Natural Number Addition is Commutative/Proof 2

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

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


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.


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