As an example, we can easily go from ##0## to ##-1/3##. Indeed, we can apply ##T## to ##0## to go to ##1##, we apply ##T## to go to ##2##, we apply ##T## to go to ##3##, and then we apply ##R## to go to ##-1/3##.

Well ##R^{-1} = R## and ##T^{-1} = RTRTR## so that shows we can invert any combination.
Secondly, observe
##T^c R T^b R T^a (0) = c - \frac{1}{b - \frac{1}{a}},##
and in general
##T^{k_1} R T^{k_2} \ldots T^{k_n}(0) = k_1 - \frac{1}{k_2 - \frac{1}{\ldots - \frac{1}{k_n}}}##
We get something similar to continued fractions, the only difference being the subtractions instead of additions. But every positive rational number can be written in this form for the same reason that every rational can be written as a continued fraction (just change "integer part" to "integer part + 1" in the algorithm). So we can map 0 to any positive rational. R'ing that means we get every negative rational.

Hmm, there's probably something left to prove. Oh well.

Posting an answer would be kind of cheating, since I've encountered it before. Someone might be interested to know that T and R are generators of the modular group [itex]\Gamma[/itex] of transformations of the upper half-plane. Answering the question amounts to proving that [itex]\Gamma[/itex] acts transitively on [itex]\mathbb{Q}[/itex].

Further insight into micro's hatred of Wikipedia and arithmetic. Clearly we have a group whose action on 0 generates Q. As pwsnafu points out we can use continued fractions. It was quite a macro oversight not to provide the reader with
$$T^{-1}=RTRTR=x-1$$
Don't you like to subtract?

The following algorithm takes any rational number to a natural number:

1. If a/b is a natural number, stop.
2. Else if a/b < 0, apply T.
3. Else if a/b > 0, apply R.
4. Repeat.

The only way this could fail to terminate is if it gets stuck in a rule or cycles continuously. Getting stuck in rule 2 is not possible by the Archimedean property of the integers. Getting stuck in rule 3 is not possible because it takes positive numbers to negative numbers. Cycling from rule 2 to 3 and back again is not possible because the first positive number after a chain of rule T's will be < 1, therefore rule R will reduce the size of the denominator. Eventually, it'll reduce it to 1. This is sufficient to show that the algorithm terminates.

To extend this, given any two rational numbers, apply the algorithm to each giving two natural numbers N < M. Apply T finitely many times to transform N into M. Then, if R and T are invertible, invert the process that generated M and voila.

Not sure why you think I should give a partial solution in a challenge question...

Anyway, very big congratulations to pwsnafu for giving such a short and elegant solution. And of course, a congratulations to verty as well for providing a more elementary solution. I hope you enjoyed it.