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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}

CompuPhysics
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    If you want actual examples, which are in some sense "smooth" I'd start here – lulu Mar 04 '23 at 15:58
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    here is a near duplicate. – lulu Mar 04 '23 at 16:02
  • To be clear: I seriously doubt that there is a "pleasant" solution. The methods discussed in the first link I provided give you something, but it sure isn't pretty. If there is something sensible, please post it as an answer here. I'd like to see it. – lulu Mar 04 '23 at 16:10
  • So, post the solution for rational $a,b$. I think that would be of interest. – lulu Mar 04 '23 at 17:09

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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.

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