# Natural Number Addition Commutes with Zero

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## 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:

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

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:

 $\ds 0 + k^+$ $=$ $\ds \left({0 + k}\right)^+$ Definition of Addition $\ds$ $=$ $\ds \left({k + 0}\right)^+$ from the induction hypothesis $\ds$ $=$ $\ds k^+$ 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:

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

$\blacksquare$