When Sets Are Not Sum-dominant
Hung Viet Chu

TL;DR
The paper provides a human-understandable proof that sets with fewer than six elements are not sum-dominant and explores how adding elements to arithmetic progressions or specific sequences affects sum-dominance.
Contribution
It offers a human-proof that sets smaller than six are not sum-dominant and analyzes how augmenting sets with certain elements influences sum-dominance.
Findings
Sets with fewer than six elements are not sum-dominant.
Adding up to two elements to an arithmetic progression does not produce a sum-dominant set.
Incorporating specific sequence elements with a few arbitrary numbers does not yield sum-dominance.
Abstract
Given a set of nonnegative integers, define the sum set and the difference set The set is said to be sum-dominant if . In answering a question by Nathanson, Hegarty used a clever algorithm to find that the smallest cardinality of a sum-dominant set is . 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 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 is not sum-dominant. Finally, we prove that if a set contains a certain number of integers from aβ¦
Peer Reviews
No public reviews on file for this paper yet. If you reviewed it on a platform where reviews are public (OpenReview, ICLR, NeurIPS, ICML), you can paste yours below so the community can read it here.
Videos
No videos yet. Explain this paper in a talk, walkthrough, or lecture? Add one.
Taxonomy
TopicsLimits and Structures in Graph Theory
**When Sets Are Not Sum-Dominant
** HΓΉng Viα»t Chu
Department of Mathematics
Washington and Lee University
Lexington, VA 24450
USA
Abstract
Given a set of nonnegative integers, define the sum set
[TABLE]
and the difference set
[TABLE]
The set is said to be sum-dominant if . In answering a question by Nathanson, Hegarty used a clever algorithm to find that the smallest cardinality of a sum-dominant set is . 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 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 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.
1 Introduction
1.1 Literature review
Given a set , define the sum set and the difference set . The set is said to be
- β’
sum-dominant, if ;
- β’
balanced, if ; and
- β’
difference-dominant, if .
As addition is commutative and subtraction is not, it was natural to conjecture that sum-dominant sets are rare. Since Nathansonβs review of the subject in 2006 [16], research on sum-dominant sets has made considerable progress: see [5, 8, 16, 19, 20, 21] for history and overview, [6, 11, 12, 17, 22] for explicit constructions , [3, 9, 24] for positive lower bound for the percentage of sum-dominant sets, [7, 14] for generalized sum-dominant sets, and [1, 2, 4, 13, 23] for extensions to other settings.
However, much less work has been done on how to determine whether or not a given set is sum-dominant. Only recently, Mathur and Wong [10] gave an algorithm for checking if a set is sum-dominant. The algorithm computes and compares all pairs of possible sums and differences among numbers of the set. This paper instead focuses on certain types of not-sum-dominant sets that we can tell without the assistance of computers.
Nathanson [16] asked What is the smallest cardinality of a sum-dominant set?. Hegarty [6] used a clever algorithm to find as the smallest sum-dominant set. Furthermore, is the unique sum-dominant set of cardinality , up to affine transformations. However, a human-understandable proof of the result has not been produced because of the complexity lurking behind the interaction of numbers in addition and subtraction. Nathanson [15, 18] asked for a human-understandable proof of the smallest cardinality of a sum-dominant set. This paper proves that a set of cardinality less than is not sum-dominant without the use of computers.
1.2 Notation
We introduce some notation.
- β’
Let . Define and . So, we can write . Define the center of to be , the point that is equidistant from the two endpoints.
- β’
For and a set , we write to mean the introduction of numbers into the set to form .
- β’
Given a set and a number , we write to mean the set .
1.3 Main results
The following is our first result.
Theorem 1**.**
A set with is not sum-dominant.
Though our proof is concise, we are unable to prove the cases of cardinality and since the level of complexity grows quite quickly. Our main technique is to argue for a lower bound for the number of equal pairs of positive differences from , which confines set to certain structures.
Our next result is about the relationship between sum-dominant sets and arithmetic progressions. Since numbers from an arithmetic progression form a balanced set, it is convenient to introduce new numbers into the set (in a clever way) and produce a sum-dominant set. Indeed, the construction of sum-dominant sets using arithmetic progressions was started by Nathanson [17] and Hegarty [6]. However, little is known about the smallest number of integers needed to turn an arithmetic progression into a sum-dominant set. 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.
Theorem 2**.**
Let be a set of numbers in an arithmetic progression and . Then is not sum-dominant.
A natural question to ask is what is the minimum number of integers to add to a set formed by numbers in an arithmetic progression and have a sum-dominant set?. Let the number be . Note that . Because , , is an arithmetic progression, we know that . Though Theorem 2 is easily stated, the proof requires a clever division into cases to reduce the complexity of interactions between numbers in addition and subtraction. As a necessary condition for a set to be sum-dominant, Theorem 2 provides a powerful tool to eliminate cases in arguing about the smallest cardinality of sum-dominant sets. For example, for a set of cardinality , if we know that at least numbers in the set form an arithmetic progression, then the set is not sum-dominant.
Our final result is another test for being sum-dominant and extends [4, Theorem 1].
Theorem 3**.**
Let be a subset of , where is a strictly increasing sequence of non-negative numbers. If there exists a positive integer such that
* for all , and* 2. 2.
set does not contain any sum-dominant set with for some , and 3. 3.
* for some ,*
Then
* is not sum-dominant and ,* 2. 2.
Let and be integers. If , then is not sum-dominant.
Example 4**.**
For the Fibonacci numbers , we have . By [4, Corollary 2], the Fibonacci numbers have no sum-dominant subsets, so we can pick any number . Let and , for example. Since satisfies , we know that a set of Fibonacci numbers and an arbitrary integer is not sum-dominant.
Example 5**.**
Let , the golden ratio. The geometric sequence has the property that . By [4, Corollary 8], this sequence has and has no sum-dominant subsets, so we can pick any number . Let and , for example. Since satisfies , we know that a set of numbers from the sequence and an arbitrary integer is not sum-dominant.
Section 2 proves several important results for the proof of our main theorems. Section 3, Section 4, and Section 5 prove Theorem 1, Theorem 2, and Theorem 3, respectively. Section 6 discusses some questions for future research.
2 Important results
We use the definition of a symmetric set in the sense of Nathanson [17]: a set is symmetric with respect to a number if . It is easy to check that any arithmetic progression is symmetric. (The sum of the two endpoints of an arithmetic progression is the point of symmetry.) The following lemma is proved by Nathanson [17].
Lemma 6**.**
A symmetric set is balanced.
Proposition 7**.**
Let and , where . Then gives new positive differences and new sums.
Proof.
We have . Clearly, all new sums are in
[TABLE]
Let , where . Then . Hence, the number of new sums is .
All new positive differences are in . Note that positive differences in are . Because . The number of new positive differences is . β
Lemma 8**.**
Let . Then is not a sum-dominant set.
Proof.
If , we are done because is an arithmetic progression, which is a symmetric set and thus, balanced
For , our set has at most 2 elements, which is symmetric and thus, not sum-dominant by Lemma 6.
We assume . Note the number of new sums is at most . We consider the following three cases.
Case I: . Arrange numbers in in increasing order. Either to the left of or to the right of , there are at least numbers. Without loss of generality, assume that is greater than numbers in ; that is, . The set of new differences has
[TABLE]
as a subset.
Subcase I.1: . Because , , implying that there are at most new sums. Since the number of new differences is at least , is not sum-dominant. 2. 2.
Subcase I.2: . Then the set of new differences has as a subset. Since the number of new differences is at least , is not sum-dominant.
Case II: . We consider two subcases.
Subcase II.1: . The set of new differences include . Therefore, the number of new differences is at least , implying that is not sum-dominant. 2. 2.
Subcase II.2: . If , then have numbers from an arithmetic progression and so, is not sum-dominant. Let for some . We have . Because , if , there are new positive differences and so, is not sum-dominant. If by Proposition 7, there are new positive differences and new sums, which shows that is not sum-dominant.
Case III: . Due to symmetry, this case is similar to Case II. This completes our proof. β
Corollary 9**.**
A set of numbers in an arithmetic progression in union with a singleton set cannot be sum-dominant.
Proof.
Let and . Our set is , where . The set is sum-dominant if and only if is sum-dominant. Note that may not be a nonnegative integer. This completes the proof. β
Remark 10*.*
Let , and . The difference set of has at least one more number than the sum set. This remark is evident from the proof of Lemma 8.
Remark 11*.*
A set of numbers in an arithmetic progression is an example of a symmetric set. Though the result of Corollary 9 holds for any set of numbers in an arithmetic progression, it is not true for symmetric sets in general. For example, the set is symmetric, and is sum-dominant.
The following proposition offers upper bounds for the cardinality of the sum set and the difference set of a set in terms of . The two inequalities are not hard to prove and are used by Hegarty [6].
Proposition 12**.**
We have the following bounds
[TABLE]
The equality in (1) is achieved if the sum of any two numbers is distinct, and the equality in (2) is achieved if the difference between any two different numbers is distinct.
The next observation plays a key role in reducing the complexity of our proof that a set of too small a cardinality cannot be sum-dominant.
Observation 13**.**
Let and consider . Let such that . If , we have another pair of equal positive differences . If , then we do not have another pair. In both cases, we have , a pair of equal sums. Hence, for pairs of equal positive differences, there exist at least pairs of equal sums.
3 Proof of Theorem 1
For clarity, we split the proof into two parts.
3.1 A set with is not sum-dominant.
Proof.
We proceed by case analysis of .
If , then and so, is not sum-dominant.
If , suppose that . Then is symmetric with respect to . By Lemma 6, is not sum-dominant.
If , then is the union of an arithmetic progression with a singleton set. By Corollary 9, is not sum-dominant.
If , by Proposition 12, , while . Let be the number of pairs of equal positive differences that has. Then, in order that is sum-dominant,
[TABLE]
The comes from Observation 13. We have . Therefore, . Denote with and . We write out all nonnegative differences in
[TABLE]
All differences in row 1 are pairwise distinct. These nonnegative differences give 7 differences in total. Because , we are not allowed to have any new differences from row 2 and row 3. Clearly, , which implies that is an arithmetic progression. By Corollary 9, is not sum-dominant. β
3.2 A set with is not sum-dominant.
Proof.
By Proposition 12, we know that , while . Let be the number of pairs of equal positive differences. Due to Observation 13, we have So, and . Denote with and . The following lists all nonnegative differences in
[TABLE]
Differences in row 1 are pairwise distinct and account for differences in . Because , we are allowed to have at most one more difference from rows 2, 3, and 4.
Case I: . Then
[TABLE]
Subcase I.1: . Because and , . Since , either or . Combine the former with to have . We have , which, by Corollary 9, makes not sum-dominant. The latter gives , which, combined with , makes symmetric and thus, not sum-dominant. 2. 2.
Subcase I.2: , implying . Because , it must be that and so, . Similarly, because and , we must have . Due to the fact that sum-dominant is preserved under affine transformations, we let and to have . This set is not sum-dominant.
Case II: . If , then . Since , is not sum-dominant due to Corollary 9. Therefore, or, equivalently, and
[TABLE]
Because and , we know . Since and , it must be that . Hence, and so, . Due to the fact that sum-dominant is preserved under affine transformations, we let and to have . This set is not sum-dominant.
We have shown that a set of cardinality is not sum-dominant.β
4 Proof of Theorem 2
Lemma 14**.**
Let and two numbers such that are not integers. Then is not sum-dominant.
Proof.
If , we know that is not sum-dominant because its cardinality is 3. So, we assume that and . The condition guarantees that . For the proof, we let our original set be and we introduce to the set; that is, . By Lemma 8, is not sum-dominant. Note that the introduction of gives at most new sums. We consider three cases.
Case I: . Similar to Case I in the proof of Lemma 8, the introduction of into gives at least new positive differences. These differences result from the interaction of and .
Subcase I.1: . Similar to Subcase I.2 in the proof of Lemma 8, either or is another new positive difference. Hence, the number of new differences is at least . Therefore, is not sum-dominant. 2. 2.
Subcase I.2: . Due to our condition that , causing and Remark 11, the difference set of has one more number than the sum set. Also, since , the number of new sums is at most . Because the number of new differences is at least , is not sum-dominant.
Case II: . We have two subcases.
Subcase II.1: . The set of new differences includes . Therefore, the number of new differences is at least , implying that is not sum-dominant. 2. 2.
Subcase II.2: . Let for some . The set of differences related to is . If , there are new positive differences and so, is not sum-dominant. If , there are at least new positive differences and at most new sums by Proposition 7.
Case III: . Due to symmetry, this case is the same as Case II. We complete the proof. β
Lemma 15**.**
Let and two numbers such that are integers. Then is not sum-dominant.
Proof.
The proof is divided into four parts, each of which deals with a specific position of and when being introduced to .
Part I.
We know that gives at most new sums.
Case I: . Then gives the following new differences . Therefore, in order that is sum-dominant, must give exactly new sums.
Subcase I.1: . To have new sums, and must be disjoint, implying that . So, is a new difference that is not in . Hence, we have new positive differences and so, is not sum-dominant. 2. 2.
Subcase I.2: .
- β’
If , then is a new positive difference not in and we have new positive differences.
- β’
If , then note that if , we have is a new difference not in . So, . Because and , . Hence, we have a pair of equal sums and so, the number of new sums is at most . Therefore, is not sum-dominant.
Case II: . Write for some and for some . Note . If , then all numbers in are new differences. Because , is another new difference. So, the number of new differences is at least . Therefore, in order that is sum-dominant, must give exactly new sums. Similar to Subcase I.1 above, we can show that is not sum-dominant. Hence, .
If , then is the set of new positive differences. In order that is sum-dominant, must gives exactly new sums. This is a contradiction because , causing . Hence, .
We have shown that . Suppose that . It can be easily checked that is not sum-dominant. So, either or .
- β’
If , we know that and . Also, because , . So, the number of new sums is at most .
- β’
If , then . Also, because , . The number of new sums is also at most .
We have shown that the number of new sums is at most . Suppose that . Because , gives new positive differences. This implies that is not sum-dominant. So, . We consider the sum set of . Note that , so the set of new sums is a subset of , which is at most due to Proposition 7. Indeed, Proposition 7 states that since and , gives new sums (including ). Also by Proposition 7, gives new positive differences. Since , is not sum-dominant.
Part II.
Case I: If , there are new positive differences from the interaction between and . Since , is another new positive difference. If , is another positive difference. Hence, we do not have a sum-dominant set because we have at most new sums. Therefore, in order that is sum-dominant, and must give exactly new sums, which implies that . So, . This inequality leads to the following two cases.
- β’
If , then and there is another new difference .
- β’
If , then . Hence, and . Because , or . Since we have a pair of equal sums, the number of new sums is less than .
Case II: If , the following are new positive differences from . Hence, the number of new differences is at least . Because and , the number of new sums is at most . In order that is sum-dominant, there must not be a new positive difference other than the differences above. So, the positive difference must be equal to for some and so, . Similarly, the positive difference must be equal to for some and so, or equivalently, . Hence, we have at least two pairs of equal sums. Therefore, the number of new sums is at most . This shows that is not sum-dominant.
Part III.
Let be the number such that .
Case I: .
Subcase I.1: . New positive differences include
[TABLE]
Because , numbers in the two rows are pairwise distinct. We have these new positive differences. Since , there exists such that . So, , implying that we have at most new sums. Therefore, is not sum-dominant. 2. 2.
Subcase I.2: . Because , we have at most new sums.
- β’
If lie on the same side of the center of , we assume that . The following are new differences
[TABLE]
Because , the numbers in these two rows are pairwise distinct. These are new positive differences. Hence, is not sum-dominant.
- β’
If lies on two sides of the center of , we assume that is closer to the center than . (If they are equidistant from the center, we have a balanced set with the same center of ). The numbers in Row 3 and Row 4 are still new positive differences. Hence, is not sum-dominant.
Case II: . We write for some .
Subcase II.1: lie on one side of the center of . Assume that . We have
[TABLE]
Because , they are not new sums and so, all new sums are , which contains numbers. The following are new positive differences
[TABLE]
So, there are at least new differences. Note that
[TABLE]
Hence, is not sum-dominant. 2. 2.
Subcase II.2: lie on two side of the center of . As above, the number of new sums is still . Without loss of generality, assume that ; that is, is closer to the center. We have
[TABLE]
Because
[TABLE]
all these numbers are distinct positive differences and we have new differences. Since ( is closer to the center), is not sum-dominant.
Part IV.
If , we have a symmetric set, which is not sum-dominant. Without loss of generality, assume that . Note gives at most new sums. If we can show that the number of new differences is at least , then the new set is not sum-dominant.
Case I: . New positive differences include
[TABLE]
Case II: . Let and . Because , .
Subcase II.1: and . We have and . Because
[TABLE]
there are at least new positive differences . Hence, is not sum-dominant. 2. 2.
Subcase II.2: and . Then . So, is not a new sum and so, we have at most new sums. As above, there are at least new positive differences . Hence, is not sum-dominant. 3. 3.
Subcase II.3: . We will find all new sums from and . We have
[TABLE]
New sums are from . (We do not count because .) So, there are at most new sums. We find a lower bound for the number of new differences by looking at
[TABLE]
The numbers in are new positive differences. Another new difference is . So, the number of new differences is bounded below by . This is bigger than . Hence, is not sum-dominant.
As these four parts consider all possible cases, we have completed the proof. β
Due to the fact that being sum-dominant is preserved under affine transformation, Theorem 2 follows immediately from Lemma 14 and Lemma 15.
Remark 16*.*
We can use Theorem 2 to simplify the proof of Theorem 1.
- β’
For , we write . Because is an arithmetic progression, adding to the set of does not give a sum-dominant set.
- β’
For , if contains an arithmetic progression of length , then we are done. We look at Section 3.2 and easily eliminate Subcase I.1 and Case II, thus shortening the proof.
The trade off is that the proof of Theorem 2 is much more computationally involved.
5 Proof of Theorem 3
Proof.
We first prove item 1. Let , where be a finite subset of . We show that with is not a sum-dominant set by induction. If , then is not sum-dominant by the second assumption of the theorem. If , consider the number of new sums and differences obtained by adding . The number of new sums is at most . We prove that the number of new differences is at least .
Since , we have . Let . Then, , which implies . The largest difference between elements in is ; we now show that we have added at least distinct differences greater than . Denote , where is an increasing function. We have
[TABLE]
The third inequality is due to the first assumption on . Since , we know that
[TABLE]
are differences greater than the greatest difference in . As we could subtract in the opposite order, the number of new differences obtained from adding to is at least
[TABLE]
We have seen that by adding to the set , the number of differences goes by at least more than the number of sums. Note that must have at least elements.
Next, we prove item 2. To form , we first add to the set, which gives at most new sums. When we add to , we have at most new sums. Continue the process until we have added all numbers to the set, the number of new sums is at most . Our original set has ; therefore, if , then . Therefore, is not sum-dominant. β
6 Future work
We list some natural topics for future research.
- β’
Is there a human-understandable proof that is not the cardinality of a sum-dominant set?
- β’
What is the smallest number of integers added to a set of numbers in an arithmetic progression to have a sum-dominant set?
- β’
Is the following conjecture true?
Conjecture 17**.**
The union of two arithmetic progressions is not a sum-dominant set.
The case analysis in this paper is unable to solve this general case because the complexity grows quickly when more numbers are added to an arithmetic progression. The case that the minimum of one arithmetic progression is bigger than the maximum of the other is easy. If this conjecture is true, our Theorem 2 follows immediately because the set of two numbers forms an arithmetic progression.
7 Acknowledgments
First, I would like to thank professor Steven Miller at Williams College for introducing me to this interesting topic. Next, many thanks to the anonymous referee for valuable comments which helped to improve the article. Thanks to professor Kevin Beanland at Washington and Lee University for proofreading this paper. Finally, I would like to dedicate this paper to my parents and my brother for their support throughout my undergraduate years at Washington and Lee University.
The reference list from the paper itself. Each links out to its DOI / PubMed record.
- 1[1] M. Asada, S. Manski, S. J. Miller, and H. Suh, Fringe pairs in generalized MSTD sets, Int. J. Number Theory 13 (2017), 2653β2675.
- 2[2] H. V. Chu, N. Luntzlara, S. J. Miller, and L. Shao, Infinite families of partitions into MSTD subsets, preprint, 2018. Available at https://arxiv.org/abs/1808.05460 .
- 3[3] H. V. Chu, N. Luntzlara, S. J. Miller, and L. Shao, Generalizations of a curious family of MSTD sets hidden by interior blocks, preprint, 2018. Available at https://arxiv.org/abs/1808.05501 .
- 4[4] H. V. Chu, N. Mc New, S. J. Miller, V. Xu, and S. Zhang, When sets can and cannot have sum-dominant subsets, J. Integer Seq , Vol. 18, 2018.
- 5[5] G. A. Freiman and V. P. Pigarev, Number Theoretic Studies in the Markov Spectrum and in the Structural Theory of Set Addition (Russian), Kalinin. Gos. Univ., Moscow, 1973.
- 6[6] P. V. Hegarty, Some explicit constructions of sets with more sums than differences, Acta Arith. 130 (2007), 61β77.
- 7[7] G. Iyer, O. Lazarev, S. J. Miller, and L. Zhang, Generalized more sums than differences sets, J. Number Theory 132 (2012), 1054β1073.
- 8[8] J. Marica, On a conjecture of Conway, Canad. Math. Bull. 12 (1969), 233β234.
