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1、AnswersandHintstoSelectedProblemsProblemsofChapter11.2Foreacha∈A,theremaynotbeoneb∈AsuchthataRb.Forexample,therelation{(b,b)}ontheset{a,b}.2A2n2n2−n1.4R⊆A.So,R∈2.NumberofsuchRis2.ReflexiveR’sare2innumber.And2n(n+1)/2symmetricR’s.1.5xR−1yiffyRx.Hencetheyhavethes
2、ameequivalenceclasses.1.6Takef(x)asthenumberofelementsinx.1.7B={[x]:x∈A}.Definef:A→Bbyf(x)=[x].1.9∪A=∅.UseDeMorgan’slawtoget∩A=A.1.12f(x)=f(y)impliesx=g(f(x))=g(f(y))=y.Hencefisinjective.Now,f(g(y))=yshowsthatf(x)=yforx=g(y).Hencefissurjective.1.13(d)g◦finjecti
3、veimpliesfisinjective;gneednotbeinjective.(e)g◦fsurjectiveimpliesgissurjective;fneednotbesurjective.1.15UseProblem1.12.1.24Supposethecollectionofallsetsisaset;denoteitbyS.Eachelementof2Sisaset,andhenceisinS.Thatis,2S⊆S.Then
4、2S
5、≤
6、S
7、.ThiscontradictsCantor’stheor
8、em.1.28LetPnbethesetofallpolynomialsofdegreenwithrationalcoefficients.Definef:P→Qn+1byf(a+ax+···+axn)=(a,a,...,a).Thisfisn01n01ninjective.HencePniscountable.Thesetofallpolynomialswithrationalcoefficientsis∪n∈Z+Pn,andhence,iscountable.EachsuchpolynomialisinsomePna
9、ndhasnroots.Therefore,allrootsofsuchpolynomialsarecountableinnumber.1.33∪r<1{x∈R:a−r≤x≤a+r}={x∈R:a−10{x∈R:a−rx2>...>xn+2.ConsiderthesetA={x1+xk,x1−xk:k=2,
10、3,...,n+2}.Bythepigeonholeprinciple,atleasttwoofthemhavethesameremainderwhendividedby2n.Iftheyareyi,yj∈A,thenconsiderthedifferenceyj−yi.Thisisintheform±(xi±xj)andisdivisibleby2n.1.41Writethen+1numbersintheformx=2qky,whereyisodd.Howmanykkkyk’sarethereatthemost?
11、1.42Forx∈√R,let√'x(denotethelargestintegerlessthanorequalto√√√√x.Considerthenumbers2−'2(,22−'22(,...,(n+1)2−'(n+1)2(.Allthesearebetween0and1.Dividethe0–1intervalintonsubintervalseachoflength1/n.Bypigeonholeprinciple,atleasttwoofthesenumberslieinthesamesubinter
12、val.√Theirdifferenceisoftheformm+n2anditliesbetween0and1/n.ProblemsofChapter2Section2.22.4bbbbaa,baaaaabaaaab∈L∗.2.7(d)Yes.(i)Yes.(j)No.2.13(c)No;e.g.,(L∗)∗=L∗L,ingeneral.(d)Yes,
