Research ArticleMidpoint-Free Subsets of the Real Numbers

Roger B Eggleton

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

Correspondence should be addressed to Roger B Eggleton rogerilstuedu

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

Academic Editor Chris A Rodger

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

A set of reals 119878 sub R is midpoint-free if it has no subset 119886 119887 119888 sube 119878 such that 119886 lt 119887 lt 119888 and 119886 + 119888 = 2119887 If 119878 sub 119883 sube R and 119878 ismidpoint-free it is amaximal midpoint-free subset of119883 if there is no midpoint-free set 119879 such that 119878 sub 119879 sube 119883 In each of the cases119883 = Z+ZQ+QR+R we determine two maximal midpoint-free subsets of 119883 characterised by digit constraints on the base 3representations of their members

1 Introduction

Let us say that a set of three real numbers 119886 119887 119888 sub R is amidpoint triple if it satisfies 119886 lt 119887 lt 119888 and 119886 + 119888 = 2119887 In thiscase 119887 is the midpoint of the set 119886 is its lower endpoint and 119888

is its upper endpoint This geometric viewpoint immediatelysuggests the main objective of this paper which is to identifyseveral significant subsets of R that contain no midpointtriple

A midpoint-free subset of R is any set that contains nomidpoint triple For instance the set 2119899 119899 isin Z of powersof 2 is midpoint-free since appropriate scaling shows that thesum of two distinct powers of 2 is never equal to a power of2 Note that if 119878 sube R and 119888119878 + 119889 = 119888119904 + 119889 119904 isin 119878 then 119888119878 + 119889

is midpoint-free for any 119888 119889 sub R with 119888 = 0 if and only if119878 is midpoint-free

If 119878 sub 119883 sube R and 119878 is midpoint-free then 119878 is amaximalmidpoint-free subset of 119883 if there is a midpoint triple in anyset 119879 such that 119878 sub 119879 sube 119883 (In this case any 119909 isin 119883 119878 isa member of a midpoint triple in 119878 cup 119909) If 119883 is a ldquonaturalrdquosubset ofR such as the nonnegative integersZ+ the rationalsQ or indeed R itself it is of considerable interest to identifymaximal midpoint-free subsets of119883

Alternatively any set 119886 119887 119888 such that 119886 lt 119887 lt 119888

and 119886 + 119888 = 2119887 can be viewed as a 3-term arithmeticprogression (AP) with first term 119886 midterm 119887 and lastterm 119888 This arithmetic viewpoint has led to many studiesof sequences of positive integers in which there is no 3-termAP raising questions such as the following How large is a

maximal subset of the first 119899 positive integers that containsno 3-term AP (See [1] for data) Given a finite set 119878 ofpositive integers with no 3-term AP what does its greedyalgorithm extension look like How does it compare witha largest possible subset of the first 119899 positive integers thatcontains 119878 and has no 3-term AP Under what conditionsmust a subset of the positive integers contain a 3-term APAmong those who have made major contributions to ourunderstanding of these questions one must list such lumi-naries as Van der Waerden Erdos Turan Rado BehrendRoth Graham and Szemeredi For a compact survey andextensive bibliography of such investigations see Guy [2]As an example of recent work in this area see Dybizbanski[3]

In contrast with the usual arithmetic viewpoint thegeometric viewpoint adopted here takes us in an apparentlynovel direction where the sets of interest are most naturallyviewed as subsets ofR

2 Notational Conventions

In what follows a key tool for discussing midpoint tripleswill be base 3 representations of the real numbers so relevantnotational conventions will now be specified If 119903 isin R 119903 gt 0

and

119903 = sum

119894isinZ

1198891198943119894 with 119889

119894isin 0 1 2 (1)

Hindawi Publishing CorporationInternational Journal of CombinatoricsVolume 2014 Article ID 214637 8 pageshttpdxdoiorg1011552014214637

2 International Journal of Combinatorics

then the two-way infinite string

d = sdot sdot sdot 119889211988911198890sdot 119889minus1119889minus2

sdot sdot sdot (2)

is a base 3 representation of 119903 and 119889119894is the digit in place 119894

of d If 119889119894isin 0 1 for infinitely many 119894 lt 0 then d is a

regular base 3 representation of 119903 otherwise d is a singularbase 3 representation of 119903 and there is an integer 119896 such that119889119894= 2 for all 119894 lt 119896 The regular base 3 representation of

119903 is unique and if 119903 has a singular base 3 representationthat is also unique If 119883 is some ldquonaturalrdquo subset of R and119863 sub 0 1 2 it will be convenient to use119883

3(119863) to denote the

set of all members of119883with a base 3 representation restrictedto119863This notation adapts to cases where119863 is a configurationof base 3 digits thus when119863 is a finite string of base 3 digits1198833([119863]) will denote the set of all members of 119883 with a base

3 representation which includes a one-way infinite string ofrecurring blocks119863

Since 119903 gt 0 at least one of the digits is nonzero Theleading digit of d is the digit 119889

ℎgt 0 such that 119889

119894= 0 for

all 119894 gt ℎ The trivial zeros of d are all the digits 119889119894= 0 with

119894 gt maxℎ 0 Conventionally these are suppressed (ie keptimplicit) when listing d If ℎ lt 0 so the leading digit 119889

ℎ

occupies a negative place the placeholder zeros of d are all thedigits 119889

119894= 0 with 0 ge 119894 gt ℎ If there is an integer 119896 le ℎ

such that 119889119896gt 0 and 119889

119894= 0 for all 119894 lt 119896 then d is a regular

representation 119889119896is its trailing digit its optional zeros are all

the digits 119889119894= 0 with 119894 lt min119896 0 and if 119896 gt 0 then its

placeholder zeros are the digits 119889119894= 0with 0 le 119894 lt 119896 If d has a

trailing digit it is conventional to suppress its optional zerosas well as its trivial zeros the finite digit string remaining isthe terminating base 3 representation of 119903

These conventions are extended to 119903 = 0 as followsAlthough the digits in this case are 119889

119894= 0 for all 119894 in this

exceptional case 1198890is defined to be both the leading digit and

the trailing digit of the representation Then its trivial zerosare all those in places 119894 gt 0 and its optional zeros are all thosein places 119894 lt 0 thus 0 is the terminating representation for119903 = 0

If d has no trailing digit it is a nonterminating represen-tation of 119903 and it may be either regular or singular If d issingular there is an integer 119896 le ℎ + 1 such that 119889

119896isin 0 1

and 119889119894= 2 for all 119894 lt 119896 the digit 119889

119896is the pivot digit of d In

this case there is an alternative regular base 3 representationof 119903 namely

e = sdot sdot sdot 119890211989011198900sdot 119890minus1119890minus2

sdot sdot sdot (3)

where 119890119894= 119889119894for all 119894 gt 119896 119890

119896= 119889119896+ 1 and 119890

119894= 0 for all 119894 lt 119896

Evidently e is a terminating representation 119890119896is its trailing

digit and 3minus119896119903 isin Z+ so 119903 isin 3

119896Z+

The regular base 3 representation of any nonnegative

rational 119902 isin Q+ has a recurring block of digits if 119902 hasa singular base 3 representation this has a recurring 2 Forbrevity any such recurring block can be enclosed in square

brackets and the subscript 3 can be used to indicate base 3notation for instance

7

3= 2 sdot 1

3= 2 sdot 1[0]

3= 2 sdot 0[2]

3

1

2= 0 sdot [1]

3

2

5= 0 sdot [1012]

3

(4)

The set of all nonnegative rationals with a terminating base 3representation is

Q+

3([0]) = ⋃

119894isinZ+

3minus119894

Z+

(5)

The set of positive rationals with a singular base 3 represen-tation is simply

Q+

3([2]) = Q

+

3([0]) 0 (6)

The set of positive rationals with base 3 representationcontaining a recurring 1 is

Q+

3([1]) = ⋃

119894isinZ+

3minus119894

(Z+

+1

2) (7)

It will later be found that this set requires particular attentionwhen members are doubled because

2Q+

3([1]) sub Q

+

3([2]) sub Q

+

3([0]) (8)

Base 3 representation of negative reals is simply achievedby writing any such number as minus119903 with 119903 isin R 119903 gt 0 sothat minus119903 has its base 3 representation adopted from that of 119903with trivial zeros suppressed and the unary minus operatorapplied to precede the leading digit If 119903 has both a regularand a singular representation so does minus119903

To distinguish between instances of base 3 and base 10representations an explicit subscript 3 is normally attachedto the former while a subscript 10 is normally kept implicitfor the latter As needed 119903

3will denote the regular base

3 representation of 119903 with 119903lowast

3reserved for the singular

representation when this exists Finally ⟦119903⟧3119894

will denote119889119894 the digit in place 119894 of the regular base 3 representation

of 119903 while ⟦119903⟧lowast

3119894will denote the corresponding digit in the

singular representation when this exists

3 Three Sparse Subsets of Z+

Let Z+3(119863) be the set of all nonnegative integers with termi-

nating base 3 representation in which the only explicit digitsare in119863 sub 0 1 2The three cases of interest are those where119863 is a 2-set They begin as follows

Z+

3(0 1)

= 0 1 3 4 9 10 12 13 27 28

30 31 36 37 39 40 81

Z+

3(0 2)

= 0 2 6 8 18 20 24 26 54 56 60 62 72 74 78 80

International Journal of Combinatorics 3

Z+

3(1 2)

= 1 2 4 5 7 8 13 14 16 17 22 23 25 26 40

(9)

Some features are obvious Clearly Z+3(0 2) = 2Z+

3(0 1)

When 119889 isin 0 1 2 and 119863 = 0 1 2 119889 all integerscongruent to 119889 (mod 3) are absent from Z+

3(119863) The first

positive integer absent from all three sets is 1023= 11 but

it is clear that an increasing proportion of integers will bemissing fromall three sets Indeed a simple calculation showsfor any positive integer 119899 thatZ+

3(0 1) andZ+

3(0 2) each have

2119899 members below 3

119899 while Z+3(1 2) has 2119899+1 minus 2 members

below 3119899 so each set is sparse and has asymptotic density 0

It turns out that these three sets are of considerableinterest when it comes to the presence or absence ofmidpointtriples so let us now address the main subject of this paper

Note that the set Z+3(1 2) contains midpoint triples such

as 1 4 7 and 2 5 8 In factZ+3(1 2) is densely packedwith

midpoint triples since eachmember 119886 isin Z+3(1 2) is the lower

endpoint of an infinite family of midpoint triples

119886 119906119899minus 119906119898+ 119886 2119906

119899minus 2119906119898+ 119886

= (119906119899minus 119906119898) 0 1 2 + 119886 sub Z

+

3(1 2)

(10)

for any integer 119899 gt 119898 where 119898 is the number of digits inthe terminating base 3 representation of 119886 (if 119889

ℎis the leading

digit in 1198863 then119898 = ℎ + 1) and

119906119899= 1 + 3 + 3

2

+ sdot sdot sdot + 3119899minus1

=3119899minus 1

2(11)

is the positive integer with terminating base 3 representationof 119899 digits all 1 Note that Z+

3(1) = 119906

119899 119899 ge 1 and

appropriate scaling shows that this set is midpoint-free (Theintegers 119906

119899are rep-units in base 3 The positive integers with

explicit decimal digits all equal to 1 are the rep-units for base10 so called by contraction of ldquorepeated unitrdquo [4] Primefactorizations of rep-units have been much studied [5 6]Clearly if 119898 | 119899 then 119906

119898| 119906119899 Thus 119906

119899can be a prime

number only if 119899 is prime but 1199065= 11111

3= 112 shows that

this condition is not sufficient Such observations generalizeto rep-units in any base)

Not only is every member of Z+3(1 2) the lower endpoint

of infinitely many midpoint triples it is also the case that anymember greater than 2 is either the midpoint or the upperendpoint of at least one midpoint triple depending on theleading digit of its base 3 representation

In contrast it turns out that Z+3(0 1) and Z+

3(0 2) are

midpoint-free To see this it suffices to show that Z+3(0 1) is

midpoint-free thenZ+3(0 2)= 2Z+

3(0 1) ensures thatZ+

3(0 2)

is also midpoint-free (Long ago Erdos and Turan [7] notedthat the nonnegative integers with a 2-less base 3 represen-tation are midpoint-free The following compact proof isincluded for completeness and to typify what follows)

Claim A The set Z+3(0 1) is midpoint-free

Proof On the contrary suppose that 119886 119887 119888 sub Z+3(0 1) is a

midpoint triple with 119886 lt 119887 lt 119888 and 119886 + 119888 = 2119887 All digits in1198863 1198873 1198883are in 0 1 and all digits in (2119887)

3are in 0 2 Then

⟦119886⟧3119894

+ ⟦119888⟧3119894

= ⟦2119887⟧3119894 (12)

for all integers 119894 and

⟦2119887⟧3119894

= 0 997904rArr ⟦119886⟧3119894

= ⟦119887⟧3119894

= ⟦119888⟧3119894

= 0

⟦2119887⟧3119894

= 2 997904rArr ⟦119886⟧3119894

= ⟦119887⟧3119894

= ⟦119888⟧3119894

= 1

(13)

Hence 119886 = 119887 = 119888 a contradiction so Z+3(0 1) contains no

midpoint triple

It will now be shown thatZ+3(0 1) is not contained in any

larger midpoint-free subset of Z+

Claim B For any positive integer 119887 isin Z+ Z+

3(0 1) there is a

midpoint triple in Z+3(0 1) cup 119887 with 119887 as its midpoint

Proof Given 119887 isin Z+ Z+

3(0 1) we seek 119886 119888 sub Z+

3(0 1) such

that 119886 lt 119887 lt 119888 and 119886 + 119888 = 2119887 Specify the base 3 digits of 119886and 119888 as follows

⟦2119887⟧3119894

= 0 997904rArr ⟦119886⟧3119894

= ⟦119888⟧3119894

= 0

⟦2119887⟧3119894

= 2 997904rArr ⟦119886⟧3119894

= ⟦119888⟧3119894

= 1

⟦2119887⟧3119894

= 1 997904rArr ⟦119886⟧3119894

= 0 ⟦119888⟧3119894

= 1

(14)

This determines integers 119886 and 119888 such that 119886 119888 sub Z+3(0 1)

and 119886 + 119888 = 2119887 since

⟦119886⟧3119894

+ ⟦119888⟧3119894

= ⟦2119887⟧3119894 (15)

for all 119894 At least one digit of (2119887)3is 1 since 119887 notin Z+

3(0 1) so 119886

3

and 1198883differ in at least one place then ⟦119886⟧

3119894lt ⟦119888⟧

3119894in each

such place ensuring that 119886 lt 119888 Put 119889 = (119888 minus 119886)2 gt 0 Then119888 = 119886+ 2119889 so 2119887 = 119886+ 119888 = 2119886+ 2119889 whence 119887 = 119886+119889 gt 119886 and119888 = 119886 + 2119889 = 119887 + 119889 gt 119887 Thus 119886 lt 119887 lt 119888

It will now be shown that extendingZ+3(0 1) by adjoining

any rational of the form 119899 + (12) with 119899 isin Z+ always yieldsa set which contains a midpoint triple

Claim C For any 119887 isin Z+

+ (12) there is a midpoint triple inZ+3(0 1) cup 119887 with 119887 as its midpoint

Proof Again we seek 119886 119888 sub Z+3(0 1) such that 119886 lt 119887 lt 119888

and 119886 + 119888 = 2119887 In the present case note that 2119887 is an oddpositive integer so (2119887)

3contains the digit 1 an odd number

of times Specifying 119886 and 119888 as in the proof of Claim B onceagain it follows that 119886 119888 sub Z+

3(0 1) and 119886 lt 119887 lt 119888

For instance the proofs of Claims B and C produce thedecompositions

12013= 100

3+ 1101

3 212

3= 101

3+ 111

3(16)

showing that 23 and 232 are the midpoints of the pairs9 37 sub Z+

3(0 1) and 10 13 sub Z+

3(0 1) respectively

4 International Journal of Combinatorics

Further doubling shows that 46 and 23 are the midpointsof the pairs 18 74 sub Z+

3(0 2) and 20 26 sub Z+

3(0 2)

respectively From Claims A B and C and the doublingmethod just illustrated we have the following

Theorem 1 The sets Z+3(0 1) and Z+

3(0 2) are maximal

midpoint-free subsets of Z+

4 Two Unexpected Results for Z

Let us now study Z+3(0 1) and Z+

3(0 2) as midpoint-free

subsets of Z the full set of integers Surprisingly the resultsfor the two sets are quite different

Calculations until now have involved sums of pairs119886 119888 sub Z

+

3(0 1) so it has been possible to work digit by digit

in base 3 without a carry over digit When the context iswidened to Z it seems that carry over digits can no longerbe avoided so we shall consider blocks of digits rather thansingle digits Let the compact notation 119889

times119899 (ldquo119889 by 119899rdquo) denotea homogeneous block of 119899 adjacent digits all equal to 119889 thus1times119899 denotes the terminating base 3 representation of rep-unit119906119899 For blocks in which the digits are not all equal we simply

abut such expressionswith119889times1 = 119889 as a natural simplificationTo illustrate consider the integer

1198863= 1222202111001121100 = 12

times4

021times3

0times2

1times2

21times2

0times2

(17)

Let us find integers 119887 119888 isin Z+

3(0 1) such that 119886 + 2119887 = 119888

Regarding 1198863as a string of homogeneous blocks we choose

matching blocks for (2119887)3so that

(2119887)3= 2000020222222202200 = 20

times4

202times7

02times2

0times2

(18)

Calculating from right to left base 3 arithmetic for 1198863+ (2119887)

3

yields successively the following blocks with each carry overdigit indicated parenthetically

0times2

+ 0times2

= 0times2

1times2

+ 2times2

= (1) 10

2 + 0 + (1) = (1) 0

1times2

+ 2times2

+ (1) = (1)1times2

0times2

+ 2times2

+ (1) = (1) 0times2

1times3

+ 2times3

+ (1) = (1) 1times3

2 + 0 + (1) = (1) 0

0 + 2 + (1) = (1) 0

2times4

+ 0times4

+ (1) = (1) 0times4

1 + 2 + (1) = (1) 1

(19)

Assembling these blocks yields

1198863+ (2119887)

3= 1times2

| 0times4

| 0 | 0 | 1times3

| 0times2

| 1times2

| 0 | 10 | 0times2

(20)

After simplifying we have

10times4

101times7

01times2

0times2

= 1000010111111101100 = 1198873

1times2

0times6

1times3

0times2

1times2

010times3

= 11000000111001101000 = 1198883

(21)

so 119886 + 2119887 = 119888 and 119887 119888 isin Z+

3(0 1) as desired This calculation

models the proof of the following

Claim D For any integer 119886 gt 0 there is a midpoint triplein Z+3(0 1) cup minus119886 with the negative integer minus119886 as its lower

endpoint

Proof Given any positive integer 119886 we seek 119887 119888 sub Z+3(0 1)

such that minus119886 lt 119887 lt 119888 and minus119886 + 119888 = 2119887 so 119886 + 2119887 = 119888 Partition1198863into its homogeneous blocks 119860

119894 so

1198863= 119860119898sdot sdot sdot 119860119894sdot sdot sdot 11986011198600

(22)

and seek corresponding blocks 119861119894and 119862

119894to construct 119887

3and

1198883If 1198863has a terminal block119860

0= 0times119899 assign 119887

3the terminal

block 1198610= 0times119899 so 119888

3has the terminal block 119862

0= 0times119899

=

1198600+ 21198610 If 1198863has a nonterminal block 119860

119894= 0times119899 for some

119894 gt 0 assume the sum of the preceding blocks 119860119894minus1

and2119861119894minus1

contributes a carry over digit of 1 and assign 1198873the

corresponding block 119861119894= 1times119899 so

119862119894= 119860119894+ 2119861119894+ (1) = 0

times119899

+ 2times119899

+ (1) = (1) 0times119899

(23)

As 2times119899 + (1) corresponds to 2119906119899+ 1 = (3

119899+1minus 1) + 1 = 3

119899+1the computation is justified

If 1198863has a block 119860

119894= 1times119899 assign 119887

3the corresponding

block 119861119894= 1times119899 so 119888

3has the corresponding block 119862

119894= 119860119894+

2119861119894+ (120575) = (1)1

times(119899minus1)

0 + (120575) where the carry over digit 120575 iseither 0 or 1 so 119862

119894= (1)1

times(119899minus1)

0 or (1)1times119899 respectively Thisholds since the sum 1

times119899+2times119899 corresponds to 119906

119899+2119906119899= 3119906119899=

119906119899+1

minus 1199061and 1times119899

+ 2times119899

+ (1) corresponds to 3119906119899+ 1 = 119906

119899+1

If 1198863has a terminal block 119860

0= 2times119899 assign 119887

3the block

1198610= 0times(119899minus1)

1 so 1198883has terminal block 119862

0= 1198600+ 21198610

=

(1)0times(119899minus1)

1 since the sum 2times119899

+ 0times(119899minus1)

2 corresponds to 2119906119899+

2 = (3119899+1

minus 1) + 2 = 3119899+1

+ 1 Similarly if 11986011198600= 2times1198990times119898

assign 1198873the corresponding blocks 119861

11198610= 0times(119899minus1)

10times119898 then

1198883has the blocks 119862

11198620= (1)0

times(119899minus1)

10times119898 Otherwise if 119886

3has

a nonterminal block 119860119894= 2times119899 for some 119894 gt 0 assume that

119860119894minus1

+ 2119861119894minus1

yields a carry over digit 1 and assign 1198873the

block 119861119894= 0times119899 so 119888

3has 119862119894= 119860119894+ 2119861119894+ (1) = (1)0

times119899For 119894 gt 0note that the only case inwhich119860

119894minus1+2119861119894minus1

doesnot contribute a carry over digit 1 to 119860

119894+ 2119861119894is when 119894 = 1

and 1198863has terminal block119860

0= 0times119899Thus all our assumptions

about carry over digits are justified a posteriori

Claim E For any integer 119899 gt 0 there is a midpoint triple inZ+3(0 2)cupminus119899 if and only if 119899 is even and thenminus119899 is the lower

endpoint of the triple

Proof By doubling it follows from Claim D that for anyinteger 119886 gt 0 there is a midpoint triple in Z+

3(0 2) cup

International Journal of Combinatorics 5

minus2119886 However we claim that there is no midpoint triple inZ+3(0 2) cup minus2119886 + 1 On the contrary suppose minus2119886 + 1 forms

a midpoint triple with the pair 119887 119888 sub Z+3(0 2) Without loss

of generality minus2119886 + 1 lt 119887 lt 119888 and (minus2119886 + 1) + 119888 = 2119887 so2119886 + 2119887 minus 1 = 119888 this is impossible since 2119886 + 2119887 minus 1 is oddand 119888 is even It follows thatZ+

3(0 2) cup minus2119886 + 1 is midpoint-

free

Claim F The set Z+3(0 2) cup minus(Z+

3(0 2) + 1) is midpoint-free

Proof The two sets forming this union are certainlymidpoint-free so any midpoint triple in the union musthave two members in one set and one member in the otherClaim E shows that there is no midpoint triple with exactlyone member in minus(Z+

3(0 2) + 1) since this set only contains

odd negative integers So suppose there is a midpoint tripleminus119886 minus119887 119888 with exactly one member 119888 isin Z+

3(0 2) Then

119886 minus 1 119887 minus 1 minus119888 minus 1 is a midpoint triple in

minus (Z+

3(0 2) cup minus (Z

+

3(0 2) + 1)) minus 1

= Z+

3(0 2) cup minus (Z

+

3(0 2) + 1)

(24)

with exactly one member in minus(Z+3(0 2) + 1) the case ruled

out by Claim E

Since Z+3(0 2) is a maximal midpoint-free subset of Z+

by Theorem 1 clearly Z+3(0 2) + 1 is a maximal midpoint-

free subset of Z+ + 1 = Z+ 0 Note that the members ofZ+3(0 2) + 1 are precisely those positive integers 119899 for which

the trailing digit of 1198993is 1 and all other digits are in 0 2

With Claims E and F it follows thatZ+3(0 2)cupminus(Z+

3(0 2)+1)

is a maximal midpoint-free subset of Z = Z+ cup minus(Z+ 0)Combined with Claim D this proves the following

Theorem 2 The sets Z+3(0 1) and Z+

3(0 2) cup minus(Z+

3(0 2) + 1)

are maximal midpoint-free subsets of Z

5 Midpoint-Free Subsets of Q+

Now let us consider the corresponding subsets Q+3(119863) of the

nonnegative rationals Q+ namely the subsets comprisingthose members with a regular base 3 representation in whicheach digit after suppression of all trivial and optional zerosis in 119863 sub 0 1 2 The three cases of interest are those where119863 is a 2-set

As expected Q+3(1 2) is densely packed with midpoint

triples This follows from the observation that each rational119902 isin Q+

3(1 2) is the lower endpoint of an infinite family of

midpoint triples

119902 119906119899minus 119906119898+ 119902 2119906

119899minus 2119906119898+ 119902

= (119906119899minus 119906119898) 0 1 2 + 119902 sub Q

+

3(1 2)

(25)

for any integer 119899 gt 119898 where 119906119899is the 119899-digit base 3 rep-unit

and 119898 = ℎ + 1 where 119889ℎis the leading digit of the regular

base 3 representation 1199023

Next consider Q+3(0 1) The details are a little more

complicated than in the integer context since allowance

must be made for singular representations when particularmembers ofQ+

3(0 1) are doubled

Claim G The setQ+3(0 1) Q+

3([1]) is midpoint-free

Proof If 119887 isin Q+3(0 1) Q+

3([1]) and 119887 gt 0 all digits of (2119887)

3

are in 0 2 They uniquely determine the digits of 1198863and 1198883

such that 119886 119888 sub Q+3(0 1) Q+

3([1]) and 119886 + 119888 = 2119887 As in

the proof of Claim A they force 119886 = 119887 = 119888 Hence there is nomidpoint triple 119886 119887 119888 sub Q+

3(0 1) Q+

3([1])

Claim H Any 119887 isin Q+3([1]) is the midpoint of a triple with

both its endpoints in the setQ+3(0 1) capQ+

3([0])

Proof We seek 119886 119888 sub Q+3(0 1) Q+

3([1]) with 119886 lt 119887 lt 119888 and

119886 + 119888 = 2119887 Then 1198863and 1198883are uniquely determined by

⟦119886⟧3119894

+ ⟦119888⟧3119894

= ⟦2119887⟧3119894 ⟦119886⟧

3119894le ⟦119888⟧

3119894(26)

for all 119894 Since 119887 isin Q+3([1]) then 3

119896119887 isin Z+ + (12) for some

119896 isin Z+ so

2119887 isin 3minus119896

(2Z+

+ 1) (27)

It follows that (2119887)3is a terminating representation with an

odd number of digits equal to 1 Then

⟦2119887⟧3119894

= 1 997904rArr ⟦119886⟧3119894

= 0 ⟦119888⟧3119894

= 1

for an odd number of integers 119894 ge minus119896

⟦2119887⟧3119894

= 0 997904rArr ⟦119886⟧3119894

= ⟦119888⟧3119894

= 0 forall119894 lt minus119896

(28)

Hence the requiredmidpoint triple 119886 119887 119888 exists and in fact119886 119888 sub Q+

3([0])

In particular if 119887 isin Q+3([1]) is such that ⟦2119887⟧

3119899= 1 for

some integer 119899 and ⟦2119887⟧3119894

= 0 for all 119894 = 119899 then 119887 = 31198992 =

119906119899+ (12) and the proof of Claim H determines 119886 = 0 and

119888 = 2119887 = 3119899 Since ⟦119887⟧

3119894= 0 if 119894 ge 119899 and ⟦119887⟧

3119894= 1 if 119894 lt 119899

we call 119887 the base 3 fractional rep-unit with offset 119899 denotedby V119899 Then V

119899isin Q+3([1]) is the midpoint of the triple with

endpoints 0 3119899 sub Q+3(0 1) capQ+

3([0])

Positive rationals not in Q+3(0 1) cup Q+

3([1]) also belong

to midpoint triples with endpoints in Q+3(0 1) Q+

3([1]) but

proving this needs care For instance

5

16= 0 sdot [0221]

3

997904rArr5

8= 0 sdot [12]

3= 0 sdot [0111]

3+ 0 sdot [1101]

3

=13

80+37

80

(29)

so 1380 516 3780 is a midpoint triple with endpoints inQ+3(0 1) Q+

3([1]) This calculation models the proof of the

following general result

Claim I Any 119887 isin Q+ Q+

3(0 1) is themidpoint of a triple with

both its endpoints in the setQ+3(0 1) Q+

3([1])

6 International Journal of Combinatorics

Proof We seek 119886 119888 sub Q+3(0 1) Q+

3([1]) with 119886 lt 119887 lt 119888 and

119886 + 119888 = 2119887 When 119887 isin Q+3([1]) the result follows from Claim

H so we may now assume that

119887 isin Q+

(Q+

3(0 1) cupQ

+

3([1])) (30)

Then (2119887)3has at least one digit 119889

119895equal to 1 and its recurring

block say [119863] has at least one digit different from 2 Werequire 119886

3and 1198883to satisfy

⟦119886⟧3119894

+ ⟦119888⟧3119894

= ⟦2119887⟧3119894

(31)

for all 119894 Requiring ⟦119886⟧3119894

le ⟦119888⟧3119894

for all 119894 ge 119895 ensures that119886 lt 119888 since

⟦119886⟧3119895

= 0 ⟦119888⟧3119895

= 1 (32)

If the recurring block [119863] of (2119887)3contains 0 then the

recurring blocks of 1198863and 1198883also contain 0 so neither 119886 nor

119888 is in Q+3([1]) Now suppose [119863] has at least one digit equal

to 1 and none equal to 0 Clearly in this case we may choosethe integer 119895 so that the digit 119889

119895= 1 occurs in [119863] If119863 has119898

digits then for all 119894 le 119895 choose

⟦119886⟧3119894

= 0 ⟦119888⟧3119894

= 1 when 119894 equiv 119895 mod 2119898

⟦119886⟧3119894

= 1 ⟦119888⟧3119894

= 0 when 119894 equiv 119895 + 119898 mod 2119898

⟦119886⟧3119894

le ⟦119888⟧3119894

otherwise

(33)

Then 1198863and 1198883are uniquely determined so that 119886 lt 119888 and

each has a recurring block containing 0 so neither 119886 nor 119888 isinQ+3([1])

If 119902 isin Q+ Q+

3(0 2) then 119902

3has at least one digit equal

to 1 so it follows that (1199022)3isin Q+ Q

+

3(0 1) Claim I shows

that there is a midpoint triple 119886 119887 119888 with 119887 = 1199022 and119886 119888 sub Q+

3(0 1) Q+

3([1]) Then 2119886 119887 119888 is a midpoint triple

with midpoint 119902 and endpoints 2119886 119888 sub Q+3(0 2) Hence

doubling applied to Claims G and I shows that Q+3(0 2) is

midpoint-free and every 119887 isin Q+ Q+

3(0 2) is the midpoint

of a triple with endpoints in Q+3(0 2) so Claims G H and I

establish the following

Theorem 3 The sets Q+3(0 1) Q+

3([1]) and Q+

3(0 2) are

maximal midpoint-free subsets ofQ+

6 Midpoint-Free Subsets of Q

Now considerQ+3(0 1) andQ+

3(0 2) as subsets ofQ The role

ofQ+3([1]) is yet more prominent in this context

Claim J The setQ+3(0 1) capQ+

3([1]) is midpoint-free

Proof Suppose 119886 119887 119888 sub Q+3(0 1) cap Q+

3([1]) satisfies 119886 lt

119887 lt 119888 and 119886 + 119888 = 2119887 There is an integer 119898 ge 0 for which3119898119886 119887 119888 sub Z+

3(0 1) + (12) so

2 sdot 3119898

119886 119887 119888 sub 2Z+

3(0 1) + 1 (34)

This contradicts the fact that 2Z+3(0 1) + 1 is midpoint-free

by Claim A

Claim K Any positive rational 119887 isin Q+3([1]) Q+

3(0 1) is the

midpoint of a triple with endpoints inQ+3(0 1) capQ+

3([1])

Proof Given 119887 we seek 119886 119888 sub Q+3(0 1) capQ+

3([1]) such that

119886 lt 119887 lt 119888 and 119886 + 119888 = 2119887 There is an integer 119898 such that3119898119887 isin Z+ + (12) so there is an integer 119861 isin Z+ Z+

3(0 1)

such that 3119898119887 = 119861 + (12) At least one digit of 1198613is 2 so at

least one digit of (2119861)3is 1 Define 119860 119862 sub Z+

3(0 1) by

⟦2119861⟧3119894

= 0 997904rArr ⟦119860⟧3119894

= ⟦119862⟧3119894

= 0

⟦2119861⟧3119894

= 1 997904rArr ⟦119860⟧3119894

= 0 ⟦119862⟧3119894

= 1

⟦2119861⟧3119894

= 2 997904rArr ⟦119860⟧3119894

= ⟦119862⟧3119894

= 1

(35)

Then ⟦119860⟧3119894

le ⟦119862⟧3119894for all 119894 with strict inequality for at least

one 119894 so 119860 lt 119862 Also ⟦119860⟧3119894

+ ⟦119862⟧3119894

= ⟦2119861⟧3119894

for all 119894 so119860 + 119862 = 2119861 Finally let

3119898

119886 119887 119888 = 119860 119861 119862 +1

2 (36)

Then 119886 + 119888 = 2119887 and 119886 119888 sub Q+3(0 1) capQ+

3([1])

Claim L The set (Q+3(0 1) Q+

3([1])) cup minus(Q+

3(0 1)capQ+

3([1]))

is midpoint-free

Proof By Claims G and J if the specified set contains amidpoint triple twomembers of the triplemust have one signand the third member must have the opposite sign Suppose119886 sub Q+

3(0 1) cap Q+

3([1]) and 119887 119888 sub Q+

3(0 1) Q+

3([1]) are

such that minus119886 lt 119887 lt 119888 and 119886 + 2119887 = 119888 Since 1198863 (2119887)3 and 119888

3

are eventually periodic there is an integer 119898 such that theirdigits in places 119894 le 119898 are purely periodic and

⟦119886⟧3119894

= 1 ⟦2119887⟧3119894

isin 0 2 ⟦119888⟧3119894

isin 0 1

forall119894 le 119898

(37)

Since 119888 notin Q+3([1]) there is a 119895 le 119898 such that ⟦119888⟧

3119895= 0 and

therefore ⟦2119887⟧3119894

cannot be equal to 0 for all 119894 le 119898 On theother hand ⟦2119887⟧

3119894cannot be equal to 2 for all 119894 le 119898 since

119887 notin Q+3([1]) Hence the recurring block [119863] of (2119887)

3contains

0 and 2 so ⟦2119887⟧3119896

= 0 ⟦2119887⟧3119896minus1

= 2 for some 119896 le 119898 But⟦119886⟧3119896

= ⟦119886⟧3119896minus1

= 1 so ⟦119886⟧3119896minus1

+ ⟦2119887⟧3119896minus1

has a ldquocarryoverrdquo of 1 regardless of whether or not the digits in place 119896minus2contribute any ldquocarry overrdquo Then

⟦119888⟧3119896

= ⟦119886⟧3119896

+ ⟦2119887⟧3119896

+ (1) = 2 notin 0 1 (38)

By this contradiction there is no midpoint triple of theproposed type A similar but simpler argument shows thatthere are no subsets 119886 119887 sub Q+

3(0 1) cap Q+

3([1]) and 119888 sub

Q+3(0 1) Q+

3([1]) such that minus119886 lt minus119887 lt 119888 and 119886 = 2119887 + 119888

ClaimM For any positive rational 119886 isin Q+ Q+3([1]) there is a

midpoint triple in (Q+3(0 1) Q+

3([1])) cup minus119886 with minus119886 as its

lower endpoint

International Journal of Combinatorics 7

Proof Given 119886 we seek 119887 119888 sub Q+3(0 1) Q+

3([1]) such that

minus119886 lt 119887 lt 119888 and 119886 + 2119887 = 119888 Note that if the triple 119886 119887 119888

has all the required properties the scaled triple 3119898119886 119887 119888

also has all the required properties for any integer 119898 Letus assume without loss of generality that the digits of 119886

3

in negative places are purely periodic and partitioning intohomogeneous blocks 119860

119894yields

1198863= 119860ℎsdot sdot sdot 11986011198600sdot [119860minus1119860minus2

sdot sdot sdot 119860minus119896] (39)

for suitable integers ℎ ge 0 119896 gt 0 As 119886 notin Q+3([1]) if 119896 = 1

then 119860minus1

= 0 so 119886 isin Z+ and a suitable set 119887 119888 exists byClaim D

Now suppose 119896 ge 2 and 1198863has at least one negative place

digit which is nonzero If 1198863has no negative place digit equal

to 2 let 119886 = 119860 + 119902 where 119860 isin Z+ and 119902 isin Q+3(0 1)

Q+3([1]) 0 lt 119902 lt 1 are the integer part and fractional

part of 119886 respectively The proof of Claim D yields integers119861 119862 isin Z+

3(0 1) such that minus119860 119861 119862 is a midpoint triple

Choose 119887 = 119861 119888 = 119862 + 119902 Then

minus119886 119887 119888 = minus119860 minus 119902 119861 119862 + 119902 (40)

is a midpoint triple with 119887 119888 sub Q+3(0 1) Q+

3([1])

Finally suppose 119860minus119895

= 2times119899 for at least one 119895 gt 0 Define

homogeneous blocks 119861119894and 119862

119894for all 119894 as follows

119860119894= 0times119899

997904rArr 119861119894= 1times119899

119862119894= 0times119899

119860119894= 1times119899

997904rArr 119861119894= 1times119899

119862119894= 1times119899

119860119894= 2times119899

997904rArr 119861119894= 0times119899

119862119894= 0times119899

(41)

Assuming a carry over digit equal to 1 note that checks as inthe proof of Claim D justify the calculations

0times119899

+ 2times119899

+ (1) = (1) 0times119899

1times119899

+ 2times119899

+ (1) = (1) 1times119899

(42)

so in all cases we have 119860119894+ 2119861119894+ (1) = (1)119862

119894 Choose

1198873= 119861ℎsdot sdot sdot 11986111198610sdot [119861minus1119861minus2

sdot sdot sdot 119861minus119896]

1198883= (1) 119862

ℎsdot sdot sdot 11986211198620sdot [119862minus1119862minus2

sdot sdot sdot 119862minus119896]

(43)

Then 119886 + 2119887 = 119888 and 119887 119888 sub Q+3(0 1) Also 119861

minus119895= 119862minus119895

= 0times119899

ensures that 119887 119888 sub Q+3(0 1) Q+

3([1]) as required

For instance beginning with

119886 = 2 sdot [201]3=

71

26isin Q+

Q+

3([1]) (44)

the proof of Claim M constructs a midpoint triple minus119886 119887 119888with

119887 = 0 sdot [011]3=

2

13isin Q+

3(0 1) Q

+

3([1])

119888 = 10 sdot [001]3=

79

26isin Q+

3(0 1) Q

+

3([1])

(45)

However if we begin with

119886 = 2 sdot [101]3=

31

13isin Q+

Q+

3([1]) (46)

the construction falls back to Claim D resulting in 119887 = 1 and119888 = 5713

It now follows from Claims I K L and M that the set(Q+

3(0 1) Q

+

3([1])) cup minus (Q

+

3(0 1) capQ

+

3([1])) (47)

is a maximal midpoint-free subset ofQThis settles midpointquestions aboutQ+

3(0 1) as a subset ofQ

What is the situation for Q+3(0 2) If 119902 isin Q+

3(0 2) then

the recurring block of 1199023must have at least one digit equal

to 0 Hence there must be at least one digit equal to 0 inthe recurring block of (1199022)

3 so 1199022 isin Q+

3(0 1) Q+

3([1]) it

follows that 2(Q+3(0 1) Q+

3([1])) = Q+

3(0 2) Again if there

is some119898 isin Z+ such that 3119898119902 isin Z+3(0 2) + 1 it follows that

1199023is a terminating representation with trailing digit 1 and all

other digits in 0 2 Hence119902

2isin Q+

3(0 1) capQ

+

3([1]) (48)

The converse also holds so2 (Q+

3(0 1) capQ

+

3([1])) = Q

+

3(0 2 1) (49)

where Q+3(0 2 1) is the set of positive rationals which have a

terminating base 3 representation with trailing digit 1 and allother digits in 0 2 so

Q+

3(0 2 1) = ⋃

119894isinZ+

3minus119894

(Z+

3(0 2) + 1) (50)

In passing note that Z+3(0 2) + 1 = Q+

3(0 2 1) cap Z+ By

doubling it now follows from Claim L that the setQ+

3(0 2) cup minusQ

+

3(0 2 1) (51)

is midpoint-free

Claim N Any rational 119887 isin Q+ Q+3(0 2) is the midpoint of a

triple with both its endpoints inQ+3(0 2)

Proof If 119887 isin Q+ Q+3(0 2) then 1198872 notin Q+

3(0 1) so 1198872 is the

midpoint of a triple with endpoints in Q+3(0 1) Q+

3([1]) by

Claim I By doubling 119887 is the midpoint of a triple with bothits endpoints in the setQ+

3(0 2)

Claim O For any rational 119886 isin Q+ Q+3(0 2 1) there is

a midpoint triple in Q+3(0 2) cup minus119886 with minus119886 as its lower

endpoint

Proof If 119886 isin Q+ Q+3(0 2 1) and 119886 gt 0 then 1198862 notin Q+

3([1])

so there is a midpoint triple in Q+3(0 1) Q+

3([1]) cup minus1198862

with minus1198862 as its lower endpoint by Claim M By doublingQ+3(0 2)cup minus119886 has a midpoint triple with lower endpoint minus119886

The preceding results now fully establish the following

Theorem 4 The two sets(Q+

3(0 1) Q

+

3([1])) cup minus (Q

+

3(0 1) capQ

+

3([1]))

Q+

3(0 2) cup minusQ

+

3(0 2 1)

(52)

are both maximal midpoint-free subsets of Q

8 International Journal of Combinatorics

7 Midpoint-Free Subsets of R+ and R

Now consider R+3(0 1) and R+

3(0 2) These sets are uncount-

able while their maximal rational subsets Q+3(0 1) and

Q+3(0 2) are countable yet no essentially new considerations

arise in extending the results of the previous two sectionsfrom Q+ to R+ and from Q to R Noting that R+

3([1]) =

Q+3([1]) and R+

3(0 2 1) = Q+

3(0 2 1) it suffices to state the

end results for this wider context

Theorem 5 The setsR+3(0 1) R+

3([1]) andR+

3(0 2) are both

maximal midpoint-free subsets of R+

Theorem 6 The two sets

(R+

3(0 1) R

+

3([1])) cup minus (R

+

3(0 1) capR

+

3([1]))

R+

3(0 2) cup minusR

+

3(0 2 1)

(53)

are both maximal midpoint-free subsets of R

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

References

[1] N Sloane OEISmdashOnline Encyclopedia of Integer SequencesNote A003002 and A005836 in particular many relevant linksare included

[2] R K GuyUnsolved Problems in NumberTheory Springer NewYork NY USA 3rd edition 2004

[3] J Dybizbanski ldquoSequences containing no 3-term arithmeticprogressionsrdquo Electronic Journal of Combinatorics vol 19 no2 pp 1ndash15 2012

[4] D Wells The Penguin Dictionary of Curious and InterestingNumbers Penguin London UK 1986

[5] S Yates ldquoPrime divisors of repunitsrdquo Journal of RecreationalMathematics vol 8 no 1 pp 33ndash38 1975

[6] J Brillhart D H Lehmer J L Selfridge B Tuckerman and SWagstaff Jr ldquoFactorizations of 119887119899 plusmn 1 119887 = 2 3 5 6 7 10 11 12

up to high powersrdquo Contemporary Mathematics vol 22 1983[7] P Erdos and P Turan ldquoOn some sequences of integersrdquo Journal

of the London Mathematical Society vol 28 pp 104ndash109 1936

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 International Journal of Combinatorics

then the two-way infinite string

d = sdot sdot sdot 119889211988911198890sdot 119889minus1119889minus2

sdot sdot sdot (2)

is a base 3 representation of 119903 and 119889119894is the digit in place 119894

of d If 119889119894isin 0 1 for infinitely many 119894 lt 0 then d is a

regular base 3 representation of 119903 otherwise d is a singularbase 3 representation of 119903 and there is an integer 119896 such that119889119894= 2 for all 119894 lt 119896 The regular base 3 representation of

119903 is unique and if 119903 has a singular base 3 representationthat is also unique If 119883 is some ldquonaturalrdquo subset of R and119863 sub 0 1 2 it will be convenient to use119883

3(119863) to denote the

set of all members of119883with a base 3 representation restrictedto119863This notation adapts to cases where119863 is a configurationof base 3 digits thus when119863 is a finite string of base 3 digits1198833([119863]) will denote the set of all members of 119883 with a base

3 representation which includes a one-way infinite string ofrecurring blocks119863

Since 119903 gt 0 at least one of the digits is nonzero Theleading digit of d is the digit 119889

ℎgt 0 such that 119889

119894= 0 for

all 119894 gt ℎ The trivial zeros of d are all the digits 119889119894= 0 with

119894 gt maxℎ 0 Conventionally these are suppressed (ie keptimplicit) when listing d If ℎ lt 0 so the leading digit 119889

ℎ

occupies a negative place the placeholder zeros of d are all thedigits 119889

119894= 0 with 0 ge 119894 gt ℎ If there is an integer 119896 le ℎ

such that 119889119896gt 0 and 119889

119894= 0 for all 119894 lt 119896 then d is a regular

representation 119889119896is its trailing digit its optional zeros are all

the digits 119889119894= 0 with 119894 lt min119896 0 and if 119896 gt 0 then its

placeholder zeros are the digits 119889119894= 0with 0 le 119894 lt 119896 If d has a

trailing digit it is conventional to suppress its optional zerosas well as its trivial zeros the finite digit string remaining isthe terminating base 3 representation of 119903

These conventions are extended to 119903 = 0 as followsAlthough the digits in this case are 119889

119894= 0 for all 119894 in this

exceptional case 1198890is defined to be both the leading digit and

the trailing digit of the representation Then its trivial zerosare all those in places 119894 gt 0 and its optional zeros are all thosein places 119894 lt 0 thus 0 is the terminating representation for119903 = 0

If d has no trailing digit it is a nonterminating represen-tation of 119903 and it may be either regular or singular If d issingular there is an integer 119896 le ℎ + 1 such that 119889

119896isin 0 1

and 119889119894= 2 for all 119894 lt 119896 the digit 119889

119896is the pivot digit of d In

this case there is an alternative regular base 3 representationof 119903 namely

e = sdot sdot sdot 119890211989011198900sdot 119890minus1119890minus2

sdot sdot sdot (3)

where 119890119894= 119889119894for all 119894 gt 119896 119890

119896= 119889119896+ 1 and 119890

119894= 0 for all 119894 lt 119896

Evidently e is a terminating representation 119890119896is its trailing

digit and 3minus119896119903 isin Z+ so 119903 isin 3

119896Z+

The regular base 3 representation of any nonnegative

rational 119902 isin Q+ has a recurring block of digits if 119902 hasa singular base 3 representation this has a recurring 2 Forbrevity any such recurring block can be enclosed in square

brackets and the subscript 3 can be used to indicate base 3notation for instance

7

3= 2 sdot 1

3= 2 sdot 1[0]

3= 2 sdot 0[2]

3

1

2= 0 sdot [1]

3

2

5= 0 sdot [1012]

3

(4)

The set of all nonnegative rationals with a terminating base 3representation is

Q+

3([0]) = ⋃

119894isinZ+

3minus119894

Z+

(5)

The set of positive rationals with a singular base 3 represen-tation is simply

Q+

3([2]) = Q

+

3([0]) 0 (6)

The set of positive rationals with base 3 representationcontaining a recurring 1 is

Q+

3([1]) = ⋃

119894isinZ+

3minus119894

(Z+

+1

2) (7)

It will later be found that this set requires particular attentionwhen members are doubled because

2Q+

3([1]) sub Q

+

3([2]) sub Q

+

3([0]) (8)

Base 3 representation of negative reals is simply achievedby writing any such number as minus119903 with 119903 isin R 119903 gt 0 sothat minus119903 has its base 3 representation adopted from that of 119903with trivial zeros suppressed and the unary minus operatorapplied to precede the leading digit If 119903 has both a regularand a singular representation so does minus119903

To distinguish between instances of base 3 and base 10representations an explicit subscript 3 is normally attachedto the former while a subscript 10 is normally kept implicitfor the latter As needed 119903

3will denote the regular base

3 representation of 119903 with 119903lowast

3reserved for the singular

representation when this exists Finally ⟦119903⟧3119894

will denote119889119894 the digit in place 119894 of the regular base 3 representation

of 119903 while ⟦119903⟧lowast

3119894will denote the corresponding digit in the

singular representation when this exists

3 Three Sparse Subsets of Z+

Let Z+3(119863) be the set of all nonnegative integers with termi-

nating base 3 representation in which the only explicit digitsare in119863 sub 0 1 2The three cases of interest are those where119863 is a 2-set They begin as follows

Z+

3(0 1)

= 0 1 3 4 9 10 12 13 27 28

30 31 36 37 39 40 81

Z+

3(0 2)

= 0 2 6 8 18 20 24 26 54 56 60 62 72 74 78 80

International Journal of Combinatorics 3

Z+

3(1 2)

= 1 2 4 5 7 8 13 14 16 17 22 23 25 26 40

(9)

Some features are obvious Clearly Z+3(0 2) = 2Z+

3(0 1)

When 119889 isin 0 1 2 and 119863 = 0 1 2 119889 all integerscongruent to 119889 (mod 3) are absent from Z+

3(119863) The first

positive integer absent from all three sets is 1023= 11 but

it is clear that an increasing proportion of integers will bemissing fromall three sets Indeed a simple calculation showsfor any positive integer 119899 thatZ+

3(0 1) andZ+

3(0 2) each have

2119899 members below 3

119899 while Z+3(1 2) has 2119899+1 minus 2 members

below 3119899 so each set is sparse and has asymptotic density 0

It turns out that these three sets are of considerableinterest when it comes to the presence or absence ofmidpointtriples so let us now address the main subject of this paper

Note that the set Z+3(1 2) contains midpoint triples such

as 1 4 7 and 2 5 8 In factZ+3(1 2) is densely packedwith

midpoint triples since eachmember 119886 isin Z+3(1 2) is the lower

endpoint of an infinite family of midpoint triples

119886 119906119899minus 119906119898+ 119886 2119906

119899minus 2119906119898+ 119886

= (119906119899minus 119906119898) 0 1 2 + 119886 sub Z

+

3(1 2)

(10)

for any integer 119899 gt 119898 where 119898 is the number of digits inthe terminating base 3 representation of 119886 (if 119889

ℎis the leading

digit in 1198863 then119898 = ℎ + 1) and

119906119899= 1 + 3 + 3

2

+ sdot sdot sdot + 3119899minus1

=3119899minus 1

2(11)

is the positive integer with terminating base 3 representationof 119899 digits all 1 Note that Z+

3(1) = 119906

119899 119899 ge 1 and

appropriate scaling shows that this set is midpoint-free (Theintegers 119906

119899are rep-units in base 3 The positive integers with

explicit decimal digits all equal to 1 are the rep-units for base10 so called by contraction of ldquorepeated unitrdquo [4] Primefactorizations of rep-units have been much studied [5 6]Clearly if 119898 | 119899 then 119906

119898| 119906119899 Thus 119906

119899can be a prime

number only if 119899 is prime but 1199065= 11111

3= 112 shows that

this condition is not sufficient Such observations generalizeto rep-units in any base)

Not only is every member of Z+3(1 2) the lower endpoint

of infinitely many midpoint triples it is also the case that anymember greater than 2 is either the midpoint or the upperendpoint of at least one midpoint triple depending on theleading digit of its base 3 representation

In contrast it turns out that Z+3(0 1) and Z+

3(0 2) are

midpoint-free To see this it suffices to show that Z+3(0 1) is

midpoint-free thenZ+3(0 2)= 2Z+

3(0 1) ensures thatZ+

3(0 2)

is also midpoint-free (Long ago Erdos and Turan [7] notedthat the nonnegative integers with a 2-less base 3 represen-tation are midpoint-free The following compact proof isincluded for completeness and to typify what follows)

Claim A The set Z+3(0 1) is midpoint-free

Proof On the contrary suppose that 119886 119887 119888 sub Z+3(0 1) is a

midpoint triple with 119886 lt 119887 lt 119888 and 119886 + 119888 = 2119887 All digits in1198863 1198873 1198883are in 0 1 and all digits in (2119887)

3are in 0 2 Then

⟦119886⟧3119894

+ ⟦119888⟧3119894

= ⟦2119887⟧3119894 (12)

for all integers 119894 and

⟦2119887⟧3119894

= 0 997904rArr ⟦119886⟧3119894

= ⟦119887⟧3119894

= ⟦119888⟧3119894

= 0

⟦2119887⟧3119894

= 2 997904rArr ⟦119886⟧3119894

= ⟦119887⟧3119894

= ⟦119888⟧3119894

= 1

(13)

Hence 119886 = 119887 = 119888 a contradiction so Z+3(0 1) contains no

midpoint triple

It will now be shown thatZ+3(0 1) is not contained in any

larger midpoint-free subset of Z+

Claim B For any positive integer 119887 isin Z+ Z+

3(0 1) there is a

midpoint triple in Z+3(0 1) cup 119887 with 119887 as its midpoint

Proof Given 119887 isin Z+ Z+

3(0 1) we seek 119886 119888 sub Z+

3(0 1) such

that 119886 lt 119887 lt 119888 and 119886 + 119888 = 2119887 Specify the base 3 digits of 119886and 119888 as follows

⟦2119887⟧3119894

= 0 997904rArr ⟦119886⟧3119894

= ⟦119888⟧3119894

= 0

⟦2119887⟧3119894

= 2 997904rArr ⟦119886⟧3119894

= ⟦119888⟧3119894

= 1

⟦2119887⟧3119894

= 1 997904rArr ⟦119886⟧3119894

= 0 ⟦119888⟧3119894

= 1

(14)

This determines integers 119886 and 119888 such that 119886 119888 sub Z+3(0 1)

and 119886 + 119888 = 2119887 since

⟦119886⟧3119894

+ ⟦119888⟧3119894

= ⟦2119887⟧3119894 (15)

for all 119894 At least one digit of (2119887)3is 1 since 119887 notin Z+

3(0 1) so 119886

3

and 1198883differ in at least one place then ⟦119886⟧

3119894lt ⟦119888⟧

3119894in each

such place ensuring that 119886 lt 119888 Put 119889 = (119888 minus 119886)2 gt 0 Then119888 = 119886+ 2119889 so 2119887 = 119886+ 119888 = 2119886+ 2119889 whence 119887 = 119886+119889 gt 119886 and119888 = 119886 + 2119889 = 119887 + 119889 gt 119887 Thus 119886 lt 119887 lt 119888

It will now be shown that extendingZ+3(0 1) by adjoining

any rational of the form 119899 + (12) with 119899 isin Z+ always yieldsa set which contains a midpoint triple

Claim C For any 119887 isin Z+

+ (12) there is a midpoint triple inZ+3(0 1) cup 119887 with 119887 as its midpoint

Proof Again we seek 119886 119888 sub Z+3(0 1) such that 119886 lt 119887 lt 119888

and 119886 + 119888 = 2119887 In the present case note that 2119887 is an oddpositive integer so (2119887)

3contains the digit 1 an odd number

of times Specifying 119886 and 119888 as in the proof of Claim B onceagain it follows that 119886 119888 sub Z+

3(0 1) and 119886 lt 119887 lt 119888

For instance the proofs of Claims B and C produce thedecompositions

12013= 100

3+ 1101

3 212

3= 101

3+ 111

3(16)

showing that 23 and 232 are the midpoints of the pairs9 37 sub Z+

3(0 1) and 10 13 sub Z+

3(0 1) respectively

4 International Journal of Combinatorics

Further doubling shows that 46 and 23 are the midpointsof the pairs 18 74 sub Z+

3(0 2) and 20 26 sub Z+

3(0 2)

respectively From Claims A B and C and the doublingmethod just illustrated we have the following

Theorem 1 The sets Z+3(0 1) and Z+

3(0 2) are maximal

midpoint-free subsets of Z+

4 Two Unexpected Results for Z

Let us now study Z+3(0 1) and Z+

3(0 2) as midpoint-free

subsets of Z the full set of integers Surprisingly the resultsfor the two sets are quite different

Calculations until now have involved sums of pairs119886 119888 sub Z

+

3(0 1) so it has been possible to work digit by digit

in base 3 without a carry over digit When the context iswidened to Z it seems that carry over digits can no longerbe avoided so we shall consider blocks of digits rather thansingle digits Let the compact notation 119889

times119899 (ldquo119889 by 119899rdquo) denotea homogeneous block of 119899 adjacent digits all equal to 119889 thus1times119899 denotes the terminating base 3 representation of rep-unit119906119899 For blocks in which the digits are not all equal we simply

abut such expressionswith119889times1 = 119889 as a natural simplificationTo illustrate consider the integer

1198863= 1222202111001121100 = 12

times4

021times3

0times2

1times2

21times2

0times2

(17)

Let us find integers 119887 119888 isin Z+

3(0 1) such that 119886 + 2119887 = 119888

Regarding 1198863as a string of homogeneous blocks we choose

matching blocks for (2119887)3so that

(2119887)3= 2000020222222202200 = 20

times4

202times7

02times2

0times2

(18)

Calculating from right to left base 3 arithmetic for 1198863+ (2119887)

3

yields successively the following blocks with each carry overdigit indicated parenthetically

0times2

+ 0times2

= 0times2

1times2

+ 2times2

= (1) 10

2 + 0 + (1) = (1) 0

1times2

+ 2times2

+ (1) = (1)1times2

0times2

+ 2times2

+ (1) = (1) 0times2

1times3

+ 2times3

+ (1) = (1) 1times3

2 + 0 + (1) = (1) 0

0 + 2 + (1) = (1) 0

2times4

+ 0times4

+ (1) = (1) 0times4

1 + 2 + (1) = (1) 1

(19)

Assembling these blocks yields

1198863+ (2119887)

3= 1times2

| 0times4

| 0 | 0 | 1times3

| 0times2

| 1times2

| 0 | 10 | 0times2

(20)

After simplifying we have

10times4

101times7

01times2

0times2

= 1000010111111101100 = 1198873

1times2

0times6

1times3

0times2

1times2

010times3

= 11000000111001101000 = 1198883

(21)

so 119886 + 2119887 = 119888 and 119887 119888 isin Z+

3(0 1) as desired This calculation

models the proof of the following

Claim D For any integer 119886 gt 0 there is a midpoint triplein Z+3(0 1) cup minus119886 with the negative integer minus119886 as its lower

endpoint

Proof Given any positive integer 119886 we seek 119887 119888 sub Z+3(0 1)

such that minus119886 lt 119887 lt 119888 and minus119886 + 119888 = 2119887 so 119886 + 2119887 = 119888 Partition1198863into its homogeneous blocks 119860

119894 so

1198863= 119860119898sdot sdot sdot 119860119894sdot sdot sdot 11986011198600

(22)

and seek corresponding blocks 119861119894and 119862

119894to construct 119887

3and

1198883If 1198863has a terminal block119860

0= 0times119899 assign 119887

3the terminal

block 1198610= 0times119899 so 119888

3has the terminal block 119862

0= 0times119899

=

1198600+ 21198610 If 1198863has a nonterminal block 119860

119894= 0times119899 for some

119894 gt 0 assume the sum of the preceding blocks 119860119894minus1

and2119861119894minus1

contributes a carry over digit of 1 and assign 1198873the

corresponding block 119861119894= 1times119899 so

119862119894= 119860119894+ 2119861119894+ (1) = 0

times119899

+ 2times119899

+ (1) = (1) 0times119899

(23)

As 2times119899 + (1) corresponds to 2119906119899+ 1 = (3

119899+1minus 1) + 1 = 3

119899+1the computation is justified

If 1198863has a block 119860

119894= 1times119899 assign 119887

3the corresponding

block 119861119894= 1times119899 so 119888

3has the corresponding block 119862

119894= 119860119894+

2119861119894+ (120575) = (1)1

times(119899minus1)

0 + (120575) where the carry over digit 120575 iseither 0 or 1 so 119862

119894= (1)1

times(119899minus1)

0 or (1)1times119899 respectively Thisholds since the sum 1

times119899+2times119899 corresponds to 119906

119899+2119906119899= 3119906119899=

119906119899+1

minus 1199061and 1times119899

+ 2times119899

+ (1) corresponds to 3119906119899+ 1 = 119906

119899+1

If 1198863has a terminal block 119860

0= 2times119899 assign 119887

3the block

1198610= 0times(119899minus1)

1 so 1198883has terminal block 119862

0= 1198600+ 21198610

=

(1)0times(119899minus1)

1 since the sum 2times119899

+ 0times(119899minus1)

2 corresponds to 2119906119899+

2 = (3119899+1

minus 1) + 2 = 3119899+1

+ 1 Similarly if 11986011198600= 2times1198990times119898

assign 1198873the corresponding blocks 119861

11198610= 0times(119899minus1)

10times119898 then

1198883has the blocks 119862

11198620= (1)0

times(119899minus1)

10times119898 Otherwise if 119886

3has

a nonterminal block 119860119894= 2times119899 for some 119894 gt 0 assume that

119860119894minus1

+ 2119861119894minus1

yields a carry over digit 1 and assign 1198873the

block 119861119894= 0times119899 so 119888

3has 119862119894= 119860119894+ 2119861119894+ (1) = (1)0

times119899For 119894 gt 0note that the only case inwhich119860

119894minus1+2119861119894minus1

doesnot contribute a carry over digit 1 to 119860

119894+ 2119861119894is when 119894 = 1

and 1198863has terminal block119860

0= 0times119899Thus all our assumptions

about carry over digits are justified a posteriori

Claim E For any integer 119899 gt 0 there is a midpoint triple inZ+3(0 2)cupminus119899 if and only if 119899 is even and thenminus119899 is the lower

endpoint of the triple

Proof By doubling it follows from Claim D that for anyinteger 119886 gt 0 there is a midpoint triple in Z+

3(0 2) cup

International Journal of Combinatorics 5

minus2119886 However we claim that there is no midpoint triple inZ+3(0 2) cup minus2119886 + 1 On the contrary suppose minus2119886 + 1 forms

a midpoint triple with the pair 119887 119888 sub Z+3(0 2) Without loss

of generality minus2119886 + 1 lt 119887 lt 119888 and (minus2119886 + 1) + 119888 = 2119887 so2119886 + 2119887 minus 1 = 119888 this is impossible since 2119886 + 2119887 minus 1 is oddand 119888 is even It follows thatZ+

3(0 2) cup minus2119886 + 1 is midpoint-

free

Claim F The set Z+3(0 2) cup minus(Z+

3(0 2) + 1) is midpoint-free

Proof The two sets forming this union are certainlymidpoint-free so any midpoint triple in the union musthave two members in one set and one member in the otherClaim E shows that there is no midpoint triple with exactlyone member in minus(Z+

3(0 2) + 1) since this set only contains

odd negative integers So suppose there is a midpoint tripleminus119886 minus119887 119888 with exactly one member 119888 isin Z+

3(0 2) Then

119886 minus 1 119887 minus 1 minus119888 minus 1 is a midpoint triple in

minus (Z+

3(0 2) cup minus (Z

+

3(0 2) + 1)) minus 1

= Z+

3(0 2) cup minus (Z

+

3(0 2) + 1)

(24)

with exactly one member in minus(Z+3(0 2) + 1) the case ruled

out by Claim E

Since Z+3(0 2) is a maximal midpoint-free subset of Z+

by Theorem 1 clearly Z+3(0 2) + 1 is a maximal midpoint-

free subset of Z+ + 1 = Z+ 0 Note that the members ofZ+3(0 2) + 1 are precisely those positive integers 119899 for which

the trailing digit of 1198993is 1 and all other digits are in 0 2

With Claims E and F it follows thatZ+3(0 2)cupminus(Z+

3(0 2)+1)

is a maximal midpoint-free subset of Z = Z+ cup minus(Z+ 0)Combined with Claim D this proves the following

Theorem 2 The sets Z+3(0 1) and Z+

3(0 2) cup minus(Z+

3(0 2) + 1)

are maximal midpoint-free subsets of Z

5 Midpoint-Free Subsets of Q+

Now let us consider the corresponding subsets Q+3(119863) of the

nonnegative rationals Q+ namely the subsets comprisingthose members with a regular base 3 representation in whicheach digit after suppression of all trivial and optional zerosis in 119863 sub 0 1 2 The three cases of interest are those where119863 is a 2-set

As expected Q+3(1 2) is densely packed with midpoint

triples This follows from the observation that each rational119902 isin Q+

3(1 2) is the lower endpoint of an infinite family of

midpoint triples

119902 119906119899minus 119906119898+ 119902 2119906

119899minus 2119906119898+ 119902

= (119906119899minus 119906119898) 0 1 2 + 119902 sub Q

+

3(1 2)

(25)

for any integer 119899 gt 119898 where 119906119899is the 119899-digit base 3 rep-unit

and 119898 = ℎ + 1 where 119889ℎis the leading digit of the regular

base 3 representation 1199023

Next consider Q+3(0 1) The details are a little more

complicated than in the integer context since allowance

must be made for singular representations when particularmembers ofQ+

3(0 1) are doubled

Claim G The setQ+3(0 1) Q+

3([1]) is midpoint-free

Proof If 119887 isin Q+3(0 1) Q+

3([1]) and 119887 gt 0 all digits of (2119887)

3

are in 0 2 They uniquely determine the digits of 1198863and 1198883

such that 119886 119888 sub Q+3(0 1) Q+

3([1]) and 119886 + 119888 = 2119887 As in

the proof of Claim A they force 119886 = 119887 = 119888 Hence there is nomidpoint triple 119886 119887 119888 sub Q+

3(0 1) Q+

3([1])

Claim H Any 119887 isin Q+3([1]) is the midpoint of a triple with

both its endpoints in the setQ+3(0 1) capQ+

3([0])

Proof We seek 119886 119888 sub Q+3(0 1) Q+

3([1]) with 119886 lt 119887 lt 119888 and

119886 + 119888 = 2119887 Then 1198863and 1198883are uniquely determined by

⟦119886⟧3119894

+ ⟦119888⟧3119894

= ⟦2119887⟧3119894 ⟦119886⟧

3119894le ⟦119888⟧

3119894(26)

for all 119894 Since 119887 isin Q+3([1]) then 3

119896119887 isin Z+ + (12) for some

119896 isin Z+ so

2119887 isin 3minus119896

(2Z+

+ 1) (27)

It follows that (2119887)3is a terminating representation with an

odd number of digits equal to 1 Then

⟦2119887⟧3119894

= 1 997904rArr ⟦119886⟧3119894

= 0 ⟦119888⟧3119894

= 1

for an odd number of integers 119894 ge minus119896

⟦2119887⟧3119894

= 0 997904rArr ⟦119886⟧3119894

= ⟦119888⟧3119894

= 0 forall119894 lt minus119896

(28)

Hence the requiredmidpoint triple 119886 119887 119888 exists and in fact119886 119888 sub Q+

3([0])

In particular if 119887 isin Q+3([1]) is such that ⟦2119887⟧

3119899= 1 for

some integer 119899 and ⟦2119887⟧3119894

= 0 for all 119894 = 119899 then 119887 = 31198992 =

119906119899+ (12) and the proof of Claim H determines 119886 = 0 and

119888 = 2119887 = 3119899 Since ⟦119887⟧

3119894= 0 if 119894 ge 119899 and ⟦119887⟧

3119894= 1 if 119894 lt 119899

we call 119887 the base 3 fractional rep-unit with offset 119899 denotedby V119899 Then V

119899isin Q+3([1]) is the midpoint of the triple with

endpoints 0 3119899 sub Q+3(0 1) capQ+

3([0])

Positive rationals not in Q+3(0 1) cup Q+

3([1]) also belong

to midpoint triples with endpoints in Q+3(0 1) Q+

3([1]) but

proving this needs care For instance

5

16= 0 sdot [0221]

3

997904rArr5

8= 0 sdot [12]

3= 0 sdot [0111]

3+ 0 sdot [1101]

3

=13

80+37

80

(29)

so 1380 516 3780 is a midpoint triple with endpoints inQ+3(0 1) Q+

3([1]) This calculation models the proof of the

following general result

Claim I Any 119887 isin Q+ Q+

3(0 1) is themidpoint of a triple with

both its endpoints in the setQ+3(0 1) Q+

3([1])

6 International Journal of Combinatorics

Proof We seek 119886 119888 sub Q+3(0 1) Q+

3([1]) with 119886 lt 119887 lt 119888 and

119886 + 119888 = 2119887 When 119887 isin Q+3([1]) the result follows from Claim

H so we may now assume that

119887 isin Q+

(Q+

3(0 1) cupQ

+

3([1])) (30)

Then (2119887)3has at least one digit 119889

119895equal to 1 and its recurring

block say [119863] has at least one digit different from 2 Werequire 119886

3and 1198883to satisfy

⟦119886⟧3119894

+ ⟦119888⟧3119894

= ⟦2119887⟧3119894

(31)

for all 119894 Requiring ⟦119886⟧3119894

le ⟦119888⟧3119894

for all 119894 ge 119895 ensures that119886 lt 119888 since

⟦119886⟧3119895

= 0 ⟦119888⟧3119895

= 1 (32)

If the recurring block [119863] of (2119887)3contains 0 then the

recurring blocks of 1198863and 1198883also contain 0 so neither 119886 nor

119888 is in Q+3([1]) Now suppose [119863] has at least one digit equal

to 1 and none equal to 0 Clearly in this case we may choosethe integer 119895 so that the digit 119889

119895= 1 occurs in [119863] If119863 has119898

digits then for all 119894 le 119895 choose

⟦119886⟧3119894

= 0 ⟦119888⟧3119894

= 1 when 119894 equiv 119895 mod 2119898

⟦119886⟧3119894

= 1 ⟦119888⟧3119894

= 0 when 119894 equiv 119895 + 119898 mod 2119898

⟦119886⟧3119894

le ⟦119888⟧3119894

otherwise

(33)

Then 1198863and 1198883are uniquely determined so that 119886 lt 119888 and

each has a recurring block containing 0 so neither 119886 nor 119888 isinQ+3([1])

If 119902 isin Q+ Q+

3(0 2) then 119902

3has at least one digit equal

to 1 so it follows that (1199022)3isin Q+ Q

+

3(0 1) Claim I shows

that there is a midpoint triple 119886 119887 119888 with 119887 = 1199022 and119886 119888 sub Q+

3(0 1) Q+

3([1]) Then 2119886 119887 119888 is a midpoint triple

with midpoint 119902 and endpoints 2119886 119888 sub Q+3(0 2) Hence

doubling applied to Claims G and I shows that Q+3(0 2) is

midpoint-free and every 119887 isin Q+ Q+

3(0 2) is the midpoint

of a triple with endpoints in Q+3(0 2) so Claims G H and I

establish the following

Theorem 3 The sets Q+3(0 1) Q+

3([1]) and Q+

3(0 2) are

maximal midpoint-free subsets ofQ+

6 Midpoint-Free Subsets of Q

Now considerQ+3(0 1) andQ+

3(0 2) as subsets ofQ The role

ofQ+3([1]) is yet more prominent in this context

Claim J The setQ+3(0 1) capQ+

3([1]) is midpoint-free

Proof Suppose 119886 119887 119888 sub Q+3(0 1) cap Q+

3([1]) satisfies 119886 lt

119887 lt 119888 and 119886 + 119888 = 2119887 There is an integer 119898 ge 0 for which3119898119886 119887 119888 sub Z+

3(0 1) + (12) so

2 sdot 3119898

119886 119887 119888 sub 2Z+

3(0 1) + 1 (34)

This contradicts the fact that 2Z+3(0 1) + 1 is midpoint-free

by Claim A

Claim K Any positive rational 119887 isin Q+3([1]) Q+

3(0 1) is the

midpoint of a triple with endpoints inQ+3(0 1) capQ+

3([1])

Proof Given 119887 we seek 119886 119888 sub Q+3(0 1) capQ+

3([1]) such that

119886 lt 119887 lt 119888 and 119886 + 119888 = 2119887 There is an integer 119898 such that3119898119887 isin Z+ + (12) so there is an integer 119861 isin Z+ Z+

3(0 1)

such that 3119898119887 = 119861 + (12) At least one digit of 1198613is 2 so at

least one digit of (2119861)3is 1 Define 119860 119862 sub Z+

3(0 1) by

⟦2119861⟧3119894

= 0 997904rArr ⟦119860⟧3119894

= ⟦119862⟧3119894

= 0

⟦2119861⟧3119894

= 1 997904rArr ⟦119860⟧3119894

= 0 ⟦119862⟧3119894

= 1

⟦2119861⟧3119894

= 2 997904rArr ⟦119860⟧3119894

= ⟦119862⟧3119894

= 1

(35)

Then ⟦119860⟧3119894

le ⟦119862⟧3119894for all 119894 with strict inequality for at least

one 119894 so 119860 lt 119862 Also ⟦119860⟧3119894

+ ⟦119862⟧3119894

= ⟦2119861⟧3119894

for all 119894 so119860 + 119862 = 2119861 Finally let

3119898

119886 119887 119888 = 119860 119861 119862 +1

2 (36)

Then 119886 + 119888 = 2119887 and 119886 119888 sub Q+3(0 1) capQ+

3([1])

Claim L The set (Q+3(0 1) Q+

3([1])) cup minus(Q+

3(0 1)capQ+

3([1]))

is midpoint-free

Proof By Claims G and J if the specified set contains amidpoint triple twomembers of the triplemust have one signand the third member must have the opposite sign Suppose119886 sub Q+

3(0 1) cap Q+

3([1]) and 119887 119888 sub Q+

3(0 1) Q+

3([1]) are

such that minus119886 lt 119887 lt 119888 and 119886 + 2119887 = 119888 Since 1198863 (2119887)3 and 119888

3

are eventually periodic there is an integer 119898 such that theirdigits in places 119894 le 119898 are purely periodic and

⟦119886⟧3119894

= 1 ⟦2119887⟧3119894

isin 0 2 ⟦119888⟧3119894

isin 0 1

forall119894 le 119898

(37)

Since 119888 notin Q+3([1]) there is a 119895 le 119898 such that ⟦119888⟧

3119895= 0 and

therefore ⟦2119887⟧3119894

cannot be equal to 0 for all 119894 le 119898 On theother hand ⟦2119887⟧

3119894cannot be equal to 2 for all 119894 le 119898 since

119887 notin Q+3([1]) Hence the recurring block [119863] of (2119887)

3contains

0 and 2 so ⟦2119887⟧3119896

= 0 ⟦2119887⟧3119896minus1

= 2 for some 119896 le 119898 But⟦119886⟧3119896

= ⟦119886⟧3119896minus1

= 1 so ⟦119886⟧3119896minus1

+ ⟦2119887⟧3119896minus1

has a ldquocarryoverrdquo of 1 regardless of whether or not the digits in place 119896minus2contribute any ldquocarry overrdquo Then

⟦119888⟧3119896

= ⟦119886⟧3119896

+ ⟦2119887⟧3119896

+ (1) = 2 notin 0 1 (38)

By this contradiction there is no midpoint triple of theproposed type A similar but simpler argument shows thatthere are no subsets 119886 119887 sub Q+

3(0 1) cap Q+

3([1]) and 119888 sub

Q+3(0 1) Q+

3([1]) such that minus119886 lt minus119887 lt 119888 and 119886 = 2119887 + 119888

ClaimM For any positive rational 119886 isin Q+ Q+3([1]) there is a

midpoint triple in (Q+3(0 1) Q+

3([1])) cup minus119886 with minus119886 as its

lower endpoint

International Journal of Combinatorics 7

Proof Given 119886 we seek 119887 119888 sub Q+3(0 1) Q+

3([1]) such that

minus119886 lt 119887 lt 119888 and 119886 + 2119887 = 119888 Note that if the triple 119886 119887 119888

has all the required properties the scaled triple 3119898119886 119887 119888

also has all the required properties for any integer 119898 Letus assume without loss of generality that the digits of 119886

3

in negative places are purely periodic and partitioning intohomogeneous blocks 119860

119894yields

1198863= 119860ℎsdot sdot sdot 11986011198600sdot [119860minus1119860minus2

sdot sdot sdot 119860minus119896] (39)

for suitable integers ℎ ge 0 119896 gt 0 As 119886 notin Q+3([1]) if 119896 = 1

then 119860minus1

= 0 so 119886 isin Z+ and a suitable set 119887 119888 exists byClaim D

Now suppose 119896 ge 2 and 1198863has at least one negative place

digit which is nonzero If 1198863has no negative place digit equal

to 2 let 119886 = 119860 + 119902 where 119860 isin Z+ and 119902 isin Q+3(0 1)

Q+3([1]) 0 lt 119902 lt 1 are the integer part and fractional

part of 119886 respectively The proof of Claim D yields integers119861 119862 isin Z+

3(0 1) such that minus119860 119861 119862 is a midpoint triple

Choose 119887 = 119861 119888 = 119862 + 119902 Then

minus119886 119887 119888 = minus119860 minus 119902 119861 119862 + 119902 (40)

is a midpoint triple with 119887 119888 sub Q+3(0 1) Q+

3([1])

Finally suppose 119860minus119895

= 2times119899 for at least one 119895 gt 0 Define

homogeneous blocks 119861119894and 119862

119894for all 119894 as follows

119860119894= 0times119899

997904rArr 119861119894= 1times119899

119862119894= 0times119899

119860119894= 1times119899

997904rArr 119861119894= 1times119899

119862119894= 1times119899

119860119894= 2times119899

997904rArr 119861119894= 0times119899

119862119894= 0times119899

(41)

Assuming a carry over digit equal to 1 note that checks as inthe proof of Claim D justify the calculations

0times119899

+ 2times119899

+ (1) = (1) 0times119899

1times119899

+ 2times119899

+ (1) = (1) 1times119899

(42)

so in all cases we have 119860119894+ 2119861119894+ (1) = (1)119862

119894 Choose

1198873= 119861ℎsdot sdot sdot 11986111198610sdot [119861minus1119861minus2

sdot sdot sdot 119861minus119896]

1198883= (1) 119862

ℎsdot sdot sdot 11986211198620sdot [119862minus1119862minus2

sdot sdot sdot 119862minus119896]

(43)

Then 119886 + 2119887 = 119888 and 119887 119888 sub Q+3(0 1) Also 119861

minus119895= 119862minus119895

= 0times119899

ensures that 119887 119888 sub Q+3(0 1) Q+

3([1]) as required

For instance beginning with

119886 = 2 sdot [201]3=

71

26isin Q+

Q+

3([1]) (44)

the proof of Claim M constructs a midpoint triple minus119886 119887 119888with

119887 = 0 sdot [011]3=

2

13isin Q+

3(0 1) Q

+

3([1])

119888 = 10 sdot [001]3=

79

26isin Q+

3(0 1) Q

+

3([1])

(45)

However if we begin with

119886 = 2 sdot [101]3=

31

13isin Q+

Q+

3([1]) (46)

the construction falls back to Claim D resulting in 119887 = 1 and119888 = 5713

It now follows from Claims I K L and M that the set(Q+

3(0 1) Q

+

3([1])) cup minus (Q

+

3(0 1) capQ

+

3([1])) (47)

is a maximal midpoint-free subset ofQThis settles midpointquestions aboutQ+

3(0 1) as a subset ofQ

What is the situation for Q+3(0 2) If 119902 isin Q+

3(0 2) then

the recurring block of 1199023must have at least one digit equal

to 0 Hence there must be at least one digit equal to 0 inthe recurring block of (1199022)

3 so 1199022 isin Q+

3(0 1) Q+

3([1]) it

follows that 2(Q+3(0 1) Q+

3([1])) = Q+

3(0 2) Again if there

is some119898 isin Z+ such that 3119898119902 isin Z+3(0 2) + 1 it follows that

1199023is a terminating representation with trailing digit 1 and all

other digits in 0 2 Hence119902

2isin Q+

3(0 1) capQ

+

3([1]) (48)

The converse also holds so2 (Q+

3(0 1) capQ

+

3([1])) = Q

+

3(0 2 1) (49)

where Q+3(0 2 1) is the set of positive rationals which have a

terminating base 3 representation with trailing digit 1 and allother digits in 0 2 so

Q+

3(0 2 1) = ⋃

119894isinZ+

3minus119894

(Z+

3(0 2) + 1) (50)

In passing note that Z+3(0 2) + 1 = Q+

3(0 2 1) cap Z+ By

doubling it now follows from Claim L that the setQ+

3(0 2) cup minusQ

+

3(0 2 1) (51)

is midpoint-free

Claim N Any rational 119887 isin Q+ Q+3(0 2) is the midpoint of a

triple with both its endpoints inQ+3(0 2)

Proof If 119887 isin Q+ Q+3(0 2) then 1198872 notin Q+

3(0 1) so 1198872 is the

midpoint of a triple with endpoints in Q+3(0 1) Q+

3([1]) by

Claim I By doubling 119887 is the midpoint of a triple with bothits endpoints in the setQ+

3(0 2)

Claim O For any rational 119886 isin Q+ Q+3(0 2 1) there is

a midpoint triple in Q+3(0 2) cup minus119886 with minus119886 as its lower

endpoint

Proof If 119886 isin Q+ Q+3(0 2 1) and 119886 gt 0 then 1198862 notin Q+

3([1])

so there is a midpoint triple in Q+3(0 1) Q+

3([1]) cup minus1198862

with minus1198862 as its lower endpoint by Claim M By doublingQ+3(0 2)cup minus119886 has a midpoint triple with lower endpoint minus119886

The preceding results now fully establish the following

Theorem 4 The two sets(Q+

3(0 1) Q

+

3([1])) cup minus (Q

+

3(0 1) capQ

+

3([1]))

Q+

3(0 2) cup minusQ

+

3(0 2 1)

(52)

are both maximal midpoint-free subsets of Q

8 International Journal of Combinatorics

7 Midpoint-Free Subsets of R+ and R

Now consider R+3(0 1) and R+

3(0 2) These sets are uncount-

able while their maximal rational subsets Q+3(0 1) and

Q+3(0 2) are countable yet no essentially new considerations

arise in extending the results of the previous two sectionsfrom Q+ to R+ and from Q to R Noting that R+

3([1]) =

Q+3([1]) and R+

3(0 2 1) = Q+

3(0 2 1) it suffices to state the

end results for this wider context

Theorem 5 The setsR+3(0 1) R+

3([1]) andR+

3(0 2) are both

maximal midpoint-free subsets of R+

Theorem 6 The two sets

(R+

3(0 1) R

+

3([1])) cup minus (R

+

3(0 1) capR

+

3([1]))

R+

3(0 2) cup minusR

+

3(0 2 1)

(53)

are both maximal midpoint-free subsets of R

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

References

[1] N Sloane OEISmdashOnline Encyclopedia of Integer SequencesNote A003002 and A005836 in particular many relevant linksare included

[2] R K GuyUnsolved Problems in NumberTheory Springer NewYork NY USA 3rd edition 2004

[3] J Dybizbanski ldquoSequences containing no 3-term arithmeticprogressionsrdquo Electronic Journal of Combinatorics vol 19 no2 pp 1ndash15 2012

[4] D Wells The Penguin Dictionary of Curious and InterestingNumbers Penguin London UK 1986

[5] S Yates ldquoPrime divisors of repunitsrdquo Journal of RecreationalMathematics vol 8 no 1 pp 33ndash38 1975

[6] J Brillhart D H Lehmer J L Selfridge B Tuckerman and SWagstaff Jr ldquoFactorizations of 119887119899 plusmn 1 119887 = 2 3 5 6 7 10 11 12

up to high powersrdquo Contemporary Mathematics vol 22 1983[7] P Erdos and P Turan ldquoOn some sequences of integersrdquo Journal

of the London Mathematical Society vol 28 pp 104ndash109 1936

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

International Journal of Combinatorics 3

Z+

3(1 2)

= 1 2 4 5 7 8 13 14 16 17 22 23 25 26 40

(9)

Some features are obvious Clearly Z+3(0 2) = 2Z+

3(0 1)

When 119889 isin 0 1 2 and 119863 = 0 1 2 119889 all integerscongruent to 119889 (mod 3) are absent from Z+

3(119863) The first

positive integer absent from all three sets is 1023= 11 but

it is clear that an increasing proportion of integers will bemissing fromall three sets Indeed a simple calculation showsfor any positive integer 119899 thatZ+

3(0 1) andZ+

3(0 2) each have

2119899 members below 3

119899 while Z+3(1 2) has 2119899+1 minus 2 members

below 3119899 so each set is sparse and has asymptotic density 0

It turns out that these three sets are of considerableinterest when it comes to the presence or absence ofmidpointtriples so let us now address the main subject of this paper

Note that the set Z+3(1 2) contains midpoint triples such

as 1 4 7 and 2 5 8 In factZ+3(1 2) is densely packedwith

midpoint triples since eachmember 119886 isin Z+3(1 2) is the lower

endpoint of an infinite family of midpoint triples

119886 119906119899minus 119906119898+ 119886 2119906

119899minus 2119906119898+ 119886

= (119906119899minus 119906119898) 0 1 2 + 119886 sub Z

+

3(1 2)

(10)

for any integer 119899 gt 119898 where 119898 is the number of digits inthe terminating base 3 representation of 119886 (if 119889

ℎis the leading

digit in 1198863 then119898 = ℎ + 1) and

119906119899= 1 + 3 + 3

2

+ sdot sdot sdot + 3119899minus1

=3119899minus 1

2(11)

is the positive integer with terminating base 3 representationof 119899 digits all 1 Note that Z+

3(1) = 119906

119899 119899 ge 1 and

appropriate scaling shows that this set is midpoint-free (Theintegers 119906

119899are rep-units in base 3 The positive integers with

explicit decimal digits all equal to 1 are the rep-units for base10 so called by contraction of ldquorepeated unitrdquo [4] Primefactorizations of rep-units have been much studied [5 6]Clearly if 119898 | 119899 then 119906

119898| 119906119899 Thus 119906

119899can be a prime

number only if 119899 is prime but 1199065= 11111

3= 112 shows that

this condition is not sufficient Such observations generalizeto rep-units in any base)

Not only is every member of Z+3(1 2) the lower endpoint

of infinitely many midpoint triples it is also the case that anymember greater than 2 is either the midpoint or the upperendpoint of at least one midpoint triple depending on theleading digit of its base 3 representation

In contrast it turns out that Z+3(0 1) and Z+

3(0 2) are

midpoint-free To see this it suffices to show that Z+3(0 1) is

midpoint-free thenZ+3(0 2)= 2Z+

3(0 1) ensures thatZ+

3(0 2)

is also midpoint-free (Long ago Erdos and Turan [7] notedthat the nonnegative integers with a 2-less base 3 represen-tation are midpoint-free The following compact proof isincluded for completeness and to typify what follows)

Claim A The set Z+3(0 1) is midpoint-free

Proof On the contrary suppose that 119886 119887 119888 sub Z+3(0 1) is a

midpoint triple with 119886 lt 119887 lt 119888 and 119886 + 119888 = 2119887 All digits in1198863 1198873 1198883are in 0 1 and all digits in (2119887)

3are in 0 2 Then

⟦119886⟧3119894

+ ⟦119888⟧3119894

= ⟦2119887⟧3119894 (12)

for all integers 119894 and

⟦2119887⟧3119894

= 0 997904rArr ⟦119886⟧3119894

= ⟦119887⟧3119894

= ⟦119888⟧3119894

= 0

⟦2119887⟧3119894

= 2 997904rArr ⟦119886⟧3119894

= ⟦119887⟧3119894

= ⟦119888⟧3119894

= 1

(13)

Hence 119886 = 119887 = 119888 a contradiction so Z+3(0 1) contains no

midpoint triple

It will now be shown thatZ+3(0 1) is not contained in any

larger midpoint-free subset of Z+

Claim B For any positive integer 119887 isin Z+ Z+

3(0 1) there is a

midpoint triple in Z+3(0 1) cup 119887 with 119887 as its midpoint

Proof Given 119887 isin Z+ Z+

3(0 1) we seek 119886 119888 sub Z+

3(0 1) such

that 119886 lt 119887 lt 119888 and 119886 + 119888 = 2119887 Specify the base 3 digits of 119886and 119888 as follows

⟦2119887⟧3119894

= 0 997904rArr ⟦119886⟧3119894

= ⟦119888⟧3119894

= 0

⟦2119887⟧3119894

= 2 997904rArr ⟦119886⟧3119894

= ⟦119888⟧3119894

= 1

⟦2119887⟧3119894

= 1 997904rArr ⟦119886⟧3119894

= 0 ⟦119888⟧3119894

= 1

(14)

This determines integers 119886 and 119888 such that 119886 119888 sub Z+3(0 1)

and 119886 + 119888 = 2119887 since

⟦119886⟧3119894

+ ⟦119888⟧3119894

= ⟦2119887⟧3119894 (15)

for all 119894 At least one digit of (2119887)3is 1 since 119887 notin Z+

3(0 1) so 119886

3

and 1198883differ in at least one place then ⟦119886⟧

3119894lt ⟦119888⟧

3119894in each

such place ensuring that 119886 lt 119888 Put 119889 = (119888 minus 119886)2 gt 0 Then119888 = 119886+ 2119889 so 2119887 = 119886+ 119888 = 2119886+ 2119889 whence 119887 = 119886+119889 gt 119886 and119888 = 119886 + 2119889 = 119887 + 119889 gt 119887 Thus 119886 lt 119887 lt 119888

It will now be shown that extendingZ+3(0 1) by adjoining

any rational of the form 119899 + (12) with 119899 isin Z+ always yieldsa set which contains a midpoint triple

Claim C For any 119887 isin Z+

+ (12) there is a midpoint triple inZ+3(0 1) cup 119887 with 119887 as its midpoint

Proof Again we seek 119886 119888 sub Z+3(0 1) such that 119886 lt 119887 lt 119888

and 119886 + 119888 = 2119887 In the present case note that 2119887 is an oddpositive integer so (2119887)

3contains the digit 1 an odd number

of times Specifying 119886 and 119888 as in the proof of Claim B onceagain it follows that 119886 119888 sub Z+

3(0 1) and 119886 lt 119887 lt 119888

For instance the proofs of Claims B and C produce thedecompositions

12013= 100

3+ 1101

3 212

3= 101

3+ 111

3(16)

showing that 23 and 232 are the midpoints of the pairs9 37 sub Z+

3(0 1) and 10 13 sub Z+

3(0 1) respectively

4 International Journal of Combinatorics

Further doubling shows that 46 and 23 are the midpointsof the pairs 18 74 sub Z+

3(0 2) and 20 26 sub Z+

3(0 2)

respectively From Claims A B and C and the doublingmethod just illustrated we have the following

Theorem 1 The sets Z+3(0 1) and Z+

3(0 2) are maximal

midpoint-free subsets of Z+

4 Two Unexpected Results for Z

Let us now study Z+3(0 1) and Z+

3(0 2) as midpoint-free

subsets of Z the full set of integers Surprisingly the resultsfor the two sets are quite different

Calculations until now have involved sums of pairs119886 119888 sub Z

+

3(0 1) so it has been possible to work digit by digit

in base 3 without a carry over digit When the context iswidened to Z it seems that carry over digits can no longerbe avoided so we shall consider blocks of digits rather thansingle digits Let the compact notation 119889

times119899 (ldquo119889 by 119899rdquo) denotea homogeneous block of 119899 adjacent digits all equal to 119889 thus1times119899 denotes the terminating base 3 representation of rep-unit119906119899 For blocks in which the digits are not all equal we simply

abut such expressionswith119889times1 = 119889 as a natural simplificationTo illustrate consider the integer

1198863= 1222202111001121100 = 12

times4

021times3

0times2

1times2

21times2

0times2

(17)

Let us find integers 119887 119888 isin Z+

3(0 1) such that 119886 + 2119887 = 119888

Regarding 1198863as a string of homogeneous blocks we choose

matching blocks for (2119887)3so that

(2119887)3= 2000020222222202200 = 20

times4

202times7

02times2

0times2

(18)

Calculating from right to left base 3 arithmetic for 1198863+ (2119887)

3

yields successively the following blocks with each carry overdigit indicated parenthetically

0times2

+ 0times2

= 0times2

1times2

+ 2times2

= (1) 10

2 + 0 + (1) = (1) 0

1times2

+ 2times2

+ (1) = (1)1times2

0times2

+ 2times2

+ (1) = (1) 0times2

1times3

+ 2times3

+ (1) = (1) 1times3

2 + 0 + (1) = (1) 0

0 + 2 + (1) = (1) 0

2times4

+ 0times4

+ (1) = (1) 0times4

1 + 2 + (1) = (1) 1

(19)

Assembling these blocks yields

1198863+ (2119887)

3= 1times2

| 0times4

| 0 | 0 | 1times3

| 0times2

| 1times2

| 0 | 10 | 0times2

(20)

After simplifying we have

10times4

101times7

01times2

0times2

= 1000010111111101100 = 1198873

1times2

0times6

1times3

0times2

1times2

010times3

= 11000000111001101000 = 1198883

(21)

so 119886 + 2119887 = 119888 and 119887 119888 isin Z+

3(0 1) as desired This calculation

models the proof of the following

Claim D For any integer 119886 gt 0 there is a midpoint triplein Z+3(0 1) cup minus119886 with the negative integer minus119886 as its lower

endpoint

Proof Given any positive integer 119886 we seek 119887 119888 sub Z+3(0 1)

such that minus119886 lt 119887 lt 119888 and minus119886 + 119888 = 2119887 so 119886 + 2119887 = 119888 Partition1198863into its homogeneous blocks 119860

119894 so

1198863= 119860119898sdot sdot sdot 119860119894sdot sdot sdot 11986011198600

(22)

and seek corresponding blocks 119861119894and 119862

119894to construct 119887

3and

1198883If 1198863has a terminal block119860

0= 0times119899 assign 119887

3the terminal

block 1198610= 0times119899 so 119888

3has the terminal block 119862

0= 0times119899

=

1198600+ 21198610 If 1198863has a nonterminal block 119860

119894= 0times119899 for some

119894 gt 0 assume the sum of the preceding blocks 119860119894minus1

and2119861119894minus1

contributes a carry over digit of 1 and assign 1198873the

corresponding block 119861119894= 1times119899 so

119862119894= 119860119894+ 2119861119894+ (1) = 0

times119899

+ 2times119899

+ (1) = (1) 0times119899

(23)

As 2times119899 + (1) corresponds to 2119906119899+ 1 = (3

119899+1minus 1) + 1 = 3

119899+1the computation is justified

If 1198863has a block 119860

119894= 1times119899 assign 119887

3the corresponding

block 119861119894= 1times119899 so 119888

3has the corresponding block 119862

119894= 119860119894+

2119861119894+ (120575) = (1)1

times(119899minus1)

0 + (120575) where the carry over digit 120575 iseither 0 or 1 so 119862

119894= (1)1

times(119899minus1)

0 or (1)1times119899 respectively Thisholds since the sum 1

times119899+2times119899 corresponds to 119906

119899+2119906119899= 3119906119899=

119906119899+1

minus 1199061and 1times119899

+ 2times119899

+ (1) corresponds to 3119906119899+ 1 = 119906

119899+1

If 1198863has a terminal block 119860

0= 2times119899 assign 119887

3the block

1198610= 0times(119899minus1)

1 so 1198883has terminal block 119862

0= 1198600+ 21198610

=

(1)0times(119899minus1)

1 since the sum 2times119899

+ 0times(119899minus1)

2 corresponds to 2119906119899+

2 = (3119899+1

minus 1) + 2 = 3119899+1

+ 1 Similarly if 11986011198600= 2times1198990times119898

assign 1198873the corresponding blocks 119861

11198610= 0times(119899minus1)

10times119898 then

1198883has the blocks 119862

11198620= (1)0

times(119899minus1)

10times119898 Otherwise if 119886

3has

a nonterminal block 119860119894= 2times119899 for some 119894 gt 0 assume that

119860119894minus1

+ 2119861119894minus1

yields a carry over digit 1 and assign 1198873the

block 119861119894= 0times119899 so 119888

3has 119862119894= 119860119894+ 2119861119894+ (1) = (1)0

times119899For 119894 gt 0note that the only case inwhich119860

119894minus1+2119861119894minus1

doesnot contribute a carry over digit 1 to 119860

119894+ 2119861119894is when 119894 = 1

and 1198863has terminal block119860

0= 0times119899Thus all our assumptions

about carry over digits are justified a posteriori

Claim E For any integer 119899 gt 0 there is a midpoint triple inZ+3(0 2)cupminus119899 if and only if 119899 is even and thenminus119899 is the lower

endpoint of the triple

Proof By doubling it follows from Claim D that for anyinteger 119886 gt 0 there is a midpoint triple in Z+

3(0 2) cup

International Journal of Combinatorics 5

minus2119886 However we claim that there is no midpoint triple inZ+3(0 2) cup minus2119886 + 1 On the contrary suppose minus2119886 + 1 forms

a midpoint triple with the pair 119887 119888 sub Z+3(0 2) Without loss

of generality minus2119886 + 1 lt 119887 lt 119888 and (minus2119886 + 1) + 119888 = 2119887 so2119886 + 2119887 minus 1 = 119888 this is impossible since 2119886 + 2119887 minus 1 is oddand 119888 is even It follows thatZ+

3(0 2) cup minus2119886 + 1 is midpoint-

free

Claim F The set Z+3(0 2) cup minus(Z+

3(0 2) + 1) is midpoint-free

Proof The two sets forming this union are certainlymidpoint-free so any midpoint triple in the union musthave two members in one set and one member in the otherClaim E shows that there is no midpoint triple with exactlyone member in minus(Z+

3(0 2) + 1) since this set only contains

odd negative integers So suppose there is a midpoint tripleminus119886 minus119887 119888 with exactly one member 119888 isin Z+

3(0 2) Then

119886 minus 1 119887 minus 1 minus119888 minus 1 is a midpoint triple in

minus (Z+

3(0 2) cup minus (Z

+

3(0 2) + 1)) minus 1

= Z+

3(0 2) cup minus (Z

+

3(0 2) + 1)

(24)

with exactly one member in minus(Z+3(0 2) + 1) the case ruled

out by Claim E

Since Z+3(0 2) is a maximal midpoint-free subset of Z+

by Theorem 1 clearly Z+3(0 2) + 1 is a maximal midpoint-

free subset of Z+ + 1 = Z+ 0 Note that the members ofZ+3(0 2) + 1 are precisely those positive integers 119899 for which

the trailing digit of 1198993is 1 and all other digits are in 0 2

With Claims E and F it follows thatZ+3(0 2)cupminus(Z+

3(0 2)+1)

is a maximal midpoint-free subset of Z = Z+ cup minus(Z+ 0)Combined with Claim D this proves the following

Theorem 2 The sets Z+3(0 1) and Z+

3(0 2) cup minus(Z+

3(0 2) + 1)

are maximal midpoint-free subsets of Z

5 Midpoint-Free Subsets of Q+

Now let us consider the corresponding subsets Q+3(119863) of the

nonnegative rationals Q+ namely the subsets comprisingthose members with a regular base 3 representation in whicheach digit after suppression of all trivial and optional zerosis in 119863 sub 0 1 2 The three cases of interest are those where119863 is a 2-set

As expected Q+3(1 2) is densely packed with midpoint

triples This follows from the observation that each rational119902 isin Q+

3(1 2) is the lower endpoint of an infinite family of

midpoint triples

119902 119906119899minus 119906119898+ 119902 2119906

119899minus 2119906119898+ 119902

= (119906119899minus 119906119898) 0 1 2 + 119902 sub Q

+

3(1 2)

(25)

for any integer 119899 gt 119898 where 119906119899is the 119899-digit base 3 rep-unit

and 119898 = ℎ + 1 where 119889ℎis the leading digit of the regular

base 3 representation 1199023

Next consider Q+3(0 1) The details are a little more

complicated than in the integer context since allowance

must be made for singular representations when particularmembers ofQ+

3(0 1) are doubled

Claim G The setQ+3(0 1) Q+

3([1]) is midpoint-free

Proof If 119887 isin Q+3(0 1) Q+

3([1]) and 119887 gt 0 all digits of (2119887)

3

are in 0 2 They uniquely determine the digits of 1198863and 1198883

such that 119886 119888 sub Q+3(0 1) Q+

3([1]) and 119886 + 119888 = 2119887 As in

the proof of Claim A they force 119886 = 119887 = 119888 Hence there is nomidpoint triple 119886 119887 119888 sub Q+

3(0 1) Q+

3([1])

Claim H Any 119887 isin Q+3([1]) is the midpoint of a triple with

both its endpoints in the setQ+3(0 1) capQ+

3([0])

Proof We seek 119886 119888 sub Q+3(0 1) Q+

3([1]) with 119886 lt 119887 lt 119888 and

119886 + 119888 = 2119887 Then 1198863and 1198883are uniquely determined by

⟦119886⟧3119894

+ ⟦119888⟧3119894

= ⟦2119887⟧3119894 ⟦119886⟧

3119894le ⟦119888⟧

3119894(26)

for all 119894 Since 119887 isin Q+3([1]) then 3

119896119887 isin Z+ + (12) for some

119896 isin Z+ so

2119887 isin 3minus119896

(2Z+

+ 1) (27)

It follows that (2119887)3is a terminating representation with an

odd number of digits equal to 1 Then

⟦2119887⟧3119894

= 1 997904rArr ⟦119886⟧3119894

= 0 ⟦119888⟧3119894

= 1

for an odd number of integers 119894 ge minus119896

⟦2119887⟧3119894

= 0 997904rArr ⟦119886⟧3119894

= ⟦119888⟧3119894

= 0 forall119894 lt minus119896

(28)

Hence the requiredmidpoint triple 119886 119887 119888 exists and in fact119886 119888 sub Q+

3([0])

In particular if 119887 isin Q+3([1]) is such that ⟦2119887⟧

3119899= 1 for

some integer 119899 and ⟦2119887⟧3119894

= 0 for all 119894 = 119899 then 119887 = 31198992 =

119906119899+ (12) and the proof of Claim H determines 119886 = 0 and

119888 = 2119887 = 3119899 Since ⟦119887⟧

3119894= 0 if 119894 ge 119899 and ⟦119887⟧

3119894= 1 if 119894 lt 119899

we call 119887 the base 3 fractional rep-unit with offset 119899 denotedby V119899 Then V

119899isin Q+3([1]) is the midpoint of the triple with

endpoints 0 3119899 sub Q+3(0 1) capQ+

3([0])

Positive rationals not in Q+3(0 1) cup Q+

3([1]) also belong

to midpoint triples with endpoints in Q+3(0 1) Q+

3([1]) but

proving this needs care For instance

5

16= 0 sdot [0221]

3

997904rArr5

8= 0 sdot [12]

3= 0 sdot [0111]

3+ 0 sdot [1101]

3

=13

80+37

80

(29)

so 1380 516 3780 is a midpoint triple with endpoints inQ+3(0 1) Q+

3([1]) This calculation models the proof of the

following general result

Claim I Any 119887 isin Q+ Q+

3(0 1) is themidpoint of a triple with

both its endpoints in the setQ+3(0 1) Q+

3([1])

6 International Journal of Combinatorics

Proof We seek 119886 119888 sub Q+3(0 1) Q+

3([1]) with 119886 lt 119887 lt 119888 and

119886 + 119888 = 2119887 When 119887 isin Q+3([1]) the result follows from Claim

H so we may now assume that

119887 isin Q+

(Q+

3(0 1) cupQ

+

3([1])) (30)

Then (2119887)3has at least one digit 119889

119895equal to 1 and its recurring

block say [119863] has at least one digit different from 2 Werequire 119886

3and 1198883to satisfy

⟦119886⟧3119894

+ ⟦119888⟧3119894

= ⟦2119887⟧3119894

(31)

for all 119894 Requiring ⟦119886⟧3119894

le ⟦119888⟧3119894

for all 119894 ge 119895 ensures that119886 lt 119888 since

⟦119886⟧3119895

= 0 ⟦119888⟧3119895

= 1 (32)

If the recurring block [119863] of (2119887)3contains 0 then the

recurring blocks of 1198863and 1198883also contain 0 so neither 119886 nor

119888 is in Q+3([1]) Now suppose [119863] has at least one digit equal

to 1 and none equal to 0 Clearly in this case we may choosethe integer 119895 so that the digit 119889

119895= 1 occurs in [119863] If119863 has119898

digits then for all 119894 le 119895 choose

⟦119886⟧3119894

= 0 ⟦119888⟧3119894

= 1 when 119894 equiv 119895 mod 2119898

⟦119886⟧3119894

= 1 ⟦119888⟧3119894

= 0 when 119894 equiv 119895 + 119898 mod 2119898

⟦119886⟧3119894

le ⟦119888⟧3119894

otherwise

(33)

Then 1198863and 1198883are uniquely determined so that 119886 lt 119888 and

each has a recurring block containing 0 so neither 119886 nor 119888 isinQ+3([1])

If 119902 isin Q+ Q+

3(0 2) then 119902

3has at least one digit equal

to 1 so it follows that (1199022)3isin Q+ Q

+

3(0 1) Claim I shows

that there is a midpoint triple 119886 119887 119888 with 119887 = 1199022 and119886 119888 sub Q+

3(0 1) Q+

3([1]) Then 2119886 119887 119888 is a midpoint triple

with midpoint 119902 and endpoints 2119886 119888 sub Q+3(0 2) Hence

doubling applied to Claims G and I shows that Q+3(0 2) is

midpoint-free and every 119887 isin Q+ Q+

3(0 2) is the midpoint

of a triple with endpoints in Q+3(0 2) so Claims G H and I

establish the following

Theorem 3 The sets Q+3(0 1) Q+

3([1]) and Q+

3(0 2) are

maximal midpoint-free subsets ofQ+

6 Midpoint-Free Subsets of Q

Now considerQ+3(0 1) andQ+

3(0 2) as subsets ofQ The role

ofQ+3([1]) is yet more prominent in this context

Claim J The setQ+3(0 1) capQ+

3([1]) is midpoint-free

Proof Suppose 119886 119887 119888 sub Q+3(0 1) cap Q+

3([1]) satisfies 119886 lt

119887 lt 119888 and 119886 + 119888 = 2119887 There is an integer 119898 ge 0 for which3119898119886 119887 119888 sub Z+

3(0 1) + (12) so

2 sdot 3119898

119886 119887 119888 sub 2Z+

3(0 1) + 1 (34)

This contradicts the fact that 2Z+3(0 1) + 1 is midpoint-free

by Claim A

Claim K Any positive rational 119887 isin Q+3([1]) Q+

3(0 1) is the

midpoint of a triple with endpoints inQ+3(0 1) capQ+

3([1])

Proof Given 119887 we seek 119886 119888 sub Q+3(0 1) capQ+

3([1]) such that

119886 lt 119887 lt 119888 and 119886 + 119888 = 2119887 There is an integer 119898 such that3119898119887 isin Z+ + (12) so there is an integer 119861 isin Z+ Z+

3(0 1)

such that 3119898119887 = 119861 + (12) At least one digit of 1198613is 2 so at

least one digit of (2119861)3is 1 Define 119860 119862 sub Z+

3(0 1) by

⟦2119861⟧3119894

= 0 997904rArr ⟦119860⟧3119894

= ⟦119862⟧3119894

= 0

⟦2119861⟧3119894

= 1 997904rArr ⟦119860⟧3119894

= 0 ⟦119862⟧3119894

= 1

⟦2119861⟧3119894

= 2 997904rArr ⟦119860⟧3119894

= ⟦119862⟧3119894

= 1

(35)

Then ⟦119860⟧3119894

le ⟦119862⟧3119894for all 119894 with strict inequality for at least

one 119894 so 119860 lt 119862 Also ⟦119860⟧3119894

+ ⟦119862⟧3119894

= ⟦2119861⟧3119894

for all 119894 so119860 + 119862 = 2119861 Finally let

3119898

119886 119887 119888 = 119860 119861 119862 +1

2 (36)

Then 119886 + 119888 = 2119887 and 119886 119888 sub Q+3(0 1) capQ+

3([1])

Claim L The set (Q+3(0 1) Q+

3([1])) cup minus(Q+

3(0 1)capQ+

3([1]))

is midpoint-free

Proof By Claims G and J if the specified set contains amidpoint triple twomembers of the triplemust have one signand the third member must have the opposite sign Suppose119886 sub Q+

3(0 1) cap Q+

3([1]) and 119887 119888 sub Q+

3(0 1) Q+

3([1]) are

such that minus119886 lt 119887 lt 119888 and 119886 + 2119887 = 119888 Since 1198863 (2119887)3 and 119888

3

are eventually periodic there is an integer 119898 such that theirdigits in places 119894 le 119898 are purely periodic and

⟦119886⟧3119894

= 1 ⟦2119887⟧3119894

isin 0 2 ⟦119888⟧3119894

isin 0 1

forall119894 le 119898

(37)

Since 119888 notin Q+3([1]) there is a 119895 le 119898 such that ⟦119888⟧

3119895= 0 and

therefore ⟦2119887⟧3119894

cannot be equal to 0 for all 119894 le 119898 On theother hand ⟦2119887⟧

3119894cannot be equal to 2 for all 119894 le 119898 since

119887 notin Q+3([1]) Hence the recurring block [119863] of (2119887)

3contains

0 and 2 so ⟦2119887⟧3119896

= 0 ⟦2119887⟧3119896minus1

= 2 for some 119896 le 119898 But⟦119886⟧3119896

= ⟦119886⟧3119896minus1

= 1 so ⟦119886⟧3119896minus1

+ ⟦2119887⟧3119896minus1

has a ldquocarryoverrdquo of 1 regardless of whether or not the digits in place 119896minus2contribute any ldquocarry overrdquo Then

⟦119888⟧3119896

= ⟦119886⟧3119896

+ ⟦2119887⟧3119896

+ (1) = 2 notin 0 1 (38)

By this contradiction there is no midpoint triple of theproposed type A similar but simpler argument shows thatthere are no subsets 119886 119887 sub Q+

3(0 1) cap Q+

3([1]) and 119888 sub

Q+3(0 1) Q+

3([1]) such that minus119886 lt minus119887 lt 119888 and 119886 = 2119887 + 119888

ClaimM For any positive rational 119886 isin Q+ Q+3([1]) there is a

midpoint triple in (Q+3(0 1) Q+

3([1])) cup minus119886 with minus119886 as its

lower endpoint

International Journal of Combinatorics 7

Proof Given 119886 we seek 119887 119888 sub Q+3(0 1) Q+

3([1]) such that

minus119886 lt 119887 lt 119888 and 119886 + 2119887 = 119888 Note that if the triple 119886 119887 119888

has all the required properties the scaled triple 3119898119886 119887 119888

also has all the required properties for any integer 119898 Letus assume without loss of generality that the digits of 119886

3

in negative places are purely periodic and partitioning intohomogeneous blocks 119860

119894yields

1198863= 119860ℎsdot sdot sdot 11986011198600sdot [119860minus1119860minus2

sdot sdot sdot 119860minus119896] (39)

for suitable integers ℎ ge 0 119896 gt 0 As 119886 notin Q+3([1]) if 119896 = 1

then 119860minus1

= 0 so 119886 isin Z+ and a suitable set 119887 119888 exists byClaim D

Now suppose 119896 ge 2 and 1198863has at least one negative place

digit which is nonzero If 1198863has no negative place digit equal

to 2 let 119886 = 119860 + 119902 where 119860 isin Z+ and 119902 isin Q+3(0 1)

Q+3([1]) 0 lt 119902 lt 1 are the integer part and fractional

part of 119886 respectively The proof of Claim D yields integers119861 119862 isin Z+

3(0 1) such that minus119860 119861 119862 is a midpoint triple

Choose 119887 = 119861 119888 = 119862 + 119902 Then

minus119886 119887 119888 = minus119860 minus 119902 119861 119862 + 119902 (40)

is a midpoint triple with 119887 119888 sub Q+3(0 1) Q+

3([1])

Finally suppose 119860minus119895

= 2times119899 for at least one 119895 gt 0 Define

homogeneous blocks 119861119894and 119862

119894for all 119894 as follows

119860119894= 0times119899

997904rArr 119861119894= 1times119899

119862119894= 0times119899

119860119894= 1times119899

997904rArr 119861119894= 1times119899

119862119894= 1times119899

119860119894= 2times119899

997904rArr 119861119894= 0times119899

119862119894= 0times119899

(41)

Assuming a carry over digit equal to 1 note that checks as inthe proof of Claim D justify the calculations

0times119899

+ 2times119899

+ (1) = (1) 0times119899

1times119899

+ 2times119899

+ (1) = (1) 1times119899

(42)

so in all cases we have 119860119894+ 2119861119894+ (1) = (1)119862

119894 Choose

1198873= 119861ℎsdot sdot sdot 11986111198610sdot [119861minus1119861minus2

sdot sdot sdot 119861minus119896]

1198883= (1) 119862

ℎsdot sdot sdot 11986211198620sdot [119862minus1119862minus2

sdot sdot sdot 119862minus119896]

(43)

Then 119886 + 2119887 = 119888 and 119887 119888 sub Q+3(0 1) Also 119861

minus119895= 119862minus119895

= 0times119899

ensures that 119887 119888 sub Q+3(0 1) Q+

3([1]) as required

For instance beginning with

119886 = 2 sdot [201]3=

71

26isin Q+

Q+

3([1]) (44)

the proof of Claim M constructs a midpoint triple minus119886 119887 119888with

119887 = 0 sdot [011]3=

2

13isin Q+

3(0 1) Q

+

3([1])

119888 = 10 sdot [001]3=

79

26isin Q+

3(0 1) Q

+

3([1])

(45)

However if we begin with

119886 = 2 sdot [101]3=

31

13isin Q+

Q+

3([1]) (46)

the construction falls back to Claim D resulting in 119887 = 1 and119888 = 5713

It now follows from Claims I K L and M that the set(Q+

3(0 1) Q

+

3([1])) cup minus (Q

+

3(0 1) capQ

+

3([1])) (47)

is a maximal midpoint-free subset ofQThis settles midpointquestions aboutQ+

3(0 1) as a subset ofQ

What is the situation for Q+3(0 2) If 119902 isin Q+

3(0 2) then

the recurring block of 1199023must have at least one digit equal

to 0 Hence there must be at least one digit equal to 0 inthe recurring block of (1199022)

3 so 1199022 isin Q+

3(0 1) Q+

3([1]) it

follows that 2(Q+3(0 1) Q+

3([1])) = Q+

3(0 2) Again if there

is some119898 isin Z+ such that 3119898119902 isin Z+3(0 2) + 1 it follows that

1199023is a terminating representation with trailing digit 1 and all

other digits in 0 2 Hence119902

2isin Q+

3(0 1) capQ

+

3([1]) (48)

The converse also holds so2 (Q+

3(0 1) capQ

+

3([1])) = Q

+

3(0 2 1) (49)

where Q+3(0 2 1) is the set of positive rationals which have a

terminating base 3 representation with trailing digit 1 and allother digits in 0 2 so

Q+

3(0 2 1) = ⋃

119894isinZ+

3minus119894

(Z+

3(0 2) + 1) (50)

In passing note that Z+3(0 2) + 1 = Q+

3(0 2 1) cap Z+ By

doubling it now follows from Claim L that the setQ+

3(0 2) cup minusQ

+

3(0 2 1) (51)

is midpoint-free

Claim N Any rational 119887 isin Q+ Q+3(0 2) is the midpoint of a

triple with both its endpoints inQ+3(0 2)

Proof If 119887 isin Q+ Q+3(0 2) then 1198872 notin Q+

3(0 1) so 1198872 is the

midpoint of a triple with endpoints in Q+3(0 1) Q+

3([1]) by

Claim I By doubling 119887 is the midpoint of a triple with bothits endpoints in the setQ+

3(0 2)

Claim O For any rational 119886 isin Q+ Q+3(0 2 1) there is

a midpoint triple in Q+3(0 2) cup minus119886 with minus119886 as its lower

endpoint

Proof If 119886 isin Q+ Q+3(0 2 1) and 119886 gt 0 then 1198862 notin Q+

3([1])

so there is a midpoint triple in Q+3(0 1) Q+

3([1]) cup minus1198862

with minus1198862 as its lower endpoint by Claim M By doublingQ+3(0 2)cup minus119886 has a midpoint triple with lower endpoint minus119886

The preceding results now fully establish the following

Theorem 4 The two sets(Q+

3(0 1) Q

+

3([1])) cup minus (Q

+

3(0 1) capQ

+

3([1]))

Q+

3(0 2) cup minusQ

+

3(0 2 1)

(52)

are both maximal midpoint-free subsets of Q

8 International Journal of Combinatorics

7 Midpoint-Free Subsets of R+ and R

Now consider R+3(0 1) and R+

3(0 2) These sets are uncount-

able while their maximal rational subsets Q+3(0 1) and

Q+3(0 2) are countable yet no essentially new considerations

arise in extending the results of the previous two sectionsfrom Q+ to R+ and from Q to R Noting that R+

3([1]) =

Q+3([1]) and R+

3(0 2 1) = Q+

3(0 2 1) it suffices to state the

end results for this wider context

Theorem 5 The setsR+3(0 1) R+

3([1]) andR+

3(0 2) are both

maximal midpoint-free subsets of R+

Theorem 6 The two sets

(R+

3(0 1) R

+

3([1])) cup minus (R

+

3(0 1) capR

+

3([1]))

R+

3(0 2) cup minusR

+

3(0 2 1)

(53)

are both maximal midpoint-free subsets of R

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

References

[1] N Sloane OEISmdashOnline Encyclopedia of Integer SequencesNote A003002 and A005836 in particular many relevant linksare included

[2] R K GuyUnsolved Problems in NumberTheory Springer NewYork NY USA 3rd edition 2004

[3] J Dybizbanski ldquoSequences containing no 3-term arithmeticprogressionsrdquo Electronic Journal of Combinatorics vol 19 no2 pp 1ndash15 2012

[4] D Wells The Penguin Dictionary of Curious and InterestingNumbers Penguin London UK 1986

[5] S Yates ldquoPrime divisors of repunitsrdquo Journal of RecreationalMathematics vol 8 no 1 pp 33ndash38 1975

[6] J Brillhart D H Lehmer J L Selfridge B Tuckerman and SWagstaff Jr ldquoFactorizations of 119887119899 plusmn 1 119887 = 2 3 5 6 7 10 11 12

up to high powersrdquo Contemporary Mathematics vol 22 1983[7] P Erdos and P Turan ldquoOn some sequences of integersrdquo Journal

of the London Mathematical Society vol 28 pp 104ndash109 1936

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 International Journal of Combinatorics

Further doubling shows that 46 and 23 are the midpointsof the pairs 18 74 sub Z+

3(0 2) and 20 26 sub Z+

3(0 2)

respectively From Claims A B and C and the doublingmethod just illustrated we have the following

Theorem 1 The sets Z+3(0 1) and Z+

3(0 2) are maximal

midpoint-free subsets of Z+

4 Two Unexpected Results for Z

Let us now study Z+3(0 1) and Z+

3(0 2) as midpoint-free

subsets of Z the full set of integers Surprisingly the resultsfor the two sets are quite different

Calculations until now have involved sums of pairs119886 119888 sub Z

+

3(0 1) so it has been possible to work digit by digit

in base 3 without a carry over digit When the context iswidened to Z it seems that carry over digits can no longerbe avoided so we shall consider blocks of digits rather thansingle digits Let the compact notation 119889

times119899 (ldquo119889 by 119899rdquo) denotea homogeneous block of 119899 adjacent digits all equal to 119889 thus1times119899 denotes the terminating base 3 representation of rep-unit119906119899 For blocks in which the digits are not all equal we simply

abut such expressionswith119889times1 = 119889 as a natural simplificationTo illustrate consider the integer

1198863= 1222202111001121100 = 12

times4

021times3

0times2

1times2

21times2

0times2

(17)

Let us find integers 119887 119888 isin Z+

3(0 1) such that 119886 + 2119887 = 119888

Regarding 1198863as a string of homogeneous blocks we choose

matching blocks for (2119887)3so that

(2119887)3= 2000020222222202200 = 20

times4

202times7

02times2

0times2

(18)

Calculating from right to left base 3 arithmetic for 1198863+ (2119887)

3

yields successively the following blocks with each carry overdigit indicated parenthetically

0times2

+ 0times2

= 0times2

1times2

+ 2times2

= (1) 10

2 + 0 + (1) = (1) 0

1times2

+ 2times2

+ (1) = (1)1times2

0times2

+ 2times2

+ (1) = (1) 0times2

1times3

+ 2times3

+ (1) = (1) 1times3

2 + 0 + (1) = (1) 0

0 + 2 + (1) = (1) 0

2times4

+ 0times4

+ (1) = (1) 0times4

1 + 2 + (1) = (1) 1

(19)

Assembling these blocks yields

1198863+ (2119887)

3= 1times2

| 0times4

| 0 | 0 | 1times3

| 0times2

| 1times2

| 0 | 10 | 0times2

(20)

After simplifying we have

10times4

101times7

01times2

0times2

= 1000010111111101100 = 1198873

1times2

0times6

1times3

0times2

1times2

010times3

= 11000000111001101000 = 1198883

(21)

so 119886 + 2119887 = 119888 and 119887 119888 isin Z+

3(0 1) as desired This calculation

models the proof of the following

Claim D For any integer 119886 gt 0 there is a midpoint triplein Z+3(0 1) cup minus119886 with the negative integer minus119886 as its lower

endpoint

Proof Given any positive integer 119886 we seek 119887 119888 sub Z+3(0 1)

such that minus119886 lt 119887 lt 119888 and minus119886 + 119888 = 2119887 so 119886 + 2119887 = 119888 Partition1198863into its homogeneous blocks 119860

119894 so

1198863= 119860119898sdot sdot sdot 119860119894sdot sdot sdot 11986011198600

(22)

and seek corresponding blocks 119861119894and 119862

119894to construct 119887

3and

1198883If 1198863has a terminal block119860

0= 0times119899 assign 119887

3the terminal

block 1198610= 0times119899 so 119888

3has the terminal block 119862

0= 0times119899

=

1198600+ 21198610 If 1198863has a nonterminal block 119860

119894= 0times119899 for some

119894 gt 0 assume the sum of the preceding blocks 119860119894minus1

and2119861119894minus1

contributes a carry over digit of 1 and assign 1198873the

corresponding block 119861119894= 1times119899 so

119862119894= 119860119894+ 2119861119894+ (1) = 0

times119899

+ 2times119899

+ (1) = (1) 0times119899

(23)

As 2times119899 + (1) corresponds to 2119906119899+ 1 = (3

119899+1minus 1) + 1 = 3

119899+1the computation is justified

If 1198863has a block 119860

119894= 1times119899 assign 119887

3the corresponding

block 119861119894= 1times119899 so 119888

3has the corresponding block 119862

119894= 119860119894+

2119861119894+ (120575) = (1)1

times(119899minus1)

0 + (120575) where the carry over digit 120575 iseither 0 or 1 so 119862

119894= (1)1

times(119899minus1)

0 or (1)1times119899 respectively Thisholds since the sum 1

times119899+2times119899 corresponds to 119906

119899+2119906119899= 3119906119899=

119906119899+1

minus 1199061and 1times119899

+ 2times119899

+ (1) corresponds to 3119906119899+ 1 = 119906

119899+1

If 1198863has a terminal block 119860

0= 2times119899 assign 119887

3the block

1198610= 0times(119899minus1)

1 so 1198883has terminal block 119862

0= 1198600+ 21198610

=

(1)0times(119899minus1)

1 since the sum 2times119899

+ 0times(119899minus1)

2 corresponds to 2119906119899+

2 = (3119899+1

minus 1) + 2 = 3119899+1

+ 1 Similarly if 11986011198600= 2times1198990times119898

assign 1198873the corresponding blocks 119861

11198610= 0times(119899minus1)

10times119898 then

1198883has the blocks 119862

11198620= (1)0

times(119899minus1)

10times119898 Otherwise if 119886

3has

a nonterminal block 119860119894= 2times119899 for some 119894 gt 0 assume that

119860119894minus1

+ 2119861119894minus1

yields a carry over digit 1 and assign 1198873the

block 119861119894= 0times119899 so 119888

3has 119862119894= 119860119894+ 2119861119894+ (1) = (1)0

times119899For 119894 gt 0note that the only case inwhich119860

119894minus1+2119861119894minus1

doesnot contribute a carry over digit 1 to 119860

119894+ 2119861119894is when 119894 = 1

and 1198863has terminal block119860

0= 0times119899Thus all our assumptions

about carry over digits are justified a posteriori

Claim E For any integer 119899 gt 0 there is a midpoint triple inZ+3(0 2)cupminus119899 if and only if 119899 is even and thenminus119899 is the lower

endpoint of the triple

Proof By doubling it follows from Claim D that for anyinteger 119886 gt 0 there is a midpoint triple in Z+

3(0 2) cup

International Journal of Combinatorics 5

minus2119886 However we claim that there is no midpoint triple inZ+3(0 2) cup minus2119886 + 1 On the contrary suppose minus2119886 + 1 forms

a midpoint triple with the pair 119887 119888 sub Z+3(0 2) Without loss

of generality minus2119886 + 1 lt 119887 lt 119888 and (minus2119886 + 1) + 119888 = 2119887 so2119886 + 2119887 minus 1 = 119888 this is impossible since 2119886 + 2119887 minus 1 is oddand 119888 is even It follows thatZ+

3(0 2) cup minus2119886 + 1 is midpoint-

free

Claim F The set Z+3(0 2) cup minus(Z+

3(0 2) + 1) is midpoint-free

Proof The two sets forming this union are certainlymidpoint-free so any midpoint triple in the union musthave two members in one set and one member in the otherClaim E shows that there is no midpoint triple with exactlyone member in minus(Z+

3(0 2) + 1) since this set only contains

odd negative integers So suppose there is a midpoint tripleminus119886 minus119887 119888 with exactly one member 119888 isin Z+

3(0 2) Then

119886 minus 1 119887 minus 1 minus119888 minus 1 is a midpoint triple in

minus (Z+

3(0 2) cup minus (Z

+

3(0 2) + 1)) minus 1

= Z+

3(0 2) cup minus (Z

+

3(0 2) + 1)

(24)

with exactly one member in minus(Z+3(0 2) + 1) the case ruled

out by Claim E

Since Z+3(0 2) is a maximal midpoint-free subset of Z+

by Theorem 1 clearly Z+3(0 2) + 1 is a maximal midpoint-

free subset of Z+ + 1 = Z+ 0 Note that the members ofZ+3(0 2) + 1 are precisely those positive integers 119899 for which

the trailing digit of 1198993is 1 and all other digits are in 0 2

With Claims E and F it follows thatZ+3(0 2)cupminus(Z+

3(0 2)+1)

is a maximal midpoint-free subset of Z = Z+ cup minus(Z+ 0)Combined with Claim D this proves the following

Theorem 2 The sets Z+3(0 1) and Z+

3(0 2) cup minus(Z+

3(0 2) + 1)

are maximal midpoint-free subsets of Z

5 Midpoint-Free Subsets of Q+

Now let us consider the corresponding subsets Q+3(119863) of the

nonnegative rationals Q+ namely the subsets comprisingthose members with a regular base 3 representation in whicheach digit after suppression of all trivial and optional zerosis in 119863 sub 0 1 2 The three cases of interest are those where119863 is a 2-set

As expected Q+3(1 2) is densely packed with midpoint

triples This follows from the observation that each rational119902 isin Q+

3(1 2) is the lower endpoint of an infinite family of

midpoint triples

119902 119906119899minus 119906119898+ 119902 2119906

119899minus 2119906119898+ 119902

= (119906119899minus 119906119898) 0 1 2 + 119902 sub Q

+

3(1 2)

(25)

for any integer 119899 gt 119898 where 119906119899is the 119899-digit base 3 rep-unit

and 119898 = ℎ + 1 where 119889ℎis the leading digit of the regular

base 3 representation 1199023

Next consider Q+3(0 1) The details are a little more

complicated than in the integer context since allowance

must be made for singular representations when particularmembers ofQ+

3(0 1) are doubled

Claim G The setQ+3(0 1) Q+

3([1]) is midpoint-free

Proof If 119887 isin Q+3(0 1) Q+

3([1]) and 119887 gt 0 all digits of (2119887)

3

are in 0 2 They uniquely determine the digits of 1198863and 1198883

such that 119886 119888 sub Q+3(0 1) Q+

3([1]) and 119886 + 119888 = 2119887 As in

the proof of Claim A they force 119886 = 119887 = 119888 Hence there is nomidpoint triple 119886 119887 119888 sub Q+

3(0 1) Q+

3([1])

Claim H Any 119887 isin Q+3([1]) is the midpoint of a triple with

both its endpoints in the setQ+3(0 1) capQ+

3([0])

Proof We seek 119886 119888 sub Q+3(0 1) Q+

3([1]) with 119886 lt 119887 lt 119888 and

119886 + 119888 = 2119887 Then 1198863and 1198883are uniquely determined by

⟦119886⟧3119894

+ ⟦119888⟧3119894

= ⟦2119887⟧3119894 ⟦119886⟧

3119894le ⟦119888⟧

3119894(26)

for all 119894 Since 119887 isin Q+3([1]) then 3

119896119887 isin Z+ + (12) for some

119896 isin Z+ so

2119887 isin 3minus119896

(2Z+

+ 1) (27)

It follows that (2119887)3is a terminating representation with an

odd number of digits equal to 1 Then

⟦2119887⟧3119894

= 1 997904rArr ⟦119886⟧3119894

= 0 ⟦119888⟧3119894

= 1

for an odd number of integers 119894 ge minus119896

⟦2119887⟧3119894

= 0 997904rArr ⟦119886⟧3119894

= ⟦119888⟧3119894

= 0 forall119894 lt minus119896

(28)

Hence the requiredmidpoint triple 119886 119887 119888 exists and in fact119886 119888 sub Q+

3([0])

In particular if 119887 isin Q+3([1]) is such that ⟦2119887⟧

3119899= 1 for

some integer 119899 and ⟦2119887⟧3119894

= 0 for all 119894 = 119899 then 119887 = 31198992 =

119906119899+ (12) and the proof of Claim H determines 119886 = 0 and

119888 = 2119887 = 3119899 Since ⟦119887⟧

3119894= 0 if 119894 ge 119899 and ⟦119887⟧

3119894= 1 if 119894 lt 119899

we call 119887 the base 3 fractional rep-unit with offset 119899 denotedby V119899 Then V

119899isin Q+3([1]) is the midpoint of the triple with

endpoints 0 3119899 sub Q+3(0 1) capQ+

3([0])

Positive rationals not in Q+3(0 1) cup Q+

3([1]) also belong

to midpoint triples with endpoints in Q+3(0 1) Q+

3([1]) but

proving this needs care For instance

5

16= 0 sdot [0221]

3

997904rArr5

8= 0 sdot [12]

3= 0 sdot [0111]

3+ 0 sdot [1101]

3

=13

80+37

80

(29)

so 1380 516 3780 is a midpoint triple with endpoints inQ+3(0 1) Q+

3([1]) This calculation models the proof of the

following general result

Claim I Any 119887 isin Q+ Q+

3(0 1) is themidpoint of a triple with

both its endpoints in the setQ+3(0 1) Q+

3([1])

6 International Journal of Combinatorics

Proof We seek 119886 119888 sub Q+3(0 1) Q+

3([1]) with 119886 lt 119887 lt 119888 and

119886 + 119888 = 2119887 When 119887 isin Q+3([1]) the result follows from Claim

H so we may now assume that

119887 isin Q+

(Q+

3(0 1) cupQ

+

3([1])) (30)

Then (2119887)3has at least one digit 119889

119895equal to 1 and its recurring

block say [119863] has at least one digit different from 2 Werequire 119886

3and 1198883to satisfy

⟦119886⟧3119894

+ ⟦119888⟧3119894

= ⟦2119887⟧3119894

(31)

for all 119894 Requiring ⟦119886⟧3119894

le ⟦119888⟧3119894

for all 119894 ge 119895 ensures that119886 lt 119888 since

⟦119886⟧3119895

= 0 ⟦119888⟧3119895

= 1 (32)

If the recurring block [119863] of (2119887)3contains 0 then the

recurring blocks of 1198863and 1198883also contain 0 so neither 119886 nor

119888 is in Q+3([1]) Now suppose [119863] has at least one digit equal

to 1 and none equal to 0 Clearly in this case we may choosethe integer 119895 so that the digit 119889

119895= 1 occurs in [119863] If119863 has119898

digits then for all 119894 le 119895 choose

⟦119886⟧3119894

= 0 ⟦119888⟧3119894

= 1 when 119894 equiv 119895 mod 2119898

⟦119886⟧3119894

= 1 ⟦119888⟧3119894

= 0 when 119894 equiv 119895 + 119898 mod 2119898

⟦119886⟧3119894

le ⟦119888⟧3119894

otherwise

(33)

Then 1198863and 1198883are uniquely determined so that 119886 lt 119888 and

each has a recurring block containing 0 so neither 119886 nor 119888 isinQ+3([1])

If 119902 isin Q+ Q+

3(0 2) then 119902

3has at least one digit equal

to 1 so it follows that (1199022)3isin Q+ Q

+

3(0 1) Claim I shows

that there is a midpoint triple 119886 119887 119888 with 119887 = 1199022 and119886 119888 sub Q+

3(0 1) Q+

3([1]) Then 2119886 119887 119888 is a midpoint triple

with midpoint 119902 and endpoints 2119886 119888 sub Q+3(0 2) Hence

doubling applied to Claims G and I shows that Q+3(0 2) is

midpoint-free and every 119887 isin Q+ Q+

3(0 2) is the midpoint

of a triple with endpoints in Q+3(0 2) so Claims G H and I

establish the following

Theorem 3 The sets Q+3(0 1) Q+

3([1]) and Q+

3(0 2) are

maximal midpoint-free subsets ofQ+

6 Midpoint-Free Subsets of Q

Now considerQ+3(0 1) andQ+

3(0 2) as subsets ofQ The role

ofQ+3([1]) is yet more prominent in this context

Claim J The setQ+3(0 1) capQ+

3([1]) is midpoint-free

Proof Suppose 119886 119887 119888 sub Q+3(0 1) cap Q+

3([1]) satisfies 119886 lt

119887 lt 119888 and 119886 + 119888 = 2119887 There is an integer 119898 ge 0 for which3119898119886 119887 119888 sub Z+

3(0 1) + (12) so

2 sdot 3119898

119886 119887 119888 sub 2Z+

3(0 1) + 1 (34)

This contradicts the fact that 2Z+3(0 1) + 1 is midpoint-free

by Claim A

Claim K Any positive rational 119887 isin Q+3([1]) Q+

3(0 1) is the

midpoint of a triple with endpoints inQ+3(0 1) capQ+

3([1])

Proof Given 119887 we seek 119886 119888 sub Q+3(0 1) capQ+

3([1]) such that

119886 lt 119887 lt 119888 and 119886 + 119888 = 2119887 There is an integer 119898 such that3119898119887 isin Z+ + (12) so there is an integer 119861 isin Z+ Z+

3(0 1)

such that 3119898119887 = 119861 + (12) At least one digit of 1198613is 2 so at

least one digit of (2119861)3is 1 Define 119860 119862 sub Z+

3(0 1) by

⟦2119861⟧3119894

= 0 997904rArr ⟦119860⟧3119894

= ⟦119862⟧3119894

= 0

⟦2119861⟧3119894

= 1 997904rArr ⟦119860⟧3119894

= 0 ⟦119862⟧3119894

= 1

⟦2119861⟧3119894

= 2 997904rArr ⟦119860⟧3119894

= ⟦119862⟧3119894

= 1

(35)

Then ⟦119860⟧3119894

le ⟦119862⟧3119894for all 119894 with strict inequality for at least

one 119894 so 119860 lt 119862 Also ⟦119860⟧3119894

+ ⟦119862⟧3119894

= ⟦2119861⟧3119894

for all 119894 so119860 + 119862 = 2119861 Finally let

3119898

119886 119887 119888 = 119860 119861 119862 +1

2 (36)

Then 119886 + 119888 = 2119887 and 119886 119888 sub Q+3(0 1) capQ+

3([1])

Claim L The set (Q+3(0 1) Q+

3([1])) cup minus(Q+

3(0 1)capQ+

3([1]))

is midpoint-free

Proof By Claims G and J if the specified set contains amidpoint triple twomembers of the triplemust have one signand the third member must have the opposite sign Suppose119886 sub Q+

3(0 1) cap Q+

3([1]) and 119887 119888 sub Q+

3(0 1) Q+

3([1]) are

such that minus119886 lt 119887 lt 119888 and 119886 + 2119887 = 119888 Since 1198863 (2119887)3 and 119888

3

are eventually periodic there is an integer 119898 such that theirdigits in places 119894 le 119898 are purely periodic and

⟦119886⟧3119894

= 1 ⟦2119887⟧3119894

isin 0 2 ⟦119888⟧3119894

isin 0 1

forall119894 le 119898

(37)

Since 119888 notin Q+3([1]) there is a 119895 le 119898 such that ⟦119888⟧

3119895= 0 and

therefore ⟦2119887⟧3119894

cannot be equal to 0 for all 119894 le 119898 On theother hand ⟦2119887⟧

3119894cannot be equal to 2 for all 119894 le 119898 since

119887 notin Q+3([1]) Hence the recurring block [119863] of (2119887)

3contains

0 and 2 so ⟦2119887⟧3119896

= 0 ⟦2119887⟧3119896minus1

= 2 for some 119896 le 119898 But⟦119886⟧3119896

= ⟦119886⟧3119896minus1

= 1 so ⟦119886⟧3119896minus1

+ ⟦2119887⟧3119896minus1

has a ldquocarryoverrdquo of 1 regardless of whether or not the digits in place 119896minus2contribute any ldquocarry overrdquo Then

⟦119888⟧3119896

= ⟦119886⟧3119896

+ ⟦2119887⟧3119896

+ (1) = 2 notin 0 1 (38)

By this contradiction there is no midpoint triple of theproposed type A similar but simpler argument shows thatthere are no subsets 119886 119887 sub Q+

3(0 1) cap Q+

3([1]) and 119888 sub

Q+3(0 1) Q+

3([1]) such that minus119886 lt minus119887 lt 119888 and 119886 = 2119887 + 119888

ClaimM For any positive rational 119886 isin Q+ Q+3([1]) there is a

midpoint triple in (Q+3(0 1) Q+

3([1])) cup minus119886 with minus119886 as its

lower endpoint

International Journal of Combinatorics 7

Proof Given 119886 we seek 119887 119888 sub Q+3(0 1) Q+

3([1]) such that

minus119886 lt 119887 lt 119888 and 119886 + 2119887 = 119888 Note that if the triple 119886 119887 119888

has all the required properties the scaled triple 3119898119886 119887 119888

also has all the required properties for any integer 119898 Letus assume without loss of generality that the digits of 119886

3

in negative places are purely periodic and partitioning intohomogeneous blocks 119860

119894yields

1198863= 119860ℎsdot sdot sdot 11986011198600sdot [119860minus1119860minus2

sdot sdot sdot 119860minus119896] (39)

for suitable integers ℎ ge 0 119896 gt 0 As 119886 notin Q+3([1]) if 119896 = 1

then 119860minus1

= 0 so 119886 isin Z+ and a suitable set 119887 119888 exists byClaim D

Now suppose 119896 ge 2 and 1198863has at least one negative place

digit which is nonzero If 1198863has no negative place digit equal

to 2 let 119886 = 119860 + 119902 where 119860 isin Z+ and 119902 isin Q+3(0 1)

Q+3([1]) 0 lt 119902 lt 1 are the integer part and fractional

part of 119886 respectively The proof of Claim D yields integers119861 119862 isin Z+

3(0 1) such that minus119860 119861 119862 is a midpoint triple

Choose 119887 = 119861 119888 = 119862 + 119902 Then

minus119886 119887 119888 = minus119860 minus 119902 119861 119862 + 119902 (40)

is a midpoint triple with 119887 119888 sub Q+3(0 1) Q+

3([1])

Finally suppose 119860minus119895

= 2times119899 for at least one 119895 gt 0 Define

homogeneous blocks 119861119894and 119862

119894for all 119894 as follows

119860119894= 0times119899

997904rArr 119861119894= 1times119899

119862119894= 0times119899

119860119894= 1times119899

997904rArr 119861119894= 1times119899

119862119894= 1times119899

119860119894= 2times119899

997904rArr 119861119894= 0times119899

119862119894= 0times119899

(41)

Assuming a carry over digit equal to 1 note that checks as inthe proof of Claim D justify the calculations

0times119899

+ 2times119899

+ (1) = (1) 0times119899

1times119899

+ 2times119899

+ (1) = (1) 1times119899

(42)

so in all cases we have 119860119894+ 2119861119894+ (1) = (1)119862

119894 Choose

1198873= 119861ℎsdot sdot sdot 11986111198610sdot [119861minus1119861minus2

sdot sdot sdot 119861minus119896]

1198883= (1) 119862

ℎsdot sdot sdot 11986211198620sdot [119862minus1119862minus2

sdot sdot sdot 119862minus119896]

(43)

Then 119886 + 2119887 = 119888 and 119887 119888 sub Q+3(0 1) Also 119861

minus119895= 119862minus119895

= 0times119899

ensures that 119887 119888 sub Q+3(0 1) Q+

3([1]) as required

For instance beginning with

119886 = 2 sdot [201]3=

71

26isin Q+

Q+

3([1]) (44)

the proof of Claim M constructs a midpoint triple minus119886 119887 119888with

119887 = 0 sdot [011]3=

2

13isin Q+

3(0 1) Q

+

3([1])

119888 = 10 sdot [001]3=

79

26isin Q+

3(0 1) Q

+

3([1])

(45)

However if we begin with

119886 = 2 sdot [101]3=

31

13isin Q+

Q+

3([1]) (46)

the construction falls back to Claim D resulting in 119887 = 1 and119888 = 5713

It now follows from Claims I K L and M that the set(Q+

3(0 1) Q

+

3([1])) cup minus (Q

+

3(0 1) capQ

+

3([1])) (47)

is a maximal midpoint-free subset ofQThis settles midpointquestions aboutQ+

3(0 1) as a subset ofQ

What is the situation for Q+3(0 2) If 119902 isin Q+

3(0 2) then

the recurring block of 1199023must have at least one digit equal

to 0 Hence there must be at least one digit equal to 0 inthe recurring block of (1199022)

3 so 1199022 isin Q+

3(0 1) Q+

3([1]) it

follows that 2(Q+3(0 1) Q+

3([1])) = Q+

3(0 2) Again if there

is some119898 isin Z+ such that 3119898119902 isin Z+3(0 2) + 1 it follows that

1199023is a terminating representation with trailing digit 1 and all

other digits in 0 2 Hence119902

2isin Q+

3(0 1) capQ

+

3([1]) (48)

The converse also holds so2 (Q+

3(0 1) capQ

+

3([1])) = Q

+

3(0 2 1) (49)

where Q+3(0 2 1) is the set of positive rationals which have a

terminating base 3 representation with trailing digit 1 and allother digits in 0 2 so

Q+

3(0 2 1) = ⋃

119894isinZ+

3minus119894

(Z+

3(0 2) + 1) (50)

In passing note that Z+3(0 2) + 1 = Q+

3(0 2 1) cap Z+ By

doubling it now follows from Claim L that the setQ+

3(0 2) cup minusQ

+

3(0 2 1) (51)

is midpoint-free

Claim N Any rational 119887 isin Q+ Q+3(0 2) is the midpoint of a

triple with both its endpoints inQ+3(0 2)

Proof If 119887 isin Q+ Q+3(0 2) then 1198872 notin Q+

3(0 1) so 1198872 is the

midpoint of a triple with endpoints in Q+3(0 1) Q+

3([1]) by

Claim I By doubling 119887 is the midpoint of a triple with bothits endpoints in the setQ+

3(0 2)

Claim O For any rational 119886 isin Q+ Q+3(0 2 1) there is

a midpoint triple in Q+3(0 2) cup minus119886 with minus119886 as its lower

endpoint

Proof If 119886 isin Q+ Q+3(0 2 1) and 119886 gt 0 then 1198862 notin Q+

3([1])

so there is a midpoint triple in Q+3(0 1) Q+

3([1]) cup minus1198862

with minus1198862 as its lower endpoint by Claim M By doublingQ+3(0 2)cup minus119886 has a midpoint triple with lower endpoint minus119886

The preceding results now fully establish the following

Theorem 4 The two sets(Q+

3(0 1) Q

+

3([1])) cup minus (Q

+

3(0 1) capQ

+

3([1]))

Q+

3(0 2) cup minusQ

+

3(0 2 1)

(52)

are both maximal midpoint-free subsets of Q

8 International Journal of Combinatorics

7 Midpoint-Free Subsets of R+ and R

Now consider R+3(0 1) and R+

3(0 2) These sets are uncount-

able while their maximal rational subsets Q+3(0 1) and

Q+3(0 2) are countable yet no essentially new considerations

arise in extending the results of the previous two sectionsfrom Q+ to R+ and from Q to R Noting that R+

3([1]) =

Q+3([1]) and R+

3(0 2 1) = Q+

3(0 2 1) it suffices to state the

end results for this wider context

Theorem 5 The setsR+3(0 1) R+

3([1]) andR+

3(0 2) are both

maximal midpoint-free subsets of R+

Theorem 6 The two sets

(R+

3(0 1) R

+

3([1])) cup minus (R

+

3(0 1) capR

+

3([1]))

R+

3(0 2) cup minusR

+

3(0 2 1)

(53)

are both maximal midpoint-free subsets of R

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

References

[1] N Sloane OEISmdashOnline Encyclopedia of Integer SequencesNote A003002 and A005836 in particular many relevant linksare included

[2] R K GuyUnsolved Problems in NumberTheory Springer NewYork NY USA 3rd edition 2004

[3] J Dybizbanski ldquoSequences containing no 3-term arithmeticprogressionsrdquo Electronic Journal of Combinatorics vol 19 no2 pp 1ndash15 2012

[4] D Wells The Penguin Dictionary of Curious and InterestingNumbers Penguin London UK 1986

[5] S Yates ldquoPrime divisors of repunitsrdquo Journal of RecreationalMathematics vol 8 no 1 pp 33ndash38 1975

[6] J Brillhart D H Lehmer J L Selfridge B Tuckerman and SWagstaff Jr ldquoFactorizations of 119887119899 plusmn 1 119887 = 2 3 5 6 7 10 11 12

up to high powersrdquo Contemporary Mathematics vol 22 1983[7] P Erdos and P Turan ldquoOn some sequences of integersrdquo Journal

of the London Mathematical Society vol 28 pp 104ndash109 1936

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

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International Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

International Journal of Combinatorics 5

minus2119886 However we claim that there is no midpoint triple inZ+3(0 2) cup minus2119886 + 1 On the contrary suppose minus2119886 + 1 forms

a midpoint triple with the pair 119887 119888 sub Z+3(0 2) Without loss

of generality minus2119886 + 1 lt 119887 lt 119888 and (minus2119886 + 1) + 119888 = 2119887 so2119886 + 2119887 minus 1 = 119888 this is impossible since 2119886 + 2119887 minus 1 is oddand 119888 is even It follows thatZ+

3(0 2) cup minus2119886 + 1 is midpoint-

free

Claim F The set Z+3(0 2) cup minus(Z+

3(0 2) + 1) is midpoint-free

Proof The two sets forming this union are certainlymidpoint-free so any midpoint triple in the union musthave two members in one set and one member in the otherClaim E shows that there is no midpoint triple with exactlyone member in minus(Z+

3(0 2) + 1) since this set only contains

odd negative integers So suppose there is a midpoint tripleminus119886 minus119887 119888 with exactly one member 119888 isin Z+

3(0 2) Then

119886 minus 1 119887 minus 1 minus119888 minus 1 is a midpoint triple in

minus (Z+

3(0 2) cup minus (Z

+

3(0 2) + 1)) minus 1

= Z+

3(0 2) cup minus (Z

+

3(0 2) + 1)

(24)

with exactly one member in minus(Z+3(0 2) + 1) the case ruled

out by Claim E

Since Z+3(0 2) is a maximal midpoint-free subset of Z+

by Theorem 1 clearly Z+3(0 2) + 1 is a maximal midpoint-

free subset of Z+ + 1 = Z+ 0 Note that the members ofZ+3(0 2) + 1 are precisely those positive integers 119899 for which

the trailing digit of 1198993is 1 and all other digits are in 0 2

With Claims E and F it follows thatZ+3(0 2)cupminus(Z+

3(0 2)+1)

is a maximal midpoint-free subset of Z = Z+ cup minus(Z+ 0)Combined with Claim D this proves the following

Theorem 2 The sets Z+3(0 1) and Z+

3(0 2) cup minus(Z+

3(0 2) + 1)

are maximal midpoint-free subsets of Z

5 Midpoint-Free Subsets of Q+

Now let us consider the corresponding subsets Q+3(119863) of the

nonnegative rationals Q+ namely the subsets comprisingthose members with a regular base 3 representation in whicheach digit after suppression of all trivial and optional zerosis in 119863 sub 0 1 2 The three cases of interest are those where119863 is a 2-set

As expected Q+3(1 2) is densely packed with midpoint

triples This follows from the observation that each rational119902 isin Q+

3(1 2) is the lower endpoint of an infinite family of

midpoint triples

119902 119906119899minus 119906119898+ 119902 2119906

119899minus 2119906119898+ 119902

= (119906119899minus 119906119898) 0 1 2 + 119902 sub Q

+

3(1 2)

(25)

for any integer 119899 gt 119898 where 119906119899is the 119899-digit base 3 rep-unit

and 119898 = ℎ + 1 where 119889ℎis the leading digit of the regular

base 3 representation 1199023

Next consider Q+3(0 1) The details are a little more

complicated than in the integer context since allowance

must be made for singular representations when particularmembers ofQ+

3(0 1) are doubled

Claim G The setQ+3(0 1) Q+

3([1]) is midpoint-free

Proof If 119887 isin Q+3(0 1) Q+

3([1]) and 119887 gt 0 all digits of (2119887)

3

are in 0 2 They uniquely determine the digits of 1198863and 1198883

such that 119886 119888 sub Q+3(0 1) Q+

3([1]) and 119886 + 119888 = 2119887 As in

the proof of Claim A they force 119886 = 119887 = 119888 Hence there is nomidpoint triple 119886 119887 119888 sub Q+

3(0 1) Q+

3([1])

Claim H Any 119887 isin Q+3([1]) is the midpoint of a triple with

both its endpoints in the setQ+3(0 1) capQ+

3([0])

Proof We seek 119886 119888 sub Q+3(0 1) Q+

3([1]) with 119886 lt 119887 lt 119888 and

119886 + 119888 = 2119887 Then 1198863and 1198883are uniquely determined by

⟦119886⟧3119894

+ ⟦119888⟧3119894

= ⟦2119887⟧3119894 ⟦119886⟧

3119894le ⟦119888⟧

3119894(26)

for all 119894 Since 119887 isin Q+3([1]) then 3

119896119887 isin Z+ + (12) for some

119896 isin Z+ so

2119887 isin 3minus119896

(2Z+

+ 1) (27)

It follows that (2119887)3is a terminating representation with an

odd number of digits equal to 1 Then

⟦2119887⟧3119894

= 1 997904rArr ⟦119886⟧3119894

= 0 ⟦119888⟧3119894

= 1

for an odd number of integers 119894 ge minus119896

⟦2119887⟧3119894

= 0 997904rArr ⟦119886⟧3119894

= ⟦119888⟧3119894

= 0 forall119894 lt minus119896

(28)

Hence the requiredmidpoint triple 119886 119887 119888 exists and in fact119886 119888 sub Q+

3([0])

In particular if 119887 isin Q+3([1]) is such that ⟦2119887⟧

3119899= 1 for

some integer 119899 and ⟦2119887⟧3119894

= 0 for all 119894 = 119899 then 119887 = 31198992 =

119906119899+ (12) and the proof of Claim H determines 119886 = 0 and

119888 = 2119887 = 3119899 Since ⟦119887⟧

3119894= 0 if 119894 ge 119899 and ⟦119887⟧

3119894= 1 if 119894 lt 119899

we call 119887 the base 3 fractional rep-unit with offset 119899 denotedby V119899 Then V

119899isin Q+3([1]) is the midpoint of the triple with

endpoints 0 3119899 sub Q+3(0 1) capQ+

3([0])

Positive rationals not in Q+3(0 1) cup Q+

3([1]) also belong

to midpoint triples with endpoints in Q+3(0 1) Q+

3([1]) but

proving this needs care For instance

5

16= 0 sdot [0221]

3

997904rArr5

8= 0 sdot [12]

3= 0 sdot [0111]

3+ 0 sdot [1101]

3

=13

80+37

80

(29)

so 1380 516 3780 is a midpoint triple with endpoints inQ+3(0 1) Q+

3([1]) This calculation models the proof of the

following general result

Claim I Any 119887 isin Q+ Q+

3(0 1) is themidpoint of a triple with

both its endpoints in the setQ+3(0 1) Q+

3([1])

6 International Journal of Combinatorics

Proof We seek 119886 119888 sub Q+3(0 1) Q+

3([1]) with 119886 lt 119887 lt 119888 and

119886 + 119888 = 2119887 When 119887 isin Q+3([1]) the result follows from Claim

H so we may now assume that

119887 isin Q+

(Q+

3(0 1) cupQ

+

3([1])) (30)

Then (2119887)3has at least one digit 119889

119895equal to 1 and its recurring

block say [119863] has at least one digit different from 2 Werequire 119886

3and 1198883to satisfy

⟦119886⟧3119894

+ ⟦119888⟧3119894

= ⟦2119887⟧3119894

(31)

for all 119894 Requiring ⟦119886⟧3119894

le ⟦119888⟧3119894

for all 119894 ge 119895 ensures that119886 lt 119888 since

⟦119886⟧3119895

= 0 ⟦119888⟧3119895

= 1 (32)

If the recurring block [119863] of (2119887)3contains 0 then the

recurring blocks of 1198863and 1198883also contain 0 so neither 119886 nor

119888 is in Q+3([1]) Now suppose [119863] has at least one digit equal

to 1 and none equal to 0 Clearly in this case we may choosethe integer 119895 so that the digit 119889

119895= 1 occurs in [119863] If119863 has119898

digits then for all 119894 le 119895 choose

⟦119886⟧3119894

= 0 ⟦119888⟧3119894

= 1 when 119894 equiv 119895 mod 2119898

⟦119886⟧3119894

= 1 ⟦119888⟧3119894

= 0 when 119894 equiv 119895 + 119898 mod 2119898

⟦119886⟧3119894

le ⟦119888⟧3119894

otherwise

(33)

Then 1198863and 1198883are uniquely determined so that 119886 lt 119888 and

each has a recurring block containing 0 so neither 119886 nor 119888 isinQ+3([1])

If 119902 isin Q+ Q+

3(0 2) then 119902

3has at least one digit equal

to 1 so it follows that (1199022)3isin Q+ Q

+

3(0 1) Claim I shows

that there is a midpoint triple 119886 119887 119888 with 119887 = 1199022 and119886 119888 sub Q+

3(0 1) Q+

3([1]) Then 2119886 119887 119888 is a midpoint triple

with midpoint 119902 and endpoints 2119886 119888 sub Q+3(0 2) Hence

doubling applied to Claims G and I shows that Q+3(0 2) is

midpoint-free and every 119887 isin Q+ Q+

3(0 2) is the midpoint

of a triple with endpoints in Q+3(0 2) so Claims G H and I

establish the following

Theorem 3 The sets Q+3(0 1) Q+

3([1]) and Q+

3(0 2) are

maximal midpoint-free subsets ofQ+

6 Midpoint-Free Subsets of Q

Now considerQ+3(0 1) andQ+

3(0 2) as subsets ofQ The role

ofQ+3([1]) is yet more prominent in this context

Claim J The setQ+3(0 1) capQ+

3([1]) is midpoint-free

Proof Suppose 119886 119887 119888 sub Q+3(0 1) cap Q+

3([1]) satisfies 119886 lt

119887 lt 119888 and 119886 + 119888 = 2119887 There is an integer 119898 ge 0 for which3119898119886 119887 119888 sub Z+

3(0 1) + (12) so

2 sdot 3119898

119886 119887 119888 sub 2Z+

3(0 1) + 1 (34)

This contradicts the fact that 2Z+3(0 1) + 1 is midpoint-free

by Claim A

Claim K Any positive rational 119887 isin Q+3([1]) Q+

3(0 1) is the

midpoint of a triple with endpoints inQ+3(0 1) capQ+

3([1])

Proof Given 119887 we seek 119886 119888 sub Q+3(0 1) capQ+

3([1]) such that

119886 lt 119887 lt 119888 and 119886 + 119888 = 2119887 There is an integer 119898 such that3119898119887 isin Z+ + (12) so there is an integer 119861 isin Z+ Z+

3(0 1)

such that 3119898119887 = 119861 + (12) At least one digit of 1198613is 2 so at

least one digit of (2119861)3is 1 Define 119860 119862 sub Z+

3(0 1) by

⟦2119861⟧3119894

= 0 997904rArr ⟦119860⟧3119894

= ⟦119862⟧3119894

= 0

⟦2119861⟧3119894

= 1 997904rArr ⟦119860⟧3119894

= 0 ⟦119862⟧3119894

= 1

⟦2119861⟧3119894

= 2 997904rArr ⟦119860⟧3119894

= ⟦119862⟧3119894

= 1

(35)

Then ⟦119860⟧3119894

le ⟦119862⟧3119894for all 119894 with strict inequality for at least

one 119894 so 119860 lt 119862 Also ⟦119860⟧3119894

+ ⟦119862⟧3119894

= ⟦2119861⟧3119894

for all 119894 so119860 + 119862 = 2119861 Finally let

3119898

119886 119887 119888 = 119860 119861 119862 +1

2 (36)

Then 119886 + 119888 = 2119887 and 119886 119888 sub Q+3(0 1) capQ+

3([1])

Claim L The set (Q+3(0 1) Q+

3([1])) cup minus(Q+

3(0 1)capQ+

3([1]))

is midpoint-free

Proof By Claims G and J if the specified set contains amidpoint triple twomembers of the triplemust have one signand the third member must have the opposite sign Suppose119886 sub Q+

3(0 1) cap Q+

3([1]) and 119887 119888 sub Q+

3(0 1) Q+

3([1]) are

such that minus119886 lt 119887 lt 119888 and 119886 + 2119887 = 119888 Since 1198863 (2119887)3 and 119888

3

are eventually periodic there is an integer 119898 such that theirdigits in places 119894 le 119898 are purely periodic and

⟦119886⟧3119894

= 1 ⟦2119887⟧3119894

isin 0 2 ⟦119888⟧3119894

isin 0 1

forall119894 le 119898

(37)

Since 119888 notin Q+3([1]) there is a 119895 le 119898 such that ⟦119888⟧

3119895= 0 and

therefore ⟦2119887⟧3119894

cannot be equal to 0 for all 119894 le 119898 On theother hand ⟦2119887⟧

3119894cannot be equal to 2 for all 119894 le 119898 since

119887 notin Q+3([1]) Hence the recurring block [119863] of (2119887)

3contains

0 and 2 so ⟦2119887⟧3119896

= 0 ⟦2119887⟧3119896minus1

= 2 for some 119896 le 119898 But⟦119886⟧3119896

= ⟦119886⟧3119896minus1

= 1 so ⟦119886⟧3119896minus1

+ ⟦2119887⟧3119896minus1

has a ldquocarryoverrdquo of 1 regardless of whether or not the digits in place 119896minus2contribute any ldquocarry overrdquo Then

⟦119888⟧3119896

= ⟦119886⟧3119896

+ ⟦2119887⟧3119896

+ (1) = 2 notin 0 1 (38)

By this contradiction there is no midpoint triple of theproposed type A similar but simpler argument shows thatthere are no subsets 119886 119887 sub Q+

3(0 1) cap Q+

3([1]) and 119888 sub

Q+3(0 1) Q+

3([1]) such that minus119886 lt minus119887 lt 119888 and 119886 = 2119887 + 119888

ClaimM For any positive rational 119886 isin Q+ Q+3([1]) there is a

midpoint triple in (Q+3(0 1) Q+

3([1])) cup minus119886 with minus119886 as its

lower endpoint

International Journal of Combinatorics 7

Proof Given 119886 we seek 119887 119888 sub Q+3(0 1) Q+

3([1]) such that

minus119886 lt 119887 lt 119888 and 119886 + 2119887 = 119888 Note that if the triple 119886 119887 119888

has all the required properties the scaled triple 3119898119886 119887 119888

also has all the required properties for any integer 119898 Letus assume without loss of generality that the digits of 119886

3

in negative places are purely periodic and partitioning intohomogeneous blocks 119860

119894yields

1198863= 119860ℎsdot sdot sdot 11986011198600sdot [119860minus1119860minus2

sdot sdot sdot 119860minus119896] (39)

for suitable integers ℎ ge 0 119896 gt 0 As 119886 notin Q+3([1]) if 119896 = 1

then 119860minus1

= 0 so 119886 isin Z+ and a suitable set 119887 119888 exists byClaim D

Now suppose 119896 ge 2 and 1198863has at least one negative place

digit which is nonzero If 1198863has no negative place digit equal

to 2 let 119886 = 119860 + 119902 where 119860 isin Z+ and 119902 isin Q+3(0 1)

Q+3([1]) 0 lt 119902 lt 1 are the integer part and fractional

part of 119886 respectively The proof of Claim D yields integers119861 119862 isin Z+

3(0 1) such that minus119860 119861 119862 is a midpoint triple

Choose 119887 = 119861 119888 = 119862 + 119902 Then

minus119886 119887 119888 = minus119860 minus 119902 119861 119862 + 119902 (40)

is a midpoint triple with 119887 119888 sub Q+3(0 1) Q+

3([1])

Finally suppose 119860minus119895

= 2times119899 for at least one 119895 gt 0 Define

homogeneous blocks 119861119894and 119862

119894for all 119894 as follows

119860119894= 0times119899

997904rArr 119861119894= 1times119899

119862119894= 0times119899

119860119894= 1times119899

997904rArr 119861119894= 1times119899

119862119894= 1times119899

119860119894= 2times119899

997904rArr 119861119894= 0times119899

119862119894= 0times119899

(41)

Assuming a carry over digit equal to 1 note that checks as inthe proof of Claim D justify the calculations

0times119899

+ 2times119899

+ (1) = (1) 0times119899

1times119899

+ 2times119899

+ (1) = (1) 1times119899

(42)

so in all cases we have 119860119894+ 2119861119894+ (1) = (1)119862

119894 Choose

1198873= 119861ℎsdot sdot sdot 11986111198610sdot [119861minus1119861minus2

sdot sdot sdot 119861minus119896]

1198883= (1) 119862

ℎsdot sdot sdot 11986211198620sdot [119862minus1119862minus2

sdot sdot sdot 119862minus119896]

(43)

Then 119886 + 2119887 = 119888 and 119887 119888 sub Q+3(0 1) Also 119861

minus119895= 119862minus119895

= 0times119899

ensures that 119887 119888 sub Q+3(0 1) Q+

3([1]) as required

For instance beginning with

119886 = 2 sdot [201]3=

71

26isin Q+

Q+

3([1]) (44)

the proof of Claim M constructs a midpoint triple minus119886 119887 119888with

119887 = 0 sdot [011]3=

2

13isin Q+

3(0 1) Q

+

3([1])

119888 = 10 sdot [001]3=

79

26isin Q+

3(0 1) Q

+

3([1])

(45)

However if we begin with

119886 = 2 sdot [101]3=

31

13isin Q+

Q+

3([1]) (46)

the construction falls back to Claim D resulting in 119887 = 1 and119888 = 5713

It now follows from Claims I K L and M that the set(Q+

3(0 1) Q

+

3([1])) cup minus (Q

+

3(0 1) capQ

+

3([1])) (47)

is a maximal midpoint-free subset ofQThis settles midpointquestions aboutQ+

3(0 1) as a subset ofQ

What is the situation for Q+3(0 2) If 119902 isin Q+

3(0 2) then

the recurring block of 1199023must have at least one digit equal

to 0 Hence there must be at least one digit equal to 0 inthe recurring block of (1199022)

3 so 1199022 isin Q+

3(0 1) Q+

3([1]) it

follows that 2(Q+3(0 1) Q+

3([1])) = Q+

3(0 2) Again if there

is some119898 isin Z+ such that 3119898119902 isin Z+3(0 2) + 1 it follows that

1199023is a terminating representation with trailing digit 1 and all

other digits in 0 2 Hence119902

2isin Q+

3(0 1) capQ

+

3([1]) (48)

The converse also holds so2 (Q+

3(0 1) capQ

+

3([1])) = Q

+

3(0 2 1) (49)

where Q+3(0 2 1) is the set of positive rationals which have a

terminating base 3 representation with trailing digit 1 and allother digits in 0 2 so

Q+

3(0 2 1) = ⋃

119894isinZ+

3minus119894

(Z+

3(0 2) + 1) (50)

In passing note that Z+3(0 2) + 1 = Q+

3(0 2 1) cap Z+ By

doubling it now follows from Claim L that the setQ+

3(0 2) cup minusQ

+

3(0 2 1) (51)

is midpoint-free

Claim N Any rational 119887 isin Q+ Q+3(0 2) is the midpoint of a

triple with both its endpoints inQ+3(0 2)

Proof If 119887 isin Q+ Q+3(0 2) then 1198872 notin Q+

3(0 1) so 1198872 is the

midpoint of a triple with endpoints in Q+3(0 1) Q+

3([1]) by

Claim I By doubling 119887 is the midpoint of a triple with bothits endpoints in the setQ+

3(0 2)

Claim O For any rational 119886 isin Q+ Q+3(0 2 1) there is

a midpoint triple in Q+3(0 2) cup minus119886 with minus119886 as its lower

endpoint

Proof If 119886 isin Q+ Q+3(0 2 1) and 119886 gt 0 then 1198862 notin Q+

3([1])

so there is a midpoint triple in Q+3(0 1) Q+

3([1]) cup minus1198862

with minus1198862 as its lower endpoint by Claim M By doublingQ+3(0 2)cup minus119886 has a midpoint triple with lower endpoint minus119886

The preceding results now fully establish the following

Theorem 4 The two sets(Q+

3(0 1) Q

+

3([1])) cup minus (Q

+

3(0 1) capQ

+

3([1]))

Q+

3(0 2) cup minusQ

+

3(0 2 1)

(52)

are both maximal midpoint-free subsets of Q

8 International Journal of Combinatorics

7 Midpoint-Free Subsets of R+ and R

Now consider R+3(0 1) and R+

3(0 2) These sets are uncount-

able while their maximal rational subsets Q+3(0 1) and

Q+3(0 2) are countable yet no essentially new considerations

arise in extending the results of the previous two sectionsfrom Q+ to R+ and from Q to R Noting that R+

3([1]) =

Q+3([1]) and R+

3(0 2 1) = Q+

3(0 2 1) it suffices to state the

end results for this wider context

Theorem 5 The setsR+3(0 1) R+

3([1]) andR+

3(0 2) are both

maximal midpoint-free subsets of R+

Theorem 6 The two sets

(R+

3(0 1) R

+

3([1])) cup minus (R

+

3(0 1) capR

+

3([1]))

R+

3(0 2) cup minusR

+

3(0 2 1)

(53)

are both maximal midpoint-free subsets of R

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

References

[1] N Sloane OEISmdashOnline Encyclopedia of Integer SequencesNote A003002 and A005836 in particular many relevant linksare included

[2] R K GuyUnsolved Problems in NumberTheory Springer NewYork NY USA 3rd edition 2004

[3] J Dybizbanski ldquoSequences containing no 3-term arithmeticprogressionsrdquo Electronic Journal of Combinatorics vol 19 no2 pp 1ndash15 2012

[4] D Wells The Penguin Dictionary of Curious and InterestingNumbers Penguin London UK 1986

[5] S Yates ldquoPrime divisors of repunitsrdquo Journal of RecreationalMathematics vol 8 no 1 pp 33ndash38 1975

[6] J Brillhart D H Lehmer J L Selfridge B Tuckerman and SWagstaff Jr ldquoFactorizations of 119887119899 plusmn 1 119887 = 2 3 5 6 7 10 11 12

up to high powersrdquo Contemporary Mathematics vol 22 1983[7] P Erdos and P Turan ldquoOn some sequences of integersrdquo Journal

of the London Mathematical Society vol 28 pp 104ndash109 1936

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 International Journal of Combinatorics

Proof We seek 119886 119888 sub Q+3(0 1) Q+

3([1]) with 119886 lt 119887 lt 119888 and

119886 + 119888 = 2119887 When 119887 isin Q+3([1]) the result follows from Claim

H so we may now assume that

119887 isin Q+

(Q+

3(0 1) cupQ

+

3([1])) (30)

Then (2119887)3has at least one digit 119889

119895equal to 1 and its recurring

block say [119863] has at least one digit different from 2 Werequire 119886

3and 1198883to satisfy

⟦119886⟧3119894

+ ⟦119888⟧3119894

= ⟦2119887⟧3119894

(31)

for all 119894 Requiring ⟦119886⟧3119894

le ⟦119888⟧3119894

for all 119894 ge 119895 ensures that119886 lt 119888 since

⟦119886⟧3119895

= 0 ⟦119888⟧3119895

= 1 (32)

If the recurring block [119863] of (2119887)3contains 0 then the

recurring blocks of 1198863and 1198883also contain 0 so neither 119886 nor

119888 is in Q+3([1]) Now suppose [119863] has at least one digit equal

to 1 and none equal to 0 Clearly in this case we may choosethe integer 119895 so that the digit 119889

119895= 1 occurs in [119863] If119863 has119898

digits then for all 119894 le 119895 choose

⟦119886⟧3119894

= 0 ⟦119888⟧3119894

= 1 when 119894 equiv 119895 mod 2119898

⟦119886⟧3119894

= 1 ⟦119888⟧3119894

= 0 when 119894 equiv 119895 + 119898 mod 2119898

⟦119886⟧3119894

le ⟦119888⟧3119894

otherwise

(33)

Then 1198863and 1198883are uniquely determined so that 119886 lt 119888 and

each has a recurring block containing 0 so neither 119886 nor 119888 isinQ+3([1])

If 119902 isin Q+ Q+

3(0 2) then 119902

3has at least one digit equal

to 1 so it follows that (1199022)3isin Q+ Q

+

3(0 1) Claim I shows

that there is a midpoint triple 119886 119887 119888 with 119887 = 1199022 and119886 119888 sub Q+

3(0 1) Q+

3([1]) Then 2119886 119887 119888 is a midpoint triple

with midpoint 119902 and endpoints 2119886 119888 sub Q+3(0 2) Hence

doubling applied to Claims G and I shows that Q+3(0 2) is

midpoint-free and every 119887 isin Q+ Q+

3(0 2) is the midpoint

of a triple with endpoints in Q+3(0 2) so Claims G H and I

establish the following

Theorem 3 The sets Q+3(0 1) Q+

3([1]) and Q+

3(0 2) are

maximal midpoint-free subsets ofQ+

6 Midpoint-Free Subsets of Q

Now considerQ+3(0 1) andQ+

3(0 2) as subsets ofQ The role

ofQ+3([1]) is yet more prominent in this context

Claim J The setQ+3(0 1) capQ+

3([1]) is midpoint-free

Proof Suppose 119886 119887 119888 sub Q+3(0 1) cap Q+

3([1]) satisfies 119886 lt

119887 lt 119888 and 119886 + 119888 = 2119887 There is an integer 119898 ge 0 for which3119898119886 119887 119888 sub Z+

3(0 1) + (12) so

2 sdot 3119898

119886 119887 119888 sub 2Z+

3(0 1) + 1 (34)

This contradicts the fact that 2Z+3(0 1) + 1 is midpoint-free

by Claim A

Claim K Any positive rational 119887 isin Q+3([1]) Q+

3(0 1) is the

midpoint of a triple with endpoints inQ+3(0 1) capQ+

3([1])

Proof Given 119887 we seek 119886 119888 sub Q+3(0 1) capQ+

3([1]) such that

119886 lt 119887 lt 119888 and 119886 + 119888 = 2119887 There is an integer 119898 such that3119898119887 isin Z+ + (12) so there is an integer 119861 isin Z+ Z+

3(0 1)

such that 3119898119887 = 119861 + (12) At least one digit of 1198613is 2 so at

least one digit of (2119861)3is 1 Define 119860 119862 sub Z+

3(0 1) by

⟦2119861⟧3119894

= 0 997904rArr ⟦119860⟧3119894

= ⟦119862⟧3119894

= 0

⟦2119861⟧3119894

= 1 997904rArr ⟦119860⟧3119894

= 0 ⟦119862⟧3119894

= 1

⟦2119861⟧3119894

= 2 997904rArr ⟦119860⟧3119894

= ⟦119862⟧3119894

= 1

(35)

Then ⟦119860⟧3119894

le ⟦119862⟧3119894for all 119894 with strict inequality for at least

one 119894 so 119860 lt 119862 Also ⟦119860⟧3119894

+ ⟦119862⟧3119894

= ⟦2119861⟧3119894

for all 119894 so119860 + 119862 = 2119861 Finally let

3119898

119886 119887 119888 = 119860 119861 119862 +1

2 (36)

Then 119886 + 119888 = 2119887 and 119886 119888 sub Q+3(0 1) capQ+

3([1])

Claim L The set (Q+3(0 1) Q+

3([1])) cup minus(Q+

3(0 1)capQ+

3([1]))

is midpoint-free

Proof By Claims G and J if the specified set contains amidpoint triple twomembers of the triplemust have one signand the third member must have the opposite sign Suppose119886 sub Q+

3(0 1) cap Q+

3([1]) and 119887 119888 sub Q+

3(0 1) Q+

3([1]) are

such that minus119886 lt 119887 lt 119888 and 119886 + 2119887 = 119888 Since 1198863 (2119887)3 and 119888

3

are eventually periodic there is an integer 119898 such that theirdigits in places 119894 le 119898 are purely periodic and

⟦119886⟧3119894

= 1 ⟦2119887⟧3119894

isin 0 2 ⟦119888⟧3119894

isin 0 1

forall119894 le 119898

(37)

Since 119888 notin Q+3([1]) there is a 119895 le 119898 such that ⟦119888⟧

3119895= 0 and

therefore ⟦2119887⟧3119894

cannot be equal to 0 for all 119894 le 119898 On theother hand ⟦2119887⟧

3119894cannot be equal to 2 for all 119894 le 119898 since

119887 notin Q+3([1]) Hence the recurring block [119863] of (2119887)

3contains

0 and 2 so ⟦2119887⟧3119896

= 0 ⟦2119887⟧3119896minus1

= 2 for some 119896 le 119898 But⟦119886⟧3119896

= ⟦119886⟧3119896minus1

= 1 so ⟦119886⟧3119896minus1

+ ⟦2119887⟧3119896minus1

has a ldquocarryoverrdquo of 1 regardless of whether or not the digits in place 119896minus2contribute any ldquocarry overrdquo Then

⟦119888⟧3119896

= ⟦119886⟧3119896

+ ⟦2119887⟧3119896

+ (1) = 2 notin 0 1 (38)

By this contradiction there is no midpoint triple of theproposed type A similar but simpler argument shows thatthere are no subsets 119886 119887 sub Q+

3(0 1) cap Q+

3([1]) and 119888 sub

Q+3(0 1) Q+

3([1]) such that minus119886 lt minus119887 lt 119888 and 119886 = 2119887 + 119888

ClaimM For any positive rational 119886 isin Q+ Q+3([1]) there is a

midpoint triple in (Q+3(0 1) Q+

3([1])) cup minus119886 with minus119886 as its

lower endpoint

International Journal of Combinatorics 7

Proof Given 119886 we seek 119887 119888 sub Q+3(0 1) Q+

3([1]) such that

minus119886 lt 119887 lt 119888 and 119886 + 2119887 = 119888 Note that if the triple 119886 119887 119888

has all the required properties the scaled triple 3119898119886 119887 119888

also has all the required properties for any integer 119898 Letus assume without loss of generality that the digits of 119886

3

in negative places are purely periodic and partitioning intohomogeneous blocks 119860

119894yields

1198863= 119860ℎsdot sdot sdot 11986011198600sdot [119860minus1119860minus2

sdot sdot sdot 119860minus119896] (39)

for suitable integers ℎ ge 0 119896 gt 0 As 119886 notin Q+3([1]) if 119896 = 1

then 119860minus1

= 0 so 119886 isin Z+ and a suitable set 119887 119888 exists byClaim D

Now suppose 119896 ge 2 and 1198863has at least one negative place

digit which is nonzero If 1198863has no negative place digit equal

to 2 let 119886 = 119860 + 119902 where 119860 isin Z+ and 119902 isin Q+3(0 1)

Q+3([1]) 0 lt 119902 lt 1 are the integer part and fractional

part of 119886 respectively The proof of Claim D yields integers119861 119862 isin Z+

3(0 1) such that minus119860 119861 119862 is a midpoint triple

Choose 119887 = 119861 119888 = 119862 + 119902 Then

minus119886 119887 119888 = minus119860 minus 119902 119861 119862 + 119902 (40)

is a midpoint triple with 119887 119888 sub Q+3(0 1) Q+

3([1])

Finally suppose 119860minus119895

= 2times119899 for at least one 119895 gt 0 Define

homogeneous blocks 119861119894and 119862

119894for all 119894 as follows

119860119894= 0times119899

997904rArr 119861119894= 1times119899

119862119894= 0times119899

119860119894= 1times119899

997904rArr 119861119894= 1times119899

119862119894= 1times119899

119860119894= 2times119899

997904rArr 119861119894= 0times119899

119862119894= 0times119899

(41)

Assuming a carry over digit equal to 1 note that checks as inthe proof of Claim D justify the calculations

0times119899

+ 2times119899

+ (1) = (1) 0times119899

1times119899

+ 2times119899

+ (1) = (1) 1times119899

(42)

so in all cases we have 119860119894+ 2119861119894+ (1) = (1)119862

119894 Choose

1198873= 119861ℎsdot sdot sdot 11986111198610sdot [119861minus1119861minus2

sdot sdot sdot 119861minus119896]

1198883= (1) 119862

ℎsdot sdot sdot 11986211198620sdot [119862minus1119862minus2

sdot sdot sdot 119862minus119896]

(43)

Then 119886 + 2119887 = 119888 and 119887 119888 sub Q+3(0 1) Also 119861

minus119895= 119862minus119895

= 0times119899

ensures that 119887 119888 sub Q+3(0 1) Q+

3([1]) as required

For instance beginning with

119886 = 2 sdot [201]3=

71

26isin Q+

Q+

3([1]) (44)

the proof of Claim M constructs a midpoint triple minus119886 119887 119888with

119887 = 0 sdot [011]3=

2

13isin Q+

3(0 1) Q

+

3([1])

119888 = 10 sdot [001]3=

79

26isin Q+

3(0 1) Q

+

3([1])

(45)

However if we begin with

119886 = 2 sdot [101]3=

31

13isin Q+

Q+

3([1]) (46)

the construction falls back to Claim D resulting in 119887 = 1 and119888 = 5713

It now follows from Claims I K L and M that the set(Q+

3(0 1) Q

+

3([1])) cup minus (Q

+

3(0 1) capQ

+

3([1])) (47)

is a maximal midpoint-free subset ofQThis settles midpointquestions aboutQ+

3(0 1) as a subset ofQ

What is the situation for Q+3(0 2) If 119902 isin Q+

3(0 2) then

the recurring block of 1199023must have at least one digit equal

to 0 Hence there must be at least one digit equal to 0 inthe recurring block of (1199022)

3 so 1199022 isin Q+

3(0 1) Q+

3([1]) it

follows that 2(Q+3(0 1) Q+

3([1])) = Q+

3(0 2) Again if there

is some119898 isin Z+ such that 3119898119902 isin Z+3(0 2) + 1 it follows that

1199023is a terminating representation with trailing digit 1 and all

other digits in 0 2 Hence119902

2isin Q+

3(0 1) capQ

+

3([1]) (48)

The converse also holds so2 (Q+

3(0 1) capQ

+

3([1])) = Q

+

3(0 2 1) (49)

where Q+3(0 2 1) is the set of positive rationals which have a

terminating base 3 representation with trailing digit 1 and allother digits in 0 2 so

Q+

3(0 2 1) = ⋃

119894isinZ+

3minus119894

(Z+

3(0 2) + 1) (50)

In passing note that Z+3(0 2) + 1 = Q+

3(0 2 1) cap Z+ By

doubling it now follows from Claim L that the setQ+

3(0 2) cup minusQ

+

3(0 2 1) (51)

is midpoint-free

Claim N Any rational 119887 isin Q+ Q+3(0 2) is the midpoint of a

triple with both its endpoints inQ+3(0 2)

Proof If 119887 isin Q+ Q+3(0 2) then 1198872 notin Q+

3(0 1) so 1198872 is the

midpoint of a triple with endpoints in Q+3(0 1) Q+

3([1]) by

Claim I By doubling 119887 is the midpoint of a triple with bothits endpoints in the setQ+

3(0 2)

Claim O For any rational 119886 isin Q+ Q+3(0 2 1) there is

a midpoint triple in Q+3(0 2) cup minus119886 with minus119886 as its lower

endpoint

Proof If 119886 isin Q+ Q+3(0 2 1) and 119886 gt 0 then 1198862 notin Q+

3([1])

so there is a midpoint triple in Q+3(0 1) Q+

3([1]) cup minus1198862

with minus1198862 as its lower endpoint by Claim M By doublingQ+3(0 2)cup minus119886 has a midpoint triple with lower endpoint minus119886

The preceding results now fully establish the following

Theorem 4 The two sets(Q+

3(0 1) Q

+

3([1])) cup minus (Q

+

3(0 1) capQ

+

3([1]))

Q+

3(0 2) cup minusQ

+

3(0 2 1)

(52)

are both maximal midpoint-free subsets of Q

8 International Journal of Combinatorics

7 Midpoint-Free Subsets of R+ and R

Now consider R+3(0 1) and R+

3(0 2) These sets are uncount-

able while their maximal rational subsets Q+3(0 1) and

Q+3(0 2) are countable yet no essentially new considerations

arise in extending the results of the previous two sectionsfrom Q+ to R+ and from Q to R Noting that R+

3([1]) =

Q+3([1]) and R+

3(0 2 1) = Q+

3(0 2 1) it suffices to state the

end results for this wider context

Theorem 5 The setsR+3(0 1) R+

3([1]) andR+

3(0 2) are both

maximal midpoint-free subsets of R+

Theorem 6 The two sets

(R+

3(0 1) R

+

3([1])) cup minus (R

+

3(0 1) capR

+

3([1]))

R+

3(0 2) cup minusR

+

3(0 2 1)

(53)

are both maximal midpoint-free subsets of R

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

References

[1] N Sloane OEISmdashOnline Encyclopedia of Integer SequencesNote A003002 and A005836 in particular many relevant linksare included

[2] R K GuyUnsolved Problems in NumberTheory Springer NewYork NY USA 3rd edition 2004

[3] J Dybizbanski ldquoSequences containing no 3-term arithmeticprogressionsrdquo Electronic Journal of Combinatorics vol 19 no2 pp 1ndash15 2012

[4] D Wells The Penguin Dictionary of Curious and InterestingNumbers Penguin London UK 1986

[5] S Yates ldquoPrime divisors of repunitsrdquo Journal of RecreationalMathematics vol 8 no 1 pp 33ndash38 1975

[6] J Brillhart D H Lehmer J L Selfridge B Tuckerman and SWagstaff Jr ldquoFactorizations of 119887119899 plusmn 1 119887 = 2 3 5 6 7 10 11 12

up to high powersrdquo Contemporary Mathematics vol 22 1983[7] P Erdos and P Turan ldquoOn some sequences of integersrdquo Journal

of the London Mathematical Society vol 28 pp 104ndash109 1936

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

International Journal of Combinatorics 7

Proof Given 119886 we seek 119887 119888 sub Q+3(0 1) Q+

3([1]) such that

minus119886 lt 119887 lt 119888 and 119886 + 2119887 = 119888 Note that if the triple 119886 119887 119888

has all the required properties the scaled triple 3119898119886 119887 119888

also has all the required properties for any integer 119898 Letus assume without loss of generality that the digits of 119886

3

in negative places are purely periodic and partitioning intohomogeneous blocks 119860

119894yields

1198863= 119860ℎsdot sdot sdot 11986011198600sdot [119860minus1119860minus2

sdot sdot sdot 119860minus119896] (39)

for suitable integers ℎ ge 0 119896 gt 0 As 119886 notin Q+3([1]) if 119896 = 1

then 119860minus1

= 0 so 119886 isin Z+ and a suitable set 119887 119888 exists byClaim D

Now suppose 119896 ge 2 and 1198863has at least one negative place

digit which is nonzero If 1198863has no negative place digit equal

to 2 let 119886 = 119860 + 119902 where 119860 isin Z+ and 119902 isin Q+3(0 1)

Q+3([1]) 0 lt 119902 lt 1 are the integer part and fractional

part of 119886 respectively The proof of Claim D yields integers119861 119862 isin Z+

3(0 1) such that minus119860 119861 119862 is a midpoint triple

Choose 119887 = 119861 119888 = 119862 + 119902 Then

minus119886 119887 119888 = minus119860 minus 119902 119861 119862 + 119902 (40)

is a midpoint triple with 119887 119888 sub Q+3(0 1) Q+

3([1])

Finally suppose 119860minus119895

= 2times119899 for at least one 119895 gt 0 Define

homogeneous blocks 119861119894and 119862

119894for all 119894 as follows

119860119894= 0times119899

997904rArr 119861119894= 1times119899

119862119894= 0times119899

119860119894= 1times119899

997904rArr 119861119894= 1times119899

119862119894= 1times119899

119860119894= 2times119899

997904rArr 119861119894= 0times119899

119862119894= 0times119899

(41)

Assuming a carry over digit equal to 1 note that checks as inthe proof of Claim D justify the calculations

0times119899

+ 2times119899

+ (1) = (1) 0times119899

1times119899

+ 2times119899

+ (1) = (1) 1times119899

(42)

so in all cases we have 119860119894+ 2119861119894+ (1) = (1)119862

119894 Choose

1198873= 119861ℎsdot sdot sdot 11986111198610sdot [119861minus1119861minus2

sdot sdot sdot 119861minus119896]

1198883= (1) 119862

ℎsdot sdot sdot 11986211198620sdot [119862minus1119862minus2

sdot sdot sdot 119862minus119896]

(43)

Then 119886 + 2119887 = 119888 and 119887 119888 sub Q+3(0 1) Also 119861

minus119895= 119862minus119895

= 0times119899

ensures that 119887 119888 sub Q+3(0 1) Q+

3([1]) as required

For instance beginning with

119886 = 2 sdot [201]3=

71

26isin Q+

Q+

3([1]) (44)

the proof of Claim M constructs a midpoint triple minus119886 119887 119888with

119887 = 0 sdot [011]3=

2

13isin Q+

3(0 1) Q

+

3([1])

119888 = 10 sdot [001]3=

79

26isin Q+

3(0 1) Q

+

3([1])

(45)

However if we begin with

119886 = 2 sdot [101]3=

31

13isin Q+

Q+

3([1]) (46)

the construction falls back to Claim D resulting in 119887 = 1 and119888 = 5713

It now follows from Claims I K L and M that the set(Q+

3(0 1) Q

+

3([1])) cup minus (Q

+

3(0 1) capQ

+

3([1])) (47)

is a maximal midpoint-free subset ofQThis settles midpointquestions aboutQ+

3(0 1) as a subset ofQ

What is the situation for Q+3(0 2) If 119902 isin Q+

3(0 2) then

the recurring block of 1199023must have at least one digit equal

to 0 Hence there must be at least one digit equal to 0 inthe recurring block of (1199022)

3 so 1199022 isin Q+

3(0 1) Q+

3([1]) it

follows that 2(Q+3(0 1) Q+

3([1])) = Q+

3(0 2) Again if there

is some119898 isin Z+ such that 3119898119902 isin Z+3(0 2) + 1 it follows that

1199023is a terminating representation with trailing digit 1 and all

other digits in 0 2 Hence119902

2isin Q+

3(0 1) capQ

+

3([1]) (48)

The converse also holds so2 (Q+

3(0 1) capQ

+

3([1])) = Q

+

3(0 2 1) (49)

where Q+3(0 2 1) is the set of positive rationals which have a

terminating base 3 representation with trailing digit 1 and allother digits in 0 2 so

Q+

3(0 2 1) = ⋃

119894isinZ+

3minus119894

(Z+

3(0 2) + 1) (50)

In passing note that Z+3(0 2) + 1 = Q+

3(0 2 1) cap Z+ By

doubling it now follows from Claim L that the setQ+

3(0 2) cup minusQ

+

3(0 2 1) (51)

is midpoint-free

Claim N Any rational 119887 isin Q+ Q+3(0 2) is the midpoint of a

triple with both its endpoints inQ+3(0 2)

Proof If 119887 isin Q+ Q+3(0 2) then 1198872 notin Q+

3(0 1) so 1198872 is the

midpoint of a triple with endpoints in Q+3(0 1) Q+

3([1]) by

Claim I By doubling 119887 is the midpoint of a triple with bothits endpoints in the setQ+

3(0 2)

Claim O For any rational 119886 isin Q+ Q+3(0 2 1) there is

a midpoint triple in Q+3(0 2) cup minus119886 with minus119886 as its lower

endpoint

Proof If 119886 isin Q+ Q+3(0 2 1) and 119886 gt 0 then 1198862 notin Q+

3([1])

so there is a midpoint triple in Q+3(0 1) Q+

3([1]) cup minus1198862

with minus1198862 as its lower endpoint by Claim M By doublingQ+3(0 2)cup minus119886 has a midpoint triple with lower endpoint minus119886

The preceding results now fully establish the following

Theorem 4 The two sets(Q+

3(0 1) Q

+

3([1])) cup minus (Q

+

3(0 1) capQ

+

3([1]))

Q+

3(0 2) cup minusQ

+

3(0 2 1)

(52)

are both maximal midpoint-free subsets of Q

8 International Journal of Combinatorics

7 Midpoint-Free Subsets of R+ and R

Now consider R+3(0 1) and R+

3(0 2) These sets are uncount-

able while their maximal rational subsets Q+3(0 1) and

Q+3(0 2) are countable yet no essentially new considerations

arise in extending the results of the previous two sectionsfrom Q+ to R+ and from Q to R Noting that R+

3([1]) =

Q+3([1]) and R+

3(0 2 1) = Q+

3(0 2 1) it suffices to state the

end results for this wider context

Theorem 5 The setsR+3(0 1) R+

3([1]) andR+

3(0 2) are both

maximal midpoint-free subsets of R+

Theorem 6 The two sets

(R+

3(0 1) R

+

3([1])) cup minus (R

+

3(0 1) capR

+

3([1]))

R+

3(0 2) cup minusR

+

3(0 2 1)

(53)

are both maximal midpoint-free subsets of R

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

References

[1] N Sloane OEISmdashOnline Encyclopedia of Integer SequencesNote A003002 and A005836 in particular many relevant linksare included

[2] R K GuyUnsolved Problems in NumberTheory Springer NewYork NY USA 3rd edition 2004

[3] J Dybizbanski ldquoSequences containing no 3-term arithmeticprogressionsrdquo Electronic Journal of Combinatorics vol 19 no2 pp 1ndash15 2012

[4] D Wells The Penguin Dictionary of Curious and InterestingNumbers Penguin London UK 1986

[5] S Yates ldquoPrime divisors of repunitsrdquo Journal of RecreationalMathematics vol 8 no 1 pp 33ndash38 1975

[6] J Brillhart D H Lehmer J L Selfridge B Tuckerman and SWagstaff Jr ldquoFactorizations of 119887119899 plusmn 1 119887 = 2 3 5 6 7 10 11 12

up to high powersrdquo Contemporary Mathematics vol 22 1983[7] P Erdos and P Turan ldquoOn some sequences of integersrdquo Journal

of the London Mathematical Society vol 28 pp 104ndash109 1936

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 International Journal of Combinatorics

7 Midpoint-Free Subsets of R+ and R

Now consider R+3(0 1) and R+

3(0 2) These sets are uncount-

able while their maximal rational subsets Q+3(0 1) and

Q+3(0 2) are countable yet no essentially new considerations

arise in extending the results of the previous two sectionsfrom Q+ to R+ and from Q to R Noting that R+

3([1]) =

Q+3([1]) and R+

3(0 2 1) = Q+

3(0 2 1) it suffices to state the

end results for this wider context

Theorem 5 The setsR+3(0 1) R+

3([1]) andR+

3(0 2) are both

maximal midpoint-free subsets of R+

Theorem 6 The two sets

(R+

3(0 1) R

+

3([1])) cup minus (R

+

3(0 1) capR

+

3([1]))

R+

3(0 2) cup minusR

+

3(0 2 1)

(53)

are both maximal midpoint-free subsets of R

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

References

[1] N Sloane OEISmdashOnline Encyclopedia of Integer SequencesNote A003002 and A005836 in particular many relevant linksare included

[2] R K GuyUnsolved Problems in NumberTheory Springer NewYork NY USA 3rd edition 2004

[3] J Dybizbanski ldquoSequences containing no 3-term arithmeticprogressionsrdquo Electronic Journal of Combinatorics vol 19 no2 pp 1ndash15 2012

[4] D Wells The Penguin Dictionary of Curious and InterestingNumbers Penguin London UK 1986

[5] S Yates ldquoPrime divisors of repunitsrdquo Journal of RecreationalMathematics vol 8 no 1 pp 33ndash38 1975

[6] J Brillhart D H Lehmer J L Selfridge B Tuckerman and SWagstaff Jr ldquoFactorizations of 119887119899 plusmn 1 119887 = 2 3 5 6 7 10 11 12

up to high powersrdquo Contemporary Mathematics vol 22 1983[7] P Erdos and P Turan ldquoOn some sequences of integersrdquo Journal

of the London Mathematical Society vol 28 pp 104ndash109 1936

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of