Given the following sequence $d(n) = x \cdot n - \lfloor x \cdot n \rfloor$ with $n \in \mathbb{N}$, $x \in \mathbb{R} \backslash \mathbb{Q}$ .

I want to show that for every $z \in [0,1)$ there is a subsequence of $d(n)$ that converges against $z$.


I am having serious problems of finding such a subsequence, first I see what the sequence does. If you take for example the subsequence $a(n) = d(10^n)$ you see that it gives you the decimal representation of $x$ starting with the $n'th$ digit. Example: let $x=\pi$, then we get $a(0)=0.141592..., a(1)=0.41592..., a(2)=0.1592...$ and so on.

Also this is a plot of $d(n)$ for $x=\pi$:

Plot of d(n)

You see that for irrational x the height of the function at integers is not constant, nor periodic.

Also this question I asked several weeks ago seems to be related: Proving that an equation is solvable, Floor function

  • $\begingroup$ Your definition needs a minus in place of the plus, I believe. $\endgroup$ Jan 2, 2011 at 21:42
  • $\begingroup$ Thanks for the correction, of course you are right. $\endgroup$
    – Listing
    Jan 2, 2011 at 21:54

1 Answer 1


If I have understood your question correctly, I believe this follows from the Equidistribution theorem. For more info refer: Equidistributed Sequence.

  • $\begingroup$ Thank you for your answer. Yes you are right but I think the fact that it is uniformly distributed is stronger than what I want to show. I hope that there is a simpler proof that can live without integrals :-) $\endgroup$
    – Listing
    Jan 2, 2011 at 22:30
  • $\begingroup$ @user3123: No idea if a simpler proof exists. Where did you get this problem? $\endgroup$
    – Aryabhata
    Jan 3, 2011 at 6:15
  • $\begingroup$ Our Analysis Teacher gave us a holiday sheet we could make if we wanted and this has prooved to be the only tough question on it. I investigated a bit further and found this -> at.yorku.ca/cgi-bin/… on a forum about topology. I guess it uses math.stackexchange.com/questions/10280/… to show that there is a sequence that has a limit in every open interval. Then you could choose a interval of length $2^{-n}$ to prove it i think? But I don't fully understand it $\endgroup$
    – Listing
    Jan 3, 2011 at 10:47
  • $\begingroup$ @user: That works, if a set A is dense in an interval, then for even real number r in that interval, there is a sequence of elements of A which converges to r. Remember the proof that for every real number there is a sequence of rationals converging to it? $\endgroup$
    – Aryabhata
    Jan 3, 2011 at 17:14

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