For $x \notin \pi\mathbb Q$, that is, a real $x$ that is not a rational multiple of $\pi$, consider the set $$\{(\cos nx,\sin nx):n = 0,1,2,...\}.$$ It is known that this set is dense in the unit circle $B(0,1)$ of $\mathbb R^2$. Could someone please give me a proof or reference for a proof?

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    $\begingroup$ Do you mean dense in the unit circle? $\endgroup$
    – paw88789
    Dec 10, 2015 at 12:26
  • $\begingroup$ Yes, I am sorry for the mistake. $\endgroup$
    – martin
    Dec 10, 2015 at 15:38
  • $\begingroup$ google irrational rotation dense, see math.stackexchange.com/questions/282102/… and en.wikipedia.org/wiki/Irrational_rotation Wikpedia gives a few references, one is math.byu.edu/~tfisher/documents/classes/2008/winter/635/… Many dynamical systems books will consider this example in detail. $\endgroup$
    – Mirko
    Dec 12, 2015 at 15:26
  • $\begingroup$ In a reference where "proof is an exercise" cannot be accepted as an answer. $\endgroup$
    – martin
    Dec 12, 2015 at 18:34
  • $\begingroup$ the above of course is not an answer (and was not meant to be), but a comment. Did you try to find a dynamical systems book? I posted a complete answer, providing references to two books (and enclosing the proof from one of them). $\endgroup$
    – Mirko
    Dec 17, 2015 at 17:57

4 Answers 4


From $x\notin \pi{\mathbb Q}$ it follows that $e^{ikx}\ne1$ for all $k\in{\mathbb Z}\setminus\{0\}$, and this implies that the numbers $e^{inx}\in S^1$ $(n\geq0)$ are all different. Assume that a point $\zeta\in S^1$ and an $\epsilon>0$ is given. Since $S^1$ has finite length we then can find two numbers $n_1<n_2$ with $|e^{in_2x}-e^{in_1 x}|< \epsilon$. Put $n':=n_2-n_1$. Then $$\bigl|e^{in'x} -1\bigr|=|e^{in_2x}-e^{in_1 x}|<\epsilon\ .$$ Put $n_k:=k\>n'$ $(k\geq0)$. Then the successive points $e^{i n_k x}\in S^1$ $(k\geq0)$ have a distance $<\epsilon$. It follows that there is a $k\geq0$ with $|e^{in_k x}-\zeta|<\epsilon$.

  • $\begingroup$ How are dense subsets related to irrational rotations of pi? $\endgroup$
    – user420360
    Jun 21, 2017 at 16:45
  • $\begingroup$ But $e^{ikx}=1$ when $k=0\in \mathbb{Z}$ $\endgroup$
    – Heisenberg
    Aug 1, 2017 at 4:11
  • $\begingroup$ @Heisenberg: Thank you for reporting the slip. Hope it's correct now. $\endgroup$ Aug 1, 2017 at 6:40
  • $\begingroup$ Yes. Could you explain why there are two numbers $n_1,n_2$ such that $|e^{in_2x}-e^{in_1 x}|< \epsilon$?\ $\endgroup$
    – Heisenberg
    Aug 1, 2017 at 6:53
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    $\begingroup$ @Heisenberg Because, to quote Christian: the numbers $e^{inx}\in S^1$ are all different, but $S^1$ is of finite length. If $N$ points are all to be at least distance $\epsilon$ from each other, then the length of $S^1$ would have to be at least $N\epsilon$ to accommodate them all. $\endgroup$ Sep 15, 2018 at 21:03

As I noted in the comments many Dynamical Systems book will have a proof. I think the book by Robert L. Devaney, An Introduction to Chaotic Dynamical Systems does, but I don't have access to it right now to confirm. The proof which I enclose below (and which I remember having seen before, perhaps on other books) is from Introduction to Dynamical Systems by Michael Brin, Garrett Stuck (link to 1-st edition at Amazon, though I know there are newer editions).

There are links to this book online, in particular there is a sample of the first chapter at a Library of Congress web page (so I assume it is legal and more or less permanent). See section 1.2, Circle Rotations, p.3-4 there. To make this answer self contained I enclose the proof from the Brin/Stuck book (with some editing, omitting the subscript $\alpha$ and a certain inequality, and comments).

For $\alpha\in\mathbb R$, let $R$ be the rotation of the circle $S^1$ by angle $2\pi\alpha$.

If $\alpha=\frac pq$ is rational, then $R^q=Id$, so every orbit is periodic. On the other hand, if $\alpha$ is irrational, then every positive semiorbit is dense in $S^1$. Indeed, the pigeon-hole principle implies that, for any $\varepsilon>0$, there are $m,n$ such that $m<n$ and $d(R^m,R^n)<\varepsilon$. Thus $R^{n-m}$ is rotation by an angle less than $\varepsilon$, so every positive semiorbit is $\varepsilon$-dense in $S^1$ (i.e., comes within distance $\varepsilon$ of every point in $S^1$). Since $\varepsilon$ is arbitrary, every positive semiorbit is dense.

Comment. To apply the pigeon-hole principle, you may represent the circle as the union of finitely many closed arcs, each of length less than $\varepsilon$. Then (for any fixed $x$) we place infinitely many $R^n(x)$, $n\ge0$, into finitely many arcs, hence at least one arc must contain both $R^n(x)$ and $R^m(x)$ for some $n\not=m$.
More precisely the book says "for any $\varepsilon>0$, there are $m,n<\frac1\varepsilon$ such that $m<n$" etc, but I do not see why this inequality would be relevant (I first thought it was a typo and should be $m,n>\frac1\varepsilon$ which is equivalent to $\frac1m,\frac1n<\varepsilon$ and implies $|\frac1m-\frac1n|<\varepsilon$), though it may have something to do with an attempt to find $n,m$ that work, with $n>m$ and smallest possible $n-m$. Or it may have something to do with the other representation of rotation (as indicated on the same page), as the sum $\mod1$ on $[0,1]$ with $0\sim1$.

It should be noted in the above proof that for irrational $\alpha$ (i.e. when the angle of rotation $2\pi\alpha$ is an irrational multiple of $\pi$), then $R^n$ is never the identity, for $n\ge1$. Indeed, if $R^n=Id$, then $n2\pi\alpha=2\pi k$ for some integer $k$, hence $\alpha=\frac kn$ would be rational.

Edit. The proof is available in Devaney's book too, e.g. in the second edition it is Example 3.12 and Theorem 3.13 (referred to as Jacobi's Theorem there) on p.22-23. He considers the forward orbit of a point and uses that any infinite set (in particular the forward orbit) must have a limit point (and hence, in the notation of the present answer, $d(R^m,R^n)$ could be made arbitrarily small, non-zero, for suitable $m,n$).


You can proove that every rational number other than 1/2 comes from an irrational angle.

  1. a = cos(pi d/z) + i sin(pi d/z) solves some equation of the form a^z + 1 = 0

  2. Prove that a can never be a rational, by supposing that where

    $$a^n = \sum^{z-1}_{e=0} x_i a^e$$

that if a were a fraction p/q, multiply through by p^n, then the LHS is not a multiple of q, and the RHS is, unless q=1.

Therefore the equation a can never be rational unless it is half-integer, which means that the rational numbers arise from irrational angles.

Since the rational numbers are dense on the circle, the irrational angles are too.

  • 1
    $\begingroup$ What do you mean by the expressions "rational number comes from an irrational angle", and "the equation $a$", and "rational numbers arise from irrational angles"**? $\endgroup$
    – Mirko
    Dec 19, 2015 at 23:12

Suppose $[a,b] \subset [0,1]$. Since $T\theta$ is ergodic, $$\lim_{N \to \infty} \frac{1}{N} \sum_{n=0}^{N-1} \chi_{[a,b)}(T_\theta ^n (t))=b-a $$

  • 2
    $\begingroup$ You need to include dollar signs around the math for it to work. $\endgroup$
    – 6005
    Dec 11, 2016 at 20:26
  • $\begingroup$ Please check whether I didn't change the meaning of your answer. Also, how does this exactly answer the question (I don't know the answer to the question, but this answer seems a bit short for this question) $\endgroup$
    – wythagoras
    Dec 11, 2016 at 20:38

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