English

An improved bound on $(A+A)/(A+A)$

Combinatorics 2016-10-13 v2

Abstract

We show that, for a finite set AA of real numbers, the size of the set A+AA+A={a+bc+d:a,b,c,dA,c+d0}\frac{A+A}{A+A} = \left\{ \frac{a+b}{c+d} : a,b,c,d \in A, c+d \neq 0 \right \} is bounded from below by A+AA+AA2+1/4A/A1/8logA.\left|\frac{A+A}{A+A} \right| \gg \frac{|A|^{2+1/4}}{|A / A|^{1/8} \log |A|}. This improves a result of Roche-Newton.

Keywords

Cite

@article{arxiv.1606.03468,
  title  = {An improved bound on $(A+A)/(A+A)$},
  author = {Ben Lund},
  journal= {arXiv preprint arXiv:1606.03468},
  year   = {2016}
}

Comments

8 pages

R2 v1 2026-06-22T14:22:51.897Z