rudin-w-solution-manual-of-principles-of-mathematical-analysis.pdf

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1、MATH413[513](PHILLIPS)SOLUTIONSTOHOMEWORK1Generally,asolution"issomethingthatwouldbeacceptableifturnedinintheformpresentedhere,althoughthesolutionsgivenareoftenclosetominimalinthisrespect.Asolution(sketch)"istoosketchytobeconsideredacompletesolutionif

2、turnedin;varyingamountsofdetailwouldneedtobe¯lledin.Problem1.1:Ifr2Qnf0gandx2RnQ,provethatr+x;rx62Q.Solution:Weprovethisbycontradiction.Letr2Qnf0g,andsupposethatr+x2Q.Then,usingthe¯eldpropertiesofbothRandQ,wehavex=(r+x)¡r2Q.Thusx62Qimpliesr+x62Q.Similar

3、ly,ifrx2Q,thenx=(rx)=r2Q.(Here,inadditiontothe¯eldpropertiesofRandQ,weuser6=0.)Thusx62Qimpliesrx62Q.Problem1.2:Provethatthereisnox2Qsuchthatx2=12.Solution:Weprovethisbycontradiction.Supposethereisx2Qsuchthatx2=12.Writex=minlowestterms.Thenx2=12impliesth

4、atm2=12n2.nSince3divides12n2,itfollowsthat3dividesm2.Since3isprime(andbyuniquefactorizationinZ),itfollowsthat3dividesm.Therefore32dividesm2=12n2.Since32doesnotdivide12,usingagainuniquefactorizationinZandthefactthat3isprime,itfollowsthat3dividesn.Wehavep

5、rovedthat3dividesbothmandn,contradictingtheassumptionthatthefractionmisinlowestterms.nAlternatesolution(Sketch):Ifx2Qsatis¯esx2=12,thenxisinQandsatis¯es¡¢2x22=3.Nowprovethatthereisnoy2Qsuchthaty=3byrepeatingthe2pproofthat262Q.Problem1.5:LetA½Rbenonempty

6、andboundedbelow.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¡b

7、isalowerboundforAisimmediatefromthefactthatbisanupperboundfor¡A.Toshowthat¡bisthegreatestlowerbound,weletc>¡bandprovethatcisnotalowerboundforA.Now¡c¡c.Then¡x2Aand¡x

8、em1.6:Letb2Rwithb>1,¯xedthroughouttheproblem.Comment:Wewillassumeknownthatthefunctionn7!bn,fromZtoR,isstrictlyincreasing,thatis,thatform;n2Z,wehavebm