Natural Number Addition is Commutative
Contents
Theorem
The operation of addition on the set of natural numbers $\N$ is commutative:
- $\forall m, n \in \N: m + n = n + m$
Proof 1
Consider the natural numbers defined as a naturally ordered semigroup.
By definition, the operation in a naturally ordered semigroup is commutative.
Hence the result.
$\blacksquare$
Proof 2
Proof by induction.
Consider the natural numbers $\N$ defined as the elements of the minimal infinite successor set $\omega$.
From the definition of addition in $\omega$, 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:
- $\forall m \in \N: m + n = n + m$
Basis for the Induction
From Natural Number Addition Commutes with Zero, we have:
- $\forall m \in \N: m + 0 = m = 0 + m$
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 $P \left({k}\right)$:
- $\forall m \in \N: m + k = k + m$
Then we need to show that $P \left({k^+}\right)$ follows directly from $P \left({k}\right)$:
- $\forall m \in \N: m + k^+ = k^+ + m$
Induction Step
This is our induction step:
\(\displaystyle k^+ + m\) | \(=\) | \(\displaystyle \left({k + m}\right)^+\) | Natural Number Addition Commutativity with Successor | ||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \left({m + k}\right)^+\) | from the induction hypothesis | ||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle m + k^+\) | by definition |
So $P \left({k}\right) \implies P \left({k^+}\right)$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall m, n \in \N: m + n = n + m$
$\blacksquare$
Proof 3
In the axiomatisation of $1$-based natural numbers, this is rendered:
- $\forall x, y \in \N_{> 0}: x + y = y + x$
Using the following axioms:
\((A)\) | $:$ | \(\displaystyle \exists_1 1 \in \N_{> 0}:\) | \(\displaystyle a \times 1 = a = 1 \times a \) | |||||
\((B)\) | $:$ | \(\displaystyle \forall a, b \in \N_{> 0}:\) | \(\displaystyle a \times \paren {b + 1} = \paren {a \times b} + a \) | |||||
\((C)\) | $:$ | \(\displaystyle \forall a, b \in \N_{> 0}:\) | \(\displaystyle a + \paren {b + 1} = \paren {a + b} + 1 \) | |||||
\((D)\) | $:$ | \(\displaystyle \forall a \in \N_{> 0}, a \ne 1:\) | \(\displaystyle \exists_1 b \in \N_{> 0}: a = b + 1 \) | |||||
\((E)\) | $:$ | \(\displaystyle \forall a, b \in \N_{> 0}:\) | \(\displaystyle \)Exactly one of these three holds:\( \) | |||||
\(\displaystyle a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y} \) | ||||||||
\((F)\) | $:$ | \(\displaystyle \forall A \subseteq \N_{> 0}:\) | \(\displaystyle \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0} \) |
Let $x \in \N_{> 0}$ be arbitrary.
For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:
- $x + n = n + x$
Basis for the Induction
From Natural Number Commutes with 1 under Addition, we have that:
- $\forall x \in \N_{> 0}: x + 1 = 1 + x$
and so $P \left({1}\right)$ holds.
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true.
So this is our induction hypothesis:
- $x + k = k + x$
Then we need to show:
- $x + \left({k + 1}\right) = \left({k + 1}\right) + x$
Induction Step
This is our induction step:
\(\displaystyle x + \left({k + 1}\right)\) | \(=\) | \(\displaystyle \left({x + k}\right) + 1\) | Natural Number Addition is Associative | ||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \left({k + x}\right) + 1\) | Induction Hypothesis | ||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle k + \left({x + 1}\right)\) | Natural Number Addition is Associative | ||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle k + \left({1 + x}\right)\) | Basis for the Induction | ||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \left({k + 1}\right) + x\) | Natural Number Addition is Associative |
So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.
$\blacksquare$
Sources
- 1951: Nathan Jacobson: Lectures in Abstract Algebra: I. Basic Concepts ... (previous) ... (next): Introduction $\S 4$: The natural numbers: $\text{A} 2$
- 1972: A.G. Howson: A Handbook of Terms used in Algebra and Analysis ... (previous) ... (next): $\S 4$: Number systems $\text{I}$: Peano's Axioms