Fix an algebraic number $\beta$ and consider a complex number $\alpha$ which admits multiple representations in base $\beta$. If one representation of $\alpha$ is ultimately periodic, must every other representation of $\alpha$ be ultimately periodic?

Bonus question: Does this depend on the choice of the set of digits?

Edit: John Bentin showed that this is false in general, even with $\beta \in \Bbb{Q}$. Is this true at least for $\beta$ algebraic integer?

Edit 2: This question came up when trying to solve this problem of mine.



All I know so far is that if $\beta$ is a Pisot integer, i.e. if it is a real algebraic integer greater than $1$ with every conjugate lying in the unit circle, then $\alpha \in \Bbb{Q}(\beta)$ if and only if it has an ultimately periodic expansion (which is stronger than my question).


Question: Is "ages" used only in British English?
Answer: In any dialect, it is informal when used hyperbolically, as in “I waited for ages”, or “It's been ages since I saw you”.  

From ngrams for ages since I saw,waited for ages in a British English corpus, and from similar ngrams in an American English corpus, it's evident that these two expressions are used slightly more frequently in BE than in AE.  I don't recall hearing an American saying something like  “I waited for ages”, but I think I have heard “It's been ages since I saw you” numerous times.