Natural Number Addition is Commutative/Proof 2

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Theorem

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

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


Proof

Proof by induction.

Consider the natural numbers $\N$ defined as the elements of the minimally inductive st $\omega$.


From the definition of addition in $\omega$‎, we have that:

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


For all $n \in \N$, let $\map P n$ 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 $\map P 0$ is seen to be true.


This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k^+}$ is true.


So this is our induction hypothesis $\map P k$:

$\forall m \in \N: m + k = k + m$


Then we need to show that $\map P {k^+}$ follows directly from $\map P k$:

$\forall m \in \N: m + k^+ = k^+ + m$


Induction Step

This is our induction step:


\(\ds k^+ + m\) \(=\) \(\ds \paren {k + m}^+\) Natural Number Addition Commutativity with Successor
\(\ds \) \(=\) \(\ds \paren {m + k}^+\) from the induction hypothesis
\(\ds \) \(=\) \(\ds m + k^+\) by definition

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


Therefore:

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

$\blacksquare$


Sources

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