Research Article | Open Access

Volume 2015 |Article ID 216475 | 14 pages | https://doi.org/10.1155/2015/216475

# Maximal Midpoint-Free Subsets of Integers

Accepted09 Dec 2014
Published20 Jan 2015

#### Abstract

A set is midpoint-free if no ordered triple satisfies and . Midpoint-free subsets of and are studied, with emphasis on those sets characterized by restrictions on the base digits of their elements when , and with particular attention to maximal midpoint-free subsets with .

#### 1. Introduction

An ordered triple of integers is a midpoint triple of if and . The midpoint of this triple is , its lower endpoint is and its upper endpoint is . For any subset , let denote the set of all midpoint triples . When , let Generically, the members of are the balance points for in . The balance points comprise the lower endpoint set , the midpoint set , and the upper endpoint set , for in . The members of are the eccentric points for in . Attention to these sets appears to be a new focus, suggested by the underlying geometrical viewpoint. There is an extensive literature associated with treating as specifying an arithmetic progression of length. A compact discussion and rich bibliography are given in Guy’s survey work [1, Section ]. For an example of recent work in this field, see Dybizbański .

If it will be convenient to say that is midpoint-free; moreover, is a maximal midpoint-free subset of if and , whenever . Hence we have for any . This implies the following characterization.

Theorem 1. If , then is a maximal midpoint-free subset of if and only if , or equivalently, if and only if and .

Note that if is infinite, any maximal midpoint-free subset must also be infinite: any pair of elements of has one midpoint and two endpoints, so precludes at most three elements of from membership of ; thus would be infinite if were finite, but then would be midpoint-free for any , contradicting maximality of .

In  the notions of midpoint triple and maximal midpoint-free subset are studied for several “natural” subsets of the real numbers, but for simplicity in the present note we restrict to and subsets, especially and . Here, the main focus will be on , , and when is a maximal midpoint-free subset of , or .

Initially has been defined to be midpoint-free if contains no midpoint triple. There are several semantic equivalents for this condition.

Theorem 2. If , then any one of the sets is empty if and only if all three are empty.

Proof. Consider the contrary. If there is a triple ; this triple ensures that and . The other two cases follow in the same way.

Take in Theorem 2. Then recovers the “natural” terminology that is midpoint-free if and only if it does not contain the midpoint of any two of its members. Equally, is midpoint-free if and only if it is lower endpoint-free, or alternatively, if and only if it is upper endpoint-free.

Corollary 3. A subset is midpoint-free if and only if

This yields another semantic equivalent: is midpoint-free exactly when it is balance point-free.

For any define the multiplicity of as a lower endpoint, midpoint, or upper endpoint for , respectively, as The multiplicities for in particular cases of will be of interest.

#### 2. An Explicit Example, Involving Base Digit Restrictions

Let us begin with an explicit example to illustrate these notions and sample some of the typical features encountered.

For any integer , let be the set of integers with regular base representation in which all digits are restricted to the set . When are distinct members of , all digits of the base representation of lie in the set , but base computation of involves no carry over, so contains the digitin each place where the base digits of and differ. Hence is impossible. It follows that is midpoint-free.

Let denote the regular base representation of . An easy base computation shows that This corresponds to the identities Note that is an even positive integer, so is a midpoint for when this set is treated as a subset of . Digit considerations make it clear that has no other solutions with , so the midpoint multiplicity of is. In summary, Again, the general base computation corresponds to the identity which implies when . Additionally, in the special case we also have Hence . There are no other solutions to with , so the multiplicities of as a lower and upper endpoint are , .

When , it can be seen that has no solutions with , so is not an endpoint for , and Now seek such that and . As , it follows that . Then it is routine to verify that when , but note that digit arguments must be sensitive to the magnitude of ; for instance, when , while for smaller we have It follows that and is not a midpoint for when . This completes the demonstration that is an eccentric point, so the midpoint-free subset is not maximal if .

We will later return to a more systematic study of and .

#### 3. New Maximal Midpoint-Free Sets from Old

If and with , then is an affine transform of . Clearly if and only if when or when , so any affine transform is midpoint-free if and only if is midpoint-free.

For example, let be the set of all integers with regular base representation in which all digits are restricted to the set . Then is midpoint-free because is midpoint-free when , and is an affine transform of . The last identity shows that . Similarly, is an affine transform of so is a midpoint-free subset of when . Then is a midpoint-free subset of , since the positive and negative components of this set are midpoint-free, and has no solution with , , and , because these conditions require to be an odd integer and to be an even integer.

Suppose is known to be a maximal midpoint-free subset of . It turns out that the principle illustrated by the example in the previous paragraph holds strongly for .

Theorem 4. If is a maximal midpoint-free subset of , then the set is a maximal midpoint-free subset of .

Proof. As is a maximal midpoint-free subset of , Theorem 1 implies But , so and ; hence and . Therefore, .
Given , there exist such that and . Let . Then , so Therefore, Conversely, suppose , so there exist such that and . But , where , so implies , whence with . But is midpoint-free, so is midpoint-free, and therefore, . Thus . Now and imply that and , so . Therefore, the reverse containments also hold: Hence
For , similar reasoning shows that , so However, this containment can in fact be proper. For instance, if , then only contains even integers, whereas .
Since is a maximal midpoint-free subset of , no integers are eccentric for , so no members of are eccentric for . Then the positive integers eccentric for satisfy where . Since it follows that Specifically, all integers eccentric for are odd, and all integers eccentric for are even. Therefore, no integer is eccentric for both and , so . Thus is a maximal midpoint-free subset of .

Corollary 5. If is a maximal midpoint-free subset of , the balance point sets for and with satisfy

It is convenient to refer to the construction in Theorem 4 as “doubling” the given set . Other constructions involving affine transforms of a set are also of interest. For example, since is midpoint-free when , it follows that the disjoint sets and are midpoint-free when . In fact, their union is midpoint-free. This turns out to be “trivial.” Multiplying a member of by simply shifts its base digits one place, and a terminalemerges to occupy the zeroth place; then addingreplaces the terminalby. Hence .

When , the disjoint midpoint-free sets and are more interesting. Let us verify that is also midpoint-free. The two component sets are midpoint-free, so any midpoint triple must have at least one member in each set. Thus in each case with . If , then for . Then and . But , so . If then , while if then . In each case and , so because . Thus , as claimed.

Generalising the latter example, a “trebling” construction which produces new maximal midpoint-free subsets of will now be studied.

Theorem 6. If is a maximal midpoint-free subset of , and all members of are endpoints for , then is a maximal midpoint-free subset of , and all members of are endpoints of .

Proof. Because is midpoint-free, each of the affine transforms and is midpoint-free. Assume . Since , there is a such that . Then , so . Thus , so no such triple exists. Hence is midpoint-free.
By hypothesis, . Also by Corollary 5, so every is an endpoint for . Suppose and . Two integers from complementary sets cannot be equal, so must be different. Also holds when . It follows that If are consecutive members of , all members of the interval are endpoints for . Then are consecutive members of and all members of are endpoints for . Also each is an endpoint for , so all members of are endpoints for . Thus , and .

Let for each , and let . Iterating the construction in Theorem 6 and combining with Theorem 4 yields the following result.

Corollary 7. If is a maximal midpoint-free subset of , and all members of are endpoints for , then the set is a maximal midpoint-free subset of , for any integer , and all members of are endpoints of . Moreover, the set is a maximal midpoint-free subset of .

#### 4. Subsets of with Base Digit Restrictions

Fix an integer . Let . Then is a digit subset for base representations of the integers or, briefly, a base digit subset. Let be the set of nonnegative integers with base representation using only digits in , and let denote the digit in position of the regular base representation of , so

Let us say that is midpoint-free as a base digit subset if and there is no ordered triple such that and .

Theorem 8. If is a midpoint-free base digit subset with and , then the set is midpoint-free.

Proof. Suppose . Then . There is no carry-over in computing this sum in base arithmetic since all its digits are less than , so for every . Since is midpoint-free, it follows that for every , so , contradicting the initial choice of . Thus , so is midpoint-free.

Three early instances of Theorem 8, the first of which was independently demonstrated earlier, are of considerable interest.

Corollary 9. Each is midpoint-free when .

Corollary 10. Each is midpoint-free when .

Corollary 11. Each is midpoint-free when .

If is a midpoint-free base digit subset, it is of interest to decide whether has any members that are eccentric for , since this is equivalent to deciding whether is a maximal midpoint-free subset of . For , let the support for , as a lower endpoint, midpoint, or upper endpoint for , be defined by The following result is useful for settling whether the eccentric set is empty in specific cases.

Theorem 12. Suppose and is a midpoint-free base digit subset. If with , then

Proof. Fix the nonnegative integer with . (For simplicity we do not explicitly require , although Theorem 8 does imply that members of have empty support sets.)
Suppose there exist such that , so . Base arithmetic for this sum takes the form for every , with carry-overs satisfying and for . But so for . In particular, so . For all , it follows that and But is midpoint-free, so for all , since otherwise yields the contradiction . Hence .
The other two cases follow simply by noting that if then , and if then .

Corollary 13. If is a midpoint-free base digit subset, then relative to the three multiplicities of any are finite.

Corollary 14. If is a midpoint-free base digit subset and with , then contains all three support sets for relative to . Moreover , where .

Corollary 15. If then is eccentric for .

Corollary 16. If then is eccentric for .

Corollary 17. If then is eccentric for .

The last three corollaries leave open the possibility that their subject sets might be maximal midpoint-free subsets of when is small enough. In the next section we shall consider and pursue other cases later.

#### 5. Greedy Midpoint-Free Subset of

The greedy midpoint-free subset of is the set in which and each with is the smallest integer satisfying such that is midpoint-free. It has long been known  that corresponding to sequence A00536 of OEIS . For brevity, let We shall attach the adjective greedy when referring to these sets. Before we prove that each of these greedy balance point sets contains all positive integers not in , let us check the example . The following basecomputations are transparent: Hence we have the midpoint triples , , , showing that . Now consider the general case.

Theorem 18. The greedy midpoint-free subset has greedy balance point sets satisfying

Proof. Let . First we show that is a midpoint for . Note that there is at least one such that . Specify by their basedigits; thus We have for all . Also , because . Hence . It is easily checked that for all , so . Hence .
Next we show that is an upper endpoint for . Let be specified by their basedigits; thus There is at least one such that , so . Also for all , so . Hence .
Finally we show that is a lower endpoint for . We need to find such that and . Basearithmetic for this sum takes the form for every , with carry-overs satisfying and Specify by their basedigits; thus
If and for then for . It follows that . Since , for every , we have .
There are integers such that and for and and for . If then , ; if then . In either case it follows that . If then so since for all . If then there is an integer such that and for while . If and then so . Hence for all . Also , and , so . Then and for all , so .
Finally, in all three cases for each , so . Hence .

Corollary 19. The greedy midpoint-free subset is maximal in .

In  it was shown that is actually a maximal midpoint-free subset of . The proof is not repeated here, but let us note that the single example follows from computing with in the form . Baseconsiderations yield corresponding to , so as claimed.

For brevity, let , , . With Theorem 18, this yields the following result.

Corollary 20. The greedy midpoint-free subset has greedy balance point sets satisfying , .

Reversing implications, this yields the following result.

Corollary 21. The greedy midpoint-free subset is maximal in .

Now the multiplicities for can be examined. First note that the sum of basedigits of any even integer is , so even integers have an even number of basedigits equal to. If , at least one digit in must be, so the total number of such digits is , and .

Theorem 22. If , then the midpoint multiplicity of is where is the number of digits equal toin and .

Proof. Suppose satisfy and . There is no carry-over in the basearithmetic for the sum, so , for all . If then we must have There is an integer such that and for all , so forces . However, when and , the requirement is satisfied if . Both possibilities are consistent with , so there are solutions in total.

To illustrate, when then so has just two triples withas midpoint: , . In contrast, when then so has eight triples withas midpoint, ranging from to , namely, for all .

Fix . The digits of the baserepresentation include at least one. An ordered pair of integers with is critical for if and , with when . Any two ordered pairs and critical for are independent if . Any set of ordered pairs critical for is independent if every two members are independent. Let denote the family of all sets of independent ordered pairs critical for , including the empty set.

Theorem 23. If , then has total endpoint multiplicity

Proof. Given , we seek such that . Any solution necessarily satisfies . Basearithmetic for this sum takes the form for every , with carry-overs satisfying and Specifying by their basedigits, the equation requires As is easily verified, these digit specifications satisfy the requirements of basearithmetic for .
Any integer is a basetransition point for the sum if . Note that the digits