Chapter1:SpeakingMathematically
Manycollegeanduniversitystudentshavedifficultyusingandinterpretinglanguageinvolvingif-then statementsandquantification.Section1.1isagentleintroductiontotherelationbetweeninformal andformalwaysofexpressingsuchstatements.Theexercisesareintendedtostarttheprocessof helpingstudentsimprovetheirabilitytointerpretmathematicalstatementsandevaluatetheirtruth orfalsity.Sections1.2-1.4areabriefintroductiontothelanguageofsets,relations,functions, andgraphs.IncludingSections1.2and1.3atthebeginningofthecoursecanhelpstudentsrelate discretemathematicstothepre-calculusorcalculustheyhavestudiedpreviouslywhileenlarging theirperspectivetoincludeagreaterproportionofdiscreteexamples.Section1.4isdesignedto broadenstudents’understandingofthewaythewordgraphisusedinmathematicsandtoshow themhowgraphmodelscanbeusedtosolvesomesignificantproblems.
Proofsofsetproperties,suchasthedistributivelaws,andproofsofpropertiesofrelationsand functions,suchastransitivityandsurjectivity,areconsiderablymorecomplexthanthoseusedin Chapter4togivestudentstheirfirstpracticeinconstructingmathematicalproofs.Forthisreason settheoryasatheoryislefttoChapter6,propertiesoffunctionstoChapter7,andpropertiesof relationstoChapter8.Bymakingslightchangesaboutexercisechoices,instructorscouldcover Section1.2justbeforestartingChapter6andSection1.3justbeforestartingChapter7.
ThematerialinSection1.4laysthegroundworkforthediscussionofthehandshaketheorem anditsapplicationsinSection4.9.Instructorswhowishtoofferaself-containedtreatmentofgraph theorycancombinebothsectionswiththematerialinChapter10.
Collegeanduniversitymathematicsinstructorsmaybesurprisedbythewaystudentsunderstand themeaningoftheterm“realnumber.”Whenaskedtoevaluatethetruthorfalsityofastatement aboutrealnumbers,itisnotunusualforstudentstothinkonlyofintegers.Thusaninformal descriptionoftherelationshipbetweenrealnumbersandpointsonanumberlineisgiveninSection 1.2toillustratethattherearemanyrealnumbersbetweenanypairofconsecutiveintegers,Examples 3.3.5and3.3.6showthatwhilethereisasmallestpositiveintegerthereisnosmallestpositivereal number,andthediscussioninChapter7,whichprecedestheproofoftheuncountabilityofthereal numbersbetween0and1,describesaprocedureforapproximatingthe(possiblyinfinite)decimal expansionforanarbitrarilychosenpointonanumberline.
Section1.1
1. a. x2 = 1(Or :thesquareof x is 1) b.arealnumber x
2. a. aremainderof2whenitisdividedby5andaremainderof3whenitisdividedby6
b. aninteger n; n isdividedby6theremainderis3
3. a. between a and b b.distinctrealnumbers a and b;thereisarealnumber c
4. a. arealnumber;greaterthan rb.realnumber r;thereisarealnumber s
5. a. r ispositive
b. positive;thereciprocalof r ispositive(Or :positive;1/r ispositive)
c. ispositive;1/r ispositive(Or :ispositive;thereciprocalof r ispositive)
6. a. s isnegative b. negative;thecuberootof s isnegative(Or : 3 √s isnegative)
c. isnegative; 3 √s isnegative(Or :thecuberootof s isnegative)
Instructor’sManual:Chapter1
7. a. Therearerealnumberswhosesumislessthantheirdifference.True.Forexample, 1+( 1)=0, 1 ( 1)=1+1=2,and0 < 2.
b. Thereisarealnumberwhosesquareislessthanitself.True.Forexample,(1/2)2 =1/4 < 1/2.
c. Thesquareofeachpositiveintegerisgreaterthanorequaltotheinteger.
True.If n isanypositiveinteger,then n ≥ 1.Multiplyingbothsidesbythepositivenumber n doesnotchangethedirectionoftheinequality(seeAppendixA,T20),andso n2 ≥ n d. Theabsolutevalueofthesumofanytwonumbersislessthanorequaltothesumoftheir absolutevalues.
True.Thisisknownasthetriangleinequality.ItisdiscussedinSection4.4.
8. a. havefoursides b. hasfoursides c. hasfoursides d. isasquare;hasfoursides e. J hasfoursides
9. a. haveatmosttworealsolutions b. hasatmosttworealsolutions c. hasatmosttwo realsolutions d. isaquadraticequation;hasatmosttworealsolutions e. E hasatmost tworealsolutions
10. a. havereciprocals b. areciprocal c. s isareciprocalfor r
11. a. havepositivesquareroots b. apositivesquareroot c. r isasquarerootfor e
12. a. realnumber;productwitheverynumberleavesthenumberunchanged b. apositivesquareroot c. rs = s
13. a. realnumber;productwitheveryrealnumberequalszero b. witheverynumberleavesthenumberunchanged c. ab =0
Section1.2
1. A = C and B = D
2. a. Thesetofallpositiverealnumbers x suchthat0islessthan x and x islessthan1
b. Thesetofallrealnumbers x suchthat x islessthanorequaltozeroor x isgreaterthan orequalto1
c. Thesetofallintegers n suchthat n isafactorof6
d. Thesetofallpositiveintegers n suchthat n isafactorof6
3. a. No, {4} isasetwithoneelement,namely4,whereas4isjustasymbolthatrepresentsthe number4
b. Three:theelementsofthesetare3,4,and5.
c. Three:theelementsarethesymbol1,theset {1},andtheset {1, {1}}
4. a. Yes: {2} isthesetwhoseonlyelementis2. b. One:2istheonlyelementinthisset c. Two:Thetwoelementsare0and {0} d. Yes: {0} isoneoftheelementslistedintheset.
e. No:Theonlyelementslistedinthesetare {0} and {1},and0isnotequaltoeitherof these.
5. Theonlysetsthatareequaltoeachotherare A and D.
A containstheintegers0,1,and2andnothingelse.
B containsalltherealnumbersthataregreaterthanorequalto 1andlessthan3.
C containsalltherealnumbersthataregreaterthan 1andlessthan3.Thus 1isin B butnotin C
D containsalltheintegersgreaterthan 1andlessthan3.Thus D containstheintegers0, 1,and2andnothingelse,andso D = {0, 1, 2} = A.
E containsallthepositiveintegersgreaterthan 1andlessthan3.Hence E containsthe integers1and2andnothingelse,thatis, E = {1, 2}.
6. T2 and T 3 eachhavetwoelements,and T0 and T1 eachhaveoneelement.
Justification: T
7. a. {1, 1}
b. T = {m ∈ Z | m =1+( 1)k forsomeinteger k} = {0, 2}. ExercisesinChapter4explorethe factthat( 1)k = 1when k isoddand( 1)k =1when k iseven.So1+( 1)k =1+( 1)=0 when k isodd,and1+( 1)k =1+1=2when k iseven.
c. thesethasnoelements
d.Z (everyintegerisintheset)
e. Therearenoelementsin W becausetherearenointegersthatarebothgreaterthan1and lessthan 3
f. X = Z becauseeveryinteger u satisfiesatleastoneoftheconditions u ≤ 4or u ≥ 1
8. a. No, B A because j ∈ B and j/ ∈ A
b. Yes,becauseeveryelementin C isin A
c. Yes,becauseeveryelementin C isin C
c..Yes,becauseitistruethateveryelementin C isin C
d. Yes, C isapropersubsetof A.Bothelementsof C arein A,but A containselements (namely c and f )thatarenotin C
9. a. Yes
b. No,thenumber1isnotasetandsoitcannotbeasubset.
c. No:Theonlyelementsin {1, 2} are1and2, and {2} isnotequaltoeitherofthese.
d. Yes: {3} isoneoftheelementslistedin {1, {2}, {3}}
e. Yes: {1} isthesetwhoseonlyelementis1.
f. No,theonlyelementin {2} isthenumber2andthenumber2isnotoneofthethree elementsin {1, {2}, {3}}
g. Yes:Theonlyelementin {1} is1,and1isanelementin {1, 2}.
h. No:Theonlyelementsin {{1}, 2} are {1} and2,and1isnotequaltoeitherofthese.
i. Yes,theonlyelementin {1} isthenumber1,whichisanelementin {1, {2}}.
j. Yes:Theonlyelementin {1} is1,whichisisanelementin {1}.Soeveryelementin {1} is in {1}.
10. a. No.Observethat( 2)2 =( 2)( 2)=4,whereas 22 = (22)= 4.So(( 2)2 , 22)= (4, 4),whereas( 22 , ( 2)2)=( 4, 4).And(4, 4) =( 4, 4)because 4 =4.
b. No:Fortwoorderedpairstobeequal,theelementsineachpairmustoccurinthesame order.Inthiscasethefirstelementofthefirstpairis5,whereasthefirstelementofthesecond
Instructor’sManual:Chapter1
pairis 5,andthesecondelementofthefirstpairis 5whereasthesecondelementofthe secondpairis5.
c. Yes.Notethat8 9= 1and 3 √ 1= 1,andso(8 9, 3 √ 1)=( 1, 1).
d. YesThefirstelementsofbothpairsequal 1 2 ,andthesecondelementsofbothpairsequal 8.
11. a. {(w,a), (w,b), (x,a), (x,b), (y,a), (y,b), (z,a), (z,b)} A × B has4 2=8elements.
b. {(a,w), (b,w), (a,x), (b,x), (a,y), (b,y), (a,z), (b,z)} B × A has4 2=8elements.
c. {(w,w), (w,x), (w,y), (w,z), (x,w), (x,x), (x,y), (x,z), (y,w), (y,x), (y,y), (y,z), (z,w), (z,x), (z,y), (z,z)} A × A has4 4=16elements.
d. {(a,a), (a,b), (b,a), (b,b)} B × B has2 · 2=4elements.
12. Allfoursetshavenineelements.
a. S × T = {(2, 1), (2, 3), (2, 5), (4, 1), (4, 3), (4, 5), (6, 1), (6, 3), (6, 5)}
b. T × S = {(1, 2), (3, 2), (5, 2), (1, 4), (3, 4), (5, 4), (1, 6), (3, 6), (5, 6)}
c. S × S = {(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}
d. T × T = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}
13. a. A × (B × C)= {(1, (u,m)), (1, (u,n)), (2, (u,m)), (2, (u,n)), (3, (u,m)), (3, (u,n))}
b. (A × B) × C = {((1,u),m), ((1,u),n), ((2,u),m), ((2,u),n), ((3,u),m), ((3,u),n)}
c. A × B × C = {(1,u,m), (1,u,n), (2,u,m), (2,u,n), (3,u,m), (3,u,n)}
14. a.R × (S × T )= {(a, (x,p)), (a, (x,q)), (a, (x,r)), (a, (y,p)), (a, (y,q)), (a, (y,r))}
b. (R × S) × T = {((a,x),p), ((a,x),q), ((a,x),r), ((a,y),p), ((a,y),q), ((a,y),r)}
c. R × S × T = {(a,x,p), (a,x,q), (a,x,r), (a,y,p), (a,y,q), (a,y,r)}
15. 0000,0001,0010,0100,1000
16. yxxxx,xyxxx, xxyxx, xxxyx, xxxxy
Section1.3
1. a. No.Yes.No.Yes.
b. R = {(2, 6), (2, 8), (2, 10), (3, 6), (4, 8)}
c. Domainof R = A = {2, 3, 4},co-domainof R = B = {6, 8, 10}
d. R 6 8 10 2 3 4
2. a. 2 S 2because 1 2 1 2 =0,whichisaninteger.
1 S 1because 1 1 1 1 =0,whichisaninteger.
3 S 3because 1 3 1 3 =0,whichisaninteger.
3 /S 3because 1 3 1 3 = 2 3 ,whichisnotaninteger.
b. S = {( 3, 3), ( 2, 2), ( 1, 1), (1, 1), (2, 2), (3, 3), (1, 1), ( 1, 1), (2, 2), ( 2, 2)}
c. Thedomainandco-domainof S areboth {−3, 2, 1, 1, 2, 3}
d. S
3. a. 3 T 0because 3 0 3 = 3 3 =1,whichisaninteger. 1 T ( 1)because 1 ( 1) 3 = 2 3 ,whichisnotaninteger.
(2, 1) ∈ T because 2 ( 1) 3 = 3 3 =1,whichisaninteger. (3, 2) / ∈ T because 3 ( 2) 3 = 5 3 ,whichisnotaninteger.
b. T = {(1, 2), (2, 1), (3, 0)}
c. Domainof T = E = {1, 2, 3},co-domainof T = F = {−2, 1, 0}
d. T –2 –1 0 1 2 3
4. a. 2 V 6because 2 6 4 = 4 4 = 1,whichisaninteger.
1 V ( 1)because ( 2) 8 4 = 6 4 ,whichisnotaninteger. ( 2) V 8because ( 2) 8 4 = 6 4 ,whichisnotaninteger.
0 V 6because 0 6 4 = 6 4 ,whichisnotaninteger.
2 V 4because 2 4 4 = 2 4 ,whichisnotaninteger.
b. V = {( 2, 6), (0, 4), (0, 8), (2, 6)}
c. Domainof V = {−2, 0, 2},co-domainof V = {4, 6, 8}
d. S –2 0 2 4 6 8 GH
5. a. (2, 1) ∈ S because2 ≥ 1.(2, 2) ∈ S because2 ≥ 2. 2 S 3because2 3.( 1)S( 2)because 1 ≥−2.
x $ y in shaded region
b. x 1
6. a. (2, 4) ∈ R because4=22
(4, 2) / ∈ R because2 =42
( 3, 9) ∈ R because9=( 3)2
(9, 3) / ∈ R because 3 =92
b.
graph of S
b. R isnotafunctionbecauseitsatisfiesneitherproperty(1)norproperty(2)ofthedefinition. Itfailsproperty(1)because(4,y) / ∈ R,forany y in B.Itfailsproperty(2)because(6, 5) ∈ R and(6, 6) ∈ R and5 =6.
S isnotafunctionbecause(5,5) ∈ S and(5, 7) ∈ S and5 =7.So S doesnotsatisfyproperty (2)ofthedefinitionoffunction.
T isnotafunctionbothbecause(5,x) / ∈ T forany x in B andbecause(6, 5) ∈ T and (6, 7) ∈ T and5 =7.So T doesnotsatisfyeitherproperty(1)orproperty(2)ofthedefinition offunction.
b. Noneof U , V ,or W arefunctions.
U isnotafunctionbecause(4,y)isnotin U forany y in B,andso U doesnotsatisfyproperty (1)ofthedefinitionoffunction.
V isnotafunctionbecause(2,y)isnotin V forany y in B,andso V doesnotsatisfyproperty (1)ofthedefinitionoffunction.
W isnotafunctionbecauseboth(2, 3)and(2, 5)arein W and3 =5,andso W doesnot satisfyproperty(2)ofthedefinitionoffunction.
9. a. Thereisonlyone: {(0, 1), (1, 1)}
b. {(0, 1)}, {(1, 1)}
10. Thefollowingsetsarerelationsfrom {a,b} to {x,y} thatarenotfunctions: {(a,x)}, {(a,y)}, {(b,x)}, {(b,y)}, {(a,x), (a,y)}, {(b,x), (b,y)}, {(a,x), (a,y), (b,x)}, {(a,x), (a,y), (b,y)}, {(b,x), (b,y), (a,x)}, {(b,x), (b,y), (a,y)}, {(a,x), (a,y), (b,x), (b,y)}
11. L(0201)=4,L(12)=2
12. C(x)= yx,C(yyxyx)= yyyxyx
13. a. Domain= A = {−1, 0, 1}, co-domain= B = {t,u,v,w}
b. F ( 1)= u,F (0)= w,F (1)= u
14. a. Domainof G = {1, 2, 3, 4}, co-domainof G = {a,b,c,d}
b. G(1)= G(2)= G(3)= G(4)= c
15. a. Thisdiagramdoesnotdetermineafunctionbecause2isrelatedtoboth2and6.
b. Thisdiagramdoesnotdetermineafunctionbecause5isinthedomainbutitisnotrelated toanyelementintheco-domain.
c. Thisdiagramdoesnotdetermineafunctionbecause4isrelatedtoboth1and2,which violatesproperty(2)ofthedefinitionoffunction.
d. Thisdiagramdefinesafunction;bothproperties(1)and(2)aresatisfied.
e. Thisdiagramdoesnotdetermineafunctionbecause2isinthedomainbutitisnotrelated toanyelementintheco-domain.
16. f ( 1)=( 1)2 =1,f (0)=02 =0,f ( 1 2 ) = ( 1 2 )2 = 1 4
17. g( 1000)= 999, g(0)=1, g(999)=1000
18. h( 12 5 )= h( 0 1 )= h( 9 17 )=2
19. Foreach x ∈ R, g(x)= 2x3 +2x x2 +1 = 2x(x2 +1) x2 +1 =2x = f (x) Therefore,bydefinitionof equalityoffunctions, f = g.
Instructor’sManual:Chapter1
20. Forall x ∈ R, K(x)=(x 1)(x 3)+1=(x2 4x +3)+1= x2 +4x +4=(x 2)2 = H(x).
Therefore,bydefinitionofequalityoffunctions, H = K.
1. V (G)= {v1,v2,v3,v4},E(G)= {e
Edge-endpointfunction:
Edge Endpoints
e
e
e3 {v3}
Section1.4
,e2,e3}
2. V (G)= {v1,v2,v3,v4}, E(G)= {e1,e2,e3,e4,e5}
Edge-endpointfunction:
Edge Endpoints
e
e
3.
4.
5. Imaginethattheedgesarestringsandtheverticesareknots.Youcanpickuptheleft-hand figureandlayitdownagaintoformtheright-handfigureasshownbelow.
7.
8. (i) e1, e2, e7 areincidenton v1
(ii) v 1, v 2,and v 3 areadjacentto v 3
(iii) e2, e8, e9,and e3 areadjacentto e1.
(iv)Loopsare e6 and e7
(v) e8 and e9 areparallel; e4 and e5 areparallel.
(vi) v 6 isanisolatedvertex.
(vii)degreeof v3 =5
9. (i) e1, e2, e7 areincidenton v1
(ii) v1 and v2 areadjacentto v3
(iii) e2 and e7 areadjacentto e1
(iv) e1 and e3 areloops.
(v) e4 and e5 areparallel.
(vi) v4 isanisolatedvertex.
(vii)degreeof v3 =2
10. a.Yes.Accordingtothegraph, SportsIllustrated isaninstanceofasportsmagazine,asports magazineisaperiodical,andaperiodicalcontainsprintedwriting.
b. Yes.Accordingtothegraph, PoetryMagazine isaninstanceofaLiteraryjournalwhichis aScholarlyjournaland,therefore,containsLongwords.
11. (vvccB/) → (vc/Bvc) → (vvcB/c) → (c/Bvvc) → (vcB/vc) → (/Bvvcc) (vvccB/) → (vv/Bcc) → (vvcB/c) → (c/Bvvc) → (vcB/vc) → (/Bvvcc) (vvccB/) → (vv/Bcc) → (vvcB/c) → (c/Bvvc) → (ccB/vv) → (/Bvvcc)
12. Tosolvethispuzzleusingagraph,introduceanotationinwhich,forexample, wc/fg means thatthewolfandthecabbageareontheleftbankoftheriverandtheferrymanandthegoat areontherightbank.Thendrawthosearrangementsofwolf,cabbage,goat,andferryman thatcanbereachedfromtheinitialarrangement(wgcf /)andthatarenotarrangementstobe avoided(suchas(wg/fc)).Ateachstageaskyourself,“WherecanIgofromhere?”anddraw linesorarrowspointingtothosearrangements.Thismethodgivesthegraphshownbelow.
13.
Examiningthediagramrevealsthesolutions
/wc) → (/wgcf )
vvvcccB/ vvvc/Bcc
vvcc/Bvc
vvvcc/Bc vvvccB/c
vvv/Bccc vvvcB/cc vc/Bvvcc
vvccB/vc cccB/vvv c/Bvvvcc
ccB/vvvc
vcB/vvcc /Bvvvccc
ccB/vvvc
Thediagramshowsseveralsolutions.Amongthemis(vvvcccB/) → (vvcc/Bvc) → (vvvccB/c) → (vvv/Bccc) → (vvvcB/cc) → (vc/Bvvcc) → (vvccB/vc) → (cc/Bvvvc) → (cccB/vvv) → (c/Bvvvcc) → (vcB/vvcc) → (/Bvvvccc),oronecanendwith(c/Bvvvcc) → (ccB/vvvc) → (/Bvvvccc),oronecanstartwith(vvvcccB/) → (vvvc/Bcc) → (vvvccB/c).
14. Representpossibleamountsofwaterinjugs A and B byorderedpairswith,say,theordered pair(1,3)indicatingthatthereisonequartofwaterinjug A andthreequartsinjug B Startingwith(0,0),drawanedgefromoneorderedpairtoanotherifitispossibletogofrom thesituationrepresentedbytheonepairtothatrepresentedbytheotherandbackbyeither
fillingajugfromthetap,emptyingajugintothedrain,ortransferringwaterfromonejug toanother.Exceptfor(0,0),onlydrawedgesfromstatesthathaveedgesincidentonthem (sincethesearetheonlystatesthatcanbereached).Theresultinggraphisshownasfollows:
(0,5)(1,5)(2,5) (3,5)
(0,4)(1,4)(2,4) (3,4)
(0,3)(1,3)(2,3) (3,3)
(0,2)(1,2)(2,2) (3,2)
(0,1)(1,1)(2,1) (3,1)
(0,0) (1,0) (2,0) (3,0)
Itisclearfromthegraphthatonesolutionis(0,
andanothersolutionis(0
→
,
→ (3, 1) → (0, 1)
Notethatitwouldbepossibletoaddarrowstotheabovegraphfromeachreachablestate toeachotherstatethatcouldbeobtainedfromiteitherbyfillingoneofthejugstothetop orbyemptyingtheentirecontentsofoneofthejugs.Forinstance,onecoulddrawanarrow from(0,3)to(0,5)orfrom(0,3)to(0,0).Becausethegraphisconnected,allsucharrows wouldpointtostatesalreadyreachablebyothermeans,sothatitisnotnecessarytoadd suchadditionalarrowstofindsolutionstotheproblem(anditmakesthediagramlookmore complicated).However,iftheproblemweretofindallpossiblesolutions,thearrowswould havetobeadded. 15.
Vertex e hasmaximaldegree,socoloritwithcolor#1.Vertex a doesnotshareanedgewith e, andsocolor#1mayalsobeusedforit.Fromtheremaininguncoloredvertices,allof d,g,and f havemaximaldegree.Chooseanyoneofthem—say, d—andusecolor#2forit.Observe thatvertices c and f donotshareanedgewith d,buttheydoshareanedgewitheachother, whichmeansthatcolor#2maybeusedforonebutnottheother.Choosetocolor f with color#2becausethedegreeof f isgreaterthanthedegreeof c.Theremaininguncolored vertices, b,c, and g,areunconnected,andsocolor#3maybeusedforallthree.
16. Representeachcommitteenamebyavertex,labeledwiththefirstletterofthenameofthe committee,andjoinverticesif,andonlyif,thecorrespondingcommitteeshaveamemberin common.Figure(a)showsonewaytocolorthegraphs.Vertex H hasmaximumdegree,so usecolor#1forit.Vertex L doesnotshareanedgewith H,andsocolor#1mayalsobe usedforit.
Fromtheremaininguncoloredvertices,eachof P , U ,and G isadjacenttothreeothervertices. Chooseanyoneofthem,say P ,andusecolor#2forit.Vertices U and C donotsharean edgewith P orwitheachother,andsocolor#2mayalsobeusedforthem. Color#3canthenbeusedfortheremainingvertex, G,atwhichpointallverticeswillbe colored.
Notethatthesameresultisobtainedif U ischoseninsteadof P instep2.However,if G is choseninsteadof P instep2,theresultisthecoloringindicatedinFigure(b).
TousetheresultsofFigure(a)toschedulethemeetings,letcolor n correspondtomeeting time n.Then
Time1:hiring,library
Time2:personnel,undergraduateeducation,colloquium
Time3:graduateeducation
UsingtheresultsofFigure(b)toschedulethemeetingsproducesthisresult:
Time1:hiring,library
Time2:graduateeducation,colloquium
Time3:personnel,undergraduateeducation
17. Inthefollowinggrapheachcoursenumberisrepresentedasavertex.Verticesarejoinedif, andonlyif,thecorrespondingcourseshaveastudentincommon.
Vertex135hasmaximumdegree,sousecolor#1forit.Allverticesshareedgeswithvertex 135,andsocolor#1cannotbeusedonanyothervertex.
Fromtheremaininguncoloredvertices,onlyvertex120hasmaximumdegree.Sousecolor#2 forit.Becausevertex100doesnotshareanedgewithvertex120,color#2mayalsobeused forit.
Fromtheremaininguncoloredvertices,allof101,102,110,and130havemaximumdegree. Chooseanyoneofthem,sayvertex101,andusecolor#3forit.Neithervertex102nor
vertex110sharesanedgewithvertex101,buttheydoshareanedgewitheachother.So color#3maybeusedforonlyoneofthem.Ifcolor#3isusedforvertex110,then,sincethe remainingvertices130and102areconnected,twoadditionalcolorswouldbeneededforthem tohavedifferentcolors.Ontheotherhand,ifcolor#3isusedforvertex102,then,sincethe remainingvertices,110and130,arenotconnectedtoeachother,color4maybeusedforboth. Therefore,tominimizethenumberofcolors,color#3shouldbeusedforvertex102andcolor #4forvertices110and130.Theresultisindicatedintheannotationsonthegraph.
Tousetheresultsforschedulingexams,letcolor n correspondtoexamtime n.Then
Time1:MCS135
Time2:MCS100andMCS120
Time3:MCS101andMCS102
Time4:MCS110andMCS130
Notethatbecause,forexample,MSC135,MSC102,MSC110,andMSC120areallconnected toeachother,theymustallbegivendifferentcolors,andsothescheduleforthesevenexams mustuseatleastfourtimeperiods.