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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?

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  • $\begingroup$ This can well be a duplicate, if so, I apologize in advance. $\endgroup$
    – Peter
    Jan 20, 2017 at 20:25
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    $\begingroup$ @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. $\endgroup$
    – Noah Schweber
    Jan 20, 2017 at 20:43
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    $\begingroup$ @Peter A household pest control device. $\endgroup$
    – Noah Schweber
    Jan 20, 2017 at 20:53
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    $\begingroup$ 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. $\endgroup$
    – Mark Bennet
    Jan 20, 2017 at 21:02
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    $\begingroup$ 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. $\endgroup$
    – Wojowu
    Jan 20, 2017 at 22:35

1 Answer 1

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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 $c_x(n)\leq 2^n$.

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  • $\begingroup$ 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 ? $\endgroup$
    – Peter
    Jan 20, 2017 at 22:02
  • $\begingroup$ 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 $n$. $\endgroup$
    – Asinomás
    Jan 20, 2017 at 22:04
  • $\begingroup$ @Peter that one wont work. The number of blocks of length $n$ is not bounded by $kn$ for any $k$. $\endgroup$
    – Asinomás
    Jan 20, 2017 at 22:06
  • $\begingroup$ 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. $\endgroup$
    – Asinomás
    Jan 20, 2017 at 22:11
  • $\begingroup$ So, the distinct blocks of length $3$ in my example would be $bbb,abb,bab,bba,aba$ ? $\endgroup$
    – Peter
    Jan 20, 2017 at 22:13

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