Dunwoody Chemistry Laboratory Final Exam Protocol

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• cours - matière potentielle : number
• cours magistral
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• cours - matière potentielle : coordinator
• cours - matière potentielle : number with the section number
• cours magistral - matière potentielle : sections
• cours - matière potentielle : coordinators

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
• scantrons
• end of the exam teach student
• red scantron
• final exam protocol
• assessment exam
• practical exam
• course coordinator
• student
• exam

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• Nomura
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• Publié par: mevang
• Nombre de lectures: 25
• Langue: English

Extrait

6.042/18.062JMathematicsforComputerScience SriniDevadasandEricLehman

ProblemSet1Solutions

Due:Monday,February7at9PM

February1,2005

Problem1.Theconnectives∧(and),∨(or),and⇒(implies)comeoftennotonlyincom puterprograms,butalsoeverydayspeech.Butdevicesthatcomputethenandoperation arepreferableincomputerchipdesigns.Hereisthetruthtablefornand: P QPnandQ T T F  T F T F T T F  FT  Foreachofthefollowingexpressions,findanequivalentexpressionusingonlynandand ¬(not).

(a)A∧B  Solution.¬(AnandB)

(b)A∨B  Solution.(¬A)nand(¬B)

(c)A⇒B Solution.Anand(¬B)

Problem2.Aselfproclaimed“greatlogician”hasinventedanewquantifier,onparwith ∃(“thereexists”)and∀(“forall”).ThenewquantifierissymbolizedbyUandread“there existsaunique”.ThepropositionPU x(x)istrueiffthereisexactlyonexforwhichP(x)is true.Thelogicianhasnoted,“Thereusedtobetwoquantifiers,butnowtherearethree!I haveextendedthewholefieldofmathematicsby50%!”

(a)WriteapropositionequivalenttoUx P(x) usingonlythe∃quantifier,=,and logicalconnectives. Solution. ∃x(P(x)∧ ¬(∃y(¬(x= y)∧P(y)))

2ProblemSet1 (b)WriteapropositionequivalenttoUx P(x) usingonlythe∀quantifier,=,and logicalconnectives. Solution. ¬∀x (¬P(x)∨ ¬∀y (x = y ∨ ¬P(y))) Problem3.AmediatycoonhasanideaforanallnewstelevisionnetworkcalledLNN: TheLogicNewsNetwork.Eachsegmentwillbeginwiththedefinitionofsomerelevant setsandpredicates.Theday’shappeningscanthenbecommunicatedconciselyinlogic notation.Forexample,abroadcastmightbeginasfollows:

“THISISLNN.LetS betheset{Bill, Monica, Ken, Linda, Betty}.LetD(x) be apredicatethatistrueifx isdeceitful.LetL(x, y)beapredicatethatistrueif x likesy.LetG(y ,x)beapredicatethatistrueifx gavegiftstoy.”

Completethebroadcastbytranslatingthefollowingstatementsintologicnotation.

(a)IfneitherMonicanorLindaisdeceitful,thenBillandMonicalikeeachother. Solution.

(¬(D(Monica)∨D(Linda))) ⇒(L(Bill, Monica)∧L(Monica, Bill))

(b)EveryoneexceptforKenlikesBetty,andnooneexceptLindalikesKen. Solution.

∀x ∈S (x = Ken∧ ¬L(x, Betty)) ∨(x �= Ken∧L(x, Betty)) ∧ ∀x ∈S (x = Linda∧L(x, Ken)) ∨(x �= Linda∧ ¬L(x, Ken))

(c)IfKenisnotdeceitful,thenBillgavegiftstoMonica,andMonicagavegiftsto someone. Solution. ¬D(Ken)⇒(G(Bill, Monica)∧ ∃x ∈G S(Monicax, )) 

(d)Everyonelikessomeoneanddislikessomeoneelse. Solution. ∀x ∈S ∃y ∈S ∃z ∈S (y�= z)∧L(, yx)∧ ¬L(x, z)

Theremainingproblemswillbegradedprimarilyontheclarityof yourproofs—notonwhetheryouhavetherightidea.Infact,ifyou can’tfigureouttherightidea,we’llgiveittoyou–justaskyourTA!

ProblemSet13 Problem4.Letn beapostiveinteger.Provethatlogn isrationalifandonlyifn is 2  apowerof2.Assumeanybasicfactsaboutdivisibilitythatyouneed;juststateyour assumptionsexplicitly. b  Solution.Assumption:Ifnisapowerof2,thenn isapowerof2. Proof.Weprovethatifn isapowerof2,thenlogn isrationalandviceversa. 2  First,weprovethatifn isapowerof2,thenlogn isrational.Assumethatn isapower 2  k k  of2.Thenn = 2forsomeintegerk ≥0.Thus,logn  == gol2k,whichisarational 2 2  number. Next,weprovethatiflogn isrational,thenn isapowerof2.Assumethatlogn is 2 2  rational.Thatmeansthereexistintegersa andb suchthat: a  logn =  2  b  Wecanrewritethisequationasfollows: a/b  n = 2(Take2topowerofeachside.) b a  n = 2(Takethebthpowerofeachside.) b  Thus,nisapowerof2.Byourassumption,n isapowerof2. Problem5.Atriangleisasetofthreepeoplesuchthateithereverypairhasshakenhands ornopairhasshakenhands.Provethatamongeverysixpeoplethereisatriangle.Sug gestion:Initially,breaktheproblemintotwocases: 1.ThereexistatleastthreepeoplewhoshookhandswithpersonX.

2.Thereexistatleastthreepeopledidn’tshakehandswithX

(Whymustatleastoneoftheseconditionshold?) Solution. Proof.Weusecaseanalysis.LetX denoteoneofthesixpeople.Therearetwopossibili ties:

1.ThereexistthreepeoplewhoshookhandswithpersonX.Nowtherearetwofur therpossibilities: (a)Amongthesethree,somepairshookhands.ThenthesetwoandX forma triangle. (b)Amongthesethree,nopairshookhands.Thenthesethreeformatriangle.

4ProblemSet1 2.Otherwise,atmosttwopeopleshookhandswithpersonX.Thus,thereexistthree peoplewhodidn’tshakehandswithX.Again,therearetwofurtherpossibilities: (a)Amongthesethree,everypairshookhands.Thenthesethreeformatriangle. (b)Amongthesethree,somepairdidn’tshakehands.ThenthesetwoandX fora triangle.

Problem6.Letx andy benonnegativerealnumbers.Thearithmeticmeanofx andy  √ isdefinedtobe(x + y)/2,andthegeometricmeanisdefinedtobexy.Provethatthe arithmeticmeanisequaltothegeometricmeanifandonlyifx = y. Solution. Proof.Weconstructachainofifandonlyifimplications.Thearithmeticmeanisequal tothegeometricmeanifandonlyif: x + y  √ √ = xy ⇔x + y = 2xy 2  2  ⇔(x + y4) =xy 2 2  ⇔x 2 +xy + y = 4xy 2 2  ⇔x −2xy + y 0 = 2  ⇔(x −y) =0  ⇔x −y =0  ⇔x = y 

Problem7.Usecaseanalysistoprovethatallintegralsolutionstotheequation 1  1 11  +  =+  m ne 2  subjecttotheseconstraints m ≥3 n ≥3  > e0 areinthistable: m n e  336  3 412 3 530 4 312 5 3 30  Theseequationsrevealsomethingfundamentalaboutthegeometryofourthreedimensional world;we’llrevisittheminaboutthreeweeks.

ProblemSet15 Solution. Proof.Weusecaseanalysis.Sincem≥3,oneoffourcasesmusthold: 1.m= 3.Therearenowfoursubcases: (a)n= 3.Rewritingtheequationintheform 1 e= 1 1 1  +− m n2  andsubtitutinginm= n= 3 impliesthate= 6,whichisthefirstsolution. (b)n=  4.Substitutingthevaluesofmandnintotheequationshowthate=  12, whichisthesecondsolution. (c)n =  5.Substitutingintotheequationshowsthate =  30,whichisthethird solution. (d)n≥6.Thisimplies: 1 1 1 1 1  + ≤+ =  m n 6 2 3 Thus,theleftsideoftheequationisstrictlylessthantherightforalle >0,so therearenosolutionsinthiscase. 2.m= 4.Therearetwosubcases: (a)n= 3.Subsitutinggivese =12,whichisthefourthsolution. (b)n≥4.Thisimplies: 1 1111  + ≤ =+ m n2 4 4 Again,theleftsideoftheequationisstrictlylessthantherightforalle >0,so therearenosolutionsinthiscase. 3.m= 5.Therearetwosubcases: (a)n= 3.Subsitutinggivese0 3=,whichisthefifthsolution. (b)n≥4.Thisimplies: 1 1111  + ≤+ <  m n5 42  Again,theequationcannothold,sotherearenosolutionsinthiscase. 4.m≥6.Thisimplies: 1 1111  + ≤ =+ m n6 32  Oncemore,theequationcannothold,sotherearenosolutionsinthiscase.