When Sets Are Not Sum-dominant

Given a set $A$ of nonnegative integers, define the sum set $$A+A = \{a_i+a_j\mid a_i,a_j\in A\}$$ and the difference set $$A-A = \{a_i-a_j\mid a_i,a_j\in A\}.$$ The set $A$ is said to be sum-dominant if $|A+A|>|A-A|$. In answering a question by Nathanson, Hegarty used a clever algorithm to find that the smallest cardinality of a sum-dominant set is $8$. Since then, Nathanson has been asking for a human-understandable proof of the result. We offer a computer-free proof that a set of cardinality less than $6$ is not sum-dominant. Furthermore, we prove that the introduction of at most two numbers into a set of numbers in an arithmetic progression does not give a sum-dominant set. This theorem eases several of our proofs and may shed light on future work exploring why a set of cardinality $6$ is not sum-dominant. Finally, we prove that if a set contains a certain number of integers from a specific sequence, then adding a few arbitrary numbers into the set does not give a sum-dominant set.Comment: 16 pages, published in J. Integer Se

Union of Two Arithmetic Progressions with the Same Common Difference Is Not Sum-dominant

Given a finite set $A\subseteq \mathbb{N}$, define the sum set $$A+A = \{a_i+a_j\mid a_i,a_j\in A\}$$ and the difference set $$A-A = \{a_i-a_j\mid a_i,a_j\in A\}.$$ The set $A$ is said to be sum-dominant if $|A+A|>|A-A|$. We prove the following results. 1) The union of two arithmetic progressions (with the same common difference) is not sum-dominant. This result partially proves a conjecture proposed by the author in a previous paper; that is, the union of any two arbitrary arithmetic progressions is not sum-dominant. 2) Hegarty proved that a sum-dominant set must have at least $8$ elements with computers' help. The author of the current paper provided a human-verifiable proof that a sum-dominant set must have at least $7$ elements. A natural question is about the largest cardinality of sum-dominant subsets of an arithmetic progression. Fix $n\ge 16$. Let $N$ be the cardinality of the largest sum-dominant subset(s) of $\{0,1,\ldots,n-1\}$ that contain(s) $0$ and $n-1$. Then $n-7\le N\le n-4$; that is, from an arithmetic progression of length $n\ge 16$, we need to discard at least $4$ and at most $7$ elements (in a clever way) to have the largest sum-dominant set(s). 3) Let $R\in \mathbb{N}$ have the property that for all $r\ge R$, $\{1,2,\ldots,r\}$ can be partitioned into $3$ sum-dominant subsets, while $\{1,2,\ldots,R-1\}$ cannot. Then $24\le R\le 145$. This result answers a question by the author et al. in another paper on whether we can find a stricter upper bound for $R$.Comment: 12 pages. arXiv admin note: text overlap with arXiv:1906.0047

On Sets with More Products than Quotients

Given a finite set $A\subset \mathbb{R}\backslash \{0\}$, define \begin{align*}&A\cdot A \ =\ \{a_i\cdot a_j\,|\, a_i,a_j\in A\},\\ &A/A \ =\ \{a_i/a_j\,|\,a_i,a_j\in A\},\\ &A + A \ =\ \{a_i + a_j\,|\, a_i,a_j\in A\},\\ &A - A \ =\ \{a_i - a_j\,|\,a_i,a_j\in A\}.\end{align*} The set $A$ is said to be MPTQ (more product than quotient) if $|A\cdot A|>|A/A|$ and MSTD (more sum than difference) if $|A + A|>|A - A|$. Since multiplication and addition are commutative while division and subtraction are not, it is natural to think that MPTQ and MSTD sets are very rare. However, they do exist. This paper first shows an efficient search for MPTQ subsets of $\{1,2,\ldots,n\}$ and proves that as $n\rightarrow \infty$, the proportion of MPTQ subsets approaches $0$. Next, we prove that MPTQ sets of positive numbers must have at least $8$ elements, while MPTQ sets of both negative and positive numbers must have at least $5$ elements. Finally, we investigate several sequences that do not have MPTQ subsets.Comment: 14 pages, to appear in Rocky Mountain Journal of Mathematic

Sets of Cardinality 6 Are Not Sum-dominant

Given a finite set $A\subseteq \mathbb{N}$, define the sum set $$A+A = \{a_i+a_j\mid a_i,a_j\in A\}$$ and the difference set $$A-A = \{a_i-a_j\mid a_i,a_j\in A\}.$$ The set $A$ is said to be sum-dominant if $|A+A|>|A-A|$. Hegarty used a nontrivial algorithm to find that $8$ is the smallest cardinality of a sum-dominant set. Since then, Nathanson has asked for a human-understandable proof of the result. However, due to the complexity of the interactions among numbers, it is still questionable whether such a proof can be written down in full without computers' help. In this paper, we present a computer-free proof that a sum-dominant set must have at least $7$ elements. We also answer the question raised by the author of the current paper et al about the smallest sum-dominant set of primes, in terms of its largest element. Using computers, we find that the smallest sum-dominant set of primes has $73$ as its maximum, smaller than the value found before.Comment: 15 pages, to appear in Integers: Electronic Journal of Combinatorial Number Theor