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.