English

If $(A+A)/(A+A)$ is small then the ratio set is large

Combinatorics 2017-05-17 v2 Number Theory

Abstract

In this paper, we consider the sum-product problem of obtaining lower bounds for 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\}, for an arbitrary finite set AA of real numbers. The main result is the bound A+AA+AA2+225A:A125logA,\left| \frac{A+A}{A+A} \right| \gg \frac{|A|^{2+\frac{2}{25}}}{|A:A|^{\frac{1}{25}}\log |A|}, where A:AA:A denotes the ratio set of AA. This improves on a result of Balog and the author (arXiv:1402.5775), provided that the size of the ratio set is subquadratic in A|A|. That is, we establish that the inequality A+AA+AA2A:AA2log25A.\left| \frac{A+A}{A+A} \right| \ll |A|^{2} \Rightarrow |A:A| \gg \frac{ |A|^2}{\log^{25}|A|} . This extremal result answers a question similar to some conjectures in a recent paper of the author and Zhelezov (arXiv:1410.1156).

Keywords

Cite

@article{arxiv.1507.07672,
  title  = {If $(A+A)/(A+A)$ is small then the ratio set is large},
  author = {Oliver Roche-Newton},
  journal= {arXiv preprint arXiv:1507.07672},
  year   = {2017}
}

Comments

In this version, Lemma 3.2 has been improved, and the new version of Lemma 3.2 is tight up to multiplicative constants. This results in a small improvement to the main result of the paper. To appear in JLMS. With thanks to Noga Alon, for providing the proof of the new and improved Lemma 3.2 via a private communication