apostol_mathematical.analysis.习题解答_cropped - 副本 (1)

《apostol_mathematical.analysis.习题解答_cropped - 副本 (1)》由会员上传分享，免费在线阅读，更多相关内容在学术论文-天天文库。

1、TheRealAndComplexNumberSystemsIntegers1.1Provethatthereisnolargestprime.Proof:Supposepisthelargestprime.Thenp!+1isNOTaprime.So,thereexistsaprimeqsuchthatq

2、p!+1⇒q

3、1whichisimpossible.So,thereisnolargestprime.Remark:Therearemanyandmanyproofsaboutit.TheproofthatwegivecomesfromArchimedes287-212B

4、.C.Inaddition,EulerLeonhard(1707-1783)findanothermethodtoshowit.Themethodisimportantsinceitdevelopstostudythetheoryofnumbersbyanalyticmethod.Thereadercanseethebook,AnIntroductionToTheTheoryOfNumbersbyLoo-KengHua,pp91-93.(ChineseVersion)1.2Ifnisapositiveinteger,provethealgebraicidentityXn−1nn

5、kn−1−ka−b=(a−b)abk=0Proof:ItsufficestoshowthatXn−1nkx−1=(x−1)x.k=01Considertherighthandside,wehaveXn−1Xn−1Xn−1kk+1k(x−1)x=x−xk=0k=0k=0XnXn−1kk=x−xk=1k=0n=x−1.1.3If2n−1isaprime,provethatnisprime.Aprimeoftheform2p−1,wherepisprime,iscalledaMersenneprime.Proof:Ifnisnotaprime,thensayn=ab,wherea>1a

6、ndb>1.So,wehaveXb−12ab−1=(2a−1)(2a)kk=0whichisnotaprimebyExercise1.2.So,nmustbeaprime.Remark:ThestudyofMersenneprimeisimportant;itisrelatedwithsocalledPerfectnumber.Inaddition,therearesomeOPENprob-lemaboutit.Forexample,isthereinfinitelymanyMersennenem-bers?Thereadercanseethebook,AnIntroducti

7、onToTheTheoryOfNumbersbyLoo-KengHua,pp13-15.(ChineseVersion)1.4If2n+1isaprime,provethatnisapowerof2.Aprimeofthe2mform2+1iscalledaFermatprime.Hint.Useexercise1.2.Proof:Ifnisanotapowerof2,sayn=ab,wherebisanoddinteger.So,2a+12ab+1and2a+1<2ab+1.Itimpliesthat2n+1isnotaprime.So,nmustbeapowerof2.R

8、emark:(1)Intheproof,weusetheidentity2Xn−2x2n−1+1=(x+1)(−1)kxk.k=02Proof:Consider2Xn−22Xn−22Xn−2(x+1)(−1)kxk=(−1)kxk+1+(−1)kxkk=0k=0k=02Xn−12Xn−2=(−1)k+1xk+(−1)kxkk=1k=02n+1=x+1.(2)ThestudyofFermatnumberisimportant;forthedetailsthereadercanseethebook,AnIntroductionToTheTheoryOfNumbersbyLoo-K

9、engHua,pp15.(ChineseVersion)1.5TheFibonaccinumbers1,1,2,3,5,8,13,...aredefinedbytherecur-sionformulaxn+1=xn+xn−1,withx1=x2=1.Provethat(xn,xn+1)=1andthatx=(an−bn)/(a−b),whereaandbaretherootsofthequadraticnequationx2−x−1=0.Proof:Letd=g.c.d.(xn,xn+1),thend

10、x