**Notation:**
$$ (0;1)\ :=\ \{x\in\mathbb R:\ 0<x<1\}$$




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There are simple rational stretches $\ f \colon (0; \, 1)\to\mathbb R,\ $ e.g. let $\ s\in(0;\, 1);\ $ then
$$ f(x)\ :=\ \frac{1-s}{1-x}-\frac sx\ $$
is an increasing bijection $\ f \colon (0;\,1)\to\mathbb R\ $ such that $\ f(s)=0.$


In the other direction, there are rational surjections, such as $\ g \colon \mathbb R\to(0; \, 1]\ $ given by
$\ g(x)=\frac 1{1+x^2}.\ $



> 
> **Question.** Does there exist a rational bijection $\ b \colon \mathbb R\to(0; \, 1)\ $?
> 
> 
> 


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EDIT: A stronger theorem holds:


**Theorem:**  Let $\ -\infty<a<b<\infty.\ $ Then there does not exist any surjective rational function $\ s:\mathbb R\to (a;b)$ onto the open interval $(a;b).$


**Proof:**  Function $s:\mathbb R\to(a;b),\ $ under the given assumptions, would admit a continuous extension defined in the projective line $ \mathbb P:=\mathbb R\cup\{\infty\},$
$$ q : \mathbb P\to[a;b] $$
Since $\mathbb P\ $ is compact, the image of it is a closed interval, actually


$$ q(\mathbb P) = [a;b] $$


However


$$ s(\mathbb R)\ \supseteq\ [a;b]\setminus \{q(\infty)\} $$


It follows that $\ a\in s(\mathbb R)\ $ or $\ b\in s(\mathbb R)\ $ -- a contradiction. **End of proof**



Rational bijections $\mathbb R\to(0;1)$
No. It should be monotone (as injective continuous function on the real line), but the limits on $\pm \infty$ are equal.


