Natural Number Addition is Commutative

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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 1

Consider the natural numbers defined as a naturally ordered semigroup.


By definition, the operation in a naturally ordered semigroup is commutative.

Hence the result.

$\blacksquare$


Proof 2

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:

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


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:

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

$\blacksquare$


Proof 3

In the axiomatisation of $1$-based natural numbers, this is rendered:

$\forall x, y \in \N_{> 0}: x + y = y + x$


Using the following axioms:

\((\text A)\)   $:$     \(\ds \exists_1 1 \in \N_{> 0}:\) \(\ds a \times 1 = a = 1 \times a \)             
\((\text B)\)   $:$     \(\ds \forall a, b \in \N_{> 0}:\) \(\ds a \times \paren {b + 1} = \paren {a \times b} + a \)             
\((\text C)\)   $:$     \(\ds \forall a, b \in \N_{> 0}:\) \(\ds a + \paren {b + 1} = \paren {a + b} + 1 \)             
\((\text D)\)   $:$     \(\ds \forall a \in \N_{> 0}, a \ne 1:\) \(\ds \exists_1 b \in \N_{> 0}: a = b + 1 \)             
\((\text E)\)   $:$     \(\ds \forall a, b \in \N_{> 0}:\) \(\ds \)Exactly one of these three holds:\( \)             
\(\ds a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y} \)             
\((\text F)\)   $:$     \(\ds \forall A \subseteq \N_{> 0}:\) \(\ds \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0} \)             


Let $x \in \N_{> 0}$ be arbitrary.

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$x + n = n + x$


Basis for the Induction

From Natural Number Commutes with 1 under Addition, we have that:

$\forall x \in \N_{> 0}: x + 1 = 1 + x$

and so $\map P 1$ holds.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is our induction hypothesis:

$x + k = k + x$


Then we need to show:

$x + \paren {k + 1} = \paren {k + 1} + x$


Induction Step

This is our induction step:

\(\ds x + \paren {k + 1}\) \(=\) \(\ds \paren {x + k} + 1\) Natural Number Addition is Associative
\(\ds \) \(=\) \(\ds \paren {k + x} + 1\) Induction Hypothesis
\(\ds \) \(=\) \(\ds k + \paren {x + 1}\) Natural Number Addition is Associative
\(\ds \) \(=\) \(\ds k + \paren {1 + x}\) Basis for the Induction
\(\ds \) \(=\) \(\ds \paren {k + 1} + x\) Natural Number Addition is Associative

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

$\blacksquare$


Sources