# Is every irrational number containing only $2$ distinct digits, transcendental?

If we have an irrational number, consisting of only $2$ distinct digits, for example:

$$0.01011011101111011111 \cdots$$

Can we conclude that the number is transcendental?

It is conjectured that every irrational algebraic number is normal in base $10$. This would imply that the answer to my question is yes. But can we prove it?

• This can well be a duplicate, if so, I apologize in advance. – Peter Jan 20 '17 at 20:25
• @DanUznanski I don't really think this constitutes killing a mosquito with a dubious howitzer; rather, the OP is observing that their question can't have an easy negative answer, which is useful data for someone trying to solve it. – Noah Schweber Jan 20 '17 at 20:43
• – Noah Schweber Jan 20 '17 at 20:53
• Well not in base $2$, and it might be that base $3$ is an easier place to start, because the Cantor Set is well studied. – Mark Bennet Jan 20 '17 at 21:02
• This problem is wide open. As far as we know, all algebraic irrationals could have expansions which eventually only use two digits, in all bases. – Wojowu Jan 20 '17 at 22:35

You may want to look at this answer in mathoverflow. Our conditions are not strong enough to use the theorem though, as we just have $c_x(n)\leq 2^n$.
• Can you explain the complexity in this context ? When can we conclude that a number is either rational or transcendental with the complexity ? In particular, is $0.ababbabbbabbbbabbbbb\cdots$ with distinct digits $a$ and $b$ always transcendental ? – Peter Jan 20 '17 at 22:02
• when there is a $k$ such that the number of distinct blocks of length $n$ that appear in the word is less than $kn$ for all $n$. – Jorge Fernández Hidalgo Jan 20 '17 at 22:04
• @Peter that one wont work. The number of blocks of length $n$ is not bounded by $kn$ for any $k$. – Jorge Fernández Hidalgo Jan 20 '17 at 22:06
• no, a block of length $n$ is a substring of length $n$. For example, the number $0.00000000$ has only one block of every length. – Jorge Fernández Hidalgo Jan 20 '17 at 22:11
• So, the distinct blocks of length $3$ in my example would be $bbb,abb,bab,bba,aba$ ? – Peter Jan 20 '17 at 22:13