# Natural Number Addition is Associative

## Theorem

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

$\forall x, y, z \in \N: x + \left({y + z}\right) = \left({x + y}\right) + z$

## Proof 1

Consider the natural numbers defined as a naturally ordered semigroup.

By definition, the operation in a semigroup is associative.

Hence the result.

$\blacksquare$

## Proof 2

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

We are to show that:

$\left({x + y}\right) + n = x + \left({y + n}\right)$

for all $x, y, n \in \N$.

From the definition of addition, we have that:

 $\displaystyle \forall m, n \in \N: \ \$ $\displaystyle m + 0$ $=$ $\displaystyle m$ $\displaystyle m + n^+$ $=$ $\displaystyle \left({m + n}\right)^+$

Let $x, y \in \N$ be arbitrary.

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

$\left({x + y}\right) + n = x + \left({y + n}\right)$

### Basis for the Induction

$P \left({0}\right)$ is the case:

 $\displaystyle \left({x + y}\right) + 0$ $=$ $\displaystyle x + y$ $\displaystyle$ $=$ $\displaystyle x + \left({y + 0}\right)$

and so $P \left({0}\right)$ holds.

This is our basis for the induction.

### Induction Hypothesis

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

So this is our induction hypothesis:

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

Then we need to show:

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

### Induction Step

This is our induction step:

 $\displaystyle \left({x + y}\right) + k^+$ $=$ $\displaystyle \left({\left({x + y}\right) + k}\right)^+$ Definition of addition $\displaystyle$ $=$ $\displaystyle \left({x + \left({y + k}\right)}\right)^+$ by the induction hypothesis $\displaystyle$ $=$ $\displaystyle x + \left({\left({y + k}\right)^+}\right)$ Definition of addition $\displaystyle$ $=$ $\displaystyle x + \left({y + k^+}\right)$ Definition of addition

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

$\blacksquare$

## Proof 3

In the Axiomatization of 1-Based Natural Numbers, this is rendered:

$\forall x, y, z \in \N_{> 0}: x + \left({y + z}\right) = \left({x + y}\right) + z$

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

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

$\paren {x + y} + n = x + \paren {y + n}$

### Basis for the Induction

From Axiom $C$, we have by definition that:

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

and so $\map P 1$ holds.

This is our basis for the induction.

### Induction Hypothesis

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

So this is our induction hypothesis:

$\paren {x + y} + k = x + \paren {y + k}$

Then we need to show:

$\paren {x + y} + \paren {k + 1} = x + \paren {y + \paren {k + 1} }$

### Induction Step

This is our induction step:

 $\displaystyle \paren {x + y} + \paren {k + 1}$ $=$ $\displaystyle \paren {\paren {x + y} + k} + 1$ Basis for the Induction $\displaystyle$ $=$ $\displaystyle \paren {x + \paren {y + k} } + 1$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle x + \paren {\paren {y + k} + 1}$ Basis for the Induction $\displaystyle$ $=$ $\displaystyle x + \paren {y + \paren {k + 1} }$ Basis for the Induction

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

$\blacksquare$