MathAudit

Paper metadata

Original submission
October 13, 2009
Claimed MSC
11J04
arXiv categories
math.NT
Comments
5 pages

Abstract

Probably we have observed a new simple phenomena dealing with approximations to two real numbers.

AI-generated audit

Audit summary

Audited against arXiv v1

Not a correctness certificate. A “Correct” result may include yellow typos or minor formal corrections that do not affect substantive soundness. It means this audit found no unresolved substantive error under the stated criteria; it does not replace expert scrutiny or formal verification.

Current report

Detailed mathematical audit

Generated August 20, 2026
01Statements1 reported findingCorrect

The sign-change theorem for the two irrationality-measure functions is correct.

Theorem 1Correct

Infinite sign changes

Page 1 · Theorem 1 · arXiv:0910.2428v1

If one difference sign were eventually fixed, the interlacing of convergent denominators forced by Lemma 2 would give either an immediate contradiction between adjacent complete quotients or eventual equality of both approximation steps. The equality case makes the consecutive denominator pairs proportional and forces α±β∈Z\alpha\pm\beta\in\mathbb Z, contrary to the hypotheses.

02Proofs4 reported findingsCorrect

The continued-fraction comparison proves the theorem after correcting two mechanical indices.

Lemmas 1–2Correct and complete

Consecutive-denominator comparison

Pages 2–3 · Lemmas 1 and 2 · arXiv:0910.2428v1

The exact complete-quotient formula gives ∥qn−1α∥qn+1>1\lVert q_{n-1}\alpha\rVert q_{n+1}>1, whereas ∥pmβ∥pm+1<1\lVert p_m\beta\rVert p_{m+1}<1. Thus qn+1≤pm+1q_{n+1}\leq p_{m+1} forces ∥qn−1α∥>∥pmβ∥\lVert q_{n-1}\alpha\rVert>\lVert p_m\beta\rVert, which is the required interlacing obstruction.

Interlacing paragraphTypo

The endpoint denominator is pm+1p_{m+1}

Page 3 · paragraph following Equation (6) · arXiv:0910.2428v1

Replace the printed pn+1p_{n+1} in qn+1≤qn+t<pn+1q_{n+1}\leq q_{n+t}<p_{n+1} by pm+1p_{m+1}, with non-strict final inequality if equality occurs. The surrounding chain is qn+t≤pm+1q_{n+t}\leq p_{m+1} and Lemma 2 needs exactly qn+1≤pm+1q_{n+1}\leq p_{m+1}.

Equation (11)Typo

The complete-quotient index is n0+j+1n_0+j+1

Pages 4–5 · derivation of Equation (11) · arXiv:0910.2428v1

Replace αn0+j−1\alpha_{n_0+j-1} by αn0+j+1\alpha_{n_0+j+1} in Equation (11). The immediately preceding expansion contains qn0+j−1/αn0+j+1q_{n_0+j-1}/\alpha_{n_0+j+1}, so the inequalities pmj+1≥qn0+jp_{m_j+1}\geq q_{n_0+j} and pmj≥qn0+j−1p_{m_j}\geq q_{n_0+j-1} yield βmj+2>αn0+j+1\beta_{m_j+2}>\alpha_{n_0+j+1}. This directly contradicts the next derived inequality βmj+2≤αn0+j+1\beta_{m_j+2}\leq\alpha_{n_0+j+1}. The intended correction is unique and restores the printed contradiction.

Equality caseCorrect and complete

Eventual equality forces the excluded rational relation

Page 5 · final two paragraphs · arXiv:0910.2428v1

Equality at consecutive steps gives two integer affine relations in α\alpha and β\beta. A nonzero determinant of their coefficient matrix would make both numbers rational, hence consecutive denominator pairs are proportional. Their coprimality and the continued-fraction recurrences then force α=±β\alpha=\pm\beta modulo integers, which the theorem excludes.

03Novelty0 reported findingsNo non-novelty findings

No non-novelty findings.

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Paper
arXiv:0910.2428v1
Authors listed
Igor D. Kan, Nikolay G. Moshchevitin
Audit date
August 20, 2026
  1. 01Establish identityMatch an authenticated scholarly identity to this paper.
  2. 02Submit evidenceIdentify the finding and give a formal mathematical response.
  3. 03Re-evaluateA separate agent checks the response and records a disposition.
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