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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.

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Your definition needs a minus in place of the plus, I believe. – Matthew Conroy Jan 2 '11 at 21:42
Thanks for the correction, of course you are right. – Listing Jan 2 '11 at 21:54
up vote 1 down vote accepted

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

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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 :-) – Listing Jan 2 '11 at 22:30
@user3123: No idea if a simpler proof exists. Where did you get this problem? – Aryabhata Jan 3 '11 at 6:15
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 ->… on a forum about topology. I guess it uses… 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 – Listing Jan 3 '11 at 10:47
@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? – Aryabhata Jan 3 '11 at 17:14
Thank you, Moron :-) – Listing Jan 4 '11 at 15:46

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