# Successor to Natural Number It has been suggested that this article or section be renamed. One may discuss this suggestion on 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:

 $(A)$ $:$ $\displaystyle \exists_1 1 \in \N_{> 0}:$ $\displaystyle a \times 1 = a = 1 \times a$ $(B)$ $:$ $\displaystyle \forall a, b \in \N_{> 0}:$ $\displaystyle a \times \paren {b + 1} = \paren {a \times b} + a$ $(C)$ $:$ $\displaystyle \forall a, b \in \N_{> 0}:$ $\displaystyle a + \paren {b + 1} = \paren {a + b} + 1$ $(D)$ $:$ $\displaystyle \forall a \in \N_{> 0}, a \ne 1:$ $\displaystyle \exists_1 b \in \N_{> 0}: a = b + 1$ $(E)$ $:$ $\displaystyle \forall a, b \in \N_{> 0}:$ $\displaystyle$Exactly one of these three holds: $\displaystyle a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y}$ $(F)$ $:$ $\displaystyle \forall A \subseteq \N_{> 0}:$ $\displaystyle \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0}$

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

Then by the definition of ordering on natural numbers:

 $\displaystyle a + x$ $=$ $\displaystyle n$ Definition of Ordering on Natural Numbers: $a < n$ $\displaystyle n + y$ $=$ $\displaystyle a + 1$ Definition of Ordering on Natural Numbers: $a < n$ $\displaystyle \implies \ \$ $\displaystyle \left({a + x}\right) + y$ $=$ $\displaystyle a + 1$ substitution for $n$ $\displaystyle \implies \ \$ $\displaystyle a + \left({x + y}\right)$ $=$ $\displaystyle a + 1$ Natural Number Addition is Associative $\displaystyle \implies \ \$ $\displaystyle x + y$ $=$ $\displaystyle 1$ 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.

$\blacksquare$