I noticed that none of the "nice" sigmoid functions I know of sends rationals to rationals.
I set out to find a sigmoid function that maps rationals into rationals. I did find one, but it is not smooth. I wonder if such a function exists, and if not, what is the connection between this non-existence and the requirement that it and its inverse map rationals to rationals.
The function
$$
x \mapsto
\begin{cases}
\frac{1}{2 (1 – x)} &, \text {if} \ \ x < 0 \\
\frac{2 x + 1}{2 (1 + x)} &, \text {if} \ \ x \geq 0
\end{cases}
$$
is a strictly increasing bijection from $\mathbb{R}$ onto $(0, 1)$, and it maps $\mathbb{Q}$ into $\mathbb{Q}\cap(0, 1)$.
Furthermore, its inverse
$$
y \mapsto
\begin{cases}
\frac{2 y – 1}{2 y} &, \text {if} \ \ 0 < y < \frac{1}{2} \\
\frac{2 y – 1}{2 (1 – y)} &, \text {if} \ \ \frac{1}{2} \leq y < 1
\end{cases}
$$
maps $\mathbb{Q}\cap(0, 1)$ into $\mathbb{Q}$.
Therefore, the restriction of this function to $\mathbb{Q}$ is a monotone bijection from $\mathbb{Q}$ onto $\mathbb{Q}\cap(0, 1)$.
The original function (over $\mathbb{R}$) is not smooth, however: its second derivative does not exist at $x = 0$.
Does there exist a smooth monotone bijection from $\mathbb{R}$ onto $(0, 1)$ whose restriction to $\mathbb{Q}$ maps $\mathbb{Q}$ onto $\mathbb{Q}\cap(0, 1)$? Is it possible to write down a formula for it?
EDIT: Below is the function one gets if one subtracts $\frac{1}{2}$ from the expressions in the original definition.
$$
x \mapsto
\begin{cases}
\frac{x}{2 (1 – x)} &, \text {if} \ \ x < 0 \\
\frac{x}{2 (1 + x)} &, \text {if} \ \ x \geq 0
\end{cases}
$$
(Informally, this "translates" the graph of the function down by $\frac{1}{2}$, so that the function's value at $0$ is $0$.)
This "translate" of the original function more clearly shows its symmetry.
EDIT2: I realized after the fact that the word "monotone" in the phrase "smooth monotone bijection" is probably redundant. I cannot envision a smooth bijection on $\mathbb{R}$ that fails to be monotone. (The sign of the first derivative must remain bounded away from zero if the function is going to be injective.)
Best Answer
First of all, by taking any diffeomorphism between ${\mathbb R}$ and $(0,1)$ and using it to send ${\mathbb Q}$ to a subset $B\subset {\mathbb R}$, the problem reduces to the following:
Given two countable dense subsets $A, B\subset {\mathbb R}$, find a diffeomorphism $f: {\mathbb R}\to {\mathbb R}$ such that $f(A)=B$.
Remark. A diffeomorphism between smooth manifolds is smooth map with smooth inverse. In the setting of maps between open intervals in ${\mathbb R}$, diffeomorphisms are bijective functions which are infinitely differentiable and have infinitely differentiable inverses.
This turns out to be always possible. One can even find a real-analytic diffeomorphism with this property. Even more, there exists an entire holomorphic function $h: {\mathbb C}\to {\mathbb C}$ which restricts to a diffeomorphism $f: {\mathbb R}\to {\mathbb R}$ such that $f(A)=B$.
This result has a long and interesting history. It is usually attributed to
Franklin, P., Analytic transformations of everywhere dense point sets., Transactions A. M. S. 27, 91-100 (1925). ZBL51.0166.01.
See also a more recent proof in
Barth, K. F.; Schneider, W. J., Entire functions mapping countable dense subsets of the reals onto each other monotonically, J. Lond. Math. Soc., II. Ser. 2, 620-626 (1970). ZBL0201.09203.
The same result holds in higher dimensions, see
Morayne, Michał, Measure preserving analytic diffeomorphisms of countable dense sets in ${\mathbb C}^n$ and ${\mathbb{R}}^n$, Colloq. Math. 52, 93-98 (1987). ZBL0627.28012.
and even for Banach manifolds:
Dobrowolski, T., On smooth countable dense homogeneity, Bull. Acad. Pol. Sci., Sér. Sci. Math. Astron. Phys. 24, 627-634 (1976). ZBL0332.58005.
When I have more time I will add a sketch of a proof for existence of $C^\infty$ diffeomorphisms from a smooth manifold $M$ to itself mapping given dense countable subset $A\subset M$ to another given dense countable subset $B\subset M$.