It is not possible.
It is consistent with set theory without choice that $\mathbb R/\mathbb Q$ has strictly larger cardinality that $\mathbb R$. (This looks counter-intuitive, since $\mathbb R/\mathbb Q$ is a quotient.)
This is the case because, using a fact that goes back to Sierpiński (Sur une proposition qui entraîne l’existence des ensembles non mesurables, Fund. Math. 34, (1947), 157–162. MR0023318 (9,338i)), in any model of $\mathsf{ZF}$ where all sets of reals have the Baire property, it is not even possible to linearly order $\mathbb R/\mathbb Q$.
(Sierpiński argues in terms of Lebesgue measure. The argument in terms of the Baire property is analogous, and has the additional technical advantage of not requiring any discussion of consistency strength issues.)
A couple of years ago, Mike Oliver gave a nice talk on this topic (How to have more things by forgetting where you put them); he is not exactly using $\mathbb R/\mathbb Q$, but the arguments easily adapt. The point of the talk is precisely to give some intuition on why we expect the quotient to be "larger".
[Of course, in the presence of choice, the two sets have the same size. The argument above shows that the use of choice is unavoidable.]
One of the tries is a bijection $f:\mathbb N\to(0,1]\cap\mathbb Q$ where $f(0)=1$ and then if $f(n)=\frac{p_k}q$ for $\gcd(p_k,q)=1$ with $p_k<q-1$ then $f(n+1)=\frac{p_{k+1}}q$ for $\gcd(p_{k+1},q)=0$ and $p_{k+1}>p_k$ over the set of proper coprimes of $q$: $\{p_k\in\mathbb N: 0<p_k<q, \gcd(p_k,q)=1\}$, and if $f(n)=\frac{q-1}{q}$ then $f(n+1)=\frac{1}{q+1}$.
I don't quite remember but this injection of $\mathbb N\to(0,1]\cap\mathbb Q$ had a few direct formulas. And it was also surjective.
This injection can be extended for $g:\mathbb N\to\mathbb Q^+$ by $g(2k)=f(k)$ and $g(2k+1)=1/f(k)$.
Best Answer
Hint: Fix $a_n$ as a sequence of irrational numbers, and write $\mathbb Z=\{z_n\mid n\in\mathbb N\}$. Define a function which sends $a_n$ to $a_{2n}$; $z_n$ to $a_{2n+1}$; and $x$ to itself otherwise.