Find bijection $\Bbb R\setminus\Bbb Q \to \Bbb R$ [duplicate]

Find bijection $$\Bbb R\setminus\Bbb Q \to \Bbb R$$
I've tried using Schröder–Bernstein theorem to show a 1-1 function in both directions. But I only succeed to prove one direction. Explicit function seems much harder to prove. thanks for any help.

• Did you mean to write $\Bbb R\setminus\Bbb Q$? Jun 27, 2021 at 18:30
• yes I'm sorry.. Jun 27, 2021 at 18:30
• Maybe you should try to find a bijection $\mathbb{R}\setminus \mathbb{N}\to\mathbb{R}$ first, and see if that helps, which I do not know, but I think it helps and should be easier. Maybe you can then generalize that bijection. Or even simpler: Start by finding a bijection $\mathbb{R}\setminus\{x\}\to\mathbb{R}$ for some arbitrary $x\in\mathbb{R}$. Jun 27, 2021 at 18:31
• I was asked to prove it specific with Schröder–Bernstein theorem. Jun 27, 2021 at 19:02

We know that all irrational numbers have a unique infinite continued fraction. So, take any irrational number $$r$$ and consider its continued fraction $$\begin{equation*} r=a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+\frac{1}{\dots}}}} \end{equation*}$$ Map $$r$$ to the real number $$0.a_1a_2a_3\dots$$. Clearly, this is an injection from $$\mathbb{R}\backslash\mathbb{Q}\to \mathbb{R}$$.
Now, consider the famous injection from $$\mathbb{R}\to \mathbb{R}\backslash\mathbb{Q}$$ which maps $$q+n\sqrt{2}$$ to $$q+(n+1)\sqrt{2}$$ for $$q\in \mathbb{Q}$$, $$n\in \mathbb{N}$$ and maps the rest to itself.
Take a countably infinite $$\Bbb S\subset \Bbb R\setminus \Bbb Q .$$ Now $$\Bbb S\cup\Bbb Q$$ and $$\Bbb S$$ are countably infinite, so take a bijection $$f:\Bbb S\cup\Bbb Q \to \Bbb S.$$ Extend the domain of $$f$$ to $$\Bbb R$$ by letting $$f(x)=x$$ for $$x\in \Bbb R\setminus (\Bbb S\cup \Bbb Q).$$ Then $$f:\Bbb R\to \Bbb R\setminus \Bbb Q$$ is a bijection.
E.g. let $$g:\Bbb Z^+\to \Bbb Q$$ be a bijection and let $$\Bbb S=\{n\sqrt 2\,:n\in\Bbb Z^+\}.$$ For $$n\in\Bbb Z^+$$ let $$f(g(n))=2n\sqrt 2$$ and let $$f(n\sqrt 2)=(2n-1)\sqrt 2.$$