In this [previous post](https://mathoverflow.net/q/412923/7113) I asked for the smallest set of continuous real functions that could generate $\mathbb Q$ by iteration starting from $0$. Surprisingly one continuous function suffices.


In the question I gave the example of three rational functions that generate $\mathbb{Q}$, $f(x)=1/x$, $g(x)=x+1$ and $h(x)=x-1$. It would be interesting to know if this is best possible and in particular whether one rational function can generate all of $\mathbb{Q}$:



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> Can $\mathbb{Q}$ be generated as the orbit of fewer than 3 rational functions?
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The question [Orbits of rational functions](https://mathoverflow.net/q/280256/7113) asks a more general question but I don't think explicitly answers it for $\mathbb{Q}$ itself.



Is $\mathbb{Q}$ the orbit of a rational function under iteration?
As was mentioned in the [comments](https://mathoverflow.net/questions/413087/is-mathbbq-the-orbit-of-a-rational-function-under-iteration#comment1058750_413087) by pregunton, it is possible to do using two rational functions. I claim it is not possible using just one. As Fedor Petrov suggests in [another comment](https://mathoverflow.net/questions/413087/is-mathbbq-the-orbit-of-a-rational-function-under-iteration#comment1058754_413087), this is because rational functions of degree higher than $1$ are never going to be surjective, which can be shown with help of [Hilbert's irreducibility theorem](https://en.m.wikipedia.org/wiki/Hilbert%27s_irreducibility_theorem). Indeed, take a rational function $\frac{f(x)}{g(x)}$ with coprime polynomials $f$, $g$ of which at least one has degree greater than $1$. The polynomial $h(x,t)=tf(x)-g(x)\in\mathbb Q[x,t]$ is irreducible then, so by Hilbert's theorem there are infinitely many values $q\in\mathbb Q$ for which $h(x,q)\in\mathbb Q[x]$ is irreducible. For all but one of these $q$, $h(x, q)$ will have degree $\max(\deg f,\deg g)>1$, so irreducibility implies it has no rational roots. Hence $q$ is not in the image of $\frac{f(x)}{g(x)}$.


The only case remaining is that of $\deg f,\deg g\leq 1$. In this case either $\deg g=1$ and the rational function has a rational pole, so its iteration can't go over all rationals, or else it is affine of the form $ax+b$ and it's easy to see explicitly its iterations do not cover all rationals.


