Can you example a Bijective correspondence from all irrational numbers to irrational numbers of any open interval (for example (0, 1) interval)? Is it even possible?
\begin{equation} f: \mathbb{Q}^{c} \rightarrow \mathbb{Q}^{c} \cap (a, b) \\ \end{equation}
This answer is only for when a and b are both rational.
Consider this function and call it f. f is defined on (0, 1).
It can be shown that the sum and product of rational and (rational/irrational) are (rational/irrational).
so define g(x) := (x - a) / (b - a) since a and b - a are both rational, we can say fog(x) is still smooth. So fog(x) is the answer for this case.