Say you picked your favorite irrational number $q$ and looking at $S = \{nq: n\in \mathbb{Z} \}$ in $\mathbb{R}$, you chopped off everything but the decimal of $nq$, leaving you with a number in $[0,1]$. Is this new set dense in $[0,1]$? If so, why? (Basically looking at the $\mathbb{Z}$-orbit of a fixed irrational number in $\mathbb{R}/\mathbb{Z}$ where we mean the quotient by the group action of $\mathbb{Z}$.)

Thanks!

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    Not that it matters terribly, but usually $q$ is used for rational numbers... – copper.hat Jan 8 '13 at 2:03
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    Here is a (an almost) duplicate of your question. – David Mitra Jan 8 '13 at 2:07
  • There is a short proof here; it’s a fairly simple application of the pigeonhole principle. – Brian M. Scott Jan 8 '13 at 14:09
up vote 35 down vote accepted

Notation: For each real number $ r $, let

  • $ \lfloor r \rfloor $ denote the largest integer $ \leq r $ and
  • $ \{ r \} $ denote the fractional part of $ r $.

Notice that $ \{ r \} = r - \lfloor r \rfloor $. Hence, $ \{ r \} $ is the ‘chopped-off decimal part’ of $ r $ that you speak of.


Most proofs begin with the Pigeonhole Principle, but we can introduce a slightly topological flavor by using the Bolzano-Weierstrass Theorem. Full detail will be provided.

Let $ \alpha $ be an irrational number. Then for distinct $ i,j \in \mathbb{Z} $, we must have $ \{ i \alpha \} \neq \{ j \alpha \} $. If this were not true, then $$ i \alpha - \lfloor i \alpha \rfloor = \{ i \alpha \} = \{ j \alpha \} = j \alpha - \lfloor j \alpha \rfloor, $$ which yields the false statement $ \alpha = \dfrac{\lfloor i \alpha \rfloor - \lfloor j \alpha \rfloor}{i - j} \in \mathbb{Q} $. Hence, $$ S := \{ \{ i \alpha \} \mid i \in \mathbb{Z} \} $$ is an infinite subset of $ [0,1] $. By the Bolzano-Weierstrass Theorem, $ S $ has a limit point in $ [0,1] $. One can thus find pairs of elements of $ S $ that are arbitrarily close.

Now, fix an $ n \in \mathbb{N} $. By the previous paragraph, there exist distinct $ i,j \in \mathbb{Z} $ such that $$ 0 < |\{ i \alpha \} - \{ j \alpha \}| < \frac{1}{n}. $$ WLOG, it may be assumed that $ 0 < \{ i \alpha \} - \{ j \alpha \} < \dfrac{1}{n} $. Let $ M $ be the largest positive integer such that $ M (\{ i \alpha \} - \{ j \alpha \}) \leq 1 $. The irrationality of $ \alpha $ then yields $$ (\spadesuit) \quad M (\{ i \alpha \} - \{ j \alpha \}) < 1. $$ Next, observe that for any $ m \in \{ 0,\ldots,n - 1 \} $, we can find a $ k \in \{ 1,\ldots,M \} $ such that $$ k (\{ i \alpha \} - \{ j \alpha \}) \in \! \left[ \frac{m}{n},\frac{m + 1}{n} \right]. $$ This is because

  • the length of the interval $ \left[ \dfrac{m}{n},\dfrac{m + 1}{n} \right] $ equals $ \dfrac{1}{n} $, while
  • the distance between $ l (\{ i \alpha \} - \{ j \alpha \}) $ and $ (l + 1) (\{ i \alpha \} - \{ j \alpha \}) $ is $ < \dfrac{1}{n} $ for all $ l \in \mathbb{N} $.

On the other hand, there is another expression for $ k (\{ i \alpha \} - \{ j \alpha \}) $: \begin{align} k (\{ i \alpha \} - \{ j \alpha \}) & = \{ k (\{ i \alpha \} - \{ j \alpha \}) \} \quad (\text{As $ 0 < k (\{ i \alpha \} - \{ j \alpha \}) < 1 $; see ($ \spadesuit $).}) \\ & = \{ k [(i \alpha - \lfloor i \alpha \rfloor) - (j \alpha - \lfloor j \alpha \rfloor)] \} \\ & = \{ k (i - j) \alpha + k (\lfloor j \alpha \rfloor - \lfloor i \alpha \rfloor) \} \\ & = \{ k (i - j) \alpha \}. \quad (\text{The $ \{ \cdot \} $ function discards any integer part.}) \end{align} Hence, $$ \{ k (i - j) \alpha \} \in \! \left[ \dfrac{m}{n},\dfrac{m + 1}{n} \right] \cap S. $$ As $ n $ is arbitrary, every non-degenerate sub-interval of $ [0,1] $, no matter how small, must contain an element of $ S $.

(Note: A non-degenerate interval is an interval whose endpoints are not the same.)


Conclusion: $ S $ is dense in $ [0,1] $.

  • @ Haskell Curry $\{k(i−j)α+k(⌊jα⌋−⌊iα⌋)\}=\{k(i−j)α\}.$ how does this step coming?please explain. – Unknown x Aug 7 '17 at 10:14
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    @N. Maneesh: It appears that Haskell Curry isn’t responding. Anyway, for all $ n \in \mathbb{Z} $ and $ x \in \mathbb{R} $, observe that \begin{align} \{ n + x \} & = (n + x) - \lfloor n + x \rfloor \\ & = (n + \lfloor x \rfloor + \{ x \}) - \lfloor n + \lfloor x \rfloor + \{ x \} \rfloor \\ & = (n + \lfloor x \rfloor + \{ x \}) - (n + \lfloor x \rfloor) \qquad (\text{As $ 0 \leq \{ x \} < 1 $.}) \\ & = \{ x \}. \end{align} In the context of Haskell Curry’s argument, we have $ n = k (\lfloor j \alpha \rfloor - \lfloor i \alpha \rfloor) $ and $ x = k (i - j) \alpha $. – Berrick Caleb Fillmore Aug 20 '17 at 20:00
  • I got the proof. Thank you – Unknown x Aug 21 '17 at 7:47

Hint: Let $ \{ z\}$ denote the fractional part of the number $z$. If $x$ is an irrational number, then for any given $n$, then there exists $1 \leq i \in \mathbb{N}$, $i \leq n+1$ such that $ 0 < \{ ix \} < \frac {1}{n}$

A bit of a late comer to this question, but here's another proof:

Lemma: The set of points $\{x\}$ where $x\in S$, (here $\{\cdot\}$ denotes the fractional part function), has $0$ as a limit point.

Proof: Given $x\in S$, Select $n$ so that $\frac{1}{n+1}\lt\{x\}\lt\frac{1}{n}$. We'll show that by selecting an appropriate $m$, we'll get: $\{mx\}\lt\frac{1}{n+1}$, and that would conclude the lemma's proof.

Select $k$ so that $\frac{1}{n}-\{x\}\gt\frac{1}{n(n+1)^k}$. Then: $$ \begin{array}{ccc} \frac{1}{n+1} &\lt& \{x\} &\lt& \frac{1}{n} - \frac{1}{n(n+1)^k} \\ 1 &\lt& (n+1)\{x\} &\lt& 1+\frac{1}{n} - \frac{1}{n(n+1)^{k-1}} \\ & & \{(n+1)x\} &\lt&\frac{1}{n} - \frac{1}{n(n+1)^{k-1}} \end{array} $$ If $\{(n+1)x\}\lt\frac{1}{n+1}$, we are done. Otherwise, we repeat the above procedure, replacing $x$ and $k$ with $(n+1)x$ and $k-1$ respectively. The procedure would be repeated at most $k-1$ times, at which point we'll get: $$ \{(n+1)^{k-1}x\}\lt\frac{1}{n} - \frac{1}{n(n+1)}=\frac{1}{n+1}. $$

Proposition: The set described in the lemma is dense in $[0,1]$.

Proof: Let $y\in[0,1]$, and let $\epsilon\gt0$. Then by selecting $x\in S$ such that $\{x\}\lt\epsilon$, and $N$ such that $N\cdot\{x\}\le y\lt (N+1)\cdot\{x\}$, we get: $\left|\,y-\{Nx\}\,\right|\lt\epsilon$.

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