Find bijection $\Bbb R\setminus\Bbb Q \to \Bbb R$ 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.
 A: 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.
Now, apply Schröder–Bernstein theorem.
Does that help?
A: 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 $f(n\sqrt 2)=(2n-1)\sqrt 2.$
