Natural Number Addition is Commutative/Proof 3
Jump to navigation
Jump to search
Theorem
The operation of addition on the set of natural numbers $\N_{> 0}$ is commutative:
- $\forall x, y \in \N_{> 0}: x + y = y + x$
Proof
Using the following axioms:
\((\text A)\) | $:$ | \(\ds \exists_1 1 \in \N_{> 0}:\) | \(\ds a \times 1 = a = 1 \times a \) | ||||||
\((\text B)\) | $:$ | \(\ds \forall a, b \in \N_{> 0}:\) | \(\ds a \times \paren {b + 1} = \paren {a \times b} + a \) | ||||||
\((\text C)\) | $:$ | \(\ds \forall a, b \in \N_{> 0}:\) | \(\ds a + \paren {b + 1} = \paren {a + b} + 1 \) | ||||||
\((\text D)\) | $:$ | \(\ds \forall a \in \N_{> 0}, a \ne 1:\) | \(\ds \exists_1 b \in \N_{> 0}: a = b + 1 \) | ||||||
\((\text E)\) | $:$ | \(\ds \forall a, b \in \N_{> 0}:\) | \(\ds \)Exactly one of these three holds:\( \) | ||||||
\(\ds a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y} \) | |||||||||
\((\text F)\) | $:$ | \(\ds \forall A \subseteq \N_{> 0}:\) | \(\ds \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0} \) |
Let $x \in \N_{> 0}$ be arbitrary.
For all $n \in \N_{> 0}$, let $\map P n$ 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 $\map P 1$ holds.
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
- $x + k = k + x$
Then we need to show:
- $x + \paren {k + 1} = \paren {k + 1} + x$
Induction Step
This is our induction step:
\(\ds x + \paren {k + 1}\) | \(=\) | \(\ds \paren {x + k} + 1\) | Natural Number Addition is Associative | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {k + x} + 1\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds k + \paren {x + 1}\) | Natural Number Addition is Associative | |||||||||||
\(\ds \) | \(=\) | \(\ds k + \paren {1 + x}\) | Basis for the Induction | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {k + 1} + x\) | Natural Number Addition is Associative |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
$\blacksquare$
Sources
- 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): $\S 2.1$: Theorem $2.3$