LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Irrational circle orbits are dense

Statement

For irrational $α$ , the subgroup ${{n α} : n \in Z}$ is dense in the circle. For every $ε > 0$ and integer $M \geq 0$ there is an integer $n > M$ with $d ({n α}, 0) < ε$ . These assertions are choice-free.

Facts & Assumptions

[F1]

Circle distance is distance to the nearest integer. The circle, rotations and the doubling map.

[F2]

N+1 points in N intervals contain a pair in one interval. The pigeonhole principle on $N$ .

Proof

Given: For irrational $α$ , the subgroup ${{n α} : n \in Z}$ is dense in the circle. For every $ε > 0$ and integer $M \geq 0$ there is an integer $n > M$ with $d ({n α}, 0) < ε$ . These assertions are choice-free.

1.1

For an integer $N \geq 2$ , place ${j α}$ , $0 \leq j \leq N$ , in the N half-open intervals $[k / N, (k + 1) / N)$ . Two indices i<j lie in one interval. Hence for q=j-i, $1 \leq q \leq N$ and $0 < d ({q α}, 0) < 1 / N$ ; positivity follows from irrationality. Change q to -q if needed to obtain a subgroup point $β \in (0, 1 / N)$ .

F1F2

2.1

The subgroup contains $0, β, 2 β, \dots, m β$ , where $m = ⌊ 1 / β ⌋$ ; the last term is interpreted modulo one if necessary. Consecutive gaps are beta and the remaining gap to 1 is at most beta. For every x in [0,1), taking $k = ⌊ x / β ⌋$ gives $0 \leq x - k β < β$ . Since N can be arbitrarily large, every circle neighborhood meets the subgroup.

step 1.1F1algebra

3.1

For fixed M>=1 let $δ = \min_{1 \leq q \leq M} d ({q α}, 0) > 0$ . Take N with $1 / N < \min (δ, ε)$ and use the positive q from step 1.1 before changing its sign. That q exceeds M and has distance less than epsilon. For M=0 any q supplied there works after taking $1 / N < ε$ . Only finite minima and finite pigeonhole choices occur.

step 1.1F1algebra∎

Depends on

Used by

False: ergodicity implies strong mixing False statement
Circle rotation is ergodic for Lebesgue measure exactly at irrational angles Theorem

Dependency tree · two levels

18 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

E–W Proposition 2.16 p.26 and Example 2.33 p.49 (standard reference, not scraped)