30数学分析原理Rudin完整答案

MATH413[513](PHILLIPS)SOLUTIONSTOHOMEWORK1Generally,a\solutionissomethingthatwouldbeacceptableifturnedinintheformpresentedhere,althoughthesolutionsgivenareoftenclosetominimalinthisrespect.A\solution(sketch)istoosketchytobeconsideredacompletesolutionifturnedin;varyingamountsofdetailwouldneedtobe¯lledin.Problem1.1:Ifr2Qnf0gandx2RnQ,provethatr+x;rx62Q.Solution:Weprovethisbycontradiction.Letr2Qnf0g,andsupposethatr+x2Q.Then,usingthe¯eldpropertiesofbothRandQ,wehavex=(r+x)¡r2Q.Thusx62Qimpliesr+x62Q.Similarly,ifrx2Q,thenx=(rx)=r2Q.(Here,inadditiontothe¯eldpropertiesofRandQ,weuser6=0.)Thusx62Qimpliesrx62Q.Problem1.2:Provethatthereisnox2Qsuchthatx2=12.Solution:Weprovethisbycontradiction.Supposethereisx2Qsuchthatx2=12.Writex=mninlowestterms.Thenx2=12impliesthatm2=12n2.Since3divides12n2,itfollowsthat3dividesm2.Since3isprime(andbyuniquefactorizationinZ),itfollowsthat3dividesm.Therefore32dividesm2=12n2.Since32doesnotdivide12,usingagainuniquefactorizationinZandthefactthat3isprime,itfollowsthat3dividesn.Wehaveprovedthat3dividesbothmandn,contradictingtheassumptionthatthefractionmnisinlowestterms.Alternatesolution(Sketch):Ifx2Qsatis¯esx2=12,thenx2isinQandsatis¯es¡x2¢2=3.Nowprovethatthereisnoy2Qsuchthaty2=3byrepeatingtheproofthatp262Q.Problem1.5:LetA½Rbenonemptyandboundedbelow.Set¡A=f¡a:a2Ag.Provethatinf(A)=¡sup(¡A).Solution:Firstnotethat¡Aisnonemptyandboundedabove.Indeed,Acontainssomeelementx,andthen¡x2A;moreover,Ahasalowerboundm,and¡misanupperboundfor¡A.Wenowknowthatb=sup(¡A)exists.Weshowthat¡b=inf(A).That¡bisalowerboundforAisimmediatefromthefactthatbisanupperboundfor¡A.Toshowthat¡bisthegreatestlowerbound,weletc¡bandprovethatcisnotalowerboundforA.Now¡cb,so¡cisnotanupperboundfor¡A.Sothereexistsx2¡Asuchthatx¡c.Then¡x2Aand¡xc.SocisnotalowerboundforA.Problem1.6:Letb2Rwithb1,¯xedthroughouttheproblem.Comment:Wewillassumeknownthatthefunctionn7!bn,fromZtoR,isstrictlyincreasing,thatis,thatform;n2Z,wehavebmbnifandonlyifmn.Similarly,wetakeasknownthatx7!xnisstrictlyincreasingwhennisDate:1October2001.12MATH413[513](PHILLIPS)SOLUTIONSTOHOMEWORK1anintegerwithn0.Wewillalsoassumethattheusuallawsofexponentsareknowntoholdwhentheexponentsareintegers.Wecan'tassumeanythingaboutfractionalexponents,exceptforTheorem1.21ofthebookanditscorollary,becausethecontextmakesitclearthatwearetoassumefractionalpowershavenotyetbeende¯ned.(a)Letm;n;p;q2Z,withn0andq0.Provethatifmn=pq,then(bm)1=n=(bp)1=q.Solution:BytheuniquenesspartofTheorem1.21ofthebook,appliedtothepositiveintegernq,itsu±cestoshowthath(bm)1=ninq=h(bp)1=qinq:Nowthede¯nitioninTheorem1.21impliesthath(bm)1=nin=bmandh(bp)1=qiq=bp:Therefore,usingthelawsofintegerexponentsandtheequationmq=np,wegeth(bm)1=ninq=hh(bm)1=niniq=(bm)q=bmq=bnp=(bp)n=hh(bp)1=qiqin=h(bp)1=qinq;asdesired.ByPart(a),itmakessensetode¯nebm=n=(bm)1=nform;n2Zwithn0.Thisde¯nesbrforallr2Q.(b)Provethatbr+s=brbsforr;s2Q.Solution:Choosem;n;p;q2Z,withn0andq0,suchthatr=mnands=pq.Thenr+s=mq+npnq.BytheuniquenesspartofTheorem1.21ofthebook,appliedtothepositiveintegernq,itsu±cestoshowthathb(mq+np)=(nq)inq=h(bm)1=n(bp)1=qinq:Directlyfromthede¯nitions,wecanwritehb(mq+np)=(nq)inq=·hb(mq+np)i1=(nq)¸nq=b(mq+np):Usingthelawsofintegerexponentsandthede¯nitionsforrationalexponents,wecanrewritetherighthandsideash(bm)1=n(bp)1=qinq=hh(bm)1=niniqhh(bp)1=qiqin=(bm)q(bp)n=b(mq+np):Thisprovestherequiredequation,andhencetheresult.(c)Forx2R,de¯neB(x)=fbr:r2Q\(¡1;x]g:Provethatifr2Q,thenbr=sup(B(r)).Solution:Themainpointistoshowthatifr;s2Qwithrs,thenbrbs.Choosem;n;p;q2Z,withn0andq0,suchthatr=mnands=pq.ThenMATH413[513](PHILLIPS)SOLUTIONSTOHOMEWORK13alsor=mqnqands=npnq,withnq0,sobr=(bmq)1=(nq)andbs=(bnp)1=(nq):Nowmqnpbecausers.Therefore,usingthede¯nitionofc1=(nq),(br)nq=bmqbnp=(bs)nq:Sincex7!xnqisstrictlyincreasing,thisimpliesthatbrbs.Nowwecanprovethatifr2Qthenbr=sup(B(r)).Bytheabove,ifs2Qands·r,thenbs·br.ThisimpliesthatbrisanupperboundforB(r).Sincebr2B(r),obviouslynonumbersmallerthanbrcanbeanupperboundforB(r).Sobr=sup(B(r)).Wenowde¯nebx=sup(B(x))foreveryx2R.WeneedtoshowthatB(x)isnonemptyandboundedabove.Toshowitisnonempty,choose(usingtheArchimedeanproperty)somek2Zwithkx;thenbk2B(x).Toshowitisboundedabove,similarlychoosesomek2Zwithkx.Ifr2Q\(¡1;x],thenbr2B(k)sothatbr·bkbyPart(c).ThusbkisanupperboundforB(x).Thisshowsthatthede¯nitionmakessense,andPart(c)showsitisconsistentwithourearlierde¯nitionwhenr2Q.(d)Provethatbx+y=bxbyforallx;y2R.Solution:Inordertodothis,wearegoingtoneedtoreplacethesetB(x)abovebythesetB0(x)=fbr:r2Q\(¡1;x)g(thatis,werequirerxratherthanr·x)inthede¯nitionofbx.(Ifyouareskeptical,readthemainpartofthesolution¯rsttoseehowthisisused.)Weshowthatthereplacementispossibleviasomelemmas.Lemma1.Ifx2[0;1)andn2Zsatis¯esn¸0,then(1+x)n¸1+nx.Proof:Theproofisbyinductiononn.Thestatementisobviousforn=0.Soassumeitholdsforsomen.Then,sincex¸0,(1+x)n+1=(1+x)n(1+x)¸(1+nx)(1+x)=1+(n+1)x+nx2¸1+(n+1)x:Thisprovestheresultforn+1.Lemma2.inffb1=n:n2Ng=1.(Recallthatb1andN=f1;2;3;:::g.)Proof:Clearly1isalowerbound.(Indeed,(b1=n)n=b1=1n,sob1=n1.)Weshowthat1+xisnot