Natural Number Addition is Associative

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Theorem

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

$\forall x, y, z \in \N: x + \paren {y + z} = \paren {x + y} + z$


Proof 1

Consider the natural numbers defined as a naturally ordered semigroup.


By definition, the operation in a semigroup is associative.

Hence the result.

$\blacksquare$


Proof 2

Consider the von Neumann construction of natural numbers $\N$, as elements of the minimally inductive set $\omega$.

We are to show that:

$\paren {x + y} + n = x + \paren {y + n}$

for all $x, y, n \in \N$.


From the definition of addition, we have that:

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

Let $x, y \in \N$ be arbitrary.

For all $n \in \N$, let $\map P n$ be the proposition:

$\paren {x + y} + n = x + \paren {y + n}$


Basis for the Induction

$\map P 0$ is the case:

\(\ds \paren {x + y} + 0\) \(=\) \(\ds x + y\)
\(\ds \) \(=\) \(\ds x + \paren {y + 0}\)

and so $\map P 0$ holds.

This is our basis for the induction.


Induction Hypothesis

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


So this is our induction hypothesis:

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


Then we need to show:

$\paren {x + y} + \paren {k^+} = x + \paren {y + \paren {k^+} }$


Induction Step

This is our induction step:


\(\ds \paren {x + y} + k^+\) \(=\) \(\ds \paren {\paren {x + y} + k}^+\) Definition of Addition in Minimally Inductive Set
\(\ds \) \(=\) \(\ds \paren {x + \paren {y + k} }^+\) induction Hypothesis
\(\ds \) \(=\) \(\ds x + \paren {\paren {y + k}^+}\) Definition of Addition in Minimally Inductive Set
\(\ds \) \(=\) \(\ds x + \paren {y + k^+}\) Definition of Addition in Minimally Inductive Set

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

$\blacksquare$


Proof 3

In the Axiomatization of 1-Based Natural Numbers, this is rendered:

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


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

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

$\paren {x + y} + n = x + \paren {y + n}$


Basis for the Induction

From Axiom $\text C$, we have by definition that:

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

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:

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


Then we need to show:

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


Induction Step

This is our induction step:


\(\ds \paren {x + y} + \paren {k + 1}\) \(=\) \(\ds \paren {\paren {x + y} + k} + 1\) Basis for the Induction
\(\ds \) \(=\) \(\ds \paren {x + \paren {y + k} } + 1\) Induction Hypothesis
\(\ds \) \(=\) \(\ds x + \paren {\paren {y + k} + 1}\) Basis for the Induction
\(\ds \) \(=\) \(\ds x + \paren {y + \paren {k + 1} }\) Basis for the Induction

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

$\blacksquare$


Sources

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