# 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?

• 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? – Qiaochu Yuan Jul 31 '10 at 1:27
• 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. – Matt Calhoun Jul 31 '10 at 1:30

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.