# Digits in an algebraic irrational number

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).