That is, are the following two properties equivalent for a real number $A \in \mathbb{R}$:

1) $A$ is a normal number

2) In any base $b$, the representation of $A$ contains every possible finite sequence of digits.

I ask because I've wondered for a while whether property 2) holds for $\pi$, and I know that the normality of $\pi$ is unknown. I don't know whether the two problems are linked or if they are two separate problems.

  • $\begingroup$ i think (2) implies transcendentalism which is a weaker claim than normality $\endgroup$ – shai horowitz Oct 27 '16 at 1:44
  • $\begingroup$ @shaihorowitz That's incorrect - the interaction between normality and transcendentality is wide open. If I recall correctly, it is conjectured that every algebraic irrational is normal, while at the same time no example of an algebraic irrational which is proved to satisfy (2) is known! $\endgroup$ – Noah Schweber Oct 27 '16 at 1:47
  • $\begingroup$ ah so trancedental is one of the few ways that a number can not be normal under that conjecture interesting $\endgroup$ – shai horowitz Oct 27 '16 at 1:49

While (1) implies (2), (2) does not imply (1). The strings could appear with the wrong frequency.

For example, let's take binary for simplicity. The finite binary strings are $$0, 1, 00, 01, 10, 11, 000, 001, 010, 100, 011, 101, 110, 111, . . .$$ (listed in an appropriate order). Consider the real gotten by putting all of these strings together, separated by "$0000$": $$0.000001000000000001000010000011000000000000010000...$$ As you can see, while every string will appear eventually (indeed, infinitely often), the string "$0$" will appear far more frequently than ${1\over 2}$ of the time, and the string "$00$" will appear far more frequently than ${1\over 4}$ of the time, etc. Indeed, if instead of always using "$0000$" we separate new strings by blocks of $0$s which (say) double in size each time, we get a number in whose binary expansion every finite string appears, but the only strings which appear with positive frequency are those consisting of all zeroes. We can even do this in all bases, simultaneously (although the full construction is a bit messy).

  • $\begingroup$ Very clear answer. I suppose you could do the same in any base, that is, construct a number that is just every sequence in order, but prepend each sequence with the same number of zeroes as digits in the sequence (if that makes sense). $\endgroup$ – Johan Oct 27 '16 at 1:46

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