Is there a bijection from a bounded open interval of $\mathbb{Q}$ onto $\mathbb{Q}$? It is easy to create a bijection between two bounded open intervals of $\mathbb{R}$, such as:
$$
\begin{align}
 f : (a,b) &\to (\alpha,\beta) \\
 x &\mapsto \alpha+(x-a)(\beta-\alpha).
\end{align}
$$
It is also possible to biject a bounded open interval of $\mathbb{R}$ onto the whole of $\mathbb{R}$, e.g.:
$$
\begin{align}
 f : (a,b) &\to \mathbb{R} \\
 x &\mapsto \tanh^{-1} x.
\end{align}
$$
Consider now the set of rational numbers $\mathbb{Q}$. The bijection between two bounded open intervals still holds, but is it possible to biject:
$$
f : (p,q) \to \mathbb{Q}
$$
where $p,q\in\mathbb{Q}$ and $(p,q) = \{x\in \mathbb{Q} : p < x < q\}$?

Start: Pick any monotonic decreasing sequence $l_i$ of positive rationals converging to zero and any monotonic increasing sequence $r_i$ of positive rationals converging to $1$, with $l_1<r_1$. Consider
$$\begin{array}{lllll} \Bbb Q= & \cdots & \sqcup & [2,-1) & \sqcup & [-1,0) & \sqcup &  [0,1) & \sqcup & [1,2) & \sqcup & [2,3) & \sqcup & \cdots \\
(0,1) = & \cdots & \sqcup & [l_3,l_2) & \sqcup & [l_2,l_1) & \sqcup & [l_1,r_1) & \sqcup & [r_1,r_2) & \sqcup & [r_2,r_3) & \sqcup & \cdots \end{array} $$
Note that $[a,b)\cong[c,d)$ are order-isomorphic for any rationals $a,b,c,d\in\Bbb Q$ with $a<b$, $c<d$.
All intervals are taken in $\Bbb Q$.
