Let $a_n=10^n \cdot \pi$. Is the set of numbers $\{a_n-\lfloor a_n \rfloor : n \in \mathbb{N}\}$ dense in [0,1]?

What is the best known result near this question?

Apparently John Nash asked this on an undergraduate analysis exam (according to an anecdote told by Seymour Haber, recounted in Sylvia Nassar's biography of Nash).

  • 3
    $\begingroup$ Surely no one doubts that it is true but no one can prove much of anything that does not apply to any irrational real. For example surely all 10 digits appear infinitely often but all we can prove is that less than 9 digits appear finitely often (or at least 2 appear infinitely often if you are not a constructivist) $\endgroup$ Feb 7, 2011 at 17:48
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    $\begingroup$ @Daniel Parry: it is believed true (it's a very weak version of normality, and $\pi$ is believed to be normal) and it has not been proved. Even the fact that the decimal expansion does not eventually consist of 0s and 1s has not been proved, and this is much stronger than that. $\endgroup$ Feb 7, 2011 at 18:34
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    $\begingroup$ Related question: mathoverflow.net/questions/51853/… $\endgroup$ Feb 7, 2011 at 18:36
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    $\begingroup$ Just to summarize what everyone else has said, it is believed to be true but is currently open. If it is false, it would imply that $\pi$ is not normal (which is open). If it is true, it would imply that all 10 digits appear infinitely often in $\pi$ (which is open). $\endgroup$
    – Tony Huynh
    Feb 7, 2011 at 19:07
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    $\begingroup$ About what is known: Furstenberg proved (Math. Systems Theory 1 (1967), 1-49) that $S\alpha$ is dense mod 1 for any $\alpha$ irrational, as long as $S$ is non-lacunary. An example is $S=\{2^m3^n\mid m,n\ge 0\]$. An example of a lacunary $S$ (so the result does not apply) is $S=\{10^n\mid n\ge0\}$. $\endgroup$ Feb 8, 2011 at 0:27

1 Answer 1


The comments have pretty much said all there is to be said about $\pi$. I'll just note that the fractional part of $10^n\alpha$ is known to be not just dense but uniformly distributed in $[0,1)$ for all real $\alpha$, except for a set of measure zero. There is no reason to think $\pi$ is in this exceptional set, and no expectation of a proof anytime soon that it isn't.

  • $\begingroup$ This may be a stupid question, but can the set of exceptions be of positive Hausdorff measure, e.g. some Cantor set? $\endgroup$
    – Mark
    Feb 8, 2011 at 0:43
  • $\begingroup$ I think the set of numbers not normal to, say, base 10 has Hausdorff dimension 1. See A D Pollington, The Hausdorff dimension of a set of normal numbers, Pac J Math 95 (1980) 193-204, also Kenji Nagasaka, On Hausdorff dimension of non-normal sets, AnnInst Stat Math 23, 515-521. $\endgroup$ Feb 8, 2011 at 0:54
  • $\begingroup$ The Hausdorff dimension of the classical Cantor set is $\log 2/\log 3$. Similarly, the Hausdorff dimension of the set of numbers in the unit interval whose decimal expansion does not use the digit 5 is $\log 9/\log 10$. Omitting longer patterns will give you a Hausdorff dimension arbitrarily close to 1. $\endgroup$
    – Goldstern
    Sep 24, 2012 at 8:58

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