Kovac 2024 eventually greedy best underapproximations egyptian fractions

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corollary_2: Deduces from the measure-zero theorem that some transcendental real has no eventually greedy best Egyptian underapproximations; the argument is non-constructive and names no such number.

lemma_3: Shows that for every integer i >= 1000 at least one thousandth of the interval (1/i, 1/(i-1)], by measure, consists of numbers whose best two-term Egyptian underapproximation is not the greedy one.

lemma_4: Shows that the measure of the set of reals in (0, H_s] whose best n-term underapproximations are nested from n = s to n = t is at most 1999/2000 times its value at t whenever t grows by two, for 100 <= s < t.

theorem_1: Proves that the set of positive reals whose best n-term Egyptian underapproximations are eventually built greedily has Lebesgue measure zero, disproving the almost-all assertion of problem 206.

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Vjekoslav Kovač, On eventually greedy best underapproximations by Egyptian fractions. J. Number Theory 268 (2025), 39--48, doi:10.1016/j.jnt.2024.09.004; arXiv:2406.07218 (2024). The arXiv record names arXiv's non-exclusive distribution license (arXiv:2406.07218), every other right reserved.

Erdos and Graham speculated that almost every positive real might have eventually greedy best Egyptian underapproximations, meaning that for all large n the best n-term sum of distinct unit fractions below x is obtained from the best (n-1)-term one by adding a further unit fraction. Theorem 1 proves the opposite extreme: the set of positive reals with this property has Lebesgue measure zero, giving a negative answer to problem 206. Corollary 2 deduces the existence of a transcendental number lacking the property, non-constructively answering a question of Nathanson, who had asked for a proof or disproof that such irrationals exist. The argument is a probabilistic-flavored recursive reduction to two-term underapproximations, resting on Lemma 3: for every integer i >= 1000, the numbers in (1/i, 1/(i-1)] whose best two-term Egyptian underapproximation is not the greedy one fill at least one thousandth of that interval, a uniform positive share that then bootstraps as more unit fractions are allowed. The paper explicitly makes no progress on the companion open questions of whether every rational, or every algebraic number, has the eventually greedy property.

Source: https://arxiv.org/abs/2406.07218.

The copy read for this card is arXiv:2406.07218v3 (26 September 2024; the arXiv comment says v3 incorporates the referee's suggestions), 7 pages; v1 is of 11 June 2024 and v2 (13 June 2024) corrected the history of the claim for rational numbers, per the same comment. The published text (J. Number Theory 268, March 2025) has not been compared with the preprint. Read status: claims checked for Theorem 1, Corollary 2 and Lemma 3, whose statements were read clause by clause on PDF p. 2, and for Lemma 4, read on PDF p. 5; the proofs (pp. 3--7) were read for structure and are sketched on Theorem 1, Corollary 2, Lemma 3 and Lemma 4; no proof is rewritten in full and none has been independently reviewed. The rational companion question that the paper leaves open is the subject of the author's 2026 preprint with Q. Tang, filed as kovac_2026_eventually_greedy_best_egyptian_underapproximations_rational.

Bears on. #206: Theorem 1 shows that the reals with eventually greedy best Egyptian underapproximations form a Lebesgue null set, the negation of the almost-all assertion; Corollary 2 shows, non-constructively, that a transcendental number without the property exists; Lemmas 3 and 4 are steps in the proof of Theorem 1. The paper states that it makes no progress on whether every rational, or every algebraic number, has the property.

Results to transcribe.

• Theorem 1: The positive reals having eventually greedy best Egyptian underapproximations form a Lebesgue null set.
• Corollary 2: There exists a transcendental real without eventually greedy best Egyptian underapproximations (non-constructive).
• Lemma 3: For every integer i >= 1000, the numbers in (1/i, 1/(i-1)] whose best two-term Egyptian underapproximation is not the greedy one have measure at least 1/1000 of the interval's length.
• Lemma 4: For all integers 100 <= s < t, |X_{s,t+2}| <= (1999/2000)|X_{s,t}|, where X_{s,t} is the set of x in (0, H_s] whose best n-term Egyptian underapproximations are partial sums of one increasing denominator list for every n from s to t.

No file of this source is held: no license on record permits its redistribution, and the card cites the edition it names above.