I have $a_n(x):=nx-\lfloor nx \rfloor$ where $x$ is real. I want to show that if $x$ is rational, then $a_n(x)$ has finitely many cluster points, if $x$ is irrational, then every real $a$ with $0\leq a \leq 1$ is cluster point of $a_n(x)$. I don't know where to start, through your help, I will understand cluster point, sequences more deeply. I am stuck how to show those 2 cases, i am still bad at proof-thinking.
Cluster points of the sequence $a_n(x):=nx-\lfloor nx \rfloor$
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sequences-and-series
irrational-numbers
rational-numbers
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1You will find some useful information in [this related question](http://math.stackexchange.com/questions/202392/cluster-points-of-multiples-of-the-fractional-part-of-an-irrational-number). – 2012-12-29
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1Don't know where to start? Start by retrieving the relevant definitions ("cluster point", "rational" etc) and write down what they mean in the particular case of your $a_n(x)$. – 2012-12-29
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0okay thanks, Henning – 2012-12-29
1 Answers
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HINT: If $x$ is rational, we can write $x=m+\frac{a}n$ for some integers $m,a$, and $n$ such that $n>0$ and $0\le a
$$\begin{align*} a_k(x)&=kx-\lfloor kx\rfloor\\ &=km+\frac{ka}n-\left\lfloor km+\frac{ka}n\right\rfloor\\ &=km+\frac{ka}n-\left(km+\left\lfloor\frac{ka}n\right\rfloor\right)\\ &=\frac{ka}n-\left\lfloor\frac{ka}n\right\rfloor\;. \end{align*}$$
Use this to find a very simple relationship between $a_k(x)$ and $a_{k+n}(x)$.
This answer to an earlier question covers the case of irrational $x$; it’s complete, but it’s pretty concise, so you may have to stare at it a bit to see just how it works.
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0Thanks Brian, great, i am studying your answer – 2012-12-29
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1@doniyor: You’re welcome; feel free to ping me if you get badly stuck. – 2012-12-29
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0i will, Brian. :). thank you so much for helping – 2012-12-29