rudin數(shù)學分析原理答案

1、MATH413[513](PHILLIPS)SOLUTIONSTOHOMEWK1Generallya“solution”issomethingthatwouldbeacceptableifturnedininthefmpresentedherealthoughthesolutionsgivenareoftenclosetominimalinthisrespect.A“solution(sketch)”istoosketchytobeco

2、nsideredacompletesolutionifturnedinvaryingamountsofdetailwouldneedtobefilledin.Problem1.1:Ifr∈Q0x∈RQprovethatrxrx?∈Q.Solution:Weprovethisbycontradiction.Letr∈Q0supposethatrx∈Q.ThenusingthefieldpropertiesofbothRQwehavex=(

3、rx)?r∈Q.Thusx?∈Qimpliesrx?∈Q.Similarlyifrx∈Qthenx=(rx)r∈Q.(HereinadditiontothefieldpropertiesofRQweuser?=0.)Thusx?∈Qimpliesrx?∈Q.Problem1.2:Provethatthereisnox∈Qsuchthatx2=12.Solution:Weprovethisbycontradiction.Supposeth

4、ereisx∈Qsuchthatx2=12.Writex=mninlowestterms.Thenx2=12impliesthatm2=12n2.Since3divides12n2itfollowsthat3dividesm2.Since3isprime(byuniquefactizationinZ)itfollowsthat3dividesm.Therefe32dividesm2=12n2.Since32doesnotdivide12

5、usingagainuniquefactizationinZthefactthat3isprimeitfollowsthat3dividesn.Wehaveprovedthat3dividesbothmncontradictingtheassumptionthatthefractionmnisinlowestterms.Alternatesolution(Sketch):Ifx∈Qsatisfiesx2=12thenx2isinQsat

6、isfies?x2?2=3.Nowprovethatthereisnoy∈Qsuchthaty2=3byrepeatingtheproofthat√2?∈Q.Problem1.5:LetA?Rbenonemptyboundedbelow.Set?A=?a:a∈A.Provethatinf(A)=?sup(?A).Solution:Firstnotethat?Aisnonemptyboundedabove.IndeedAcontainss

7、omeelementxthen?x∈AmeoverAhasalowerboundm?misanupperboundf?A.Wenowknowthatb=sup(?A)exists.Weshowthat?b=inf(A).That?bisalowerboundfAisimmediatefromthefactthatbisanupperboundf?A.Toshowthat?bisthegreatestlowerboundweletc?bp

8、rovethatcisnotalowerboundfA.Now?c?c.Then?x∈A?x1fixedthroughoutthement:Wewillassumeknownthatthefunctionn?→bnfromZtoRisstrictlyincreasingthatisthatfmn∈Zwehavebm0sobr=(bmq)1(nq)bs=(bnp)1(nq).Nowmqx.Ifr∈Q∩(?∞x]thenbr∈B(k)sot

9、hatbr≤bkbyPart(c).ThusbkisanupperboundfB(x).ThisshowsthatthedefinitionmakessensePart(c)showsitisconsistentwithourearlierdefinitionwhenr∈Q.(d)Provethatbxy=bxbyfallxy∈R.Solution:Indertodothiswearegoingtoneedtoreplacetheset

10、B(x)abovebythesetB0(x)=br:r∈Q∩(?∞x)(thatiswerequirer1N=123....)Proof:Clearly1isalowerbound.(Indeed(b1n)n=b1=1nsob1n1.)Weshowthat1xisnotalowerboundwhenx0.If1xwerealowerboundthen1x≤b1nwouldimply(1x)n≤(b1n)n=bfalln∈N.ByLemm

11、a1wewouldget1nx≤bfalln∈NwhichcontradictstheArchimedeanpropertywhenx0.Lemma3.supb?1n:n∈N=1.Proof:Part(b)showsthatb?1nb1n=b0=1whenceb?1n=(b1n)?1.Sinceallnumbersb?1narestrictlypositiveitnowfollowsfromLemma2that1isanupperbou