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
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Differential EquationsInternational Journal of
Volume 2014
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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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
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Differential EquationsInternational Journal of
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
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
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
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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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
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Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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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
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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
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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
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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
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
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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
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