Dunwoody Chemistry Laboratory Final Exam Protocol
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Dunwoody Chemistry Laboratory Final Exam Protocol

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5 pages
English

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Tuesday, April 19, 2011 Dunwoody Chemistry Laboratory Final Exam Protocol General Instructors are to fully supervise the exams and should walk around the benches periodically. All lab safety rules shall apply during final exams having practical portions. Principals of Chemistry Labs: CHEM 1211L & CHEM 1212L Written exam (20 questions CHEM 1211L and 10 questions CHEM 1212L) 60% of the grade. 1. Students should not write on the exam.
  • masses for the practical part of the exam
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  • final exam protocol
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Nombre de lectures 25
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6.042/18.062JMathematicsforComputerScienceSriniDevadasandEricLehman
ProblemSet1Solutions
Due:Monday,February7at9PM
February1,2005
Problem1.Theconnectives(and),(or),and(implies)comeoftennotonlyincom puterprograms,butalsoeverydayspeech.Butdevicesthatcomputethenandoperationarepreferableincomputerchipdesigns.Hereisthetruthtablefornand:P QPnandQT T F  T F T F T T F  FT  Foreachofthefollowingexpressions,findanequivalentexpressionusingonlynandand¬(not).
(a)ASolution.¬(AnandB)
(b)ASolution.(¬A)nand(¬B)
(c)AB Solution.Anand(¬B)
Problem2.Aselfproclaimed“greatlogician”hasinventedanewquantifier,onparwith(“thereexists”)and(“forall”).ThenewquantifierissymbolizedbyUandread“thereexistsaunique”.ThepropositionPU x(x)istrueiffthereisexactlyonexforwhichP(x)istrue.Thelogicianhasnoted,“Thereusedtobetwoquantifiers,butnowtherearethree!Ihaveextendedthewholefieldofmathematicsby50%!”
(a)WriteapropositionequivalenttoUx P(xusingonlythequantifier,=,andlogicalconnectives.Solution.x(P(x)∧ ¬(y(¬(xy)P(y)))
2ProblemSet1(b)WriteapropositionequivalenttoUx P(xusingonlythequantifier,=,andlogicalconnectives.Solution.¬∀(¬P(x)∨ ¬∀(∨ ¬P(y))) Problem3.AmediatycoonhasanideaforanallnewstelevisionnetworkcalledLNN:TheLogicNewsNetwork.Eachsegmentwillbeginwiththedefinitionofsomerelevantsetsandpredicates.Theday’shappeningscanthenbecommunicatedconciselyinlogicnotation.Forexample,abroadcastmightbeginasfollows:
“THISISLNN.Letbetheset{BillMonicaKenLindaBetty}.LetD(xbeapredicatethatistrueifisdeceitful.LetL(x, y)beapredicatethatistrueiflikesy.LetG(y ,x)beapredicatethatistrueifgavegiftstoy.”
Completethebroadcastbytranslatingthefollowingstatementsintologicnotation.
(a)IfneitherMonicanorLindaisdeceitful,thenBillandMonicalikeeachother.Solution.
(¬(D(Monica)D(Linda))) (L(BillMonica)L(MonicaBill))
(b)EveryoneexceptforKenlikesBetty,andnooneexceptLindalikesKen.Solution.
(Ken∧ ¬L(x, Betty)) (KenL(x, Betty)) (LindaL(x, Ken)) (Linda∧ ¬L(x, Ken))
(c)IfKenisnotdeceitful,thenBillgavegiftstoMonica,andMonicagavegiftstosomeone.Solution.¬D(Ken)(G(BillMonica)∧ ∃G S(Monicax, )) 
(d)Everyonelikessomeoneanddislikessomeoneelse.Solution.(yz)L(, yx)∧ ¬L(x, z)
Theremainingproblemswillbegradedprimarilyontheclarityofyourproofs—notonwhetheryouhavetherightidea.Infact,ifyoucan’tfigureouttherightidea,we’llgiveittoyou–justaskyourTA!
ProblemSet13Problem4.Letbeapostiveinteger.Provethatlogisrationalifandonlyifisapowerof2.Assumeanybasicfactsaboutdivisibilitythatyouneed;juststateyourassumptionsexplicitly.Solution.Assumption:Ifnisapowerof2,thenisapowerof2.Proof.Weprovethatifisapowerof2,thenlogisrationalandviceversa.First,weprovethatifisapowerof2,thenlogisrational.Assumethatisapowerk k  of2.Then= 2forsomeinteger0.Thus,log == gol2k,whichisarational2 2  number.Next,weprovethatiflogisrational,thenisapowerof2.Assumethatlogis2 2  rational.Thatmeansthereexistintegersandsuchthat:logWecanrewritethisequationasfollows:a/b  = 2(Take2topowerofeachside.)b a  = 2(Takethebthpowerofeachside.)Thus,nisapowerof2.Byourassumption,isapowerof2.Problem5.Atriangleisasetofthreepeoplesuchthateithereverypairhasshakenhandsornopairhasshakenhands.Provethatamongeverysixpeoplethereisatriangle.Sug gestion:Initially,breaktheproblemintotwocases:1.ThereexistatleastthreepeoplewhoshookhandswithpersonX.
2.Thereexistatleastthreepeopledidn’tshakehandswithX
(Whymustatleastoneoftheseconditionshold?)Solution.Proof.Weusecaseanalysis.Letdenoteoneofthesixpeople.Therearetwopossibili ties:
1.ThereexistthreepeoplewhoshookhandswithpersonX.Nowtherearetwofur therpossibilities:(a)Amongthesethree,somepairshookhands.Thenthesetwoandformatriangle.(b)Amongthesethree,nopairshookhands.Thenthesethreeformatriangle.
4ProblemSet12.Otherwise,atmosttwopeopleshookhandswithpersonX.Thus,thereexistthreepeoplewhodidn’tshakehandswithX.Again,therearetwofurtherpossibilities:(a)Amongthesethree,everypairshookhands.Thenthesethreeformatriangle.(b)Amongthesethree,somepairdidn’tshakehands.Thenthesetwoandforatriangle.
Problem6.Letandbenonnegativerealnumbers.Thearithmeticmeanofandisdefinedtobe(y)/2,andthegeometricmeanisdefinedtobexy.Provethatthearithmeticmeanisequaltothegeometricmeanifandonlyify.Solution.Proof.Weconstructachainofifandonlyifimplications.Thearithmeticmeanisequaltothegeometricmeanifandonlyif:√ √ xy = 2xy (y4) =xy 2 2  2 +xy = 4xy 2 2  2xy 0 = (y) ==0 
Problem7.Usecaseanalysistoprovethatallintegralsolutionstotheequation1  1 1+  =m nsubjecttotheseconstraints > e0 areinthistable:m n 336  3 412 3 530 4 312 5 3 30  Theseequationsrevealsomethingfundamentalaboutthegeometryofourthreedimensionalworld;we’llrevisittheminaboutthreeweeks.
ProblemSet15Solution.Proof.Weusecaseanalysis.Sincem3,oneoffourcasesmusthold:1.m= 3.Therearenowfoursubcases:(a)n= 3.Rewritingtheequationintheform1e=1 1 +m nandsubtitutinginmn= 3 impliesthate= 6,whichisthefirstsolution.(b)n=  4.Substitutingthevaluesofmandnintotheequationshowthate=  12,whichisthesecondsolution.(c)=  5.Substitutingintotheequationshowsthat=  30,whichisthethirdsolution.(d)n6.Thisimplies:1 1 1 1 1  + =  m n 6 2 3 Thus,theleftsideoftheequationisstrictlylessthantherightforalle >0,sotherearenosolutionsinthiscase.2.m= 4.Therearetwosubcases:(a)n= 3.Subsitutinggivese =12,whichisthefourthsolution.(b)n4.Thisimplies:1 1111   =+ m n4 4 Again,theleftsideoftheequationisstrictlylessthantherightforalle >0,sotherearenosolutionsinthiscase.3.m= 5.Therearetwosubcases:(a)n= 3.Subsitutinggivese0 3=,whichisthefifthsolution.(b)n4.Thisimplies:1 1111  m n5 4Again,theequationcannothold,sotherearenosolutionsinthiscase.4.m6.Thisimplies:1 1111   =+ m n6 3Oncemore,theequationcannothold,sotherearenosolutionsinthiscase.
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