International Journal of CombinatoricsVolume 2014 (2014), Article ID 214637, 8 pageshttp://dx.doi.org/10.1155/2014/214637

Research Article

Midpoint-Free Subsets of the Real Numbers

Roger B. Eggleton

School of Mathematical and Physical Sciences, University of Newcastle, Callaghan, NSW 2308, Australia

Received 22 May 2014; Accepted 29 July 2014; Published 26 August 2014

Academic Editor: Chris A. Rodger

Copyright © 2014 Roger B. Eggleton. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

A set of reals is midpoint-free if it has no subset such that and . If and is midpoint-free, it is a maximal midpoint-free subset of if there is no midpoint-free set such that . In each of the cases , we determine two maximal midpoint-free subsets of characterised by digit constraints on the base 3 representations of their members.

1. Introduction

Let us say that a set of three real numbers is a midpoint triple if it satisfies and . In this case is the midpoint of the set, is its lower endpoint, and is its upper endpoint. This geometric viewpoint immediately suggests the main objective of this paper, which is to identify several significant subsets of that contain no midpoint triple.

A midpoint-free subset of is any set that contains no midpoint triple. For instance, the set of powers of 2 is midpoint-free, since appropriate scaling shows that the sum of two distinct powers of 2 is never equal to a power of 2. Note that if and , then is midpoint-free, for any with , if and only if is midpoint-free.

If and is midpoint-free, then is a maximal midpoint-free subset of if there is a midpoint triple in any set such that . (In this case, any is a member of a midpoint triple in .) If is a “natural” subset of , such as the nonnegative integers , the rationals , or indeed itself, it is of considerable interest to identify maximal midpoint-free subsets of .

Alternatively, any set such that and can be viewed as a 3-term arithmetic progression (A.P.) with first term , midterm , and last term . This arithmetic viewpoint has led to many studies of sequences of positive integers in which there is no 3-term A.P., raising questions such as the following. How large is a maximal subset of the first positive integers that contains no 3-term A.P.? (See [1] for data.) Given a finite set of positive integers with no 3-term A.P., what does its greedy algorithm extension look like? How does it compare with a largest possible subset of the first positive integers that contains and has no 3-term A.P.? Under what conditions must a subset of the positive integers contain a 3-term A.P.? Among those who have made major contributions to our understanding of these questions, one must list such luminaries as Van der Waerden, Erdős, Turán, Rado, Behrend, Roth, Graham, and Szemerédi. For a compact survey and extensive bibliography of such investigations, see Guy [2]. As an example of recent work in this area, see Dybizbański [3].

In contrast with the usual arithmetic viewpoint, the geometric viewpoint adopted here takes us in an apparently novel direction, where the sets of interest are most naturally viewed as subsets of .

2. Notational Conventions

In what follows, a key tool for discussing midpoint triples will be base 3 representations of the real numbers, so relevant notational conventions will now be specified. If and then the two-way infinite string is a base 3 representation of and is the digit in place of . If for infinitely many , then is a regular base 3 representation of ; otherwise, is a singular base 3 representation of , and there is an integer such that for all . The regular base 3 representation of is unique, and if has a singular base 3 representation, that is also unique. If is some “natural” subset of and , it will be convenient to use to denote the set of all members of with a base 3 representation restricted to . This notation adapts to cases where is a configuration of base 3 digits; thus, when is a finite string of base 3 digits, will denote the set of all members of with a base 3 representation which includes a one-way infinite string of recurring blocks .

Since , at least one of the digits is nonzero. The leading digit of is the digit , such that for all . The trivial zeros of are all the digits with . Conventionally these are suppressed (i.e., kept implicit) when listing . If , so the leading digit occupies a negative place, the placeholder zeros of are all the digits with . If there is an integer such that and for all , then is a regular representation, is its trailing digit, its optional zeros are all the digits with , and if then its placeholder zeros are the digits with . If has a trailing digit, it is conventional to suppress its optional zeros as well as its trivial zeros; the finite digit string remaining is the terminating base 3 representation of .

These conventions are extended to , as follows. Although the digits in this case are for all , in this exceptional case is defined to be both the leading digit and the trailing digit of the representation. Then its trivial zeros are all those in places , and its optional zeros are all those in places ; thus, 0 is the terminating representation for .

If has no trailing digit, it is a nonterminating representation of and it may be either regular or singular. If is singular, there is an integer such that and for all ; the digit is the pivot digit of . In this case there is an alternative, regular base 3 representation of , namely, where for all , , and for all . Evidently is a terminating representation, is its trailing digit, and , so .

The regular base 3 representation of any nonnegative rational has a recurring block of digits; if has a singular base 3 representation, this has a recurring 2. For brevity, any such recurring block can be enclosed in square brackets, and the subscript 3 can be used to indicate base 3 notation; for instance, The set of all nonnegative rationals with a terminating base 3 representation is The set of positive rationals with a singular base 3 representation is simply The set of positive rationals with base 3 representation containing a recurring 1 is It will later be found that this set requires particular attention when members are doubled, because

Base 3 representation of negative reals is simply achieved by writing any such number as , with , so that has its base 3 representation adopted from that of , with trivial zeros suppressed and the unary minus operator applied to precede the leading digit. If has both a regular and a singular representation, so does .

To distinguish between instances of base 3 and base 10 representations, an explicit subscript 3 is normally attached to the former, while a subscript 10 is normally kept implicit for the latter. As needed, will denote the regular base 3 representation of , with reserved for the singular representation when this exists. Finally, will denote , the digit in place of the regular base 3 representation of , while will denote the corresponding digit in the singular representation when this exists.

3. Three Sparse Subsets of

Let be the set of all nonnegative integers with terminating base 3 representation in which the only explicit digits are in . The three cases of interest are those where is a 2-set. They begin as follows:

Some features are obvious. Clearly . When and , all integers congruent to (mod 3) are absent from . The first positive integer absent from all three sets is , but it is clear that an increasing proportion of integers will be missing from all three sets. Indeed, a simple calculation shows for any positive integer that and each have members below , while has members below , so each set is sparse and has asymptotic density 0.

It turns out that these three sets are of considerable interest when it comes to the presence or absence of midpoint triples, so let us now address the main subject of this paper.

Note that the set contains midpoint triples, such as and . In fact, is densely packed with midpoint triples, since each member is the lower endpoint of an infinite family of midpoint triples: for any integer , where is the number of digits in the terminating base 3 representation of (if is the leading digit in , then ), and is the positive integer with terminating base 3 representation of digits, all 1. Note that , and appropriate scaling shows that this set is midpoint-free. (The integers are rep-units in base 3. The positive integers with explicit decimal digits all equal to 1 are the rep-units for base 10, so called by contraction of “repeated unit” [4]. Prime factorizations of rep-units have been much studied [5, 6]. Clearly, if , then . Thus can be a prime number only if is prime, but shows that this condition is not sufficient. Such observations generalize to rep-units in any base.)

Not only is every member of the lower endpoint of infinitely many midpoint triples, it is also the case that any member greater than 2 is either the midpoint or the upper endpoint of at least one midpoint triple, depending on the leading digit of its base 3 representation.

In contrast, it turns out that and are midpoint-free. To see this, it suffices to show that is midpoint-free: then ensures that is also midpoint-free. (Long ago, Erdős and Turán [7] noted that the nonnegative integers with a 2-less base 3 representation are midpoint-free. The following compact proof is included for completeness and to typify what follows.)

Claim A. The set is midpoint-free.

Proof. On the contrary, suppose that is a midpoint triple, with and . All digits in are in and all digits in are in . Then for all integers , and Hence, , a contradiction, so contains no midpoint triple.

It will now be shown that is not contained in any larger midpoint-free subset of .

Claim B. For any positive integer , there is a midpoint triple in with as its midpoint.

Proof. Given , we seek such that and . Specify the base 3 digits of and as follows: This determines integers and such that and since for all . At least one digit of is 1, since , so and differ in at least one place: then in each such place, ensuring that . Put . Then , so , whence and . Thus .

It will now be shown that extending by adjoining any rational of the form , with , always yields a set which contains a midpoint triple.

Claim C. For any , there is a midpoint triple in with as its midpoint.

Proof. Again we seek , such that and . In the present case, note that is an odd positive integer, so contains the digit 1 an odd number of times. Specifying and as in the proof of Claim B, once again it follows that and .

For instance, the proofs of Claims B and C produce the decompositions showing that 23 and 23/2 are the midpoints of the pairs and , respectively. Further, doubling shows that 46 and 23 are the midpoints of the pairs and , respectively. From Claims A, B, and C and the doubling method just illustrated, we have the following.

Theorem 1. The sets and are maximal midpoint-free subsets of .

4. Two Unexpected Results for

Let us now study and as midpoint-free subsets of , the full set of integers. Surprisingly, the results for the two sets are quite different.

Calculations until now have involved sums of pairs , so it has been possible to work digit by digit in base 3 without a carry over digit. When the context is widened to , it seems that carry over digits can no longer be avoided, so we shall consider blocks of digits rather than single digits. Let the compact notation (“ by ”) denote a homogeneous block of adjacent digits all equal to ; thus, denotes the terminating base 3 representation of rep-unit . For blocks in which the digits are not all equal we simply abut such expressions with as a natural simplification. To illustrate, consider the integer Let us find integers such that . Regarding as a string of homogeneous blocks, we choose matching blocks for so that Calculating from right to left, base 3 arithmetic for yields successively the following blocks, with each carry over digit indicated parenthetically: Assembling these blocks yields After simplifying we have so and as desired. This calculation models the proof of the following.

Claim D. For any integer , there is a midpoint triple in with the negative integer as its lower endpoint.

Proof. Given any positive integer , we seek , such that and , so . Partition into its homogeneous blocks , so and seek corresponding blocks and to construct and .If has a terminal block , assign the terminal block , so has the terminal block . If has a nonterminal block for some , assume the sum of the preceding blocks and contributes a carry over digit of 1, and assign the corresponding block , so As corresponds to , the computation is justified.If has a block , assign the corresponding block , so has the corresponding block , where the carry over digit is either 0 or 1, so or , respectively. This holds since the sum corresponds to and corresponds to .If has a terminal block , assign the block , so has terminal block , since the sum corresponds to . Similarly if , assign the corresponding blocks ; then has the blocks . Otherwise, if has a nonterminal block for some , assume that yields a carry over digit 1, and assign the block, so has .For note that the only case in which does not contribute a carry over digit 1 to is when and has terminal block . Thus, all our assumptions about carry over digits are justified a posteriori.

Claim E. For any integer , there is a midpoint triple in if and only if is even, and then is the lower endpoint of the triple.

Proof. By doubling, it follows from Claim D that, for any integer , there is a midpoint triple in . However, we claim that there is no midpoint triple in . On the contrary, suppose forms a midpoint triple with the pair . Without loss of generality, and , so : this is impossible, since is odd and is even. It follows that is midpoint-free.

Claim F. The set is midpoint-free.

Proof. The two sets forming this union are certainly midpoint-free, so any midpoint triple in the union must have two members in one set and one member in the other. Claim E shows that there is no midpoint triple with exactly one member in , since this set only contains odd negative integers. So suppose there is a midpoint triple with exactly one member . Then is a midpoint triple in with exactly one member in , the case ruled out by Claim E.

Since is a maximal midpoint-free subset of by Theorem 1, clearly is a maximal midpoint-free subset of . Note that the members of are precisely those positive integers for which the trailing digit of is 1, and all other digits are in . With Claims E and F, it follows that is a maximal midpoint-free subset of . Combined with Claim D, this proves the following.

Theorem 2. The sets and are maximal midpoint-free subsets of .

5. Midpoint-Free Subsets of

Now let us consider the corresponding subsets of the nonnegative rationals , namely, the subsets comprising those members with a regular base 3 representation in which each digit, after suppression of all trivial and optional zeros, is in . The three cases of interest are those where is a 2-set.

As expected, is densely packed with midpoint triples. This follows from the observation that each rational is the lower endpoint of an infinite family of midpoint triples, for any integer , where is the -digit base 3 rep-unit and , where is the leading digit of the regular base 3 representation .

Next, consider . The details are a little more complicated than in the integer context, since allowance must be made for singular representations when particular members of are doubled.

Claim G. The set is midpoint-free.

Proof. If and , all digits of are in . They uniquely determine the digits of and such that and . As in the proof of Claim A, they force . Hence, there is no midpoint triple .

Claim H. Any is the midpoint of a triple with both its endpoints in the set .

Proof. We seek with and . Then and are uniquely determined by for all . Since , then for some , so It follows that is a terminating representation with an odd number of digits equal to 1. Then Hence, the required midpoint triple exists, and in fact .

In particular, if is such that for some integer and for all , then and the proof of Claim H determines and . Since if and if , we call the base 3 fractional rep-unit with offset , denoted by . Then is the midpoint of the triple with endpoints .

Positive rationals not in also belong to midpoint triples with endpoints in , but proving this needs care. For instance, so {13/80, 5/16, 37/80} is a midpoint triple with endpoints in . This calculation models the proof of the following general result.

Claim I. Any is the midpoint of a triple with both its endpoints in the set .

Proof. We seek with and . When , the result follows from Claim H, so we may now assume that Then has at least one digit equal to 1 and its recurring block, say , has at least one digit different from 2. We require and to satisfy for all . Requiring for all ensures that , since If the recurring block of contains 0, then the recurring blocks of and also contain 0, so neither nor is in . Now suppose has at least one digit equal to 1 and none equal to 0. Clearly, in this case we may choose the integer so that the digit occurs in . If has digits, then for all choose Then and are uniquely determined, so that and each has a recurring block containing 0, so neither nor is in .

If then has at least one digit equal to 1, so it follows that . Claim I shows that there is a midpoint triple with and . Then is a midpoint triple with midpoint and endpoints . Hence, doubling applied to Claims G and I shows that is midpoint-free, and every is the midpoint of a triple with endpoints in , so Claims G, H, and I establish the following.

Theorem 3. The sets and are maximal midpoint-free subsets of .

6. Midpoint-Free Subsets of

Now consider and as subsets of . The role of is yet more prominent in this context.

Claim J. The set is midpoint-free.

Proof. Suppose satisfies and . There is an integer for which so This contradicts the fact that is midpoint-free, by Claim A.

Claim K. Any positive rational is the midpoint of a triple with endpoints in .

Proof. Given , we seek , such that and . There is an integer such that , so there is an integer such that . At least one digit of is 2, so at least one digit of is 1. Define by Then for all , with strict inequality for at least one , so . Also for all , so . Finally, let Then and .

Claim L. The set is midpoint-free.

Proof. By Claims G and J, if the specified set contains a midpoint triple, two members of the triple must have one sign and the third member must have the opposite sign. Suppose and are such that and . Since , , and are eventually periodic, there is an integer such that their digits in places are purely periodic, and Since , there is a such that and therefore cannot be equal to 0 for all . On the other hand, cannot be equal to 2 for all since . Hence the recurring block of contains 0 and 2, so for some . But , so has a “carry over” of 1, regardless of whether or not the digits in place contribute any “carry over”. Then By this contradiction, there is no midpoint triple of the proposed type. A similar but simpler argument shows that there are no subsets and such that and .

Claim M. For any positive rational there is a midpoint triple in , with as its lower endpoint.

Proof. Given , we seek such that and . Note that if the triple has all the required properties, the scaled triple also has all the required properties, for any integer . Let us assume, without loss of generality, that the digits of in negative places are purely periodic, and partitioning into homogeneous blocks yields for suitable integers . As , if , then so and a suitable set exists, by Claim D.Now suppose and has at least one negative place digit which is nonzero. If has no negative place digit equal to 2, let , where and are the integer part and fractional part of , respectively. The proof of Claim D yields integers , such that is a midpoint triple. Choose . Then is a midpoint triple with .Finally, suppose for at least one . Define homogeneous blocks and for all as follows: Assuming a carry over digit equal to 1, note that checks as in the proof of Claim D justify the calculations so in all cases we have . Choose Then and . Also ensures that , as required.

For instance, beginning with the proof of Claim M constructs a midpoint triple with However, if we begin with the construction falls back to Claim D, resulting in and .

It now follows from Claims I, K, L, and M that the set is a maximal midpoint-free subset of . This settles midpoint questions about as a subset of .

What is the situation for ? If then the recurring block of must have at least one digit equal to 0. Hence, there must be at least one digit equal to 0 in the recurring block of , so : it follows that . Again, if there is some , such that , it follows that is a terminating representation with trailing digit 1 and all other digits in . Hence, The converse also holds, so where is the set of positive rationals which have a terminating base 3 representation with trailing digit 1 and all other digits in , so In passing, note that . By doubling, it now follows from Claim L that the set is midpoint-free.

Claim N. Any rational is the midpoint of a triple with both its endpoints in .

Proof. If , then , so is the midpoint of a triple with endpoints in , by Claim I. By doubling, is the midpoint of a triple with both its endpoints in the set .

Claim O. For any rational , there is a midpoint triple in with as its lower endpoint.

Proof. If and , then , so there is a midpoint triple in with as its lower endpoint, by Claim M. By doubling, has a midpoint triple with lower endpoint .

The preceding results now fully establish the following.

Theorem 4. The two sets are both maximal midpoint-free subsets of .

7. Midpoint-Free Subsets of and

Now consider and . These sets are uncountable, while their maximal rational subsets and are countable, yet no essentially new considerations arise in extending the results of the previous two sections from to and from to . Noting that and , it suffices to state the end results for this wider context.

Theorem 5. The sets and are both maximal midpoint-free subsets of .

Theorem 6. The two sets are both maximal midpoint-free subsets of .

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

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