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I would like to ask you about the following question. It is conjectured that every algebraic irrational number is normal (absolutely normal). I know the result by Bugeaud and Adamczewski about the non-linearity of the complexity function. Unfortunately, this was not enough to solve the question I have in mind. But, my question is the following:

Any algebraic irrational number contains at least one 0 in its expansion in basis $b$ for all $b\geq 2$ sufficiently large (or at least for $b$ in the form $10^s$, for every $s$ sufficiently large)?

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  • $\begingroup$ Do you really mean base $b$ rather than $b$-adic? $\endgroup$ – Robert Israel Jan 8 at 4:48
  • $\begingroup$ @RobertIsrael thanks! $\endgroup$ – Jeremy Jan 8 at 5:01
  • $\begingroup$ Are you asking whether this assertion has been proved? To my knowledge, it has not. $\endgroup$ – Greg Martin Jan 8 at 8:44
  • $\begingroup$ Do you know this article by Michel Waldschmidt: arxiv.org/abs/0908.4034? Look at Conjecture 1.1. $\endgroup$ – Kurisuto Asutora Jan 8 at 8:48
  • $\begingroup$ I'm pretty sure it is not known whether, say, $\sqrt{2}$ contains infinitely many zeros in its decimal expansion. If not, you could very easily make a quadratic irrational with no zeros past the decimal point. $\endgroup$ – Wojowu Jan 8 at 12:26

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