How Irrational Numbers Fill the Circle?

August 23, 2026 · #number theory · #real analysis

Let $𝛼 > 0$ be an irrational number. In this post, we are interested in the sequence ${({𝑛 𝛼})}_{𝑛 \geq 0}$ where ${𝑥}$ denotes the fractional part of a real number $𝑥$ . Clearly, this is a sequence of real numbers contained in the unit interval $[0, 1)$ . As $𝛼$ is irrational, all elements in this sequence are distinct. A natural first question is whether this sequence is dense in the unit interval? To answer this, we introduce the concept of equidistribution.

Definition 1. We say a sequence of numbers ${(𝑠_{𝑛})}_{𝑛 \geq 0}$ is equidistributed modulo 1 if for any $0 \leq 𝑎 < 𝑏 < 1$

lim_{𝑛 \to \infty} \frac{| {𝑠_{0}, \dots, 𝑠_{𝑛}} \cap [𝑎, 𝑏] |}{𝑛} = 𝑏 - 𝑎 .

Theorem 2. Let ${(𝑠_{𝑛})}_{𝑛 \geq 0}$ be a sequence of real numbers in $[0, 1)$ . Then the following are equivalent:

this sequence is equidistributed modulo 1,
for any Riemann-integrable function $𝑓$ , then
$lim_{𝑛 \to \infty} \frac{1}{𝑛} \sum_{𝑗 = 0}^{𝑛} 𝑓 (𝑠_{𝑗}) = \int_{0}^{1} 𝑓 (𝑥) d 𝑥,^{1}$ We actually have the converse from the de Bruijin–Post Theorem: if for a function $𝑓$ this identity holds for every equidistributed sequence in $[0, 1)$ , then $𝑓$ is Riemann-integrable.
(1)
for any $𝑘 \in ℤ_{+}$ , it holds that
$lim_{𝑛 \to \infty} \frac{1}{𝑛} \sum_{𝑗 = 0}^{𝑛} e^{2 𝜋 𝑖 𝑘 𝑠_{𝑗}} = 0 .$
(2)

Proof. For $1 \Rightarrow 2$ , we note from the definition of equidistribution that Equation (1) holds for any indicator function $𝑓 (𝑥) = 1_{[𝑎, 𝑏]} (𝑥)$ . Moreover, for any step function $ℎ$ , Equation (1) holds. Since $𝑓$ is Riemann-integrable, its Darboux upper and lower sums converge to the same value. In particular, we can find a step function $ℎ \leq 𝑓$ such that $\int_{0}^{1} ℎ d 𝑥 \geq \int_{0}^{1} 𝑓 d 𝑥 - 𝜀$ . This gives us

\underset{𝑛 \to \infty}{lim inf} \frac{1}{𝑛} \sum_{𝑗 = 0}^{𝑛} 𝑓 (𝑠_{𝑗}) \geq lim_{𝑛 \to \infty} \frac{1}{𝑛} \sum_{𝑗 = 0}^{𝑛} ℎ (𝑠_{𝑗}) = \int_{0}^{1} ℎ (𝑥) d 𝑥 \geq \int_{0}^{1} 𝑓 (𝑥) d 𝑥 - 𝜀 .

Passing $𝜀 \to 0$ and by symmetry, we derive Equation (1).

For $2 \Rightarrow 3$ , we note that for any $𝑘 \in ℤ_{+}$ , $𝑓 (𝑥) = e^{2 𝜋 𝑖 𝑘 𝑥}$ is Riemann-integrable and $\int_{0}^{1} e^{2 𝜋 𝑖 𝑘 𝑥} d 𝑥 = 0 .$

For $3 \Rightarrow 1$ , by the Stone–Weierstrass theorem, any continuous function on $[0, 1]$ can be uniformly approximated by trigonometric polynomials. Therefore, Equation (1) holds for all continuous functions. By a similar approximation argument from the first part, indicator functions can be approximated from below/above by continuous functions, and we show that the sequence ${(𝑠_{𝑛})}_{𝑛 \geq 0}$ is equidistributed.⁠∎

The next proposition answers the question we posed at the beginning. The sequence ${({𝑛 𝛼})}_{𝑛 \geq 0}$ is not only dense but also fills the circle in a uniform way.

Proposition 3. The sequence ${({𝑛 𝛼})}_{𝑛 \geq 0}$ is equidistributed modulo 1.

Proof. We only need to verify Equation (2). Fix $𝑘 \in ℤ_{+}$ . We note that

| \frac{1}{𝑛} \sum_{𝑗 = 0}^{𝑛} e^{2 𝜋 𝑖 𝑘 {𝑗 𝛼}} | = | \frac{1}{𝑛} \sum_{𝑗 = 0}^{𝑛} e^{2 𝜋 𝑖 𝑘 𝑗 𝛼} | = | \frac{1 - e^{2 𝜋 𝑖 𝑘 (𝑛 + 1) 𝛼}}{𝑛 (1 - e^{2 𝜋 𝑖 𝑘 𝛼})} | \leq \frac{2}{𝑛 | 1 - e^{2 𝜋 𝑖 𝑘 𝛼} |} .

Therefore, by Theorem 2 we conclude the proof.⁠∎

Although the empirical distribution of ${({𝑛 𝛼})}_{𝑛 \geq 0}$ converges to the uniform distribution, its finite configurations behave very differently from independent random samples. Let $𝑆_{𝑘} = {({𝑛 𝛼})}_{0 \leq 𝑛 \leq 𝑘}$ . The gap between any two neighboring elements in $𝑆_{𝑘}$ can take at most 3 different values.

Proposition 4. We write $𝑆_{𝑘} = {𝑠_{0} < 𝑠_{1} < \dots < 𝑠_{𝑘}}$ . Then there exists a set $𝐺_{𝑘}$ with at most three elements such that for any $0 \leq 𝑗 \leq 𝑘$ , we have $𝑠_{𝑗 + 1} - 𝑠_{𝑗} \in 𝐺_{𝑘}$ , where we set $𝑠_{𝑘 + 1} = 1$ .

Proof. Set $𝐴 = {min}_{1 \leq 𝑛 \leq 𝑘} {𝑛 𝛼}, 𝐵 = {min}_{1 \leq 𝑛 \leq 𝑘} {- 𝑛 𝛼},$ and choose $𝑎, 𝑏 \in {1, \dots, 𝑘}$ such that

𝐴 = {𝑎 𝛼}, 𝐵 = {- 𝑏 𝛼} = 1 - {𝑏 𝛼} .

Thus $𝐴$ is the clockwise distance from $0$ to the first point of $𝑆_{𝑘}$ , while $𝐵$ is the distance from the last point of $𝑆_{𝑘}$ back to $0$ . We first claim that $𝑎 + 𝑏 > 𝑘 .$ Otherwise $𝑎 + 𝑏 \leq 𝑘$ . If $𝐴 \geq 𝐵$ , then

{(𝑎 + 𝑏) 𝛼} = 𝐴 - 𝐵 < 𝐴 .

Similarly, if $𝐴 < 𝐵$ , then

{- (𝑎 + 𝑏) 𝛼} = 𝐵 - 𝐴 < 𝐵,

contradicting the definition of $𝐵$ .

Now consider a point ${𝑟 𝛼} \in 𝑆_{𝑘}$ .

If $𝑟 + 𝑎 \leq 𝑘$ , then the next point clockwise is ${(𝑟 + 𝑎) 𝛼}$ , and the corresponding gap has length $𝐴$ .
If $𝑟 \geq 𝑏$ , then the next point clockwise is ${(𝑟 - 𝑏) 𝛼}$ , and the corresponding circular gap has length $𝐵$ .
It remains to consider $𝑘 - 𝑎 < 𝑟 < 𝑏 .$ Since $𝑎 + 𝑏 > 𝑘$ , the integer $𝑟 + 𝑎 - 𝑏$ lies in ${0, \dots, 𝑘}$ . The next point clockwise is then
${(𝑟 + 𝑎 - 𝑏) 𝛼},$
and the corresponding gap has length $𝐴 + 𝐵$ .

Hence, every gap has length in

𝐺_{𝑘} = {𝐴, 𝐵, 𝐴 + 𝐵} .

Therefore there are at most three distinct gap lengths, and whenever all three occur, the largest one is $𝐴 + 𝐵$ , the sum of the other two.⁠∎

In the following, we discuss the upper and lower bounds for the gap sizes. For a real number $𝑥$ , we set $‖ 𝑥 ‖ = 𝑑 (𝑥, ℤ)$ .

Proposition 5. Let $𝐺_{𝑘}$ be the set of gaps as above. We have

min (𝐺_{𝑘}) = min_{1 \leq 𝑗 \leq 𝑘} ‖ 𝑗 𝛼 ‖ \leq \frac{1}{𝑘 + 1} .

Proof. The proof is an immediate consequence of the pigeonhole principle. As we have

(𝑘 + 1)

distinct points on the circle, the shortest arc must have length no greater than

\frac{1}{𝑘 + 1}

.⁠∎

To obtain a lower bound, irrationality alone is not enough: some irrational numbers admit extraordinarily accurate rational approximations. A very different picture appears when the rotation angle is algebraic.

Theorem 6 (Liouville). Assume that $𝛼$ is a solution to a $𝑑$ -degree integer coefficient polynomial $𝑝 (𝑥)$ . Then, there exists $𝐶_{𝛼}$ such that

min (𝐺_{𝑘}) = min_{1 \leq 𝑗 \leq 𝑘} ‖ 𝑗 𝛼 ‖ \geq \frac{𝐶_{𝛼}}{𝑘^{𝑑 - 1}} .^{2}

Roth’s theorem sharpens this estimate: for every

𝜀 > 0

, there exists some constant

𝐶_{𝛼, 𝜀} > 0

such that

min (𝐺_{𝑘}) \geq \frac{𝐶_{𝛼, 𝜀}}{𝑘^{1 + 𝜀}} .

Proof. Without loss of generality, we may assume that $𝑝$ is irreducible. Fix $1 \leq 𝑗 \leq 𝑘$ , and choose $𝑙 \in ℤ$ such that $‖ 𝑗 𝛼 ‖ = | 𝑗 𝛼 - 𝑙 | .$ Since $𝛼$ is irrational, we have $𝑝 (\frac{𝑙}{𝑗}) \neq 0 .$ Because $𝑝$ has integer coefficients, $𝑗^{𝑑} 𝑝 (\frac{𝑙}{𝑗})$ is a nonzero integer. Therefore,

| 𝑝 (\frac{𝑙}{𝑗}) | \geq \frac{1}{𝑗^{𝑑}} .

On the other hand, by the mean value theorem,

| 𝑝 (\frac{𝑙}{𝑗}) | = | 𝑝 (\frac{𝑙}{𝑗}) - 𝑝 (𝛼) | \leq 𝐶_{𝛼} | \frac{𝑙}{𝑗} - 𝛼 |

for some constant $𝐶_{𝛼}$ depending only on $𝛼$ . Combining the above estimates, we derive

| 𝛼 - \frac{𝑙}{𝑗} | \geq \frac{1}{𝐶_{𝛼} 𝑗^{𝑑}},

and therefore

‖ 𝑗 𝛼 ‖ = | 𝑗 𝛼 - 𝑙 | \geq \frac{1}{𝐶_{𝛼} 𝑗^{𝑑 - 1}} \geq \frac{1}{𝐶_{𝛼} 𝑘^{𝑑 - 1}} .

⁠∎