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;varyingamountsofdetailwouldneedtobeˉlledin.Problem1.1:Ifr2Qnf0gandx2RnQ,provethatr+x;rx62Q.Solution:Weprovethisbycontradiction.Letr2Qnf0g,andsupposethatr+x2Q.Then,usingtheˉeldpropertiesofbothRandQ,wehavex=(r+x)?r2Q.Thusx62Qimpliesr+x62Q.Similar

3、ly,ifrx2Q,thenx=(rx)=r2Q.(Here,inadditiontotheˉeldpropertiesofRandQ,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):Ifx2Qsatisˉesx2=12,thenxisinQandsatisˉes?￠2x22=3.Nowprovethatthereisnoy2Qsuchthaty=3byrepeatingthe2pproofthat262Q.Problem1.5:LetA?Rbenonempty

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,ˉxedthroughouttheproblem.Comment:Wewillassumeknownthatthefunctionn7!bn,fromZtoR,isstrictlyincreasing,thatis,thatform;n2Z,wehavebm