Natural Number Addition Commutativity with Successor/Proof 1
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Theorem
Let $\N$ be the natural numbers.
Then:
- $\forall m, n \in \N: m^+ + n = \paren {m + n}^+$
Proof
Proof by induction:
From definition of addition:
\(\ds \forall m, n \in \N: \, \) | \(\ds m + 0\) | \(=\) | \(\ds m\) | |||||||||||
\(\ds m + n^+\) | \(=\) | \(\ds \paren {m + n}^+\) |
For all $n \in \N$, let $\map P n$ be the proposition:
- $\forall m \in \N: m^+ + n = \paren {m + n}^+$
Basis for the Induction
From definition of addition:
\(\ds \forall m \in \N: \, \) | \(\ds m^+ + 0\) | \(=\) | \(\ds m^+\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {m + 0}^+\) |
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 $\map P k$:
- $\forall m \in \N: m^+ + k = \paren {m + k}^+$
Then we need to show that $\map P {k^+}$ follows directly from $\map P k$:
- $\forall m \in \N: m^+ + k^+ = \paren {m + k^+}^+$
Induction Step
This is our induction step:
\(\ds m^+ + k^+\) | \(=\) | \(\ds \paren {m^+ + k}^+\) | Definition of Addition | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {m + k}^+}^+\) | induction hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {m + k^+}^+\) | Definition of Addition |
So $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall m, n \in \N: m^+ + n = \paren {m + n}^+$
$\blacksquare$
Sources
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 13$: Arithmetic