Let $I \subseteq \mathbb{R}$ be an interval and $f: I \to \mathbb{R}$ a continuous function. We’ll say that $f$ is totally rational if the following propositions are true for any $x\in I$:

  1. If $x \in \mathbb{Q}$ then $f(x) \in \mathbb{Q}$

  2. If $f(x) \in \mathbb{Q}$ then $x \in \mathbb{Q}$

A simple example of such a function is the identity function $f(x)=x$. More generally any function of the form $f(x)=ax + b$ with $a,b\in \mathbb{Q}$ will do. Another class of functions that are totally rational are those of the form $$f(x)=\frac{ax + b}{cx + d}\qquad \text{with}\ a,b,c,d\in \mathbb{Q} \ \text{and}\ x\neq-\frac{d}{c}.$$

Besides functions of these kinds (and piecewise combinations thereof) I cannot find any other examples of such functions. It is easy to see, for instance, that any higher-order polynomial or rational function will fail condition (2).

But do other totally rational functions exist?


Here's a different type of example with the property:

$f(n + 0.a_1 a_2 a_3 \dots) = 0.0 a_1 0 a_2 0 a_3 \dots$


$f(n + \sum_{i \ge 1}{a_i \cdot 10^{-i}}) = \sum_{i \ge 1}{a_i \cdot 100^{-i}}$

In other words, the function takes the decimal expansion of the fractional part and inserts a $0$ between every digit. Or it writes it in base $10$ and reads it again in base $100$.

The function maps numbers with eventually periodic expansions (rational numbers) to numbers with eventually periodic expansions, and the converse is also true. Also it is continuous.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.