Natural Number Addition Commutes with Zero
Jump to navigation
Jump to search
Theorem
Let $\N$ be the natural numbers.
Then:
- $\forall n \in \N: 0 + n = n = n + 0$
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:
- $0 + n = n = n + 0$
Basis for the Induction
By definition, we have:
- $0 + 0 = 0 = 0 + 0$
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 $\map P k$:
- $0 + k = k = k + 0$
Then we need to show that $\map P {k^+}$ follows directly from $\map P k$:
- $0 + k^+ = k^+ = k^+ + 0$
Induction Step
This is our induction step:
\(\ds 0 + k^+\) | \(=\) | \(\ds \paren {0 + k}^+\) | Definition of Addition | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {k + 0}^+\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds k^+\) | Definition of Addition: $k + 0 = k$ |
By definition:
- $k^+ + 0 = k^+$
So $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \N: 0 + n = n = n + 0$
$\blacksquare$
Sources
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 13$: Arithmetic