Proving that there exists no natural number between 2 consecutive natural numbers using the Axiomatic description of $\mathbb{R}$

Question

From an axiomatic description of $\Bbb R$ as given in (for example) Calculus by Apostol, which assumes the field axioms, order relations and the completeness axiom.

The Definition of natural numbers being the set of all real numbers that belong to every inductive set.

Prove that there exists no natural number between $n$ and $n+1$ where $n$ is a natural number.

An inductive set is a subset of $\Bbb R$ with the following properties:

1. $1$ belongs to this set
2. if $x$ belongs to this set, them $x+1$ belongs to this set.

From the comments:

i proved the induction principle that if A is a subset of N and A is inductive then A=N. Then i could prove that 1 is the least element of N by considering A as the set of all natural numbers greater than or equal to 1

Answer 1

By the definition of the natural numbers, given a number $n<x<n+1$, it is enough to find any inductive set that doesn't contain $x$ to prove that $x\notin \mathbb N$.

By induction on $n$:

for the case where $n=1$, take the inductive set $\{1,2,...\}$, there is no $1<x<2$ in that set.

For $n>1$: given $x$ such that $n<x<n+1$, by the induction hypothesis we know that $x-1$ isn't a natural number because $n-1<x-1<n$, therefore there is an inductive set $S$ such that $x-1\notin S$. Then by the definition of an inductive set, the set $S-\{x\}$ obtained by removing $x$ from $S$ is also an inductive set, and doesn't contain $x$.