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First let us establish that

  1. There exists a bijection between rationals, $\mathbb{Q}$ and natural numbers $\mathbb{N}$.
  2. Using the Peano Axiom's (and considering Gödel's incompleteness theorem), we can fundamentally describe the nature of $\mathbb{N}$.

Let us define $f : \mathbb{N} \to \mathbb{Q}$ as the bijection from naturals to rationals, and $q : \mathbb{Q} \to \mathbb{N}$ as the bijection from rationals to naturals.

Consider Peano's axioms (Also $s : \mathbb{N} \to \mathbb{N}$)

  1. $0 \in \mathbb{N}$
  2. For any $n \in \mathbb{N}$, $n = n$
  3. For any $a, b \in \mathbb{N}$, $a = b \iff b = a$
  4. For any $a, b, c \in \mathbb{N}$, if $a = b$ and $b = c$ then $a = c$
  5. For any $a \in \mathbb{N}$ if $a = b$ then $b \in \mathbb{N}$
  6. For any $a \in \mathbb{N}$, $s(a) \in \mathbb{N}$
  7. For any $a, b \in \mathbb{N}$, $s(a) = s(b) \iff a = b$
  8. There exists no $a \in \mathbb{N}$ such that $s(a) = 0$
  9. Let $N \le \mathbb{N}$ be a set. If $0 \in N$ and for all $n \in N$, then $s(n) \in N$, then we can say that $\mathbb{N} = N$.

Now consider that for all $n \in \mathbb{N}$ we can say that under $f$ there is a $q \in \mathbb{Q}$ such that $n \mapsto q$.

Considering all of the above, can a Peano-like system be constructed for rationals?

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    I really don't understand what this is asking – 2017-01-02
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    @mercio I'm trying to ask if an axiomatic system, similar to Peano's axioms can be created to fundamentally describe rationals. – 2017-01-02
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    When you talk about an axiom system for $\mathbb{Q}$ what operations, and structures do you want to have ? $+, \times, <, $ ? – 2017-01-02
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    These are not Peano's axioms. e.g. 2,3,4 are poperties of $=$ and 5 is also. For 6, need to say $s$ is a map from N into N. Missing is also the induction axiom. – 2017-01-02
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    but then why would a set-theory bijection between set-theory models of $\Bbb Q$ and $\Bbb N$ be important ? (also I am not sure if I would say that Peano axioms *fundamentally* represent $\Bbb N$ when it's an incomplete (hopefully) system) – 2017-01-02
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    @coffeemath (9) is the induction axiom, and defining equality over $\mathbb{N}$ is a part of Peano's axioms. – 2017-01-02
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    @EliSadoff Yes it is 9, for induction. I missed that one.... – 2017-01-02
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    @mercio My idea for the set theory bijection was $\forall n \in \mathbb{N}$ we can say if $f : \mathbb{N} \to \mathbb{Q}$ and if $n$ is described by Peano's axioms then $f(n) \in \mathbb{Q}$ would then be able to be described by a rational form of Peano's axioms. – 2017-01-02
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    @ReneSchipperus The operation would be $s\prime : \mathbb{Q} \to \mathbb{Q}$ which is equivalent to $f \circ s$ where $s : \mathbb{N} \to \mathbb{N}$ and $f : \mathbb{N} \to \mathbb{Q}$ – 2017-01-02
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    how is $17$ described by Peano's axioms and what does this say about $23/3$ ? – 2017-01-02
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    @mercio We can say $s^{17}(0) = 17$ using Peano's axioms (where $s^{n} $ is a shorthand for $s \circ s \circ s \cdots \circ s$, applied $n$ times). – 2017-01-02
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    @mercio That is my fundamental question. Because there must exist some function $s\prime : \mathbb{Q} \to \mathbb{Q} \equiv f \circ s$ where $f: \mathbb{N} \to \mathbb{Q}$ and $s : \mathbb{N} \to \mathbb{N}$. – 2017-01-02
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    well yeah you can make $\Bbb Q$ into a model of Peano arithmetic via any $f$ and so you can have $s'(23/3) = -17/5$ but what's the point ? – 2017-01-02
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    @mercio I realized in both of those comments I incorrectly described the type of $s\prime$, it should be $s\prime : \mathbb{N} \to \mathbb{Q}$. – 2017-01-02
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    @mercio The point would be to create a fundamental (yet incomplete) description of the rationals. – 2017-01-02
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    But for what? And there is no reason to since the bijection between naturals and rationals will not respect the arithmetic operations. – 2017-01-03

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