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# Introduction to Mathematical Thought MATH 124

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GPA 3.72

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This 45 page Class Notes was uploaded by Breanne Schaden PhD on Saturday October 3, 2015. The Class Notes belongs to MATH 124 at Boise State University taught by Tommy Conklin Jr in Fall. Since its upload, it has received 5 views. For similar materials see /class/218000/math-124-boise-state-university in Mathematics (M) at Boise State University.

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Date Created: 10/03/15

3 5 ANALYZING ARGUMENTS WITH EULER DIAGRAMS 57 39 The sinienieni quotThe square nrn Iwnd iiniis digit is 5 will end in 25quot becomes wordig 1 lie 39 is 39 u dh39 39iv39 eni i 9quot 139 quot d g 5 39 L a 39 58 Answcrsw lvai39y Oneexnnipleis ieniiileiKeibeiis inz V 3 years nlii leiiniier Kurhei is less ihen 3 years iihL nil mnnher whose 57 Answers will Vary One example is Tliiii man is Ktm ii Merrill Thzll man sells hunks quotIfil no The sinieineni An inieger whose nniis digil is o in s is iiiyisi lo y 5quot hcunics lfan inwgcrhas n unilsdigil or leis ihen ii i iiiiisihle by 5quot 35 EXERCISES il 0 iiiinDisiheenswersineeu ineessn finsquot V i W 1 Draw nn Enleminernm whenlhcregmnrepmscnung remnantshecunvse8uniullolmnmhcr V V y hmiers niiisi he inside ihe ieeinn iepieseninig ihose Sl lellklnl quot A whii wear iiiinks V iliai ihe ii i prenusc 5 line 4 Wrilingcxercisc 4 M ng emeise mwhnwwmk 44 wiiiing xercisc 45 The iienieni quot5 9 gt1 ifund only ire 2 v 10quot is m sinus his is a hicondiiinnnl composed of iwo true Swmmm xmpnsems Chris Made 39 V 4 Lei e Icprescnl ciis Mnder13y ieniise 2 e iniisi lie in W Th WW 3 T 1 f G quot quotd gt quot r h A ihe quotboxersquot I39ezilln since ihis inncs iIie eenclusionio sinee iliis is n hicundllmn l en iiiig ofa iiiie and a be 1m be when is m false sinienieni i 47 The sldlcmclll x i i 15 i and nnl ii393 x 5 3s 9quot is 239 E E lg f 5 s f2225639f e em Sin quot5 is i bim m39mm39 0 Sihm39g 01 a 5 cpres39cming loca nnx ihsi have ihrill micsquot se ihai ihe 3quot n WC 5W1 I iiisi premise is inie 4x Tllcsmiemcnl39fl x 2 Hi nd un ir 9 7 e 11239quot is m xillue ihis is e hieonilniininl callsisllng er lwa l39nlse hllcmums leniuus hm M will rides 49 The sluemClquotBiHC1illmn wns piesidem new only if Jiiniiiy csiiei was iini pieshleniquot is since iliis is ii mm 9 bimndiiicllal collsisi39lng ilf A mic and a false slaicmem so Burger King sells Bi Miiesii iinii nnlyil lBM newuenMirwllmcrim ninnuieeiin s ennipnie is since ihi 39 n e Lei in rel39 mll ihe nniiisenieni lurk areiimiimicn39 n musi Iii in me amuseman 31 The sinieinenis quotElvis is alivequot ilnd Elvis iienir an Parks mng Sim his mm m wmmmn m be quotW since bulk canilm be we in mi some lim mg umumcm is 53 The simmems quotGeorge W Bush s u Democratquot a 1 Drilw nn Eniei L hlgmm where iiie region iepieeeniing quotGWTSC W BuSh is 1 Rtimbliclm39 MC 414 60 W WW quotSoiiihernersquot lies inside ihe reginii ieprcseniiiig quot ihose bolh cannm hc true I ma same rm whn sneak wiih nii aeeeni Lei e iepreseni Bill Lennard 5 The sinieinenis quotThat linininl lins rein l quot and Thm By K39 mud P WM Bi L00 1quot W L quotWide 1 nniinni isaiiiig39 neenn 39sieiii Smutbolhsmlemcnlsmm r0250 01 m whoanea I lnnmenmud hum may 1ic inside ihe inner or ilie niiiei rcgluni Bui ii i e 39 conclusion m be iine he mm he only inside ilie inner 54 The sinienienis Thin hunk is noniieiienquot and quotThai hook egm Thumm gnmcn sN ki e mine i inn s70 nre gtimzc in niulcmcnb L39Lm b mic 39ninsc who Speak Mill in acceni 55 The sinieinenis quotThis nuinhei is nn integerquot and This x number is irrmiunulquot zlrc cnu rurmsmce bmh cannot be inie ni ihe snnie iiine 56 The sinieiinnis This number is pi v quot ilnd Thih xmprzsems Bill mm liiiiiilier is ii iihiiiinl number are cilnsisleni since hoili sialemunis Cim be lrue 8 CHAPTER 3 INTRODUCTION TO LOGIC Draw an Euler diagram where the region representing politicians must be inside the region representing those who lie cheat and steal so that the rst premise is truc Those who lie cheat and steal x Politicians X it represents that man Let l39 represent that man By the second premise a must lie in the those who lie cheat and steal region Thus he could be inside or outside the inner region Since this allows for a false conclusion he doesn39t have to be in the politicians region for both premises to be true the argument is invalid Draw an Euler diagram where the region representing contractors must be inside the region representing those who use cell phonesquot so that the rst premise is true Those who use cell phones Contractors it represents Doug Boyle Let m represent Doug Boyle By the second premise I must lie outside the region representing those who use cell phonesquot Since this forces the conclusion to be true the argument is yam Draw an Euler diagram where the region representing dogs must be inside the region representing creatures that love to bury bones so that the rst premise is true Creatures that love to bury bones Dogs x represents Archie Let 1 represent Archie By the second premise 0 must lie outside the region representing creatures that love to bury bones Since this forces the conclusion to be true the argument is valid Draw an Euler diagram where the region representing people who apply for a loan must be inside the region representing people who pay for a title search so that the first premise is true People who apply for a title search it People who apply for a loan it 1 represents Cindy Herring Let w represent Cindy Herringquot By the second premise 1 must lie in the people who pay for a title searc region Thus she could be inside or outside the inner region Since this allows for a false conclusion she doesn39t have to be in the people who apply for a loan region for both premises to be true the argument is invalid Draw an Euler diagram where thc region representing residents of Minnesota must be inside the region representing those who know how to live in freezing temperatures so that the rst premise is true 0 he know how to live in freezing temperatures at Residents of Minnesota X 1 represents Wendy Rockswold Let 6 represent Wendy Rockswold By the second premise I must lie in the those who know how to live in freezing temperaturesquot region Thus she could be inside or outside the inner region Since this allows for false conclusion she doesn t have to be in the residen of Minnesotaquot region for both premises to be true the argument is invalid Draw an Euler diagram where the region representing philosophers intersects the region representing thost who are absent mindedquot This keeps the rst premise true Philosophers x 1 represents Amanda Pardarls Those who are absennuinded Let or represent Amanda Perdaris By the second premise a must lie in the region representing philosophers Thus she could be inside or outside t region representing people who are absent minded Since this allows for a false conclusion the argument invalid 59 mi DI39JW an Euler iliiigrzim whom iiio region rcpi39cscnling me an Euler Aims nni whore me region represeniing lnDS ulSquot inrei eois Hm region reiiieseiiiing quotpinrir inunii ircesquot niiisi he inside U18 region roprescniing cliursquotTl s keups ilie rsi prem e ime 39Things ihni iiiive groan lcuvcs so ml me iiisi promise i irii bank Dlnuaxuls P39 x Things wilh green have x mprlzscms Danny ioi i reprcsoiiinoannyt By iiic sucoiid prciillsc1 imisi liein hciwun rcpicscnlngquotplunl leis 39 Thus he Louiii hui iriooroii ihii IHHWS ror ii i39iilso im iiio iegiun dino nuo conclusion lhc iirgnmeni is since Imprcsmu onion Lei x i cpi cscnl mini plantquot By miisi inside riie region represen V as ihc cunc D ii premise B 4 ml irees Druw un Enler diagram where irie wgion mpwscnling iiiicks lnwracc s ilie region rcpresoniins whicies wih sound ysiems quotThis keeps iiie iirsi premiseirue There 7 V m me my m rewmm m in 1mm 35 W 11 The valid ai39guxiiuni oi Exuiiiplc4 in ihe mu is One way is as below EXZKI i g e dIA mm v we 1 apparenrihaiihe conclusion is lalsc All cxpcnsnc things oro iiesirnoio All desirable ihings make you l39eei good All I mnk ml BI mind In W All expen we iliings mnko you live longer Aiiuiher possihle cunulusion which Will keep ihe quotAll ex an veih V I gt nrgiiiricrii Vallde gs nioke you feel Since ihe diagram shows lrue premises hul lm a inlse gaudy Tim migumm mm wild since the p es cunclusmn the 5 summit xs Nola Umle ways 10 dmgmmmed I gum l3 in me BM iorce mis conclusion a v 39l Euler in am represeniirig Hm premises musi io he ime Liko a5u yltld ii Lruc Conulusiun lor iiie ngiinieni io he validv 3quot U G IW iin Euloi iliiigriiin whim ht region roprcsoriiing s ilie region rcprcscming quotiilose who 39 39h s keeps ihe l iisi premix irue P P L Who haw quotN 3 none Dc soquot The ioiiownig is ii valid argulncm which can be eonsiriieioii i39roni iho given Euler diagram or surgery mUS go in me h him i m srv iiiliiinne Pm on rimsi g0 io me hosp JL lo o luwing i n id urguinunl which can he consimoieii rroni iho g vun Elllcl diagram x epmienis icini Flilgoul All pimple wiili blui eyes more blond Imir Lei represeni Kim Foigoui By hc second premise Julie W I this nm L blond ha i iniisi lic in iho mgiun reprosoniinn nurse Thu iniie wind docs nor have blue eyes she cuuld he in as or oulx id ihe iegion quotHm wlm weor mire iinii39oims quot Since ihis aiiom ior a misc 7 V mncwun m mum i L l7 Thu lolluwmg Eillcrdizigrurn rcprcsenis irue premise V liiici hanging ihe secomi PI CH inii ihe wnulu on ul Examplc 3 in H ext yields he following iirgunwni ningi um nine iubbcr Ail honmin irees have green leaves nrngs ulqihzvciims Th I lum x h n lll Thni pl n iins gic ri loaves Sincl ilie diagram forces me cnnCIuSion io hr iroe also ilio m39gumem i iii CHAPTER 3 INTRODUCTION TO LOGIC as a l ill5 picniises sc ihal ihizy are iruc however ihe arugmeni is invalid since xccurdiiig 0 the diagram all birds are planes which is false cvcn ihough he sialed cnlnclusicn is LINE x Izpi39escm s Oimwa 35 m V u x can reside inside UT oumirie uflhe Ciiicsl12ilme nor lhizzisl of Toronto diagram ln ihe one case M a insidcj in conclusion is true in lhb ulher case ince lrue premises he argument IS a c m 5 D lt n a 9 E 5 19 The fulkiwiiig Euler diagram lepieseiils lnle premises Cmmres ihat have a bank Birds Figm win lougesi sides The arguiiiunl is iiiv even innugh me conclusion is 39 39 inquot mi mi hnwr Pr39 39 4 L r a false statem m lclniionship between the largest angle and the iongesi sidc The argiimeni is invalid even ihough iiic 3950 ms 3 WWW W be quotW 24 The following Eulenliagruni mpresenis ilie iwD prcmises is being rruc and we are breed inlu u mic conclusion x x rcprnsmls 4 Aninuis wiiii beaks Thus lhl argument in v Observe ihal the diagram is he only way to show lruc premises m pmmi39xm market A B and c m39efullawed b y veveml n posxi39ble concluxiunx Exercises 25 30 Take each 39 J the mu 39 Tiius he arguiiieni is valid also forces he conclusion to be true is valid Dr lIlWllldv A All people who drive conirihuie in air piilluiion B All people who conu ihuic lo iiir pallulion make liii Li lilile worse C Sn 0 people who live in a suburb make life i lillle worse Diagiziiii ilic ilircc premisss 1 be iriie Ciucs soilih oi Mammy Cliiss SollLi l orTarnpicn x x repmsenis Vmcriiz Thus ihe ulgumiml in valid Observe ihiii ilie diagram is he only way in show lruu prcmises Tm Wm cunnibinr in air pollulioii C n Then who live in in suburbs EXTENSION LOGIC PUZZLES 61 N Ur N as N 3 N 92 N O 394 0 We are not forced into the conclusion Some people who live in a suburb drive since diagrams A or B represent true premises where this conclusion is false Thus the argument is invalid We are not forced into the conclusion Some people who live in a suburb contribute to air pollutionquot since option A represents true premises and a false conclusion Thus the argument is invalid We are not forced into the conclusion Some people who contribute to air pollution live in a suburb since option A represents true premises and a false conclusion Thus the argument is invalid We are not forced into the conclusion Suburban residents never drive since diagrams C represents true premises where this conclusion is false Thus the argument is invalid The conclusion All people who drive make life a little worsequot yields a argument since all three options A C repreSent true premises and force this conclusion to be true The conclusion Some people who make life a little worse live in a suburb yields a lLd argument since all three options A C39 represent true premiSes and force this conclusion to be true 31732 No answers EXTENSION LOGIC PUZZLES Draw charts as indicated and complete using the initial information given Use 0 for Yes and x for No For any cell that is assigned a mark x in the remaining unmarked cells in that row and column I Neither Lauren nor Zach was the child accompanied by Ms Reed Tam was more interested in shopping at the country store than in picking pumpkins Mark x s in the boxes for Lauren and Zach under Ms Reed s column Place a into the intersection of Tara and country store since that was her interest Note remember to always mark x39s in remaining cells of a row or column when marking a cell with a 2 Xander and his father Mr Morgan didn39t go on the hay ride Mr Fedor39s child who isn39tZuci or Lauren was fascinated by the cidermaking process Place 0 into the cell representing the intersection of Xander and Mr Morgan since you know that they are father and son Place x into the cell representing the intersection of Xantler and hay ride as well as the cell for his father Mr Morgan and hay ride Place 0 into the cell at the intersection of Ms Fetlor and cidet39 making Place x39s in the cells representing Lauren and Zach in Ms Fedor s column Since neither Lauren nor Zach is Ms Fedor s child neither can be interested in cidermaking Mark x s in the cells for cidermaking in Zach and Lauren39s rows 3 Zach is neither the child who went on the hay ride nor the one who wanted to go ripple picking Mr Hanson s child didn 39I go on the hay ride Place x in the aspect cells for apple picking and hay ride in Zach39s row This leaves feeding animals as Zach39s favorite aspect Place I in the feeding animal cell for Zach Mark it in the aspect cell for the hay ride in Mr Hanson39s column Tara and Raven are the only two children that could still belong to Ms Fedor Since Ms Fedor39s child liked cider making check the aspect columns and notice Tara liked the country store This leaves Raven as Ms Fedor39s child Place 0 in the cell for Raven in Ms Fedor39s column and o in the cidermaking cell in Raven s row The only aspect now available to Xander is apple picking Place u in that cell The last aspect the hay ride is now Lauren39s only possible aspect Place u in the appropriate cell At this point Chart 1 will have been completed Churl I Tara is now the only child left in Ms Reed39s column Place 0 at the intersection of Lauren and Ms Reed Notice that Xander s favorite aspect is apple picking and we know that Mr Morgan is Xander39s father Thus We can place a in Mr Morgan39s column in the apple picking cell This leaves Lauren and Zach with the possibility oer Hanson and Mr Maier for parents By comparing Lauren39s favorite aspect the hay ride to Mr Hanson and Mr Maier s aspect columns we nd that Mr Hanson s child did NOT like the hay ride Thus Mr Maier must be Lauren39s father Mr Hanson is left as Zach39s father Now it is a simple matter of matching each chi ld s interest with the aspect cell for his or her parent Zach likes feeding animals mark 0 in the feeding animals cell for Mr Hanson Lauren likes the hay ride mark o in the hay ride cell for Mr Maier The country store is now the only available interest in Ms Reed s column Check to see that Ms Reed s child Tara is in fact the child that likes the country store 62 CHAPTER 3 INTRODUCTION TO LOGIC lquot Plate is in lhc uppropn39ula cell and the logic nunnle is complete Chan 2 Char 2 Thus Lauren Mr Muir hay rldc Raven Ms Fedor Filler making Tam Ms Reed country store Xuntler Mr Morgan apple picking 2 ch Mr llunson feuding animals Draw churts as indicated and cnmplele using the mitiul 39 uliun aiven Use quotcquot for Yes turn x for o For nssi ed mark x in lhc remaining unmarked cells in ha row and column 1 Aliens nan her mime imprinted on Iler new omnge bowling bnlL Dnvon bongII n bowling bugfur his new ball which is lighter than Sl lm39s Arlenes ball is uran and ere orecun e another color Deyrrns balli lighter than Silas s Tllus Devon39s on 1 cannot weigh the most ls 1 Mark x uppmprialcly 2 Tina bowl mm n Hrpauml bull The pink bullmg halll39 utly 6 pounds ligluerlhml he ntrqnnne me Tin 3 ball is 14 lbs amt thereiore Can39l be any other weigh ull olher cells x 8 lb Cell in m red column Thu red ball which is 18 lbs is no Rosellu s so x the 8 lb and red columns in Rusella s row Alsu mark x in lhc 16 lb cell in Rosella s low The smallest ball is IL 35 and the largcsl bull is 18 lbs Hav ng already dclcrmined lhzn he 8 lb ball is rell the turquoise ball which 39s 6 pounds heavier than the pink ball musl be If lb and Lhc pink hull 10 1h Pul Xs in all rows and Columns that conlain a known value l this polnl yott will have complc ed the following chnrl Chatr l The orange hall cannul he ll 16 or 8 lbs By transferring mesa xl39s lo Arlene s row Arlene owns the alarm bull we determine Ihul Arlene s ball is 12 lb chart should now be compleled as in CIlarIZ below Chart 2 Thus Ar one orange 12 Devan turquoise l6 39n a Sila l Rosella ll k l sred 18 Tina gray l4 Drnw c 3 d5 illtllculcd and cunlpltle using hc tmtml int ormniinn givntl Us 4 lot Yr and mm For nny uc that is assigned 3quot marl xquot in he rmnnining unmarked cells in that rnw and column t ll 39rliu lie with llquot grinning epl39EIIltlulu quotmil n nmmlihm n dungHell Mark K in it call m lhe illlcl39seclilln oldzlughler nnu grinning leprechaun I lit 39 CI LIl lie fmmw IlL lIIl l39 119 dancing leilltle39el39 llur Illz yrllrlltl Imp rm x he dnnumg leilldcer nan the yullnw hllppy 1 le l onl Ml39 me s mw 3 Will Spuiglzr39 lie ttrrmt l a rmml from llt39 Place x in ma llnclc cell in ML Speigler n rtiw m39lc 4 Tim lie with l lle yulltm39 lumpy39th lt mll39l n gllflom tl S39l39Allfl Mark it in lhc cell at lite InlmSCCllOn ofyclluw happy litres lnd sislcr lil Mr El lllli mil Mr Spel39glzr won lite lit wt39lh 1 l pl esenl lihm r Biz The lcprccltmtn lit could s l EXTENSION LOGIC PUZZLES 63 Ctnn E 55 ch in on 39lltns Mr Crow cupids daughter Mr 1 nm lcprechuuns uncle Mr Hurley reindeer Slslcr Mr Speigler happy ftlccs lalherrinlaw Draw rs39 lldicul and uumplel usmg lheiniliul inlormlllinnk veil Usu uquot for man orNoFnr nny cell lhal is assigned x mark x in he remaining unmarked coils in mm mw and column I The l eal ivw iJll39I the tremiult veilher Juanita or puceslti t 39 have received 1 ill mm his l39tllhtr law so X Lnlll llL39ilwr IL llE lrl Ihe illJail in a p I ur 00L ll l nlhcril39llllw crlls for Mr Crow and Mr Hmley Thu In39p amitlltl Iii ltlorltl all a magic Il nill Platelx in h 39 39 39L 39 Fr LCM l l 39 L seal httr Mrl 39 39 lmmu 1nd LL x In nppmprlall ccllx Mr me39s Unly ttt opllml is l well Ill 19 Cirrus Place X in the cell or uly bear n inlerscclion ol39nllpidx and Mr Cluw i Juannh39s mw Mark 0 at lhi ittlcrsccllon ul Juanne and 39 I 39 I l5Mrny 39 39 39 at forlhc cell for sister in Mr Hutlcy s inn T lh unly choice for Mr Hurley39s row cl lhc L mindccr tn Mrl Hurley39s row an to lllc cell for rcimlm in lhe sislcr mw VBL39IHISC Mr Hurluy received his lie from lei m Ill apprupl39iale cell in the 39 zly hear column l3 Willllle39x llllaglmtlyjl and is it zebra Placc I or zcbm in Wi 39 row Joanne is lull will the mouse as her clltll39tlcler st plate l in mm cell This leaves the grizzly bear l39ur Lou and once lhul is markctl lhe sen lcl l for erl 11 10mm 5 39 place u in ht cell I ur circus seal39s unly ildvtnlurc chuice is now me rock band so i i L I SinceMrCrtlwtithOT 39 39 39 l39 39L inan or i m i 39 The lhlllglrinrlaw coulrl nuw only llavc given me i it so he ilpprllpnult cell This lenvr hum l 393 lllL cupid lie for the n lit for mellncle Sinualhc unptdlt b w place u in lhu dilughlcrccll in ML Crows mw ll39 Speigler s lie mull nnw only have come tnnn his l ulhcrrlnr w luavlng lilt uncle39s lie for Mr Evansv Now nullcc lllul Mr Evan s lit came from his unclc and the uncle pulcllnscd lhc lcpl39echzlun tie n Mr Emits rel ved l 2 lie wllh lepruuhulllis and Mr Slvgigltl lcceivctl lhe mil rcmuining tic lllL lie with htlmly faces Your Conlplcmd cllnll should look like Ih following 4 Thu bear rlidlt39l lmml quotimamhip 0 1h minim Mark K in lhe cell at the intersection of grizzly llcal39 nnd spacesl 39 The grizzly bear must have lakcn the trnin sn t Ihl lmin cell in lhc grimy bearcalulllll This llm ubra Inu39s character lhc gtr ly a a lit rain 30 mark 0 in he lrain cell in Ltlu s low This leaves lhc spaceship l39or Winnie39s character he hill The Complcled ulla waultl look likr Ihc ll 0 3 64 CHAPTER 3 INTRODUCTION TO LOGIC Chart Grizzly bear Rock band Spaceship Thus Joanne moose circus Lou grizzly bear train Ralph seal rock band Winnie zebra spaceship 36 EXERCISES 1 Let p represent you use binoculars q represent you get a glimpse of the comet quot and 739 represent you will be amazed The argument is then represented symbolically by P 4 q q 7 p gt r This is the valid argument form reasoning by transitivity 2 Let 1 represent Billy Joel comes to town 1 represent 1 will go to the concert and 7quot represent I39ll call in for work The argument is then represented symbolically by P 4 w p r This is the valid argument form reasoning by transitivity 3 Let 1 represent Frank Steed sells his quota and 1 represent He will get a bonusquot The argument is then represented symbolically by P gt C 1 q This is the valid argument form modus ponens 4 Let p represent Amy McRee works hard enough and q represent she will get a promotion The argument is then represented symbolically by P q P G This is the valid argument form modus ponens Let p represent She buys another pair of shoes and q represent her closet will over ow The argument is then represented symbolically by P q 1 I Since this is the form fallacy of the converse it is invalid and considered a My Let p represent He doesn39t have to get up at 530 am and q represent he is ecstaticquot The argument is then represented symbolically by P q I p Since this is the form fallacy of the converse it is invalid and considered a fallacy Let p represent Patrick Roy playsquot and 1 represent the opponent gets shutout The argument is then represented symbolically by P q 1 Np This is the valid argument form modus tollens Let p represent Pedro Martinez pitches and 1 represent the Red Sox wins The argument is then represented symbolically by P 394 fl quot 11 NP This is the valid argument form modus tollens Let 17 represent we evolved a race of Isaac Newtonsquot and q represent that would not be progress The argument is then represented symbolically by Z7 quot q i Nq Note that since we let q represent that would not be progress then q represents thath progress Since this is the form fallacy ofthe inverse it is invalid and considered a fallacy Social Choice The Impossible Dream Exercises Majority Rule and Condorcet s Method 1 In a few sentences explain why minority rule the voting procedure for two alternatives that is described on page 341 satisfies conditions 1 and 2 on page 341 but not 3 2 In a few sentences explain why imposed rule the voting procedure for two alternatives that is described on page 341 satisfies conditions 1 and 3 on page 341 but not 2 3 In a few sentences explain why a dictatorship the voting procedure for two alternatives that is described on page 341 satisfies conditions 2 and 3 on page 341 but not 1 4 Find or invent a voting rule for two alternatives that satis es a condition 1 on page 341 but neither 2 nor 3 b condition 2 on page 341 but neither 1 nor 3 c condition 3 on page 341 but neither 1 nor 2 5 In a sentence ortwo explain why it39s impossible with an odd number of voters to have two distinct candidates win the same election using Condorcet39s method 6 Construct a realworld example perhaps involving yourself and two friends where the individual preference lists for three alternatives are as in the voting paradox of Condorcet Other Voting Systems for 3 or 4 Candidates 7 Plurality voting is illustrated by the 1980 Senate race in New York among Alfonse D Amato D conservative Elizabeth Holtzman H liberal and Jacob Javits l liberal Reasonable estimates based on exit polls suggest that voters ranked the candidates according to the following table 22 23 15 29 7 4 D D H H J J H J D J H D J H J D D H a Is there a Condorcet winner b Who won using plurality voting 8 Everyone wins Considerthe following set of preference lists Number of Voters9 Rank 3 1 1 1 1 1 1 First A A B B C C D Second D B C C B D C Third B C D A D B B Fourth c D A D A A A Note that the rst list is held by three voters not just one Calculate the winner using a plurality voting b the Borda count c the Hare system d sequential painNise voting with the agenda A B C D 9 Consider the following set of preference lists Number of Votes 7 1 1 1 Rank First C D C B A Second A A D D D Third B C A A B Fourth D B B C C Calculate the winner using a plurality voting b the Borda count c the Hare system d sequential painNise voting with the agenda B DCA 10 Consider the following set of preference lists Number of Voters 8 Rank 2 2 1 1 1 1 First A E A B C D Second B B D E E E Third C D C C D A Fourth D C B D A B Fifth E A E A B C Calculate the winner using a plurality voting b the Borda count c the Hare system d sequential painNise voting with the agenda B D C A E 11 Considerthe following set of preference lists Number of Voters 5 Rank 1 1 1 First A B C D Second B C B C D 11 Third E A E A C Fourth D D D E A Fifth C E A B B Calculate the winner using a plurality voting b the Borda count c the Hare system d sequential painNise voting with the agenda A B CDE 12 Considerthe following set of preference lists Number of Voters 7 Rank 2 2 1 1 1 First A B A C D Second D D B B B Third C A D D A Fourth B C C A C Calculate the winner using a plurality voting b the Borda count c the Hare system d sequential painNise voting with the agenda B D CA 13 Considerthe following set of preference lists Number of Votes 7 2 Rank First C E C D A Second E B A E E Third D D D A C Fourth A C E C D Fifth B A B B B Calculate the winner using a plurality voting b the Borda count c the Hare system d sequential painNise voting with the agenda A B CDE 14 Considerthe following set of preference lists Number of Voters 7 Rank 1 1 1 1 1 1 1 First C D C B E D C Second A A E D D E A Third E E D A A A E Fourth B C A E C B B Fifth D B B C B C D Calculate the winner using a plurality voting b the Borda count c sequential painNise voting with the agenda A B C D E d the Hare system 15 operates exactly as does the Hare system but instead of deleting alternatives with the fewest rstplace votes it deletes those with the most lastplace votes a Use the Coombs procedure to nd the winner if the ballots are as in Exercise 14 b Show that for two voters and three alternatives it is possible to have ballots that result in one candidate winning if the Coombs procedure is used and a tie between the othertwo ifthe Hare system is used 16 In a few sentences explain why Condorcet39s rule satis es a the Pareto condition b monotonicity 17 In a few sentences explain why plurality voting satisfies a the Pareto condition b mono tonicity 18 In a few sentences explain why the Borda count satis es a the Pareto condition b monotonicity 19 In a few sentences explain why sequential painNise voting satisfies a the Condorcet winner criterion b monotonicity 20 In a few sentences explain why the Hare system satis es the Pareto condition 21 In a few sentences explain why the plurality runoff method satis es the Pareto condition 22 Use the following ballots to show that the plurality runoff method does not satisfy the Condorcet winner criterion Number of Voters 5 Rank 2 2 1 First A B C Second C C B Third B A A 23 Use the following ballots to show that the plurality runoff method does not satisfy monotonicity Number of Voters 13 Rank 4 3 2 1 First A C D E Second B A A B D Third C C B C C Fourth D D D A B Fifth E E E E A 24 Considerthe following two elections among candidates A B and C Number of Voters 4 1 1 Rank 1 1 First A A A A Second B B C B Third C C A A Number of Voters 4 Rank 1 1 1 First A A B B Second B B C C Third C C A A a Use these two elections to show that plurality voting does not satisfy independence of irrelevant alternatives b Use these two elections to show that the Hare system does not satisfy independence of irrelevant alternative 25 Construct ballots for the alternatives A B and C to show that the Borda count does not satisfy the Condorcet winner criterion 26 Show that the nonmonotonicity of the Hare system can also be demonstrated by the following 17voter 4 alternative election In a number of recent books this example is used to show the nonmonotonicity of the Hare system The easier 13voter 3alternative example given in the text was pointed out to us by Matt Gendron an undergraduate at Union College Number of Voters 17 Rank 7 4 1 First A C B D Second D A C B Third B B D A Fourth C D A C 27 The following example illustrates how badly the Hare system can fail to satisfy monotonicity Consider the following sequence of preference lists Number of Voters 21 Rank 7 6 5 3 First A B C D Second B A B C Third C C A B Fourth D D D A a Show that A is the unique winner if the Hare system is used b Find the winner using the Hare system in the new election wherein the three voters on the right all move A from last place on their preference lists to first place on their preference lists 28 In a few sentences explain why with an odd number of voters a sequential painNise voting always yields a unique winner b we can never have exactly two winners with the Hare system 29 In a few sentences explain why the plurality runoff method can never elect a candidate ranked last on a majority of ballots assuming there are no ties for rst or second place in the voting 30 Produce ballots showing that plurality voting can in fact elect a candidate ranked last on a majority of the ballots lnsurrnountable Difficulties Arrow s Impossibility Theorem 31 Complete the proof ofthe version of Arrow39s theorem from the text by showing that neither B nor C can be a winner in the situation described Your argument will be almost word for word the same as the proofs in the text A Better Approach 32 Ten board members vote by approval voting on eight candidates for new positions on their board as indicated in the following table An X indicates an approval vote For example Voter 1 in the rst column approves of candidates A D E F and G and disapproves of B C and H Voters Candidate 1 2 3 4 5 6 7 8 9 10 A X X X X X X X B X X X X X X X X C X X D X X X X X X X X X E X X X X X F X X X X X X X X G X X X X X X H X X X X X a Which candidate is chosen forthe board ifjust one of them is to be elected b Which candidates are chosen ifthe top four are selected c Which candidates are elected if 80 approval is necessary and at most four are elected d Which candidates are elected if 60 approval is necessary and at most four are elected 33 The 45 members ofa school39s football team vote on three nominees A B and C by approval voting for the award ofquotmost improved playerquot as indicated in the following table An X indicates an approval vote Number of Voters 45 Nominee 7 8 9 9 6 3 1 2 A X X X X B X X X X C X X X X a Which nominee is selected for the award b Which nominee gets announced as runnerup for the award c Note that two of the players quotabstainedquot that is approved of none of the nominees Note also that one person approved ofall three ofthe nominees What would be the difference in the outcome if one were to quotabstainquot or quotapprove of everyonequot Writing Projects 1 In the 2000 presidential election in Florida the nal results were as follows of of Candidate Votes Votes Bush 2911872 4904 Gore 2910942 4902 Nader 97419 164 Buchanan 17472 029 5937705 Making reasonable assumptions about voters39 preference schedules give a onepage discussion of how the election might have turned out under the different voting methods discussed in this chapter 2 Frequently in presidential campaigns the winner ofthe rst few primaries is given front runner status that can lead to the nomination of his or her party Moreover there are often several candidates running in early primaries such as New Hampshire In one page consider a recent election and discuss how the nominating process might have proceeded through the campaign if approval voting had been used to decide primary winners 78 CHAPTER 4 NUMERATION AND MATHEMATICAL SYSTEMS 16x900 51 7d 1 1 900 52 0 1 1 2 1800 V 4 3600 53 Writing exerctse 8 7200 54 Writing exercise gt 16 111400 77000 in Egyptian numerals is f fi 1 737gtlt103x17x1013x100 441600 in Egyptian numerals is nenesg wwgwww 2 9259x1002x105gtlt1 9 x 102 2 X1015 X100 14 1 I 400 1 Emma mm 5 3 3774 3 X 1000 7 X 100 7 x 10 f 9999 4X1 Then 3 x 103 7 x 102 7 x 101 4 x 10 M f f e30 994 c 999999 4 123981gtlt100002gtlt10003gtlt100 9gtlt108x1 397 M 9999 1gtlt10 2gtlt1033gtlt102 332 99999 1 0 303 WWW rum 99999 9X108xl0 5 4024 4 x 1000 9 x 100 2 x 10 Regroup len scrolls to make one lotus ower 4 X 1 4 x1039 x 102 2 x101 5563353213 0 4 x 10 ff f R a 1 H k f I 6 52118 5 x 10000 2 x 1000 1 x 100 euroup ten otus owers to ma eone pomtmg mger 1 X 10 8 X 1 99531399 WWW 9 5gtlt10quot2gtlt1031gtlt102 MMM39 1x1018x10 Regroup ten pomting ngers to make one burbolhsh 7 14206040 1 x 10000000 4 X LOOOIOC egepeeebee My 2x10070000gtlt10000 This is equivalent to 533000shekeis 6 X 1000 0 X 100 4X100gtlt1 41 wining exercrse 1 x 1074 x 105 2 X 105 42 Writing exercise 0 X 10 6 gtlt 103 43 Writing exercise 0 X 102 4 x 101 44 Writing exercise 0 X 100 45 99999 Five distinct symbols allows only ve positions 8 212011916 2 gtlt 100000000 1 gtlt 10000000 2 x 1000000 4639 9 999 999 99939 0 x 100000 1 X 10000 47 The largest number is 44444 w which is equivalent to 1 x 1000 0 x 100 3124 1 X 10 6 x 1 8 7 6 48 The largest number is 4444444444im which is 2 X 10 51 x 10 X 10 equivalent to 9765624 0 x 10 l X 10 3 2 49 10d 1 Examine Exercise 45 to see that 1 X 101 9 X 100 105 1100000 199999 1X 106X 10 5d1 9 4X142 N5 b T 5 42 ARITHMETIC IN THE HINDUARABIC SYSTEM 79 3 x 100 5 x 10 350 6 x 1000 2 X 100 9 X 1 6209 5 X 100 000 3 x 1000 5 x 100 6 x 10 5 8gtlt 1 503 7 x 10000 000 4 x 100 000 1 x 1000 9 X 1 70401009 3 x 100 000000 8 X 10 000 000 2 x 100 3 x 1 380000201 54 5 X 101 4 x10 35 3 x1015 x10quot 8x1019gtlt10 809 89 782 7 x 103 8 x 101 2 x10quot 413 4 x1031x1013 x10quot 11 x 102 9 x 101 5 x10 1 X 10 1 x 102 9 x10 5 x10 1000 100 90 5 1195 85 8 x 101 5 x10quot 53 5 X113 x10 3 x10 2 X10 302 32 784 7 X 102 8 X1014 x100 523 5 x 102 2 x 101 3 x100 2 X 102 6 X1011gtlt100 200 01 261 757gtlt1015 x10 34 3 gtlt 101 4 x10quot 10 x 101 9 x10 2 1 x 1039x 100 1009 109 537 5x102 3x1017x100 P 7x 10 8 x10 7 x 102 10 X10115 X 10 7 gtlt102 10 X1011gtlt101 5 x10 7 x1021gtlt103921gtlt101 5 x10quot 8 x 102 l x 101 5 X100 800105 815 2782x102 434 4 X 101 3 x 101 4 x 10 299 2 X 102 9 x 101 9 x10quot 2 6 x 102 12 X 101 13 x10 6 x10212 gtlt 101 1 x10 3 x10 0 x 102 13 x 101 3 x10 6 x1021gtlt1023 x101 3 x 10 7 x1023 x 101 3 x10 700 30 3 733 6755 039 X 103 7 gtlt 103 5 x101 5 x 10J 4827 4 x 103 8 x 102 2 x101 7 x 10quot 10 x 103 15 X 102 7 X101 12 x 10quot 10 x 10 15 x 102 7 X10 1 gtlt 1012 x10 11 X 103 5 x 102 8 x101 2 x10quot 2 10 x 10 1 x 103 5 X10 8 X 101 2 x10 1 x 10quot 1 X 103 5 X102 8 x1012 X100 10000 1000 500 80 2 11582 54 5 x 10 4 X10 48 4 x 101 8 x10 Since in the units position we czmnot subtract 8 from 4 we use the distributive property to modify the top expansion as follows 4 gtlt10391gtlt1014gtlt10 4 x 101 10 x 100 4 x 10 54 4 X 101 14 x10 48 4 x 101 8 x 100 0 x1016 X10quot 6 304 3 X 103 6 x 101 4 X 10 59 0 x 103 5 x 101 9 X10 Since in the units position we cannot subtract 9 from 4 we use the distributive properly to modify the top expansion as follows 5 CHAPTER 4 mumquot 3 x 102 5 x 10 1 X 10 1 x 10 3 x1015 x 10 10 X10quot1 x100 304 3 x 103 5 X 10 1 1 x in is 2 0 x 10quot 5 X 10 9 X10 3 x 103 0 x10 5 x10 3005 305 25 645 6 x107l x10 5 X 10quot 4397 41x102 3x10l 9x10quot Since in Ihe uniis position we cannol sublracl 9 iiuni 5 e use he distributive properly i0 modify the up expansion as rniinws 6 x 101 3 x1011 x 10 5 x 111 6 x 101 3 x 10 10 x10quot s x 10 015 6 x 103 3 x 10 15 x10quot 7419 4 x 102 3 x10 9 x10quot 2 x 102 0 X 10 6 x 10 200 6 200 la Fquot Since in he lens position we cnnnni submacl 3 from i we use the disinbuiive properly a mndify he mp expansion as follows 7 gtlt 102 1 x 10 1gtlt1016 x10 7 x 102 10 x 10 1 x 10 6 x 10 7 gtlt 10 11 x10 6 X 10 310 7 X 10 11 x 10 6 x10 305 a x 102 3 X 10 a x10quot 4 x1028 gtlt 1n i x10 400 80 1 481 N ending lh abacus from lhe righi The number represemcd by unis abacus i 1x51 gtlt 11 x502x 100 50 200 256 28 Reading the abacus from the righi 3x 12 X1032023 EMATICAL SYSTEMS 29 The number mpicscmcd by ihis abacus is lx 514 1 1x50 2x 100 3 x 1000 1 x 50000 1 x 10 000 50 200 3000 50 000 1 101000 I 63 259 1x5H 1 gtlt 13x 101x500 4 00 6 30 500 4000 2 4536 31 383gtlt101x53x1 32 18311X1001X503x103xl 33 2517 2 x 1000 1 x 5011 4 x 111 lt1x52gtlt11 34 70103 1 x 50000 2 x 10000 1 x1001x 501X103x 1 E 55 x 29 is wriilen around the up and right side 3992 V o 8 01min me numbers inside cach box by nding the products of all the pairs ofdigiis a up and side S X 2 12 5 x 2 10 etc T an Add diagonally 39 he sums nulsk F 39 read the emswe 1885 c a 5 n E n m Starling from the bottom r a For example 0 4 4 8 aund he Icfl side and 111 bollom as E 5 52 42 ARITHMETIC IN THE HINDUARABIC SYSTEM 81 35 32 x 71115 wrillen around lhe lop and righlside 39 723 gtlt 1198 is wriusn aruund lhe lnp and righl sidu onlnin Lhu nunibcis inside cash box by nding he nnnlnnls ofall inc pairsofdighson in up nnd side 39 1 5 391 7 x 3 014 x ri 12 Then add diagonally idc the boxes by nding me L Find each number i n i39 ii L 39 iii For exumvlv 3 U 5 A W 1 1 0W quotW Then add diagonally beginning mini hB hononl riglll hilx and cnny he one in he next diagonal linnliyicnd Formmple G 9 7 5 We he 5msidcmc 39 gt Y i 7 V mun w 7 uuu ma 1 Nowanlil6ls 24s Agni can39y the 2 m Aha nuxL diagonal above the answer around he on ids and 1h bollom as g 37 525 X 73 is written alound the up and right sidc 2 Find cacll number insnle he hum by nding me prndncl ol39 nil in pniis ul39dlglls on me mp and sin Then Add diagonally beginning l iom inc bollom righl 6 1quot Wrile IhC 2 mliside he Read 3 15 Wriling exercise Select the rods in 5 and 2 ml place will side by side Use lhc index 0 locate the raw or level for 1 nlulliplier of 8 Flirmrn box and nnny ht I Now add 1 a nin carry 10 me nex diagonal above Ran me answer 5 Ag umuml he ld l side nnil lhe bottom as 3832 1 f n I 38 912 x 1183 is written around the up and riglll Sid 9 1 2 i1 3 G 4 7 Z n 2 7 This resulung lulllce ls shown hclnw 4 4 VP Find such number inside inn hum by nding the producl ni39ull 1h uirn ofdigils on the lop and side 9 Then mm In anally he inning from he hollom rinhl i For uxalllplmgb l i 7 4 in me 4 onEide Th Wm quot 8 d 2 5 49 quot lhc box and curry lhc 2 Nuw add 2 391 2 2 10 0 he nexl diugnnul zlhovr Read It answer illmind 11 111 sidc and he lmllmn as 4404 E 539 2 82 CHAPTER 4 NUMERATION AND MATHEMATICAL SYSTEMS 42 Select the rods for 7 and 3 and place them side by side 4 394 Use the index to locate the row or level for multipliers of 3 and 2 Index To create the table below write the multiplicand on the top row Write the digits of the multiplier in reverse order in the right hand column as shown Insert the product of 2 and 73 as the rst entry Insert the product of 3 and 73 as the second entry shifted one column to the left because it is actually 30 x 73 The nal answer is found by addition 23 X 73 2336 73 146 2 219 3 2336 Select the rods for 8 3 5 and 4 and place them side by side Use the index to rst locate the row or level for multipliers of 2 and 6 4 Index 8 3 5 4 The product 6 x 8354 50124 Find the product of 2 x 8354 in a similar way but usin the index for a multiplier of 2 I To create the table below write the multiplicand on th lop row and the multiplier in the right hand column as shown Insert the product of 6 and 8354 as the rst entry Insert the product of 2 and 8354 as the second entry shifted one column to the left because it is actu 20 X 8354 The nal answer is found by addition 26 X 8354 217204 8354 50124 6 16708 2 217204 526 x 4863 Select the rods for 4 8 6 and 3 and place them sidcl side Use the index to first locate the row or level for multipliers 015 2 and 6 42 ARITHMETIC IN THE HINDUARABIC SYSTEM 83 2zvm 0 7 8 The product 6 x 4863 291781 Find the product ofl X 4863 in a similar way but using the index fora multiplier of The product 2 X 4863 9726 Find the product 015 X 4863 in a similar way but using the index fora multiplier ot 5 2anan 4 5 The product 5 X 1863 24315 To create the table below write the multiplicand on the top row and the multiplier in the right hand column as shown Insert the product of 6 and 4863 as the first entry Insert the product of2 and 4863 as the second entry shifted one column to the left because it is actually 20 X 4863 Insert the product ofS and 4863 as the third entry shifted two columns to the left because it is actually 500 X 4863 The nal answer is found by addition 526 x 4863 2 557 938 39 45 Complete missing place value with 0 283 0111 Replace digits in subtrahend 041 with the nines complement of each and add 283 958 1241 Delete the rst digit on left and add that l to the remaining part of the sum 241 1 242 46 536 425 Replace digits in subtrahend 425 with the nines complement of each and add 536 574 1110 Delete the rst digit on left and add that l to the remaining part of the sum 110 1 111 4 9 Complete missing place values with 0 50000 00199 Replace digits in subtrahend 00199 with the nines complement of each and add 50000 99800 149800 Delete the rst digit on left and add that l to the remaining part of the sum 49 800 l 1 49 801 t 06 Complete missing place value with 0 40002 04846 Replace digits in subtrahend 04846 with the nines complement of each and add 40002 95153 135155 Delete the rst digit on left and add that l to the remaining part of the sum 35155 1 35156 g D To multiply 5 and 92 using the Russian peasant method write each number at the top of a column 92 HMOquot gt 99 u Divide me lirsl column by 2 unll double lhe second column unlil l is obtained in lhe rsl enlumn Ignore lire remainders when dlvldlng Add llre numbers in lhe n at orrcspond 1o lire urlrl numbers in lile 0 50 To mulllply 41 and 53 using me Russian peasani mellmd wrne each number m lhe up of a column Divide tilt F1151 caiumn by 2 and double he second cululnl39l 1111 il l is obiained in lhe rst column Iguana he rcmzunders when dividing Add the numbers in the L 1 1 urn lirsr 53 1124 1696 2173 1 To mu a 529 using me Russian p021 ml nrelhud wrue eLIc number a lire mp or u column 62 5 25 Av 31 1058 t 15 2116 o v 7 4232 4 3 8464 e a 1 15928 39 Dividcihe rst unlumn by 2 21nd dauhle he seuuud column unlil l is obtained in Ihe rst column Ignore the 1 u 39 39 439 L 39 h second column 11ml correspond in me odd numbers in 1h rsl column 1068 2116 11232 8464 16928 32 798 84 CHAPTER 4 NUMEHATION AND MATHEMATICAL SYSTEMS v 63 145 1 3 l 290 v 9 15 580 7 1 160 e 2320 v 1 4010 7 Divide 11w l39lrsl column by 2 and double he sccond column unlil l is obtained in lhl rSI column Ignore 1 remainders when dividing Add me numbers in i a second enlumn lhal correspond in the odd numbers in i1 rsl column 145 290 580 1160 2320 4640 9135 43 EXERCISES l 1 2 3 4 5 and 5 an me rnsl six digils To represenr I A 911 39 l x 7 7 The hem six numbers would he ll 1213 1 IS and 16 To express rhe numherfrlurleen10 is use meaning 2 x 7 0 x 7 Cunlinue in lhls pallerr 21 22 23 24 25 26 2 l 2 3 4 5 6 and 7 are lhe rsl seven uigirs To represenl the number eiglll 10 is used menning 1 x 8 0 x 8 The Hex seven numbers would ll 12131415la l7 Tncxpresslhu num er sixiccn2015 used menning 2 x 8 0 x a Conrinue in this pariern 21 22 1 3 12 34 5 67 and s are he lirsreighl digirs To k nine 10 is used w ieli means 1 x 9 10 x 9 The nexl eighr numbers are 11 13 1415 I6 17 113 Tn cxpressihe nurnhereiglrm 20 is used which means 2 X 9 u x 9 Conliu in lhis puliem 21 22 Te multiply 62 and 529 using me Russiun neusam mahodv i 4 1n base sinleen we need sixleen digits Since me 11 nluul 39 b 39 52 529 suf cienl we add me digits A lhrough F 10 represe 31 was k l nu erx len 1hmugli 15 Note 111211 10 represems a 15 me P number sixreen The rst lwcmy numbers in base 39 7 12312 P sixlcen are 2 3 4 5 6 7 8 9 A B C D E F a 3 464 k 1121314 1 16928 5 13M is me number ills before and 205 is lhe num jusi aller me given number Dime quot 3 sl 39m by 2 3quot mm m cm 554x is the number jusr hel ure and 0001ix3uslnfi column unul 1 is ohlnrnerl in he rst column lgnnre ihe quotWequot number remainders when dividing Add me numbers in he 7 39 39ecund Culumn li l i correspond to me odd numbers in the 7 136E an is the numberjusl before 11nd B70mm i rm column numberjusl after me given numhe 8 1058 2116 4232 81611 16928 32 798 101 mm is me numberjusl hernre and 110110m is numllerjusl afler llie given number 52 To mulliply 63 and 39 g melhod wrirs each number or me mp 0139 n culurnn 10 Seven disiincl symbols are needed 93 PERIMETER AREA AND CIRCUMFERENCE 289 EXTENSION GEOMETRIC CONSTRUCTIONS With the raditts of the compasses greater than one half the length PQ place the point of the compasses at P and swing arcs above and below liner Then with the same radius and the point of the compasses at Q swing two more arcs above and below line 139 Locate the two points of intersections of the ar above and below and call them A and B With a straiglitedgejoin A and B AB is the perpendicular hiseetor of PQ With raditts of the compasses greater than onehalf the length PQ place the point of the compasses at P and swing arcs to the left and right of line 139 Then with the same radius and the point of the compasses at Q swing two more arcs to the left and right of line 739 Locate the two points of intersections of the arcs to the left and right and call them A and B With a straightedgejoin A and B AB is the perpendicular hisector of PQ With the radius of the compasses greater than the distance from P to 739 place the point of the compasses at P and swing an arc intersecting line r in two points Call these points A and B Swing arcs of equal radius to the left of line 139 with the point of the compasses at A and at B intersecting at point Q With a stt39aightedgejoiti P and Q PQ is the perpendicular from P to 139 With the radius of the compasses greater than the distance from P to 139 place the point of the compasses at P and swing an arc intersecting line 1 in two points Call these points A and B Swing arcs of equal radius above line 139 with the point of the compasses at A and at B intersecting at point Q With a straightedge join P and Q PQ is the perpendicular from P to line r With any radius place the point of the compasses at P and swing ares to the left and right intersecting line 1 in two points Call these points A and B With an arc of sufficient length place the point of the compasses first at A and then at B and swing arcs either both above or both below liner intersecting at point Q With a straiglttedgejoin P and Q PQ is perpendicular to line r at P With any radius place the point of the compasses at P and swing arcs above and below intersecting litte r in two points Call these points A and B With an arc of sufficient length place the point of the compasses first at A and then at B and swing arcs either both to the left or both to the right of line 139 intersecting at pointQ With a straightedgc join P and Q PQ is perpendicular to line I at P With any radius place the point of the compasses at vertex A and swing an arc intersecting both sides of the angle Call these points ot intersection P and Q With 3 an arc of sufficient length place the point of the compasses rst at P and then at Q and swing arcs in the interior of the angle intersecting each other at point B With a slrttightedgejoin A and B AB is the hisector of the angle With any radius place the point of the compasses at vertex P and swing an arc intersecting both sides of the angle Call these points of intersection A and B With an arc of sufficient length place the point of the compasses rst at A and then at B and swing arcs in the interior of the angle intersecting each other at point Qt With a straightedgeJoin P and Q PQ is the bisector of the angle Use Construction 3 to construct a perpendicular to a line at a point Then use Construction 4 to hisect one of the right angles formed This yields a 45 angle Writing exercise 93 EXERCISES to The perimeter of an equilateral triangle with side length equal to l2 inches is the same as the perimeter of a rectangle with length IO inches and width 8 inches The perimeter of the rectangle is 2 10 2 8 3639 The perimeter of the triangle is also 36 fall three sides must have the same length then one side has length 30 3 12 inches A square with an area 16 square cm has perimeterlj cm The area ofu square is found by the formula A 52 if the area is 16 the length of one side is 4 cm Perimeter is found by P 45 then P 4 4 1039 cm If the area ofa certain triangle is 24 square inches and the base measures 8 inches then the height must measure 9 inches The formula for the area of a triangle is A b L Substitute the given values into the formula and solve for It If the radius of a circle is doubled then its area is multiplied by a factor of g The formula for the area ofzt circle it39 1 7rr2 Replace 739 with 2 and compute the new area A 7r2r2 7r Ali2 Ila392 The area of an equilateral triangle with side length 6 inches is square inches Use Heron39s area formula withabca Thens66j9and 290 CHAPTERS GEOMETRY A 55 alis be C 15 99 69 69 6 v 9333 243 81 3 saxE Ifth length and the width ofa rectangle are doubled 1639 then the area is multiplied by a factor of A In the formula for the area of a rectangle replace 1 and w by El and 2111 Formula for area A lw Replace l and 111 with 21 and 211 A 21 2111 Area is multiplied by 4 A 4111 A 111 A 4 3 17 A 12 cm2 A 32 A 32 A 9 cm2 A w 18 1 2 A 22 A 501112 A In A 1 3 A 3cm2 A bh 19 A 4 2 A in2 A bh 1 20 A 4 25 4 5 A 1 2 2 A 10m 2 A bh A 3 15 A 45 cm2 A bh 1 22 A 52 36 A 936 mm2 1 14 51111 1 9 A 2 2 38 ABE 2 1 1 A2418mm2 1 bh 1 A 2 5 3 1 5 3 A 5 1391 15 A 2 A75ml A 13 1 A3 235 A 8 A8cm2 AhbB A345 3 9l 5 2 A1350m2 A7rr2 A 31z112 A314cm l A7rr2 A 314152 A 314225 A 7065 cm2 The diameter is 36 so the radius is 18 m A m 2 A 314182 A 314324 A 101736 1112 The diameter is 12 so the radius is 6 m A 7r39r2 A 31462 A 31436 A 11304 n12 23 24 m 9quot 26 N 9 Let s length of a side of the window Use the formula P 4s Replace P with 75 12 and solve for s P45 75 124s 75 75 1243 7s 12 3s 12 3s 73 45 The length of a side of the window is 4 In Use the formula P 21 211 and l 90 11 Replace P with I76 and replace 1 with 20 a Solve for n P 91 2w 176 220 39u1 2w 176 410 l 2w 2a 176 40 4111 130 41w 136 a 4m T Ti 3411 The width is 34 inches the length is 20 34 54 inches The formula for perimeter ol39a triangle is P a l l c Translating the problem let a be the shortest side Then b 100 u and c 200 rt Replace a l and c in the l39orntula with these expressions replace P with 1200 and solve for a P a b c 1200 a 100 a 200 a 1200 3n 300 900 3a 300 a Side a is 30011 100 300 400 ft side c 200 300 500 It Let a the length of the shorter side 0139 the triangle Then the other two sides each have a length Ola 18 Substitute these values into the formula for the perimeter ol a triangle P 511 inches qu l bl c 54uo18a18 5423a36 183o 6rt Side a is 6 in the other two equal sides each have a length ot G 18 221 inches One formula for circumference is C 21m Translating the second sentence at the problem C 6139 l 1288 Equate these two expressions for C and solve the equation lbr I 93 PERIMETER AREA AND CIRCUMFERENCE 291 6r 1288 277739 67 l 1288 r31Llr 67 1288 6281 67 6r 1288 628739 67 1288 028r 1288 g 0287 028 028 116 739 The radius is 46 ft 28 Use the formula C 27rr Translating the problem indicates C 3r 82 Set these expressions equal to each other and solve for r 3739 82 277139 Br 82 2314v 3739 82 6287 3r 3739 82 6287quot 3r 82 3281 82 3281 328 328 25 39739 The radius is 25 cm 29 The formula for the area of a trapezoid is A ltl B Substitute the nttmerical values given in the problem and compute to nd area I A Ell B 1 A 51659711580 l 17100 1 A 3165972868 A 2 417600196 A 23 800098 Rounded to the nearest hundredth the area is 23800 0 sq ft 30 The formula for the area ot39a trapezoid is A hb l B Substitute the numerical values given in the problem and compute to nd area 1 A tb B 1 A 165972684 l 8205 1 A 1659710889 0724 A 18 73 A 90362365 Rounded to the nearest hundredth the area is 903624 sq ft 31 You would need to use perimeter Fencing is sold by linear measure 292 CHAPTER9 GEOMETRY N 32 You would need to use area 4 Table zr Exercises 33 40 7 43 33 d2r2612in C27rr27r6127rin A7rr27rA62367rin2 34 d2r2918in C27rr27r9187rin A7rrz7r92817rin2 44 35 39r 105t t C7rd1r10101rft A7rr27r52257rft2 36 r r4020tt C7rd7r40 401rl39l A7rr27r202400739rft2 C 121r 37 d T120m 4539 39r126cm A7rt 2337rcm2 38 d 18 1r18cm 17r 1r r21890n1 46 A1r92817rcm2 39 r2 5 100 Then 39I 100 10 in 71 d2r21020in C27rr27r 10207rin 40 r2 25a Thenr25616in d27392 1632in C2rr139 2 7r16327rin 4 Use the formula P 4s replacing s with x and P with PIlL39 5843 ea 4 4 145te Use the formula P a b c replacing a b and c with the expressions in e and replacing P with 42 42za2z7 zt23c9 3331 333m 3 quot 113 Use the formula P 2 21 211 replacing l and to with the expressions in x and replacing P with 38 38 22x 3 2x 1 384z 62m2 3863 4 426 42Ga F 731 Use the formula P 21 211 replacing l and 111 with the expressions in 139 and P with 278 278 251 1 2139 278 102322x 278 12x 2 276129 mm 12 12 23z Use the formula A 32 replacing s with a and A with 260 2601 x2 2601 E 51 1 Use the formula A lw replacing l and to with the expressions in a and replacing A with 28 28wm3 m w 023z 28 Now factor the Lrinomial and set each factor equal to zero 077z 4 m70 or 17 4 CB 7 x 0 4 The solution 7 is not meaningful because length or width of a rectangle must be positive numbers The answer then is 4 47 Use the formula A 711 replacing b and IL with the expressions in w and replacing A with 15 1 15 a cc 1 2 2 1 239 r 1 15 1 2LL 30 17139 1 30m2z 0rlr2m 30 New factor the trinomial and set each factor equal to zero 0 z6tv 5 I60 0r m 5 6 as 0 G The solution 6 is not meaningful because the base ot a triangle must be positive a number The answer then is 1 DC with the expressions in 139 and replacing A with 30 30 3 wzl LL 2 30 213 1 1 1 2 LL l39 60 3w e4 3lw 4 3 3 20 z39 4 202m4 1621 89 49 Use the formula C 2m replacing C with 3768 and 739 with the expression in 17 3768 2314w 1 3768 628 1 3768 6281 1 m 628 G 0 1 5 139 Ln 53 C with 5495 5495 3143m 5 5495 31zt3m 5 314 314 175 3139 5 225 33c 75 139 Use the formula A hb B replacing h l and B Use the formula C 7rd replacing d with 7 5 and 51 u 4 Ln LII 93 PERIMETER AREA AND CIRCUMFEHENCE 293 Use the formula A 7172 replacing A with 80864 and r with 180864 314372 180864 314112 3 314 H a 763 241 in Use the formula A m2 replacing A with 2826 The diameter is 49 so this expression must be divided in half for 7quot 2726 314m2 a A4520cm b A 810 2 80cm2 e A 1215 180 cm2 d A 16 20 320 cm2 e The rectangle in part b had sides twice as long as the sides of the rectangle in part a Divide the larger area by the smaller 80 20 4 By doubling the sides the area increased A times I39 To get the rectangle in part c each side ol the rectangle of part a was multiplied by 3 This made the larger area 9 times the smaller area 180 20 9 g To get the rectangle of part d each side of the rectangle of part a was multiplied by 4 This made the area increase to lg times what it was originally 320 20 13 h In general if the length of each side 01 a rectangle is multiplied by in the area is multiplied by M Because each measurement is multiplied by 2 the area will increase by 22 4 Then 4 e 60 2210 5 Because each measurement is multiplied by 2 the area will increase by 22 4 Then 4200 800 6 Because each measurement is multiplied by 3 the area will increase by 3391 z 9 Then 9 v 80 720 294 U l Ur 00 VI 0 6 0 6 6 IQ CHAPTER 9 GEOMETRY If the radius of a circle is multiplied by n then the area of the circle is multiplied by It If the height of a triangle is multiplied by n and the base length remains the same then the area of the triangle is multiplied by u Only one dimension is multiplied by n so the area is multiplied by 72 only once Find the area of the parallelogram and the area of the triangle Then add the two area values Parallelogram A 6 10 60 Triangle A 00M 20 Total area 60 20 80 Find the area of the triangle rectangle and parallelogram Then add the three area values 1 A 109 45 Rectangle A 4 10 40 Parallelogram A 10 3 30 Total area 45 40 30 115 There are 2 semicircles or equivalently 1 full circle with radius of3 Find the area of this circle and of the rectangle Triangle Rectangle A 86 48 Circle A 314 32 2826 Total area 48 2826 7626 The four semicircles create two full circles each with a radius of 4 Find the sum of the areas of the two circles and the square Square A8864 Circle A 314 42 5024 Total area 64 25024 16448 Find the area of the trapezoid that surrounds the triangle Subtract the area of the triangle Trapezoid A 1218 11 174 A 127 42 Shaded area 174 42 132 112 Triangle Find the area 01 the trapezoid that surrounds the triangle Subtract the area of the triangle The length of the lower base of the trapezoid is 19 38 57 1 Trapezoid A 524 28 57 1020 1 Triangle A 51916 152 Shaded area 1020 152 868 cm392 6 6 l 6 00 as gt0 Fquot Find the area of the rectangle that surrounds the triangles Subtract the areas of the triangles The length of the rectangle is 48 48 96 Rectangle A 74 96 7104 1 One triangle A 54836 864 Shaded area 7104 2864 5376 cm2 Find the area of the rectangle that surrounds the semicircle Subtract the area of the semicircle The diameter is 21 therefore the radius is 21 2 105 A 21 23 483 1 A 53141052 1730925 Shaded area 483 1730925 3099075 Rounded to the nearest hundredth the shaded area is 30991 it Find the area of the square that surrounds the circle Subtract the area of the circle The diameter is 26 therefore the radius is 26 2 3 A 262 676 Circle A 314132 53066 Shaded area 676 53066 14534 in Rectangle Semicircle Square Find the area of the large circle that surrounds the two smaller circles Subtract the area of the smaller circles The radius of each of the smaller circles is 4 the radius of the large circle is 8 A 314s2 20096 Small circle A 3144 3 5024 Shaded area 20096 25024 10048 cm2 Large circle The best buy is the pizza with the lowest cost per square inch or unit price if you have enough money and you can eat all of it 10quot pizza A 31457 785 in2 599 U 391 39 m 076 nl prlce 785 12quot pizza A 3146 2 11304 in2 7 9 9 N Unit prlce 11304 3071 l4quot pizza A 314 72 or 15386 in2 899 Unlt prlce 058 The best buy is the 14 pizza 93 PERIMETER AREA AND CIRCUMFEHENCE 295 70 The bcsi buy i m pizza wiih ihc Iuwesi cnsl pci aqu l39e By subsiiiuiinn ihc pariniiziur of AAEB is inch or unii price 20 34 54 in ii pizza A 314m 785 in2 75 Thu key is o Conslruui TV and U W 0 cream more U39m Price 39 z 03 triangles By inspcuiioir all the mail li i ng es are Equal WW A avHJW lmmni P Rs39hussiriungics TUVWliiis4triiingles Therefore TUVi 39 has halfihc men of gas which is i Unii price 2 was gt25 1 Oiiiciwisc nd the area by I39irsl mlving mi mi 1 3 0 1 lenglh of one side using ihc Pylhugnriaun theorem 14quot plzzil 5117 or 15386 iii mm 76 Ill iccuiligle4BCDl 2hr Since he pciimclcr 0139 Unii price m 56 m 3071 ABC39D i5 96 inches The best buy is he 14quotpizzii P 1 2 06 72m 2m 7 The hen buy is he pizza wiih ihc iowesi cosi pcr quaia inch or unii price 0quot pizza A 3i1r153 755 in2 w 16 1iiandl 2w 32 A midpiiini divides a Segmcnl imo VO parts prcqhui iihir prim x 3127 lengih SinczP Q R and s are niidpninmufihe sides 7 AP PB D and RC are each equiviilunl in ii which I2quot pizza 314 113011 In is I6 Also AS SD BQ and QC are each erfuivaiciii uhir price m iuu 1 5 Wh39ch 5 i w mm The aria of rizcmnglc ABQD m 16 512 inch ulice ihni ilic iriangles iire right lriuiigles so Unii prim he lags can hi used as riieir bases and heighis 39 1 Thu best buy is Ill 14quot pizza Awnm AAPS 510mg 6 72A l39heliesih uy isi1c pizza wiih he Inwesi cusi per square Am 01 APBS 16 8 4 inch or umi pncc 10 pizza A iii 5 723r in39 Aw 0quot ASDR 06 64 Unil price 171839 m 3153 AmofAQCR 06105 64 I2quot pizm A f ii14W 1130 1i3912 Thi arc i ofquzidriluierul PQ R iq uqual lo ihp area or ABC D minus ll sum ol ihe an nl39llic loui39 triangle Uiii ric L99 P 1137 5124mm 1 z IAquot pm A 311720r15386in1 uiiii price his a s 097 77 Tlickcy is loconslrllct u pcrpcndicularlinc from E iu r sidepa Lui ihu puinl oi ihiciscczioh ni39side DC he 1quot w my 5 quot 3 394 W labeled puihi F By inspecliou mere aic Wu seisor Q 3 The key is in consuiiui OB and m realize that the diagonal pi ll rcclungli are equal in lcngm sh hy inspeulimi 0 AC 13 in GB is n ruiliiix Therefore the dininelci 9 13 26 in In he gure I ED and F3 2 DC Thu pim39mz ler of AAEB is equal lriangkxx ADAE and AEFD 39and AEFC Then the 2mm of the shaded mgion Li half the m pr he sqiarc Therefore 36171296 7 2 Aica 648 in2 P A r 78 Yes Ihe perimeiercaiibcmupi Drawiirm n cniiiplaic A EB quotw AL EF PB 4 AB39 ihe ling muniiglc These lines N cquill in measure L0 Because EF ED ihc shnrl veriical and horizontal lines Ihuicrehii ihi uiiiuui pnrliull of the large reullingle AE iEFAEEDAD20in Also bbciluse FB BC FB 139 AB BC AB 31 in 296 CHAPTER 9 GEOMETHY l w 0 La Thcrcforc the perimeter i 4 STATEMENTS AS P2132740in 1 301 53quot L I 2 2 Given he key is lo construct a square using lwn radii from 0 3 441350 4 3 Wlemms 0 equal and bounding lhe shaded reginn The uren el uie small 3398 film W each other unre is rquot and lhe urea uflhe qunrler eirele is uni4 4 DB PB 4 Rc cch properly Therefore lhe urea oilhe shaded region is 5 AUBC 2 ABBA 5 SAS congruence property 2 S STATEMENTS REASON 1 139 1 l l Given 2 2 Given First nd me Icngm 0MB ln me right mangle ABE 3 A0 A0 3 Re exive PTOPWY u a ns 5 We and 4 AABC39 a AADC 4 ASA congruence properly height we have 6 1 BO CE 2 0393 is perpen 2 Given dicular in El 230 7 3 OHE is perpen 3 Given 6 diculur U 13739quot 10 A3 4 ABO AFEO 4 Born are right nngles b Now nd lhe areu of lhe uapczcid wilh Il 6 b 10 dclmllw 0139 PSYPEHGWUMHIY and 3 l4 5 mos MOE Variicnlznglcsareequ 1 e AAOB 6 ASAC ngmence property A 2 7 lr 13 measures 46quot than 24 measures Miami LC 4 me 57 In an isosceies lrinngle lhe angles A opstiIe eequalsirlesareulnuequul in measure Thus A 1A C The sum of lhe angles at lhe lriangle is 180 A Then 180 e 46 134 and l34 67 8 Because 10 2 AA 52 heenuse lhe angles uppusile lhe equal sides ere also equal in lucasurc Then 94 EXERC39SES because me sum of ihe llnee angles is ISOquot STATEMENTS REASONS 18 52 52 75 l 4 BD l Given 2 AD BC I GM 9 The length ufsidu AB 12 hecuuseil as lllesame 3 A3 AB 3 Re uxiw properly measurc as BC Then 30 7 2 12 6 ill 4 AABD E ABAC 4 SSS Congruence Prowl ln Suhueer 10 mm 40 in nd lhe uIaI length ul lhe lv0 TATEM reinnlningsieles Then heeuuse lhey urccquzll in lenglh 5 AC Eggs fr gz x divide by 2 40 e 10 so und so 2 l i 2 ACD 1301 2 Given ll Writing exercise 3 CD D 3 Re exive properly a y 4 AADC 2 ABDU 4 SAS Cungrueuce Properly 12 wquot Wm 39 in mm llr STATEMENTS kmsalvs M d 0 an 1 DB IS I pan 1 Given B and 1Q dlcular L0 AC KC and m 2 AB BC 2 Gwen 3 also 433 3 delinilion orperpenaieulurily I 4 DB DB 4 Rerlerlve properly if 1 ff 5 AABD g AGED 5 SAS congmcncc properly AC and PH CB and it 39Jt L11 57 114 USING PASCAL39S TRIANGLE AND THE BINOMIAL THEOREM 387 c There is only 1 way that the groom can stand in line with the bride next to him There are 6 ways the remaining people can stand in line Thus by the fundamental counting principle the total number of arrangements is l6 720 Five percent of the 60 students is 0560 or 3 students Five percent of the 40 students is 0540 or 2 students The number of ways that he can assign Agrades to the rst class is C60 3 The number of ways that he can assign Agrades to the second class is C40 2 Apply the fundamental counting principle to nd the total number of ways the professor may assign A grades to his students C60 3 C402 26 691 500 21 There are only two sets of distinct digits that add to 12 They are 1 2 3 6 and 1 2 4 5 Try to nd others There are 4 distinct permutations which lead to a different counting number for each set of digits Thus using the fundamental counting principle the total number of counting numbers whose sum of digits is 12 is 2 4 48 b There are only three sets of distinct digits that add to 13 They are 1 2 3 7 1246 and 1 34 5 Try to nd others There are 4 distinct permutations which lead to a different counting number for each set of digits Therefore using the fundamental counting principle the total number of counting numbers whose sum ofdigits is 13 is 3472 This question is the same as asking how many 5 element subsets can be formed from a set with 30 elements The number of different samples is thus C30 5 142506 21 0029 9112 399 12 220 12 7 312 3 12 220 Thus C12 9 C12 3 0023 58 5 U3 0 By the factorial formula for combinations n 0quotquot l and Cn n r 7 n H n n Thus CnJ Cnn r q n 21 Since P7zr n P n 0 n 7 0 n H m 1 b Writing exercise n a Since Cnr TL Cn O 0ln 0 n 2 l n 1 b Writing exercise 114 EXERCISES Read the following combination values directly from Pascal39s triangle F or exercises 18 refer to Table 5 in the text 1 To find from the value of C4 3 from Pascal s triangle read entry number 3 in row 4 remember that the top row is row 0 and that in row 4 the I is entry 0 C39 4 3 4 To find the value of 05 2 from Pascal39s triangle read entry number 2 in row 5 C5 2 10 To nd the value of C 6 4 from Pascal39s triangle read entry number 4 in row 6 C64 15 385 CHAPTER 11 COUNTING METHODS gt9 Fquot To nd the value of C7 5 from Pascal39s triangle read entry number 5 in row 7 C75 21 To nd the value of C 8 2 from Pascal s triangle read entry number 2 in row 8 C8 2 28 To nd the value of C 9 4 from Pascal39s triangle read entry number 4 in row 9 C9 4 126 To nd the value of CQ 7 from Pascal s triangle read entry number 7 in row 9 C9 7 36 To nd the value of CIO 6 from Pascal39s triangle read entry number 6 in row 10 010 6 210 Selecting the committee is a two part task There are C7 1 ways of choosing the one Democrat and C3 3 way of choosing the remaining 3 Republicans The combination values can be read from Pascal s triangle By the fundamental counting principle the total number of ways is 071 C33 71 7 A committee with exactly two Democrats will consist of two Democrats and two Republicans Selecting the committee is a twopart task The two Democrats can be selected from seven Democrats in C7 2 ways while the two Republicans can be selected from three Republicans in C3 2 ways The combination values can be read from Pascal39s u39iangle Using the fundamental counting principle the number of ways to select the committee is 072 C32 21 3 63 A committee with exactly three Democrats will consist of three Democrats and one Republican Selecting the committee is a two part task There are C 7 3 ways of choosing three Democrats and C3 1 ways to choose the one remaining Republican Hence there are C7 3 C3 1 353 105 ways in total The number of ways to select four Democrats and no Republicans is C74C3035A1 3 9 0 N O E P U 9 gt N H N N There are C39 8 3 56 ways to choose three dif positions for heads Using Pascal39s triangle nd to entry 3 Remember to count rst row and rst entr 0 The remaining positions will automatically be ta The number of ways of obtaining exactly four head thus exactly four tails can be found from entry 4 i 8 of Pascal39s triangle that is C84 70 There are C 8 5 56 ways to choose exactly different positions for heads Using Pascal39s triang would be found in row 8 entry 5 The number of ways of obtaining exactly six beads C 86 28 Using Pascal39s triangle this would be found in row 6n The number of selections for four rooms is given b C 9 4 126 Using Pascal s triangle this would be found in row try 4 Since only one of the classrooms contains the econ class there are eight classrooms that do not The n of ways to select four classrooms that do not conta class is C84 70 The number of selections that succeed in locating t class is given by total number of selections Exerc 17 minus the number of ways which will fail to lo the classroom Exercise 18 or C94 CS 4 126 70 56 ways From Exercise 17 there is a total of CQ 4 126 different selections From Exercise 18 CS 4 selections will fail to locate the class From Exerc the number of selections which succeed in locating class is 126 7 70 56 Therefore the fraction of possible selections that will lead to success is 56 z 444 126 The number of O element subsets for a set of ve elements is entry 0 the rst entry in row 5 of Pasl triangle This number is 1 The number of 1element subsets for a set of ve elements is entry 1 the second entry in row 5 of Pascal39s triangle This number is 5 tfferent row 8 ntry as tails ads and 4 in row ve angle this eads is row 8 ten by n row 9 economics The number contain the min g the Exercise iii to locate 5 ways 126 84 70 n ExerciSe 19 locating the ction of the ssquot is of ve 5 of Pascal39s t of ve row 5 of 114 USING PASCAL39S TRIANGLE AND THE BINOMIAL THEOREM J DJ is 5 N UI N gt1 NM pee m 32 The number of 2element subsets for a set of ve elements is entry 2 the third entry in row 5 ot l ascal39s triangle This numberis 10 The number of 3element subsets for a set of ve elements is entry 3 the fourth entry in row 5 This number is 10 The number of 4 element subsets for a set of five elements is entry 4 the fth entry in row 5 This number is 5 The number of S element subsets of a set of ve elements is entry 5 the last entry in row 3 This number 151 The total number of subsets is given by C50 C5 1 Cs 2 C5 3 05 4 C55 3s 1510105132 This is the sum of elements in the fth row of Pascal39s triangle Writing exercise a All are multiples of the row number b The same pattern holds c Row 1 I 1 11 55 165 330 462 462 330 165 55 11 1 All are multiples of ll Thus the same pattern holds Name the next ve numbers ofthe diagonal sequence indicated in the gure shown in the text What special name applies to the numbers of this sequence 15 21 28 3645 numbers These are the triangular Following the indicated sums 1 1 2 3 5 the sequence continues 8 13 21 34 A number in this sequence comes from the sum of die two preceding terms This is the Fibonacci sequence In rows 2 and 4 every entry except for the beginning and ending 139s is O This is because the corresponding entries in the original triangle were all even 33 Row 8 would be the next row to begin and end with l with all other entries 0 each internal entry in row 8 of Pascal39s triangle is even 34 From Exercises 32 and 33 we see a pattern In Pascal39s triangle every entry except for the beginning and endin 139s will be even whenever the row number is a power 0 2 Since 256 28 is a power of 2 this pattern Will app39 to row number 256 In row number 256 there are 256 4 1 257 entries The rst and last will be l39s while all the others will bt even Thus the number of even numbers will be 257 2 255 The sum of the squares of the entries across the top roV equals the entry at the bottom vertex Choose for example the second triangle from the bottom 1 9 9 1 20 the vertex value The rows ot Tartaglia39s rectangle correspond to the diagonals of Pascal39s triangle Ln 9 37 The sum N Any entry in the array equals the sum the two entries immediately above it and immediatel39 its left 38 The sum N Any entry in the array equals the sun the column of entries from its immediate left upwart the top of the array 39 The sum N any entry in the array equals the sun the row of entries from the cell immediately above i the left boundary of the array 40 The sum N 1 Any entry in the array equals 1 than the sum of all entries whose cells comprise the largest rectangle entirely to the left and above that r 4i Reading the coef cients from row 6 of Pascal39s trir and applying the binomial theorem we obtain w l y6 16 6151 15274112 20333313 15222311 63145 ye 42 Reading the coef cients from row 8 of Pascal s tri and applying the binomial theorem we obtain 1 y8 m8 8371 2826312 5617513 7 565315 2812116 81117 ya 43 Reading the coef cients om row 3 of Pascal39s n and applying the binomial theorem we obtain e2fz1mhm3q 2362212z 8 385 390 CHAPTER 11 COUNTING METHODS 44 Reading the coef cients from row 5 of Pascal39s triangle 1 VI 4 Ch 4 I p 00 and applying the binomial theorem we obtain in 35 1115 5w 3 10w3 3 10111283 5w34 35 39 w5 15114 90w3 270102 40511 243 Reading the coef cients from row 4 of Pascal39s triangle and applying the binomial theorem we obtain 20 514 2a 42a35b 62az5b2 42a 51 5194 16a4 16003b 600a2b2 ioooaba 625124 Reading the coef cients from row 4 of Pascal39s triangle and applying the binomial theorem we obtain 3d 5f4 3d 43d35 f 63d25f2 43d5f3 W aid4 540113 f 135Dd2f2 150iatf3 625 f4 Reading the coef cients from row 6 of Pascal39s triangle and applying the binomial theorem we obmin b m b Hm b6 6b5h 151 h2 201943 15bz hd 6bh5 106 b6 6b5h 151M2 203 15b2h4 6bh5 h Reading the coef cients from row 5 of Pascal39s triangle and applying the binomial theorem we obtain 27 4m5 2n 4m15 2n5 52n 4m 102n3 4m2 102n2 4m3 52n 4m4 4m5 32715 320n4m 1280n3m2 e 2560n2m3 2560nm 1024m5i For the expansion 1 yquot there will be H 1 terms The exponent on y is r 1 Since the sum of the exponents on c and y must be n the exponent on 1 is n e r 1 n r 1 Thus the variable part of the rth term is nrlyr 1 Each term of the expansion is the product of a coef cient and its corresponding variable part Thus the rth term the general term is 7 n r1 7 1 n T1TA1 y 51 52 53 Heren 14and7395 Therm 1 4and n 7 1 10 Substituting these values into the result of Exercise 50 we find that the 5th term of w my1 is 141 10 4 10 4 Tome y 1001z y Here n 18 and 7quot 16 Then 1quot 1 15 and n e r 1 3 Substituting these values into the result of Exercise 50 we find that the 16th term of a b18 is I i 1839 3115 816L135 315 Prove Cn739 Cn 12 1 Cn 1r Cn 1r 1C n 1r 721 n1l r 1n 1 r 1 rn1 7quot rt 1 r 1nr rlnr139 n n41 39r 77 n l n7 i nr lln rlnrln r 1iii r nlr n n r nrln T nrln r i n139nn r n rln r n39rn4r nTln r 1M yirln r n T n Tl Cnr 115 EXERCISES l 2 3 Writing exercise Writing eXercise The total number of subsets is 24 for a set with 4 elements The only subset which is not a proper subset is the given set itself Thus by the complements principle the number of proper subsets is 24 16 1215 The total number of subsets is 25 The only subset which is not a proper subset is the given set Thus by the complements principle the number of proper subsets is 25 1232 1231 64 CHAPTER 3 H LOGIC Uitll l P E r s 2 q E x 395 5 This is the l argument torm mod us poncnsquot 5 Let p represent quotShe liuys another pair nfshnesquot anti q represent her closet will overflowquot The nrgumenl is then represented symbolieally by p m q ll p Sincet s the fami fallacy of the conversequot it is Thus inanne moose circus Lou griz39 b r train 3min and cansidemd 9 gm Ralph seal nick hand Winnie zebra sp p 6 Let p represent He doesn t have 0 get up nt510 amquot nnd 4 represent ht is ecstaticquot The argumcnl is then 35 EXERCISES represented symtuilleally by F 1 1 Let p represent you use blnoculars q represent quotyou q gct a glimpse of the comet quot and 139 represent quotyou will tie If amazedquot The argumcnl is then represented symhotienliy P b Since this is the term quotfallacy of the conversequot ir is I e invalid and cansitteretl n l nilecy w 7 Let p represent Patrick Roy playsquot and q represent lhe p a r opponent gets shut outquot The argument is V V represented symbolically by Th s the vnlld argumenlfoml quotreasoning by transitivity 2 Let 1 represent Billy inel comes 0 tnwn q represent 1 will go to the concertquot and i represent I ll cull in tin work The argumenl 39 39 by Hluuu toilettequot 8 Let prepresent quotPedro Marlinez pitchesquot and q represent P T 1 quotlhl Red Sox wi quot The argument is rhen represented 91 symbolically by p 39r p a a This is the valid argument form reasoning by q lransilivityquot T 3 rem quotpl39 iff wi l l39gg gmgx iqlhm This is the EM ingumenl intm mndus tollens represented symbolically by 9 Let prepresent we evolved a race of Isaac Newtonsquot I q and 1 represent lhul would not be regressquot The p argument is then represented symbolically by39 q p r a P This is the valid zlrgunmnt rnrm modus penens 4 Let p represent Amy Mch works hrtrd enoughquot Lind 11 represent she will get a prumollonquot The nrgumenl is then represented symbolically by Nola hm since we let 1 represent that would no he progressquot then q represenls that is progress Since lhl it lhr form fallile of ie inversequot i is 36 ANN WITH TRUTHTAELES 65 o m rcillquot L nl Ihal csccnllirlhcrdmn ollml39s undq H Furlquotlllhcundiliunulhuncmcnl icpissunlquotilisiassnnscliinvssiuoil ihssnouldersnl U A M 1 M ginnisquot l39hc nrgunicnlisiiisnrsnrcssnisilsymhnlicnlly 1 q 11 l 39 i39roin he a plsis n mnh 11quot F q llA vq T T F ii N T F T 1 F T F F F F Since his is he l39oi in quotl allncy d l39 ni lln inverse il is invlilidiiml considers a 39 v Lu 2 mpiescni quotAlison lieinningsquot and n rcpwbum quotK Tnylnr pumps iron The argument is than rcprcsenleu syniinilicniiy by pv q oiiiv 1 q 9 SINCE liiis is he loi39m llisjnncnve syllogismquot ii is a v algnnis l B Let rcpresenl She uses ccommeruequot nnu i ispisseni gnu pays by credil cardquot The argument is inn reprcxenlcd symbolically by I V a Or a V 1 L ii Since iins is me lonn disjum llv syllugismquot ii is n valid argument To show valldiiyfor ilie argillllmts in llwfolhiwing mercixex we mini nllmv 1111 he miin39ililciion ofilie premisln EmpliL39x ln mm llli39iall Thle is he 39Uml iulmlsluts112 iPl A P s Pl 7 C must by n qululugy Far smcim l and l4 We will Ile Iwm undurd langfonnml I0 deVL IDp ilie wrrexpmlllmg mil 1 ll les r 39 v a v i s will ins ills nliennns Sllol lfamml to new I m it1 tables 1339 Flinn Lhc cundiliunul Simemcnl 12 V q A P rl l min lhc alrguman Cvmpltlc n ll39ulh lulilc Sines llu Clnllilionnl i39onnuu by lhu cuiiiunclion of picniiss i iplying Illc conclusion is n0 n laulology the argumcnl is invalid Since lhc cundiliunal formed by me conjunction m premises implying me conclusion is n luumlngy mt nignmeni i inl 5 Form lhe condilional Sli bmcnl ll w N1 P min ills argulhcnl Conlplele n ll39ulh mble Since Lhc cilmlilinnal formed by the conjunction of piuiiiiscs implying in concluninn is n muiolugY he urgumcnl is valid Form he cundiliunal slalemcnl KINquot3911 A p r1 5 from it ul gllKHEnL Cumplcm n Lruih table ll WWII All A q TTF TT F 3913 39TI ii lrJ TI l39n I TT FF FF T T T 1 wamam 3 2 Since he condiualmL armed by he Cinjunclinn if 39emiscs implying ht conclusion is nui n mulology mt argumcnl is im Form llle Clndiiional slalcmcnl p v a A ii A l0 M q i Ma A39 Hlvll N39 39 39 39 66 CHAPTER 3 INTRODUCTION TO LOGIC 1r Sincc the euntlition tuned by me cunjunclitm nr at ptemtses implying the conclusion is not atutltoiogY lhc argu nl is inyoliol 8 Farm the conditional statement MD gt r A p q min the argument Complete a truth tablet Since the conditional formed by the conjunction of premises implying the conclusmtt is not a tautulngy tht quot quot uterit39 wk truth table tilting rows rather than down columns you could sing after cumpleting tne rsl row knowing that with a false conditional the sl lemmll will not he a iaulology 19 Form the conditional statement til 21 A 11 A p from 118 argument Complete a truth table Since the conditional rnnned by the conjunction of premises implying the conclusion is n tnutnlugy ihc argument is vn 39d 20 Form the conditional stutetttenl P 39 ltt A 7111 a 41 rrom the argument Complete n trmh table Since lne Conditional united by the nnjunetinn of premises implying tlie conclusion is not n lmlmlog llie argument is inv id N N lquot Form the conditional statement WNWWith 4MP gt q from the argument Complete tt truth table P a erll A Mm q q T T T 39nll l l nvl Since the conditional formed by the conjunction of premises implying the conclusion is not a iaulology the argument is 39 valid Form Lhe conditional statement P 391IIP1PquotPV4 Since the conditional tortured by the conjunction of premises implying tit conclusion is a titutotttgy the argument is M Furnt the canditional statement Hm quotJ a quH T H 72 a q 39T from the arguman in r v a s TT FTT T T T TT FFT T F F l F FTT T T T TF FFT T T T FT TTT T T T FT FFT T T F FF 39I TF F T T FF FFT F T T at 3 2 s 4 The F in the nal cttlumtt 5 shows us that the statement is not a tnutoiogy and hence the argument is invg id 36 ANALYZING ARGUMENTS WITH TFlUTH TABLES 67 14 Form the conditional statement 1 q r 7 A q I A qr r T T T T F T F F F F T T T T lllAptVrlllAqPl IVpl T T F TF TFF F F TF T F from the argument T F T T T F T F T T T T T T T F F T T F T F F T F F T F p q r 39I39p 39I39th qp 39v I39Vp F T T F F T T T F F T T T T TTT TTTTTTTT TTT T TTT FT F FF TTT FFTF TF TTF FFT TFTT T TTT T FTT F F T FF FTT F TTT T T T F T TTT TTTT F FFT T TTT F F F FF FTT F T FF T F T F F FFT T FTT F FFT T FTT 21 12 l 4 gt32 5 t FTT TFF TTTF F TFF T TTF F T F FFF T FFF F TFF T FFF Since the conditional hunted by the conjunction of F F T TFF T TTF F FFF T TTF premises implying theconclusion isatautology the F F F FFF T FFF F FFF T FFF argumentisLlid 2 l 3 12 l 4 2 3 2 5 3 l 3 28 Let p represent that tree is infested with pine bark beetles q represent quotit will die and 7 represent people plant trees on Arbor Dayquot The argument is then represented symbolically by Since the conditional formed by the conjunction 0139 premises implying the conclusion is a tautology the argument is 25 Writing exercise 17 y 1 26 Every time something squeaks I use WD 40 139 q Every time I use WD 40 must go to the hardware store 7 73 Every time something squeaks I go to the hardware V 510m Construct the truth table lor 1 q r A 1ll H 7 I P 1 39 itJ I rA r1l 397 39 P T T T T T T F TF F T T F F T T F T T T F FF F T F T F T F T T F F F TT T T T F F T F I39 T F F F FF T T F T F F T T F T T F TF F T T T T F T F F T T F FF F T F T T F F T F T F T TT T T T T T 27 Let 1 represent Jel l loves to play goll39quotq represent F F F F T F F FF T T F T Joan likes to sew and 39r represent Brad sings in the l 2 l 3 l 2 I 4 2 3 2 choitzquot The argument is then represented symbolically by Since the conditional formed by the conjunction of premises implying the conclusion is a tautology the p argument is valid I 39 1 WI r 29 Let p represent the Bobble head doll craze continues q represent Beanie Babies will remain popular and r Tquot represent Barbie dolls continue to be favoritesquot The Conslrucl he mm able for argument is then represented symbolically by c lim 0 II 11 mot 12 7 F1 139 V q 7 Np 68 CHAPTER 3 INTRODUCTION TO LOGIC Construct lhc truth tuhle lin 13 a q A TV q A rll P with a lttlse Ctlndilimtal the stuletncnl will not b a tnutriltrgy 31 Let p represent quotI39ve gal you under my skinquot q represen time u you I I wt V r n r me columns to crcatc the trutlt value far catch connective p a I 1141 A TVA A i39 v in 39I quotT T T T T T T F F T F qVT T T F T T T T T F F p rr T F T F F T F F T F 1 F F F F F F T 1 F Construct the truth table for FTT T TTFFTT AAV FTP TITTTTT Pqill1 FFTTTTFFTT pqr par F F F T F F F T T T T T T T T 1 2 l 3 2 4 3 T T F F F T F T T T Sincelhecnndilinnttlfonnedbytheconjunclion of T F F T F premises nnpiyingtnecnncinnnn ix nntnrnutningy llie F T T T T axgumentisinvnlid Note Uytitiarecumplelinglhe F T F T T truthtnhlenlong rowsratticrthandowncolumns you F F T T T could stopnirercompletingmesecondmwknowing that F F F T T wiih a false conditional the statement will not be u 5 4 tautnlngy 30 Let 11 represent ChristinaAguilcm singsquot 1 represent Sim1hc conditional fomicd by the conjunc on of quotRicky Martin is a teen idolquot and r represent Britney premises implying the conclusion is not a rauwlogy mt Spams winxan American Music wardquot The argument at umemis 39 m Ifyuume comFlelingthe ii men represented symbolically by truin rnhle nlnng rows rather than down columns you cuultt step after compleling the smmd row knowing th P V q with 2 false conditional the statement will nnl be a 394 39 739 tautology 32 p Cumlruct the truth table far 12 V t A 41 W A T a 17 39H I39I Tl rl l ltltl39s 39nti39rl lTltl39rl li Since the conditional fanned hy the mnjunctitin trl premises implying the conclusion is not a tnutnlagy the are ent rsinvultd Note If you are completing the truth table along rows ra her mnn down coiu could stop alter utimpleti t mus you rig the rsl mw knowing that Let p represent quotThe Cult will be in the ptnynt rsquot q repritvenl Pcyltin lends the league Ill passing and r mprmenl Marv loves Lhe COINquot Thu argument is lhen represented symbolically by it M q r V q r Np Construct the truth ttthlt for 14 it A r V q A 4 39 p p q r qJArVqAdrlp TTTTTTFFTF l TF TTTTT FF TFT FFTFFTF TFF FFFFTTF FTT FFTFFTT FTF FFTFTTT FFT TTTFFTT FFF TFFFTTT I 2132 43 t39TJ l le hl39l39339Tl39quotl39139i3939lt quotn rl U H li ii39t 39n rtrll 9 Since the conditional formed by the conjunction of premises implying the conclusion is not a tautology the argument is invalid Note Il39you are completing the truth table along rows rather than down columns you could stop alter completing the second row knowing that with a false conditional the statement will not be a Let 1 represent Otis is a disc jockeyquot q represent he lives in Lexington and 739 represent he is a history buff The argument is then represented symbolically by P HI qr 739 p Construct the truth table for p a q q A gt P quot NP towwqmttactwt T T T T FTF T F F T TFF F F F T FTF F F F T TFF T T T T FTF T F F T TTT T F F T FTT T F F T TTT 2 l 3 13 Since the conditional formed by the conjunction ol premises implying the conclusion is a tuutology the argument is valid Let p represent 1 am your women 1 represent you are my man and 139 represent quot1 stop loving you argument is then represented symbolically by p A q r r pVq Construct the truth table for lP r1 39 r A 739 p v 11 39H39TJ39U TlHFi i l39ti mma imm iaa mam am rm is S lltMu i r Ar i NWL F FT 39Tl Tl39l l39Tl Tl lt l 3 3 1 7 1 t F rr 11 F F T T T T T T quotl 39TJ 3 1 JRivl iH EH TH roHPl i l39 39 rl rj i 1n ngt gt quotnrt 4 m to 36 ANALYZING AHGUMENTS WITH TRUTH TABLES 69 Since the conditional formed by the conjunction of premises implying the conclusion is u tautology the argument is valid The following exercises involve Quanti ed arguments and can be analyzed is such by Euler diagrams owever t e quantified statements can be represented as conditional statements as well This allows us to use a truth table or recognize it valid argument farm 10 attaiyze the validity of the cirgtnnent 35 Let 12 represent you are a manquot 1 represent you are created equal and 391 represent you are a women The argument is then represented symbolically by P I 1 t 739 p a 72 This is a Reasoning by Transitivity argument form and hence is 36 Let 1 represent you are a man I represent you are mortal and r represent you are Socrates The argument is then represented symbolically by P 39 Q r p 39r gtq By interchanging the first premise with the second premise the argument becomes a reasoning by transitivity form and hence is valid 37 We apply reasoning by repeated transitivity to the six premises A conclusion from this reasoning which makes the argument valid is reached by linking the rst antecedent to the last consequent This conclusion is If I tell you the time then my life will be miserablequot 38 Writing exercise Answers in Exercises 39 46 may be replaced by their contrapusitives 39 The statement All my poultry are ducks becomes If it is my poultry then it is a duckquot 40 The statement None of your sons can do logic becomes Ifhe is your son then he can39t do logic 4 The statement Guinea pigs are hopelessly ignorant of music becomes If it is a Guinea pig then it is hopelessly ignorant 01 music quot 42 The statement No teetotalers are pawnbrokers becomes 1139 the person is a teetotztler then the person is not a pawnbroker 43 The statement No teachable kitten has green eyes becomes If it is it teachable kitten then it does not have green eyesquot u s 1 a 9 CHAPTER 3 INTRODUCTION TO LOGIC The sialcnlenl n 39 39 quot D I ilhe becomes commandquot The slzlemlflll 1 have not led any or lhem lhal i can readquot becomes quotll39l can read ils than l have not led llquot The slaien lenl All ofihexn wrillcn on blue paper are l quot ucumes Ifil is written on blue paper hen ii is ledquot a No ducks are willing to waltz bccumes quotll duck men ii is noi willing lo wallzl 7 Nu unicers ever decline in waluquot becomes ifonc is an nrlicer lhen one IS willing I0 wallzquot a All my poullry are ducksquot bemmcsquotifil is my poullly lhen il 15 a uck d In symbols lhe lhree premises an p s 7 a s I 439 7 Begin Wilh 4 which only appears once Replacing r a 5 Mill iLs conlmpusilive s a 72 rearrange lhe ihree premises 1 P p a s s i r By repealed usc orleasnning by lransilivlly lhc conclusion which provides a valid argumenl is r in words lril is my poultry than iris nor an of cerquot or none of my poullry are uf cers a Everyone who is sane can do logicquot becomes if one is snne lhen one is able lo do lngicquot b No lunatic are iii in serve on a juryquot becomes if one is a lunnlic or nol sane Lhen cm is nul til 0 serve on a jury c Noni ofynursons can do logicquot ecomcs ifhs is yourson lhcn lhen he can nor do logicquot 1 ln symhnls he lhree premises are r p Replacing 7 p with ils cunlrnposilivc p a 1 reaming lhe three premises s s p a w r t q n n lull 39 p 5 a q In words quotIl he is your son lhcn leis nul lil lu serve on aJury or your sons are nol t lo serve in ajuryl a Promisebreakers an untruslwurlhy becomes if one is a promisebreaker him one is nol lruslwonhyquot b Winedrinksrs are very communicativequot become if He is a wincedrinker him one is very communicalive c A person who keeps a promise is honeslquot becomes if on is nnl a pmmlscbrcaker man one is honest d Na leelmalers are pawnbrokcrsquot becomes ifonc is nul n winc drinker men one is no 1 pawnblokerquot c One can always ir sl a very cammunicmive personquot becomes limit is very communicalive lhen one is iruslwnnhy quot i In symbols he smtcnlenls are Begin wiih g which nnly appears once Using lhe conlrnposluve of u o 11 l a u and r s s 139 rearrange rhe ve premises as follows 1 a u 11 a t t a s s r i a p By Cpcalw use ufreasoning by unnsilivily lhe conclusion which provides a valid argumenl is 4 39 P In words his conclusion can be slated as if one is a pawnbroker zhen one is honeslquot or all pawnbmkers are honeslquot Lel p be quotii is a guinea pigquot L be n is hopelessly ignorant ol39rnuslcquot r be ii keeps silent while he conlighl Snnnla 39s bcing playedquot and s be ll apprccialcs Eealhuven n Nobody who Cally nppreeinles Beellloven fails ll keep silenl while he Moonligln Sunala is being playedquot comes Hon really nppreclnlcs Beelhoven men one fails to keep silenl while ihe Moonligln Sallalu is being playedquot 11 Guinea pigs are hnpelessly lglloranl nl musir ecnmos If you are a guinea pig lllen you are hopelessly ignomnl of musicquot u ci No int who is hopcicssly ignomnl of music cvcr k p ileni w ilc am is being pitiycd hccumes Mom is hopelessly ignomnl 0139 m sic him one fails to keep silent while the Mouniighl Sunzim is being playedquot 5 5 E 2 m 2 d in symbuls Ihi blulcmenis an a a a fbi 11 q C II u 39 g the uoiilriipmilive iil39ihu premix s 4 r H 5 rcarrunge the premises in l tiiinwa By uprated use oi rctlsoning by transiiivity the conclusion which pmviiiei ivaiiti urgumcni is 7 In wordsv thh conclusion cun be 5 mi is ii ynu m u guinea pig ihen you do nni nppi39ecim Beeihoven m equivaienily minim pigs don39t nppi39cciuu Beethovenquot Begin by changing each quitniil39iiztl premix It a condiiional sitiiemeni a The statumcm quotAll the dalcd letters in this room are written on blue paper becomes Ii39it is dated men it is an hhie puperquot h The Sluicmcni Noni ni thcm am in black ink e lhnSi ihat arc wriiten in the third person becomes quot not in thc third person then it is not in black 39nk quot pt Ii i c The slalenizm quot have nUl iiled my 0139 ihem mm 1 can readquot bucomcs if can mud it then it is not lcdvquot id The smlemcnt None or them mm are written on one sheul in iindulcdquot htCUmES quotil39il is on uric sheet then it is datedquot All of hem thil ilm nni cms ed are in it is iml crusscd than it is in e The niitnnieni black in r hecumcn I The slhlemcnl All of them Wrilten hy Brawn begin with 39Dezlr DoctorquotY becomes quotIfil is wriilen by Brown then it hegins wiih Deur Sir iquot g The statcmcnt AII or lhum WrillCn on blue paper are led hccnmcs is on hills paper hen iI is iiiutiquot it The slulnmenl quotNiin oi lhtnl Wi39illcn in mm than quot heuoiiies on sheet than il is not crossedquot 36 ANAI w u IHUTH TABLES 71 i The Sullcmcnl quot one of them that begin with quotDear ir will an in the lhird personquot becomes If it begins wilh Dcar Sir llicn ii is not wriitcn in the third person i In nymhnls the stutemcnis 1m a i h ii t 1 ii a s id 1 it r E I a 139 y P 3 w h 11 g 41 I I u Begin with g which appears only once Us cunlrap iiivesui39v t 3s 11 11 a rm v n 1 1 0 reun39ange the nine xttnememn U I p a u u a t t q q a r q 7 i 7 w u a s a u By renamed use ni i eusuning by ll39ansllivily the conclusion ihal makes ha argument valid is y U 5 written in words ihe conclusion can be sIztlud by Br 39 I can 1 us own then I can t read it or equivalently read any of Brown s ieilerxiquot a N0 on who is going in a party ever i aiis m brush his hair becomes quotif one is going to a party him he brusth his hair 1 No Om looks l39ascindling if he is untidyquot becomes If one is untidy then he does not look fascinatingquot c Opiumeaters have nu seilluommandquot bewmzx H39 mm is an upium eiiler him he h39 nn sell39cnnimiquot d quotEveryone who has brushed his hair looks fascinatingquot becomes Ii39one has brushed his hair Ihcn hninnk mnnhngquot c No one wcurs whitc kid gloves unless he is going to a partyquotbccomcsquot1fhe wears white gloves than he is I A man is always untidy ifih has no selfcommandquot bewmcs becomes If ti man has no selfcommand than a man is no iidy 72 CHAPTER 3 INTRODUCTION TO LOGIC g In symbols the statements are a peg M W gt s c t 2r 7 4H3 e 39t gt p t r gt u Begin with t which only occurs once Using the contrapositives of q t s s v q p q q gt p7 and 39u gt p 11 11 rearrange the ve premises as follows t gt 1 r t u u gt s S q quotI quot P p gt u By repeated use of reasoning by transitivity the conclusion which provides a valid argument is t t t I In words this conclusion can be stated as If he is an opiumeater then he doesn39t wear white gloves or equivalently Opiumeaters do not wear white kid gloves Chapter 3 Test 1 The negation of 6 3 3 is 6 3 75 3 2 The negation 01 All men are created equal is Some men are not created equal 3 The negation of Some members of the class went on the eld trip is No members of the class went on the field tripquot An equivalent answer would be All members of the class did not go on the eld trip 4 The negation of If that39s the way you feet then will accept itquot is That39s the way you feel and I won39t accept it Remember that p q E pA 11 5 The negation of She passed GO and collected 200 is She did not pass GO or did not collect 200 Remember that p q E 11 V q Let 1 represent You will love mequot and let q represent quot1 will love you quot 6 The symbolic form 01quot If you won39t love me then I will love you is p qt 7 The symbolic form of I will love you itquot you will love me or equivalently if you will love me then I will love you is p e q 8 The symbolic form of I won39t love you if and only if you won39t love me is q t gt p 9 Writing the symbolic form p q in words we get You won39t love me and I will love you 10 Writing the symbolic form p V q in words we get It is not the case that you will love me orI won39t love you or equivalently by DeMorgan s you won39t love me and I will love youquot Assume that p is true aml that q and 39r arefalxefor Exercises 1 l Replacing q and 139 with the given truth values we have F F T T T The compound statement 11 139 is true 12 Replacing p q and r with the given truth values we have F V T F F v T T F V T T The compound statement r V p q is true 13 Replacing 139 with the given truth value 3 not known we ave F gt s V F F not known T The compound statement r gt s V 739 is true 14 Replacing p and q with the given truth values we have T HT gtF TH F F The compound statement 3 gt p i q is false 15 Writing exercise 16 The necessary condition for a a conditional statement to be false is that the antecedent must be true and the consequent must be false b a conjunction to be Lrue is that both component statements must be true 0 a disjunction to be false is that both component statements must be false

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