Countability of $\mathbb{Q}$ I have seen a few proofs which shows the countability of rationals (denoted $\mathbb{Q}$). But they always involve picking a representation for the rationals (like fractions,Calkin-Wilf trees) and then showing a bijection between $\mathbb{N}$ and that representation. Is there some sort of canonical bijection between $\mathbb{Q}$ and $\mathbb{N}$?
P.S By canonical bijection, I mean one which doesn't rely on a specific representation of $\mathbb{Q}$.
 A: If you really want to define the rationals without using fractions, you could define them as the smallest subset of $\mathbb R$ that contains $1$ and is closed under the four basic arithmetic operations.
(Here we need to explicitly ignore the circular feeling that we need the rationals in order to be sure $\mathbb R$ exists in the first place. If we put our minds to it, we could probably find some way to side-step that -- for example by defining $\mathbb R$ as the completion of the ordered ring of terminating decimal fractions, and then proving separately that what we get is a complete ordered field).
Then we can construct $\mathbb Q$ as follows: Let $f$ be a function that maps finite subsets of $\mathbb R$ to other finite subsets of $\mathbb R$ as follows:
$$ \begin{align} f(A) = A & \cup \{ a+b \mid a,b\in A \} \cup \{ a-b \mid a,b \in A \}
\\& \cup \{ a\times b \mid a,b\in A \} \cup \{ a\div b \mid a,b \in A, b\ne 0 \} 
\end{align}$$
Then by our definition above we can show
$$ \mathbb Q  = f(\{1\}) \cup f(f(\{1\})) \cup \cdots = \bigcup_{n\ge 1} f^n(\{1\}) $$
Since a countable union of finite sets is countable, $\mathbb Q$ must be countable.
However, one might very well object that this is really just a thin disguise of choosing arithmetic expressions as "representations" of the rationals.
