# Conjecture on irrational algebraic numbers

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:

1. Has this conjecture been proved, refuted or neither?

2. 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 $q_b$.

3. 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

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For the last point: yes, as you seem to know, it fails for transcendental numbers: for example, Liouville's constant is transcendental, but only has $1$'s and $0$'s in its decimal expansion. For more information, read about general Liouville numbers. –  Geoff Robinson Jul 6 at 9:00
@GeoffRobinson: 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, it does contain all the digits)? –  barak manos Jul 6 at 9:04
Assuming you are talking about initial sequences 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 at 9:07
The Liouville's constant I refer to is $\sum_{n=1}^{\infty}10^{-n!}.$ –  Geoff Robinson Jul 6 at 9:55
I just want to remark since no one has mentioned it yet: there is a much stronger conjecture that every irrational algebraic real number is normal in every base. –  Bill Mance Jul 6 at 19:46

The conjecture has been neither refuted nor proved. The state of the art, as far as I know, is contained in the papers of Adamczewski and Bugeaud, in which they show that anything with a very low complexity decimal expansion cannot be an algebraic irrational. The complexity is the function $c_x(n)$ giving the number of blocks of length $n$ in the decimal expansion of $x$ (or any base). They show that if there exists a $k$ such that $c_x(n)\le kn$ for all $n$, then $x$ is either rational or transcendental. Of course, it's conjectured that $c_x(n)=10^n$ for all algebraic irrationals $x$. Your condition would be implied by the conjecture $c_x(n)>9^n$ for all algebraic irrationals $x$.

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You might also look at questions 114905, 114758, 99039, where very similar questions are asked. –  Anthony Quas Jul 6 at 13:12
Thanks. Did you mean "cannot be irrational algebraic"? –  barak manos Jul 6 at 13:14
right - fixed now. –  Anthony Quas Jul 6 at 13:17
For this to be relevant to the conjecture, the number you constructed would have to be algebraic, but it is transcendental because it is $2/9 - \sum_{n=1}^\infty (1/\sqrt{10})^{n(n+1)}$. The latter term is related to a theta function value and was proved to be transcendental. projecteuclid.org/… –  Douglas Zare Jul 6 at 9:49
Has this been done for $\sqrt D$ in arbitrary base? –  Lev Borisov Jul 6 at 12:20