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Conjecture:

For every irrational algebraic number $q$ and natural number $b$, the representation of $q$ on base $b$ contains all the digits $[0,\dots,b-1]$.

Questions:

  1. Has this conjecture been proved, refuted or neither?

  2. If proved:

    Is there an estimate of the minimum length of $q_b$ containing all the digits?

    For example, I would expect something like $2b$ or $b^2$ for any given $q_b$.

  3. If not refuted:

    I suppose that it is not true for transcendental numbers. Is that correct?

    How can we construct a transcendental number $q_b$ which does not contain all the digits?

Thanks

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    $\begingroup$ For the last point: yes, as you seem to know, it fails for transcendental numbers: for example, Liouville's constant is transcendental, but only has $1$'s and $0$'s in its decimal expansion. For more information, read about general Liouville numbers. $\endgroup$
    – Geoff Robinson
    Commented Jul 6, 2014 at 9:00
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    $\begingroup$ Assuming you are talking about initial sequences of digits: for the second point, as for any $b$ and any $n$ there are irrational algebraic numbers whose representation in base $b$ starts with $n$ zeros, there is no such bound. $\endgroup$
    – Stefan Kohl ♦
    Commented Jul 6, 2014 at 9:07
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    $\begingroup$ @StefanKohl: You mean, I can simply take any irrational algebraic number $q$ and any natural number $b$, and divide $q$ by $b$ over and over? Hmmmm... good one, thanks. $\endgroup$
    – barak manos
    Commented Jul 6, 2014 at 9:10
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    $\begingroup$ The Liouville's constant I refer to is $\sum_{n=1}^{\infty}10^{-n!}.$ $\endgroup$
    – Geoff Robinson
    Commented Jul 6, 2014 at 9:55
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    $\begingroup$ I just want to remark since no one has mentioned it yet: there is a much stronger conjecture that every irrational algebraic real number is normal in every base. $\endgroup$
    – Bill Mance
    Commented Jul 6, 2014 at 19:46

2 Answers 2

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The conjecture has been neither refuted nor proved. The state of the art, as far as I know, is contained in the papers of Adamczewski and Bugeaud, in which they show that anything with a very low complexity decimal expansion cannot be an algebraic irrational. The complexity is the function $c_x(n)$ giving the number of blocks of length $n$ in the decimal expansion of $x$ (or any base). They show that if there exists a $k$ such that $c_x(n)\le kn$ for all $n$, then $x$ is either rational or transcendental. Of course, it's conjectured that $c_x(n)=10^n$ for all algebraic irrationals $x$. Your condition would be implied by the conjecture $c_x(n)>9^n$ for all algebraic irrationals $x$.

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  • $\begingroup$ You might also look at questions 114905, 114758, 99039, where very similar questions are asked. $\endgroup$
    – Anthony Quas
    Commented Jul 6, 2014 at 13:12
  • $\begingroup$ Thanks. Did you mean "cannot be irrational algebraic"? $\endgroup$
    – barak manos
    Commented Jul 6, 2014 at 13:14
  • $\begingroup$ right - fixed now. $\endgroup$
    – Anthony Quas
    Commented Jul 6, 2014 at 13:17
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What about the number in base 10 that has decimal expansion using only digits 1 and 2 in the following pattern: 0.121221222122221 ... That is the digit 1 occurs always alone: 2's occur in blocks of increasing length. This has no periodicity and is not a rational number (and misses many digits of the base 10 system)

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    $\begingroup$ For this to be relevant to the conjecture, the number you constructed would have to be algebraic, but it is transcendental because it is $2/9 - \sum_{n=1}^\infty (1/\sqrt{10})^{n(n+1)}$. The latter term is related to a theta function value and was proved to be transcendental. projecteuclid.org/… $\endgroup$
    – Douglas Zare
    Commented Jul 6, 2014 at 9:49
  • $\begingroup$ I did not know this was transcendental. Then this answers 3rd part of the question by OP. Thanks, Douglas Zare for such a detailed information and pointing out the reference. But I am still not able to figure out how to see that given number is the sum of the series you have described, also not able to see the connection to the Theta function paper you have cited. I'll think about it. $\endgroup$
    – P Vanchinathan
    Commented Jul 6, 2014 at 10:16
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    $\begingroup$ Thanks. From the comments to my question, I realized that both section #2 and section #3 were pretty easy to answer (kind of dumb questions to begin with I suppose). @Geoff Robinson gave a good example answering section #3, similar to yours I think. $\endgroup$
    – barak manos
    Commented Jul 6, 2014 at 10:42
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    $\begingroup$ Has this been done for $\sqrt D$ in arbitrary base? $\endgroup$
    – Lev Borisov
    Commented Jul 6, 2014 at 12:20

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