# Natural Number Addition Commutes with Zero

## Contents

## Theorem

Let $\N$ be the natural numbers.

Then:

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

## Proof

Proof by induction:

From the definition of addition, we have that:

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

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

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

- $0 + n = n = n + 0$

### Basis for the Induction

By definition, we have:

- $0 + 0 = 0 = 0 + 0$

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

- $0 + k = k = k + 0$

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

- $0 + k^+ = k^+ = k^+ + 0$

### Induction Step

This is our induction step:

\(\displaystyle 0 + k^+\) | \(=\) | \(\displaystyle \left({0 + k}\right)^+\) | Definition of Addition | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({k + 0}\right)^+\) | from the induction hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle k^+\) | Definition of Addition: $k + 0 = k$ |

By definition:

- $k^+ + 0 = k^+$

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

Therefore:

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

$\blacksquare$

## Sources

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