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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

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  • $\begingroup$ 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. Now i could prove that the induction principle applies to N-{1}, but i am not sure about extending it. When i tried extending it using sets made through inequalities i had to use what i was trying to prove, at least the way i tried it. $\endgroup$ – Random guy Aug 15 '17 at 16:48
  • $\begingroup$ I proved that 1 is the least elements because i also tried to asuume $n<k<n+1$ which implies $0<k-n<1$ , which further implies k-n is not a natural number but i couldnt prove that it has to be one if i assume k and n to be natural numbers $\endgroup$ – Random guy Aug 15 '17 at 16:52
  • $\begingroup$ You are missing some of the def'n of an inductive set. Conditions 1. and 2. are satisfied by $\mathbb R$ itself. $\endgroup$ – DanielWainfleet Aug 15 '17 at 17:17
  • $\begingroup$ It was there before , but now i made it a separate paragraph. $\endgroup$ – Random guy Aug 15 '17 at 17:25
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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 $x$.

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  • $\begingroup$ Thanks, Although i was trying not to consider things like $\{1,2,...\}$ (ie. things with ellipsis), your argument especially the last part will work for me. $\endgroup$ – Random guy Aug 15 '17 at 17:01
  • $\begingroup$ Glad to help :) $\endgroup$ – Kessem Clein Aug 15 '17 at 17:06

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