If $f$, for $x>0$, is a continuous function such that for all rationals $m=p/q$ with $\gcd(p,q)=1$, $f(m)$ is equal to the number of digits of $q$ (either base-2 or base-10), what is $f(\pi)$?
And 2) If $f(m)=q$, then what is $f(\pi)$?
If $f$, for $x>0$, is a continuous function such that for all rationals $m=p/q$ with $\gcd(p,q)=1$, $f(m)$ is equal to the number of digits of $q$ (either base-2 or base-10), what is $f(\pi)$?
And 2) If $f(m)=q$, then what is $f(\pi)$?
Such a function cannot possibly exist. To see this, first observe
$f\left(\frac{n+1}{n}\right)=\left\lfloor \log_b \, n \right\rfloor \xrightarrow{n\to\infty}\infty.$
All the while, $(n+1)/n\xrightarrow{n\to\infty}1$ but $f(1)=1\ne\infty$, which contradicts continuity.
This same example applies to case (2) as well.