Multiples of an irrational number forming a dense subset

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!

• Not that it matters terribly, but usually $q$ is used for rational numbers... Commented Jan 8, 2013 at 2:03
• Here is a (an almost) duplicate of your question. Commented Jan 8, 2013 at 2:07
• There is a short proof here; it’s a fairly simple application of the pigeonhole principle. Commented Jan 8, 2013 at 14:09

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.
– user464147
Commented Aug 7, 2017 at 10:14
• @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 $x = k (i - j) \alpha$. Commented Aug 20, 2017 at 20:00
• I got the proof. Thank you
– user464147
Commented Aug 21, 2017 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: $\left|\,y-\{Nx\}\,\right|\lt\epsilon$.