Natural Number Addition Commutes with Zero
Contents
Theorem
Let $\N$ be the natural numbers.
Then:
- $\forall n \in \N: 0 + n = n = n + 0$
Proof
Proof by induction:
From the definition of addition, we have that:
\(\displaystyle \forall m, n \in \N: \ \ \) | \(\displaystyle m + 0\) | \(=\) | \(\displaystyle m\) | ||||||||||
\(\displaystyle m + n^+\) | \(=\) | \(\displaystyle \left({m + n}\right)^+\) |
For all $n \in \N$, let $P \left({n}\right)$ be the proposition:
- $0 + n = n = n + 0$
Basis for the Induction
By definition, we have:
- $0 + 0 = 0 = 0 + 0$
Thus $P \left({0}\right)$ is seen to be true.
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k^+}\right)$ is true.
So this is our induction hypothesis $P \left({k}\right)$:
- $0 + k = k = k + 0$
Then we need to show that $P \left({k^+}\right)$ follows directly from $P \left({k}\right)$:
- $0 + k^+ = k^+ = k^+ + 0$
Induction Step
This is our induction step:
\(\displaystyle 0 + k^+\) | \(=\) | \(\displaystyle \left({0 + k}\right)^+\) | Definition of Addition | ||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \left({k + 0}\right)^+\) | from the induction hypothesis | ||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle k^+\) | Definition of Addition: $k + 0 = k$ |
By definition:
- $k^+ + 0 = k^+$
So $P \left({k}\right) \implies P \left({k^+}\right)$ 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