Given $x,y\in\mathbb R$ is there a "formulaic" way to obtain a $q\in\mathbb Q$ with $a<q<b?$

Question

Is there an assignment of reals $x,y$ to a rational number $q(x,y)$ for which

$$\forall_{\mathbb R} x.\forall_{\mathbb R}(x<y).\left(x<q(x,y)<y\right)\hspace{.2cm}?$$

For computable reals, using the definition, I can show existence of $F$ by averaging the upper and lower bounds for $a$ and $b$, respectively. In fact I can do it practically by truncating the first decimal where $a$ and $b$ differ and trail with a lot of zeroes and then a $1$.

Maybe the answer is no due to undecidability issues of $\le$ for $a,b$ in the first place?

Answer 1

If you want a definite arithmetic-looking formula you could do something like:

Let $n = 1+\left\lfloor \frac1{b-a} \right\rfloor$ -- then there will be at least one rational with denominator $n$ strictly between $a$ and $b$.

Now $\displaystyle\frac{1+\lfloor n a \rfloor}{n} $ will be the smallest such rational.