We just had another pi day, and once again people talk about how infinite pi is, and that it contains everything.

It seems to me completely irrational to expect that just because something is infinite and non-repeating it means that it contains all possible messages. I could easily (uhm, with infinite time and effort) construct an irrational number by rolling dice and adding the value to the end of the base 10 string. That would be infinite and non-repeating, but never contain 0,7,8 or 9.

Is this just a problem of base conversion, which means that every irrational number contains "everything", or is there a pi-specific proof that it does contain " everything"? Of course we can read the dice number in base 6 and get all possible values, but that's a bijection, which i suspect is more powerful than a base conversion.

Filip Haglund
  • 295
  • 1
  • 12
  • 1
    It's widely imagined that $\pi$ is a ["normal number"](http://mathworld.wolfram.com/NormalNumber.html) which would mean that it's digits contained every finite string of digits. But this is not known to be true. – lulu Mar 15 '17 at 11:43

1 Answers1


This has come up before I think, but the business about a number 'containing all possible strings of any length' is a consequence of a number being normal, which very approximately says that no finite block appears more or less often than any other its sequence of digits.

There is a result of Borel (though its proof was fixed by Faber and then Hausdorff), that 'almost all' real numbers are normal ('almost all' here means the complement has Lebesgue measure zero). Of course there are infinitely (in fact uncountably) many irrational numbers whose representation in a given base is not normal, as you described I could decide to only pick digits from a small range of possible digits. However in some sense you are still doing something 'non-general'.

It's possibly worth noting that $\pi$ is not actually known to be normal, though as the Wikipedia page suggests - it is expected.

  • 106
  • 3