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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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    If there was a natural number between 0 and 1, what interesting consequences would there be?2016-09-03

3 Answers 3

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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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    @Charles: In any model, it is still true every natural is either $0$ or $s(n)$ (by induction for formula $\phi(n) = (n=0) \vee \exists m. n=s(m)$), remainder of the proof is unchanged.2011-12-21
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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 '$\in$'.2011-12-21