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Is there a natural number between $0$ and $1$?

A proof, s'il vous plaît, not your personal opinion. (Assume the Peano Postulates.)

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Can't make a comment - but I would say this would depend on the ordering. Peano postulates don't define 1 either. – Mark Bennet Jun 24 '11 at 12:06
I suppose that with 1, Prof means S(0). – user6701 Jun 24 '11 at 12:09
@Tim: No opinions, please! $$$$ :) – The Chaz 2.0 Jun 24 '11 at 12:48
up vote 8 down vote accepted

Every natural number $m$ is either $0$ or $s(n)$, where $n$ is a natural number.

Proof: It can't be both, because $s(n)$ can't be $0$. Set of all natural numbers which are either $0$ or $s(n)$ for some $n$ satisfies induction principle, so it contains all natural numbers.

Direct consequence: Every natural number is either $0$, or $s(0)$ or $s(s(n))$ for some natural number $n$.

Suppose there is $m$ such that $0 < m < s(0)$. Either $m$ is $0$, $s(0)$ or $s(s(n))$. First two cannot hold, so you have $s(s(n)) < s(0)$, i.e., $s(n) < 0$.

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How to prove that there are no natural numbers that are neither 0 or s(n)? Is the set defined so that "0 and S(n), and only those, are natural number"? – user6701 Jun 24 '11 at 12:14
The set is defined A = {x : x = 0 or there exists natural n such that x=s(n)}. It is easy to prove it satisfies $0 \in A$ and $n \in A \implies s(n) \in A$. It follows that $A = \mathbb{N}$. – sdcvvc Jun 24 '11 at 12:15
Well, so s(n) < 0. Why not? – Prof Duck Jun 24 '11 at 14:00
@Prof Duck: It depends on how you define "<". For example, $x<y$ could mean there exists $z$ that $x+z=y$ and $x \neq y$. If $s(n) < 0$ then there would be $z$ such that $s(n) + z = 0$, but $s(n) + z = s(n + z) = 0$, but 0 is not a successor. – sdcvvc Jun 24 '11 at 14:12
Charles: The proof works for them too. – sdcvvc Dec 20 '11 at 22:37

HINT $\rm\ \ S\:n\ =\ S\:0\ \Rightarrow\ n\: =\: 0\: \ne\: S\: m$

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One is defined as $\{\emptyset\}$. That a number n is between 0 and 1 means that $0\in n$ and $n\in 1$. Since $n\in 1$, it follows that $n=0$, a contradiction.

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In Peano arithmetic $1$ is defined as $S(0)$. There is not even the notion of a set $\{\emptyset\}$, nor of the binary relation '$\in$'. – George Lowther Dec 21 '11 at 2:35

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