Peano Equivalent for Rationals

Question

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?