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Is the countability of the set of irrational algebraic numbers somehow reflected in a characteristic property of their decimal expansions?

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  • $\begingroup$ I don't understand either of the last two paragraphs. What countable subsets of irrational numbers? What countable sets of permutations? And what does this have to do with algebraic numbers? $\endgroup$
    – Qiaochu Yuan
    Commented Jul 31, 2010 at 1:27
  • $\begingroup$ Sorry, i deleted that part because it didn't make sense. I;d like to just ask the question as it stands and I can explain my motivation later I suppose. $\endgroup$
    – Matt Calhoun
    Commented Jul 31, 2010 at 1:30

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The answer is not known, and it is a conjecture of Borel that the answer is no. See Words and Transcendence by M. Waldschmidt for references, in particular your previous claim that every digit in irrational algebraic numbers occurs infinitely often is an open problem. We don't know if there are irrational algebraic numbers in the Cantor set.

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One weak answer to this question is that every algebraic number is computable in the sense that there is a computer program that outputs as many digits as desired of the decimal expansion of the number. There are countably many computer programs and thus countably many computable numbers (and hence countably many algebraic numbers).

The vague upshot of this is that any uncountable set of real numbers has to contain some very strange numbers that aren't in some sense "accessible to computers".

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Well, something may be said for sure: decimal expression of irrational algebraic number can not have too many small periodic blocks in a row (like 0.1239239239239...), since it would imply the existence of too good rational approximations and so contradict the Roth's theorem.

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  • $\begingroup$ If you can give a precise statement of "not too many small periodic blocks in a row" that would apparently be a counterexample to Borel's conjecture. But Borel's conjecture may have a more precise statement that rules this out. $\endgroup$
    – SixWingedSeraph
    Commented Jul 31, 2010 at 16:28
  • $\begingroup$ No counterexamples may be gotten on this way, since almost all numbers also satisfy Roth's theorem. I have basically no doubt that Borel conjecture is true. $\endgroup$
    – Fedor Petrov
    Commented Aug 1, 2010 at 11:52
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I'm not an expert on such topics, but I would say "probably not". The wikipedia page on normal numbers says $\sqrt{2}$ is "widely believed" to be normal, though this hasn't been proven. Normality basically means the decimal (and also base $n$ for all $n$) expansions look random, which is true of almost all real numbers. It's conceivable that such decimal expansions could still have some structure in the case of algebraic numbers, so it depends what you're looking for, but "probably not". Either no such structure is known, or the known structure is weak enough not to interfere with normality.

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