Question on the decimal expansion of algebraic numbers Is the countability of the set of irrational algebraic numbers somehow reflected in a characteristic property of their decimal expansions?  
 A: 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.
A: 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".
A: 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.
A: 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.
