[joshuamorton]

> However, proving such results about _specific_ numbers is notoriously hard [3]. As far as I am aware, there has not been a single irrational algebraic number proven to be normal. Normality of well-known constants like pi and e is also an open problem! I would not be surprised if proving distributional results for continued fraction expansions of pi is also very hard.

Is it known if algebraic numbers can be normal? I'm not a mathematician, but almost all numbers are normal, and almost all numbers are non-algebraic (or even non-computable!). Something akin to "most of the non-algebraics and non-computables are normal, and none of the algebraics are normal" is feasible, right? It contradicts the commonly held idea that e or sqrt(2) or pi are normal, but we don't even have a (non-constructuve) proof that there exist irrational algebraic normals, do we?

[arutar]

You're absolutely correct---countable sets have probability zero (with probability 1 a uniformly random number in [0,1] will irrational), and it would not be contradictory for all algebraic numbers to be non-normal. I am pretty sure it is unknown if there exist irrational algebraic normal numbers, even non-constructively.

One reason for believing that irrational algebraics, or pi, or e are normal, is a crude heuristic which is that "there is nothing special about base b". You can also take a computer and compute digit frequencies up to some very large precision, and see what happens. Generally speaking, it feels like "naturally defined numbers" should be normal unless there is a good reason for them not to be (and this is entirely independent of the fact that uniformly random numbers are normal). Proving this is a very different matter!

[joshuamorton]

I don't know that I follow

> is a crude heuristic which is that "there is nothing special about base b

In this context. Imagine the stronger statement, for all bases b, sqrt(2) is not normal in base b. This is base-independent, and we know one base for which it is true (the irrational base sqrt(2), where it's 10). I assume this has something do with restricting b. This argument sounds reasonable for integer or rational bases, but not for noncomputable bases (where presumably sqrt(2) becomes non-computable, how does that work?)

Obviously if we restrict to integer/rational bases, I follow that argument.