Perpendicularity in an Abelian Group

Hindawi Publishing Co po a ion In e na ional Jou nal o Ma hema ics and Ma hema ical Sciences Volume 2013, A icle ID 983607, 8pages h p://dx.doi.o g/10.1155/2013/983607 Resea ch A icle Pe pendicula i y in an Abelian G oup Pen i Haukkanen,1Mika Ma ila,1Jo ma K. Me ikoski,1and Timo Tossa ainen2 1School o In o ma ion Sciences, Uni e si y o Tampe e, 33014, Finland 2School o Applied Educa ional Science and Teache Educa ion, Uni e si y o Eas e n Finland, P.O. Box 86, 57101 Sa onlinna, Finland Co espondence should be add essed o Pen i Haukkanen; pen [email p o ec ed] Recei ed 12 Janua y 2013; Accep ed 19 Ma ch 2013 Academic Edi o : Pe u Jebelean Copy igh © 2013 Pen i Haukkanen e al. This is an open access a icle dis ibu ed unde he C ea i e Commons A ibu ion License, which pe mi s un es ic ed use, dis ibu ion, and ep oduc ion in any medium, p o ided he o iginal wo k is p ope ly ci ed. We gi e a se o axioms o es ablish a pe pendicula i y ela ion in an Abelian g oup and hen s udy he exis ence o pe pendicula i ies in (Z𝑛,+)and (Q+,⋅)and in ce ain o he g oups. Ou app oach p o ides a jus i ica ion o he use o he symbol ⊥deno ing ela i e p imeness in numbe heo y and ex ends he domain o his con en ion o some deg ee. Rela ed o ha , we also conside pa allelism om an axioma ic pe spec i e. 1. In oduc ion In [1,page115],G ahame al.made he ollowingsugges ion: When gcd(𝑚,𝑛)=1, he in ege s 𝑚and 𝑛ha e no p ime ac o s in common and we say ha hey a e ela i ely p ime. This concep is so impo an in p ac ice, we ough o ha e a special no a ion o i ; bu alas, numbe heo is s ha e no ag eed on a e y good one ye . The e o e we c y: hea us, o ma hema icians o he wo ld! le us no wai any longe ! we can make many o mulas clea e by adop ing a new no a ion now! le us ag ee o w i e “𝑚⊥𝑛”, a n d o s ay “ 𝑚is p ime o 𝑛,” i 𝑚and 𝑛a e ela i ely p ime. Like pe pendicula lines do no ha e a common di ec ion, pe pendicula numbe s do no ha e common ac o s. In ac , hisc yhadbeenanswe ede enbe o ei was made. Namely, in s udying 𝑙-g oups (i.e., g oups wi h a la ice s uc u e), Bi kho [2,page295]de ines ha woposi i e elemen s 𝑎and 𝑏o an 𝑙-g oup a e disjoin i 𝑎∧𝑏=0 and uses he no a ion 𝑎⊥𝑏 o disjoin elemen s. He also ema ks ha disjoin ness specializes o ela i e p imeness in he 𝑙-g oup o posi i e in ege s. Amo i a ion o hep esen pape is os udyhow jus i ied ul ima ely i is o use he symbol o pe pendicula i y odeno e ela i ep imeness.Does hisp ac ice elyonly on he analogy be ween ha ing no common di ec ion and ha ing no common ac o o is he e a deepe linkage o en i le hiscon en ion?Thisques ionleadsus oaskwhich p ope ies essen ially es ablish he no ion o pe pendicula i y in he algeb aic con ex and wha he mos sui able algeb aic con ex o he axioma iza ion o pe pendicula i y ac ually is; we ha e ecen ly s udied he axioms o pe pendicula i y om an elemen a y geome ic poin o iew [3]. In an inne p oduc space, pe pendicula i y ob iously aces back o he inne p oduc being ze o. Howe e , ce ain ea u es o his pe pendicula i y can be shi ed down o simple algeb aic s uc u es. We will de ine pe pendicula i y in an Abelian g oup and examine i in Sec ion 2.InSec ion 3, we will ocus on pe pendicula i y in (Z𝑛,+).Da is[4] de ined pe pendicula i y in an Abelian g oup di e en ly. In Sec ion 4, we will in oduce his app oach and compa e i wi h ou s. The ea e , we will conside di isibili y in (Q+,⋅)in Sec ion 5 and pa allelism in an Abelian g oup in Sec ion 6.Wewill conclude ou pape wi h a b ie discussion and a supplemen o hesugges ionci edp e iously. 2. Axioms and P ope ies o Pe pendicula i y Th oughou his pape , 𝐺 = (𝐺,+)is an Abelian g oup so ha 𝐺 ={0}. Unless o he wise s a ed, ⊥is a bina y ela ion in 𝐺sa is ying (A1)∀𝑎∈𝐺:∃𝑏∈𝐺:𝑎⊥𝑏, (A2)∀𝑎∈𝐺 {0}:𝑎 ⊥𝑎, (A3)∀𝑎,𝑏∈𝐺:𝑎⊥𝑏⇒𝑏⊥𝑎, 2 In e na ional Jou nal o Ma hema ics and Ma hema ical Sciences (A4)∀𝑎,𝑏,𝑐∈𝐺:𝑎⊥𝑏∧𝑎⊥𝑐⇒𝑎⊥(𝑏+𝑐), (A5)∀𝑎,𝑏∈𝐺:𝑎⊥𝑏⇒𝑎⊥−𝑏. We call ⊥ape pendicula i y in 𝐺.Thisconcep canbe de ined also in weake s uc u es by changing hese axioms app op ia ely. Fo example, i 𝐺is an Abelian monoid, hen we simply omi (A5). Since he i ial pe pendicula i y 𝑥⊥𝑦⇐⇒𝑥=0∨𝑦=0 (1) always exis s, we a e mainly in e es ed in non i ial pe pen- dicula i ies. We call ⊥maximal i i is no a sub ela ion o any o he pe pendicula i y in 𝐺. The e always exis s a maximal pe pendicula i y. This is ob ious i 𝐺is ini e and, o he wise, i ollows om Zo n’s lemma. P oposi ion 1 eco ds some elemen a y p ope ies o pe - pendicula i y; we lea e he p oo o he eade . P oposi ion 1. Pe pendicula i y ⊥has he ollowing p ope - ies: (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 ollowing cha ac e iza ion is use ul in p o ing ha a gi en ela ion is pe pendicula i y. P oposi ion 2. Abina y ela ion⊥in 𝐺is pe pendicula i y i and only i i sa is ies (A1) and (A2) and (A6)∀𝑎,𝑏,𝑐∈𝐺:𝑎⊥𝑏∧𝑎⊥𝑐⇒(𝑏−𝑐)⊥𝑎. P oo . The“onlyi ”-pa is i ial.Top o e he“i ”-pa ,we i s show ha ou assump ions imply P oposi ion 1(a). Le 𝑎∈𝐺.By(A1), he eis𝑏∈𝐺such ha 𝑎⊥𝑏.Pu ing𝑐:=𝑏 in (A6) implies 0⊥𝑎;inpa icula 0⊥0. Fu he , (A6) wi h 𝑎:=0,𝑏:=𝑎and 𝑐:=0gi es (𝑎−0)⊥0, ha is,𝑎⊥0.Now we can e i y he emaining axioms. (A3) Assume 𝑎⊥𝑏. Apply (A6) wi h 𝑐:=0; hen𝑏⊥𝑎. (A5) Assume 𝑎⊥𝑏. Apply (A6) wi h 𝑏:=0and 𝑐:=𝑏. Then (−𝑏)⊥𝑎,andso,by(A3),𝑎⊥−𝑏. (A4) Assume 𝑎⊥𝑏and 𝑎⊥𝑐; hen𝑎⊥−𝑐by (A5). Now (A6) wi h 𝑐:=−𝑐implies (𝑏−(−𝑐)) ⊥ 𝑎, ha is, (𝑏+𝑐)⊥𝑎.Hence,by(A3),𝑎⊥(𝑏+𝑐). Is he e a simple condi ion unde which (A5) ollows om (A1)–(A4)? The answe is posi i e. P oposi ion 3. I all elemen s o 𝐺ha e ini e o de and i ⊥ sa is ies (A1)–(A4), hen i sa is ies (A5). I 𝐺has a leas one elemen o in ini e o de , hen he e exis s a ela ion ⊥which sa is ies (A1)–(A4) bu no (A5). P oo . Fo he i s pa , assume ha 𝑎,𝑏∈𝐺sa is y 𝑎⊥𝑏, and le he o de o 𝑏be 𝑛.Then𝑎⊥(𝑛−1)𝑏by (A4). Bu (𝑛−1)𝑏=−𝑏and (A5) ollows. Fo he second pa , le 𝑎∈ 𝐺ha e in ini e o de . Then he subg oup {0,±𝑎,±2𝑎,...}is isomo phic o Z.The ela ion⊥de ined by 𝑥⊥𝑦⇐⇒(∃𝜇,]∈Z:𝑥=𝜇𝑎∧𝑦=]𝑎∧𝜇]<0) ∨𝑥=0∨𝑦=0 (2) sa is ies (A1)–(A4) bu no (A5). I 0 =𝐴⊆𝐺,wede ine hepe pendicula complemen o ⊥-complemen o 𝐴as ollows: 𝐴⊥={𝑦∈𝐺|𝑦⊥𝐴}= ⋃ 𝐺⊇𝐵⊥𝐴𝐵. (3) He e 𝑦⊥𝐴means ha 𝑦⊥𝑥 o all 𝑥∈𝐴,and𝐵⊥𝐴 means ha 𝑦⊥𝐴 o all 𝑦∈𝐵.Thus𝐴⊥is he maximal se pe pendicula o 𝐴.Inpa icula ,𝐺⊥={0}and {0}⊥=𝐺.We also de ine 0⊥=𝐺. P oposi ion 4. I 𝐴⊆𝐺, hen𝐴⊥is a subg oup o 𝐺.I 𝐺is cyclic, hen 𝐴⊥is cyclic. P oo . The i s pa ollows by applying he subg oup es and P oposi ion 2.Thesecondpa ollows om he ac ha any subg oup o a cyclic g oup is cyclic. The nex heo em ells when 𝐺has a non i ial pe pen- dicula i y. Theo em 5. The ollowing condi ions a e equi alen : (a) 𝐺has a non i ial pe pendicula i y ⊥, (b) 𝐺has non i ial cyclic subg oups 𝐻and 𝐾sa is ying 𝐻∩𝐾={0}, (c) 𝐺has non i ial subg oups 𝐻and 𝐾sa is ying 𝐻∩ 𝐾={0}. P oo . (a)⇒(b). Since ⊥is non i ial, he e exis 𝑥,𝑦∈𝐺 {0} such ha 𝑥⊥𝑦.Then𝐻=⟨𝑥⟩and 𝐾=⟨𝑦⟩apply. He e ⟨𝑎⟩ s ands o he cyclic g oup gene a ed by 𝑎. (b)⇒(c). T i ial. (c)⇒(a). De ine ⊥by 𝑥⊥𝑦⇐⇒(𝑥∈𝐻∧𝑦∈𝐾)∨(𝑥∈𝐾∧𝑦∈𝐻) ∨𝑥=0∨𝑦=0. (4) Nex we conside he maximal pe pendicula i y in some examples o g oups. In Examples 6–9, heg oupope a ionis addi ion. Example 6. Le 𝐺=Z6.ByLag ange’s heo em[5,page 130, Theo em 2], he smalles 𝑛such ha Z𝑛has a non i ial pe pendicula i y is 6=2⋅3because 𝑛mus ha e a leas wo di e en p ime ac o s. The non i ial subg oups o Z6a e 𝐻=⟨3⟩={0,3}and 𝐾=⟨2⟩={0,2,4}.Since𝐺=𝐻⊕𝐾, In e na ional Jou nal o Ma hema ics and Ma hema ical Sciences 3 i has exac ly one non i ial pe pendicula i y, de ined by 0⊥ 0,1,2,3,4,5and 3⊥2,4and ice e sa.Consequen ly, his pe pendicula i y is maximal. Example 7. Le 𝐺=Z2×Z2, he Klein ou g oup. Deno e 0 = (0,0),𝑎 = (0,1),𝑏 = (1,0),𝑐 = (1,1).The non i ial subg oups a e 𝐴={0,𝑎},𝐵={0,𝑏},𝐶={0,𝑐}. So, he e a e h ee non i ial pe pendicula i ies ob ained as ollows: Choose wo elemen s o 𝑎,𝑏,and𝑐.De ine ha hey a e pe pendicula o each o he and o 0. De ine ha he emaining elemen is pe pendicula o 0 only. All pe pendicula i ies a ising in his way a e clea ly maximal. We also no e ha 𝐺=𝐴⊕𝐵=𝐵⊕𝐶=𝐶⊕𝐴. Example 8. Le 𝐺=Z.Fo each𝑛≥2, hesubg oup⟨𝑛⟩= 𝑛Zis non i ial and he e a e no o he non i ial subg oups han hose ound in his way. Because 𝑚𝑛∈⟨𝑚⟩∩⟨𝑛⟩, he eis no pai o non i ial subg oups wi h in e sec ion {0}.Hence 𝐺has only he i ial pe pendicula i y. Example 9. Le 𝐺=R.SinceRhas in ini ely many pai s o non i ial subg oups wi h in e sec ion {0},i hasin ini ely many non i ial pe pendicula i ies. Fo example, le 𝐻=Q and 𝐾={𝑥√2|𝑥∈Q}and de ine ⊥by (4). To see ha his pe pendicula i y is no maximal, le 𝐻1={𝑥√3|𝑥∈Q}and 𝐾1={𝑥√5|𝑥∈Q}and de ine ⊥󸀠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. Le 𝐺=(Q+,⋅),whe eQ+deno es he se o posi i e a ional numbe s. E e y 𝑐∈Q+can be uniquely exp essed as 𝑐=∏ 𝑝∈P𝑝]𝑝(𝑐),(6) whe e ]𝑝(𝑐)∈Z o each 𝑝∈Pandonlya ini enumbe o hema enonze o.ThesymbolPs ands o hese o p imes. Fo example, i 𝑐=8/25, hen]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 o he wo ds, i 𝑎=𝑚 𝑢,𝑏= 𝑛 V,𝑚,𝑢,𝑛,V∈Z+, gcd (𝑚,𝑢)=gcd (𝑛,V)=1, (8) hen 𝑎⊥𝑏⇐⇒gcd (𝑚𝑢,𝑛V)=1. (9) Hence, o example, 8/9⊥7/5.Inpa icula , o 𝑚,𝑛∈Z+, applying (9) o𝑚/1and 𝑛/1yields ha 𝑚⊥𝑛⇐⇒gcd (𝑚,𝑛)=1. (10) So, i seems ha G aham e al. we e p ophe ically qui e igh wi h hei sugges ion—and no o ge ing Bi kho ei he ! We will discuss he pe pendicula i y o posi i e a ional numbe s in mo e de ail in Sec ion 5. 3. Pe pendicula i y in Z𝑛 S udying pe pendicula i ies equi es ha we know he s uc- u e o 𝐺. Nex we ake a mo e ho ough look a pe pen- dicula i y in Z𝑛. To ha end, we begin by in oducing a sui able no a ion o discuss he s uc u e o Z𝑛and eco d wo lemmas which a e use ul in he sea ch o he maximal pe pendicula i y. We will also use he no a ions in oduced in Theo em 11 and he ollowing lemmas h oughou he nex sec ions. Theo em 11. I 𝑛=𝑝𝛼1 1⋅⋅⋅𝑝𝛼𝑟 𝑟,(11) whe e 𝑝1,...,𝑝𝑟∈Pa e dis inc and 𝛼1,...,𝛼𝑟>0, hen Z𝑛=𝐻1⊕⋅⋅⋅⊕𝐻𝑟,(12) whe e 𝐻𝑖=⟨𝑒𝑖⟩, 𝑒𝑖=𝑛 𝑝𝛼𝑖 𝑖, 𝑖=1,...,𝑟. (13) The decomposi ion (12)isunique(up o heo de o subg oups). P oo . The claim (12) ollows om[5,page399,Co olla y1] and om he ac s ha Z𝑝𝛼𝑖 𝑖≅𝐻 𝑖and 𝐻𝑖∩𝐻𝑗={0} o all 𝑖,𝑗=1,...,𝑟,𝑖 =𝑗. Uniqueness ollows om [5,page399, Co olla y 2]. Al hough we conside Z𝑛mainly as an Abelian g oup, i is now use ul o wo k wi h Z𝑛as a ing. Lemma 12. Fo all 𝑖,𝑗=1,...,𝑟,𝑖 =𝑗, 𝑒2 𝑖=0, 𝑒𝑖𝑒𝑗=0. (14) P oo . I is enough o conside 𝑖=1,𝑗=2. Rega ding 𝑒1and 𝑒2as in ege s, we ha e 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) ollows. 4 In e na ional Jou nal o Ma hema ics and Ma hema ical Sciences Lemma 13. Le ⊥be a pe pendicula i y in Z𝑛.Then ∀𝑎,𝑏,𝑐,𝑑∈Z𝑛:𝑎⊥𝑏󳨐⇒𝑐𝑎⊥𝑑𝑏. (16) P oo . Le 𝑐=𝛾1,𝑑=𝛿1,whe e𝛾and 𝛿a e in ege s wi h 0≤𝛾,𝛿<𝑛.Since𝑐𝑎=(𝛾1)𝑎=𝛾(1𝑎)=𝛾𝑎and, simila ly, 𝑑𝑏=𝛿𝑏,P oposi ion 1(d) implies (16). Now we a e eady o in oduce a pe pendicula i y which u ns ou o be maximal in Z𝑛.Le 𝑥=𝑥1+⋅⋅⋅+𝑥𝑟,𝑦=𝑦 1+⋅⋅⋅+𝑦𝑟∈Z𝑛,(17) whe e 𝑥𝑖,𝑦𝑖∈𝐻𝑖,𝑖=1,...,𝑟.The ela ion⊥0, de ined in Z𝑛 by 𝑥⊥0𝑦⇐⇒∀𝑖∈{1,...,𝑟}:𝑥𝑖=0∨𝑦𝑖=0, (18) is clea ly a pe pendicula i y. Theo em 14. The pe pendicula i y ⊥0is maximal and e e y o he pe pendicula i y in Z𝑛is con ained in i . P oo . Le ⊥be ano he pe pendicula i y in Z𝑛.Ou claimis ha 𝑥⊥𝑦⇒𝑥⊥0𝑦.By(12), we can exp ess 𝑥=𝜉1𝑒1+⋅⋅⋅+𝜉𝑟𝑒𝑟,𝑦=𝜂 1𝑒1+⋅⋅⋅+𝜂𝑟𝑒𝑟,(19) whe e he in ege s 𝜉𝑖,𝜂𝑖∈{0,...,𝑝𝛼𝑖 𝑖−1}and he esidueclass 𝑒𝑖=𝑛/𝑝𝛼𝑖 𝑖,𝑖=1,...,𝑟. Supposeagains heclaimo heo em ha he eexis 𝑥,𝑦∈Z𝑛such ha 𝑥⊥𝑦bu 𝑥 ⊥0𝑦.Then𝜉𝑖,𝜂𝑖=0 o some 𝑖. Reo de ing he indices so ha 𝑖=1and applying (16), we ha e 𝑥𝑒1⊥𝑦𝑒1which implies ha 𝜉1𝑒2 1⊥𝜂1𝑒2 1(20) by (14). Hence, by P oposi ion 1(d), 𝜂1 gcd (𝜉1,𝜂1)𝜉1𝑒2 1⊥𝜉1 gcd (𝜉1,𝜂1)𝜂1𝑒2 1,(21) ha is, lcm (𝜉1,𝜂1)𝑒2 1⊥lcm (𝜉1,𝜂1)𝑒2 1.(22) Consequen ly, lcm (𝜉1,𝜂1)𝑒2 1=0by (A2). In o he wo ds, ega ding also 𝑒1as an in ege , lcm (𝜉1,𝜂1)𝑒2 1=lcm (𝜉1,𝜂1)𝑛2 𝑝2𝛼1 1 =lcm (𝜉1,𝜂1)𝑝2𝛼2 2⋅⋅⋅𝑝2𝛼𝑟 𝑟 ≡0 (mod 𝑛), (23) and 𝑝𝛼1 1di ides lcm(𝜉1,𝜂1). Howe e , since i di ides nei he 𝜉1no 𝜂1, his is a con adic ion. Hence, 𝑥⊥0𝑦. Conside ing he di ec sum (12) ex e nal, we can iden i y 𝑥and 𝑦in (17) wi h ec o s (𝑥1,...,𝑥𝑟)and (𝑦1,...,𝑦𝑟), espec i ely. So, i is na u al o de ine hei “inne p oduc ” by ⟨𝑥,𝑦⟩=𝑥1𝑦1+⋅⋅⋅+𝑥𝑟𝑦𝑟.(24) P oposi ion 15 shows ha his ope a ion coincides wi h he o dina y mul iplica ion in Z𝑛. P oposi ion 15. Gi en 𝑥,𝑦∈Z𝑛, ⟨𝑥,𝑦⟩=𝑥𝑦. (25) P oo . We ha e 𝑥𝑦=( 𝑟 ∑ 𝑖=1𝑥𝑖)(𝑟 ∑ 𝑖=1𝑦𝑖)=𝑟 ∑ 𝑖=1𝑥𝑖𝑦𝑖+𝑟 ∑ 𝑖,𝑗=1 𝑖 =𝑗𝑥𝑖𝑦𝑗.(26) Bu , ecalling (19)and(14), 𝑟 ∑ 𝑖,𝑗=1 𝑖 =𝑗𝑥𝑖𝑦𝑗=𝑟 ∑ 𝑖,𝑗=1 𝑖 =𝑗𝜉𝑖𝜂𝑗𝑒𝑖𝑒𝑗=0. (27) The claim ollows. Theo em 16. Le ⊥be a pe pendicula i y in Z𝑛.Then ∀𝑥,𝑦∈Z𝑛:𝑥⊥𝑦󳨐⇒𝑥𝑦=0. (28) P oo . I 𝑥⊥𝑦, hen𝑥⊥0𝑦by Theo em 14.So,𝑥𝑦=0by (18) and (25). Does he con e se o Theo em 16 hold i ⊥=⊥0?And, ela ed o P oposi ion 15,is⟨𝑥,𝑦⟩ = 𝑥𝑦 ap ope inne p oduc ? Namely, an inne p oduc in a eal ec o space is symme ic and bilinea and i sa is ies ⟨𝑥,𝑥⟩=0⇒𝑥=0. The ope a ion ⟨𝑥,𝑦⟩ = 𝑥𝑦in Z𝑛has clea ly he i s and second p ope ies bu wha abou he hi d one? The answe s o bo h ques ions a e con ained in Theo em 17. Theo em 17. The ollowing condi ions a e equi alen : (a) 𝛼1=⋅⋅⋅=𝛼𝑟=1, (b) ∀𝑥,𝑦∈Z𝑛:𝑥𝑦=0⇒𝑥⊥0𝑦, (c) ∀𝑥∈Z𝑛:𝑥2=0⇒𝑥=0. P oo . (a)⇒(b). Assume ha 𝑥 ⊥0𝑦.Exp ess𝑥and 𝑦as in (19). We can ea ange he indices so ha , o 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) I 𝜉1𝜂1𝑒2 1+⋅⋅⋅+𝜉𝑠𝜂𝑠𝑒2 𝑠=0, hen he in ege 𝜉1𝜂1𝑒2 1+⋅⋅⋅+𝜉𝑠𝜂𝑠𝑒2 𝑠≡ 0(mod 𝑛), ha is, 𝜉1𝜂1𝑛2 𝑝2 1+⋅⋅⋅+𝜉𝑠𝜂𝑠𝑛2 𝑝2 𝑠≡0 (mod 𝑛=𝑝1⋅⋅⋅𝑝𝑟). (31) In e na ional Jou nal o Ma hema ics and Ma hema ical Sciences 5 Howe e , his is impossible because none o 𝑝1,...,𝑝𝑠di ides hele -handside.(Namely,𝑝𝑖di ides e e y o he summand excep he 𝑖 h one.) The e o e, 𝑥𝑦 =0and ou claim ollows by con adic ion. (b)⇒(c). I 𝑥2=0, hen𝑥⊥0𝑥by (b) and 𝑥=0by (A2). (c)⇒(a). Suppose ha (a) does no hold. Then, say, 𝛼1>1. Le 𝑥=𝑛/𝑝1. Since he in ege 𝑥2=𝑛2 𝑝2 1=𝑝2𝛼1 1⋅⋅⋅𝑝2𝛼𝑟 𝑟 𝑝2 1 =𝑝2𝛼1−2 1𝑝2𝛼2 2⋅⋅⋅𝑝2𝛼𝑟 𝑟≡0 (mod 𝑛),(32) he esidue class 𝑥2=0.Bu 𝑥 =0and hence (c) does no hold. Again, ou claim ollows now by con adic ion. Co olla y 18. I and only i he condi ions o Theo em 17 a e sa is ied, hen ∀𝑥,𝑦∈Z𝑛:𝑥⊥0𝑦⇐⇒𝑛|(𝑥𝑦),(33) whe e 𝑥𝑦is he p oduc o in ege s 𝑥and 𝑦. Example 19. Le 𝐺=Z30.Since30=2⋅3⋅5, hedecomposi- ion (12)is Z30 =⟨30 2⟩⊕⟨30 3⟩⊕⟨30 5⟩ ={0,15}⊕{0,10,20}⊕{0,6,12,18,24}.(34) Fo example, since 2=0⋅15+2⋅10+2⋅6and 15=1⋅15+0⋅ 10+0⋅6,weha e2⊥015. Gene ally, (33)implies ha 𝑥⊥0𝑦 i and only i he co esponding in ege s sa is y 30|(𝑥𝑦). Example 20. Le 𝐺=Z360.Since360=23⋅32⋅5,weha e Z360 =⟨360 23⟩⊕⟨360 32⟩⊕⟨360 5⟩ ={45,90,...,315}⊕{40,80,...,320} ⊕{72,144,...,288}. (35) Fo example, 5⊥072because 5=1⋅45+8⋅40+0⋅72and 72=0⋅45+0⋅40+1⋅72.Now(33)isonlynecessa y o ⊥0 bu no su icien . Fo example, 10 ⊥036due o he ac ha 10=2⋅45+7⋅40+0⋅72and 36=4⋅45+0⋅40+3⋅72. Howe e , 360|(10⋅36). 4. Ano he De ini ion o Pe pendicula i y Da is [4] de ined pe pendicula i y as a bina y ela ion ⊥in 𝐺 sa is ying (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 ha 𝐺is an Abelian g oup, bu he de ini ion applies mo e gene ally o an Abelian monoid, oo. I is easy o see ha (D1)–(D4) a e equi alen o (A1)–(A4). Axiom (D5) a ises om in oducing he concep o “disjoin ness” on a ec o la ice; see [2,page295],[6]. In ac , ⇔canbe eplaced wi h ⇐in (D5) due o he ollowing obse a ion. P oposi ion 21. Assume ha ⊥sa is ies (D1)–(D3) (o , equi - alen ly, (A1)–(A3)). Then ∀𝑎,𝑏∈𝐺:𝑎⊥𝑏⇒{𝑎}⊥⊥ ∩{𝑏}⊥⊥ ={0}.(36) P oo . We show i s ha i 0 =𝐴⊆𝐺, hen 𝐴∩𝐴⊥={0}.(37) I 𝑥∈𝐴∩𝐴⊥, hen𝑥⊥𝑦 o all 𝑦∈𝐴.Inpa icula ,𝑥⊥𝑥, and hence 𝑥=0by (D3) and (37) ollows. Assume nex ha 𝑎⊥𝑏and le 𝑥∈{𝑎}⊥⊥ ∩{𝑏}⊥⊥.Since 𝑥⊥{𝑏}⊥and 𝑎∈{𝑏}⊥,weha e𝑥⊥𝑎implying ha 𝑥∈{𝑎}⊥. Thus 𝑥∈{𝑎}⊥∩{𝑎}⊥⊥.Bu (37)applied o𝐴={𝑎}⊥implies ha {𝑎}⊥∩{𝑎}⊥⊥ ={0}and 𝑥=0 ollows. How a e hese wo pe pendicula i ies ela ed? We gi e a pa ial answe . Le us deno e by 𝐴and 𝐷 he axioms (A1)– (A5) and (D1)–(D5), espec i ely. P oposi ion 22. I all elemen s o 𝐺ha e ini e o de , hen 𝐷⇒𝐴.I 𝐺hasa leas oneelemen o in ini eo de , hen he e exis s a ela ion ⊥sa is ying 𝐷bu no 𝐴. P oo . The i s claim ollows om P oposi ion 3.Conce ning he second one, ⊥de ined by (2) es ablishes a ela ion sa is ying 𝐷bu no (A5). P oposi ion 23. Assume ha 𝐺has elemen s 𝑎1,𝑎2,𝑎3,𝑎4=0 such ha ⟨𝑎𝑖⟩∩⟨𝑎𝑗⟩={0}whene e 𝑖 =𝑗. Then he e exis s a ela ion ⊥sa is ying 𝐴bu no 𝐷. P oo . The ela ion ⊥de ined by (5)wi h𝐻=⟨𝑎1⟩,𝐾=⟨𝑎2⟩, 𝐻1=⟨𝑎3⟩,and𝐾1=⟨𝑎4⟩sa is ies 𝐴.Since{𝑎1}⊥⊥ ∩{𝑎3}⊥⊥ = ⟨𝑎1⟩∩⟨𝑎3⟩={0}and 𝑎1 ⊥𝑎3, i does no sa is y (D5). 5. Di isibili y in Q+ I will u n ou ha pe pendicula i y has go some hing o do also wi h di isibili y in Q+.To ha end,webeginbyno icing ha e e y 𝑏∈Q+canbesaid obea a ionaldi iso o e e y 𝑎∈Q+because 𝑎=𝑐𝑏 o some 𝑐∈Q+. So, his di isibili y is i ial. In o de o be able o discuss non i ial di isibili ies in Q+, we ha e o conside which p ope ies essen ially es ablish his ela ion. The ollowing h ee ones seem qui e ob ious. Le |be a ela ion in Q+sa is ying (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 |adi isibili y in Q+. In o he wo ds, di isibili y is a e lexi e and ansi i e ela ion (i.e., a p eo de ) sa is ying (ii). 6 In e na ional Jou nal o Ma hema ics and Ma hema ical Sciences I 𝑏|𝑎, henwesay ha 𝑏is a di iso o 𝑎and ha 𝑎is di isible by 𝑏.I 𝑑|𝑎,𝑑|𝑏and 𝑐|𝑎∧𝑐|𝑏⇒𝑐|𝑑, hen𝑑is a g ea es common di iso o 𝑎and 𝑏, deno ed by gcd|(𝑎,𝑏).All hese no ions a e meaning ul also in any Abelian monoid. Le us ecall ha e e y 𝑐∈Q+canbeexp essedas 𝑐=∏ 𝑝∈P𝑝]𝑝(𝑐),(38) whe e ]𝑝(𝑐)∈Z o each 𝑝∈P, and only a ini e numbe o hem a e nonze o. I ]𝑝(𝑐) =0, hen𝑝is a p ime ac o o 𝑐. Conside he se 𝑆o all sequences (𝑛2,𝑛3,...,𝑛𝑝,...), whe e he index uns h ough P, each 𝑛𝑝∈Z,andonlya ini e numbe o hem a e nonze o. The mapping 𝑓(𝑐)=(]2(𝑐),]3(𝑐),...,]𝑝(𝑐),...) (39) is an isomo phism om (Q+,⋅)on o (𝑆,+)whe e addi ion is de ined e mwise. Fo 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) Gi en 𝑎,𝑏 ∈ Q+, we de ine hei “inne p oduc ” being he Euclidean inne p oduc o he ec o s 𝑓(𝑎)and 𝑓(𝑏): ⟨𝑎,𝑏⟩=⟨𝑓(𝑎),𝑓(𝑏)⟩=∑ 𝑝∈P ]𝑝(𝑎)]𝑝(𝑏).(41) Since only a ini e numbe o summands a e nonze o, his sum is ini e. Fo example, ⟨45, 8 25⟩=0⋅3+2⋅0+1⋅(−2)+0+0+⋅⋅⋅=−2. (42) Nex we de ine |𝑐|by se ing ]𝑝(|𝑐|)=|]𝑝(𝑐)| o all 𝑝∈ Po , equi alen ly, |𝑐|=𝑓−1((]2(|𝑐|),]3(|𝑐|),]5(|𝑐|),...)).Fo example, i 𝑐=40/63=23⋅3 −2 ⋅5 1⋅7 −1, hen|𝑐|=23⋅ 32⋅51⋅71=2520.Le ing⊥1be hesame ela ionas heone de ined by (7),i canbecha ac e izednowby 𝑎⊥1𝑏⇐⇒⟨|𝑎|,|𝑏|⟩=0. (43) Also he ela ion ⊥2in Q+, de ined by 𝑎⊥2𝑏⇐⇒⟨𝑎,𝑏⟩=0, (44) is a pe pendicula i y. We will in oduce one mo e non i ial pe pendicula i y using di isibili y. Fo ha pu pose, we i s no ice ha he ela ion 𝛿de ined by 𝑏𝛿𝑎⇐⇒∀𝑝∈P:]𝑝(𝑏)≤]𝑝(𝑎)(45) is a di isibili y, gcd𝛿(𝑎,𝑏)exis s and is unique o all 𝑎,𝑏∈Q+, and gcd𝛿(𝑎,𝑏)=∏ 𝑝∈P𝑝min(]𝑝(𝑎),]𝑝(𝑏)).(46) Assume now ha 𝑚,𝑛,𝑢,V∈Z+so ha gcd(𝑚,𝑢) = gcd(𝑛,V)=1. An al e na i e exp ession o (45)is 𝑛 V𝛿𝑚 𝑢⇐⇒ 𝑛|𝑚∧𝑢|V,(47) and ha o (46)is gcd𝛿(𝑚 𝑢,𝑛 V)=gcd (𝑚,𝑛) lcm (𝑢,V).(48) Fo example, i 𝑎=45/14=2−1 ⋅32⋅51⋅7−1 and 𝑏=33/100= 2−2 ⋅31⋅5−2 ⋅111, hengcd 𝛿(𝑎,𝑏)=2−2 ⋅31⋅5−2 ⋅7−1 =3/700. Al e na i ely, gcd𝛿(𝑎,𝑏)=gcd (45,33) lcm (14,100)=3 700.(49) Since gcd𝛿(|𝑚/𝑢|,|𝑛/V|)=gcd(𝑚𝑢,𝑛V),weha eby(9) 𝑎⊥1𝑏⇐⇒gcd𝛿(|𝑎|,|𝑏|)=1. (50) This ela ion gene alizes (10) and answe s he c y o G aham e al. in a sligh ly wide con ex han wha hey, pe haps, had hough . Eugeni and Rizzi [7, Sec ion 2] de ined di isibili y in Q+ by se ing he ela ion 𝛾so ha 𝑛 V𝛾𝑚 𝑢⇐⇒ 𝑛|𝑚∧V|𝑢. (51) Then gcd𝛾(𝑎,𝑏)always exis s and is unique, and gcd𝛾(𝑚 𝑢,𝑛 V)=gcd (𝑚,𝑛) gcd (𝑢,V).(52) Fo example, gcd𝛾(45 14,33 100)=gcd (45,33) gcd (14,100)=3 2.(53) We de ine now he co esponding pe pendicula i y by w i - ing 𝑎⊥ER 𝑏⇐⇒gcd𝛾(𝑎,𝑏)=1 ⇐⇒ gcd (𝑚,𝑛)=gcd (𝑢,V)=1. (54) Summing up, we ha e a leas h ee non i ial pe pendic- ula i ies in Q+. Le us see how hey ela e o one ano he . ⊥1 e sus ⊥2.Clea ly⊥1⇒⊥2(i.e., 𝑥⊥1𝑦⇒𝑥⊥2𝑦). The con e se does no hold. Fo example, 6⊥22/3bu 6 ⊥12/3. ⊥1 e sus ⊥ER.Clea ly⊥1⇒⊥ER.Thecon e sedoesno hold. Fo example, 2/3⊥ER3/2bu 2/3 ⊥13/2. ⊥2 e sus ⊥ER. These pe pendicula i ies a e independen . Fo example, 6⊥22/3 bu 6 ⊥ER 2/3. On he o he hand, 2/3⊥ER3/2bu 2/3 ⊥23/2. Howe e , ega ding (Z+,⋅)as a submonoid o (Q+,⋅),i is ob ious ha ⊥1=⊥ 2=⊥ ER in Z+.Mo eo e ,inZ+, hey yield he e y pe pendicula i y p oposed by G aham e al. In e na ional Jou nal o Ma hema ics and Ma hema ical Sciences 7 6. Pa allelism Pa allelism is closely ela ed o pe pendicula i y. Conside ing di e en geome ic con ex s we no ice soon ha , in gene al, pa allelism does no ha e any o he p ope ies excep hose o equi alence. Howe e , any equi alence ela ion canno be said o s and o pa allelism in any easonable way. This leads us o ask whe he i is possible o no o de ine pa allelism in Abelian g oups ha ing a pe pendicula i y so ha i makes sense. Le 𝐺ha e a pe pendicula i y ⊥and le 𝑎,𝑏 ∈ 𝐺.We say ha 𝑎and 𝑏a e pa allel and w i e 𝑎‖𝑏i {𝑎}⊥= {𝑏}⊥. The ela ion ‖is clea ly an equi alence. I 𝑎 =0, hen𝑎∦0, since {0}⊥=𝐺by P oposi ion 1(a) bu {𝑎}⊥=𝐺by (A2). All nonze o elemen s a e pa allel i and only i ⊥is i ial. I 𝐺=Z𝑛and ⊥=⊥0, hen, ecalling(19), 𝑥‖𝑦⇐⇒(∀𝑖∈{1,...,𝑟}:𝜉𝑖=0⇐⇒𝜂𝑖=0) ⇐⇒ {𝑥}⊥={𝑦}⊥=𝐻𝑖𝑖⊕⋅⋅⋅⊕𝐻𝑖𝑡,(55) whe e 𝜉𝑖=𝜂 𝑖=0⇔𝑖∈{𝑖1,...,𝑖𝑡}. Fo example, conside Z30 (see Example 19). Since 2=0⋅15+2⋅10+2⋅6and 16=0⋅15+1⋅10+1⋅6,weha e{2}⊥={16}⊥={0,15},and so 2‖16. Now, le 𝐺=Q+and le ⊥1,⊥2,and⊥ER be as be o e. Deno e he co esponding pa allelisms by ‖1,‖2,and‖ER, espec i ely. Then 𝑎‖1𝑏i and only i 𝑎and 𝑏ha e he same p ime ac o s. Fu he , 𝑚/𝑢‖ER𝑛/Vi and only i 𝑚and 𝑛ha e hesamep ime ac o sand𝑢and Vha e hesamep ime ac o s. Le us s udy how hese pa allelisms ela e o one ano he . ‖1 e sus ‖2.Weshow ha ‖2⇒‖ 1. Assume i s ha 𝑎‖2𝑏.I 𝑎∦1𝑏, hen he e exis s 𝑝0∈Psuch ha , say, ]𝑝0(𝑎) = 0and ]𝑝0(𝑏) =0.Bu now𝑝0⊥2𝑎and 𝑝0 ⊥2𝑏,and so {𝑎}⊥2={𝑏}⊥2con adic ing he assump ion. The con e se does no hold. Fo example, le 𝑎=6and 𝑏=12; hen𝑎‖1𝑏. I 𝑥=2/3, hen𝑥⊥2𝑎bu 𝑥 ⊥2𝑏, and hence {𝑎}⊥2={𝑏}⊥2.In o he wo ds, 𝑎∦2𝑏. ‖1 e sus ‖ER.Clea ly‖ER ⇒‖ 1.Thecon e sedoesno hold. Fo example, 2/3‖13/2bu 2/3∦ER3/2. ‖2 e sus ‖ER.Weshow ha ‖2⇒‖ ER.Gi en𝑝1,...,𝑝𝑡∈ P, deno e by 𝑁(𝑝1,...,𝑝𝑡) he se o such posi i e in ege s ha a eno di isiblebyany𝑝𝑖,𝑖=1,...,𝑡.Le 𝑎=𝑚/𝑢∈Q+, gcd(𝑚,𝑢)=1.Fac o ize 𝑚=𝑝𝛼1 1⋅⋅⋅𝑝𝛼ℎ ℎ,𝑢=𝑞 𝛽1 1⋅⋅⋅𝑞𝛽𝑘 𝑘,(56) whe e 𝑝1,...,𝑝ℎ,𝑞1,...,𝑞𝑘∈Pa e dis inc and 𝛼1,..., 𝛼ℎ,𝛽1,...,𝛽𝑘>0.(I 𝑚=1o 𝑢=1, hen he co esponding “emp y p oduc ” 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 “emp y sum” is ze o.) Assume ha 𝑏=𝑛/V∈Q+, gcd(𝑛,V)=1, sa is ies 𝑎‖2𝑏, ha is,{𝑎}⊥2= {𝑏}⊥2.Then,by (57), necessa ily 𝑛=𝑝𝜌1 1⋅⋅⋅𝑝𝜌ℎ ℎ,V=𝑞𝜎1 1⋅⋅⋅𝑞𝜎𝑘 𝑘,(58) whe e 𝜌1,...,𝜌ℎ,𝜎1,...,𝜎𝑘>0.Hence𝑎‖ER𝑏,and heclaim ollows. The con e se is no alid. Fo example, 2/3‖ER4/3bu 2/3∦24/3. 7. Discussion This pape began wi h a ci a ion by h ee es ablished ma he- ma icians and compu e scien is s who showed a ema kable in ui ion by p omo ing he use o he symbol o pe pendicu- la i y in numbe heo y. Indeed, we ha e p e iously seen how his no ion se les com o ably in his se ing and gains new meanings a a mo e gene al le el in he con ex o Abelian g oup heo y. We conclude his pape wi h he ollowing supplemen o hei p oposal. Le pe pendicula i y and pa allelism mean he e ⊥1and ‖1, espec i ely. Conside he “di ec ion ec o ” o 𝑐∈Q+ by (𝑐(2),𝑐(3),...,𝑐(𝑝),...),whe e𝑐(𝑝)=0i ]𝑝(𝑐)=0and 𝑐(𝑝) = 1o he wise. Fo example, he di ec ion ec o s o 45,1,and8/25a e, espec i ely (0,1,1,0,0,...),(0,0,...),and (1,0,1,0,0,...). Now, like he di ec ions o pe pendicula lines a e as di e en as possible, he p ime ac o s o pe pendicula (posi i e a ional) numbe s a e as di e en as possible; ha is, such numbe s do no ha e common p ime ac o s. In o he wo ds, he di ec ion ec o s o pe pendicula numbe s a e as di e en as possible in he sense ha hey ha e no common elemen o alue one. Like pa allel lines ha e he same di ec ion, pa allel numbe s ha e he same p ime ac o s. In o he wo ds, hei di ec ion ec o s a e equal. Finally, we no e ha pe pendicula i y can be axioma ized in a na u al way also in many o he algeb aic s uc u es. Da is [8] did ha in a ing. In a ec o space, pe pendic- ula i y is cus oma ily de ined based on an inne p oduc . Ano he possible app oach is o supplemen (A1)–(A5) wi h sui able axioms conce ning he mul iplica ion o a ec o by a scala . I migh be in e es ing o s udy unde which addi ional condi ions he e exis s an inne p oduc inducing his pe pendicula i y. Acknowledgmen The au ho s would like o hank he e e ees o ca e ully eading he pape and kind commen s. Re e ences [1] R. L. G aham, D. E. Knu h, and O. Pa ashnik, Conc e e Ma h- ema ics: A Founda ion o Compu e Science, Addison-Wesley, Reading, Mass, USA, 2nd edi ion, 1994. [2] G. Bi kho , La ice Theo y, Ame ican Ma hema ical Socie y, P o idence, RI, USA, 3 d edi ion, 1993. [3] P.Haukkanen,J.K.Me ikoski,andT.Tossa ainen,“Axioma iz- ing pe pendicula i y and pa allelism,” Jou nal o Geome y and G aphics, ol.15,no.2,pp.129–139,2011. 8 In e na ional Jou nal o Ma hema ics and Ma hema ical Sciences [4] G. Da is, “O hogonali y ela ions on abelian g oups,” Jou nal o he Aus alian Ma hema ical Socie y. Se ies A, ol.19,pp.173– 179, 1975. [5] W.K.Nicholson,In oduc ion o Abs ac Algeb a,JohnWiley & Sons, New Yo k, NY, USA, 2nd edi ion, 1999. [6] A. I. Veksle , “Linea spaces wi h disjoin elemen s and hei con e sion in o ec o la ices,” Lening adski˘ ıGosuda s enny ˘ ı Pedagogiˇ ceski˘ ı Ins i u imeni A. I. Ge cena. Uˇ cenye Zapiski, ol. 328, pp. 19–43, 1967 (Russian). [7] F. Eugeni and B. Rizzi, “An incidence algeb a on a ional numbe s,” Rendicon i di Ma ema ica, ol.12,no.3-4,pp.557– 576, 1979. [8]G.Da is,“Ringswi ho hogonali y ela ions,”Bulle in o he Aus alian Ma hema ical Socie y, ol.4,pp.163–178,1971.