Is there a rational surjection $\Bbb N\to\Bbb Q$?

Question

The question is in the title. Is there a one-dimensional rational function $f\in\Bbb R(X)$ which restricts to $\Bbb N\to\Bbb Q$, which is a surjection onto $\Bbb Q$? My guess is no.

Expanding the scope a little (at the expense of precision), are there any "nice" functions that enumerate $\Bbb Q$? Here "nice" is meant to exclude the floor function or absolute value function and related trickery. At first I thought it might work to use analytic functions here, but there is an analytic function taking any chosen values on $\Bbb N$ subject to a mild growth hypothesis (I think $f(n)\in o(n)$), by defining $f(z)=\sum_{n\in\Bbb N}a_n{\rm sinc}(\frac{z-n}\pi)$, where ${\rm sinc}(z)=\frac{\sin(z)}z$ continuously extended over zero. I will let answerers supply their own definitions of "nice" if they want to tackle the broader question.

Answer 1

At least for the question of a rational function $\Bbb N\to\Bbb Q$, the answer is no. In fact, we can make a stronger claim:

If $f:\Bbb N\to\Bbb R$ is a rational function, then the range of $f$ is either bounded above or bounded below.

Proof: Separate the highest order term of $f$, writing it as $f(x)=ax^n+g(x)$ where $g\in o(x^n)$ for some $n\in\Bbb Z$. If $a\ge 0$, then we claim $f$ is bounded below (and by symmetry, $f$ is bounded above if $a\le 0$).

By the definition of little-O, there is some $N\in\Bbb N$ such that for all $x>N$, $|g(x)|\le ax^n$. Then $f(x)\ge0$, so $f(x)\ge\min\{f(0),f(1),\dots,f(N)\}$.

Edit: Alternatively, we can use Arnaud's suggestion: the derivative of $f$ is also a real rational function, which only changes sign at the roots of the numerator and denominator. Thus $f$ is monotone above the largest root of the derivative (let's say $f$ is increasing, the other case is covered by symmetry), so $f$ is eventually bounded below, and hence bounded below by the same argument as above.

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The analytic function example in the OP can be extended to functions with any polynomial growth rate. Define ${\rm sincz}(z)={\rm sinc}(\frac z\pi)=\frac\pi z\sin(\frac z\pi)$. This function has the property ${\rm sincz}(0)=1$ and ${\rm sincz}(n)=0$ for all other $n\in\Bbb Z$. Additionally, since $|\sin(x)|\le 1$ on $\Bbb R$, $|{\rm sinc}(x)|\le\frac 1x$ and $|{\rm sincz}(x)|\le\frac\pi x$. To achieve even higher convergence rates, we can take ${\rm sincz}^n(x)$, which also has the same behavior on integers but is $O(x^{-n})$.

The convergence rate is necessary to ensure that $\sum_{n\in\Bbb N}a_n{\rm sincz}^m(z-n)$ converges. If $a_n$ is $O(n^k)$, then $a_n{\rm sincz}^m(z-n)$ is $O(n^{k-m})$, so the sum converges for $m>k+1$. Then as a uniformly convergent sum (uniformly because it is using a global bound on ${\rm sincz}$) of analytic functions, it is itself analytic, and takes the value $f(n)=a_n$ at each integer. Needless to say, most constructed bijections $\Bbb N\to\Bbb Q$ have polynomial growth, usually $O(n^{1/2})$, so this shows that there is an analytic function whose restriction to the natural numbers is a bijection to $\Bbb Q$, although it is quite contrived and does not really "simplify" the expression.

Answer 2

This is an alternative argument based on Arnaud's suggestion. Without loss of generality, you can assume the rational function $f$ is $\frac{p}{q}$ where $n = \deg(p) \ge \deg(q) = m$ (otherwise trade $f$ in for $\frac{1}{f}$). But then (using $(\frac{p}{q})' = \frac{p'q - pq'}{q^2}$) you have: $$ f'(x) = \frac{(n-m)p_nx^{m+n-1} + \mbox{terms of lower degree}}{x^{2m} + \mbox{terms of lower degree}} $$ where, without loss of generality, I have assumed $q$ is monic and where $p_n \neq 0$ and $n > 0$ (if $n = 0$, $f$ is constant). So as $x$ tends to $\pm\infty$, $f'(x)$ tends to $\pm\infty$, if $n > m+ 1$, and to $p_n \neq 0$ if $n = m + 1$. So $f(x)$ is monotonic whenever $|x|$ is sufficiently large if $n > m$. If $m = n$, then as $x$ tends to $\pm\infty$, $f(x)$ tends to $\pm p_n$, so $f(x)$ is bounded. But if $f(x)$ is either monotonic for large enough $|x|$ or bounded, then $f$ cannot map $\Bbb{N}$ onto $\Bbb{Q}$.

Answer 3

If nice is taken to mean analytic function there sure is. There exists a result from complex analysis that states that if $\lim_{n\to\infty} c_n=\infty$ and $a_n$ is a sequence of complex number then there's an analytic functionn $f$ such than $f(c_n) = a_n$. Now let $c_n = n$ and $a_n$ be an enumeration of $\mathbb Q$ then this result guarantees the existence of such a function.