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Introduction to DiscreteMathematics

Course OutlineProf. G. Kesidis, instructor

CourseOutline• Introductiontologic• Introductiontomethodsofproof• Sequencesandinduction• Introductiontosettheory• Countingandprobability– discreterandomvariables(seeslidedeck on

undergraduateprobability)• Functionsandthepigeonholeprinciple• Recursions• IntroductiontoComplexity• Relations• Introductiontomodulararithmeticandpublic-keycryptography• Introductiontographs

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Introduction to DiscreteMathematics

Introduction to Logic© S.S. Epp (2006)

This revision G. Kesidis (2013)This document revision is copy-left 2013.

It is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License

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Intro to Logic – Outline• LogicalFormandEquivalence• ConditionalStatements• ValidandInvalidArguments• IntrotoPredicatesandQuantifiedStatements• StatementswithMultipleQuantifiers• ArgumentswithQuantifiedStatements

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Concept of Logical Form

Howdoyouknowtodayisn'tEaster?

IftodayisEaster,thenitisSunday.ItisnotSundayTherefore,todayisnotEaster.

Theformofthisargumentiscalled“modustollens”.

Ifp,thenq.Notq.Thereforenotp.

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Modus tollens exampleExample: Argumentintheformofmodustollens.

IfSueisguilty,thenSuewasinRioonAug.18.(Ifp,thenq.)

SuewasnotinRioonAug.18.(Notq.)

Therefore,Sueisnotguilty.(Therefore,notp.)

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Modus tollens: premises, conclusionCommonform(modustollens):

Ifp,thenq.(premise)Notq.(premise)Thereforenotp.(conclusion)

Apremiseisastatementpresumed trueforpurposesofanargument.

Modustollensisvalid inthesensethat• ifyouhaveanargumentofthisforminwhichthefirsttwo

statements(thepremises)arebothtrue,• thenyoucanbe100%surethatthefinalstatement(theconclusion)

willbetrue.

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Mathematical Communication

• Mustbeunambiguous

• Mustexpressthingssothateveryoneunderstandsexactlywhatwearesaying

• Consequence: Carefuldefinitionsarenecessary

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Statements

Statement: Adeclarativesentencethatiseithertrueor

falsebutnotboth.

Example: Whichofthefollowingarestatements?

§ 1+1=2 isastatement– truesentence

§ 1+1=5 isastatement– falsesentence

§ 1+x=5 isnotastatement;trueforsomexand

falseforothers(thiskindofsentenceis

calleda“predicate”).

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Logical Connectives

Usedtoputsimplestatementstogethertomakecompoundstatements

not ~ negationand, but ⋀ conjunctionor (inclusive) ⋁ disjunctionif-then → conditional

if-and-only-if ↔ biconditional

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Precedence Rules1.~

2.⋁and⋀(withleft-to-rightprecedenceamongthesetwooperations,butbettertouseparenthesestoavoidambiguityinsteadoftheleft-to-rightprecedence)

3.→and↔(ditto)

4.Parenthesesmaybeusedtooverriderules1-3

Example:• ~p⋀q=(~p)⋀q:checkthatthisisnotthesameas~(p⋀q)• p⋀ q⋁ rmaybeambiguous.Byleft-to-rightprecedence,thisisthe

sameas(p⋀ q)⋁ r.Bettertojustspecifywhether(p⋀ q)⋁ rorp⋀(q⋁ r)isintended.

Recallleft-to-rightprecedencefromarithmetic:6/3�9=18(not0.5).

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Logical connectives - examplesLetpbe“Itiswinter”,qbe“Itiscold”,andrbe“Itisraining”.

Writethefollowingstatementssymbolically.

• Itiswinterbutitisnotcold=p⋀~q=p⋀(~q)

• Neitherisitwinternorisitcold=~p⋀~q=(~p)⋀(~q)

• Itisnotwinterifitisnotcold=~q→~p

• Itisnotwinterbutitisrainingoritiscold=(~p⋀r)⋁q

• Itisnotwinterbutitisrainingorcold=~p⋀(r⋁q)?

Notetheuseofleft-to-rightprecedenceinthesecond-lastexampleabove.

Noteinthelastexamplethatthelackofseparatesubjectandverb(“itis”)withtheclause“cold”mayimplythedifferentanswergiven.

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Truth Values for Compound Statement Forms

• NotStatements(Negations): Thenegationofastatementisastatementthatexactlyexpresseswhatitwouldmeanforthegivenstatementtobefalse.

• Thenegationofastatementhasoppositetruthvaluefromthestatement.

p ~pT F

F T

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TruthTablefor⋁ (“inclusiveor”)• Notehowallpossiblecombinationsofveracityofthetwocomponent

statements(independentvariables)areconsideredinthefirsttwocolumnsofthetruthtableforlogical or:

• Theonlytimeanor statementisfalseiswhenbothcomponentstatementsarefalse– cf.DeMorgan’slawsonnegatinglogicalor statements:~(p⋁q)=~p⋀~q

p q p⋁q

T T T

T F TF T TF F F

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TruthTablefor⋀(“and”,“but”)

• Theonlytimeanand statementistrueiswhenbothcomponentsaretrue– againcf.DeMorgan’slawonnegatinglogicalandstatements:~(p⋀q)=~p⋁ ~q

p q p⋀q

T T TT F FF T FF F F

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De Morgan’s laws for negating ⋀

• Notehowthe~(p⋀q)and ~p⋁ ~q columnsareidentical,i.e.,thetwocompoundstatementsarelogicallyequivalent).

• Thisproofthat~(p⋀q)≡ ~p⋁ ~q isdirectlybytakingallcases(eachcaseisarow ofthetruthtable)ontheveracity(truthvalues)ofcomponentstatementspandq.

p q p⋀q ~(p⋀q) ~p ~q ~p⋁ ~q

T T T F F F FT F F T F T TF T F T T F TF F F T T T T

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De Morgan’s laws for negating ⋁• Exercise:repeatthisfornegatingor statements,i.e.,show~(p⋁q)≡ ~p⋀~q

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De Morgan’s laws - examples• Question:Writethenegationof,“Sueleftthedoorunlockedorsheleftthewindowopen.”

• Answer:“Suelockedthedoorandsheclosedthewindow.”• But isthenegationof“Sueleftthedoorunlocked”really“Suelockedthedoor”?

• Question:Writethenegationof,“HalgotanAonthemidtermandHalgotanAonthefinalexam.”

• Answer:“Haldidnot getanAonthemidtermor HaldidnotgetanAonthefinalexam.”

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Truth Table for → (implies)• Theonlytimeastatementoftheformifpthenqisfalseiswhenthe

hypothesis(p )istrueandtheconclusion(q)isfalse,i.e.,~(p→q)≡p⋀~q(Exercise:checkthisbytruthtable)

• Whenthehypothesisofanif-thenstatementisfalse,wesaythattheif-thenstatementis�vacuouslytrue� or�truebydefault.� Inotherwords,itistruebecauseitisnotfalse.

• p→q≡ifpthenq≡qifp≡ponlyifq≡qisanecessaryconditionforp≡pisasufficientconditionforq

≡~porq

p q p→ qT T T

T F FF T T

F F T

p q

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If-then Statements (Conditionals)• Imagine thatIpromiseyou:

"IfyougetanAonthefinalexam,thenyouwillgetanAinthecourse.”• Formofconditional:“Ifhypothesis,thenconclusion”,equivalently

“hypothesis→conclusion”.• Jumpforwardtooneweekaftertheendofthesemester.Findingout

yourfinalexamgradeandyourcoursegrade,youexclaim:”Helied.”• WhatwouldhavetobetruetoleadyoutosaythatIlied?• YouwouldhavetohaveearnedanAonthefinalexam andnotreceived

anAforthecourse.InallothercasesitwouldnotbefairtosaythatIlied.

• Note:Theexamplestatementaboveislogicallyequivalentto”YougetanAonthefinalexamonlyifyougetanAinthecourse.”

If-then statements (cont)• Question:Writethenegationof“IfJimgottherightanswer,thenhesolvedtheproblemcorrectly.”

• Answer:“Jimgottherightanswerand hedidnot solvetheproblemcorrectly.”

• Notethatonecoulduse“but” insteadof“and” intheanswer.

• ~(p→q)≡p⋀ ~q

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If-and-Only-If (iff) Statements (Biconditionals)• Astatementoftheformpifandonlyifq istruewhenbothpandqhavethesametruthvalue(veracity).

• pifandonlyifq isfalsewhenpandqhaveoppositetruthvalues.

• Exercise:Checkbytruthtablethatp↔ q≡(q→p)⋀(p→q),i.e.,pifandonlyifq≡(pifq)and(ponlyifq)

p q p↔ qT T T

T F F

F T F

F F T

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Some Potentially Ambiguous TermsOr

• Example:Asyourprize,youmaychoosetheTVor themicrowaveoven.

• Question:Isittobeunderstoodthatyoucan'tchooseboththeTVand

themicrowave?

• Answer:YES.Inthissentence,theintendedmeaningof“or”iscalled

“exclusiveor”,i.e.,

p⊕q=falsewhenp=trueandq=true.

If-then

• Example:If youeatyourdinner,then youwillgetdessert.

• Question:Doesthisimplythatifyoudon'teatyourdinner,thenyou

won'tgetdessert?Inotherwords,doesitmeanthatifyougetdessert,

youwillhaveeatenyourdinner?

• Answer:Momwantsyoutothinkthis,butitisnotimpliedbythelogical

meaningofif-then.

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Moral

• Inordinaryspeech,wordslike“or” and“if-then”mayhavemultiplemeanings,

• butintechnicalsubjects,wemustallunderstandwordsinthesameway.

• Sowechooseonemeaningforeachword.

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Compound Statement FormsExample :§ p:DePaulUniversityisinChicago§ q:1+1=5§ r:ChicagoisnexttoLakeMichigan

Question:Knowingthatp isT,q isFandr isT,whatisthetruthvalueofthefollowingcompoundstatement?

(p⋀~q)⋁~r

Answer:(T⋀~F)⋁~T=(T⋀T)⋁F=T⋁F=Ti.e.,theanswerisTrue.

Notehowwefollowedprecedenceorder.

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Truth Table for Exclusive Or

• logicalexpressionforastatementoftheformp orq butnotboth:(p⋁q)⋀~(p⋀q)≝p⊕q

p q p⋁q p⋀q ~(p⋀q) (p⋁q)⋀~(p⋀q)

T T T T F FT F T F T TF T T F T TF F F F T F

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Arguments and Argument Forms • Argument:Sequenceofstatements.Thefinalstatementinthesequenceisthe(final)conclusion;theprecedingstatementsarepremises. Somestatementsfollowingthe“initial”premisesmaythemselvesbeconclusions,i.e.,formingcomponentarguments.

• ArgumentForm:Obtainedbyreplacingcomponentstatementsintheargumentbyvariables.

• Example(modustollens):Ifpthenq. (premise)Notq. (premise)Therefore,notp. (∴conclusion)

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Valid Arguments• Formofargumentisvalidwhen: Everyargumentofthatformthathas

truepremiseshasatrueconclusion.(Amoreformalversionofthedefinitionisinthebook)

• Claim:Modustollensisavalidformofargument.

• Proof:Bylastrowofthefollowingtruthtable,whereinbothpremises(columns3and4)aretrueand(rightmost)column5isthetrueconclusion.

p q p→q ~q ~p

T T T F FT F F T FF T T F TF F T T T

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Invalid Arguments• Aformofargumentisinvalid ifthereisatleastoneargumentofthatformthathastruepremisesandafalseconclusion.

• Example:Determinewhetherthefollowingargumentform(withthreepremises)isvalidorinvalid:

~p⋁qp→rq→p∴r

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Valid or Invalid? Example (cont)

p q r ~p ~p⋁q p→r q→p rT T T F T T T TT T F F T F T FT F T F F T T TT F F F F F T FF T T T T T F TF T F T T T F FF F T T T T T TF F F T T T T F

• Onlyrows1,7,8aregermaneasallpremises(cols.5,6,7)aretruejustforthosecases.

• The8th (bottom)rowshowsthatitispossibleforanargumentofthisformtohavetruepremisesandafalseconclusion(col.8).

• Sothisformofargumentisinvalid.

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More Valid and Invalid Forms of Argument

• Modusponens(valid):

IfTedisaCSmajor,thenTedhastotakeCSC211.

TedisaCSmajor.

Therefore,TedhastotakeCSC211.

• Form: Ifpthenq

p

∴q

• Itisobviouslynotpossibleforthetwopremisestobetrueandtheconclusiontobefalse,sothisformofargumentisvalid.

• Foraproof,seethefirstrowofthetruthtableforp→q

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Invalid Argument with a “Converse Error”• Example:IftodayisEaster,thenitisSunday.ItisSunday.Therefore,todayisEaster.

• Form: Ifpthenqq∴p

• Exercise:Checkbytruthtablethatthisisnot avalidformofargument(3rd rowofthetruthtableforp→q).

• Thestatementq→pisthe converseofthestatementp→q.• Theconverseofatruestatementmaybetrueorfalse,soemploying

theconverseasaboveresultsinaninvalidargument.

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Invalid Argument with an “Inverse Error” • Example:IfTedisamathmajor,thenTedhastotakeMAT152.Tedisnotamathmajor.Therefore,TeddoesnothavetotakeMAT152.

• Form: Ifpthenq~p∴~q

• Exercise:Checkbytruthtablethatthisisnot avalidformofargument.

• Thestatement~p→~qistheinverse ofthestatementp→q.• Again,ifastatementistrueitsinversemaybetrueorfalseso

employingtheinverseasaboveresultsinaninvalidargument.

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�Sound Argument�: Valid Argument with True Premises (and thus a true conclusion)

• Aninvalidargumentmayhaveatrueconclusionandavalidargument(withafalsepremise)mayhaveafalseconclusion,thoughthesedefinitionsareoftenusedinterchangeablyinordinaryspeech.

• Thevalidityoftheargumentdoesnotdependonthetruthvaluesofthepremises.

• Example:Invalidargument(converseerror)withtrueconclusion.IfDickCheneywasvicepresident,thenD.C.workedforHalliburton.D.C.workedforHalliburton.Therefore,D.C.wasvicepresident.

• Example:Validargument(modusponens)withfalseconclusion.Ifallhumansaremicethenallmicearesixfeettall.Allhumansaremice.Therefore,allmicearesixfeettall.

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Exercise: Spot the logical errorsParaphrasedfromMontyPythonandTheHolyGrail:Note:Shewas awitchintheendbecauseshedidweighasmuchasaduck((i.e.,trueconclusionthoughargumentisinvalid)

SIRBEDEMIR:Therearewaysoftellingwhethersheisawitch.VILLAGER#1:Arethere?BEDEMIRTellme.Whatdoyoudowithwitches?CROWD:Burn!Burnthemup!Burn!...BEDEMIR:Andwhatdoyouburnapartfromwitches?VILLAGER#1:Wood!BEDEMIR:So,whydowitchesburn?VILLAGER#3:B--...'causethey'remadeof...wood?BEDEMIR:Good!So,howdowetellwhethershe’smadeofwood?

VILLAGER#1:Buildabridgeoutofher.BEDEMIR:Ah,butcanyounotalsomakebridgesoutofstone?VILLAGER#1:Oh,yeah.BEDEMIR:Doeswoodsinkinwater?VILLAGER#2:No,itfloats!Itfloats!BEDEMIR:Whatalsofloatsinwater?(Villagersdon’tknow)KINGARTHUR:Aduck!BEDEMIR:Exactly.So,logically...VILLAGER#1:If...she...weighs...thesameasaduck,...she'smadeofwood.BEDEMIR:Andtherefore?VILLAGER#2:Awitch!

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Logical Equivalence: e.g., De Morgan

• Recallhowthe~(p⋀q)and ~p⋁ ~q columnsareidentical,i.e.,thesetwocompoundstatementsarelogicallyequivalent.

• Thisproofthat~(p⋀q)≡ ~p⋁ ~q isbytakingallcases(rows ofthetruthtable)ontheveracity(truthvalues)ofcomponentstatementspandq.

p q p⋀q ~(p⋀q) ~p ~q ~p⋁~q

T T T F F F FT F F T F T TF T F T T F TF F F T T T T

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Definition of Logical Equivalence

• Definition: Twostatementformsarelogicallyequivalentif,andonlyif,itisimpossibleforoneofthemtobetrueandtheotherfalse.

• Logicalequivalencemakesitconvenienttoexpressstatementsinmorethanoneway.

• Thesymbolforlogicalequivalenceis≡

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De Morgan�s Laws, Negation of �If p Then q�• DeMorgan’slaws:

~(p⋀q)≡ ~p⋁~q ,~(p⋁q)≡ ~p⋀ ~q

• Example:Thenegationof

-4<x≤7(read:-4<xandx≤7)

is

-4≥ xorx>7.

• Negationof→:~(p→q)≡ p⋀~q

• Example:Thenegationof

IfTomisAnn’sfather,thenLeoisheruncle.

is

Tomis Ann’sfatherandLeoisnot heruncle.

• So,byDeMorgan,p→q≡ ~(p⋀~q)≡ ~p⋁qandrecall

howp→qisvacuouslytrue.

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Contrapositive,Converse,Inverseofp →q

Fortheconditionalstatement“ifpthenq”:• theconverse is:ifqthenp• thecontrapositiveis: ifnotqthennotp• theinverse is:ifnotpthennotq.

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ConditionalStatement:IfSamlivesinChicagothenSamlivesinIllinois.

Converse:IfSamlivesinIllinois,thenSamlivesinChicago.

Contrapositive:IfSamdoesnotliveinIllinois,thenSamdoesnotliveinChicago.

Inverse:IfSamdoesnotliveinChicago,thenSamdoesnotliveinIllinois.

Example– Converse,InverseandContrapositive

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Which are Logically Equivalent?

• p→q≡~q→~p(conditionalstatementis logicallyequivalenttoitscontrapositive,recallmodustollens)

• p→q≢ q→p(conditionalstatementis notlogicallyequivalenttoitsconverse)

• p→q≢ ~p→~q(conditionalstatementisnot logicallyequivalenttoitsinverse)

• inverseiscontrapositiveofconverse,soinverseandconversearelogicallyequivalent:q→p≡~p→~q

• Proofbytruthtable(directlybyexhaustivecases):p q ~q ~p p→q ~q→~p q→p ~p→~ qT T F F T T T TT F T F F F T TF T F T T T F FF F T T T T T T

Converse (of an if-then statement) • Note from the truth table that “p→q” truedoes notimplythatitsconverse“q→p” istrue,i.e.,itsconversemaybetrueorfalse.

• Thatis,theconverseofatrueconditional(if-then)statementisnotnecessarilytrue.

• Ontheotherhand,when“p→q” isfalse(i.e.,pand~qaretrue),theconverse“q→p” is(vacuously)true.

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p q ~q ~p p→q ~q→~p q→p ~p→~ qT T F F T T T TT F T F F F T TF T F T T T F FF F T T T T T T

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Sets

• SetNotation:x∈ Ameansthat“xisanelementofthesetA,”or“xisinA.”

• Importantsetsofrealnumbers:ℝ,thesetofallrealnumbersℚ,thesetofallrationalnumbersℤ,thesetofallintegersℝ>0,thesetofallstrictlypositiverealnumbersℤ≥0=G,thesetofallnonnegativeintegers(wholenumbers)ℤ>0= ℕ,thesetofallstrictlypositiveintegers(naturalnumbers),sometimealsodenotedℤ+etc.

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Quantified Statements§ Apredicate isasentencethatisnotastatementbutcontainsoneor

morevariablesandbecomesastatementifspecificvaluesare

substitutedforthevariables.

§ Thedomainofapredicatevariable isthesetofallowablevaluesfor

thevariable.

§ Example: LetP(x)bethesentence“x2 >4” wherethedomainofxis

understoodtobethesetofallrealnumbers.ThenP(x)isapredicate.

§ Question: ForwhatnumbersxisP(x)true?

Answer:Thesetofallrealnumbersforwhichx>2orx<-2.

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Truth Set of a Predicate• Thetruthsetofapredicate P(x)isthesetofelementsinthedomainD

ofxforwhichP(x)istrue.

• WewritetruthsetofP(x)as{x∈D|P(x)}

andwereadthisas“thesetofallxinDsuchthatPofx.”

• Note:Theverticallinedenotesthewords“suchthat”fortheset-bracketnotationonly.

• Thewords“suchthat”mayalsobesymbolizedby“s.t.”or“s.th.” or“∋”or“:”

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Truth set of a predicate - example• LetP(x)bethepredicate“x2 <x”,withdomainthesetℝofallrealnumbers.

• Whichofthestatements P(0),P(1/2),P(2)andP(-3)aretrueandwhicharefalse?

• WhatisthetruthsetofP(x)?P(0): 02 <0 FALSEP(1/2): (½)2 <½ TRUEP(2): 22 <2 FALSEP(-3): (-3)2 <-3 FALSETruthset={x∈ℝ|0<x<1}

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UniversalQuantification∀• Thesymbol ∀,standingforthewords“forall”iscalledtheuniversal

quantifier.• Question:Rephrasethefollowinguniversalstatementinlessformal

language.∀MAT140studentsx,x hasstudiedcalculus.

• Answers:– AllMAT140studentshavestudiedcalculus.– EveryMAT140studenthasstudiedcalculus.– IfapersonisaMAT140student,thenthatpersonhasstudiedcalculus.

– etc.

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Definition of Counterexample• Definition: Givenauniversalstatementoftheform

“∀x∈D,P(x)”,acounterexample forthestatement

isavalueofx∈DforwhichP(x)isfalse.

• Notethat:∀x∈D,P(x)≡∀x,x∈D⇒P(x)wherethecontextisthat

xmaybelongtosomesupersetcontainingD(⇒is“implies”for

predicates).

• Alsonotethat“∀x∈D,P(x)”isastatementifDisdisclosedor

understood,otherwiseit’sapredicatewithvariableD.

• Example:Trueorfalse?

∀MAT140studentsx,xhasstudiedcalculus.

• Answer:ThestatementisfalsebecauseyouhappentoknowthatFred

isacounterexampletothestatement,i.e., becauseFredisaMAT140

studentandFredhasnot studiedcalculus.

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ExistentialQuantification∃• Thesymbol∃,standingforthewords“thereexists”,iscalledthe

existentialquantifier

• Example:Rephrasethefollowingexistentialstatementinlessformallanguage,anddeterminewhetheritistrueorfalse.

∃aMAT140studentx suchthatx hasstudiedcalculus.

• Answers:– ThereisaMAT140studentwhohasstudiedcalculus.– SomeMAT140studenthasstudiedcalculus.– SomeMAT140studentshavestudiedcalculus.– AtleastoneMAT140studenthasstudiedcalculus.– etc.

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Truth Values of Universal and Existential Statements, and their negations

• Universalstatement:“∀x∈D,Q(x)”istrue if,andonlyif,Q(x)istrueforeachindividualx∈D.

• Itisfalse if,andonlyif,Q(x)isfalseforatleastonex∈D.• AvalueofxforwhichQ(x)isfalseiscalledacounterexample tothe

statement“∀x∈D,Q(x)”.• So,~(∀x∈D,Q(x))≡∃x∈Dsuchthat~Q(x)

• Existentialstatement:“∃x∈DsuchthatQ(x)”istrue if,andonlyif,Q(x)istrueforatleastonex∈D.

• Itisfalse if,andonlyif,Q(x)isfalseforeachindividualx∈D.• So,~(∃x∈DsuchthatQ(x))≡∀x∈D,~Q(x)

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Universal/Existential Statement Examples• Determinewhetherthefollowingstatementsaretrueandrewritethem

formallyusingaquantifierandavariable:

• Allevenintegersarepositive.– False:-2isacounterexample;itisevenbutnotpositive.– ∀evenintegersn,nispositive.– ∀integersn,ifniseventhennispositive.– ∀n,ifnisanevenintegerthennispositive.– Noteinthepreviousanswer,thereisanimpliedsupersetoftheevenintegerstowhichnbelongs

• Someintegershaveintegersquareroots.– True:Thenumber4hasasquarerootof2,whichisaninteger.– ∃ anintegerxsuchthatxhasanintegersquareroot.– ∃xsuchthatxisanintegerandxhasanintegersquareroot.

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• Indicate,ifpossible,whichofthefollowingstatementsaretrueandwhicharefalse,andrewritethemformallyusingaquantifierandavariable:

• Noprimenumbersareeven.– False:Thenumber2isevenandprime.– ∀ primenumbersp,pisnoteven.– ∀ integersp,ifpisprimethenpisnoteven.– Notethattheintegersareasupersetoftheprimes.

• IfapersonisastudentinMAT140,thenthatpersonisatleast18yearsold.– True?– ∀ studentsPinMAT140,Pisatleast18yearsold.– ∀ peopleP,ifPisastudentinMAT140thenPisatleast18yearsold.– Notethatpeopleareasupersetofstudents.

Universal/Existential Statement Examples

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Negating Quantified Statements - Examples

1.∀studentsxinMAT140,xisatleast30yearsold.Negation:∃ astudentxinMAT140suchthatxislessthan30yearsold.

2.∃astudentxinMAT140suchthatxisatleast70yearsold.Negation:∀studentsxinMAT140,xislessthan70yearsold.

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Negating Quantified Statements - Examples

3.∀realnumbersx,x2 >0.Negation:∃arealnumberxsuchthatx2 ≤0.

4.∃arealnumberxsuchthatx2 =-1.Negation:∀realnumbersx,x2 ≠-1.

Notethattheoriginalstatementinbothexamplesabovehappenstobefalse.

Negating Quantified Statements and De Morgan’s Laws

• InSummary:~(∀x∈D,Q(x))≡∃x∈Dsuchthat~Q(x)…i.e.,acounterexample~(∃x∈DsuchthatQ(x))≡∀x∈D,~Q(x)• ButthesearejustDeMorgan’slawsonthe(potentiallyuncountablyinfinite)setD,becausewehavethefollowinglogicalequivalents:∀x∈D,Q(x)≡⋀x∈DQ(x)∃x∈DsuchthatQ(x)≡⋁x∈DQ(x)

• Forexample,ifthesetD={a,b,c}(hasthreeelements),then∀x∈D,Q(x)≡Q(a)⋀Q(b)⋀Q(c)∃x∈DsuchthatQ(x)≡Q(a)⋁Q(b)⋁Q(c)

• SobyDeMorgan,~(∀x∈D,Q(x))≡⋁x∈D~Q(x)55

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Universal Conditional Statements

• Definition:Anystatementofthefollowingformiscalledauniversalconditionalstatement:

∀x∈D,ifP(x)thenQ(x)

• Equivalentnotationforif-thenstatementsinvolvingpredicates:

∀x∈D,P(x)⇒Q(x)

i.e.,⇒means“implies”forpredicates.

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Universal Conditional Statement - Example• ∀ x∈ℝ,x>2⇒ x2 >4.• Read:Forallxinthesetofrealnumbers,ifxisgreaterthan2thenx-squaredisgreaterthan4.

• Lessformalversions:• Allrealnumbersthataregreaterthan2havesquaresthataregreaterthan4.

• Ifarealnumberisgreaterthan2,thenitssquareisgreaterthan4.

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Disproving a Universal Statement

CommonMethod:Findacounterexample!

Example:

• Isthefollowingstatementtrueorfalse?Explain.

∀x∈ℝ,ifx2 >25thenx>5.

• Solution:

– Letx=-6.Thenx2 =(-6)2 =36,and36>-6but-6≤5.

– So(forthisx),x2 >25andx≤5.

– So,thestatementisfalsebycounterexample.

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Negate a universal-conditional statement

• ~(∀x∈D,ifP(x)thenQ(x))≡∃x∈Ds.t.P(x)⋀~Q(x)

• Example:Writeanegationforthefollowingstatement∀x∈ℝ,ifx2 >4thenx>2.

• Negation:∃ x∈ℝs.t.x2 >4andx≤2.

• Notethatthefactthatthenegationistrue(andoriginalstatementisfalse)isimmaterialtowhatisbeingasked.

60

Example: Vacuous truth

• Supposethat:Therearenoelephantsonthisdesk.

• IsthefollowingstatementTrueorFalse?

Alltheelephantsonthisdeskhavetwoheads.(*)

• Thenegationofthisstatementis:

Thereisanelephantonthisdeskwithouttwoheads.

whichisclearlyfalse.

• So,theoriginalstatement(*)hastobetrue!

• Thisisanotherexampleof“vacuoustruth”or“truthbydefault”

becauseitinvolvesanempty-setquantifier:

∀x∈{elephantsonthisdesk},xhastwoheads,

wherethesetofelephantsonthedeskisempty.

• Thisisrelatedvacuoustruthofp→qwhenthepremisestatementp

isfalse;forpredicatestatementswewrite,

x∈{elephantsonthisdesk}⇒xhastwoheads

61

Arguments with Quantified Statements• Universalinstantiation:Ifapropertyistrueforalltheelementsinaset,

thenitistrueforeachindividualelementoftheset.• Example(multiplicationdistributesoveraddition):

∀a,b,c∈ℝ,a(b+c)=ab+ac.

Thus:7∙106=7(100+6)=7∙100+7∙6=742

And:Ifk∈ℝ,then2k∙3+2k ∙5=2k ∙(3+5) (=2k∙8=2k ∙23 =2k+3)

62

Universal Modus Ponens and Modus Tollens• Exampleuniversalfact(truestatement)

Allhumansaremortal.• Equivalently,

∀x,ifxishumanthenxismortal.• xisimplicitlyinthesupersetofelementsforwhichmortalityor

immortalityareattributes,i.e.,asupersetoflivingbeingsincludinggods,butnotinanimateobjectslikerocks.

• Byuniversalinstantiationthefollowingistrue:IfSocratesishuman,thenSocratesismortal.

• Sowecanargue(modusponens):Socratesishuman.∴Socratesismortal.

• Byuniversalinstantiation,thefollowingistrue:IfZeusishumanthenZeusismortal.

• Sowecanargue(modustollens):Zeusisnotmortal.∴Zeusisnothuman.

63

Necessary and Sufficient ConditionsExample:Whatdothefollowingsentencesmean?

• PassingallthetestsisasufficientconditionforJontopassthecourse.Answer:IfJonpassesallthetests,thenJonwillpassthecourse.

• PassingallthetestsisanecessaryconditionforJontopassthecourse.Answer:IfJondoesn’tpassallthetests,thenJonwon’tpassthecourse.Or:IfJonpassesthecourse,thenJonwillhavepassedallthetests.

• Notethatreplacing“sufficient”with“necessary”createsaconversestatement.

64

Necessary and sufficient conditions -Definitions

• AisasufficientconditionforB– meansifAthenB.– i.e.,theoccurrenceofAimplies/guaranteestheoccurrenceofB.

• AisanecessaryconditionforB– means if~Athen~B.– i.e.,ifAdidn’toccur,thenBdidn’toccureither.– Or,equivalently, itmeansthecontrapositive,ifBthenA.

– i.e.,ifBoccurredthenAalsohadtooccur.

65

Necessary and sufficient conditions -Examples

Writeeachofthefollowingusingif-thenstatements:

1. Beingatleast35yearsoldisanecessaryconditionforAnntobecomepresidentoftheU.S.IfAnnislessthan35yearsold,thenAnncannotbecomepresidentoftheU.S.IfAnnbecomespresidentoftheU.S.,thenAnnisatleast35yearsold.

2. Beingappropriatelydressedforajobinterviewisnecessary(butnotsufficient)forTomtogetthejob.IfTomisnotappropriatelydressedforajobinterview,thenTomwon’tgetthejob,butitcanhappenthatTomisappropriatelydressedandstilldoesn’tgetthejob.

66

Necessary and sufficient conditions –Examples (cont)

3. GettingallA’sissufficient(butnotnecessary)forSuetograduatewithhonors.

IfSuegetsallA’sthenshewillgraduatewithhonors,butit’spossibleforSuetograduatewithhonorsevenifshedoesn’tgetallA’s.

4. Supposeateachersays:Getting100%correctonalltheexamsisbothnecessaryandsufficientforyoutoearnanAinthecourse.Whatdoesthismean?IfyouearnanAinthecourse,thenyougot100%correctonalltheexams,andifyougot100%correctonalltheexams,thenyougotanAinthecourse.

67

Only if and the Biconditional

Example: Cinderella’sstepmothermakesthefollowingstatement.Writeitasanif-thenstatement:

Youhavepermissiontogototheballonlyifyoufinishallyourwork.

Answer:Ifyoudonotfinishallyourwork,thenyoudonothavepermissiontogototheball.

Or:Ifyouhavepermissiontogototheball,thenyouwillhavefinishedallyourwork.

68

Definition: Only If

• ronlyifsmeans if~sthen~r• Thatis,ifsdidn’toccur,thenrdidn’toccureither.• Or,equivalently,thecontrapositive, ifrthens• Thatis,ifristruethensalsohastobetrue.

69

Interpretation of If and Only If• So,ronlyifsmeans ifrthens

• and rifsmeansifsthenr(theconverseof:ifrthens)

• Thus,rifandonlyifs

– Means ronlyifs and rifs

– Whichmeans ifrthens and ifsthenr

• Symbolically,

r↔s≡(r→s)⋀(s→r)≡s↔r

≡risanecessaryandsufficientconditionfors

≡sisanecessaryandsufficientconditionforr

• Example:

Anemployeeisanon-profitcollege’sfootballcoachifandonlyif

theyarethehighestpaidemployeeoftheirnon-profitcollege.

r=“employeeisfootballcoachofnon-profitcollege”

s=“employeeishighestpaidinnon-profitcollege”

70

Introduction to DiscreteMathematics

Introduction to Methods of Proof© S.S. Epp (2006)

This revision G. Kesidis (2013)This document revision is copy-left 2013.

It is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License

71

Intro to Methods of Proof - Outline

• Directproofandcounterexample– Introduction– ExampleswithRationalNumbers– ExamplesonDivisibility– MethodofDivisionintoCases– ExampleswithFloorandCeilingOperators

• IndirectArguments:contradictionandcontraposition• Examples:√2isirrational,andthatthereareinfinitelymanyprimes

72

Elementary Number Theory and Methods of Proof

Assumptions:• Knowledgeofpropertiesoftherealnumbers(AppendixA),“basic” algebra,andlogic

• Propertiesofequality:A=A (=isreflexive)IfA=B,thenB=A. (=iscommutative)IfA=BandB=C,thenA=C. (=istransitive)

• Integersℤ={0,1,2,3,…,-1,-2,-3,…}={…-3,-2,-1,0,1,2,3,…}

• Anysum,difference,orproductofintegersisaninteger.

73

Definition of Even and Odd• Anintegeriseven↔itcanbeexpressedas2timessomeinteger.

• Anintegerisodd↔itcanbeexpressedas2timessomeintegerplus1.

Clarifyingtheroleof↔(ifandonlyif)inthesedefinitions:• Ifaninteger,sayn,iseventhenthereisanintegerksuchthatn=2k.

ANDtheconverse,• Ifaninteger,sayn,canbeexpressedas2k,forsomeintegerk,thenniseven.

Yes!

74

Exercise

• Question:Ifkisaninteger,is2k– 1anoddinteger?• Referencefact/definition:Anintegerisodd↔itcanbeexpressedas2timessomeintegerplus 1.

• Alsocanusebasicfacts,e.g.re.arithmeticwithintegers.• Answer:Yes(sotheaboveQuestionisaLemma).• Explanation/Proof:– Notethatk– 1isanintegerbecauseitisadifferenceofintegers.

– And2(k– 1)+1=2k– 2+1=2k– 1– Q.E.D.

75

Exercise

• Somepeoplesaythatanintegerisevenifitequals2k.Aretheyright?

• Is1anevennumber?

• Does1=2k?

• Yes:1=2(0.5)

• Soit’sprettyimportantthat“somepeople”requirektobeaninteger(therebyprecludingk=0.5),

• otherwisewehaveadisproofofwhat“somepeoplesay”bycounter-example,i.e.,takek=0.5.

76

Directly Proving a Universal Statement• MethodofGeneralizingfromtheGenericParticular: Ifaproperty

canbeshowntobetrueforaparticularbutarbitrarilychosenelementofaset,thenitistrueforeveryelementoftheset.

• Theorem: Picka(finite!)number.Add3.Multiplyby6.Subtract10.Divideby2.Subtract3timestheoriginalnumbertoarriveatyourfinalanswer,whichisalways4.

• Proof:Considerparticularbutarbitrary finitenumberx.Thefinalansweris(6(x+3)-10)/2-3x=3(x+3)–5-3x

=3x+9-5-3x=4.Q.E.D.

• Notethattheelementaryarithmeticoperationsintheproofholdforandarbitrarynumberx!

77

Direct Proving a Universal Statement Over a Finite Set

• MethodofExhaustion:Ifatheoremstatementconcernselementsofa

finiteset,thenthismethodprovesthetheoremseparatelyforeach

elementoftheset.

• Notethatourproofsoflogicalequivalencebytruthtablewere

exhaustive:checkingtheequivalenceforeverycombinationoftruth-

valuesoftheindependentvariables.

• Theorem:Provethateveryevenintegerfrom2through10canbe

expressedasasumofatmost3perfectsquares.

• Proof: 2=12 +12

4=22

6=22 +12 +12

8=22 +22

10=32 +12 Q.E.D.

78

Direct Proof - Example

• Question:Isthesumofanevenintegerplusanoddintegeralwayseven?alwaysodd?sometimesevenandsometimesodd?

• Theorem:∀x,y ∈ℤ,ifxisevenandyisodd,thenx+yisodd.• Proof:Takeparticularbutarbitraryx,y ∈ℤ.Byhypothesis,thereare

k,i ∈ℤ suchthatx=2kandy=2i+1.Thus,x+y =2k+2i+1=2(k+i)+1whichisoddbecausek+i ∈ℤ (ℤ isclosedunderaddition).Q.E.D.

• Notethatanexhaustiveproofwouldnotworkbecausetheintegersℤ isnotafiniteset.

79

Directproofgeneralizingfromthegenericparticular:form

Theorem:∀ xinD,if<hypothesis>then<conclusion>

Proof:• supposexisaparticularbutarbitrarilychosenelementinDsuchthat<hypothesis>istrue,and

• showthat<conclusion>istrueforx(notrelyingonanythingaboutxthefactthatx∈D).

• Q.E.D.(quoderat demonstrandum,i.e.,“whichwastobeshown).

80

Directions for Writing Proofs1. Copythestatementofthetheoremtobeprovedontoyourpaper

(exceptonexams!).

2. Clearlymarkthebeginningofyourproofwiththeword“Proof:”.

3. Writeyourproofincompletesentences.

4. Makeyourproofself-contained.(E.g.,introduceallvariables,but

donotrenamevariablesalreadydefinedthere!)

5. Giveareasonforeachassertioninyourproof.

6. Includethe“littlewords” thatmakethelogicofyourarguments

clear.(E.g.,then,thus,therefore,so,hence,because,since,Notice

that,etc.)

7. Makeuseofdefinitionsbutdonotincludethemverbatiminthe

bodyofyourproof.

81

Common Proof-Writing Mistakes1. Arguingfromnon-arbitraryexamples.

2. Usingthesamelettertomeantwodifferentthingsortwodifferentvariablestomeanthesamething.

3. Jumpingtoaconclusion.

4. Beggingthequestion.

5. Misuseoftheword “if.”

82

Rational Numbers• Definition: Arealnumberisrational if,andonlyif,itcanbewrittenasa

ratioofintegerswithanonzerodenominator.• Thatis:r∈ℚ⇔ ∃a,b∈ℤsuchthatr=a/bandb≠0.• (*)Numberswithfiniteorrepeatingdecimalexpansionsarerational.• Forexample,

43.205=43205/10002.3333…=7/3-1.2=(-6)/5=6/(-5)0=0/121.34343434…=:xwhere100x-x=2134-21sox=(2134-21)/99

• Theseexamplesdonotformaproofof(*).• Exercise:Buttheideasfortheproofarethere…

83

Example: ℚ closed under addition

• SimpleFact(ZeroProductProperty): Ifanytwononzerorealnumbersaremultiplied,theproductisnonzero.

• Theorem: Ifm,n∈ℤ≠0 (arenonzerointegers), then(n/m)+(m/n)∈ℚ(isrational).

• Proof:– Byexpressingwithacommondenominator(nm),wegetthat

(m/n)+(n/m)=(m2+n2)/(nm),where:– m2+n2 andmn∈ℤ(sinceintegersareclosedunderbothmultiplicationandaddition)and

– mn≠0bytheabovesimplefact.– Q.E.D.

• Exercise:Proveℚisclosedunderadditionandmultiplication.

84

Disproof of a universal statement by counter- example

Theorem:Therationals arenotclosedunderdivision.Proof:• Notethatweneedtoprovethefollowingstatementisfalse:∀x,y∈ℚ,x/y∈ℚ

• Equivalentlythatthefollowingstatementistrue:~(∀x,y∈ℚ,x/y∈ℚ )≡∃x,y∈ℚ s.t. x/y∉ℚ

• Considertherationalnumbersx=1=1/1andy=0=0/1.

• Theirquotientx/y=1/0isnotrational(notevenreal).

• Q.E.D.

85

Note regarding the last example

• Readthetheoremstatementcarefullyanddon’tconfusedefinitions.

• Theprevioustheoreminvolvesaquotientofrationalnumbers,not(directly)aquotientofintegers(asfactorsinthedefinitionofarationalnumber).

• Alsonotethatthetheoremstatementneedsitsconditionsbecausethequotientofanytworealnumbers,e.g.,20.5/3,isobviouslynotrational.

86

Divisibility

Definition: Givenanyintegersnandd≠0,thefollowingstatementsareequivalent.

• nisdivisibleby d• nisamultipleof d• disafactorof n• disadivisorof n• ddivides n• d| n• nequalsdtimessomeinteger• ∃k∈ℤ suchthatn=dk.

87

Examples

1.Is18divisibleby6?Answer:Yes,18=6∙3.

2.Does3divide15?Answer:Yes,15=3∙5.

3.Does5|30?Answer:Yes,30=5∙6.

4.Is32amultipleof8?Answer:Yes,32=8∙4.

5.Ifkisanyinteger,doeskdivide0?Answer:Yes,0=k∙0,except0ł 0.

88

Examples (cont)

• Fact:Ifaandbarepositiveintegersanda|b,thena≤b.

6.Whichintegersdivide1?Answer:Only1and-1.

7.Ifmandnareintegers,is10m+25ndivisibleby5?Answer:Yes.10m+25n=5(2m+5n)and2m+5nisaninteger

becauseitisasumofproductsofintegers(thesetofintegersisclosedundersumsandproducts).

89

Divisibility stated formally

nisdivisiblebyd(d≠0)impliesddividesn:d|n⇒∃k∊ℤ suchthatn=dk.

Thecontrapositiveofthisstatementis: ∀k∀k∊ℤ,n≠dk ⇒d ł n

Example:Does5|12?Solution: Nobecause12/5isnotaninteger.

Note:• 12/5isanumber:(twelfthfifths)=2.4• 5|12isastatement:“5divides12”.

90

Divisibility is transitive

Theorem:∀a,b,c∈ℤ,ifa|bandb|cthena|c.

Proof:• Takearbitrarya,b,c∈ℤ suchthata|bandb|c.• So,∃n,m ∈ℤ suchthatb=na andc=mb.• Bysubstitution,c=m(na)=(mn)a,wherethesecondequalityis

theassociativepropertyofmultiplicationofrealnumbers.• Sincetheintegersareclosedundermultiplication,mn ∈ℤ.• So,a|c.• Q.E.D.

91

Disproof of a universal statement by counter-example

Theorem:Thefollowingstatementisfalse:∀a,b ∈ℤ,ifa|b2 thena|b.

Proof:• Wearerequiredtoprovethatthefollowingstatementistrue:

~(∀a,b ∈ℤ,ifa|b2 thena|b)≡∃a,b ∈ℤ s.t. a|b2 anda ł b

• Fora=20andb=10,clearly:• 20,10∈ℤ• 20|102 where102=100• 20 ł 10• Q.E.D.

92

48

4 12

2 2 4 3

2 2

Prime and Composite Numbers• Definition:Anintegernisprime if,andonlyif,n>1andtheonly

positivefactorsofnare1andn.• Definition:Anintegerniscomposite if,andonlyif,n>1andn=rs

forsomepositiveintegersrandswhereneitherrnorsis1.• Note:Anintegerniscomposite if,andonlyif,n>1andn=rsfor

somepositiveintegersrandswhere1<r<nand1<s<n.• Also,anintegern>1iseitherprimeorcomposite.• Theorem (DivisibilitybyaPrime): Givenanyintegern>1,thereis

aprimenumberpsothatp| n.• Recallingthatdivisibilityistransitive:Tracingdownanybranchofa

binaryfactortreeleadstoaprimefactor(attheleaf)ofthenumber(attheroot),e.g.,

93

Unique Factorization Theorem(a.k.a. Fundamental Theorem of Arithmetic)

• UniqueFactorizationTheoremfortheIntegers: Givenanyintegern>1,eithernisprimeorncanbewrittenasaproductofprimenumbersinawaythatisunique,except,possibly,fortheorderinwhichthenumbersarewritten

• Ex.1:500=5�100=5�25�4=5�5�5�2�2=2�5�5�2�5=2253(standardfactoredform)

• Ex.2:5003 =(2253)3 =(2253)(2253)(2253)=2659

• Note:If1wereprime,thenprimefactorizationswouldnotbeunique.Forinstance,10=2�5=1�2�5=1�1�2�5.

• Note: theprevioustheoremsimplyimpliedby(acorollaryof)theUFT.

94

Examplere.8!=8�7�6�5�4�3�2�1

Ex.3:Howmany0’sareattheendof8!?

Solution:• 8!=8�7�6�5�4�3�2�1=23�7� (2�3)�5� (2�2)�3�2=27�32�5�7

• Eachfactorof10givesrisetoonezeroattheendofanumber.• 10=2�5.• Thefactorizationof8!containsonlyonepair2�5.• Sotheansweristhatthereisonlyonezeroattheendof8!

95

Quotient-Remainder Theorem • Example:Suppose14objectsaredividedintogroupsof3?

• Thatis:xxxxxxxxxxxxxx

• Theresultis4groupsof3eachwith2leftover.

• Wewrite14=4�3+2where– 4isthequotientand

– 2istheremainder(theremainderisalwaysnonnegativeandlessthanthedivisor,3).

• Equivalently,14/3=4+2/3

• Andbylongdivision:

4 ⃪quotient

divisor→3)14⃪dividend

-12

2⃪remainder

96

Quotient-Remainder Theorem• Forallintegersnandpositive integersd,thereexist

uniqueintegersqandrsuchthat

n=dq +rand 0≤r<d.

• Ex:Findqandrifn=23andd=6.

Answer:q=3andr=5

• Ex:Findqandrifn=–23andd=6.

Answer:q=– 4andr=1

97

Consequences: every integer is either even or odd

• Applyingthequotient-remaindertheoremwithd=2givesthatthere

existuniqueintegersqandrsuchthat

n=2q+rand0≤r<2.

• Whatarepossiblevaluesforr?

Answer:r=0orr=1

• Consequence:Nomatterwhatintegeryoustartwith,iteitherequals

2q+0(=2q)or2q+1forsomeintegerq.

• So: Everyintegeriseitherevenorodd.

98

Consequence: dividing by 3• Similarly:Givenanyintegern,applythequotient-remaindertheorem

withd=3.Theresultisthatthereexistuniqueintegersqandrsuchthat

n=3q+rand0≤r<3.• Whatarepossiblevaluesforr?Answer:r=0orr=1orr=2

• Consequence:Givenanyintegern,thereisanintegerqsothatncanbewritteninoneofthefollowingthreeforms:

n=3q, n=3q+1,n=3q+2.

• Similarlyforothervaluesofintegerdivisorsd>0:onecandefineddifferentgroupsofintegersbasedontheirremaindersafterdividingbyd,cf.equivalenceclasses.

99

Example Direct Proof by Division into CasesTheorem: Givenanyintegern,thereisanintegerksothatn2 =3korn2 =

3k+1.OutlineofProof:• Supposenisany integer.• Bythequotient-remaindertheoremwithd=3,thereisanintegerqso

thatn=3qORn=3q+1ORn=3q+2.

• Wewillshowthatregardlessofwhichofthesehappenstobethecase,theconclusionofthetheoremfollows.

• Case1,n=3qforsomeintegerq:(exercise:fillin)• Case2,n=3q+1forsomeintegerq:(exercise:fillin)• Case3,n=3q+2forsomeintegerq:(exercise:fillin)• Hence,ineverycasethereexistsanintegerksothat

n2 =3korn2 =3k+1.• Q.E.D.

100

Example Direct Proof by Division into Cases (cont)HowwouldwefillinCase2,forexample?

• Ifn=3q+1forsomeintegerq,thenn2 =(3q+1)2 bysubstitution

=(3q+1)(3q+1)

=9q2 +6q+1

=3(3q2 +2q)+1.

• Letk=3q2 +2q.• kisanintegerbecauseitisasumofproductsofintegers.

• Thusthereisanintegerksuchthatn2 =3k+1.

101

Proof by Division into Cases

• Howdoyouprove:IfAor BistruethenCisalsotrue.• Technique: ProveifAistruethenCistrueand ifBistruethenCistrue.• Thisisacorrectformofargumentbecause

(A⋁B)→C≡(A→C)⋀(B→C)• Exercise:checkthisbytruthtable,i.e.,thatthisformofproofisvalid.

Also,drawtherelatedVenndiagram.• SomesignificantthoughtmayberequiredtodividethestatementA⋁B

intothecases(hereA,B)whichresultinthesimplestproofsforeachcase(A→C,B→C).

• Obviously,morethantwocasesmaybeinplayandeachcasemayhavesub-cases,andeachsub-casehavesub-sub-cases,etc.

• Example:Allproofsoflogicalequivalencebytruthtableemploydivisionintocases,whereeachcasecorrespondstoarowinourtableformat.

102

IdeaofProof:

• Supposenisanarbitraryintegersuchthatn>1.Taketwocases.

• Ifnisprime,wearedone(sincen|n).

• Ifnot(i.e.,ifniscomposite),thenbydef’n n=rsforsomeintegersrands

satisfying

1<r<nand1<s<n.

• Ifeitherrorsisprime,wearedone.

• Ifnot,r=r1s1,wherer1 ands1 areintegerswith

1<r1 <rand1<s1 <r.

• Ifeitherr1 ors1 isprime,wearedone(asthesearefactorsofnbytransitivityofdivisibility).

• Ifnot,repeatwithr1 inplaceofr.Etc.

• Thisprocessmusteventuallyterminateataprimefactorofnbecause:each

successivefactorisapositivefactorofn(transitivityofdivisibility)andn<∞sothatthereareonlyafinitenumberofsuccessivelystrictlysmallerfactorsr(all>1bydefinition).

• Q.E.D.

Theorem:∀n∊ℤ,ifn>1then∃primeps.t.p|n.

103

Floor and Ceiling -mapping Reals to integers

• Definition:Givenanyrealnumberx,thefloorofx,denoted⌊x⌋(or[x]),istheuniqueintegernsuchthat

n≤x<n+1.• Definition:Theceilingofx,denoted⌈x⌉,istheuniqueintegernsuch

thatn– 1<x≤n.

• Examples:⌊2.8⌋=2;⌈2.8⌉=3;⌊–2.8⌋ =-3;⌈–2.8⌉=-2• Lemma:Forallintegersk,⌊k+1/2⌋=k.• Proof:k≤k+1/2<k+1,so⌊k+1/2⌋=kbydefinition.

104

Example theorem involving floor functionwith a direct proof not involving cases

Theorem:Ifnisodd,then⌊ n/2⌋=(n– 1)/2.

Proof:• Letnbeanarbitraryoddinteger.• Bydefinitionofodd,thereisanintegerksothatn=2k+1.

• ThentheLHSoftheequationis• ⌊ n/2 ⌋ =⌊ k+1/2 ⌋ bysubstitution

=k bylemmaofpreviousslide=(n-1)/2 bysubstitution

• Q.E.D.

105

Proof by Contradiction: Introduction• Definition:Acontradictionisastatementthatis�always� false.• Ifcisacontradiction,thenthefollowingisavalidformofargumentthatp

istrue:~p (premise,i.e.,presumedtrue)~p →c (bysomevalidargument)∴c (modusponens)butacontradictioncisalwaysfalsesothepremisepmustbefalse∴p (validconclusionbyproofbycontradiction)

• Notethatif~p →c≡p⋁cistrue(byavalidargument)andcisfalsethenpmustbetrue.

106

Method of Proof by Contradiction

• Suppose thestatementtobeprovedisnottrue.

• Show thatthissuppositionleadslogicallytoacontradiction.

• Conclude:Thesuppositionisfalse.Thatis,concludethatthegivenstatementistrue.

107

Example 1 – Proof by ContradictionTheorem:Thereisnolargestrealnumber.

Proof:• Supposetherewerealargestrealnumber,N.• ConsiderthenumberN+1.• N+1isarealnumber(realsareclosedunderaddition),andN+1islargerthanN.

• ThisresultcontradictsthesuppositionthatNisthelargestrealnumber.

• Hencethesuppositionisfalse.• Q.E.D.

108

Example 2 – Proof by ContradictionDefinition:Anirrationalnumber isarealnumberthatisnotrational.

Theorem: Thenegativeofanyirrationalnumberisirrational(∊ℚ*).Proof:

• Restatingthetheoremstatementformally:

∀x∊ℚ*,–x∊ℚ*

• Foraproofbycontradiction,assume thenegationofwhatistobe

proved:

∃x∊ℚ* s.t –x∊ℚ

• Takeanysuchx.

• Since–x∊ℚ,∃a,b ∈ℤ suchthatb≠0and–x=a/b.

• Thusx=(-a)/b.

• Sincea∈ℤ and-a∈ℤ.

• So,x∊ℚ.

• Thiscontradictsthepreviousassumption,whichmustthereforebe

false.

• Q.E.D.

109

Review of elementary definitions

• Anintegernisevenif,andonlyifnisequaltotwicesomeinteger.

• Anintegernisoddif,andonlyifnisequaltotwicesomeintegerplus1.• Arealnumberrisrationalif,andonlyifitisequaltoaquotientof

integers(withnonzerodenominatorotherwiserwouldnotbereal).• Givenintegersnandd≠0,ddividesn(d|n)if,andonlyif,nequalsd

timessomeinteger.• Anintegernisprimeif,andonlyif,n>1andtheonlypositiveinteger

divisorsofnare1andn.• Givenarealnumberx,thefloorofxistheintegern=⌊x⌋suchthat

n≤ x<n+1(nisthelargestinteger≤ x)• Givenarealnumberx,theceilingofxistheintegern=⌈x⌉suchthat

n– 1<x≤n(nisthesmallestinteger≥x)• Note:Thereareotherequivalentdefinitions.

110

Recall• Thequotient-remainder theorem:Forallintegersnandpositive

integersd,thereexistuniqueintegersqandrsuchthatn=dq +rand0≤r<d.

• The“transitivityofdivisibility” theorem:∀a,b,c∈ℤ,ifa|bandb|c,thena|c.

• Anintegerniscomposite ⇔n>1andnisaproductofpositiveintegers,neitherofwhichis1.

• Theuniquefactorization theorem:Givenanyintegern>1,eithernisprimeorncanbewrittenasaproductofprimenumbersinawaythatisunique,except,possibly,fortheorderinwhichthenumbersarewritten.

111

Proof by Contraposition: Outline

Toproveauniversalconditionalstatement(i.e.,oftheform“∀ xinD,ifP(x)thenQ(x)” bycontraposition:

1) Firstformulatethelogicallyequivalentcontrapositiveofthestatement,i.e.,“∀ xinD,if~Q(x)then~P(x)”

2) Useadirectproofforthecontrapositivestatement.

112

Example proof by contraposition

Lemma:∀ integersn,ifn2 iseventhenniseven.Proof:• The(logicallyequivalent)contrapositiveoftheconditionalstatement

inthelemmais∀ integersn,ifnisoddthenn2 isodd.

• Toprovethiscontrapositive,takeanarbitraryoddintegern.• Sincenisodd,∃anintegerksuchthatn=2k+1.• Thus,n2=(2k+1)2 =2(2k2+2k)+1.• Since2k2 +2kisaninteger(beingthesumofproductsofintegers),

n2 isodd.• Q.E.D.

Recall:Alemmaisastatementusedtoproveamoreimportantresult.Exercise:Trytoprovethistheoremdirectly.

113

Proving√2IsIrrational:Background

§ Foranynonnegativerealnumbern,thesquarerootofnisthenonnegativenumbern0.5 =√nwhich,whensquared(multipliedbyitself),equalsn.

§ Examples:√4=2,(25)0.5 =5,(16/9)0.5 =4/3

§ Recall:Anirrationalnumberisarealnumberthatisnotrational.

§ Notethat1/0isnotrationalasitisnotreal.

§ Recall:Everyrationalnumbercanbewrittenin“lowestterms” (withsmallestdenominatorinmagnitude)bycancellingoutcommonfactorsinthenumeratoranddenominator,i.e.,sothatthey’re“co-prime.”

§ Example:18/24=3/4(cancelledout6),20/(-10)=-2/1(cancelledout-10)

114

Theorem:√2IsIrrationalProofbycontradiction:• Supposenot.Thatis,assume√2isrational.• So,thereexistsintegersaandbwithb≠0suchthat√2=a/b

whereaandbareassumedtohavenocommonfactors.• Thus,a2 =2b2.• So,a2 iseven,andhencealsothataisevenbythepreviouslemma.• So,thereexistsanintegerksuchthata=2k.• Aftersubstitutingfora,weget4k2 =2b2 ,so2k2 =b2.• Thusb2 iseven,andhencebisevenbythepreviouslemma.• Sinceaandbarebotheven,theyhaveacommonfactorof2.• Thiscontradictstheassumptionthataandbhavenocommon

factors.• Thustheassumptionthat√2isrationalisfalse.• Q.E.D.

115

Theorem: There are infinitely many prime numbers

Proofbycontradiction:• Supposenot.Thatis,assumethereisafinitesetofprimenumbers

p1 ,p2 …,pn withn<∞.• Considertheproductoftheseprimesplus1,i.e.,theinteger

k=1+p1�p2�…�pn• Notethatkisanintegerlargerthanallprimes(obviouslyk>1

becauseweknowthereexistsomeprimeslike2,3,5,7,11,…).• Also,noprimenumberdividesk,indeeddivisionbyanyprime

leadstoaremainderof1(i.e.,kmodp=1forallprimesp).• Thuskisnotacompositenumberasithasnoprimefactors.• Thuskisprime,whichisacontradiction.• Q.E.D.

Euclid’s algorithm for greatest common divisor (gcd)

Def’n:gcd(A,B)isthegreatestcommondivisor(>0)ofintegersAandB.0. InitialdataA,BwithA≥B≥0.Also,initiallynotethatgcd(A,A)=Aand

gcd(0,0)isnotdefined(inlattercasereturn0orerrorstatement).1. IfB=0,thenreturngcd=AandSTOP.2. Elsefinduniqueintegersq,r suchthatA=Bq+r,where0≤r<B(i.e.,

quotient-remaindertheorem).3. SetA=BandthenB=r(i.e.,(A,B)→(B,r)),andgotostep1.Notes:• Exercisere.step2:provethatgcd(A,B)=gcd(B,r);i.e.,foreachiteration

ofsteps2and3,gcd(A,B)doesnotchange.• Asaresultofsteps2and3,B(≥0)isstrictly reduced(r<B).• Alsoinstep2,r=0impliesB=gcd(A,B)(i.e.,B|A).• So,Euclid’salgorithmwillconvergetofindthegcd inafinitenumberof

iterations(cf.wellorderingprinciple).• 1845theoremofLamegivescomplexityofEAusingFibonaccisequence,

seehttp://wwwdh.cs.fau.de/IMMD8/Lectures/THINF3/Texte/euclid4.pdf 116

Euclid’s algorithm - example

• Findgcd(284,16)byEuclid’salgorithm.• Euclid’salgorithmproceedsbythefollowingiterationssince,initially,A=284>B=16>0.

1. 284=16∙17+122. 16 =12∙1+43. 12=4∙3+0(remainderisnow0)4. B=0soreturnA=4– i.e.,Euclid’salgorithm

stopshere.• So,gcd(284,16)=4.

117

118

Introduction to DiscreteMathematics

Sequences and InductionThis revision G. Kesidis (2013)

This document revision is copy-left 2013. It is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License

Note: these slides adapted in part from those of S.J. Lomonaco Jr., available at

http://www.csee.umbc.edu/~lomonaco/f05/203/Slides203.html

119

Sequences and Induction- Outline

• Sequences• Induction• StrongInduction

120

Sequences• Sequences representorderedlists ofelements.

• Asequence isdefinedasafunctionfromasubsetofℕ (i.e.,fhascountabledomain)toasetS.

• Weusethenotationan =f(n)todenotetheimageoftheintegern.• Wecallan aterm(thenth term)ofthesequence.

• Example:f(n)=2nforn∈ℕ (hereSmustcontainallevenpositiveintegers);sinceherefisdefinedforallnaturalnumbers(positiveintegers),wecanwritef:ℕ→S.

121

Sequences• Weusethenotation{an}n∈ℕ todescribeasequence.• Whenpossibleitisconvenienttodescribeasequencewithaformula.• Forexample,thesequence{2,4,6,8,…}ofthepreviousslidecanbe

specifiedmorepreciselyas{an}n=1∞ ={an}n∈ℕ wherean=2n.

• Findingaformula,whenoneexistsandisunique,foragivennumericalsequence(orportionthereof)canbeverytricky.

122

Formulas for simple example sequencesWhataretheformulasthatdescribethefollowingsequencesa1,a2,a3,…?

• 1,3,5,7,9,… answer:an =2n– 1

• -1,1,-1,1,-1,… answer:an =(-1)n

• 2,5,10,17,26,… answer:an =n2 +1

• 0.25,0.5,0.75,1,1.25… answer:an =0.25n

• 3,9,27,81,243,… answer:an =3n

123

Series summations of numeric sequences• ∑k=mn ak represents the sum am + am+1 + am+2 + … + an.

• This sum has n-m+1 terms when n≥m (otherwisethesumissometimesdefinedtobezero).

• The variable k is called the index of summation, running from its lower limit m to its upper limit n.

• We could as well have used any other letter to denote this index, i.e., k is a “dummy” variable.

124

Summations - examples

• Expressincompactsummationnotationthesumofthefirst1000termsofthesequence{an}withan=n2 forn∈ℕ.

Answer:∑k=11000 k2

• Whatisthevalueof∑j=16 j?

Answer:1+2+3+4+5+6=21

125

Summations

• Itissaidthatasachild,Gausscameupwiththefollowingformula:

∑k=1n k=1+2+…+n=n(n+1)/2

• Forexample:1+2+…+100=100(101)/2=5050

• Gaussusedthefollowingargument:

• LetS(n)=∑k=1n k.

• Writingthesumforwardandbackward,weget:

S(n)=1+2+3+…+n-1+n

S(n)=n+n-1+n-2+…+2+1

• Notethatwehavealignednnumbersincolumns,e.g.,1andn,2andn-1,

3andn-2,etc.,eachcolumnwithsumn+1.

• Thus,byaddingthecolumnsweget

2S(n)=(n+1)+(n+1)+…+(n+1),

wheretherearenidenticalterms(n+1)beingadded.

• Thus,2S(n)=(n+1)n.

• Exercise:Writetheaboveasatheoremstatementanddirectproof.

Arithmetic sequences of real numbers• Anarithmeticsequenceisonewherethedifferencebetween

consecutivetermsisconstant,whentheyarearrangedineitherincreasingordecreasingorder.

• Letdbethisdifferenceanda1 bethe“initial”termofanarithmeticsequence.

• So,thekth termofthisseriesisak=a1+(k-1)d

• Forexample,thesequenceofnaturalnumbersisanarithmeticserieswithd=1anda1=1.

• Notethatifd>0thena1 isthesmallesttermofthesequence,butifd<0thena1isthelargesttermofthesequence.

• Alsonotethatthesetofalloddintegersisanarithmeticsequencewithoutalargestorsmallestterm(withd=2).

• Exercise:UseGauss’sdirectargumenttodirectlyprovetheidentity∑k=1

n (a1+(k-1)d)=n(2a1+(n-1)d)/2=n(a1+an)/2∀a1,d∊ℝ,n∈ℕ,i.e.,ntimestheaveragenumberinthesequence.

126

Arithmetic series - Example• Findthesumofthearithmeticseries7+4.5+…-18.

Solution:

• Fromthegivenfacts(includingthatwearedealingwithanarithmeticsequence),weimmediatelydeducea1=7andd=-2.5.

• Thefinal(nth)termisan=a1+(n-1)d=-18;so,n=11.

• Finally,applythepriorformulatogetthatthesumis

n(a1+an)/2=11(7– 18)/2=-60.5

• Exercise: Showthatthesamesumisobtainedbyaddingthearithmeticseriesinreverseorder,i.e.,witha1=-18andd=2.5.

127

Geometric sequences of real numbers• Ageometricsequenceisonewheretheratiobetweenconsecutive

termsisconstant.

• Letrbethisratioanda1 bethe“initial”termofageometricsequence.

• So,thekth termofthissequenceis

ak=a1rk-1

• Notethatif|r|>1thena1 isthesmallesttermofthesequenceinmagnitude(absolutevalue),butif|r|<1thena1isthelargesttermofthesequenceinmagnitude.

• Foraformulaforthesumofageometricseries,S:=∑k=1n a1rk-1,first

notethatrS=∑k=2n+1 a1rk-1

• Thus,(r-1)S=rS – S=a1rn – a1 aftern-2commontermscanceloutfromeachseriesrS andS.

• Andso,

∑k=1n a1rk-1=a1(rn-1)/(r-1)∀a1,r∊ℝ withr≠1.

• Exercise: Writetheaboveasatheoremstatementanddirectproof.128

Geometric series - Example• Findthesumofthegeometricseries3-3/5+…+3/625.Solution:• Fromthegivenfacts(includingthatwearedealingwithangeometric

sequence),weimmediatelydeducea1=3,r=-1/5.• Thefinal(nth)termisan=a1rn-1 =3/625;so,n=5.• Finally,applythepriorformulatogetthatthesumis

a1(1-rn)/(1-r)=…• Exercise: Showthatthesamesumisobtainedbyaddingthegeometric

seriesinreverseorder,i.e.,witha1=3/625andr=-5.

129

130

Two Other Useful Series

• ∑k=1n k=n(n+1)/2 …arithmetic

• ∑k=1n rk =(1-rn)/(1-r) …geometric

• ∑k=1n k2 =n(n+1)(2n+1)/6

• ∑k=1n k3 =n2(n+1)2/4

• We’llreturntotheproofsofthelasttwoidentitieslater.

131

Multiple Summations

• ∑j=ab ∑k=m(j)

n(j) f(j,k)= ∑j=abg(j)whereg(j):=∑k=m(j)

n(j) f(j,k)• Obviously,multiplesummationsarenumericallycomputedbynested

loops.• Example:

∑j=1b ∑k=1

j k= ∑j=1b j(j+1)/2=0.5∑j=1

b j2 +0.5∑j=1b j

=0.5b(b+1)(2b+1)/6+0.5b(b+1)/2

• Note:Thoughthesummationorderand“splitting”thesumupasintheexampleaboveisgenerallyOKforsumsoverfinitedomains,equivalenceafterswitchingtheorderofsummation(Fubini’s theorem)overinfinitedomainsrequiresuniformconvergence.

• Exercise: Workoutthepreviousexamplebyfirstswitchingtheorderofsummations,i.e.,∑k=1

b ∑j=kb k=∑k=1

b (b-k+1)(b+k)/2=…

Induction for proof of predicate over a countably infinite domain

• Supposewewanttoprove∀n∈D,Q(n)

wherethesetSisofcountablyinfinitesize,i.e.,Scanbeenumerated

(mapped)bythenaturalnumbersℕ orsubsetthereof.

• Clearly,wecannotprovebydividingintoindividualcasesifthe

numberofcasesisinfinite.

• Wemighttrytodirectlyprovethestatementbeginningwithan

arbitraryn∈S.

• Alternatively,aproofbyinductionmaybepossible.

• ForD=ℕ,theinductiveproofof“∀n∈ℕ,Q(n)”isoftheform:

1. ProveQ(1),i.e.,basecase.

2. Forarbitraryn∈ℕ,assumeQ(n),i.e.,inductiveassumption.

3. ProveQ(n+1),i.e.,inductivestep.

4. Q.E.D.

132

Induction for proof of predicate over a countably infinite domain (cont)

• Theinductiveassumptionistrueforthelowestindexn=1(basecase)ofthepredicatetobeproved.

• So,theinductivestepiteratively(ascounting)provesQ(n)forarbitrarilylargen∈ℕ startingfrom1.

• Moresuccinctly,theinductiveargumentonℕ is∀n∈ℕ,Q(n)≡Q(1)⋀(∀n∈ℕ,ifQ(n)⋀Q(1)thenQ(n+1))

133

Induction Example

Theorem:n<2n forallpositiveintegersn.

Proof:

• LetQ(n)bethepredicate“n<2n.”

• Q(1)istruebecause1<21 =2.

• Forarbitraryintegern≥1,assumeQ(n).

• Nownotethat:

n+1<2n +1(bytheinductiveassumption)

<2n +2n (sincen≥1,2n ≥21>1)

=2n+1

i.e.,wehaveestablishedQ(n+1).

• Q.E.D.

• Notethatinsteadwecouldhavearguedasn+1≤ 2n<2n+1 wherethesecondinequalityistheinductiveassumptiontimes2.

134

Induction Example

Theorem:∀n∈ℕ, ∑k=1n k2 =n(n+1)(2n+1)/6

Proof:

• LetQ(n)bethepredicate“∑k=1n k2 =n(n+1)(2n+1)/6”.

• Forn=1,wehavethat12=1(1+1)(2∙1+1)/6,i.e.,Q(1)istrue.

• Forarbitraryintegern≥1,assumeQ(n).

• ToestablishQ(n+1),beginwith:

∑k=1n+1 k2 =(n+1)2+∑k=1

n k2

=(n+1)2+ n(n+1)(2n+1)/6(inductiveassumption)=(n+1)(n+2)(2n+3)/6

• Q.E.D.Exercise:Checkthelastequality

Exercise:Provetheformulafor∑k=1n k3

135

Induction Example (cont)• In the previous example, the inductive step can instead be argued as:Q(n+1)≡ ∑k=1

n+1 k2 =(n+1)(n+2)(2n+3)/6

⇔ (n+1)2+∑k=1n k2 =(n+1)(n+2)(2n+3)/6

⇔ ∑k=1n k2 =(n+1)(n+2)(2n+3)/6- (n+1)2 …algebra…

⇔ ∑k=1n k2 =n(n+1)(2n+1)/6 ≡Q(n)

• SinceQ(n)istruebytheinductiveassumptionandeachstepaboveis“if

andonlyif”(⇔),Q(n+1)istrue.

• Exercise:Spotthelogicalflawinthefollowingargumentthat1=0

leveragingthefactthat1=1:

1=0

0=1

1+0=0+1

1=1

136

Strong Induction

• Theprincipleofstrongmathematicalinductioninvolvesa“stronger”inductiveassumption.

• Inordertoprove“∀n∈ℕ,Q(n)”,bystronginduction:1. ProveQ(1)(basecase)2. Forarbitraryintegern≥1,assumeQ(k)forall

k∈{1,2,…,n}(strong inductiveassumption)3. ProveQ(n+1).(inductivestep)Q.E.D.

137

Strong Induction Example 1Theorem:Ifs(0)=1,s(1)=2,ands(k+1)=2s(k-1)forallintegersk>0,

thens(k)=2⌊0.5k+0.5⌋ forallintegersk≥0.Proof:

• LetQ(k)bethepredicate“s(k)=2⌊0.5k+0.5⌋”.

• Q(0)istruebecause2⌊0.5∙0+0.5⌋ =2⌊0.5⌋ =20=1=s(0),wherethelastequalityisthetheorem’shypothesis.

• Forarbitraryintegerk≥0,assumeQ(n)foralln ∈{0,1,2,3,…,k}.

• ToproveQ(k+1),wetaketwocases.

– Ifk=0,then2⌊0.5∙1+0.5⌋ =2⌊1⌋ =21=2,establishingQ(1)bytheorem’shypothesis.

– Ifk>0,thens(k+1) =2s(k-1)…bytheorem’shypothesis

=2∙2⌊0.5(k-1)+0.5⌋ …bystronginductiveassump.

=2⌊0.5(k-1)+0.5⌋+1=2⌊0.5(k-1)+0.5+1⌋=2⌊0.5(k+1)+0.5⌋

• Q.E.D.138

Strong Induction Example 1 (cont)• Theproofofthepreviousexamplecanbereworkedas:1. ProveQ(0)andQ(1)areestablishedinthebasecase2. Forarbitraryintegerk≥1,assumeQ(k)andQ(k-1)

(inductiveassumption).3. ProveQ(k+1)byusingthedifferenceequation

(inductivestep).• Thisisavalidargumentbecause:

ifQ(0)⋀Q(1)and∀k≥1, Q(k)⋀Q(k-1) ⇒Q(k+1),then∀k≥0, Q(k).

• Thefollowingtheoremrequiresastronginductionproof.

139

Strong Induction Example 2

Theorem:Everyintegergreaterthanoneiseitherprimeorcanbewrittenasaproductofprimes.

Proof:• LetQ(n)bethepredicate“nisprimeorcanbewrittenasaproduct

ofprimes.”• Q(2)istruebecause2isprime.• Forarbitraryintegern≥2,assumeQ(k)forallk ∈{2,3,…,n}.• ToproveQ(n+1),wetaketwocases.

– Ifn+1isprime,Q(n+1)istruebydefinition.– Ifn+1isnotprime(composite),

• thenthereexistsintegersn≥a,b ≥2suchthatn+1=ab;• applytheinductiveassumptiontobothaandb,and• substitutetheirprimefactorizationsinton+1=abtoshown+1is

aproductofprimes.• Q.E.D.

140

Induction - Comments

• Notethattheinitialindexinthisexamplewas2,not1,andthattheinductiveassumptionwasadjustedaccordingly.

• Exercise:Caninductionbeusedtoproveapredicatetrueoverafinitedomain?Ifso,how(ifatall)doestheformoftheinductiveproofchange?

141

More general �bootstrapping� examples

• ConsiderapredicateP.• SupposethatwehaveprovedP(3).• FindthecompletesetofintegerskforwhichP(k)istrueifwealso

provedthat:∀k≥1,P(k+1)⇒P(k)⋀P(k+3).(*)

• Answer:The�bootstrap� (*)worksforP(k+1)onlywithk+1≥2.So:P(3)⇒P(2)⋀P(5)P(2)⇒P(1)⋀P(4)P(4)⇒P(3)⋀P(6)

• Sofar,weseehowP(3)canbebootstrappedtoshowP(k)fork∈{1,2,3,4,5,6}

• Continuinginthisway,weseethattheP(k)istrueforallintegersk≥1.• Exercise: WhyisP(0)notnecessarilytrue?

142

More general �bootstrapping� examples

• FindthecompletesetofintegerskforwhichP(k)istrueifP(5)istrueandwealsoprovedthat:

∀k>0,P(k)⇒P(2k)⋀P(k-1).(*)• Answer:

P(5)⇒P(10)⋀P(4)P(4)⇒P(8)⋀P(3)P(10)⇒P(9)⋀P(20)

• Sofar,weseehowP(5)canbebootstrappedtoshowP(k)fork∈{3,4,5,8,9,10,20}

• Continuinginthisway,weseethattheP(k)istrueforallintegersk≥0(notethat(*)cannotbeappliedwithk=0).

• NotethatP(5)and∀k>0,P(k)⇒P(2k)impliesthat∀k≥0,P(5∙2k),and∀k>0,P(k)⇒P(2k)”fillsthegaps”betweenthesenumbersandgivesP(1),P(2),P(3),P(4).

143

144

Remarks:InductionandCountability

• Asetiscountable ifeachofitselementcanbeplacedinone-to-onecorrespondencewithasubsetofthepositiveintegers,ℕ=ℤ>0

• Asetiscountablyinfinite ifitiscountable andhasaninfinitenumberofelements,e.g.,

– thesetofall integers,ℤ– thesetofallrationalnumbers,ℚ

• Toseethatℚ+ iscountable,constructthematrixwhose(i,j)th entryisi/jandcounttheseentriesalongtheanti-diagonalsstartingfrom1=1/1.

• Toseethatℚ iscountable,countzerothenproceedasforℚ+ butcounteachentryi/jtwice(onceforthepositiveandtheotherforthenegative).

• Uncountable setsinclude:

– thesetofrealnumbers,ℝ

– thesetofirrationalnumbers,ℚ*=ℝ-ℚ

• Inductioncannotbeusedtoprove“∀x∈D,Q(x)”foranuncountablesetD.

THE WELL-ORDERING PRINCIPLE -background

• Definition: LetBbeasetofintegers.AnintegermiscalledaleastelementofBifmisanelementofB,andforeveryjinB,m≤j.

• Example: 3isaleastelementoftheset{4,3,5,11}.• Example: IfAbethesetofallpositiveoddintegers,then1isaleastelementofA.

• Example: LetUbethesetofalloddintegers.ThenUhasnoleastelement.

• Exercise:Giveaproofofthepreviousstatementbycontradiction.

145

WELL-ORDERING PRINCIPLE -definition

• TheWell-OrderingPrinciple: IfBisanon-emptysetofintegersandthereexistsanintegercsuchthateveryelementjofBsatisfiesc<j,thenthereexistsaleastelementofB.

• Forexample:IfBisanonemptysetofnonnegativeintegersthenthereexistsaleastelementinB(sinceBhasthewell-orderingpropertywhencisanynegativenumber).

146

WELL-ORDERING - existence of quotient-remainder theorem

Theorem:Foreveryintegeraandpositiveintegerd>0thereexist

integersqandrsuchthat0≤r<danda=qd +r.

Proof:

• LetAbethesetofallvaluesa– kd forintegersksuchthat

a– kd ≥0,i.e.,A={a-kd |k∊ℤ,a-kd≥0}.

• Toapplythewell-orderingprinciple,weneedtoshowthatAisnotempty:

– Ifa ≥0,thena-kd=a ≥0fork=0,i.e.,a∊A;

– else(a<0),letk=asothata-kd=a-ad=a(1-d) ≥0 because

1-d≤0(sinced>0)anda<0,i.e.,a ∊A.

• Bythewell-orderingprinciple,Ahasaleastelement.

• Letthatleastelementber=a– qd whereq∊ℤ

• Soa=qd +rand0≤r.

• So,thetheoremisprovedifr<d,whichcanbeshownbycontradiction…

147

WELL ORDERING – existence of quotient remainder theorem (cont)

Proof(continued):• Supposethatd≤r.• Notethata=qd +r=qd +d+(r-d)=(q+1)d+r’wherer’:=(r-d)≥0.

• Alsor’=r-d<r,becaused>0.• Soa– (q+1)d=r’<r=a– qd,anda– (q+1)d=r’≥0.• Sor’=a– (q+1)disinA,andr’<r,thuscontradictingthefactthatristheleastelementofA.

• Q.E.D.

Recallthatforeacha,d,suchq,r inthetheoremstatementmustalsobeunique,andthatthisisquicklyprovedbycontradiction.

148

149

Introduction to DiscreteMathematics

Introduction to Set TheoryThis revision G. Kesidis (2013)

This document revision is copy-left 2013.

It is licensed under a Creative Commons Attribution-

Noncommercial-Share Alike 3.0 United States License

Note: these slides adapted in part from those of

© S.S. Epp and S.J. Lomonaco Jr., the latter available at

http://www.csee.umbc.edu/~lomonaco/f05/203/Slides203.html

150

• Setscontainelementsandarecompletelydeterminedbytheelementstheycontain.

• So:twosetsareequalifandonlyiftheyhaveexactlythesameelements.

• Notation:x∈Aisread�xisanelementofA� (or�xisinA�)• x∉Aisread�xisnotanelementofA� (or�xisnotinA�)

• Ex:Let A={1,3,5}B={5,1,3}C={1,1,3,3,5}D={x∈ℤ | xisanoddintegerand0<x<6}

HowareA,B,C,andDrelated?Answer:Theyareallequal.

A Glimpse into Set Theory

151

Sets – some notation and examples• {a,b}isthesetcontainingelements�a� and�b�• {a,{a}}isthesetcontainingelements�a� and{a},thelatterbeingtheset

containingtheelement�a�• So,setscontainelementsandelementsthemselvesmaybesets,cf.the

powersetofaset.• Cannotlistelementsas{a,b,…}foranuncountableset.• Fora,b∊ℝ,thefollowingareintervals(�connected� subsets)ofℝ (an�ordered� set):• (a,b]={x∊ℝ |a<x≤b}• [a,b)={x∊ℝ |a≤x<b}• [a,b]={x∊ℝ |a≤x≤b},a�closed� interval• (a,b)={x∊ℝ |a<x<b},an�open� interval

152

Subsets• Definition:GivensetsAandB,A⊆ B(read�AisasubsetofB”)

meanseveryelementinAisalsoinB.• Equivalently:A⊆ B⇔∀x,ifx∈Athenx∈B

• Note:A⊄ B⇔∃xsuchthatx∈Aandx∉B.• Note:A= B⇔A⊆ BandB⊆ A.

• Ex:LetA={2,4,5}andB={1,2,3,4,6,7}.IsA⊆B?• Answer:No,because5isinAbut5isnotinB.

• Ex:LetC={2,4,7}andB={1,2,3,4,6,7}.IsC⊆B?• Answer:Yes,becauseeveryelementinCisinB.

153

• GivensetsAandBthataresubsetsofa�universalset”Ω.• TheirunionisA⋃B={x∈Ω | x∈Aor x∈B}

where�or� isinclusive.• TheirintersectionA⋂B={x∈Ω | x∈A and x∈B}• ThesetdifferenceA– B={x∈Ω | x∈Aandx ∉ B}• ThecomplementofAisAc ={x∈Ω | x∉A}=U– A.• Notethatonlycomplementarityrequiresspecifyingtheuniversalset.• Venndiagrams:

Definitions of Basic Set Operations

A È B

A Ç B

A – B

Ac

154

The Empty Set• Example:Assumepeopleliveinonlyoneplace.LetAbethesetof

allthepeopleintheroomwholiveinChicagoandBbethesetofallpeopleintheroomwholiveoutsideChicago.

• WhatisA⋂B=A⋂Bc ? NoteB=AcAnswer:Thissetcontainsnoelementsatall.

• Notation:Thesymbolϕ denotesasetwithnoelements.• Wecallittheemptysetorthenullset.• ToshowasetAisempty,onemightfirstassumethereisanelement

ofAandproveacontradictionresults.• Exercise:Show(5,3]=ϕ.• Notethatϕ ⊆AforanysetA(vacuouslytruebythe�if…then…�

definitionofthesubsetrelation,⊆).

155

Example

• LetA={1,2,3}andB={3,4,5}andlettheuniversalsetbeΩ ={1,2,3,4,5,6,7,8}.

• A⋃B={1,2,3,4,5}• A⋂B={3}• A– B={1,2}=A– (A⋂B)=A⋂Bc• Ac ={4,5,6,7,8}

156

Venn Diagrams• TheVenndiagrambelowrepresentssetsA,B,andCsothat

A⊆C,A⋂B=ϕ (i.e.,AandBaredisjoint),andB⋂ C≠ϕ.• AVenndiagramrepresentsinfinitelymanydifferentdiscreteinstances,

e.g.,A={1},C={1,2},andB={2,3},sothatA⊆C,A⋂B=ϕ,andB⋂ C={2}≠ϕ.

• WeallowVenndiagramstocontainemptyspaces,sowemayplaceadotinsidetheregionB⋂Ctoexplicitlyindicateotherwise.

• Notethataccordingtothe3facts- A⊆C,A⋂B=ϕ,B⋂ C≠ϕ – it’spossiblethatB⊆C(astrongercondition)butit’salsopossiblethatB⊄C.

Reviewofsettheory

• SupposefollowingsetsaresubsetsofsomeuniversalsetΩ.• “xisanelementofA”isdenotedx∈A• AisasubsetofB,denotedA⊆B,whenthefollowingstatementholds:

∀(forall)x∈Ω,ifx∈Athenx∈B• TheintersectionoftwosetsAandBistheset

AB=A∩B={s∈Ω |s∈Aand s∈B}={s∈Ω |s∈A, s∈B}=BAwherethesymbol“|”denotes“suchthat”(asdoes“:”)

• TheunionoftwosetsAandBisthesetA∪B={s∈Ω |s∈Aor s∈B}=B∪A

where“or”isinclusiveallowings∈AB• Thedifferenceoftwosetsisdefinedas

A- B=A\ B={s∈Ω |s∈Aands∉ B}=A∩Bc =A-AB• Theexclusiveunionoftwosetsisdefinedas

A⊕B={s∈Ω |s∈Aor s∈Bbut s∉ AB}=A∪B- AB• ThecomplementofasetAc ={s∈Ω |s∉ A}=Ω –A(hereneedtospecifyΩ)• NotethatΩc=ϕ,theemptyset,andϕc=Ω.• Forexample,ifA={1,2}andB={2,3,4},thenA∪B={1,2,3,4},AB={2},

A-B={1},B-A={3,4},A⊕B{1,3,4}.157

Venndiagramsforsetoperations/relations,whereeverythinginsidetheboxisΩ

158

E

F

(h)E⊆F

Ω

DeMorgan’stheorem- adirectproofTheorem:∀setsA,B,(A∩ B)c =Ac ∪ Bc

Proof:• Plan:showtwosetsareequalbyshowingtheycontaineachother.• ChoosearbitrarysetsA,B(addressingtheuniversalqualifier∀).• FirstshowAc ∪ Bc ⊆ (A∩ B)c :• Ifx∈Ac ∪ Bc thenx∉Aorx∉Bbydefinitionofunion,soconsiderthe

twocases.1.Ifx∉Athen x∉A∩ BbecauseA∩ B⊆A.2.Ifx∉B then x∉A∩ BbecauseA∩ B⊆B.

• Thus,x∉A∩ B(asthisisaconsequenceofbothpossiblecases).• Nextshow(A∩ B)c ⊆ Ac ∪Bc.• Ifx∈A∩ Bthenx∈B andx∈A,i.e.,xisanelementofboth.• So,ifx∈(A∩ B)c then:xcannotbeinbothAandB,i.e.,eitherx∉Aor

x∉B,equivalentlyx∈Ac ∪Bc.• Q.E.D.

159

DeMorgan’stheoremforlogic–

Directproofbyexhaustivecases

• Statementspandq,e.g.,foranarbitraryx∈ Ω,p=x∈Aandq=x∈B.

• p,qareeitherTrue=1orFalse=0.

• Thelogicalnegationoperatoris~,e.g.,~p=x∉ A.

• Notethatx∈AB≡p∧q(i.e.,pandq),x∈A∪B≡p∨q(i.e.,porq).

• ProofofDeMorgan’stheorembytruthtable(allcombinationsofp,q):

• Notethelogicalequivalence(bycolumn)~(p∧q)≡~p∨~q.Q.E.D.

• Exercise:Show(A∪ B)c=A

c∩B

cforallsetsA,B,asanimmediateconsequence

(corollary)oftheabovehalfofDeMorgan’stheorembytruthtable.

• Exercise:Showthemorebasic“contrapositive”:A⊆BifandonlyifBc⊆Ac.

p q ~p ~q ~p�~q p�q ~(p�q)0 0 1 1 1 0 10 1 1 0 0 1 01 0 0 1 0 1 01 1 0 0 0 1 0

160

pq ~(p∧ q)≡~p∨ ~q

p ~p∨ ~qq

DeMorgan’stheoremforequivalentAND(∧)andOR(∨) gateconfigurations

161

DeMorgan’stheorem–

Proofbycontradiction:background• SupposewanttoshowX≝ (A∩B)

c⊆ Y≝ A

c∪ B

cforarbitraryA,B⊆Ω.

• Definethestatementsp=z∈Xandq=z∈Yforarbitraryz∈Ω.

• So,wewanttoshowp→q,i.e.,“ifpthenq”.

• Thenegationofthisstatement,~(p→q)≡p∧ ~q

• Toprovep→qistrue,wewillshowp∧ ~qresultsinacontradiction

(andsoisfalse).

p q p→q p � ~q

1 1 1 01 0 0 10 1 1 00 0 1 0

162

DeMorgan’stheorem–Proofbycontradiction (cont)

• Assumep∧ ~q,i.e., z∈X=(A∩B)c butz∉Y=Ac ∪ Bc.

• Ifz∉Ythenz∉ Ac andz∉ Bc otherwisezwouldbeintheirunion(Y).

• So,z ∈Aandz∈B.

• So,z ∈A∩B,whichcontradictsz∈X.

• Thus,thefirstassumptionaboveisfalseandsoX⊆Y.

• Similarlyshowq∧ ~pisfalsesothatY⊆ X.

• Thus,X=Y.

• Q.E.D.

Inthiscase,thisargumentisverysimilartothedirectproofabove.

163

DeMorgan’stheoremforarbitrarynumberofsets-Proofbyinduction

Theorem:(A1∪A2 ∪…∪An)c =A1c∩ A2c ∩ … ∩ Anc foranysetsAk andanyn∈{2,3,4,…}.Proof:• DefinepredicateQ(n)=“(A1∪A2 ∪…∪An)c =A1c∩A2c ∩ … ∩ Anc ”• WehavealreadyestablishedQ(2).• Stronginductiveassumption:Forarbitraryintegern≥2,assumeQ(k)

forallk∈{2,3,…,n}.• Tocompletetheinductiveproof,weneedtoshowQ(n+1):• Tothisend,letB=A1∪A2 ∪…∪An• ByQ(2),(B∪An+1)c=Bc∩ An+1candsubstitutetheinductive

assumptionQ(n)appliedto Bc.• Q.E.D.

Note: Boththebasecaseandinductiveassumptionwereusedtoprovetheinductivestep.Also,wecouldhaveusedweakinduction.

164

Miscellaneoussimplesetidentities• Commutativepropertyof∪

A∪B=B∪AforallsetsA,B• Associativepropertyof∪

(A∪B)∪C=A∪(B∪C)forallsetsA,B,C• Similarly,associativeandcommutativepropertiesof∩• Distributivepropertyof∩over∪:

(A∪B)∩C=(AC) ∪(BC)forallsetsA,B,C.• Distributivepropertyof∪ over∩:

(AC) ∪B=(A∪B)∩(C∪B)forallsetsA,B,C.• Exercise:provetheseidentities.

165

PartitionsandDisjointCoverings• Ann-partitionofasetBisacollectionofnsetsΔ1,Δ2,…,Δn suchthat:

– Noneareempty,Δk≠ ϕ forallk– Covering,B=Δ1 ∪ Δ2 ∪…∪ Δn– Disjoint(non-overlapping), Δk∩Δj=ϕ forallk ≠ j

• Wecanwriteanobviousinclusion-exclusionformulafortheunionoftwoarbitrarysetsastheunionof3disjointsets(adisjointcovering):A∪B=[A-B]∪[B-A]∪[AB]

• Similarly,wecanwritetheunionofthreearbitrarysetsasdisjoint-coveringof7sets:A∪B∪C=[A-(B∪C)]∪[B-(A∪C)]∪[C-(A∪B)]∪[AB-C]∪ [BC-A]

∪ [CA-B] ∪[ABC]=ABcCc ∪AcBCc∪AcBcC ∪ABCc∪AcBC ∪ABcC∪ABC

• Ignoringtheemptysets,these“inclusion-exclusion”decompositions(disjoint-coverings)formpartitionsoftheleft-hand-sideunionsofarbitrarysets.

• NotethatAcBcCc⊄A∪B∪CbyDeMorgan’slaw. 166

Partitioning∪i=13 Ei intosevendisjointregions(Δ1,...,Δ7)byinclusion/exclusion

Exercise:WriteeachΔsetintermsoftheEsets.

167

Disjointcoveringsformulaforunionofan

arbitrarynumberofsetsbyinduction

• WriteA1∪A2 ∪…∪Ak =B∪Ak whereB=A1 ∪…∪Ak-1

• Findthe3-disjoint-coveringofB∪Ak byinclusion-exclusion.

• NotethatbyDeMorgan,thetermtoAk –B=Ak ∩Bc simplifiestoa

singleelementoftheinclusion-exclusionexpansion.

• Applytheinductivelyassumedknowninclusion-exclusionexpansion

forB=A1 ∪…∪Ak-1 toeachoftheothertwoterms.

• Eachoftheseothertermswillproducen-1elementsafterdistribution

of-Ak (∩Akc)or∩Ak

• Notethatifg(k-1)isthenumberofsetsinvolvedintheinclusion-

exclusionformulafork-1sets,theng(k)=2g(k-1)+1.

• Exercise:Fillinthedetailsofthisinductiveargument.

• Exercise:Doestheargumentworkifyoudistribute∪Ak overthe

expansionofBfirst,andthenapplythe3-partitionineachofthe

resultingterms?

168

AnInclusion-ExclusionFormulawithNot-DisjointSets

• ConsiderfoursetsA,B,C,Dnotnecessarilydisjoint.• Aninclusion-exclusionformulafor

P(A⋃B⋃C⋃D)=P(A)+P(B)+P(C)+P(D)-P(AB)-P(AC)-P(AD)-P(BC)-P(BD)-P(CD)+P(ABC)+P(ABD)+P(ACD)+P(BCD)-P(ABCD)

• Note:P(ABCD)isadded4=C(4,1)timesinP(A)+…+P(D),6=C(4,2)timesin–P(AB)-…-P(CD),andadded4=C(4,3)timesinP(ABC)+…+P(BCD),andthusmustsubtractedoncesothatitcontributesonlyoncetotheuniononLHS.

• Exercise:Stateandprovethisinclusion-exclusionformulaforarbitraryfinitenumberofsetsnbyinduction.

169

Cartesian products• Supposewewanttodefinethesetof“outcomes”aftertossingtwo

different6-sideddice(i.e.,thenumbersontheirupturnedfacesaftertheystopmoving).

• Eachdiehasoutcomesbelongingtotheset{1,2,3,4,5,6}.• So,thepairofdicehasoutcomesbelongingtotheCartesianproduct,

{1,2,3,4,5,6}� {1,2,3,4,5,6}={1,2,3,4,5,6}2={(n,m)|n,m∈{1,2,3,4,5,6}},wherenistheindexofthefirstdie,andmisthatofthesecond.

• Thetotalnumberofpossibleoutcomesofthepairis6⨯6=36.• Similarly,ℝ�ℝ�ℝ=ℝ3 ={(x,y,z)|x,y,z∊ℝ}.• TheCartesianproductofintervalsofℝis

(a,b]�[c,d]�(e,f)={(x,y,z)|a<x≤b,c≤y≤d,e<z<f}⊆ℝ3

170

Power Set

• ThepowersetofasetX,denoted2X,isthesetofallsubsetsofX,includingtheemptyset,ϕ,andXitself.

• Forexample,ifX={1,2},then2X={ϕ,{1},{2},X}

• Again,notehowtheelementsofthepowerset2X arethemselvessets.

• IfthenumberofelementsinXisfinite,i.e.,|X|<∞,thenthesizeofitspowersetis2|X|,i.e.,|2X|=2|X|.

• Exercise:Provethisbyinductionon|X|.

• Cf.thebinomialtheorem.

171

172

Introduction to DiscreteMathematics

Introduction to Probability TheoryThis revision G. Kesidis (2013)

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3.0 United States License

See slidedeck on undergraduate probability

173

Introduction to DiscreteMathematics

Functions & Pigeonhole PrincipleThis revision G. Kesidis (2013)

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Share Alike 3.0 United States License

Note: these slides adapted in part from those of S.J. Lomonaco Jr.,available at

http://www.csee.umbc.edu/~lomonaco/f05/203/Slides203.html

174

Functions - definition• AfunctionffromasetAtoasetBisanassignment

– ofexactlyone elementofB

– toeachandevery elementofA.

• Wewrite

f(x)=y

ifyistheuniqueelementofBassignedbythefunctionftox∈A.

• So,iffisafunctionthen:∀x∈A,∃!y∈B s.t. f(x)=y

• IffisafunctionisamappingfromsetAtoB,wewrite

f:A→B

175

Functions – domain, co-domain and range

• Iff:A→B,wesaythatAisthedomainoffandBistheco-domainoff(theco-domainissometimescalledtherange).

• Iff(x)=y,wesaythatyistheimageofxunderfandxisapre-imageofyunderf.

• Therangeoff:A→B isthesetofallimagesofelementsofA(this�range� issometimescalledthestrictrange),whichisdenoted

f(A):={f(x)|x∈A}⊆ B

176

Functions –example where range = co-domain

Let’stakealookatthefunctionf:P→Cwith• P={Linda,Max,Kathy,Peter},and• C={Boston,NewYork,HongKong,Moscow}.

Suppose:• f(Linda)=Moscow• f(Max)=Boston• f(Kathy)=HongKong• f(Peter)=NewYork

Here,therangeoffisC,i.e.,f(P)=C.

177

Functions –example where range ⊊ co-domain

Letusre-specifyfasfollows:

• f(Linda)=Moscow• f(Max)=Boston• f(Kathy)=HongKong• f(Peter)=Boston

• Isfstillafunction? Answer:Yes.• Therangeoffisjust{Moscow,Boston,HongKong}

178

Functions – table or bipartite-graph representations

• Otherwaystorepresentafunctionf:

BostonPeter

Hong KongKathy

BostonMax

MoscowLinda

f(x)x Linda

Max

Kathy

Peter

Boston

New York

Hong Kong

Moscow

179

FunctionsIfthedomainofourfunctionfislarge,itisconvenienttospecifyfwithaformula,e.g.,

f:ℝ→ℝ withf(x)=2x,∀x∈ℝ

Thisleadsto,e.g.,• f(1)=2• f(3)=6• f(-2.5)=-5

180

Sums and products of functions• Letf1 andf2 befunctionsfromAtoℝ.• Thenthesumandtheproductoff1 andf2 arealsofunctions

fromAtoℝ definedby:(f1 +f2)(x)=f1(x)+f2(x),∀x∈ℝ(f1f2)(x)=f1(x)f2(x),∀x∈ℝ

• Examples:∀x∈ℝ:f1(x)=3x,f2(x)=x+5(f1 +f2)(x)=f1(x)+f2(x)=3x+x+5=4x+5(f1f2)(x)=f1(x)f2(x)=3x(x+5)=3x2 +15x

181

Functions – image of a domain subset

• Recalltherangeofafunctionf:A→Bisthesetofallimagesf(x)ofelementsx∈A,denotedf(A).

• IfweonlyregardasubsetS⊆A,thesetofallimagesofelementss∈S iscalledtheimageofS.

• WesimilarlydenotetheimageofSbyf(S):

f(S)={f(s)|s∈S}⊆ B

A BS f(S)f

182

FunctionsConsiderthefollowingfunction:• f(Linda)=Moscow• f(Max)=Boston• f(Kathy)=HongKong• f(Peter)=Boston

• WhatistheimageofS={Linda,Max}?• Answer:f(S)={Moscow,Boston}

• WhatistheimageofS={Max,Peter}?• Answer:f(S)={Boston}

183

Properties of Functions –one-to-one/injective

• Afunctionf:A→B issaidtobeone-to-one(orinjective),if∀x,y ∈A,f(x)=f(y)⇒ x=y.

• Sincefisafunction,wecanstatethatit’sone-to-oneif∀x,y ∈A,f(x)=f(y)⇔ x=y.

• Inotherwords:fisone-to-oneifandonlyifitdoesnotmaptwodistinctelementsofAontothesameelementofB.

• Thatis,theabove(first)definitionin(logicallyequivalent)contrapositiveformis:Afunctionf:A→B issaidtobeone-to-one(orinjective),if

∀x,y ∈A,x≠ y⇒ f(x)≠ f(y).

184

Properties of Functions –one-to-one: example

Considerthefunctionsfandg:• f(Linda)=Moscow• f(Max)=Boston• f(Kathy)=HongKong• f(Peter)=Boston

• Isfone-to-one?

• No,MaxandPeteraremappedontothesameelementoftheimage.

• g(Linda)=Moscow• g(Max)=Boston• g(Kathy)=HongKong• g(Peter)=NewYork

• Isgone-to-one?

• Yes,eachelementisassignedauniqueelementoftheimage.

185

Properties of Functions –disproving one-to-one property

• Howcanweprovethatafunctionfisone-to-one?

• Wheneveryouwanttoprovesomething,firsttakealookattherelevant

definition(s);here,

∀x,y ∈A,x≠ y⇒ f(x)≠ f(y)

• Example:

f:ℝ→ℝ withf(x)=x2

Note~[∀x,y ∈A,x≠ y⇒ f(x)≠ f(y)]≡∃x,y ∈As.t. x≠y ⋀ f(x)=f(y)

Disproofofone-to-onestatementbycounterexample:x=3,y=-3,i.e.,

3≠-3butf(3)=f(-3),sofisnotone-to-one.

• Forℝ-valuedfunctions,thereisthehorizontallinetest:i.e.,ifyoucan

drawahorizontallinethatmeetsthegraphofthefunctionatmorethan

onepoint,thenthefunctionisnotone-to-one.

• Ifbipartite-graphrepresentationoff(withfinite-sizeddomainandco-

domain)hastwoarrowsthatmeetatthesameco-domainelement,thenf

isnotone-to-one.

186

Properties of Functions –one-to-one example

• …andyetanotherexample:f:ℝ→ℝ with f(x)=3x

• Def’n:f:A→B isone-to-one,orinjective,if:∀x,y ∈A,f(x)=f(y)⇒ x=y.

• Exercise: Writethisdefinitionincontrapositiveform.

• Toshowfisone-to-oneconsiderarbitraryx,y ∈Asuchthatf(x)=f(y)⇒ 3x=3y⇒x=y.

187

Increasing and decreasing functions are one-to-one

• Considerafunctionf:A→B withA,B⊆ ℝ (bothorderedsets).

• fiscalledstrictlyincreasing,if

∀x,y∈A,x<y⇔ f(x)<f(y),

• fisstrictlydecreasing,if

∀x,y∈A,x>y⇔ f(x)<f(y).

• Exercise: Provethatafunctionthatiseitherstrictlyincreasingorstrictly

decreasingisone-to-one

– byusingtheoriginaldefinitionofone-to-oneandcontradiction,and

– byusingthecontrapositivefprm andadirectproof.

188

Properties of Functions –onto/surjective, bijective

• Afunctionf:A→B iscalledonto,orsurjective,ifandonlyifforeveryelementy∈B thereisanelementx∈A withf(x)=y.

• Inotherwords,fisontoifandonlyifitsrangeisitsentireco-domain,i.e.f(A)=B.

• RecallthatasetSiscountableifthereisasurjective fsuchthatf:ℕ→S.

• Afunctionf:A→Bisbijective ifandonlyifitisbothone-to-oneandonto.

• Exercise:Provebycontractionthatiff:A→B isabijection andAandBarefinite-sizedsets,then|A|=|B|.

189

Properties of Functions - examples

• Inthefollowingexamples,weusethebipartite-graph(arrows)representationtoillustratefunctionsf:A→B.

• Ineachexample,thecompletesetsAandBareshown.

190

Properties of Functions - example

• Isfinjective?• No.• Isfsurjective?• No.• Isfbijective?• No.

Linda

Max

Kathy

Peter

Boston

New York

Hong Kong

Moscow

191

Properties of Functions - example

• Isfinjective?• No.• Isfsurjective?• Yes.• Isfbijective?• No.

Linda

Max

Kathy

Peter

Boston

New York

Hong Kong

Moscow

Paul

192

Properties of Functions - example

• Isfinjective?• Yes.• Isfsurjective?• No.• Isfbijective?• No.

Linda

Max

Kathy

Peter

Boston

New York

Hong Kong

Moscow

Lübeck

193

Properties of Functions - example

• Isfinjective?• No.• fisnotevenafunction

becausex=ydoesnotimplythatf(x)=f(y)forthecasewherex=Peter,i.e.,fistwovaluedatPeter:f(Peter)={NewYork,Lubeck}.

Linda

Max

Kathy

Peter

Boston

New York

Hong Kong

Moscow

Lübeck

194

Properties of Functions - example

• Isfinjective?• Yes.• Isfsurjective?• Yes.• Isfbijective?• Yes.

Linda

Max

Kathy

Peter

Boston

New York

Hong Kong

Moscow

LübeckHelena

195

Inversion

• Aninterestingpropertyofbijectionsisthattheyhaveaninversefunction.

• Theinversefunction ofthebijectionf:A→B isalsoafunctiondenotedf-1:B→Awith

f-1(y)=xwheneverf(x)=y.• Toseewhy,notethat

– fisontoimpliesthatf-1(y)isdefined∀y∈B– fisone-to-oneimpliesthatf-1(y)isaunique∈A

• Sometimesbijectionsarecalledinvertiblefunctions.

196

Inversion - exampleExample:

f(Linda)=Moscowf(Max)=Bostonf(Kathy)=HongKongf(Peter)=Lübeckf(Helena)=NewYork

Clearly,fisbijective.

Theinversefunctionf-1 isgivenby:

f-1(Moscow)=Lindaf-1(Boston)=Maxf-1(HongKong)=Kathyf-1(Lübeck)=Peterf-1(NewYork)=Helena

197

Inversion - exampleLinda

Max

Kathy

Peter

Boston

New York

Hong Kong

Moscow

LübeckHelena

f

f -1

f-1:C→Pisnotafunctionbecause

• itisnotdefinedforallelementsofC(i.e.,forBoston),and

• itassignstwoimagestoitspre-imageNewYork.

Relations

• Abroaderfamilyofmappingsf:A→2B arerelations fromAtoB,i.e.,frelates elementsx∈A tosubsetsofB.

• It’spossiblethatforsomex∈A,f(x)=ϕ,i.e.,noelementsofBarerelatedtox∈A underf.

• If∀x∈A,f(x)isasingletonsubsetofB,thenrelationfcorrespondstoafunctionfromAtoB.

• So,ifafunctionfisnotabijection,f-1 isonlyarelation.• Forexample,iff:ℝ→ℝ isafunctionsuchthatf(x)=x2 thenf-1:ℝ→ℝis

relationsuchthatf-1(x)=ϕ forx<0andf-1(x)={�√x}forx≥0.

198

199

Composition• Thecompositionoftwofunctionsg:A→B andf:B→C,denotedby

f∘g ,isdefinedby

(f∘g)(x)=f(g(x))

• Thismeansthat:• First,functiongisappliedtoelementx∈A mappingitontoan

elementy=g(x)ofB,• Then,functionfisappliedtothiselementofy∈B,mappingit

ontoanelementz=f(y)=f(g(x))ofC.• Therefore,thecompositefunctionf∘g:A→C (mapsfromAtoC).

• Notethatg∘f iswelldefinediftherangef(B)⊆A,i.e.,f(B)⊆A⋂C

Ax g

By

f

Cz

200

Composition - example

• Example:

• f(x)=7x– 4,g(x)=3x,• f:ℝ→ℝ,g:ℝ→ℝ

• (f∘g)(5)=f(g(5))=f(15)=105– 4=101

• (f∘g)(x)=f(g(x))=f(3x)=21x– 4

• Exercise:Showg∘f≠f ∘g

201

Composition with inverse

• Compositionofabijection anditsinverse:

(f-1 ∘f)(x)=f-1(f(x))=x,∀x.

• Thecompositionofafunctionanditsinverseistheidentityfunction

f-1 ∘f=f∘f-1 =IwhereI(x)≡x.

• Whenf:A →B isnotabijection(i.e.,f-1 isarelation),then(f-1 ∘f)(x)={a

202

Graphs of functions

• Thegraph ofafunctionf:A→B isthesetoforderedpairs

{(x,y)|x∈A andy=f(x)}={(x,f(y))|x∈A }⊆A�B.

• Thegraphcanbeusedtovisualizefinatwo-dimensionalcoordinatesystem.

• Plottingthegraphwiththedomainasahorizontalaxis:

• Allhorizontallinescutthegraphofaone-to-one(injective)functionlessthan2places.

• Allhorizontallinescutthegraphofanonto(surjective)functioninatleast1place.

203

Recall Floor and Ceiling Functions• Thefloor andceiling functionsmaptherealnumbersontothe

integers:⌊∙⌋,⌈∙⌉:ℝ→ℤ.

• Thefloor functionassignstor∈ℝ thelargestz∈ℤ withz≤r,denotedby ⌊r⌋

• Examples: ⌊2.3⌋ =2,⌊2⌋ =2,⌊0.5⌋=0,⌊-3.5⌋ =-4

• Theceiling functionassignstor∈ℝ thesmallestz∈ℤ withz≥r,denotedby⌈r⌉.

• Examples: ⌈2.3⌉ =3,⌈2⌉ =2,⌈0.5⌉ =1,⌈-3.5⌉ =-3

204

The Pigeonhole Principle• Thepigeonholeprinciple: If(k+1)ormoreobjectsareplacedintok

boxes,thenthereisatleastoneboxcontainingtwoormoreoftheobjects.

• Example1: Ifthereare11playersinasoccerteamthatwins12-0,theremustbeatleastoneplayerintheteamwhoscoredatleasttwice.

• Example2: Ifyouhave6classesfromMondaytoFriday,theremustbeatleastonedayonwhichyouhaveatleasttwoclasses.

• Wenowprovethepigeonholeprinciplebycontradiction:– Assume∀k∈ℤ+,allkboxeshave<2objects,i.e.,has1or0objects.– Thusthetotalnumberofobjects≤k <k+1 (acontradiction).– Q.E.D.

205

The Generalized Pigeonhole Principle• Thegeneralizedpigeonholeprinciple: IfNobjectsareplacedintokboxes,

thenthereisatleastoneboxcontainingatleast⌈N/k⌉ oftheobjects.

• Example1: Ina62-studentclass,atleast13=⌈62/5⌉ studentswillgetthesamelettergrade,ofwhichthereare5differentones(A,B,C,D,orF).

• Exercise: Provethegeneralizedpigeonholeprinciplebycontradiction.Hint:∀x∈ℝ,⌈x⌉-1<x

206

The Pigeonhole Principle• Example2: Assumeyouhaveadrawercontainingarandomdistributionof

adozenidenticalbrownsocksandadozenidenticalblacksocks,alllooseinthedrawer.Itisdark,sohowmanysocksdoyouhavetopicktobesurethatamongthemthereisamatchingpair?

• Therearetwotypesofsocks,soifyoupickatleast3socks,theremustbeeitheratleasttwobrownsocksoratleasttwoblacksocks.

• Thatis,bythegeneralizedpigeonholeprinciple,wewanttofindtheintegernsuchthat⌈n/2⌉ =2.Thesolutionisn=3.

Pigeonhole principle example: Dirichlet’s Approximation Theorem

• Theorem:∀a∈ℝand∀positiveN∈ℤ,∃p,q∈ℤsuchthat

1≤q≤Nand|qa-p|<1/N.

• Note:Thisimpliesthatforanyirrationala,thereisarationalp/qsuchthat|a-p/q|<1/(qN)≤1/N,i.e.,anarbitrarilycloseapproximationbyarational.

• Thus,ℚisadensesubsetofℝ,andsinceℚiscountable,wesaythat(uncountable!)ℝisseparable.

• Dirichlet,whoemployedthepigeonholeprincipletoprovethistheorem,alsocoinedthenameforthismethodofargument.

• Notethatyoucanintuitivelydosuchapproximations(givenadecimalexpansionofanirrational)bytruncatingitsdecimalexpansion,thetruncationbeingrational:ifonetruncatesatthekthdecimaldigit,thentheresultingapproximationerroris<10-k

207

Proof of Dirichlet’s Approximation Theorem

• Ifaisrational,wearedonebydefinition.Sosupposeaisirrational.• Definef(a)=a- ⌊a⌋ asthefractionalpartofa∈ℝ,sothatf:ℝ→(0,1).• Dividetheinterval[0,1)intoNsubintervalsofequallength,1/N.• Considerthesetofnumbers{f(an)|n∈{1,2,…,N+1}}.• Exercise:OnecaneasilyshowthatthereareN+1differentnumbersin

thisset(otherwisethereisacontradictionoftheirrationalityofa).• So,bythepigeonholeprinciple,theremustbetwodifferentnumbers

fromthissetthatareinthesamesubdivision,sayk/N≤f(an),f(am)<(k+1)/Nforsomek∈{0,1,2,…,N-1}withn>m.

• So,1/N>|f(an)-f(am)|=|an- ⌊an⌋- (am-⌊am⌋)|• Wecantakeq=n-mandp=⌊an⌋-⌊am⌋.• Q.E.D.

208

Introduction to Discrete Mathematics

RecursionsThis revision G. Kesidis (2013)

This document revision is copy-left 2013. It is licensed under a Creative Commons

Attribution-Noncommercial-Share Alike 3.0 United States License

Outline• Definitionofarecurrencerelationandinitialconditionsdefininga

numericalsequence.• Numericallygeneratingthesequencebyiterativesubstitutionfrominitial

conditions.• Verifyinganexplicitformula(solution)totherecurrencerelationby

stronginduction.• Heuristictechniquesforsolvingrecursions.• Systematicsolutionofalinearandtime-invariantrecurrencerelation.• Examples.

210

Recursively defined sequences• Considerasequencesofnumbersx0,x1,x2,x3,...• Sequencescanbeexplicitlyspecifiedwithafunctionf:ℤ≥0 →ℝ,i.e.,

xn =f(n),foralln∈{0,1,2,3,...}.• Forexample,theharmonicsequencexn =1/(n+1),foralln≥0.• Alternatively,wecandefinethesequence“recursively”byexpressing

xn asafunctionofxn-1,xn-2,xn-3,...,,x0 ,foralln∈{1,2,3,...}.• Thatis,inarecursivespecification,asequenceelementisdefinedinterms

ofpastelements.• Formanyproblemsofinterest,thereisafinite“order” or“degree”,sayk,of

therecursioninthateachsequenceelementcanbespecified(recursively)onlyintermsofthepastkvaluesofthesequence,i.e.,

xn asafunctionofxn-1,xn-2,xn-3,...,,xn-k ,foralln∈{k,k+1,k+2,...}where

x0,x1,x2,...xk-1arethek“initialconditions” ofthesequencefortherecursion.

211

Recursively defined sequences –computing terms

• Givenarecursionoffiniteorderk,xn =gn(xn-1,xn-2,xn-3,...,,xn-k )forn≥k,

(i.e.,givengn)andgiventhekinitialconditionsx0,x1,x2,...xk-1 ,

wecaniterativelysubstitutetocomputexk,xk+1,xk+2,...

• Forexample,supposek=2,xn =gn(xn-1,xn-2)=xn-1+3nxn-2+7forn≥2,x0=1,x1 =0.

• Bysubstitution:x2=g2(x1,x0)=x1+3∙2∙x0+7=0 +3∙2∙1+7=13x3=g3(x2,x1)=x2+3∙3∙x1+7=13 +3∙3∙0+7=20x4=g4(x3,x2)=x3+3∙4∙x2+7=20 +3∙4∙13+7=183etc.

• Notehowboththerecursiongn andtheinitialconditionsaffecttheoutcome

• Note:Thisexampleistime-varyingbecausethecoefficientofxn-2,3n,dependsonindexn(time).

212

Recursively defined sequences –checking a solution

• Supposeaputativeexplicitformula(�solution�)xn =f(n)foragivenrecursionxn =gn(xn-1,xn-2,xn-3,...,,xn-k ),forn≥k,andi.c.�sx0,x1,x2,...xk-1 .

• Toverifythesolution,weneedtocheckf(n)=xn forall0≤n≤k-1(i.e.,theinitialconditions)andf(n)=gn(f(n-1),f(n-2))forarbitraryn≥k …bystronginduction!

• Example:Theorem:xn= f(n)=2(-2)n -2.5n(-2)n isthesolutiontoxn=-4xn-1-4xn-2 forn≥2,x0=2,x1 =1.

• Proof:Toverifythissolution,firstcheckthek=2initialconditions:f(0)=2(-2)0 -2.5∙0∙(-2)0 =2=x0f(1)=2(-2)1 -2.5∙1∙(-2)1 =-4+5=1=x1

• Foranarbitraryn≥1,assumexm= f(m)uptom=n.• Toprovetheclaimforn+1:

xn+1 =-4xn-4xn-1 (bythegivenrecursion)=-4f(n)-4f(n-1) (bytheinductiveassumption)=-4[2(-2)n -2.5n(-2)n ]-4[2(-2)n-1 -2.5(n-1)(-2)n-1 ]=f(n+1).Q.E.D.

• Exercise:verifythelastequality.• Exercise:Explainwhyn≥1intheinductiveassumptionandnotn≥0orn≥2.

213

Solving simple first-order recursions:Arithmetic and geometric progressions

• Recallthe(firstorder)arithmeticprogression,xn =xn-1 +dforn≥1,withinitialconditionx0

• Thus,forn≥0,xn =(xn-2 +d)+d(substitutingforxn-1 )

=xn-2 +2d=xn-n +nd (afternsuchsubstitutions)=x0 +nd (=f(n))forn≥0

• Similarly,forageometricprogression,xn =rxn-1 forn≥1,withinitialconditionx0 ,

thesolutionisxn =rn x0 (=f(n))forn≥0

214

Solving simple first-order recursions:Arithmetic and geometric progressions

• Wecan,e.g.,sumasequencetocreateanothersn =∑i=0n xi i.e.,sn =sn-1 +xn =sn-1 +x0 +nd

• sisafirstorderrecursion.• Recallthatifxisarithmeticthenthesolutionforsis:

sn =(n+1)(xn +x0 )/2=(n+1)(2x0 +nd )/2• Ifxisgeometric,thentheseriesrecursionis

sn =sn-1 +xn =sn-1 +x0rnandrecallthatthesolutionforsis:

sn =x0 (rn+1 -1)/(r-1).

215

Solving linear and time-invariant recurrence relations of finite degree• Supposethattherecurrencerelationisoffinitedegree/orderkandis

linearandtime-invariant:xn =∑i=1k aixn-i =g(xn-1,xn-2,xn-3,...,,xn-k), (*)

wheregislinear(theai areallscalars)anddoesnotdependonindex/time(n).

• Giventhekinitialconditions,oneemploythetheoryofZ-transformstoderiveanexplicitformulaforxn foralln≥k.

• Equivalently,findtherootsofthe(characteristic)polynomialQ(z)=zk - ∑i=1k aizk-i .

• Bythefundamentaltheoremofalgebra,thereareksuchroots(orcharacteristicmodes),r1,r2,...rk

• Exercise:Verifyxn =rn satisfiestherecursion(*)above∀ rootsrofQ.• Ingeneral,therootsrmayberepeatedandmaybecomplex,butif

complextheycomeincomplex-conjugatepairsbecauseallcoefficientsaiofQareassumedreal.

216

Linear and time-invariant recurrence relations of finite degree• AssumingnoneoftherootsrofQarenotrepeated(i.e.,eachisjusta

singlerootofQ),wecanwriteanexplicitformulafor xn (i.e.,“solve” xn ):xn =∑i=1k cirin foralln≥0

wherethekscalarsci arefoundbyusingthekgiveninitialconditions,i.e.,solvethekequations,

xn =∑i=1k cirin k-1≥n≥0,forthekunknownsc1,c2,...ck

• IftherootrofQisofmultiplicitym≤k,thenusethefollowingm“linearlyindependent”characteristicmodesinthesolutionabove(insteadofminstancesofrin ):

rin ,nrin ,n2rin ,...nm-1rin ,seethepreviousexampleforrootr=-2oforderm=2.

217

Linear and time-invariant recurrence relations of finite degree• Forexample,supposeandxn=gn(xn-1,xn-2)=3xn-1-2xn-2 forn≥2,and

x0=1,x1 =-1.

• Here,k=2andQ(z)=z2 -3z +2=(z-2)(z-1)

• So,therootsofQare2and1.

• Thus,thesolutionxn= f(n)=c22n +c1 1n =c22n +c1 forn≥0.

• Tosolveforthescalarsc2,c1 ,usethegiveninitialconditions:

f(0)=c2 +c1 =1=x0f(1)=2c2 +c1 =-1=x1

leadingto

c2 =-2andc1 =3.

• Thus,xn=-2n+1 +3forn≥0.

218

Solving more general forms ofrecurrence relations

• SystematicapproachestotheLTIrecursioncanbeextendedtoconsidergiven“input”(orforcing)function,sothatthesolutioncanbewrittenasasumof– asolutiondependingonlyontheinitialconditions(zeroinput

response)and– anothersolutiondependingonlyonthegiveninputfunction(zero

initialstateresponse);– e.g.,theconstantinput(7)ofthepreviousaffineexample.

• Systematicapproachestothesolutionofmoregeneral(e.g.,nonlinear)formsofrecurrencerelationsmaynotexist.

• Sometimes,iterativesubstitutionornumericallyevaluatingthesequenceforseveralinstanceswillgivesomeinsightandbethebasisofaguessaboutthesolution.

• Anyguessshouldbecheckedbystronginductiveproof,ofcourse.

219

The Tower of Hanoi• Inthe3-pegTowerofHanoiproblem,astackofdisksinitiallyresideona

singlepeg,whiletheothertwopegsareempty.• Thedisksareallofdifferentwidths.• Initially,theyarestackedinwidth-ordersuchthatthelargestdiskisatthe

bottom,i.e.,theyformaconeshapeonthepeg.• Theproblemistomoveallthedisksfrompegtopegsothat

– theyformthesameconeonadifferentpeg,and– alargediskisneversetdownuponasmallerdiskonapeg.

220

The Tower of Hanoi - 3 pegs• Letxn =betheminimum numberofmoves/stepsrequiredtomovendisks.• Thinking“inductively”,supposewehaven+1disksonpeg1.• Leavethelargestdiskatbottomalone,andmovetheremainingndisksto

peg2inxn steps.• Movethelargest(n+1)st topeg3,requiring1step.• Finally,movethendisksfrompeg2topeg3(atopthelargestdisk),again

requiringxn steps.• So,xn+1 = xn +1+ xn =2xn +1steps.

221

The Tower of Hanoi (cont)• Toseewhythisrecursiongivestheminimalnumberofmoves,notethatone

simplycannotmovethelargestdiskuntilonoftheothertwopegsisempty.• Whenthisoccurs,onlythelargestdiskisononepeg,onepegisempty,and

thethirdpegmusthaveallnotherdisksstackedbywidthorder.• Thatis,togettothepointwhereonecanallowablymovethelargedisk

requiresaminimumofxn steps/movesofthensmallerdisks.• Nowhowthisargumentdoesnotholdiftherearefourormorepegs

becauseneednotstackallnsmallerdisksonasinglepostbeforewecanmovethelargerone.

• So,inthecaseof4ormoreposts,wecanusetheaboveargumentonlytoconcludeisthat

xn+1 ≤ 2xn +1.

222

The Tower of Hanoi (cont)• Forthe3-pegproblem,notethat

xn+1 =2xn +1forn>1withx1 =1isafirstorderrecursionwithconstantinput1.

• Numericallyevaluatingthesequence,weget1,3,7,15,31,…

fromwhichwecanguessxn =2n-1forn≥1.• Thezeroinputresponse(tojusttheinitialconditions)is

xn =2n-1x1 =2n-1 forn≥1• Thezero(initial)stateresponse(tojusttheinput1)is

xn =2n-1-1forn≥0;(thiscanbesystematicallyfoundbyconvolutionorZ-transform).

• So,thetotalresponseisxn =2n-1 +2n-1-1=2n-1forn≥1.

223

Example:Choosingyournoodles• Considerabowlwithnnoodlestrands,eachwith2ends.• Twonoodle-endsareindependentlychosenuniformlyatrandomfrom

thebowl;ifbothendsbelongtothesamenoodle,thenoodleisremovedfromthebowl,otherwisetheendsarejoinedtoformalongernoodlewhichisreturnedtothebowl.

• Thisprocessisrepeateduntilthebowlisempty(innsteps).• FindtheexpectednumberXn ofnoodlesEXn drawnfromthebowl.• Clearly1≤Xn≤n.

224

Example:Choosingyournoodles (cont)• Toobtaintheanswer,considerwhathappensinthefirst choice:

– withprobabilityn/C(2n,2)=1/(2n-1)theendsofthesamenoodlearechosen,i.e.,Xn =1+Xn-1

– withprobability1- 1/(2n-1)theendsofthedifferentnoodlesarechosen,i.e.,Xn =Xn-1

• So,by conditioningonthefirstchoice,EXn =(1+EXn-1)∙1/(2n-1)+(EXn-1)∙(1-1/(2n-1))

=1/(2n-1)+EXn-1wherewehaveobviouslyusedthelinearitypropertyofexpectation.

• Byrecursivesubstitution,EXn =1/(2n-1)+1/(2n-3)+…+1/3+1,

whereEX1 =X1 =1,i.e.,X1 isconstant.

• ThisquestionwasaskedduringQualcommjobinterviewsin2011.225

Finding the maximum of a unimodal function by sectional search

• Afunctionf:[a,b]→ℝ isstrictlyunimodaloveraninterval[a,b]⊆ℝ if– ithasauniquemaximizingpointz∈[a,b],– fisstrictlyincreasingon[a,z],and– fisstrictlydecreasingon[z,b].

• Theobjectiveistofindzbysamplingfin[a,b].• Butassumef(x)isverycostlytoevaluateforallpointsx∈[a,b].• Initially,for“sectional”search,wehave

– intervalI(0)=[a,b],withendpointsa(0)=aandb(0)=b– Twoinitialintermediatepointsx(0),y(0)∈I(0)suchthat

a(0)<x(0)<y(0)<b(0).– Thefunctionfevaluatedatallofthesefourpoints.

226

Finding the maximum of a unimodal function by sectional search

• Sincefisunimodal:If f(x(0))≥f(y(0)) thenz∈[a(0),y(0)],elseiff(x(0))≤f(y(0))thenz∈[x(0),b(0)]

• Thatis,wecandeterminethesub-intervalcontainingthef-maximizingpointzifwehavetwointermediatepoints.

• Notethatonecannotmakethisdeterminationifonlyoneintermediatepointisused.

• Theplanistoshrinktheintervaldowntothesub-intervalcontainingzinthenextiteration,andtodosobyaconstantfactorforeveryiteration.

227

a

f

y bxy' b'a' x'

Finding the maximum of a unimodal function by sectional search

So,forthenthintervalI(n)=[a(n),b(n)],n≥0:• iff(x(n))≥f(y(n))then:

• a(n+1)=a(n)andb(n+1)=y(n), (sincez∈[a(n),y(n)])• selectapointv∈[a(n+1),b(n+1)],• evaluatef(v),and• take{x(n+1),y(n+1)}={x(n),v}s.t. x(n+1)<y(n+1)

• else:• a(n+1)=x(n)andb(n+1)=b(n), (sincez∈[x(n),b(n)])• selectapointv∈[a(n+1),b(n+1)],• evaluatef(v),and• take{x(n+1),y(n+1)}={y(n),v}s.t. x(n+1)<y(n+1)

• Iffisalsocontinuousnearitsmaximum,anaturalstoppingruleforsearchiswhentheintervalbecomessmallerthanapredefinedthreshold,otherwisen++andrepeattheabove.

228

Finding the maximum of a unimodal function by golden-section search

• Notethat,beyondpreservinga(n)<x(n)<y(n)<b(n),

wehaven’tyetpreciselyspecifiedhowsectionalsearchworks.• Supposeourobjectiveis:foreachiterationn:

– evaluatefatjustonenewpointv(exceptattwopointsinitially(n=0)),and– reducethewidthoftheintervalbythesamefactor,g.

• Thatis,letd(n)=b(n)-a(n)betheintervalwidth,andsupposethat,foralln,wewishtofixtheproportions

g:=[y(n)-a(n)]/d(n)=[b(n)-x(n)]/d(n)⇒[b(n)-y(n)]/d(n)=1-[y(n)-a(n)]/d(n)=1-g.

• So,ifa(n+1)=x(n)andb(n+1)=b(n),sothatd(n+1)=gd(n),wethenassignx(n+1)=y(n)andtakev=y(n+1)>x(n+1)suchthatg=[v-a(n+1)]/d(n+1).

• Thusalsog=[b(n+1)-x(n+1)]/d(n+1)whichimpliesg=[b(n)-y(n)]/[gd(n)]=(1-g)/g

229

Finding the maximum of a unimodal function by golden-section search

• So,g=(1-g)/g,i.e.,g2 +g-1=0.

• Therefore,g=(-1+50.5)/2≈0.618

• 1+gisthe“goldenratio”ofancientGreece.• Ifthestoppingthresholdispositiveθ≪1thenbecausethelengthofthe

intervalssatisfiestherecursiond(n+1)=gd(n),thenumberofiterationsNrequiredbygolden-sectionsearchisthesmallestintegerNs.t.d(N)=d(0)gN <θ ,i.e.,N=é log(θ/d(0))/log(g)ù .

• So,thecomputationalcomplexityofgolden-sectionsearchisΘ(N).

230

231

Introduction to DiscreteMathematics

Introduction to ComplexityThis revision G. Kesidis (2013)

This document revision is copy-left 2013. It is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License

Note: these slides adapted in part from those of S.J. Lomonaco Jr.,available at

http://www.csee.umbc.edu/~lomonaco/f05/203/Slides203.html

232

Algorithms

• Whatisanalgorithm?

• Analgorithmisafinitesetofpreciseinstructionsforperformingacomputationorforsolvingaproblem.

• Thisisarathervaguedefinition.Youwillgettoknowamorepreciseandmathematicallyusefuldefinition.

• Butthisoneisgoodenoughfornow…

233

Algorithms

• Analgorithmhasaninput fromaspecifiedset,

andproducesanoutput fromaspecifiedset(solution).

• Propertiesofalgorithms:• Definiteness ofeverystepinthecomputation,• Correctness ofoutputforeverypossibleinput,• Finiteness ofthenumberofcalculationsteps,• Effectiveness ofeachcalculationstepand• Generality foraclassofproblems.

234

Algorithm Example:Find the Maximum

• Wewilluseapseudocodetospecifyalgorithms,whichslightlyremindsusofBasicandPascal.

• Example:

procedure max(a1,a2,…,an:integers)max:=a1for i :=2to n

if max<ai then max:=ai

• Notethattheprocedurefinds�max� =largestelementof{a1,a2,…,an}

235

Algorithm Example: Linear Search of Unordered List

• Anotherexample:alinearsearchalgorithm,thatis,analgorithmthat

linearlysearchesasequenceforaparticularelement.

procedure linear_search(x:integer;a1,a2,…,an:integers)

i :=1

while(i £ nandx¹ ai)i :=i +1

if i £ nthen location:=ielse location:=0

• Notethatlocationisthesubscriptofthetermthatequalsx,oriszeroifx

isnotfound

• Also,theworst-casenumberofiterationsisn-1.

236

Algorithm Example:Binary Search of Ordered List

• Ifthetermsinasequenceareordered,abinarysearchalgorithmismoreefficientthanlinearsearch.

• Thebinarysearchalgorithmiterativelyrestrictstherelevantsearchintervaluntilitclosesinonthepositionoftheelementtobelocated.

237

Algorithm Example:Binary Search of Ordered List (cont)

procedure binary_search(x:integer;a1,a2,…,an:integers)i :=1{i isleftendpointofsearchinterval}j:=n{jisrightendpointofsearchinterval}while(i <j)begin

m:=⌊(i +j)/2 ⌋if x>am then i :=m+1else j:=m

endif x=ai then location:=ielse location:=0

• Note:locationisthesubscriptofthetermthatequalsx,oriszeroifxisnotfound

• Exercise:Showthatintheworstcase,thenumberofiterationsi requiredsatisfies 1=n/2i ,i.e.,i =log2(n)

• Comparetheproblemaddressedbythisalgorithmtothatofgolden-sectionsearch(assumingunimodalfunction).

238

Complexity

• Wearetypicallynotsomuchinterestedinthetimeandspace(memorysize)complexityforverysmallinputssizes.

• Forexample,whilethedifferenceintimecomplexitybetweenlinearandbinarysearchismeaninglessforasequencewithn=10elements,itisgiganticforn=230.

239

Complexity - example

• Forexample,letusassumetwoalgorithmsAandBthatsolvethesameclassofproblemsof�size� n,wherenreflectstheamountofinputdata.

• ThetimecomplexityofAis5,000n,theoneforBisé1.1nùforaninputwithnelements.

• Forn=10,Arequires50,000steps,butBonly3,soBseemstobesuperiortoA.

• Forn=1000,however,Arequires5,000,000steps,whileBrequires2.5�1041 steps.

240

Complexity - example (cont)

• ThismeansthatalgorithmBcannotbeusedforlargeinputs,whilealgorithmAisstillfeasible.

• Sowhatisimportantisthegrowth ofthecomplexityfunctions.

• Thegrowthoftimeandspacecomplexitywithincreasinginputsizenisasuitablemeasureforthecomparisonofalgorithms.

241

Complexity - example (cont)

• Comparison:timecomplexityofalgorithmsAandB

AlgorithmA AlgorithmBInputSize

n

10

100

1,000

1,000,000

5,000n

50,000

500,000

5,000,000

5�109

é1.1nù

3

2.5�104113,781

4.8�1041392

242

The Growth of Functions• Thegrowthoffunctionsisusuallydescribedusingthebig-Onotation.

• Definition: Letfandgbefunctionsfromtheintegersortherealnumberstotherealnumbers.

• Wesaythatfisoforderatmost g(orjustoforderg),i.e.,f(x)isO(g(x))(orjustf=O(g))iftherearepositiveconstantsC,k <∞ suchthat

|f(x)|£ C|g(x)|wheneverx>k,

i.e.,theinequalityholdsforallsufficientlylargex.

243

The Growth of Functions

• Whenweanalyzethegrowthofcomplexityfunctions,f(x)andg(x)arealwayspositive.

• Therefore,wecansimplifythebig-Orequirementto

f(x)£ C×g(x)wheneverx>k,

i.e.,absolutevaluesarenotnecessary.

• Ifwewanttoshowthatf=O(g),weonlyneedtofindone pair(C,k),neverunique,thatmaketheinequalitywork.

244

The Growth of Functions• Theideabehindthebig-Onotationistoestablishanupperbound forthe

growthofafunctionf(x)forlargex.

• Thisboundisspecifiedbyafunctiong(x)thatisusuallymuchsimplerthanf(x).

• WeaccepttheconstantCintherequirement

f(x)£ C×g(x)wheneverx>k,

becauseCdoesnotgrowwithx.

• Weareonlyinterestedinlargex,soitisOKiff(x)>C×g(x)forx£ k.

245

The Growth of Functions –Polynomial Example

• Example:Showthatf(x)=x2 +5x+1isO(x2).

• Forx>1wehave:

x2 +5x+1£ x2 +5x2 +x2Þ x2 +5x+1£ 7x2

• Therefore,forC=7andk=1,

f(x)£ Cx2 wheneverx>k

Þ f(x)=O(x2).

246

Complexity Example• Whatdoesthefollowingalgorithmcompute?

procedure who_knows(a1,a2,…,an:integers)m:=0for i :=1ton-1

for j:=i +1tonif |ai – aj|>mthen m:=|ai – aj|

• Answer:(thereturnedvalueof)misthemaximumdifferencebetweenanytwonumbersintheinputsequence,maxi<j |ai - aj|

• Numberofcomparisonsmadeintheprocedurearen-1+n-2+n-3+…+1=(n– 1)n/2=0.5n2 – 0.5n=C(n,2)

• So,timecomplexityoftheprocedureisO(n2).

247

Complexity Example• Anotheralgorithmsolvingthesameproblem:

procedure max_diff(a1,a2,…,an:integers)min:=a1max:=a1for i :=2ton

if ai <minthen min:=aielse ifai >maxthen max:=ai

m:=max- min

• Thatis,maxi<j |ai - aj|=maxiai - miniai• Thenumberofcomparisonsinthisprocedureis2n- 2

• So,thetimecomplexityisO(n).

248

The Growth of Functions –common simple functions

• Common“simple”functionsg(n)are1,2n,n2,n!,n,n3,log(n),nlog(n),etc.

• Exercise:DoweneedtospecifythebaseoftheloginO(logn)?• Listedfromslowesttofastestgrowth:

1log(n) …binarysearchoforderedlistn …linearsearchofunorderedlistisO(n)nlog(n) …somesortingalgorithmsareO(nlogn)n2

n3

2n

n!nn

• Note:costofsortingn-listmaybeamortizedovermsearches:nlogn +mlogn =m((nlogn)/m+logn)<mn forsufficientlylargem

249

Simple Properties for Big-O• Foranypolynomial f(x)=anxn +an-1xn-1 +…+a0,wherea0,a1,…,an are

realnumbers,f(x)isO(xn),i.e.,∃C,k>0s.t. |f(x)|≤C∀x≥k

• Toseewhynotethatbythetriangleinequality|f(x)|≤∑k=0

n |akxk |=∑k=0n |ak|xk ≤(∑k=0

n |ak|)xn forx≥max{1,k}

• Similarly,onecanusethetriangleinequalitytoshowthenexttwostatements.

• Iff1 =O(g)andf2=O(g),thenf1 +f2=(g);forexample,seepolynomialfabove.

• Moregenerally,iff1 =O(g1)andf2 =O(g2),thenf1 +f2=O(max{g1,g2});orderoftandemalgorithmsismostcomplexone.

• Iff1=O(g1)andf2=O(g2),thenf1f2=O(g1g2);eachiterationofonealgorithmcallsanother.

250

The Growth of Functions –Imprecision of big-O

• IfCisverylargeinorderfortheupperboundtohold,thenthebig-Oboundmaybecrude.

• Also,ifkismuchlargerthanthejobsizesxseeninpractice,thenthebig-Oboundmaynotberelevant.

• Foralgorithmsofcomplexityfonlargeinputsx,wecanaskthefollowingtypeofquestion:

Iff(x)isO(x2),isitalsoO(x3)?• TheanswerisYes: x3 growsfasterthanx2,sox3 alsogrowsfasterthan

f(x).• Therefore,wealwayswanttofindthesmallest simple-functiongfor

whichf=O(g).

Ω and Θ notation• Wesayfisoforderatleast g,i.e.,f=Ω(g),if∃positivec,k<∞suchthat

|f(x)|≥c|g(x)|∀x>k.

• Wesayfisoforderexactly g(orjustoforderg),i.e.,

f=Θ(g)if∃positiveC,c,k<∞suchthat

C|g(x)|≥|f(x)|≥c|g(x)|∀x>k.

• Clearly,f= Θ(g)⇔f=O(g)andf=Ω(g).

• ThesimplepropertiesforBig-OmentionedabovehaveanalogsforΩ

andΘ.

• Notethatwhenauthorssayjust�oforderg� theyoftenmean=O(g).

251

252

The Growth of Functions –Rational Polynomial Example

Example:Showthatf(x)=(x5 +2x+1)/(5x2+6x)isΘ(x3).• Asabove,wehavex5 +2x+1=O(x5)withk=1andC=4,and

5x2+6x=O(x2)withk=5andC=11.• Butalso,x5 +2x+1≥ x5 forallx>1,sox5 +2x+1=Ω(x5)withk=1

(alsowithk=0)andC=1.• Similarly,5x2+6x=Ω(x2)withk=1andC=5.• Thus,byusingtheupperboundonthenumeratorandlowerboundon

thedenominator,wegetf(x)=O(x3)becausef(x)£ Cx3 wheneverx>kwithk=1andC=4/5.

• Orjustf(x)=O(x5)/Ω(x2)=O(x3).• Similarly,byusingthelowerboundonthenumeratorandthelower

boundonthedenominator,wegetf(x)=Ω(x3)becausef(x)≥Cx3 wheneverx>kwithk=1andC=1/11.

• Orjust,f(x)=Ω(x5)/O(x2)= Ω(x3).• Thus,f(x)=Θ(x3).• Notehowweestablishedthesimplerresultthatx5 +2x+1=Θ(x5).• Exercise:Provethisresultbyfirstusingpolynomiallongdivisiononf.• Exercise: Prove5x2-6x=Ω(x2).Hint:consider5x2(1-6/(5x))forlargex.

253

Problem tractability and solvability

• Aproblemthatcanbesolvedwithpolynomialworst-casecomplexityiscalledtractable.

• Problemsofhighercomplexityarecalledintractable.

• Therearesomewell-knownproblemsforwhichthequestionoftractabilityisnotknown,e.g.,thetravellingsalesmanproblem.

• Problemsthatnoalgorithmcansolvearecalledunsolvable.

• Moreonthissubjectinlatercourses…

254

Introduction to DiscreteMathematics

RelationsThis revision G. Kesidis (2013)

This document revision is copy-left 2013. It is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License

Note: these slides adapted in part from those of S.J. Lomonaco Jr., available at

http://www.csee.umbc.edu/~lomonaco/f05/203/Slides203.html

255

Relations• IfwewanttodescribearelationshipbetweenelementsoftwosetsAandB,

wecanuseorderedpairs withtheirfirstelementtakenfromAandtheirsecondelementtakenfromB.

• Sincethisisarelationbetweentwosets,itiscalledabinaryrelation.• Definition: LetAandBbesets.AbinaryrelationfromAtoBisasubsetof

A´B.• Inotherwords,forabinaryrelationRwehave

RÍ A´B={(a,b)|aÎA,bÎB}.• WemayalsousethenotationaRb todenotethat(a,b)ÎR.

256

Relations• When(a,b)belongstoR,aissaidtoberelated tobbyR.• Example: LetPbeasetofpeople,Cbeasetofvehicles,andDbethe

relationdescribingwhichpersondriveswhichvehicle(s).• P={Carl,Suzanne,Peter,Carla},• C={Mercedes,BMW,tricycle},• D={(Carl,Mercedes),(Suzanne,Mercedes),(Suzanne,BMW),

(Peter,tricycle)}• ThismeansthatCarldrivesaMercedes,SuzannedrivesaMercedesanda

BMW,Peterdrivesatricycle,andCarladoesnotdriveanyofthesevehicles.• Thatis,CarlDMercedes.• NotehowCarladoesnotdriveanythingandhowSuzannedrivesmore

thanonevehicle- foreitherofthesereasons,Risnotafunction.

257

Functions as Relations• Youmightrememberthatafunction ffromasetAtoasetB(f:A→B)

assignsasingleelementbÎ BtoeachelementaÎ A,i.e.,fisafunctionfromdomainAtoco-domainBif

∀a Î A, ∃uniquebÎ Bs.t. b=f(a).• Recallthatwedonot meanthatthereisasingleelementofBtowhichf

mapsallelementsofA,i.e.,noticethedifferencebetweentheabovedefinitionofafunctionand

∃uniquebÎ Bs.t. ∀a Î A,b=f(a).• So,wecandefinetherelationcorrespondingtofunctionfas

{(a,f(a))|aÎA}.• Conversely,ifRisarelationfromAtoBsuchthateveryelementinAisthe

firstelementofexactlyoneorderedpairofR,thenafunctioncanbedefinedusingR.

• ThisisdonebyassigningtoeachelementaÎA theuniqueelementbÎBsuchthat(a,b)ÎR.

258

Relations on a (single) set

• Definition: ArelationonthesetAisarelationfromAtoA.

• Inotherwords,arelationonthesetAisasubsetofA´A=A2.

• Example: LetA={1,2,3,4}.WhichorderedpairsareintherelationR={(a,b)|a<b}?

• Solution: R={(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)}

259

Relations – Bipartite Graph Depiction(both domain and co-domain depicted)

• Example:R={(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)}

1 1

2

3

4

2

3

4

260

Relations on a single set –directed graph (digraph) depiction

• Example: R={(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)}

1

2

3

4

261

Relations – adjacency matrix• ExampleSolution: R={(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)}• Note:rowisdomainelementandcolumnsareco-domainelements

R 1 2 3 4

1

2

3

4

X X X

X X

Xúúúú

û

ù

êêêê

ë

é

=

0000100011001110

R

262

Number of Relations on a SetHowmanydifferentrelationscanwedefineonasetAwithnelements?

• ArelationonasetAisasubsetofA´A.• HowmanyelementsareinA´A?

• Therearen2 elementsinA´A,sohowmanysubsets(=relationsonA)doesA´Ahave?

• Thenumberofsubsetsthatwecanformoutofasetwithmelementsis2m.

• Therefore,2n2 =2n∙n subsetscanbeformedoutofA´A.

263

Properties of Relations on a Set - Reflexivity• Wewillnowlookatsomeusefulwaystoclassifyrelations.

• Definition: ArelationRonasetAiscalledreflexive if(a,a)ÎRforeveryelementaÎA,i.e.:∀aÎA,aRa.

• Arethefollowingrelationson{1,2,3,4}reflexive?

• R={(1,1),(1,2),(2,3),(3,3),(4,4)}…No,missing(2,2)

• R={(1,1),(2,2),(2,3),(3,3),(4,4)}…Yes.

• R={(1,1),(2,2),(3,3)}…No,missing(4,4)

• Definition: ArelationonasetAiscalledirreflexive if(a,a)ÏRforeveryelementaÎA,i.e.:∀aÎA,(a,a)∉R.

• Notethatareflexiverelationwillhaveanadjacencymatrixwithdiagonal

all1s(diag(R)=1),whileanirreflexiverelation�sadjacencymatrixwillhavediagonalall0s(diag(R)=0).

• Also,allelementswillhaveselfloopsinthedirected-graphdepictionofa

reflexiverelation,whileanirreflexiverelationwillhavenoself-loops.

• Noneofthethreeexamplesaboveareirreflexive.

• Previous�<� relationwasirreflexive,while�≤� isreflexive.

264

Properties of Relations on a Set - Symmetry

• ArelationRonasetAiscalledsymmetric if:(b,a)ÎRwhenever(a,b)ÎR; equivalently,∀a,bÎA,if(a,b)ÎRthen(b,a)ÎR.

• NotethatifRissymmetric,itsadjacencymatrixR=RT (itstranspose).• ArelationRonasetAiscalledantisymmetric if:

∀a,bÎA,if(a,b)ÎRand(b,a)ÎRthena=b≡ ∀a,bÎA,ifa≠b then(a,b)∉Ror(b,a) ∉R

• ArelationRonasetAiscalledasymmetric if:∀a,bÎA,if(a,b)ÎRthen(b,a)ÏR.

• Notethatif(a,a)ÎRforsomeaÎA,thenRmaybeantisymmetricbutnotasymmetric.

• Exercise:ArguethatRisasymmetricifandonlyifitisirreflexiveandantisymmetric.

265

Properties of Relations –Symmetry Examples

• Arethefollowingrelationson{1,2,3,4}symmetric,antisymmetric,asymmetric,reflexive,orirreflexive?

• R={(1,1),(1,2),(2,1),(3,3),(4,4)}issymmetric• R={(1,1)}issymmetricandantisymmetric• R={(1,3),(3,2),(2,1)}isantisymmetric,irreflexiveandasymmetric• R={(4,4),(3,3),(1,4)}isantisymmetric

266

Counting Relations with Properties

• HowmanydifferentreflexiverelationscanbedefinedonasetAcontainingnelements?• Areflexiverelationmust containthenelements(a,a)foreveryaÎA.• Consequently,wecanonlychooseamongtheremaining

n2 – n=n(n– 1)elementstogeneratedifferentreflexiverelations.• So,thereare2n(n– 1) ofthem.

• HowmanydifferentsymmetricrelationscanbedefinedonasetAcontainingnelements?• Asymmetricrelationmay containthen=C(n,1)elements(a,a)for

everyaÎA.• Asymmetricrelationmay containC(n,2)=n(n-1)/2elementpairs

(a,b)and(b,a)fora≠b.• So,thereare2n2n(n– 1)/2 =2n(n+1)/2 ofthem,includingthenullrelation.

• Exercise:ShowthatthenumberofantisymmetricrelationsonasetAcontainingnelementsis2n 3n(n– 1)/2 ,andfindthenumberofasymmetricrelations.Hint:∑k=1

N C(N,k)2k =3N whereN=C(n,2)=n(n-1)/2

267

Properties of Relations - Transitivity

• ArelationRonasetAiscalledtransitive if∀ a,b,cÎA,if (a,b)ÎRand(b,c)ÎR,then(a,c)ÎR.

• Thedirected-graphdepictionoftransitiverelationshavethepropertythatalltwo-hoppathsconnectingelementsalsohavesingle-hoppathsconnecting(relating)thoseelements.

• Forexample,�<� and�≤� arebothtransitive.

• Arethefollowingrelationson{1,2,3,4}transitive?

• R={(1,1),(1,2),(2,2),(2,1),(3,3)} …Yes• R={(1,3),(3,2),(2,1)} …No• R={(2,4),(4,3),(2,3),(4,1)} …No

268

Combining Relations

• Relationsaresets,andtherefore,wecanapplytheusualsetoperations tothem.

• IfwehavetworelationsR1 andR2,andbothofthemarefromasetAtoasetB,thenwecancombinethemtoR1 ⋂ R2,R1 ⋃ R2,orR1 – R2.

• Ineachcase,theresultwillbeanotherrelationfromAtoB.

Closure of Relations to Achieve Properties• SupposearelationRonasetAdoesnot possessapropertyP.• TherelationSonAiscalledtheP-closureofRif

– R⊆S,– ShaspropertyP,and– S-R(therelationsaddedtoRtoobtainS)isminimal insize.

269

Transitive Closure - Example• Forexample,recallthatR={(1,3),(3,2),(2,1)}onA={1,2,3,4}was

nottransitive.• LetS=R.• Fortransitiveclosure,weneed(1,2)because(1,3),(3,2)∈S,soset

S=S⋃{(1,2)}.• Similarly,weneed(3,1)because(3,2),(2,1)∈S,soset

S=S⋃{(3,1)}.• Butadding(1,2)meanswealsoneedtoadd(1,1)and(2,2)since

(2,1)∈S,sosetS= S⋃{(1,1),(2,2)}.• Now,(2,1)and(1,3)∈S,soweneedtoadd(2,3)toSand,because

(3,2)∈S,wealsoneedtoadd(3,3).• VerifythatS={(a,b)|a,b ∈{1,2,3}},i.e.,the�completelyconnected�

relationon{1,2,3} ⊆A,isthetransitiveclosureofRonA.

270

271

Combining Relations by Composition• Definition: LetRbearelationfromasetAtoasetBandSarelationfromB

toasetC.Thecomposite ofRandSistherelationconsistingoforderedpairs(a,c),whereaÎA,cÎC,andforwhichthereexistsanelementbÎBsuchthat(a,b)ÎRand(b,c)ÎS:

• WedenotethiscompositeofRandSbyS°R.

• Again,If(a,c)Î S°Rthen∃bÎB s.t. (a,b)Î Rand(b,c)Î S

• Also:If(a,b)Î Rand(b,c)Î S,then(a,c)Î S°R

272

Combining Relations - Example• Example: LetDandSberelationsonA={1,2,3,4}.• D={(a,b)|b=5- a}�bequals(5– a)�• S={(a,b)|a<b}�aissmallerthanb�• Equivalently,

D={(1,4),(2,3),(3,2),(4,1)}S={(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)}

• S °D={(2,4),(3,3),(3,4),(4,2),(4,3),(4,4)}• Dmapsanelementatotheelement(5– a),andafterwardsSmaps(5– a)to

allelementslargerthan(5– a),resultinginS °D={(a,b)|b>5– a},equivalentlyS°D={(a,b)|a+b>5}.

• Exercise:FindD ° SandcomparewithS °D.

273

Combining a Relation with Itself

• Definition: LetRbearelationonthesetA.ThepowersRn,n=1,2,3,…,aredefinedinductivelybyR1 =R,andRn+1 =Rn°Rforn≥1.

• Inotherwords:Rn =R°R° …°R(nrelationsRcomposed)

274

Self-Combining Transitive RelationsTheorem: TherelationRonasetAistransitiveifandonlyifRn Í Rforallpositiveintegersn.Proof:• RecallthatArelationRonasetAiscalledtransitiveifwhenever(a,b)ÎR

and(b,c)ÎR,then(a,c)ÎRforalla,b,cÎA.• Sobydefinition,transitivityofR⇔ R°RÍ R.• WhatremainstoproveisthatifRistransitive,thenRn Í Rforallintegers

n>2.• Thiscanbeeasilyshownbyinduction,i.e.,compositionofRn-1 by

(transitive)RdoesnotaddrelationsnotalreadyÎR.• Q.E.D.

Partial-order relations (PORs)

• ArelationRonAisapartial-orderrelation(POR)ifitisreflexive,

antisymmetric,andtransitive.

• Forexample,ifAisasetofsetsandRisthesubsetrelation⊆,thenRis

aPORbecause⊆isclearlyreflexiveandtransitive,andforanytwo

different subsetsa,b ∈A,ifa⊆bthenb⊄a (i.e.,⊆isantisymmetric).

• Exercise:Checkthatthedivides“|”relationonℤ isaPOR.

• Notethattherelations>and<onℝ arenotPORsbecausetheyarenot

reflexive.

• Orderrelationsaresometimescalledstrongorweakdependingon

whethertheyarereflexive-and-antisymmetric(as≥and≤)or

asymmetric(as<and>)– wewillnotconsidersuchqualifications

here.

275

Total-order relations (TORs)

• ArelationRonAistotal if

∀a,b∈A,ifa≠b then(a,b)∈Ror(b,a)∈Rbutnotboth.

• Thatis,RonAistotalifeverypairofelementsinAiscomparable/relatable.

• ArelationRonAisatotal-orderrelation(TOR)ifitisatotalPOR.

• Forexample,ifAisasetofrealnumbersthenanyR∈{≤,≥}isaTOR.

• WithatotalorderrelationR,allelementsofacountablesetAcanbeenumeratedas{an}=Asothat(an,an+1)∈Rforalln.

• Notethat⊆maynotbetotal,andhencenotaTOR,becauseit’spossibletohavetwodifferentsetsa,b∈A suchthatb⊄a anda⊄b,e.g.,

a={1,2,3}andb={2,3,4}.

276

Chains and extrema in PORs

• AchainofaPORRonAisasubsetB⊆AsuchthatPisaTORonB.

• Forexample,forthedivides�|� relationonA={3k4i :k,i ∈{0,1,2,3,4}},thesets{3,9,27}and{1,4,12,36}arechains.

• IfweinterpretaRb asbis�greaterthanorequalto� foraPORR,thenwehavethefollowingnotions:

• Amaximalelementisthelargestelementofachain,ifitexists.• Agreatestelementistheuniquemaximalelement,ifitexists.

• Aminimalelementisthesmallestelementofachain,ifitexists.

• Aleastelementistheuniquemaximalelement,ifitexists.

• Fortheaboveexample,theleastelementis1andthegreatest3444.

• Butforthesamedividesrelationon{3,4,12,16,18,24,32},theminimalelementsare3,4,themaximalonesare18,24,32,andthereisneitheragreatestnorleastelement.

277

HasseDiagramsforPORs– Example:divideson{3,4,12,16,18,24,32}

278

43

1612

322418

279

Equivalence Relations - Definition

• Equivalencerelations areusedtorelateobjectsthataresimilarinsomeway.

• Definition: ArelationonasetAiscalledanequivalencerelationifitisreflexive,symmetric,andtransitive.

• TwoelementsthatarerelatedbyanequivalencerelationRarecalledequivalent.

280

Equivalence Relations –Definition (cont)

• SinceRissymmetric,aisequivalenttobwheneverbisequivalenttoa.

• SinceRisreflexive,everyelementisequivalenttoitself.

• SinceRistransitive,ifaandbareequivalentandbandcareequivalent,thenaandcareequivalent.

• Obviously,thesethreepropertiesarenecessaryforareasonabledefinitionofequivalence.

281

Equivalence Relations - Example Example:SupposethatRistherelationonthesetofstringsthatconsistofEnglishletterssuchthataRb ifandonlyifl(a)=l(b),wherel(x)isthelengthofthestringx.IsRanequivalencerelation?Solution:• Risreflexive,becausel(a)=l(a)andthereforeaRa foranystringa.• Rissymmetric,becauseifl(a)=l(b)thenl(b)=l(a),soifaRb thenbRa.• Ristransitive,becauseifl(a)=l(b)andl(b)=l(c),thenl(a)=l(c),soaRb

andbRc impliesaRc.• Risanequivalencerelation.

282

Equivalence Classes • Definition:LetRbeanequivalencerelationonasetA.Thesetofall

elementsthatarerelatedtoanelementa∈ Aiscalledtheequivalenceclassofa.

• TheequivalenceclassofawithrespecttoRisdenotedby[a]R.• Whenonlyonerelationisunderconsideration,wewilldeletethesubscript

Randwrite[a] forthisequivalenceclass.• IfbÎ[a]R,biscalledarepresentative ofthisequivalenceclassandwecan

write[a]R =[b]R .

283

Equivalence Classes - Example

• Example:Inthepreviousexample(stringsofidenticallength),whatistheequivalenceclassofthewordmouse,denotedby[mouse]?

• Solution: [mouse]isthesetofallwordscontainingfiveEnglishletters.

• Forexample,“horse” wouldbearepresentativeofthisequivalenceclass.

284

Equivalence Classes and Set Partitions • Theorem:IfRisanequivalencerelationonasetA,thenthefollowing

statementsareequivalent:• aRb• [a]=[b]• [a]Ç [b]¹ Æ

• Definition: ApartitionofasetAisacollectionofdisjointnonemptysubsetsofAthathaveAastheirunion(i.e.,that“cover”A).

• Inotherwords,thecollectionofsubsetsSi,iÎI,formsapartitionofAifandonlyif– Si ¹ Æ foriÎI– SiÇ Sj =Æ ifi ¹ j– ÈiÎI Si =A

285

Equivalence Classes - Examples Examples:LetAbetheset{u,m,b,r,o,c,k,s}.DothefollowingcollectionsofsetspartitionA?• {{m, o, c, k}, {r, u, b, s}} yes.• {{c, o, m, b}, {u, s}, {r}} no (k is missing).• {{c, o, m, t}, {b,u, s}, {r}} no (t ∉ A).• {{u, m, b, r, o, c, k, s}} yes (A is the trivial 1-partition of itself).• {{b, o, o, k}, {r, u, m}, {c, s}} yes ({b,o,o,k} = {b,o,k}).• {{u, m, b}, {r, o, c, k, s}, Æ} no (Æ is not allowed as a partition element).

286

Equivalence Classes and Set Partitions (cont)

Theorem:• IfRisanequivalencerelationonasetA,thentheequivalenceclasses ofR

formapartition ofA.• Conversely,givenapartition{Si |iÎI}ofthesetA,thereisanequivalence

relationRthathasthesetsSi,iÎI,asitsequivalenceclasses.

Proof:• Theproofofthefirststatementfollowsdirectlybydefinition(theprevious

Theorem).• GivenapartitionofA,onecanconstructthecorrespondingequivalence

relationbysimply�relating� (ineveryway)elementsofAthatarepartofthesamepartitionelement,andbynotrelating(inanyway)elementsofAfromdifferentpartitions.

287

Equivalence Classes - Example • Example:LetusassumethatFrank,SuzanneandGeorgeliveinBoston,

StephanieandMaxliveinLübeck,andJenniferlivesinSydney.• LetRbetheequivalencerelation {(a,b)|aandbliveinthesamecity}on

thesetP={Frank,Suzanne,George,Stephanie,Max,Jennifer}.• ThenR={(Frank,Frank),(Frank,Suzanne),

(Frank,George),(Suzanne,Frank),(Suzanne,Suzanne),(Suzanne,George),(George,Frank),(George,Suzanne),(George,George),(Stephanie,Stephanie),(Stephanie,Max),(Max,Stephanie),(Max,Max),(Jennifer,Jennifer)}.

• Exercise:CanoneconstructanequivalencerelationonPifSuzannealsolivedinSydney?

288

Equivalence Classes – Example (cont) • Thentheequivalenceclasses ofRare:

{{Frank,Suzanne,George},{Stephanie,Max},{Jennifer}}.• ThisisapartitionofP.• TheequivalenceclassesofanyequivalencerelationRdefinedonasetS

constituteapartitionofS,becauseeveryelementinSisassignedtoexactlyone oftheequivalenceclasses.

• IftheelementsofasetParelistedbyequivalenceclass,thentheadjacencymatrixoftheequivalencerelationisblockdiagonal.

• ForthepreviousexamplewithP={F,Su,G,St,M,J},

úúúúúúúú

û

ù

êêêêêêêê

ë

é

=

100000011000011000000111000111000111

R

289

Equivalence Classes – Modulo Relation • Anotherexample:LetRbetherelation

{(a,b)|aº b(mod3)}onthesetofintegers.• IsRanequivalencerelation?• Yes,Risreflexive,symmetric,andtransitive.

• WhataretheequivalenceclassesofR?• [0]={{…,-6,-3,0,3,6,…},

[1]={…,-5,-2,1,4,7,…},[2]={…,-4,-1,2,5,8,…}}

Introduction to Discrete Mathematics

Introduction to Modular Arithmetic and Public Key Cryptography

This revision G. Kesidis (2013)This document revision is copy-left 2013. It is licensed under a Creative Commons

Attribution-Noncommercial-Share Alike 3.0 United States LicenseThisslidedeckwasrevisedfromthoseavailableat

http://www.cs.kent.edu/~rothstei/modular_arith.ppthttp://homepages.gold.ac.uk/rachel/CIS326week5lession1.ppt

Definition of modulo operator• Theremainderafterdividingintegerkbyn≠0 denoted

kmodn or k%n• Thatisr=kmodnmeansn|(k-r)or∃q∊ℤ s.t. k=nq+r.• Thisremainderistypicallyrestrictedtorange

{0,1,2,…,n-1}• So,thesetofintegersℤ isdividedintonequivalenceclassesbythemodn

operator,i.e.,twointegerskandmareinthesameequivalenceclassiftheyarecongruentmodn,i.e.,ifk≡mmodn,equivalently:kmodn≡mmodn,0≡(k-m)modn,k-m≡0modn.

• Congruence(equivalence)modn(i.e.,equalremaindersafterdivisionbyn)isoftensignifiedby≡,thoughwewillusejust=inthefollowing.

• Forexample,5=40mod7=-9mod7,so5,40and-9areinthesameequivalenceclass(congruent)mod7– thisequivalenceclassisdenoted“5”because0≤5<7,i.e.,here“5”=[5]={…,-9,-2,5,12,17,…}

291

What is modular arithmetic?

• Modulararithmeticisarithmeticwiththeremaindersupondivisionbyafixednumbern.

• Itisbasedupontheideathattheremainderofthesum/difference/productoftwonumbersistheremainderofthesum/difference/productoftheremainders.

• Foranexampleofhowmoddistributesover+:ifn=5,– (34+7)%5=41%5=1,and– (34%5+7%5)%5=(4+2)%5=1too

292

Properties of Modular Arithmetic

Theorem:∀n≠0,a,b∈ℤ,1. [(amod n)+(bmod n)]mod n=(a+b)mod n2. [(amod n)- (bmod n)]mod n=(a- b)mod n3. [(amod n)� (bmod n)]mod n=(a� b)mod n

Proof of3.: Forarbitraryn≠0,a,b∈ℤ,• Let Ra=amod nand Rb =bmod n.• So ∃j,k∈ℤ s.t. a=Ra +jn and b=Rb +kn.• Thus,the right-hand side ofstatement 3.is

(ab)mod n =(Ra Rb +n(jRb +kRa +nkj)) mod n=Ra Rb mod n=[(amod n)(bmod n)]mod n

• Q.E.D.

293

So, what is arithmetic mod n?

• Our“numbers” areequivalenceclasses0,1,2,...,n-1.• Weadd,subtractasusual,butsubtractoraddnasnecessarytogetananswerbetween0andn-1.

• Formultiplication,theprocessissimilar;multiplythetwonumberstogether,andthentaketheremainderafterdividingbyn.

294

Some examples mod n for n= 6

• (28+3)%6=31%6=1=7%6=4%6+3%6=28%6+3%6• (13– 15)%6=-2%6=4=13%6-15%6• (8�7)%6=56%6=2=2�1=8%6�7%6• Toefficientlycompute1119 %6:

§ 11=5%6§ Thus,112 =52 %6=1%6§ Thus,1119 =11� (112)9 =5� 19 %6=5%6

§ Moregenerally,sincemoddistributesoverproduct:ifak =1%mthenan =an%k %m

• Notethat1=19%2intheaboveexample.• Wewillreturnefficientmodularexponentiationlater.

295

Division mod n for n=6

• Todivide%6,allnumbersneedamultiplicativeinverse%6.• Buttheprimefactorsof6donotpossessmultiplicativeinverses%6.• Suppose otherwise,i.e.,that2and3dohavemultiplicativeinverses%6.• Notethat1isthemultiplicativeidentityalsoformodarithmetic,sothatz

isthemultiplicativeinverseofxmod6ifxz=1mod6.• So,wearesupposingthat

– thereisazsuchthatz� 2=1%6,and– thereisysuchthaty� 3=1%6.

• Thus,z� 2� y� 3=1�1 %6=1%6.• However,z� 2� y� 3=6zy=0%6.• Thisisacontradiction,soourassumptionthattheprimefactorsof6have

amultiplicativeinversemod6isfalse.

296

We will see that we can divide mod p if p is a prime.

• Fromnowon,ourmodulus(divisor)willbeaprimep.• Wewillshowhowtodivideinarithmeticmodpbyusingtheextended Euclideanalgorithm.

• RecallthattheEuclideanalgorithmfindsthegreatestcommondivisor(gcd)oftwointegersaandb.

297

Recall basic Euclidean Algorithm to find gcd

1. Initialdata:integersn0 ,n1 s.t.0≠|n0|≥|n1|2. seta⃪|n0|3. setb⃪|n1|(≤a)4. whileb>0do

r⃪amodb,with0≤r<ba⃪bb⃪r

5. returna=gcd(n0,n1)• Again,asthewhileloopexecutes,b≥0strictlydecreasesandgcd(a,b)does

notchange- recallthewell-orderingprinciplere.convergenceoftheEuclideanalgorithminfinitetime.

• TheworkingofEuclid’sAlgorithmismadeclearerbyconsideringtheextended Euclideanalgorithmthatwillgiveintegerssandtsuchthat

sn0 +tn1 =gcd(n0,n1)

298

Extended Euclidean Algorithm1. Initialdata:integersn0 ,n1 s.t. n0 ≥n1>0

2. indexk=1

3. Ifn1 |n0 then{Returnn1 =gcd(n0,n1)=n0 -(-1+n0/n1)n1 }else:

4. whilenk >0do

nk+1 =nk-1 modnk with0≤nk+1 <nk

ifnk+1 >0then

nk+1 =nk-1- qk nk withquotientqk >0

=skn0 +tkn1 byrecursivesubstitution(seebelow)

k++

5. Returnnk-1 =gcd(n0,n1)and

6. Returnintegers(sk-2 ,tk-2)s.t. sk-2n0 +tk-2n1 =gcd(n0,n1)

• Notethats1=1andt1=-q1

• Exercise:writethepseudo-codefortherecursivesubstitution(seethe

followingexamples).

299

Extended Euclidean Algorithm – Example 1

• Ifn0 =21andn1 =15thentheEEAwillreturns=-2andt=3,sothat

sn0 +tn1 =3∙15- 2∙21=3=gcd(n0 ,n1)• ThestepsfortheEEAofthisexampleare:

1. 6 =21mod15 ⇒6 =-1∙15+1∙212. 3=15 mod6 ⇒3=-2∙6 +1∙15=3∙15– 2∙21(sub6fromstep1)3. 0=6mod34. Stop

5. Returngcd(21,15)=3=3∙15– 2∙21

300

Extended Euclidean Algorithm – Example 2

• Whenn0 =256andn1 =100,thestepsfortheEEAare:

1. 56 =256mod100 ⇒56 =-2∙100+1∙2562. 44=100 mod56 ⇒44 =1∙100- 1∙56 =3∙100– 1∙256(sub56ofstep1)3. 12=56mod44 ⇒12 =1∙56– 1∙44=-5∙100+2∙256(sub56,44ofsteps1,2)4. 8=44mod12 ⇒8=1∙44 – 3∙12 =18∙100– 7∙256(sub44,12ofsteps2,3)5. 4=12mod8 ⇒4=1∙12– 1∙8=-23∙100+9∙256(sub12,8ofsteps3,4)6. 0=8mod47. Stop8. Returngcd(256,100)=4=-23∙100+9∙256

301

Division mod p for prime p –multiplicative inverse mod p

• ByproofofconvergenceoftheExtendedEuclideanAlgorithminafinitenumberofiterations:ifaandbhavenocommonfactors>1(i.e.,theyare“relativelyprime”),thenthereexistintegerss,tsuchthat

as+bt =1.• Again,givena,b,theintegerss,t arefoundbytheEEA.• So,forallintegersbwithoutprimefactorp,wecanfindintegerss,t such

thatps +bt =1,

wheretiscalledthemultiplicativeinverse ofb(modulop).• Thatis,1=(bt)%p=(b%p)(t%p)%p.• Themultiplicativeinversetcanbereducedtoauniqueelementof

{1,2,…,p-1} bysuitablyselectings.• Restrictingourattentiontotherange{1,2,...,p-1},wehaveproved:

• Theorem:Forallprimepandb∈{1,2,…,p-1},thereisauniquemultiplicativeinverset∈{1,2,…,p-1} ofbsuchthat

1=(bt)%p

302

Division mod p for prime p

Theorem:Forallprimepandintegersa andywithoutfactorpbutwithfactorx,ifa%p=ythen(a/x)%p=(y/x).

Proof:• Notethatxalsodoesnothavefactorpbytransitivityofdivisibility.• Lets ∈{1,2,…,p-1}bethemodpmultiplicative-inverseofx,i.e.,

1=sx%p=(s%p)(x%p)%p• So,(as)%p=(a/x)%pand(ys)%p=(y/x)%p.• Ifamodp=ythenbymultiplicativepropertyofmoduloarithmetic:

(as)%p=ys.• Q.E.D.

303

Modular division - example

• Whatis5� 3mod11?• Note11isprime.• Weneedtomultiply5bytheinverse of3mod11• Bysimplearithmetic,ifyoumultiplyanumberbyits(multiplicative)inverse

thentheansweris1(themultiplicativeidentity),e.g.,themultiplicativeinverseof2is½since2�½=1

• So,theinverseof3mod11is4since3�4=12=1mod11• Thus5� 3mod11=5� 4mod11=9

304

Fermat’s Little Theorem

Theorem:Ifadoesnothaveprimefactorp,thenap-1 modp=1.Proof:• Thenumbersa,2a,3a,…,(p-1)ahavedifferentremaindersmodp(easilyproven

bycontradiction).• So,{1,2,…,p-1}and{a,2a,…,(p-1)a}arethesamesetofnumbersmodp.• So,theproductsarethesamemodp,i.e.,

(p-1)!=a∙2a∙3a…∙(p-1)amodp=ap-1 (p-1)!modp

• Finally,sincepisprime,wecandividethisequationby(p-1)!.• Q.E.D.

305

Chinese Remainder Theorem

• Iftheintegersm1,m2,m3,....mk anda1,a2,a3,....ak aresuchthat

• themi arepositive,pairwiserelativelyprime,and• theai <miarealsopositiveintegers,

• thenthereexistsanintegerr suchthat∀i,mi|r-ai ,i.e.,∀i ,ai =r%mi

• Ifwerequirethatr<∏imi,thenthisrisunique.

306

Proof of Chinese Remainder Theorem

• Sufficestotakek=2becausethegeneralk≥2istheneasilyproved byinduction.

• Thus,weneedtoprovethat,for0<a1 <m1 and0<a2 <m2 ifm1 andm2arerelativelyprime,thereexistsauniquerbetween0andm1m2 suchthat

r%m1 =a1 andr%m2 =a2• Sincem1 andm2 arerelativelyprime,thereexistintegerssandtsuchthat

sm1 +tm2 =1(*).• r:=a2sm1 +a1tm2 %m1m2 satisfiesalloftheconditions.• Inparticular,notethatbydefinitionofrthereisanintegerqsuchthat

a2sm1 +a1tm2 =qm1m2 +r;• so,

r%m1 =a1tm2%m1 =a1(1-sm1)%m1 =a1 ,whereweused(*)forthesecondequality.

• Q.E.D.

Exercise:CheckthatthedefinedrsatisfiestheotherrequiredpropertiesoftheCRT,andprovethemoregeneralresultfork>2.

307

The RSA TheoremTheorem:Ifpandqaretwodifferentprimesandtheintegeraisrelatively

primetopandqsuchthat1≤a≤pq -1,thenforallpositiveintegersk,

a1+k(p-1)(q-1) =amod(pq)

Proof:

• ByFermat’sLittleTheorem,ap-1 =1modp.

• Thus,ak(p-1)(q-1) =(ap-1)k(q-1) =1modp(sincemoddistributesover

product).

• Thus,a1+k(p-1)(q-1) =amodp(ditto).

• Similarly,a1+k(p-1)(q-1) =amodq.

• Thatis,q|a1+k(p-1)(q-1) - aandp|a1+k(p-1)(q-1) – a.

• So,(pq)|a1+k(p-1)(q-1) – a.

• Q.E.D.

308

How RSA works to send confidential messages• AnindividualAwishestobeablesecurelycommunicatewithothers(even

potentiallyuntrustworthystrangers)overaninsecurecommunicationmedium.

• Taketwolargeprimes,pandq.• Letn=pq.• Choseanintegere,relativelyprimeto(p-1)(q-1).• UsingtheextendedEuclideanalgorithm,findadsuchthat

de– k(p-1)(q-1)=1.• Thatis,de=1+k(p-1)(q-1).• “Publish” (n,e)asapublickey,i.e.,securelydisseminatethepublickey

throughatrustworthyKeyDistributionCenter (KDC),acriticalcomponentofthepublickeyinfrastructure(PKI).

• Thatis,individualsarereliably“bound”totheirpublickeysbytheKDC.• Retain(n,d)asaprivatekey.• Now,anotherindividualBmayencryptaplain-textbinarymessage�m� to

Abyraisingm totheeth power,andtransmittingthecypher-textme mod(pq).

• Uponreceipt,Awilldecryptthecypher-textbyraisingittothedth powerandtherebyrecoverm=(me mod(pq))d =med mod(pq).Ex.:Prove.

309

How RSA works to sign messages• SupposeBwishestoreceiveamessagemfromAthatissignedby

AinawaythatcannotbelaterrepudiatedbyAoreasilyforgedbyanotherparty.

• Using its privatekey(d,n=pq),AsignsthemessagebysendingtoBmdmodn

together with the messagem(oritscypher)• Having received this signaturefromA,Bcan“check”itby

decryptingitusingA’spublic keyandcomparingtheresulttotheplaintextmessagem,i.e.,

(mdmodn)emodn=m• BcannotforgeA’ssignaturewithoutfactoringn=pq andAcannot

laterrepudiateitssignatureforthesamereason.

310

Strength of RSA

• ThestrengthofRSApublic-keycryptographyliesinthedifficultyinfactoringntofindpandq.

• Thatis,athirdpartyinterceptingthecypher-textae willalsohavethepublickeyofA,particularlyn,butwillneedtofactorntofindpandqandhenced.

• So,typically,pandqareverylargeprimessothatfactoringnisdifficult,thoughrecentlyprovedtobeofpolynomialcomplexityinthesizeofn.

• Asofthiswriting,thelargestprimenumberyetfoundisoftheform2k-1(aMersenneprime)withk=57,885,161.

• Fewerthan50Mersenneprimesareknown.

• Exercise:Showthatthisprimenumberwillbe17,425,170decimaldigitslong.

311

“Efficient” exponentiationtocomputean

• power(a,n)• setr:=1• ifn=0,returnr• whilen>1do

– Ifnisodd,setr:=ar andn:=n-1– n:=n/2anda:=a2

• returnar

Exercise:Provethecorrectnessofthisalgorithm,i.e.,thatitreturnsan,by(strong?)induction.Intheprocess,observethestrategyusedforexponentiation.

Notethattheearlierexampleofefficientcomputationofakmodnrequireddiscoveringanon-negativeintegeri≪k suchthat1=aimodn;thus,

akmodn=akmodimodn.

312

Another Crypto-system: Diffie-Hellman key exchange

• Letpbealargeprime,sanumberbetween2andp-2;pandsare“publiclyknown”.

• Supposetwopeoplewishtosecurelycommunicateandeachpersonhasaprivatekey,respectivelyaandb.

• Theyfirstsendeachothernumbers,respectivelyx=sa modpandy=sb modp,

i.e.,employingtheirownprivatekeys.• Theyraisethenumbertheyreceivetotheirprivatekeypowermodp,andtherebyhaveacommonexchangekeykforasymmetriccrypto-system,i.e.,

k = ya modp=xb modp.

313

Another Crypto System: El-Gamal

• Asbefore,letpbealarge(publiclyknown)primenumber,ssomenumberbetween2andp-2.

• Eachpersonchoosesaprivatekeyeand“publishes”E=se modp.

• Tosendmessagem,wefirstgeneratea“sessionkey” k,andsendt=sk modpandc =(Ek m)modp

• Wedecryptbycomputingt-e c=mmodp

314

Secure communication protocols

• See Kurose & Ross, Chapter 8, available athttp://www.pearsonhighered.com/educator/product/PowerPoint-Slides-for-

Computer-Networking-6E/9780132856225.page

315