Dunwoody Chemistry Laboratory Final Exam Protocol

pages

English

Documents

Lire

Le téléchargement nécessite un accès à la bibliothèque YouScribe Tout savoir sur nos offres

pages

English

Ebook

Lire

Le téléchargement nécessite un accès à la bibliothèque YouScribe Tout savoir sur nos offres

Publié par

mevang

Nombre de lectures

Langue

English

cours - matière potentielle : number
cours magistral
cours - matière potentielle : number with section
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

Voir

Publié par

mevang

Nombre de lectures

Langue

English

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)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“thereexistsaunique”.ThepropositionPU x(x)istrueiffthereisexactlyonexforwhichP(x)istrue.Thelogicianhasnoted,“Thereusedtobetwoquantiﬁers,butnowtherearethree!Ihaveextendedthewholeﬁeldofmathematicsby50%!”

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

2ProblemSet1(b)WriteapropositionequivalenttoUx P(x) usingonlythe∀quantiﬁer,=,andlogicalconnectives.Solution.¬∀x (¬P(x)∨ ¬∀y (x = y ∨ ¬P(y))) Problem3.AmediatycoonhasanideaforanallnewstelevisionnetworkcalledLNN:TheLogicNewsNetwork.Eachsegmentwillbeginwiththedeﬁnitionofsomerelevantsetsandpredicates.Theday’shappeningscanthenbecommunicatedconciselyinlogicnotation.Forexample,abroadcastmightbeginasfollows:

“THISISLNN.LetS betheset{Bill, Monica, Ken, Linda, Betty}.LetD(x) beapredicatethatistrueifx isdeceitful.LetL(x, y)beapredicatethatistrueifx 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,andMonicagavegiftstosomeone.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)

Theremainingproblemswillbegradedprimarilyontheclarityofyourproofs—notonwhetheryouhavetherightidea.Infact,ifyoucan’tﬁgureouttherightidea,we’llgiveittoyou–justaskyourTA!

ProblemSet13Problem4.Letn beapostiveinteger.Provethatlogn isrationalifandonlyifn is2 apowerof2.Assumeanybasicfactsaboutdivisibilitythatyouneed;juststateyourassumptionsexplicitly.b Solution.Assumption:Ifnisapowerof2,thenn isapowerof2.Proof.Weprovethatifn isapowerof2,thenlogn isrationalandviceversa.2 First,weprovethatifn isapowerof2,thenlogn isrational.Assumethatn isapower2 k k of2.Thenn = 2forsomeintegerk ≥0.Thus,logn == gol2k,whichisarational2 2 number.Next,weprovethatiflogn isrational,thenn isapowerof2.Assumethatlogn is2 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.Atriangleisasetofthreepeoplesuchthateithereverypairhasshakenhandsornopairhasshakenhands.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 formatriangle.(b)Amongthesethree,nopairshookhands.Thenthesethreeformatriangle.

4ProblemSet12.Otherwise,atmosttwopeopleshookhandswithpersonX.Thus,thereexistthreepeoplewhodidn’tshakehandswithX.Again,therearetwofurtherpossibilities:(a)Amongthesethree,everypairshookhands.Thenthesethreeformatriangle.(b)Amongthesethree,somepairdidn’tshakehands.ThenthesetwoandX foratriangle.

Problem6.Letx andy benonnegativerealnumbers.Thearithmeticmeanofx andy √ isdeﬁnedtobe(x + y)/2,andthegeometricmeanisdeﬁnedtobexy.Provethatthearithmeticmeanisequaltothegeometricmeanifandonlyifx = y.Solution.Proof.Weconstructachainofifandonlyifimplications.Thearithmeticmeanisequaltothegeometricmeanifandonlyif: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.Usecaseanalysistoprovethatallintegralsolutionstotheequation1 1 11 + =+ m ne 2 subjecttotheseconstraintsm ≥3 n ≥3 > e0 areinthistable:m n e 336 3 412 3 530 4 312 5 3 30 Theseequationsrevealsomethingfundamentalaboutthegeometryofourthreedimensionalworld;we’llrevisittheminaboutthreeweeks.

ProblemSet15Solution.Proof.Weusecaseanalysis.Sincem≥3,oneoffourcasesmusthold:1.m= 3.Therearenowfoursubcases:(a)n= 3.Rewritingtheequationintheform1e=1 1 1 +−m n2 andsubtitutinginm= n= 3 impliesthate= 6,whichistheﬁrstsolution.(b)n= 4.Substitutingthevaluesofmandnintotheequationshowthate= 12,whichisthesecondsolution.(c)n = 5.Substitutingintotheequationshowsthate = 30,whichisthethirdsolution.(d)n≥6.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)n≥4.Thisimplies:1 1111 + ≤ =+ m n2 4 4 Again,theleftsideoftheequationisstrictlylessthantherightforalle >0,sotherearenosolutionsinthiscase.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.

Voir

Univers
Ebooks
Livres audio
Presse
Podcasts
BD
Documents

Dunwoody Chemistry Laboratory Final Exam Protocol

YouScribe

Le catalogue

Le service

Les conditions