Is it true, that $\mathbb{R} \times \mathbb{Q} \sim \mathbb{R}$? Just like in title, is it true that $\mathbb{R} \times \mathbb{Q} \sim \mathbb{R}$?
My answer would be yes, since $\mathbb{R}^2 \sim \mathbb{R}$.
 A: If we suppose that $\Bbb R \sim \Bbb R^2$, then by Cantor-Schröder-Bernstein: it suffices to find an injection from $\Bbb R$ to $\Bbb R \times \Bbb Q$, and an injection from $\Bbb R \times \Bbb Q$ to $\Bbb R \times \Bbb R$.
A: If by $\sim$ you mean "there is a bijection between", then the answer is yes, and your argument works fine through Cantor Bernstein's theorem : there is an obvious injection $\mathbb{Q}\to \mathbb{R}$ and so a sequence of injections as follows :
$\mathbb{R}\to \mathbb{R}\times \mathbb{Q} \to \mathbb{R}\times \mathbb{R} \to \mathbb{R}$.  
By Cantor-Bernstein, all these injections "can become" (I'll let you make that precise) bijections
A: Without going through $\mathbb R\times\mathbb R$:


*

*Inject $\mathbb R$ into $(0,1)$ with $f(x)=\frac{1}{2}+\frac{1}{\pi}\arctan x$

*Inject $(x,p/q)\in\mathbb R\times\mathbb Q$, with $\gcd(p,q)=1$ and $q>0$ into $2^p(2q+1)+f(x)$ (if $p<0$, then inject it into $-2^{|p|}(2q+1)+f(x)$


So you get an injection $\mathbb R\times\mathbb Q\to\mathbb R$. The other injection is trivial. By Cantor-Bernstein, they have the same cardinality.
