forall a, b \in \N_{> 0}:$ $\displaystyle$Exactly one of these three holds: $\displaystyle a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y}$ $(F)$ $:$ $\displaystyle \forall A \subseteq \N_{> 0}:$ $\displaystyle \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0}$

Let $x \in \N_{> 0}$ be arbitrary.

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:

$x + n = n + x$

### Basis for the Induction

From Natural Number Commutes with 1 under Addition, we have that:

$\forall x \in \N_{> 0}: x + 1 = 1 + x$

and so $P \left({1}\right)$ holds.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is our induction hypothesis:

$x + k = k + x$

Then we need to show:

$x + \left({k + 1}\right) = \left({k + 1}\right) + x$

### Induction Step

This is our induction step:

 $\displaystyle x + \left({k + 1}\right)$ $=$ $\displaystyle \left({x + k}\right) + 1$ Natural Number Addition is Associative $\displaystyle$ $=$ $\displaystyle \left({k + x}\right) + 1$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle k + \left({x + 1}\right)$ Natural Number Addition is Associative $\displaystyle$ $=$ $\displaystyle k + \left({1 + x}\right)$ Basis for the Induction $\displaystyle$ $=$ $\displaystyle \left({k + 1}\right) + x$ Natural Number Addition is Associative

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

$\blacksquare$