A set of reals S⊂R is midpoint-free if it has no subset a,b,c⊆S such that a<b<c and a+c=2b. If S⊂X⊆R and S is midpoint-free, it is a maximal midpoint-free subset of X if there is no midpoint-free set T such that S⊂T⊆X. In each of the cases X=Z+,Z,Q+,Q,R+,R, we determine two maximal midpoint-free subsets of X characterised by digit constraints on the base 3 representations of their members.
1. Introduction
Let us say that a set of three real numbers {a,b,c}⊂R is a midpoint triple if it satisfies a<b<c and a+c=2b. In this case b is the midpoint of the set, a is its lower endpoint, and c is its upper endpoint. This geometric viewpoint immediately suggests the main objective of this paper, which is to identify several significant subsets of R that contain no midpoint triple.
A midpoint-free subset of R is any set that contains no midpoint triple. For instance, the set {2n:n∈Z} of powers of 2 is midpoint-free, since appropriate scaling shows that the sum of two distinct powers of 2 is never equal to a power of 2. Note that if S⊆R and cS+d≔{cs+d:s∈S}, then cS+d is midpoint-free, for any {c,d}⊂R with c≠0, if and only if S is midpoint-free.
If S⊂X⊆R and S is midpoint-free, then S is a maximal midpoint-free subset of X if there is a midpoint triple in any set T such that S⊂T⊆X. (In this case, any x∈X∖S is a member of a midpoint triple in S∪{x}.) If X is a “natural” subset of R, such as the nonnegative integers Z+, the rationals Q, or indeed R itself, it is of considerable interest to identify maximal midpoint-free subsets of X.
Alternatively, any set {a,b,c} such that a<b<c and a+c=2b can be viewed as a 3-term arithmetic progression (A.P.) with first term a, midterm b, and last term c. This arithmetic viewpoint has led to many studies of sequences of positive integers in which there is no 3-term A.P., raising questions such as the following. How large is a maximal subset of the first n positive integers that contains no 3-term A.P.? (See [1] for data.) Given a finite set S of positive integers with no 3-term A.P., what does its greedy algorithm extension look like? How does it compare with a largest possible subset of the first n positive integers that contains S and has no 3-term A.P.? Under what conditions must a subset of the positive integers contain a 3-term A.P.? Among those who have made major contributions to our understanding of these questions, one must list such luminaries as Van der Waerden, Erdős, Turán, Rado, Behrend, Roth, Graham, and Szemerédi. For a compact survey and extensive bibliography of such investigations, see Guy [2]. As an example of recent work in this area, see Dybizbański [3].
In contrast with the usual arithmetic viewpoint, the geometric viewpoint adopted here takes us in an apparently novel direction, where the sets of interest are most naturally viewed as subsets of R.
2. Notational Conventions
In what follows, a key tool for discussing midpoint triples will be base 3 representations of the real numbers, so relevant notational conventions will now be specified. If r∈R,r>0 and
(1)r=∑i∈Zdi3iwithdi∈{0,1,2},
then the two-way infinite string
(2)d≔⋯d2d1d0·d-1d-2⋯
is a base 3 representation of r and di is the digit in place i of d. If di∈{0,1} for infinitely many i<0, then d is a regular base 3 representation of r; otherwise, d is a singular base 3 representation of r, and there is an integer k such that di=2 for all i<k. The regular base 3 representation of r is unique, and if r has a singular base 3 representation, that is also unique. If X is some “natural” subset of R and D⊂{0,1,2}, it will be convenient to use X3(D) to denote the set of all members of X with a base 3 representation restricted to D. This notation adapts to cases where D is a configuration of base 3 digits; thus, when D is a finite string of base 3 digits, X3([D]) will denote the set of all members of X with a base 3 representation which includes a one-way infinite string of recurring blocks D.
Since r>0, at least one of the digits is nonzero. The leading digit of d is the digit dh>0, such that di=0 for all i>h. The trivial zeros of d are all the digits di=0 with i>max{h,0}. Conventionally these are suppressed (i.e., kept implicit) when listing d. If h<0, so the leading digit dh occupies a negative place, the placeholder zeros of d are all the digits di=0 with 0≥i>h. If there is an integer k≤h such that dk>0 and di=0 for all i<k, then d is a regular representation, dk is its trailing digit, its optional zeros are all the digits di=0 with i<min{k,0}, and if k>0 then its placeholder zeros are the digits di=0 with 0≤i<k. If d has a trailing digit, it is conventional to suppress its optional zeros as well as its trivial zeros; the finite digit string remaining is the terminating base 3 representation of r.
These conventions are extended to r=0, as follows. Although the digits in this case are di=0 for all i, in this exceptional case d0 is defined to be both the leading digit and the trailing digit of the representation. Then its trivial zeros are all those in places i>0, and its optional zeros are all those in places i<0; thus, 0 is the terminating representation for r=0.
If d has no trailing digit, it is a nonterminating representation of r and it may be either regular or singular. If d is singular, there is an integer k≤h+1 such that dk∈{0,1} and di=2 for all i<k; the digit dk is the pivot digit of d. In this case there is an alternative, regular base 3 representation of r, namely,
(3)e≔⋯e2e1e0·e-1e-2⋯,
where ei=di for all i>k, ek=dk+1, and ei=0 for all i<k. Evidently e is a terminating representation, ek is its trailing digit, and 3-kr∈Z+, so r∈3kZ+.
The regular base 3 representation of any nonnegative rational q∈Q+ has a recurring block of digits; if q has a singular base 3 representation, this has a recurring 2. For brevity, any such recurring block can be enclosed in square brackets, and the subscript 3 can be used to indicate base 3 notation; for instance,
(4)73=2·13=2·1[0]3=2·0[2]3,12=0·[1]3,25=0·[1012]3.
The set of all nonnegative rationals with a terminating base 3 representation is
(5)Q3+([0])=⋃i∈Z+3-iZ+.
The set of positive rationals with a singular base 3 representation is simply
(6)Q3+([2])=Q3+([0])∖{0}.
The set of positive rationals with base 3 representation containing a recurring 1 is
(7)Q3+([1])=⋃i∈Z+3-i(Z++12).
It will later be found that this set requires particular attention when members are doubled, because
(8)2Q3+([1])⊂Q3+([2])⊂Q3+([0]).
Base 3 representation of negative reals is simply achieved by writing any such number as -r, with r∈R,r>0, so that -r has its base 3 representation adopted from that of r, with trivial zeros suppressed and the unary minus operator applied to precede the leading digit. If r has both a regular and a singular representation, so does -r.
To distinguish between instances of base 3 and base 10 representations, an explicit subscript 3 is normally attached to the former, while a subscript 10 is normally kept implicit for the latter. As needed, r3 will denote the regular base 3 representation of r, with r3* reserved for the singular representation when this exists. Finally, ⟦r⟧3,i will denote di, the digit in place i of the regular base 3 representation of r, while ⟦r⟧3,i* will denote the corresponding digit in the singular representation when this exists.
3. Three Sparse Subsets of Z+
Let Z3+(D) be the set of all nonnegative integers with terminating base 3 representation in which the only explicit digits are in D⊂{0,1,2}. The three cases of interest are those where D is a 2-set. They begin as follows:
(9)Z3+(0,1)={0,1,3,4,9,10,12,13,27,28,30,31,36,37,39,40,81,…},Z3+(0,2)={0,2,6,8,18,20,24,26,54,56,60,62,72,74,78,80,…},Z3+(1,2)={1,2,4,5,7,8,13,14,16,17,22,23,25,26,40,…}.
Some features are obvious. Clearly Z3+(0,2)=2Z3+(0,1). When d∈{0,1,2} and D={0,1,2}∖{d}, all integers congruent to d (mod 3) are absent from Z3+(D). The first positive integer absent from all three sets is 1023=11, but it is clear that an increasing proportion of integers will be missing from all three sets. Indeed, a simple calculation shows for any positive integer n that Z3+(0,1) and Z3+(0,2) each have 2n members below 3n, while Z3+(1,2) has 2n+1-2 members below 3n, so each set is sparse and has asymptotic density 0.
It turns out that these three sets are of considerable interest when it comes to the presence or absence of midpoint triples, so let us now address the main subject of this paper.
Note that the set Z3+(1,2) contains midpoint triples, such as {1,4,7} and {2,5,8}. In fact, Z3+(1,2) is densely packed with midpoint triples, since each member a∈Z3+(1,2) is the lower endpoint of an infinite family of midpoint triples:
(10){a,un-um+a,2un-2um+a}=(un-um){0,1,2}+a⊂Z3+(1,2)
for any integer n>m, where m is the number of digits in the terminating base 3 representation of a (if dh is the leading digit in a3, then m=h+1), and
(11)un≔1+3+32+⋯+3n-1=3n-12
is the positive integer with terminating base 3 representation of n digits, all 1. Note that Z3+(1)={un:n≥1}, and appropriate scaling shows that this set is midpoint-free. (The integers un are rep-units in base 3. The positive integers with explicit decimal digits all equal to 1 are the rep-units for base 10, so called by contraction of “repeated unit” [4]. Prime factorizations of rep-units have been much studied [5, 6]. Clearly, if m∣n, then um∣un. Thus un can be a prime number only if n is prime, but u5=111113=112 shows that this condition is not sufficient. Such observations generalize to rep-units in any base.)
Not only is every member of Z3+(1,2) the lower endpoint of infinitely many midpoint triples, it is also the case that any member greater than 2 is either the midpoint or the upper endpoint of at least one midpoint triple, depending on the leading digit of its base 3 representation.
In contrast, it turns out that Z3+(0,1) and Z3+(0,2) are midpoint-free. To see this, it suffices to show that Z3+(0,1) is midpoint-free: then Z3+(0,2)=2Z3+(0,1) ensures that Z3+(0,2) is also midpoint-free. (Long ago, Erdős and Turán [7] noted that the nonnegative integers with a 2-less base 3 representation are midpoint-free. The following compact proof is included for completeness and to typify what follows.)
Claim A. The set Z3+(0,1) is midpoint-free.
Proof.
On the contrary, suppose that {a,b,c}⊂Z3+(0,1) is a midpoint triple, with a<b<c and a+c=2b. All digits in a3,b3,c3 are in {0,1} and all digits in (2b)3 are in {0,2}. Then
(12)⟦a⟧3,i+⟦c⟧3,i=⟦2b⟧3,i,
for all integers i, and
(13)⟦2b⟧3,i=0⟹⟦a⟧3,i=⟦b⟧3,i=⟦c⟧3,i=0,⟦2b⟧3,i=2⟹⟦a⟧3,i=⟦b⟧3,i=⟦c⟧3,i=1.
Hence, a=b=c, a contradiction, so Z3+(0,1) contains no midpoint triple.
It will now be shown that Z3+(0,1) is not contained in any larger midpoint-free subset of Z+.
Claim B. For any positive integer b∈Z+∖Z3+(0,1), there is a midpoint triple in Z3+(0,1)∪{b} with b as its midpoint.
Proof.
Given b∈Z+∖Z3+(0,1), we seek {a,c}⊂Z3+(0,1) such that a<b<c and a+c=2b. Specify the base 3 digits of a and c as follows:
(14)⟦2b⟧3,i=0⟹⟦a⟧3,i=⟦c⟧3,i=0,⟦2b⟧3,i=2⟹⟦a⟧3,i=⟦c⟧3,i=1,⟦2b⟧3,i=1⟹⟦a⟧3,i=0,⟦c⟧3,i=1.
This determines integers a and c such that {a,c}⊂Z3+(0,1) and a+c=2b since
(15)⟦a⟧3,i+⟦c⟧3,i=⟦2b⟧3,i,
for all i. At least one digit of (2b)3 is 1, since b∉Z3+(0,1), so a3 and c3 differ in at least one place: then ⟦a⟧3,i<⟦c⟧3,i in each such place, ensuring that a<c. Put d≔(c-a)/2>0. Then c=a+2d, so 2b=a+c=2a+2d, whence b=a+d>a and c=a+2d=b+d>b. Thus a<b<c.
It will now be shown that extending Z3+(0,1) by adjoining any rational of the form n+(1/2), with n∈Z+, always yields a set which contains a midpoint triple.
Claim C. For any b∈Z++(1/2), there is a midpoint triple in Z3+(0,1)∪{b} with b as its midpoint.
Proof.
Again we seek {a,c}⊂Z3+(0,1), such that a<b<c and a+c=2b. In the present case, note that 2b is an odd positive integer, so (2b)3 contains the digit 1 an odd number of times. Specifying a and c as in the proof of Claim B, once again it follows that {a,c}⊂Z3+(0,1) and a<b<c.
For instance, the proofs of Claims B and C produce the decompositions
(16)12013=1003+11013,2123=1013+1113
showing that 23 and 23/2 are the midpoints of the pairs {9,37}⊂Z3+(0,1) and {10,13}⊂Z3+(0,1), respectively. Further, doubling shows that 46 and 23 are the midpoints of the pairs {18,74}⊂Z3+(0,2) and {20,26}⊂Z3+(0,2), respectively. From Claims A, B, and C and the doubling method just illustrated, we have the following.
Theorem 1.
The sets Z3+(0,1) and Z3+(0,2) are maximal midpoint-free subsets of Z+.
4. Two Unexpected Results for Z
Let us now study Z3+(0,1) and Z3+(0,2) as midpoint-free subsets of Z, the full set of integers. Surprisingly, the results for the two sets are quite different.
Calculations until now have involved sums of pairs {a,c}⊂Z3+(0,1), so it has been possible to work digit by digit in base 3 without a carry over digit. When the context is widened to Z, it seems that carry over digits can no longer be avoided, so we shall consider blocks of digits rather than single digits. Let the compact notation d×n (“d by n”) denote a homogeneous block of n adjacent digits all equal to d; thus, 1×n denotes the terminating base 3 representation of rep-unit un. For blocks in which the digits are not all equal we simply abut such expressions with d×1=d as a natural simplification. To illustrate, consider the integer
(17)a3≔1222202111001121100=12×4021×30×21×221×20×2.
Let us find integers b,c∈Z3+(0,1) such that a+2b=c. Regarding a3 as a string of homogeneous blocks, we choose matching blocks for (2b)3 so that
(18)(2b)3≔2000020222222202200=20×4202×702×20×2.
Calculating from right to left, base 3 arithmetic for a3+(2b)3 yields successively the following blocks, with each carry over digit indicated parenthetically:
(19)0×2+0×2=0×2,1×2+2×2=(1)10,2+0+(1)=(1)0,1×2+2×2+(1)=(1)1×2,0×2+2×2+(1)=(1)0×2,1×3+2×3+(1)=(1)1×3,2+0+(1)=(1)0,0+2+(1)=(1)0,2×4+0×4+(1)=(1)0×4,1+2+(1)=(1)1.
Assembling these blocks yields
(20)a3+(2b)3=1×2∣0×4∣0∣0∣1×3∣0×2∣1×2∣0∣10∣0×2.
After simplifying we have
(21)10×4101×701×20×2=1000010111111101100=b3,1×20×61×30×21×2010×3=11000000111001101000=c3,
so a+2b=c and b,c∈Z3+(0,1) as desired. This calculation models the proof of the following.
Claim D. For any integer a>0, there is a midpoint triple in Z3+(0,1)∪{-a} with the negative integer -a as its lower endpoint.
Proof.
Given any positive integer a, we seek {b,c}⊂Z3+(0,1), such that -a<b<c and -a+c=2b, so a+2b=c. Partition a3 into its homogeneous blocks Ai, so
(22)a3=Am⋯Ai⋯A1A0
and seek corresponding blocks Bi and Ci to construct b3 and c3.
If a3 has a terminal block A0=0×n, assign b3 the terminal block B0=0×n, so c3 has the terminal block C0=0×n=A0+2B0. If a3 has a nonterminal block Ai=0×n for some i>0, assume the sum of the preceding blocks Ai-1 and 2Bi-1 contributes a carry over digit of 1, and assign b3 the corresponding block Bi=1×n, so
(23)Ci=Ai+2Bi+(1)=0×n+2×n+(1)=(1)0×n.
As 2×n+(1) corresponds to 2un+1=(3n+1-1)+1=3n+1, the computation is justified.
If a3 has a block Ai=1×n, assign b3 the corresponding block Bi=1×n, so c3 has the corresponding block Ci=Ai+2Bi+(δ)=(1)1×(n-1)0+(δ), where the carry over digit δ is either 0 or 1, so Ci=(1)1×(n-1)0 or (1)1×n, respectively. This holds since the sum 1×n+2×n corresponds to un+2un=3un=un+1-u1 and 1×n+2×n+(1) corresponds to 3un+1=un+1.
If a3 has a terminal block A0=2×n, assign b3 the block B0=0×(n-1)1, so c3 has terminal block C0=A0+2B0=(1)0×(n-1)1, since the sum 2×n+0×(n-1)2 corresponds to 2un+2=(3n+1-1)+2=3n+1+1. Similarly if A1A0=2×n0×m, assign b3 the corresponding blocks B1B0=0×(n-1)10×m; then c3 has the blocks C1C0=(1)0×(n-1)10×m. Otherwise, if a3 has a nonterminal block Ai=2×n for some i>0, assume that Ai-1+2Bi-1 yields a carry over digit 1, and assign b3 the blockBi=0×n, so c3 has Ci=Ai+2Bi+(1)=(1)0×n.
For i>0 note that the only case in which Ai-1+2Bi-1 does not contribute a carry over digit 1 to Ai+2Bi is when i=1 and a3 has terminal block A0=0×n. Thus, all our assumptions about carry over digits are justified a posteriori.
Claim E. For any integer n>0, there is a midpoint triple in Z3+(0,2)∪{-n} if and only if n is even, and then -n is the lower endpoint of the triple.
Proof.
By doubling, it follows from Claim D that, for any integer a>0, there is a midpoint triple in Z3+(0,2)∪{-2a}. However, we claim that there is no midpoint triple in Z3+(0,2)∪{-2a+1}. On the contrary, suppose -2a+1 forms a midpoint triple with the pair {b,c}⊂Z3+(0,2). Without loss of generality, -2a+1<b<c and (-2a+1)+c=2b, so 2a+2b-1=c: this is impossible, since 2a+2b-1 is odd and c is even. It follows that Z3+(0,2)∪{-2a+1} is midpoint-free.
Claim F. The set Z3+(0,2)∪-(Z3+(0,2)+1) is midpoint-free.
Proof.
The two sets forming this union are certainly midpoint-free, so any midpoint triple in the union must have two members in one set and one member in the other. Claim E shows that there is no midpoint triple with exactly one member in -(Z3+(0,2)+1), since this set only contains odd negative integers. So suppose there is a midpoint triple {-a,-b,c} with exactly one member c∈Z3+(0,2). Then {a-1,b-1,-c-1} is a midpoint triple in
(24)-(Z3+(0,2)∪-(Z3+(0,2)+1))-1=Z3+(0,2)∪-(Z3+(0,2)+1)
with exactly one member in -(Z3+(0,2)+1), the case ruled out by Claim E.
Since Z3+(0,2) is a maximal midpoint-free subset of Z+ by Theorem 1, clearly Z3+(0,2)+1 is a maximal midpoint-free subset of Z++1=Z+∖{0}. Note that the members of Z3+(0,2)+1 are precisely those positive integers n for which the trailing digit of n3 is 1, and all other digits are in {0,2}. With Claims E and F, it follows that Z3+(0,2)∪-(Z3+(0,2)+1) is a maximal midpoint-free subset of Z=Z+∪-(Z+∖{0}). Combined with Claim D, this proves the following.
Theorem 2.
The sets Z3+(0,1) and Z3+(0,2)∪-(Z3+(0,2)+1) are maximal midpoint-free subsets of Z.
5. Midpoint-Free Subsets of Q+
Now let us consider the corresponding subsets Q3+(D) of the nonnegative rationals Q+, namely, the subsets comprising those members with a regular base 3 representation in which each digit, after suppression of all trivial and optional zeros, is in D⊂{0,1,2}. The three cases of interest are those where D is a 2-set.
As expected, Q3+(1,2) is densely packed with midpoint triples. This follows from the observation that each rational q∈Q3+(1,2) is the lower endpoint of an infinite family of midpoint triples,
(25){q,un-um+q,2un-2um+q}=(un-um){0,1,2}+q⊂Q3+(1,2)
for any integer n>m, where un is the n-digit base 3 rep-unit and m=h+1, where dh is the leading digit of the regular base 3 representation q3.
Next, consider Q3+(0,1). The details are a little more complicated than in the integer context, since allowance must be made for singular representations when particular members of Q3+(0,1) are doubled.
Claim G. The set Q3+(0,1)∖Q3+([1]) is midpoint-free.
Proof.
If b∈Q3+(0,1)∖Q3+([1]) and b>0, all digits of (2b)3 are in {0,2}. They uniquely determine the digits of a3 and c3 such that {a,c}⊂Q3+(0,1)∖Q3+([1]) and a+c=2b. As in the proof of Claim A, they force a=b=c. Hence, there is no midpoint triple {a,b,c}⊂Q3+(0,1)∖Q3+([1]).
Claim H. Any b∈Q3+([1]) is the midpoint of a triple with both its endpoints in the set Q3+(0,1)∩Q3+([0]).
Proof.
We seek {a,c}⊂Q3+(0,1)∖Q3+([1]) with a<b<c and a+c=2b. Then a3 and c3 are uniquely determined by
(26)⟦a⟧3,i+⟦c⟧3,i=⟦2b⟧3,i,⟦a⟧3,i≤⟦c⟧3,i
for all i. Since b∈Q3+([1]), then 3kb∈Z++(1/2) for some k∈Z+, so
(27)2b∈3-k(2Z++1).
It follows that (2b)3 is a terminating representation with an odd number of digits equal to 1. Then
(28)⟦2b⟧3,i=1⟹⟦a⟧3,i=0,⟦c⟧3,i=1foranoddnumberofintegersi≥-k;⟦2b⟧3,i=0⟹⟦a⟧3,i=⟦c⟧3,i=0∀i<-k.
Hence, the required midpoint triple {a,b,c} exists, and in fact {a,c}⊂Q3+([0]).
In particular, if b∈Q3+([1]) is such that ⟦2b⟧3,n=1 for some integer n and ⟦2b⟧3,i=0 for all i≠n, then b=3n/2=un+(1/2) and the proof of Claim H determines a=0 and c=2b=3n. Since ⟦b⟧3,i=0 if i≥n and ⟦b⟧3,i=1 if i<n, we call b the base 3 fractional rep-unit with offset n, denoted by vn. Then vn∈Q3+([1]) is the midpoint of the triple with endpoints {0,3n}⊂Q3+(0,1)∩Q3+([0]).
Positive rationals not in Q3+(0,1)∪Q3+([1]) also belong to midpoint triples with endpoints in Q3+(0,1)∖Q3+([1]), but proving this needs care. For instance,
(29)516=0·[0221]3⟹58=0·[12]3=0·[0111]3+0·[1101]3=1380+3780,
so {13/80, 5/16, 37/80} is a midpoint triple with endpoints in Q3+(0,1)∖Q3+([1]). This calculation models the proof of the following general result.
Claim I. Any b∈Q+∖Q3+(0,1) is the midpoint of a triple with both its endpoints in the set Q3+(0,1)∖Q3+([1]).
Proof.
We seek {a,c}⊂Q3+(0,1)∖Q3+([1]) with a<b<c and a+c=2b. When b∈Q3+([1]), the result follows from Claim H, so we may now assume that
(30)b∈Q+∖(Q3+(0,1)∪Q3+([1])).
Then (2b)3 has at least one digit dj equal to 1 and its recurring block, say [D], has at least one digit different from 2. We require a3 and c3 to satisfy
(31)⟦a⟧3,i+⟦c⟧3,i=⟦2b⟧3,i
for all i. Requiring ⟦a⟧3,i≤⟦c⟧3,i for all i≥j ensures that a<c, since
(32)⟦a⟧3,j=0,⟦c⟧3,j=1.
If the recurring block [D] of (2b)3 contains 0, then the recurring blocks of a3 and c3 also contain 0, so neither a nor c is in Q3+([1]). Now suppose [D] has at least one digit equal to 1 and none equal to 0. Clearly, in this case we may choose the integer j so that the digit dj=1 occurs in [D]. If D has m digits, then for all i≤j choose
(33)⟦a⟧3,i=0,⟦c⟧3,i=1wheni≡jmod2m,⟦a⟧3,i=1,⟦c⟧3,i=0wheni≡j+mmod2m,⟦a⟧3,i≤⟦c⟧3,iotherwise.
Then a3 and c3 are uniquely determined, so that a<c and each has a recurring block containing 0, so neither a nor c is in Q3+([1]).
If q∈Q+∖Q3+(0,2) then q3 has at least one digit equal to 1, so it follows that (q/2)3∈Q+∖Q3+(0,1). Claim I shows that there is a midpoint triple {a,b,c} with b=q/2 and {a,c}⊂Q3+(0,1)∖Q3+([1]). Then 2{a,b,c} is a midpoint triple with midpoint q and endpoints 2{a,c}⊂Q3+(0,2). Hence, doubling applied to Claims G and I shows that Q3+(0,2) is midpoint-free, and every b∈Q+∖Q3+(0,2) is the midpoint of a triple with endpoints in Q3+(0,2), so Claims G, H, and I establish the following.
Theorem 3.
The sets Q3+(0,1)∖Q3+([1]) and Q3+(0,2) are maximal midpoint-free subsets of Q+.
6. Midpoint-Free Subsets of Q
Now consider Q3+(0,1) and Q3+(0,2) as subsets of Q. The role of Q3+([1]) is yet more prominent in this context.
Claim J. The set Q3+(0,1)∩Q3+([1]) is midpoint-free.
Proof.
Suppose {a,b,c}⊂Q3+(0,1)∩Q3+([1]) satisfies a<b<c and a+c=2b. There is an integer m≥0 for which 3m{a,b,c}⊂Z3+(0,1)+(1/2) so
(34)2·3m{a,b,c}⊂2Z3+(0,1)+1.
This contradicts the fact that 2Z3+(0,1)+1 is midpoint-free, by Claim A.
Claim K. Any positive rational b∈Q3+([1])∖Q3+(0,1) is the midpoint of a triple with endpoints in Q3+(0,1)∩Q3+([1]).
Proof.
Given b, we seek {a,c}⊂Q3+(0,1)∩Q3+([1]), such that a<b<c and a+c=2b. There is an integer m such that 3mb∈Z++(1/2), so there is an integer B∈Z+∖Z3+(0,1) such that 3mb=B+(1/2). At least one digit of B3 is 2, so at least one digit of (2B)3 is 1. Define {A,C}⊂Z3+(0,1) by
(35)⟦2B⟧3,i=0⟹⟦A⟧3,i=⟦C⟧3,i=0,⟦2B⟧3,i=1⟹⟦A⟧3,i=0,⟦C⟧3,i=1,⟦2B⟧3,i=2⟹⟦A⟧3,i=⟦C⟧3,i=1.
Then ⟦A⟧3,i≤⟦C⟧3,i for all i, with strict inequality for at least one i, so A<C. Also ⟦A⟧3,i+⟦C⟧3,i=⟦2B⟧3,i for all i, so A+C=2B. Finally, let
(36)3m{a,b,c}={A,B,C}+12.
Then a+c=2b and {a,c}⊂Q3+(0,1)∩Q3+([1]).
Claim L. The set (Q3+(0,1)∖Q3+([1]))∪-(Q3+(0,1)∩Q3+([1])) is midpoint-free.
Proof.
By Claims G and J, if the specified set contains a midpoint triple, two members of the triple must have one sign and the third member must have the opposite sign. Suppose {a}⊂Q3+(0,1)∩Q3+([1]) and {b,c}⊂Q3+(0,1)∖Q3+([1]) are such that -a<b<c and a+2b=c. Since a3, (2b)3, and c3 are eventually periodic, there is an integer m such that their digits in places i≤m are purely periodic, and
(37)⟦a⟧3,i=1,⟦2b⟧3,i∈{0,2},⟦c⟧3,i∈{0,1}∀i≤m.
Since c∉Q3+([1]), there is a j≤m such that ⟦c⟧3,j=0 and therefore ⟦2b⟧3,i cannot be equal to 0 for all i≤m. On the other hand, ⟦2b⟧3,i cannot be equal to 2 for all i≤m since b∉Q3+([1]). Hence the recurring block [D] of (2b)3 contains 0 and 2, so ⟦2b⟧3,k=0,⟦2b⟧3,k-1=2 for some k≤m. But ⟦a⟧3,k=⟦a⟧3,k-1=1, so ⟦a⟧3,k-1+⟦2b⟧3,k-1 has a “carry over” of 1, regardless of whether or not the digits in place k-2 contribute any “carry over”. Then
(38)⟦c⟧3,k=⟦a⟧3,k+⟦2b⟧3,k+(1)=2∉{0,1}.
By this contradiction, there is no midpoint triple of the proposed type. A similar but simpler argument shows that there are no subsets {a,b}⊂Q3+(0,1)∩Q3+([1]) and {c}⊂Q3+(0,1)∖Q3+([1]) such that -a<-b<c and a=2b+c.
Claim M. For any positive rational a∈Q+∖Q3+([1]) there is a midpoint triple in (Q3+(0,1)∖Q3+([1]))∪{-a}, with -a as its lower endpoint.
Proof.
Given a, we seek {b,c}⊂Q3+(0,1)∖Q3+([1]) such that -a<b<c and a+2b=c. Note that if the triple {a,b,c} has all the required properties, the scaled triple 3m{a,b,c} also has all the required properties, for any integer m. Let us assume, without loss of generality, that the digits of a3 in negative places are purely periodic, and partitioning into homogeneous blocks Ai yields
(39)a3=Ah⋯A1A0·[A-1A-2⋯A-k]
for suitable integers h≥0,k>0. As a∉Q3+([1]), if k=1, then A-1=0 so a∈Z+ and a suitable set {b,c} exists, by Claim D.
Now suppose k≥2 and a3 has at least one negative place digit which is nonzero. If a3 has no negative place digit equal to 2, let a=A+q, where A∈Z+ and q∈Q3+(0,1)∖Q3+([1]),0<q<1 are the integer part and fractional part of a, respectively. The proof of Claim D yields integers B,C∈Z3+(0,1), such that {-A,B,C} is a midpoint triple. Choose b=B,c=C+q. Then
(40){-a,b,c}={-A-q,B,C+q}
is a midpoint triple with {b,c}⊂Q3+(0,1)∖Q3+([1]).
Finally, suppose A-j=2×n for at least one j>0. Define homogeneous blocks Bi and Ci for all i as follows:
(41)Ai=0×n⟹Bi=1×n,Ci=0×nAi=1×n⟹Bi=1×n,Ci=1×nAi=2×n⟹Bi=0×n,Ci=0×n.
Assuming a carry over digit equal to 1, note that checks as in the proof of Claim D justify the calculations
(42)0×n+2×n+(1)=(1)0×n,1×n+2×n+(1)=(1)1×n,
so in all cases we have Ai+2Bi+(1)=(1)Ci. Choose
(43)b3=Bh⋯B1B0·[B-1B-2⋯B-k],c3=(1)Ch⋯C1C0·[C-1C-2⋯C-k].
Then a+2b=c and {b,c}⊂Q3+(0,1). Also B-j=C-j=0×n ensures that {b,c}⊂Q3+(0,1)∖Q3+([1]), as required.
For instance, beginning with
(44)a=2·[201]3=7126∈Q+∖Q3+([1]),
the proof of Claim M constructs a midpoint triple {-a,b,c} with
(45)b=0·[011]3=213∈Q3+(0,1)∖Q3+([1]),c=10·[001]3=7926∈Q3+(0,1)∖Q3+([1]).
However, if we begin with
(46)a=2·[101]3=3113∈Q+∖Q3+([1]),
the construction falls back to Claim D, resulting in b=1 and c=57/13.
It now follows from Claims I, K, L, and M that the set
(47)(Q3+(0,1)∖Q3+([1]))∪-(Q3+(0,1)∩Q3+([1]))
is a maximal midpoint-free subset of Q. This settles midpoint questions about Q3+(0,1) as a subset of Q.
What is the situation for Q3+(0,2)? If q∈Q3+(0,2) then the recurring block of q3 must have at least one digit equal to 0. Hence, there must be at least one digit equal to 0 in the recurring block of (q/2)3, so q/2∈Q3+(0,1)∖Q3+([1]): it follows that 2(Q3+(0,1)∖Q3+([1]))=Q3+(0,2). Again, if there is some m∈Z+, such that 3mq∈Z3+(0,2)+1, it follows that q3 is a terminating representation with trailing digit 1 and all other digits in {0,2}. Hence,
(48)q2∈Q3+(0,1)∩Q3+([1]).
The converse also holds, so
(49)2(Q3+(0,1)∩Q3+([1]))=Q3+(0,2;1),
where Q3+(0,2;1) is the set of positive rationals which have a terminating base 3 representation with trailing digit 1 and all other digits in {0,2}, so
(50)Q3+(0,2;1)=⋃i∈Z+3-i(Z3+(0,2)+1).
In passing, note that Z3+(0,2)+1=Q3+(0,2;1)∩Z+. By doubling, it now follows from Claim L that the set
(51)Q3+(0,2)∪-Q3+(0,2;1)
is midpoint-free.
Claim N. Any rational b∈Q+∖Q3+(0,2) is the midpoint of a triple with both its endpoints in Q3+(0,2).
Proof.
If b∈Q+∖Q3+(0,2), then b/2∉Q3+(0,1), so b/2 is the midpoint of a triple with endpoints in Q3+(0,1)∖Q3+([1]), by Claim I. By doubling, b is the midpoint of a triple with both its endpoints in the set Q3+(0,2).
Claim O. For any rational a∈Q+∖Q3+(0,2;1), there is a midpoint triple in Q3+(0,2)∪{-a} with -a as its lower endpoint.
Proof.
If a∈Q+∖Q3+(0,2;1) and a>0, then a/2∉Q3+([1]), so there is a midpoint triple in Q3+(0,1)∖Q3+([1])∪{-a/2} with -a/2 as its lower endpoint, by Claim M. By doubling, Q3+(0,2)∪{-a} has a midpoint triple with lower endpoint -a.
The preceding results now fully establish the following.
Theorem 4.
The two sets
(52)(Q3+(0,1)∖Q3+([1]))∪-(Q3+(0,1)∩Q3+([1])),Q3+(0,2)∪-Q3+(0,2;1)
are both maximal midpoint-free subsets of Q.
7. Midpoint-Free Subsets of R+ and R
Now consider R3+(0,1) and R3+(0,2). These sets are uncountable, while their maximal rational subsets Q3+(0,1) and Q3+(0,2) are countable, yet no essentially new considerations arise in extending the results of the previous two sections from Q+ to R+ and from Q to R. Noting that R3+([1])=Q3+([1]) and R3+(0,2;1)=Q3+(0,2;1), it suffices to state the end results for this wider context.
Theorem 5.
The sets R3+(0,1)∖R3+([1]) and R3+(0,2) are both maximal midpoint-free subsets of R+.
Theorem 6.
The two sets
(53)(R3+(0,1)∖R3+([1]))∪-(R3+(0,1)∩R3+([1])),R3+(0,2)∪-R3+(0,2;1)
are both maximal midpoint-free subsets of R.
Conflict of Interests
The author declares that there is no conflict of interests regarding the publication of this paper.
SloaneN.OEIS—Online Encyclopedia of Integer Sequences, Note A003002 and A005836 in particular; many relevant links are includedGuyR. K.20043rdNew York, NY, USASpringerDybizbańskiJ.Sequences containing no 3-term arithmetic progressions2012192115MR2928630WellsD.1986London, UKPenguinYatesS.Prime divisors of repunits1975813338MR0429722BrillhartJ.LehmerD. H.SelfridgeJ. L.TuckermanB.WagstaffS.Jr.Factorizations of bn±1,b=2,3,5,6,7,10,11,12 up to high powers198322ErdősP.TuránP.On some sequences of integers193628104109