Successor to Natural Number

It has been suggested that this page be renamed. To discuss this page in more detail, feel free to use the talk page.

Theorem

Let $\N_{> 0}$ be the 1-based natural numbers:

$\N_{> 0} = \left\{{1, 2, 3, \ldots}\right\}$

Let $<$ be the ordering on $\N_{> 0}$:

$\forall a, b \in \N_{>0}: a < b \iff \exists c \in \N_{>0}: a + c = b$

Let $a \in \N_{>0}$.

Then there exists no natural number $n$ such that $a < n < a + 1$.

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:\( \)
					\(\ds a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y} \)
\((\text F)\)	$:$			\(\ds \forall A \subseteq \N_{> 0}:\)	\(\ds \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0} \)

Suppose that $\exists n \in \N_{>0}: a < n < a + 1$.

Then by the definition of ordering on natural numbers:

\(\ds a + x\)

\(=\)

\(\ds n\)

Definition of Ordering on Natural Numbers: $a < n$

\(\ds n + y\)

\(=\)

\(\ds a + 1\)

Definition of Ordering on Natural Numbers: $a < n$

\(\ds \implies \ \ \)

\(\ds \left({a + x}\right) + y\)

\(=\)

\(\ds a + 1\)

substitution for $n$

\(\ds \implies \ \ \)

\(\ds a + \left({x + y}\right)\)

\(=\)

\(\ds a + 1\)

Natural Number Addition is Associative

\(\ds \implies \ \ \)

\(\ds x + y\)

\(=\)

\(\ds 1\)

Addition on $1$-Based Natural Numbers is Cancellable

By Axiom $D$, either:

$y = 1$

or:

$y = t + 1$ for some $t \in \N_{>0}$

Then either:

$x + 1 = 1$ when $y = 1$

or:

$x + \left({t + 1}\right) = \left({x + t}\right) + 1 = 1$ when $y = t + 1$

Both of these conclusions violate Natural Number is Not Equal to Successor.

Hence the result.

$\blacksquare$

Sources

• 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): $2.2$: Theorem $2.15$