13
$\begingroup$

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?

  • 0
    This can well be a duplicate, if so, I apologize in advance. – 2017-01-20
  • 0
    Killing a mosquito with a howitzer you're not even sure exists. Welcome to mathematics, I guess. – 2017-01-20
  • 1
    @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. – 2017-01-20
  • 0
    @NoahSchweber What is a howitzer ? – 2017-01-20
  • 1
    @Peter [A household pest control device](https://en.wikipedia.org/wiki/Howitzer). – 2017-01-20
  • 1
    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. – 2017-01-20
  • 0
    Metaphors aside, is there a specific algebraic number you're giving as an example? I tried a few things in Robert Munafo's RIES but couldn't find something like your number, e.g., $\sqrt{e^{-9}}$. – 2017-01-20
  • 1
    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. – 2017-01-20

1 Answers 1

2

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

  • 0
    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 ? – 2017-01-20
  • 0
    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 $n$. – 2017-01-20
  • 0
    @Peter that one wont work. The number of blocks of length $n$ is not bounded by $kn$ for any $k$. – 2017-01-20
  • 0
    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. – 2017-01-20
  • 0
    So, the distinct blocks of length $3$ in my example would be $bbb,abb,bab,bba,aba$ ? – 2017-01-20
  • 0
    If the number of consecutive $a$'s and that of consecutive $b$'s are both limited, can we then apply the theorem ? – 2017-01-20
  • 0
    no, we wan't apply it even then, it seems to be pretty limited really. There is also a theorem about the number of digit-changes. see here: https://arxiv.org/abs/0709.1560 – 2017-01-20