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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 $f(\pi)$?

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What makes you think such a function can even exist? –  anon Feb 3 '12 at 3:02
at least not a continuous one... –  yoyo Feb 3 '12 at 3:07
"and 2) If f(m)=2", what does it mean? –  user21436 Feb 3 '12 at 3:14
I dont know. It says "And 2) If f(m)=q" –  user24283 Feb 3 '12 at 3:18

1 Answer 1

up vote 11 down vote accepted

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.

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