Perpendicularity in an Abelian Group

Hindawi Publishing Corporation International Journal of Mathematics and Mathematical Sciences Volume 2013, Article ID 983607, 8pages http://dx.doi.org/10.1155/2013/983607 Research Article Perpendicularity in an Abelian Group Pentti Haukkanen,1Mika Mattila,1Jorma K. Merikoski,1and Timo Tossavainen2 1School of Information Sciences, University of Tampere, 33014, Finland 2School of Applied Educational Science and Teacher Education, University of Eastern Finland, P.O. Box 86, 57101 Savonlinna, Finland Correspondence should be addressed to Pentti Haukkanen; pent[email protected] Received 12 January 2013; Accepted 19 March 2013 Academic Editor: Petru Jebelean Copyright © 2013 Pentti Haukkanen et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We give a set of axioms to establish a perpendicularity relation in an Abelian group and then study the existence of perpendicularities in (Z𝑛,+)and (Q+,⋅)and in certain other groups. Our approach provides a justification for the use of the symbol ⊥denoting relative primeness in number theory and extends the domain of this convention to some degree. Related to that, we also consider parallelism from an axiomatic perspective. 1. Introduction In [1,page115],Grahametal.madethefollowingsuggestion: When gcd(𝑚,𝑛)=1, the integers 𝑚and 𝑛have no prime factors in common and we say that they are relatively prime. This concept is so important in practice, we ought to have a special notation for it; but alas, number theorists have not agreed on a very good one yet. Therefore we cry: hear us, o mathematicians of the world! let us not wait any longer! we can make many formulas clearer by adopting a new notation now! let us agree to write “𝑚⊥𝑛”, a n d t o s ay “ 𝑚is prime to 𝑛,” i f 𝑚and 𝑛are relatively prime. Like perpendicular lines do not have a common direction, perpendicular numbers do not have common factors. Infact,thiscryhadbeenansweredevenbeforeitwas made. Namely, in studying 𝑙-groups (i.e., groups with a lattice structure), Birkhoff [2,page295]definesthattwopositive elements 𝑎and 𝑏of an 𝑙-group are disjoint if 𝑎∧𝑏=0 and uses the notation 𝑎⊥𝑏for disjoint elements. He also remarks that disjointness specializes to relative primeness in the 𝑙-group of positive integers. Amotivationforthepresentpaperistostudyhow justified ultimately it is to use the symbol of perpendicularity todenoterelativeprimeness.Doesthispracticerelyonly on the analogy between having no common direction and having no common factor or is there a deeper linkage to entitlethisconvention?Thisquestionleadsustoaskwhich properties essentially establish the notion of perpendicularity in the algebraic context and what the most suitable algebraic context for the axiomatization of perpendicularity actually is; we have recently studied the axioms of perpendicularity from an elementary geometric point of view [3]. In an inner product space, perpendicularity obviously traces back to the inner product being zero. However, certain features of this perpendicularity can be shifted down to simpler algebraic structures. We will define perpendicularity in an Abelian group and examine it in Section 2.InSection 3, we will focus on perpendicularity in (Z𝑛,+).Davis[4] defined perpendicularity in an Abelian group differently. In Section 4, we will introduce his approach and compare it with ours. Thereafter, we will consider divisibility in (Q+,⋅)in Section 5 and parallelism in an Abelian group in Section 6.Wewill conclude our paper with a brief discussion and a supplement tothesuggestioncitedpreviously. 2. Axioms and Properties of Perpendicularity Throughout this paper, 𝐺 = (𝐺,+)is an Abelian group so that 𝐺 ={0}. Unless otherwise stated, ⊥is a binary relation in 𝐺satisfying (A1)∀𝑎∈𝐺:∃𝑏∈𝐺:𝑎⊥𝑏, (A2)∀𝑎∈𝐺\{0}:𝑎 ⊥𝑎, (A3)∀𝑎,𝑏∈𝐺:𝑎⊥𝑏⇒𝑏⊥𝑎, 2 International Journal of Mathematics and Mathematical Sciences (A4)∀𝑎,𝑏,𝑐∈𝐺:𝑎⊥𝑏∧𝑎⊥𝑐⇒𝑎⊥(𝑏+𝑐), (A5)∀𝑎,𝑏∈𝐺:𝑎⊥𝑏⇒𝑎⊥−𝑏. We call ⊥aperpendicularity in 𝐺.Thisconceptcanbe defined also in weaker structures by changing these axioms appropriately. For example, if 𝐺is an Abelian monoid, then we simply omit (A5). Since the trivial perpendicularity 𝑥⊥𝑦⇐⇒𝑥=0∨𝑦=0 (1) always exists, we are mainly interested in nontrivial perpendicularities. We call ⊥maximal if it is not a subrelation of any other perpendicularity in 𝐺. There always exists a maximal perpendicularity. This is obvious if 𝐺is finite and, otherwise, it follows from Zorn’s lemma. Proposition 1 records some elementary properties of perpendicularity; we leave the proof for the reader. Proposition 1. Perpendicularity ⊥has the following properties: (a) ∀𝑎∈𝐺: 𝑎⊥0, (b) ∀𝑎∈𝐺\{0}: 𝑎 ⊥−𝑎, (c) ∀𝑎,𝑏1,...,𝑏𝑘∈𝐺,𝛾 1,...,𝛾𝑘∈Z:𝑎⊥𝑏 1,...,𝑏𝑘⇒ 𝑎⊥(𝛾1𝑏1+⋅⋅⋅+𝛾𝑘𝑏𝑘), (d) ∀𝑎,𝑏∈𝐺,𝜇,]∈Z:𝑎⊥𝑏⇒𝜇𝑎⊥]𝑏. The following characterization is useful in proving that a given relation is perpendicularity. Proposition 2. Abinaryrelation⊥in 𝐺is perpendicularity if and only if it satisfies (A1) and (A2) and (A6)∀𝑎,𝑏,𝑐∈𝐺:𝑎⊥𝑏∧𝑎⊥𝑐⇒(𝑏−𝑐)⊥𝑎. Proof. The“onlyif”-partistrivial.Toprovethe“if”-part,we first show that our assumptions imply Proposition 1(a). Let 𝑎∈𝐺.By(A1),thereis𝑏∈𝐺such that 𝑎⊥𝑏.Putting𝑐:=𝑏 in (A6) implies 0⊥𝑎;inparticular0⊥0. Further, (A6) with 𝑎:=0,𝑏:=𝑎and 𝑐:=0gives (𝑎−0)⊥0,thatis,𝑎⊥0.Now we can verify the remaining axioms. (A3) Assume 𝑎⊥𝑏. Apply (A6) with 𝑐:=0;then𝑏⊥𝑎. (A5) Assume 𝑎⊥𝑏. Apply (A6) with 𝑏:=0and 𝑐:=𝑏. Then (−𝑏)⊥𝑎,andso,by(A3),𝑎⊥−𝑏. (A4) Assume 𝑎⊥𝑏and 𝑎⊥𝑐;then𝑎⊥−𝑐by (A5). Now (A6) with 𝑐:=−𝑐implies (𝑏−(−𝑐)) ⊥ 𝑎,thatis, (𝑏+𝑐)⊥𝑎.Hence,by(A3),𝑎⊥(𝑏+𝑐). Is there a simple condition under which (A5) follows from (A1)–(A4)? The answer is positive. Proposition 3. If all elements of 𝐺have finite order and if ⊥ satisfies (A1)–(A4), then it satisfies (A5). If 𝐺has at least one element of infinite order, then there exists a relation ⊥which satisfies (A1)–(A4) but not (A5). Proof. For the first part, assume that 𝑎,𝑏∈𝐺satisfy 𝑎⊥𝑏, and let the order of 𝑏be 𝑛.Then𝑎⊥(𝑛−1)𝑏by (A4). But (𝑛−1)𝑏=−𝑏and (A5) follows. For the second part, let 𝑎∈ 𝐺have infinite order. Then the subgroup {0,±𝑎,±2𝑎,...}is isomorphic to Z.Therelation⊥defined by 𝑥⊥𝑦⇐⇒(∃𝜇,]∈Z:𝑥=𝜇𝑎∧𝑦=]𝑎∧𝜇]<0) ∨𝑥=0∨𝑦=0 (2) satisfies (A1)–(A4) but not (A5). If 0 =𝐴⊆𝐺,wedefinetheperpendicular complement or ⊥-complement of 𝐴as follows: 𝐴⊥={𝑦∈𝐺|𝑦⊥𝐴}= ⋃ 𝐺⊇𝐵⊥𝐴𝐵. (3) Here 𝑦⊥𝐴means that 𝑦⊥𝑥for all 𝑥∈𝐴,and𝐵⊥𝐴 means that 𝑦⊥𝐴for all 𝑦∈𝐵.Thus𝐴⊥is the maximal set perpendicular to 𝐴.Inparticular,𝐺⊥={0}and {0}⊥=𝐺.We also define 0⊥=𝐺. Proposition 4. If 𝐴⊆𝐺,then𝐴⊥is a subgroup of 𝐺.If𝐺is cyclic, then 𝐴⊥is cyclic. Proof. The first part follows by applying the subgroup test and Proposition 2.Thesecondpartfollowsfromthefactthatany subgroup of a cyclic group is cyclic. The next theorem tells when 𝐺has a nontrivial perpendicularity. Theorem 5. The following conditions are equivalent: (a) 𝐺has a nontrivial perpendicularity ⊥, (b) 𝐺has nontrivial cyclic subgroups 𝐻and 𝐾satisfying 𝐻∩𝐾={0}, (c) 𝐺has nontrivial subgroups 𝐻and 𝐾satisfying 𝐻∩ 𝐾={0}. Proof. (a)⇒(b). Since ⊥is nontrivial, there exist 𝑥,𝑦∈𝐺\{0} such that 𝑥⊥𝑦.Then𝐻=⟨𝑥⟩and 𝐾=⟨𝑦⟩apply. Here ⟨𝑎⟩ stands for the cyclic group generated by 𝑎. (b)⇒(c). Trivial. (c)⇒(a). Define ⊥by 𝑥⊥𝑦⇐⇒(𝑥∈𝐻∧𝑦∈𝐾)∨(𝑥∈𝐾∧𝑦∈𝐻) ∨𝑥=0∨𝑦=0. (4) Next we consider the maximal perpendicularity in some examples of groups. In Examples 6–9,thegroupoperationis addition. Example 6. Let 𝐺=Z6.ByLagrange’stheorem[5,page 130, Theorem 2], the smallest 𝑛such that Z𝑛has a nontrivial perpendicularity is 6=2⋅3because 𝑛must have at least two different prime factors. The nontrivial subgroups of Z6are 𝐻=⟨3⟩={0,3}and 𝐾=⟨2⟩={0,2,4}.Since𝐺=𝐻⊕𝐾, International Journal of Mathematics and Mathematical Sciences 3 it has exactly one nontrivial perpendicularity, defined by 0⊥ 0,1,2,3,4,5and 3⊥2,4andviceversa.Consequently,this perpendicularity is maximal. Example 7. Let 𝐺=Z2×Z2, the Klein four group. Denote 0 = (0,0),𝑎 = (0,1),𝑏 = (1,0),𝑐 = (1,1).The nontrivial subgroups are 𝐴={0,𝑎},𝐵={0,𝑏},𝐶={0,𝑐}. So, there are three nontrivial perpendicularities obtained as follows: Choose two elements of 𝑎,𝑏,and𝑐.Define that they are perpendicular to each other and to 0. Define that the remaining element is perpendicular to 0 only. All perpendicularities arising in this way are clearly maximal. We also note that 𝐺=𝐴⊕𝐵=𝐵⊕𝐶=𝐶⊕𝐴. Example 8. Let 𝐺=Z.Foreach𝑛≥2,thesubgroup⟨𝑛⟩= 𝑛Zis nontrivial and there are no other nontrivial subgroups than those found in this way. Because 𝑚𝑛∈⟨𝑚⟩∩⟨𝑛⟩,thereis no pair of nontrivial subgroups with intersection {0}.Hence 𝐺has only the trivial perpendicularity. Example 9. Let 𝐺=R.SinceRhas infinitely many pairs of nontrivial subgroups with intersection {0},ithasinfinitely many nontrivial perpendicularities. For example, let 𝐻=Q and 𝐾={𝑥√2|𝑥∈Q}and define ⊥by (4). To see that this perpendicularity is not maximal, let 𝐻1={𝑥√3|𝑥∈Q}and 𝐾1={𝑥√5|𝑥∈Q}and define ⊥󸀠by 𝑥⊥󸀠𝑦⇐⇒(𝑥∈𝐻∧𝑦∈𝐾)∨(𝑥∈𝐾∧𝑦∈𝐻) ∨(𝑥∈𝐻1∧𝑦∈𝐾1)∨(𝑥∈𝐾1∧𝑦∈𝐻1) ∨𝑥=0∨𝑦=0. (5) Then 𝑥⊥𝑦⇒𝑥⊥󸀠𝑦. Example 10. Let 𝐺=(Q+,⋅),whereQ+denotes the set of positive rational numbers. Every 𝑐∈Q+can be uniquely expressed as 𝑐=∏ 𝑝∈P𝑝]𝑝(𝑐),(6) where ]𝑝(𝑐)∈Zfor each 𝑝∈Pandonlyafinitenumberof themarenonzero.ThesymbolPstandsforthesetofprimes. For example, if 𝑐=8/25,then]2(𝑐)=3,]3(𝑐)=0,]5(𝑐)=−2, ]7(𝑐)=]11(𝑐)=⋅⋅⋅=0. Assign now 𝑎⊥𝑏⇐⇒∀𝑝∈P:]𝑝(𝑎)=0∨]𝑝(𝑏)=0. (7) In other words, if 𝑎=𝑚 𝑢,𝑏= 𝑛 V,𝑚,𝑢,𝑛,V∈Z+, gcd (𝑚,𝑢)=gcd (𝑛,V)=1, (8) then 𝑎⊥𝑏⇐⇒gcd (𝑚𝑢,𝑛V)=1. (9) Hence, for example, 8/9⊥7/5.Inparticular,for𝑚,𝑛∈Z+, applying (9)to𝑚/1and 𝑛/1yields that 𝑚⊥𝑛⇐⇒gcd (𝑚,𝑛)=1. (10) So, it seems that Graham et al. were prophetically quite right with their suggestion—and not forgetting Birkhoff either! We will discuss the perpendicularity of positive rational numbers in more detail in Section 5. 3. Perpendicularity in Z𝑛 Studying perpendicularities requires that we know the structure of 𝐺. Next we take a more thorough look at perpendicularity in Z𝑛. To that end, we begin by introducing a suitable notation to discuss the structure of Z𝑛and record two lemmas which are useful in the search for the maximal perpendicularity. We will also use the notations introduced in Theorem 11 and the following lemmas throughout the next sections. Theorem 11. If 𝑛=𝑝𝛼1 1⋅⋅⋅𝑝𝛼𝑟 𝑟,(11) where 𝑝1,...,𝑝𝑟∈Pare distinct and 𝛼1,...,𝛼𝑟>0,then Z𝑛=𝐻1⊕⋅⋅⋅⊕𝐻𝑟,(12) where 𝐻𝑖=⟨𝑒𝑖⟩, 𝑒𝑖=𝑛 𝑝𝛼𝑖 𝑖, 𝑖=1,...,𝑟. (13) The decomposition (12)isunique(uptotheorderofsubgroups). Proof. The claim (12)followsfrom[5,page399,Corollary1] and from the facts that Z𝑝𝛼𝑖 𝑖≅𝐻 𝑖and 𝐻𝑖∩𝐻𝑗={0}for all 𝑖,𝑗=1,...,𝑟,𝑖 =𝑗. Uniqueness follows from [5,page399, Corollary 2]. Although we consider Z𝑛mainly as an Abelian group, it is now useful to work with Z𝑛as a ring. Lemma 12. For all 𝑖,𝑗=1,...,𝑟,𝑖 =𝑗, 𝑒2 𝑖=0, 𝑒𝑖𝑒𝑗=0. (14) Proof. It is enough to consider 𝑖=1,𝑗=2. Regarding 𝑒1and 𝑒2as integers, we have 𝑒2 1=𝑛2 𝑝2𝛼1 1=𝑝2𝛼2 2⋅⋅⋅𝑝2𝛼𝑟 𝑟≡0 (mod 𝑛), 𝑒1𝑒2=𝑛 𝑝𝛼1 1𝑛 𝑝𝛼2 2 =𝑝𝛼2 2⋅⋅⋅𝑝𝛼𝑟 𝑟𝑝𝛼1 1𝑝𝛼3 3⋅⋅⋅𝑝𝛼𝑟 𝑟 =𝑝𝛼3 3⋅⋅⋅𝑝𝛼𝑟 𝑟𝑛≡0 (mod 𝑛), (15) and (14) follows. 4 International Journal of Mathematics and Mathematical Sciences Lemma 13. Let ⊥be a perpendicularity in Z𝑛.Then ∀𝑎,𝑏,𝑐,𝑑∈Z𝑛:𝑎⊥𝑏󳨐⇒𝑐𝑎⊥𝑑𝑏. (16) Proof. Let 𝑐=𝛾1,𝑑=𝛿1,where𝛾and 𝛿are integers with 0≤𝛾,𝛿<𝑛.Since𝑐𝑎=(𝛾1)𝑎=𝛾(1𝑎)=𝛾𝑎and, similarly, 𝑑𝑏=𝛿𝑏,Proposition 1(d) implies (16). Now we are ready to introduce a perpendicularity which turns out to be maximal in Z𝑛.Let 𝑥=𝑥1+⋅⋅⋅+𝑥𝑟,𝑦=𝑦 1+⋅⋅⋅+𝑦𝑟∈Z𝑛,(17) where 𝑥𝑖,𝑦𝑖∈𝐻𝑖,𝑖=1,...,𝑟.Therelation⊥0, defined in Z𝑛 by 𝑥⊥0𝑦⇐⇒∀𝑖∈{1,...,𝑟}:𝑥𝑖=0∨𝑦𝑖=0, (18) is clearly a perpendicularity. Theorem 14. The perpendicularity ⊥0is maximal and every other perpendicularity in Z𝑛is contained in it. Proof. Let ⊥be another perpendicularity in Z𝑛.Ourclaimis that 𝑥⊥𝑦⇒𝑥⊥0𝑦.By(12), we can express 𝑥=𝜉1𝑒1+⋅⋅⋅+𝜉𝑟𝑒𝑟,𝑦=𝜂 1𝑒1+⋅⋅⋅+𝜂𝑟𝑒𝑟,(19) where the integers 𝜉𝑖,𝜂𝑖∈{0,...,𝑝𝛼𝑖 𝑖−1}andtheresidueclass 𝑒𝑖=𝑛/𝑝𝛼𝑖 𝑖,𝑖=1,...,𝑟. Supposeagainsttheclaimoftheoremthatthereexist 𝑥,𝑦∈Z𝑛such that 𝑥⊥𝑦but 𝑥 ⊥0𝑦.Then𝜉𝑖,𝜂𝑖=0for some 𝑖. Reordering the indices so that 𝑖=1and applying (16), we have 𝑥𝑒1⊥𝑦𝑒1which implies that 𝜉1𝑒2 1⊥𝜂1𝑒2 1(20) by (14). Hence, by Proposition 1(d), 𝜂1 gcd (𝜉1,𝜂1)𝜉1𝑒2 1⊥𝜉1 gcd (𝜉1,𝜂1)𝜂1𝑒2 1,(21) that is, lcm (𝜉1,𝜂1)𝑒2 1⊥lcm (𝜉1,𝜂1)𝑒2 1.(22) Consequently, lcm (𝜉1,𝜂1)𝑒2 1=0by (A2). In other words, regarding also 𝑒1as an integer, lcm (𝜉1,𝜂1)𝑒2 1=lcm (𝜉1,𝜂1)𝑛2 𝑝2𝛼1 1 =lcm (𝜉1,𝜂1)𝑝2𝛼2 2⋅⋅⋅𝑝2𝛼𝑟 𝑟 ≡0 (mod 𝑛), (23) and 𝑝𝛼1 1divides lcm(𝜉1,𝜂1). However, since it divides neither 𝜉1nor 𝜂1, this is a contradiction. Hence, 𝑥⊥0𝑦. Considering the direct sum (12) external, we can identify 𝑥and 𝑦in (17) with vectors (𝑥1,...,𝑥𝑟)and (𝑦1,...,𝑦𝑟), respectively. So, it is natural to define their “inner product” by ⟨𝑥,𝑦⟩=𝑥1𝑦1+⋅⋅⋅+𝑥𝑟𝑦𝑟.(24) Proposition 15 shows that this operation coincides with the ordinary multiplication in Z𝑛. Proposition 15. Given 𝑥,𝑦∈Z𝑛, ⟨𝑥,𝑦⟩=𝑥𝑦. (25) Proof. We have 𝑥𝑦=( 𝑟 ∑ 𝑖=1𝑥𝑖)(𝑟 ∑ 𝑖=1𝑦𝑖)=𝑟 ∑ 𝑖=1𝑥𝑖𝑦𝑖+𝑟 ∑ 𝑖,𝑗=1 𝑖 =𝑗𝑥𝑖𝑦𝑗.(26) But, recalling (19)and(14), 𝑟 ∑ 𝑖,𝑗=1 𝑖 =𝑗𝑥𝑖𝑦𝑗=𝑟 ∑ 𝑖,𝑗=1 𝑖 =𝑗𝜉𝑖𝜂𝑗𝑒𝑖𝑒𝑗=0. (27) The claim follows. Theorem 16. Let ⊥be a perpendicularity in Z𝑛.Then ∀𝑥,𝑦∈Z𝑛:𝑥⊥𝑦󳨐⇒𝑥𝑦=0. (28) Proof. If 𝑥⊥𝑦,then𝑥⊥0𝑦by Theorem 14.So,𝑥𝑦=0by (18) and (25). Does the converse of Theorem 16 hold if ⊥=⊥0?And, related to Proposition 15,is⟨𝑥,𝑦⟩ = 𝑥𝑦 aproperinner product? Namely, an inner product in a real vector space is symmetric and bilinear and it satisfies ⟨𝑥,𝑥⟩=0⇒𝑥=0. The operation ⟨𝑥,𝑦⟩ = 𝑥𝑦in Z𝑛has clearly the first and second properties but what about the third one? The answers to both questions are contained in Theorem 17. Theorem 17. The following conditions are equivalent: (a) 𝛼1=⋅⋅⋅=𝛼𝑟=1, (b) ∀𝑥,𝑦∈Z𝑛:𝑥𝑦=0⇒𝑥⊥0𝑦, (c) ∀𝑥∈Z𝑛:𝑥2=0⇒𝑥=0. Proof. (a)⇒(b). Assume that 𝑥 ⊥0𝑦.Express𝑥and 𝑦as in (19). We can rearrange the indices so that, for some 𝑠∈ {1,...,𝑟},𝜉𝑖,𝜂𝑖=0, 𝑖=1,...,𝑠, 𝜉𝑖=0∨𝜂𝑖=0, 𝑖=𝑠+1,...,𝑟. (29) By (25), 𝑥𝑦=𝜉1𝜂1𝑒2 1+⋅⋅⋅+𝜉𝑠𝜂𝑠𝑒2 𝑠.(30) If 𝜉1𝜂1𝑒2 1+⋅⋅⋅+𝜉𝑠𝜂𝑠𝑒2 𝑠=0, then the integer 𝜉1𝜂1𝑒2 1+⋅⋅⋅+𝜉𝑠𝜂𝑠𝑒2 𝑠≡ 0(mod 𝑛),thatis, 𝜉1𝜂1𝑛2 𝑝2 1+⋅⋅⋅+𝜉𝑠𝜂𝑠𝑛2 𝑝2 𝑠≡0 (mod 𝑛=𝑝1⋅⋅⋅𝑝𝑟). (31) International Journal of Mathematics and Mathematical Sciences 5 However, this is impossible because none of 𝑝1,...,𝑝𝑠divides theleft-handside.(Namely,𝑝𝑖divides every other summand except the 𝑖th one.) Therefore, 𝑥𝑦 =0and our claim follows by contradiction. (b)⇒(c). If 𝑥2=0,then𝑥⊥0𝑥by (b) and 𝑥=0by (A2). (c)⇒(a). Suppose that (a) does not hold. Then, say, 𝛼1>1. Let 𝑥=𝑛/𝑝1. Since the integer 𝑥2=𝑛2 𝑝2 1=𝑝2𝛼1 1⋅⋅⋅𝑝2𝛼𝑟 𝑟 𝑝2 1 =𝑝2𝛼1−2 1𝑝2𝛼2 2⋅⋅⋅𝑝2𝛼𝑟 𝑟≡0 (mod 𝑛),(32) the residue class 𝑥2=0.But𝑥 =0and hence (c) does not hold. Again, our claim follows now by contradiction. Corollary 18. If and only if the conditions of Theorem 17 are satisfied, then ∀𝑥,𝑦∈Z𝑛:𝑥⊥0𝑦⇐⇒𝑛|(𝑥𝑦),(33) where 𝑥𝑦is the product of integers 𝑥and 𝑦. Example 19. Let 𝐺=Z30.Since30=2⋅3⋅5,thedecomposition (12)is Z30 =⟨30 2⟩⊕⟨30 3⟩⊕⟨30 5⟩ ={0,15}⊕{0,10,20}⊕{0,6,12,18,24}.(34) For example, since 2=0⋅15+2⋅10+2⋅6and 15=1⋅15+0⋅ 10+0⋅6,wehave2⊥015. Generally, (33)impliesthat𝑥⊥0𝑦 if and only if the corresponding integers satisfy 30|(𝑥𝑦). Example 20. Let 𝐺=Z360.Since360=23⋅32⋅5,wehave Z360 =⟨360 23⟩⊕⟨360 32⟩⊕⟨360 5⟩ ={45,90,...,315}⊕{40,80,...,320} ⊕{72,144,...,288}. (35) For example, 5⊥072because 5=1⋅45+8⋅40+0⋅72and 72=0⋅45+0⋅40+1⋅72.Now(33)isonlynecessaryfor⊥0 but not sufficient. For example, 10 ⊥036due to the fact that 10=2⋅45+7⋅40+0⋅72and 36=4⋅45+0⋅40+3⋅72. However, 360|(10⋅36). 4. Another Definition of Perpendicularity Davis [4] defined perpendicularity as a binary relation ⊥in 𝐺 satisfying (D1)∀𝑎,𝑏∈𝐺:𝑎⊥𝑏⇒𝑏⊥𝑎, (D2)∀𝑎∈𝐺:0⊥𝑎, (D3)∀𝑎∈𝐺:𝑎⊥𝑎⇒𝑎=0, (D4)∀𝑎,𝑏,𝑐∈𝐺:𝑏⊥𝑎∧𝑐⊥𝑎⇒(𝑏+𝑐)⊥𝑎, (D5)∀𝑎,𝑏∈𝐺:𝑎⊥𝑏⇔{𝑎}⊥⊥ ∩{𝑏}⊥⊥ ={0}. He assumes that 𝐺is an Abelian group, but the definition applies more generally to an Abelian monoid, too. It is easy to see that (D1)–(D4) are equivalent to (A1)–(A4). Axiom (D5) arises from introducing the concept of “disjointness” on a vector lattice; see [2,page295],[6]. In fact, ⇔canbereplaced with ⇐in (D5) due to the following observation. Proposition 21. Assume that ⊥satisfies (D1)–(D3) (or, equivalently, (A1)–(A3)). Then ∀𝑎,𝑏∈𝐺:𝑎⊥𝑏⇒{𝑎}⊥⊥ ∩{𝑏}⊥⊥ ={0}.(36) Proof. We show first that if 0 =𝐴⊆𝐺,then 𝐴∩𝐴⊥={0}.(37) If 𝑥∈𝐴∩𝐴⊥,then𝑥⊥𝑦for all 𝑦∈𝐴.Inparticular,𝑥⊥𝑥, and hence 𝑥=0by (D3) and (37) follows. Assume next that 𝑎⊥𝑏and let 𝑥∈{𝑎}⊥⊥ ∩{𝑏}⊥⊥.Since 𝑥⊥{𝑏}⊥and 𝑎∈{𝑏}⊥,wehave𝑥⊥𝑎implying that 𝑥∈{𝑎}⊥. Thus 𝑥∈{𝑎}⊥∩{𝑎}⊥⊥.But(37)appliedto𝐴={𝑎}⊥implies that {𝑎}⊥∩{𝑎}⊥⊥ ={0}and 𝑥=0follows. How are these two perpendicularities related? We give a partial answer. Let us denote by 𝐴and 𝐷the axioms (A1)– (A5) and (D1)–(D5), respectively. Proposition 22. If all elements of 𝐺have finite order, then 𝐷⇒𝐴.If𝐺hasatleastoneelementofinfiniteorder,then there exists a relation ⊥satisfying 𝐷but not 𝐴. Proof. The first claim follows from Proposition 3.Concerning the second one, ⊥defined by (2) establishes a relation satisfying 𝐷but not (A5). Proposition 23. Assume that 𝐺has elements 𝑎1,𝑎2,𝑎3,𝑎4=0 such that ⟨𝑎𝑖⟩∩⟨𝑎𝑗⟩={0}whenever 𝑖 =𝑗. Then there exists a relation ⊥satisfying 𝐴but not 𝐷. Proof. The relation ⊥defined by (5)with𝐻=⟨𝑎1⟩,𝐾=⟨𝑎2⟩, 𝐻1=⟨𝑎3⟩,and𝐾1=⟨𝑎4⟩satisfies 𝐴.Since{𝑎1}⊥⊥ ∩{𝑎3}⊥⊥ = ⟨𝑎1⟩∩⟨𝑎3⟩={0}and 𝑎1 ⊥𝑎3, it does not satisfy (D5). 5. Divisibility in Q+ It will turn out that perpendicularity has got something to do also with divisibility in Q+.Tothatend,webeginbynoticing that every 𝑏∈Q+canbesaidtobearationaldivisorofevery 𝑎∈Q+because 𝑎=𝑐𝑏for some 𝑐∈Q+. So, this divisibility is trivial. In order to be able to discuss nontrivial divisibilities in Q+, we have to consider which properties essentially establish this relation. The following three ones seem quite obvious. Let |be a relation in Q+satisfying (i) ∀𝑎∈Q+:𝑎|𝑎, (ii) ∀𝑎,𝑏,𝑐∈Q+:𝑐|𝑎∧𝑐|𝑏⇒𝑐|(𝑎𝑏), (iii) ∀𝑎,𝑏,𝑐∈Q+:𝑐|𝑏∧𝑏|𝑎⇒𝑐|𝑎. We call |adivisibility in Q+. In other words, divisibility is a reflexive and transitive relation (i.e., a preorder) satisfying (ii). 6 International Journal of Mathematics and Mathematical Sciences If 𝑏|𝑎,thenwesaythat𝑏is a divisor of 𝑎and that 𝑎is divisible by 𝑏.If𝑑|𝑎,𝑑|𝑏and 𝑐|𝑎∧𝑐|𝑏⇒𝑐|𝑑,then𝑑is a greatest common divisor of 𝑎and 𝑏, denoted by gcd|(𝑎,𝑏).Allthese notions are meaningful also in any Abelian monoid. Let us recall that every 𝑐∈Q+canbeexpressedas 𝑐=∏ 𝑝∈P𝑝]𝑝(𝑐),(38) where ]𝑝(𝑐)∈Zfor each 𝑝∈P, and only a finite number of them are nonzero. If ]𝑝(𝑐) =0,then𝑝is a prime factor of 𝑐. Consider the set 𝑆of all sequences (𝑛2,𝑛3,...,𝑛𝑝,...), where the index runs through P, each 𝑛𝑝∈Z,andonlya finite number of them are nonzero. The mapping 𝑓(𝑐)=(]2(𝑐),]3(𝑐),...,]𝑝(𝑐),...) (39) is an isomorphism from (Q+,⋅)onto (𝑆,+)where addition is defined termwise. For example, 𝑓(45)+𝑓(8 25)=(0,2,1,0,0,...)+(3,0,−2,0,0,...) =(3,2,−1,0,0,...), 𝑓(45⋅8 25)=𝑓(72 5)=𝑓(23⋅32⋅5−1) =(3,2,−1,0,0,...).(40) Given 𝑎,𝑏 ∈ Q+, we define their “inner product” being the Euclidean inner product of the vectors 𝑓(𝑎)and 𝑓(𝑏): ⟨𝑎,𝑏⟩=⟨𝑓(𝑎),𝑓(𝑏)⟩=∑ 𝑝∈P ]𝑝(𝑎)]𝑝(𝑏).(41) Since only a finite number of summands are nonzero, this sum is finite. For example, ⟨45, 8 25⟩=0⋅3+2⋅0+1⋅(−2)+0+0+⋅⋅⋅=−2. (42) Next we define |𝑐|by setting ]𝑝(|𝑐|)=|]𝑝(𝑐)|for all 𝑝∈ Por, equivalently, |𝑐|=𝑓−1((]2(|𝑐|),]3(|𝑐|),]5(|𝑐|),...)).For example, if 𝑐=40/63=23⋅3 −2 ⋅5 1⋅7 −1,then|𝑐|=23⋅ 32⋅51⋅71=2520.Letting⊥1bethesamerelationastheone defined by (7),itcanbecharacterizednowby 𝑎⊥1𝑏⇐⇒⟨|𝑎|,|𝑏|⟩=0. (43) Also the relation ⊥2in Q+, defined by 𝑎⊥2𝑏⇐⇒⟨𝑎,𝑏⟩=0, (44) is a perpendicularity. We will introduce one more nontrivial perpendicularity using divisibility. For that purpose, we first notice that the relation 𝛿defined by 𝑏𝛿𝑎⇐⇒∀𝑝∈P:]𝑝(𝑏)≤]𝑝(𝑎)(45) is a divisibility, gcd𝛿(𝑎,𝑏)exists and is unique for all 𝑎,𝑏∈Q+, and gcd𝛿(𝑎,𝑏)=∏ 𝑝∈P𝑝min(]𝑝(𝑎),]𝑝(𝑏)).(46) Assume now that 𝑚,𝑛,𝑢,V∈Z+so that gcd(𝑚,𝑢) = gcd(𝑛,V)=1. An alternative expression for (45)is 𝑛 V𝛿𝑚 𝑢⇐⇒ 𝑛|𝑚∧𝑢|V,(47) and that for (46)is gcd𝛿(𝑚 𝑢,𝑛 V)=gcd (𝑚,𝑛) lcm (𝑢,V).(48) For example, if 𝑎=45/14=2−1 ⋅32⋅51⋅7−1 and 𝑏=33/100= 2−2 ⋅31⋅5−2 ⋅111,thengcd 𝛿(𝑎,𝑏)=2−2 ⋅31⋅5−2 ⋅7−1 =3/700. Alternatively, gcd𝛿(𝑎,𝑏)=gcd (45,33) lcm (14,100)=3 700.(49) Since gcd𝛿(|𝑚/𝑢|,|𝑛/V|)=gcd(𝑚𝑢,𝑛V),wehaveby(9) 𝑎⊥1𝑏⇐⇒gcd𝛿(|𝑎|,|𝑏|)=1. (50) This relation generalizes (10) and answers the cry of Graham et al. in a slightly wider context than what they, perhaps, had thought. Eugeni and Rizzi [7, Section 2] defined divisibility in Q+ by setting the relation 𝛾so that 𝑛 V𝛾𝑚 𝑢⇐⇒ 𝑛|𝑚∧V|𝑢. (51) Then gcd𝛾(𝑎,𝑏)always exists and is unique, and gcd𝛾(𝑚 𝑢,𝑛 V)=gcd (𝑚,𝑛) gcd (𝑢,V).(52) For example, gcd𝛾(45 14,33 100)=gcd (45,33) gcd (14,100)=3 2.(53) We define now the corresponding perpendicularity by writing 𝑎⊥ER 𝑏⇐⇒gcd𝛾(𝑎,𝑏)=1 ⇐⇒ gcd (𝑚,𝑛)=gcd (𝑢,V)=1. (54) Summing up, we have at least three nontrivial perpendicularities in Q+. Let us see how they relate to one another. ⊥1versus ⊥2.Clearly⊥1⇒⊥2(i.e., 𝑥⊥1𝑦⇒𝑥⊥2𝑦). The converse does not hold. For example, 6⊥22/3but 6 ⊥12/3. ⊥1versus ⊥ER.Clearly⊥1⇒⊥ER.Theconversedoesnot hold. For example, 2/3⊥ER3/2but 2/3 ⊥13/2. ⊥2versus ⊥ER. These perpendicularities are independent. For example, 6⊥22/3 but 6 ⊥ER 2/3. On the other hand, 2/3⊥ER3/2but 2/3 ⊥23/2. However, regarding (Z+,⋅)as a submonoid of (Q+,⋅),itis obvious that ⊥1=⊥ 2=⊥ ER in Z+.Moreover,inZ+,they yield the very perpendicularity proposed by Graham et al. International Journal of Mathematics and Mathematical Sciences 7 6. Parallelism Parallelism is closely related to perpendicularity. Considering different geometric contexts we notice soon that, in general, parallelism does not have any other properties except those of equivalence. However, any equivalence relation cannot be said to stand for parallelism in any reasonable way. This leads us to ask whether it is possible or not to define parallelism in Abelian groups having a perpendicularity so that it makes sense. Let 𝐺have a perpendicularity ⊥and let 𝑎,𝑏 ∈ 𝐺.We say that 𝑎and 𝑏are parallel and write 𝑎‖𝑏if {𝑎}⊥= {𝑏}⊥. The relation ‖is clearly an equivalence. If 𝑎 =0,then𝑎∦0, since {0}⊥=𝐺by Proposition 1(a) but {𝑎}⊥=𝐺by (A2). All nonzero elements are parallel if and only if ⊥is trivial. If 𝐺=Z𝑛and ⊥=⊥0,then,recalling(19), 𝑥‖𝑦⇐⇒(∀𝑖∈{1,...,𝑟}:𝜉𝑖=0⇐⇒𝜂𝑖=0) ⇐⇒ {𝑥}⊥={𝑦}⊥=𝐻𝑖𝑖⊕⋅⋅⋅⊕𝐻𝑖𝑡,(55) where 𝜉𝑖=𝜂 𝑖=0⇔𝑖∈{𝑖1,...,𝑖𝑡}. For example, consider Z30 (see Example 19). Since 2=0⋅15+2⋅10+2⋅6and 16=0⋅15+1⋅10+1⋅6,wehave{2}⊥={16}⊥={0,15},and so 2‖16. Now, let 𝐺=Q+and let ⊥1,⊥2,and⊥ER be as before. Denote the corresponding parallelisms by ‖1,‖2,and‖ER, respectively. Then 𝑎‖1𝑏if and only if 𝑎and 𝑏have the same prime factors. Further, 𝑚/𝑢‖ER𝑛/Vif and only if 𝑚and 𝑛have thesameprimefactorsand𝑢and Vhavethesameprime factors. Let us study how these parallelisms relate to one another. ‖1versus ‖2.Weshowthat‖2⇒‖ 1. Assume first that 𝑎‖2𝑏.If𝑎∦1𝑏, then there exists 𝑝0∈Psuch that, say, ]𝑝0(𝑎) = 0and ]𝑝0(𝑏) =0.Butnow𝑝0⊥2𝑎and 𝑝0 ⊥2𝑏,and so {𝑎}⊥2={𝑏}⊥2contradicting the assumption. The converse does not hold. For example, let 𝑎=6and 𝑏=12;then𝑎‖1𝑏. If 𝑥=2/3,then𝑥⊥2𝑎but 𝑥 ⊥2𝑏, and hence {𝑎}⊥2={𝑏}⊥2.In other words, 𝑎∦2𝑏. ‖1versus ‖ER.Clearly‖ER ⇒‖ 1.Theconversedoesnot hold. For example, 2/3‖13/2but 2/3∦ER3/2. ‖2versus ‖ER.Weshowthat‖2⇒‖ ER.Given𝑝1,...,𝑝𝑡∈ P, denote by 𝑁(𝑝1,...,𝑝𝑡)the set of such positive integers thatarenotdivisiblebyany𝑝𝑖,𝑖=1,...,𝑡.Let𝑎=𝑚/𝑢∈Q+, gcd(𝑚,𝑢)=1.Factorize 𝑚=𝑝𝛼1 1⋅⋅⋅𝑝𝛼ℎ ℎ,𝑢=𝑞 𝛽1 1⋅⋅⋅𝑞𝛽𝑘 𝑘,(56) where 𝑝1,...,𝑝ℎ,𝑞1,...,𝑞𝑘∈Pare distinct and 𝛼1,..., 𝛼ℎ,𝛽1,...,𝛽𝑘>0.(If𝑚=1or 𝑢=1, then the corresponding “empty product” is one.) Now {𝑎}⊥2={𝑝𝜉1 1⋅⋅⋅𝑝𝜉ℎ ℎ 𝑞𝜂1 1⋅⋅⋅𝑞𝜂𝑘 𝑘𝑥 𝑦|𝛼1𝜉1+⋅⋅⋅+𝛼ℎ𝜉ℎ+𝛽1𝜂1 +⋅⋅⋅+𝛽𝑘𝜂𝑘=0, 𝑥,𝑦∈𝑁(𝑝1,...,𝑝ℎ,𝑞1,...,𝑞𝑘)}. (57) (The “empty sum” is zero.) Assume that 𝑏=𝑛/V∈Q+, gcd(𝑛,V)=1, satisfies 𝑎‖2𝑏,thatis,{𝑎}⊥2= {𝑏}⊥2.Then,by (57), necessarily 𝑛=𝑝𝜌1 1⋅⋅⋅𝑝𝜌ℎ ℎ,V=𝑞𝜎1 1⋅⋅⋅𝑞𝜎𝑘 𝑘,(58) where 𝜌1,...,𝜌ℎ,𝜎1,...,𝜎𝑘>0.Hence𝑎‖ER𝑏,andtheclaim follows. The converse is not valid. For example, 2/3‖ER4/3but 2/3∦24/3. 7. Discussion This paper began with a citation by three established mathematicians and computer scientists who showed a remarkable intuition by promoting the use of the symbol of perpendicularity in number theory. Indeed, we have previously seen how this notion settles comfortably in this setting and gains new meanings at a more general level in the context of Abelian group theory. We conclude this paper with the following supplement to their proposal. Let perpendicularity and parallelism mean here ⊥1and ‖1, respectively. Consider the “direction vector” of 𝑐∈Q+ by (𝑐(2),𝑐(3),...,𝑐(𝑝),...),where𝑐(𝑝)=0if ]𝑝(𝑐)=0and 𝑐(𝑝) = 1otherwise. For example, the direction vectors of 45,1,and8/25are, respectively (0,1,1,0,0,...),(0,0,...),and (1,0,1,0,0,...). Now, like the directions of perpendicular lines are as different as possible, the prime factors of perpendicular (positive rational) numbers are as different as possible; that is, such numbers do not have common prime factors. In other words, the direction vectors of perpendicular numbers are as different as possible in the sense that they have no common element of value one. Like parallel lines have the same direction, parallel numbers have the same prime factors. In other words, their direction vectors are equal. Finally, we note that perpendicularity can be axiomatized in a natural way also in many other algebraic structures. Davis [8] did that in a ring. In a vector space, perpendicularity is customarily defined based on an inner product. Another possible approach is to supplement (A1)–(A5) with suitable axioms concerning the multiplication of a vector by a scalar. It might be interesting to study under which additional conditions there exists an inner product inducing this perpendicularity. Acknowledgment The authors would like to thank the referees for carefully reading the paper and kind comments. References [1] R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics: A Foundation for Computer Science, Addison-Wesley, Reading, Mass, USA, 2nd edition, 1994. [2] G. Birkhoff, Lattice Theory, American Mathematical Society, Providence, RI, USA, 3rd edition, 1993. [3] P.Haukkanen,J.K.Merikoski,andT.Tossavainen,“Axiomatizing perpendicularity and parallelism,” Journal for Geometry and Graphics,vol.15,no.2,pp.129–139,2011. 8 International Journal of Mathematics and Mathematical Sciences [4] G. Davis, “Orthogonality relations on abelian groups,” Journal of the Australian Mathematical Society. Series A,vol.19,pp.173– 179, 1975. [5] W.K.Nicholson,Introduction to Abstract Algebra,JohnWiley & Sons, New York, NY, USA, 2nd edition, 1999. [6] A. I. Veksler, “Linear spaces with disjoint elements and their conversion into vector lattices,” Leningradski˘ ıGosudarstvenny ˘ ı Pedagogiˇ ceski˘ ı Institut imeni A. I. Gercena. Uˇ cenye Zapiski,vol. 328, pp. 19–43, 1967 (Russian). [7] F. Eugeni and B. 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