**Conjecture:**

For every irrational algebraic number $q$ and natural number $b$, the representation of $q$ on base $b$ contains all the digits $[0,\dots,b-1]$.

**Questions:**

Has this conjecture been proved, refuted or neither?

If proved:

Is there an estimate of the minimum length of $q_b$ containing all the digits?

For example, I would expect something like $2b$ or $b^2$ for any given $q_b$.

If not refuted:

I suppose that it is not true for transcendental numbers. Is that correct?

How can we construct a transcendental number $q_b$ which does not contain all the digits?

Thanks

Liouville's constant? Is that from the proof of the fact that some problems cannot be solved on a Turing machine? Isn't that number given in base $2$ (in which case, itdoescontain all the digits)? – barak manos Jul 6 '14 at 9:04initialsequences of digits: for the second point, as for any $b$ and any $n$ there are irrational algebraic numbers whose representation in base $b$ starts with $n$ zeros, there is no such bound. – Stefan Kohl Jul 6 '14 at 9:07