Is there a bijection between $(0,1)$ and $\mathbb{R}$ that preserves rationality?

Solution 1:

$(1/x)-2$ on $(0,1/2]$ and $2-(1/(x-1/2))$ on $(1/2,1)$.

Solution 2:

$$ f(x) = \frac{2x - 1}{1 - |2x - 1|}. $$

Solution 3:

With the axiom of choice we can find well orderings of $\mathbb{R}$ and of $(0,1)$ such that the first $\omega$ elements are all the rationals of the set, then we can define our map to go from one well ordering to another by preserving the index (that is $a_\alpha\mapsto b_\alpha$, for $\alpha<2^{\aleph_0}$)

As discussed in the comments below by Colin and Jason (and myself), one does not need the axiom of choice for that. Using the Cantor-Schroeder-Bernstein theorem one can have two bijections, one from $(0,1)\setminus\mathbb{Q}$ to $\mathbb{R}\setminus\mathbb{Q}$ and one from $(0,1)\cap\mathbb{Q}$ to $\mathbb{Q}$, and define a bijection as needed without the use of the axiom of choice.