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:

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

$\blacksquare$


Sources

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