Natural Number Commutes with 1 under Addition
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Theorem
Let $n \in \N_{> 0}$ be a natural number.
Then $n$ commutes with $1$ under the operation of addition:
- $\forall n \in \N_{> 0}: n + 1 = 1 + n$
Proof
Using the axiomatization:
\((\text A)\) | $:$ | \(\ds \exists_1 1 \in \N_{> 0}:\) | \(\ds a \times 1 = a = 1 \times a \) | ||||||
\((\text B)\) | $:$ | \(\ds \forall a, b \in \N_{> 0}:\) | \(\ds a \times \paren {b + 1} = \paren {a \times b} + a \) | ||||||
\((\text C)\) | $:$ | \(\ds \forall a, b \in \N_{> 0}:\) | \(\ds a + \paren {b + 1} = \paren {a + b} + 1 \) | ||||||
\((\text D)\) | $:$ | \(\ds \forall a \in \N_{> 0}, a \ne 1:\) | \(\ds \exists_1 b \in \N_{> 0}: a = b + 1 \) | ||||||
\((\text E)\) | $:$ | \(\ds \forall a, b \in \N_{> 0}:\) | \(\ds \)Exactly one of these three holds:\( \) | ||||||
\(\ds a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y} \) | |||||||||
\((\text F)\) | $:$ | \(\ds \forall A \subseteq \N_{> 0}:\) | \(\ds \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0} \) |
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
- $n + 1 = 1 + n$
Basis for the Induction
Setting $n = 1$ we have that:
- $1 + 1 = 1 + 1$
and so $\map P 1$ holds trivially.
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:
- $k + 1 = 1 + k$
Then we need to show:
- $\paren {k + 1} + 1 = 1 + \paren {k + 1}$
Induction Step
This is our induction step:
\(\ds 1 + \paren {k + 1}\) | \(=\) | \(\ds \paren {1 + k} + 1\) | Axiom $C$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {k + 1} + 1\) | Induction Hypothesis |
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 $2.3$