MAD3105 Worksheet1Solutions
SupposeR,SarerelationsonasetA.
1.
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IfR,Sarebothreflexive,thenR∩Sisreflexive.
Proof:Supposea∈A.SinceR,SarebothreflexiveonA,(a,a)∈Rand(a,a)∈S.
Since(a,a)isinbothRandS,(a,a)∈R∩S,soR∩Sisreflexive.
2.
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IfR,Sarebothirreflexive,thenR∩Sisirreflexive.
Proof:Supposea∈A.SinceR,SarebothirreflexiveonA,(a,a)∉Rand(a,a)∉S.
Since(a,a)isinneitherRnorS,(a,a)∉R∩S,soR∩Sisirreflexive.
3.
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IfR,Sarebothsymmetric,thenR∩Sissymmetric.
Proof:Suppose(a,b)∈R∩S.
Because(a,b)∈R∩S,(a,b)∈Rand(a,b)∈S.
SinceR,Sarebothsymmetric(b,a)∈Rand(b,a)∈S.
Since(b,a)∈Rand(b,a)∈S,(b,a)∈R∩S,soR∩Sissymmetric.
4.
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IfR,Sarebothantisymmetric,thenR∩Sisantisymmetric.
Proof:Suppose(a,b)∈R∩Sand(b,a)∈R∩S.Weneedtoshowthata=b.
Because(a,b)∈R∩Sand(b,a)∈R∩S,weknowand(a,b)∈Rand(b,a)∈R,andsinceRis
antisymmetric,a=b.
Wehaveshownthatwhenever(a,b)∈R∩Sand(b,a)∈R∩S,a=b,soR∩Sisantisymmetric.
5.
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IfR,Sarebothasymmetric,thenR∩Sisasymmetric.
Proof:Suppose(a,b)∈R∩S.Weneedtoshowthat(a,b)∉R∩S.
Because(a,b)∈R∩S,weknowand(a,b)∈R,and,sinceRisasymmetric,weknowthat(b,a)∉R.
Since(b,a)∉R,weknowthat(b,a)∉R∩S.
Wehaveshownthatwhenever(a,b)∈R∩S,(b,a)∉R∩S,soR∩Sisasymmetric.
6.
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IfR,Sarebothtransitive,thenR∩Sistransitive.
Proof:Suppose(a,b)∈R∩Sand(b,c)∈R∩S.Weneedtoshowthat(a,c)∈R∩S.
Because(a,b)∈R∩S,weknowand(a,b)∈Rand(a,b)∈S.
Because(b,c)∈R∩S,weknowand(b,c)∈Rand(b,c)∈S.
Since(a,b)∈Rand(b,c)∈RandRistransitiveweknowthat(a,c)∈R.
Since(a,b)∈Sand(b,c)∈SandSistransitiveweknowthat(a,c)∈S.
Since(a,c)∈Rand(a,c)∈S,(a,c)∈R∩S.
Wehaveshownthatwhenever(a,b)∈R∩S,and(b,c)∈R∩S,(a,c)∈R∩S,soR∩Sistransitive.
SupposeR,SarerelationsonasetA.
7.
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IfR,Sarebothreflexive,thenR∪Sisreflexive.
Proof:Supposea∈A.
SinceRisreflexive,(a,a)∈R.
Since(a,a)∈R,(a,a)∈R∪S,soR∪Sisreflexive.
8.
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IfR,Sarebothirreflexive,thenR∪Sisirreflexive.
Proof:Supposea∈A.
SinceRisirreflexive,(a,a)∉R.
SinceSisirreflexive,(a,a)∉S.
Since(a,a)∉R,and(a,a)∉S,(a,a)∉R∪S,soR∪Sisirreflexive.
9.
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IfR,Sarebothsymmetric,thenR∪Sissymmetric.
Proof:(a,b)∈R∪S.Wemustshow(b,a)∈R∪S.
Then(a,b)∈Ror(a,b)∈S.
So,(b,a)∈Ror(b,a)∈S,becauseRandSarebothsymmetric.
Since(b,a)∈Ror(b,a)∈S,(a,b)∈R∪S.
Wehaveshownthatwhenever(a,b)∈R∪S,(b,a)∈R∪S,so∈R∪Sissymmetric.
10. T
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IfR,Sarebothantisymmetric,thenR∪Sisantisymmetric.
Counterexample:LetA={1,2,3},letR={(1,2)},letS={(2,1)}.
NotethatRandSarebothantisymmetric.
ThenR∪S={(1,2),(2,1)},whichisnotantisymmetric.
11. T
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IfR,Sarebothasymmetric,thenR∪Sisasymmetric.
Counterexample:LetA={1,2,3},letR={(1,2)},letS={(2,1)}.
NotethatRandSarebothasymmetric.
ThenR∪S={(1,2),(2,1)},whichisnotasymmetric.
12. T
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IfR,Sarebothtransitive,thenR∪Sistransitive.
Counterexample:LetA={1,2,3},letR={(1,2)},letS={(2,1)}.
NotethatRandSareboth(vacuously)transitive.
ThenR∪S={(1,2),(2,1)},whichisnottransitive,because,forinstance,1isrelatedto2and2isrelated
to1but1isnotrelatedto1.
SupposeR,SarerelationsonasetA.
13. T
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IfR,Sarebothreflexive,thenR–Sisreflexive.
Counterexample:LetA={1,2},R={(1,1),(2,2),(1,2)},S={(1,1),(2,2)}.RandSarebothreflexive,but
R–S={(1,2)},whichisnotreflexive.
14. T
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IfR,Sarebothirreflexive,thenR–Sisirreflexive.
Proof:SupposeR,Sarebothirreflexive,andleta∈A.SinceRisirreflexive,(a,a)∉R,andso(a,a)∉R–S.
15. T
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IfR,Sarebothsymmetric,thenR–Sissymmetric.
Proof:Suppose(a,b)∈R–S.Then(a,b)∈Rbut(a,b)∉S.
Since(a,b)∈RandRissymmetric,wehave(b,a)∈R.
Weneedtoshow(b,a)∉S.
Suppose(b,a)∈S.
Then,becauseSissymmetric,wehave(a,b)∈S.Butthen(a,b)∈Rand(a,b)∈S,contradicting
assumptionthat(a,b)∈R–S.
Wehaveshownthat(b,a)∈R–Swhenever(a,b)∈R–S,soR–Sissymmetric.
16. T
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IfR,Sarebothantisymmetric,thenR–Sisantisymmetric.
Proofbycontraposition:supposeR–Sisnotantisymmetric.Thentherearedistincta,bsuch(a,b)∈R–S
and(b,a)∈R–S.Butthen(a,b)∈Rand(b,a)∈R,soRisnotantisymmetric.
17. T
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IfR,Sarebothasymmetric,thenR–Sisasymmetric.
Proofbycontraposition:supposeR–Sisnotasymmetric.Thentherearea,bsuch(a,b)∈R–S
and(b,a)∈R–S.Butthen(a,b)∈Rand(b,a)∈R,soRisnotasymmetric.
18. T
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IfR,Sarebothtransitive,thenR–Sistransitive.
Counterexample:LetA={1,2,3},letR={(1,2),(2,3),(1,3)},letS={(1,3)}.
NotethatRistransitiveandSis(vacuously)transitive.
ThenR–S={(1,2),(2,3)},whichisnottransitive.
LetRbearelationonasetA.
19. T
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IfRisreflexive,thenR–1isreflexive.
Proof:SupposeRisreflexiveanda∈Aisarbitrary.SinceRisreflexive,(a,a)∈R,andthenswitchingthe
twocoordinatesweget(a,a)∈R–1,soR–1isreflexive.
20. T
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IfRisirreflexive,thenR–1isirreflexive.
Proof:SupposeRisnotirreflexive.Thenthereisa∈Asuchthat(a,a)∈R;so,switchingthetwo
coordinates,weget(a,a)∈R,soRisnotirreflexive.
21. T
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IfRissymmetric,thenR–1issymmetric.
Proof:SupposeRissymmetricand(x,y)isanarbitraryelementofR–1.ThenbydefintionofR–1,(y,x)∈R,
andsinceRissymmetric(x,y)∈R;since(x,y)∈R,bydefinition(y,x)∈R–1,soR–1issymmetric.
22. T
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IfRisantisymmetric,thenR–1isantisymmetric.
Proofbycontraposition:SupposeR–1isnotantisymmetric.Thentherearedistinctx,ysuchthat
(x,y)∈R–1and(y,x)∈R–1.Butthismeansthat(y,x)∈Rand(x,y)∈R,x≠y,soRisnotantisymmetric.
23. T
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IfRisasymmetric,thenR–1isasymmetric.
Proofbycontraposition:SupposeR–1isnotasymmetric.Thentherearex,ysuchthat
(x,y)∈R–1and(y,x)∈R–1.Butthismeansthat(y,x)∈Rand(x,y)∈R,soRisnotasymmetric.
24. T
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IfRistransitive,thenR–1istransitive.
Proof:Suppose(x,y)∈R–1and(y,z)∈R–1forsomearbitraryx,y.Weneedtoshowthatthat(x,z)∈R–1.
Since(x,y)∈R–1and(y,z)∈R–1,wehavethat(y,x)∈Rand(z,y)∈R,and,sinceRistransitive,(z,x)∈R.
Since(z,x)∈R,wehave(x,z)∈R–1,whichiswhatweneededtoprove.
25.
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IfRisreflexive,then!R isreflexive.
( )
Counterexample:LetA={1}andletR={(1,1)};then 1,1 ∉R ,so!R isnotreflexive.
!
26.
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IfRisirreflexive,then!R isirreflexive.
( )
Counterexample:LetA={0,1}andletR={(1,0)};notethatRisirreflexive,but 1,1 ∈R ,so!R isnot
!
irreflexive.
27.
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IfRissymmetric,then!R issymmetric.
( )
Proofbycontraposition:Suppose!R isnotsymmetric.Thentherearex,ysuchthat x , y ∈R but
!
( y,x ) ∉R .Thismeans,however,that!( y,x ) ∈R but!( x , y ) ∉R ,soRisnotsymmetric.
!
28.
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IfRisantisymmetric,then!R isantisymmetric.
( )
( )
Counterexample:LetA={1,2,3},R={(1,3)}.NotethatRisantisymmetric, 1,2 ∈R ,and 2,1 ∈R ,so!R !
!
isnotantisymmetric.
29. T
F
IfRisasymmetric,thenR–1asymmetric.
( )
( )
Counterexample:LetA={1,2,3},R={(1,3)}.NotethatRisasymmetric, 1,2 ∈R ,and 2,1 ∈R ,so!R is
!
!
notasymmetric.
30.
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IfRistransitive,then!R istransitive.
( )
Counterexample:LetA={1,2},R={(1,1)}.NotethatRisvacuouslytransitive.Notethat 1,2 ∈R ,and
!
2,1 ∈R ,but 1,1 ∉R ,soisnot!R transitive.
!
!
( )
( )
31.Usethefactthatcompositionisassociative,andinduction,toprove R n ! R = R ! R n forn=1,2,3,
…
Proofofbasis:Weneedtoprove R1 ! R = R ! R1 . Bydefinition, R1 = R ,so R1 ! R = R ! R = R ! R1 .
Proofofinductive:Assume R k ! R = R ! R k forsomek≥1.Then
R k+1 ! R = R k ! R ! R (definitionofRn)
(
)
= ( R ! R ) ! R = R ! ( R ! R ) k
(inductivehypothesis)
k
(becausecompositionisassociative)
= R ! R k+1 (definitionofRn)
32. Provethat(x,y)∈RniffonthedigraphforRthereisadirectedpathoflengthnfromxtoy.
Proofbyinduction
Basis:Note(x,y)∈R1ifandonlyifthereisadirectededge(thatis,adirectedpathoflength1)fromxtoy.
Inductive:Assumethat(x,y)∈Rkifandonlyifthereisadirectedpathoflengthkfromxtoy.
Supposethatthereisadirectedpathoflengthk+1fromxtoy.Ifwedeletethelastedge,thenwehavea
directedpathoflengthkfromxtosomevertexv,andthedeletededgeisadirectedpathoflengthone
fromvtoy.
Sincewehaveadirectedpathoflengthkfromxtov,ourinductivehypothesisstatesthattheorderedpair
(x,v)∈Rk,andthedirectededgefromvtoymeansthat(v,y)∈R,sobycomposition,(x,y)∈Rk+1.
Conversely,suppose(x,y)∈Rk+1.Then,bydefinitionofRn,thereisvsuchthat(x,v)∈Rkand(v,y)∈R.
Since(x,v)∈Rk,ourinductivehypothesisstatesthatthereisadirectedpathoflengthkfromxtov.
Since(v,y)∈R,thedirectededgefromvtoyisadirectedpathoflength1fromvtoy.Concatenating
thosepathsgivesusadirectedpathoflengthk+1fromxtoy.
SupposeR,SarerelationsonasetA.
33. T
F
IfR,Sarebothreflexive,then!S !R isreflexive.
Proof:Leta∈A.SinceRisreflexive,(a,a)∈R;sinceSisreflexive,(a,a)∈S;so,(a,a)∈!S !R .
34. T
F
IfR,Sarebothirreflexive,then!S !R isirreflexive.
Counterexample:LetA={1,2},R={(1,2)},S={(2,1)}.NotethatR,Sareirrefelxive,but(1,1)∈!S !R ,so
!S !R isnotirreflexive.
35. T
F
IfR,Sarebothsymmetric,then!S !R issymmetric.
Counterexample:LetA={1,2,3},R={(1,2),(2,1)},S={(2,3),(3,2)}.
NotethatRandSarebothsymmetric,but!S !R ={(1,3)}isnotsymmetric.
36. T
F
IfR,Sarebothantisymmetric,then!S !R isantisymmetric.
Counterexample:LetA={1,2,3},R={(1,2),(3,2)},S={(2,3),(2,1)}.
ThenR,Sarebothantisymmetricbut!S !R ={(1,3),(3,1)}isnotantisymmetric.
37. T
F
IfR,Sarebothantisymmetric,then!S !R isasymmetric.
Counterexample:LetA={1,2,3},R={(1,2),(3,2)},S={(2,3),(2,1)}.
ThenR,Sarebothasymmetricbut!S !R ={(1,3),(3,1)}isnotasymmetric.
38. T
F
IfR,Sarebothtransitive,then!S !R istransitive.
Counterexample:LetA={1,2,3,4,5},letR={(1,5),(2,4)},letS={(5,2),(4,3)}.
NotethatRandSarevacuouslytransitivebut!S !R ={(1,2),(2,3)}isnottransitive.
SupposeR,SarerelationsonasetA.
39. T
F
IfR,Sarebothreflexive,then!S ⊕ R isreflexive.
Counterexample:LetA={1,2},R={(1,1),(2,2),(1,2)},S={(1,1),(2,2),(2,1)}.
RandSarereflexivebut!S ⊕ R isn’treflexive.
40. T
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IfR,Sarebothirreflexive,then!S ⊕ R isirreflexive.
Proof:SupposeR,Sareirreflexiveanda∈A.Wehave(a,a)∉Rand(a,a)∉S,so(a,a)∉R⊕S.
41. T
F
IfR,Sarebothsymmetric,then!S ⊕ R issymmetric.
Proof:SupposeR,Saresymmetricand(a,b)∈R⊕S.Then(a,b)∈Rand(a,b)∉Sor(a,b)∈Sand(b,a)∉R.
Incase(a,b)∈Rand(a,b)∉S,notethat(b,a)∈RbecauseRissymmetric,and(b,a)∉Sbecause,sinceSis
symmetric,if(b,a)wereinS,(a,b)mustalsobeinS;thisshowsthatif(a,b)∈Rand(a,b)∉Sthen
(b,a)∈R⊕S.Incase(a,b)∈Sand(a,b)∉R,asimilarproofshows(b,a)∈R⊕S.Therefore,R⊕Sis
symmetric.
42. T
F
IfR,Sarebothantisymmetric,then!S ⊕ R isantisymmetric.
Counterexample:LetA={1,2},R={(1,2)},S={(2,1)}.
ThenR,Sarebothantisymmetricbut!S ⊕ R ={(1,2),(2,1)}isnotantisymmetric.
43. T
F
IfR,Sarebothasymmetric,then!S ⊕ R isasymmetric.
Counterexample:LetA={1,2},R={(1,2)},S={(2,1)}.
ThenR,Sarebothasymmetricbut!S ⊕ R ={(1,2),(2,1)}isnotasymmetric.
44. T
F
IfR,Sarebothtransitive,then!S ⊕ R istransitive.
Counterexample:LetA={1,2},R={(1,2)},S={(2,1)}.
ThenR,Sareboth(vacuously)transitive,but!S ⊕ R ={(1,2),(2,1)}isnottransitive.
45. T
F
Ristransitive,ifandonlyifRn⊆Rforn=1,2,3,…
First,weproveRn⊆RimpliesRistransitive:
SupposeRn⊆Rforn=1,2,3,…Let(a,b)∈Rand(b,c)∈R.Then(a,c)∈R2accordingtothedefinitionof
R2.ButourhypothesissaysthatR2⊆R,so(a,c)∈R.ThisshowsthatRistransitive.
Toprovetheconverse,wewilluseinduction.
Hereistheproofofinductivestep:SupposewehavetransitiveR,andRk⊆Rforsomek.Toshowthat
Rk+1⊆RweshowthatanyorderedpairinRk+1isalsoinR.
Let(a,b)∈Rk+1.Thenthereiscsuchthat(a,c)∈Rand(c,b)∈Rk.OurinductivehypothesissaysRk⊆R,
so(c,b)∈R.ButRistransitive,so(a,c)∈Rand(c,b)∈Rimplies(a,b)∈Randtheproofiscomplete.
SupposeRisarelationonasetA.
46. T
F
IfRisreflexive,thenRnisreflexiveforn≥2.
Proofofinductivestep:AssumeRkisreflexiveforsomeforsomek≥2,andleta∈A.SinceRisreflexive,
(a,a)∈R;sinceRkisreflexive,(a,a)∈Rk;so,(a,a)∈Rk+1=!Rk !R .
47. T
F
IfRisirreflexive,thenRnisirreflexiveforn≥2.
Counterexample:LetA={1,2},R={(1,2),(2,1)}.ThenR2={(1,1),(2,2)}.
48. T
F
IfRissymmetric,thenRnissymmetricforn≥2.
Proofofinductivestep:SupposeRkissymmetricforsomesymmetricrelationR,andlet(a,b)∈Rk+1.
BydefinitionofRk+1,thereiscsuch(a,c)∈Rand(c,b)∈Rk.SinceRandRkarebothsymmetric,(c,a)∈R
and(b,c)∈Rk,respectively,so b,a ∈R!Rk = Rk !R = Rk+1 !
Notethatthisproofmadeuseofatheoremprovedearlier:!R!Rn = Rn !R forallrelationsRonasetA.
49. T
F
IfRisantisymmetric,thenRnisantisymmetricforn≥2.
Counterexample:LetA={1,2,3,4},R={(1,2),(2,3),(3,4),(4,1)}.
ThenRisantisymmetricbutR2={(1,3),(2,4),(3,1),(4,2)}isnotantisymmetric.
50. T
F
IfR,Sarebothasymmetric,then!S ⊕ R isasymmetric.
Counterexample:LetA={1,2,3,4},R={(1,2),(2,3),(3,4),(4,1)}
ThenRisasymmetricbutR2={(1,3),(2,4),(3,1),(4,2)}isnotasymmetric.
51. T
F
IfRistransitive,thenRnistransitiveforn≥2.Usethistheoreminyourproof:
RistransitiveifandonlyifRn⊆Rforn=1,2,3,…
First,weuseinductiontoprovethatifRistransitive,thenRnistransitive.
Basisstep:Nothingtoprove,becauseR1=Rwhichistransitivebyhypothesis.
Proofofinductivestep.SupposeRkistransitiveforsometransitiverelationR,andlet(a,b)∈Rk+1.
(b,c)∈Rk+1.Weneedtoshow(a,c)∈Rk+1.
First,since(a,b)∈Rk+1andRk+1⊆R,wehave(a,b)∈R.
Next,since(b,c)∈Rk+1thereisdsuchthat(b,d)∈Rand(d,c)∈Rk.
Also,since(a,b)∈R,(b,d)∈RandRistransitive,wehave(a,d)∈R.
Finally,since(a,d)∈Rand(d,c)∈Rk,wehave(a,c)∈Rk+1.
Conversely,wemustprovethatifRn⊆ Rforalln≥1,thenRistransitive.
Thishalfoftheproofdoesn’trequireinduction.
Proof:SupposeRn⊆Rforalln≥1,andlet(a,b)∈R,(b,c)∈R.Weneedtoshow(a,c)∈R.
AccordingtothedefinitionofR2,since(a,b)∈R,(b,c)∈Rweknow(a,c)∈R2.
OurhypothesissaysifRn⊆Rforalln≥1,so,inparticular,ifR2⊆R.
Since(a,c)∈R2andR2⊆Rwehave(a,c)∈R,whichiswhatweneedtoprove.
( )
52. T
F
IfRisasymmetricrelationonA,thenr(R)issymmetric.
Proof:Suppose(x,y)∈r(R)=R∪Δ.
Then(x,y)∈Ror(x,y)∈Δ.
If(x,y)∈Rthen(y,x)∈RbecauseRissymmetric,andso(y,x)∈R∪Δ=r(R).
If(x,y)∈Δthenx=yso(x,y)=(x,x)=(y,x)∈Δandso(y,x)∈R∪Δ=r(R).
53. T
F
IfRisasymmetricrelationonA,thent(R)issymmetric.
Proof:Suppose(x,y)∈t(R).Then,bydefinitionoft(R),(x,y)∈Rnforsomen≥1.SinceRissymmetric,Rn
isalsosymmetricforalln≥1(weprovedthisawhileago),so(y,x)∈Rn.Since(y,x)∈Rn,thisimpliesthat
(y,x)∈t(R),andsot(R)issymmetric.
54. T
F
IfRisatransitiverelationonA,thenr(R)istransitive.
Proof:Suppose(x,y)∈r(R)=R∪Δand(y,z)∈r(R)=R∪Δ.
Case1:(x,y)∈Rand(y,z)∈R.Inthiscase(x,z)∈RsinceRistransitive,andso(x,z)∈R∪Δ=r(R).
Case2:(x,y)∈Δand(y,z)∈Δ.Inthiscasex=y=zandso(y,z)∈Δ⊆r(R).
Case3:(x,y)∈Rand(y,z)∈Δ.Inthiscasey=zandso(x,z)=(x,y)∈R⊆r(R).
Case4:(x,y)∈Δand(y,z)∈R.Inthiscasex=yandso(x,z)=(y,z)∈R⊆r(R).
55. T
F
IfRisatransitiverelationonA,thens(R)istransitive.
Counterexample:LetA={1,2,3}andletRbethevacuouslytransitiverelation{(1,2)}.Thens(R)={(1,2),
(2,1)},whichisnottransitivebecause(1,2)∈s(R)and(2,1)∈s(R)but(1,1∉)∈s(R).
56. T
F
IfR⊆SthenRn⊆Snforn=1,2,3,…
Proofofinductivestep:SupposeR⊆SandRk⊆Skforsomek.
Let(x,y)∈Rk+1.Thenthereisasuchthat(x,a)∈Rand(a,y)∈Rk.
But,sinceR⊆S,wehave(x,a)∈S,andsinceRk⊆Skwehave(a,y)∈Sk,so,accordingtothedefinitionofSn,
wehave(x,y)∈Sk+1.
57. T
F
IfR⊆SandT⊆Uthen!R!T ⊆ S !U .
( )
Proof:Suppose x , y ∈R!T .Thenthereisasuchthat(x,a)∈Rand(a,y)∈T.ButsinceR⊆S,wehave
!
(x,a)∈S,andlikewisesinceT⊆Uwehave(a,y)∈U,so, x , y ∈S !U .
!
( )
58.SupposeS,RarerelationsfromAtoB.Prove:
a.(R–1)–1=R
Proof:(x,y)∈(R–1)–1⇔(y,x)∈R–1⇔(x,y)∈R.
b.(R∪S)–1=R–1∪S–1
Proof:(x,y)∈(R∪S)–1
⇔(y,x)∈R∪S
⇔(y,x)∈Ror(y,x)∈S
⇔(x,y)∈R–1or(x,y)∈S–1
(x,y)∈R–1∪S–1
c.(R∩S)–1=R–1∩S–1
Proof:(x,y)∈(R∩S)–1
⇔(y,x)∈R∩S
⇔(y,x)∈Rand(y,x)∈S
⇔(x,y)∈R–1and(x,y)∈S–1
⇔(x,y)∈R–1∩S–1
d.(A×B)–1=B×A
Proof:(x,y)∈(A×B)–1
⇔(y,x)∈A×B
⇔y∈Aandx∈B
⇔(x,y)∈B×A
e.∅–1=∅
Proof:Weuseproofbycontradictiontoshowthat∅–1isempty.
Supposethereisanorderedpair(x,y)in∅–1.Thentheorderedpair(y,x)isin∅.
Thiscontradictsthefactthat∅isempty.Hence∅–1isempty,so∅–1=∅.
()
f. R
!
−1
= R −1 Proof
( x , y ) ∈( R)
!
−1
( )
( )
( )
( )
⇔ y,x ∈R ⇔ y,x ∉R ⇔ x , y ∉R −1 ⇔ x , y ∈R −1 (
g. R − S
!
Proof:
)
−1
= R −1 − S −1 ( x , y ) ∈( R − S )
⇔ ( y,x ) ∈R − S
⇔ ( y,x ) ∈R ∧ ( y,x ) ∉S
⇔ ( x , y ) ∈R ∧ ( x , y ) ∉S
⇔ ( x , y ) ∈R − S
!
−1
−1
−1
−1
−1
(
59.IfR,SarerelationsonasetAthen R! S
!
Proof:
( ) (
)
)
−1
= S −1 !R −1 .
( )
−1
(
)
Notethat x , y ∈ R! S ifandonlyif y,x ∈R! S ,ifandonlyifthereiswsuchthat y,w ∈S and
!
!
!
w,x ∈R. !
Butthisisequivalentto w, y ∈S −1 and x ,w ∈R −1 ,so x , y ∈S −1 !R −1 .
!
!
!
60.IfR,SarerelationsonasetAandR⊆S,thenR–1⊆S–1.
Proof:SupposeR⊆Sand(x,y)∈R–1.Then(y,x)∈R.Since(y,x)∈RandR⊆Swehave(y,x)∈S.Since
(y,x)∈Swehave(x,y)∈S–1,andtheproofiscomplete.
61.RefertothisdigraphofarelationRonthesetA={a,b,c,d,e}
a.TF(c,d)∈R3
YoucanconfirmthiseitherbyfindingthematrixforR3,orby
observingthatthereisnopathoflength3fromctod.
b.TF(e,a)∈R5
YoucanconfirmthiseitherbyfindingthematrixforR5,orby
observingthate,c,e,c,e,aisapathoflength5frometoa.
c.TF(b,a)∈R4
YoucanconfirmthiseitherbyfindingthematrixforR5,orby
observingthatb,e,c,e,aisapathoflength4frombtoa.
d.TF(c,c)∈R5
YoucanconfirmthiseitherbyfindingthematrixforR5,orby
observingthatisnopathoflength5fromctoc.
( )
(
)
( )
( )