I am trying to solve a problem and I got a conditional result related to normality of algebraic irrational numbers (Borel conjecture).

I know that by using Ridout theorem or Schmidt subspace theorem is possible to find a good lower bound for the number of nonzero "digits" in the $g$-ary expansion of an algebraic irrational number (for any basis $g\geq 2$).

However, my question is in the opposite direction:

Is it possible to prove that every algebraic irrational number has at least one 0 in its $g$-ary expansion, for all sufficiently large $g\geq 2$?

Of course, if this statement is true, then it is possible to prove that, in fact, there are infinitely many $0$'s in its $g$-ary expansion (by multiplying the algebraic number for some convenient power of $10$).

Any help will be welcomed.


What is known is that every real irrational has a $0$ in its $g$-ary expansion for infinitely many $g$. WLOG take $0 < x < 1$. Taking an even-numbered convergent of the continued fraction of $x$ gives us a rational $p/q$ such that $$\frac{p}{q} < x < \frac{p}{q} + \frac{1}{q^2}$$ so that the first two digits in the base-$q$ expansion of $x$ are $p$ and $0$.

Of course, this is a far cry from all sufficiently large $g$ (which is surely true, but I very much doubt it's provable in the current state of the art).


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