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Very related, but not the same, to this question Multiples of an irrational number forming a dense subset, is the next one:

Is the sequence $(\{10^n\pi\})_{n=1}^\infty$ dense in the interval $[0,1]$? (where $\{x\}=x\ mod\,1=x-\lfloor x\rfloor$, is de decimal part of $x$)

I tried to extend the proof of HAskell in the comments of the prvious post, but I wasn't able.

EDIT: I change the question to any normal (in base 10) irrational

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Tito Eliatron
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1 Answers1

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Not enough is known about the decimal expansion of $\pi$ to give a definitive answer to the question. For all anyone knows, there are no sevens in the decimal expansion of $\pi$ after some point, which would mean no terms in your sequence between .7 and .8 after some point.

It is widely believed that your sequence is not just dense but uniformly distributed but, as I said, no one can prove even far weaker results.

Gerry Myerson
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  • there are no sevens in the decimal expansion of π after some point? That wouldn't imply that $\pi$ is not normal in base 10? – Tito Eliatron Apr 25 '16 at 07:36
  • Sure it would – but no one has been able to prove that $\pi$ is normal in base 10. For all we know, $\pi$ could be all twos and threes after some point. No one has yet been able to prove anything to the contrary. – Gerry Myerson Apr 25 '16 at 11:19
  • @char, please don't make things up. What I wrote was, "FOR ALL ANYONE KNOWS, there are no sevens in the decimal expansion of $\pi$ after some point." No one has been able to prove that there are infinitely many sevens in the decimal expansion of $\pi$, and no one has been able to prove that there aren't. Pretty nearly the only thing anyone has ever been able to prove about the decimal expansion of $\pi$ is that it isn't periodic or terminating. – Gerry Myerson Apr 25 '16 at 13:12
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    Thank you. I think I have missunderstand your comment. Apologize for a spanish speaker. – Tito Eliatron Apr 26 '16 at 22:10