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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.2011-06-24
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    I suppose that with 1, Prof means S(0).2011-06-24
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    @Tim: No opinions, please! $$$$ :)2011-06-24
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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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    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"?2011-06-24
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    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 $A = \mathbb{N}$.2011-06-24
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    Well, so s(n) < 0. Why not?2011-06-24
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    @Prof Duck: It depends on how you define "<". For example, $x 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.2011-06-24
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    What about nonstandard models?2011-12-20
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    Charles: The proof works for them too.2011-12-20
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    @Charles: Any nonstandard model of PA begins with a copy of the standard model. In fact it is enough for the model to satisfy just a certain finite number of axioms of PA to ensure this. The remainder of the model will look like some number of copies of $\mathbb{Z}$, which will themselves be densely ordered. But there is always a gap between any number and its successor.2011-12-21
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    @CarlMummert: I know the order type of nonstandard PA models, so I agree. I just wasn't sure if the proof covered this. (Despite sdcvvc's assertion, it's still not obvious to me that it does.)2011-12-21
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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