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Just like in title, is it true that $\mathbb{R} \times \mathbb{Q} \sim \mathbb{R}$?

My answer would be yes, since $\mathbb{R}^2 \sim \mathbb{R}$.

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    What does the $\sim$ mean in this instance? – 2017-02-08
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    What relation is $\sim$ here? Are you asking whether the two sets have the same cardinality? In that case the answer is yes (but it is also yes in the category of $\Bbb Q$-vector spaces). – 2017-02-08
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    Isomorphic I think – 2017-02-08
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    Yes, $\sim$ here means, that both sets have the same cardinality. – 2017-02-08
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    This is also answered (without proof) [on wikipedia](https://en.wikipedia.org/wiki/Cardinal_number#Cardinal_multiplication). – 2017-02-08

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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 $\Bbb R \times \Bbb R$.

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    @TryingToGraduate But the injections are totally trivial! $x\mapsto(x,0)$ and $(x,y)\mapsto(x,y)$. – 2017-02-08
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    @TryingToGraduate a good hint teaches more than a complete answer – 2017-02-08
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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

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Without going through $\mathbb R\times\mathbb R$:

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