Natural Number Addition is Associative/Proof 2
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
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.
$\blacksquare$
Sources
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 13$: Arithmetic