1
SOLUTION MANUAL FOR A FIRST COURS v v v v v
E IN ABSTRACT ALGEBRA, WITHAPPLIC
v v v v v
ATIONS8TH EDITION BY JOSEPH v
v
v v
J. ROTMAN LATEST 2024
v v v
Exercises for Chapter 1
v v v
1.1 Truevorvfalsevwithvreasons.
(i) Therevisvavlargestvintegervinveveryvnonemptyvsetvofvnegativevinte-
gers.
v
Solution.vTrue.vIfvCv isvavnonemptyvsetvofvnegativevintegers,vthen
−Cv =v {−nv :v nv ∈v Cv}
isvavnonemptyvsetvofvpositivevintegers.vIfv−avisvthevsmallestvelementv
ofv−Cv,vwhichvexistsvbyvthevLeastvIntegervAxiom,vthenv−av≤v−cvf
orvallvcv ∈vC,vsovthatvav ≥vcvforvallvcv ∈vC.
(ii) Therevisvavsequencevofv13vconsecutivevnaturalvnumbersvcontainingv
exactlyv2vprimes.
Solution.vTrue.vThevintegersv48vthroughv60vformvsuchvavsequence;v
onlyv53vandv59varevprimes.
(iii) Therev arev atv leastv twov primesv inv anyv sequencev ofv 7v consecutivev
naturalvnumbers.
Solution.v False.v Thev integersv 48v throughv 54v arev 7v consecutivev
naturalvnumbers,vandvonlyv53visvprime.
(iv) Ofvallvthevsequencesvofvconsecutivevnaturalv numbersvnotvcontainingv
2vprimes,vtherevisvavsequencevofvshortestvlength.
Solution.vTrue.vThevsetvCv consistingvofvthevlengthsvofvsuchv(finite)v
sequencesvisvavnonemptyvsubsetvofvthevnaturalvnumbers.
(v) 79visvavprime.
√v √v
Solution.vTrue.v 79v< 81v =9,vandv79visvnotvdivisiblevbyv2,v3,
5,vorv7.
(vi) Therev existsv av sequencev ofv statementsv S(1),vS(2),v... v withv S(2n)vt
ruevforvallvnv≥v1vandvwithvS(2nv−v1)vfalsevforveveryvnv≥v1.vSoluti
on.vTrue.vDefinevS(2nv−v1)vtovbevthevstatementvnv/=vn,vandvdefinev S(
2n)vtovbevthevstatementvnv =vn.
(vii) Forv allv n ≥v 0,v wev havev nv ≤v Fnv,v wherev Fnv isv thev nthv Fibonacci
number.
,2
Solution.vTrue.v Wev havev 0v =v F0,v 1v =v F1,v 1v=v F2,v andv 2v =vF
3.v Usevthevsecondvformvofvinductionvwithvbasevstepsvnv =v2vandvnv
=v3v(verifyingvthevinductivevstepvwillvshowvwhyvwevchoosevthesev
numbers).v Byvthevinductivevhypothesis,vnv—v2v≤vFn−2v andvnv—
1v Fn≤1.vHence,v2nv 3v Fnv.−Bu≤
tvnv 2nv 3vforval≤lvnv 3,− ≥
−
asvdesired.
(viii) Ifvmv andvnv arevnaturalvnumbers,vthenv(mn)!= vm!n!.
Solution.v False.v Ifvmv =v2v=vn,vthenv(mn)!= v24vandvm!n!= v4.
1.2 (i) Forvanyvnv ≥v0vandvanyvrv /=v1,vprovevthat
1v+vrv +vrv2v +vrv3v +v·v·v·v +vrnv =v (1v −vrn+1)/(1v −vr).
Solution.v Wevusevinductionvonvnv ≥v 1.v Whenvnv =v 1,v bothvsidesve
qualv1v+vrv.v Forvthev inductivev step,vnotev that
[1v +vrv +vrv2v +vrv3v +v·v·v·v +vrnv]+ v rn+1v =v (1v −vrn+1)/(1v −vr)v +vrn+1
1v −vrn+1v +v(1v −vr)rn+1
=
1v−vr
1v−v rn+2
=v
.
1v−vr
(ii)v Provevthat
1v +v2v +v22v +v·v·v·v +v2nv =v 2n+1v −v1.
Solution.v Thisv isv thev specialv casev ofv thev geometricv seriesv whenvr
v =v2;vhence,vthevsumvisv(1v−v2
n+1)/(1v−v2)v=v2n+1v−v1.vOnevcanval
sovprovevthisvdirectly,vbyvinductionvonvnv ≥v0.
1.3 Show,vforvallvnv≥v1,vthatv10nvleavesvremainderv1vaftervdividingvbyv9.vSolut
ion.v Thisv mayv bev rephrasedv tov sayv thatv therev isv anv integerv qnv withv10nv =v
9qnv+v1.vIfvwevdefinevq1v =v1,vthenv10v=vq1v+v1,vandvsovthevbasevstepvisvtrue.
Forvthevinductivevstep,vtherevisvanvintegervqnvwith
10n+1v =v10v×v10nv =v10(9qnv +v1)
=v90qnv +v10v =v9(10qnv +v1)v+v1.
Definev qn+1v =v10qnv +v1,v whichv isv anv integer.
1.4 Provevthatvifv0v≤vav ≤vb,vthenvanv ≤vbnv forvallvnv ≥v0.
Solution.vBasevstep.va0v =v1v=vb0,vandvsova0v ≤vb0.
Inductivevstep.vThevinductivevhypothesisvis
anv≤vbn.
, 3
Sincevavisvpositive,vTheoremv1.4(i)vgivesvan+1v=vaanv≤vabnv;vsincevbvisvpo
sitive,vTheoremv1.4(i)vnowvgivesvabnv ≤vbbnv =vbn+1.
1.5v Provev thatv 12v +v22v +v·v·v·v +vn2v =v 1v6n(nv +v1)(2nv +v1)v =v 1vn33v +v 1vn22v +v 1vn.6
Solution.vThevproofvis1byvin1vduc1tionvonvnv ≥v1.vWhenvnv =v1,vthevleftvsidevis
1vandvthevrightvsidevis + + =v1.
3 2 6
Forvthevinductivevstep,
[12v +v22v +v··· v +vn2v]+v (nv +v1)2v =v 13vn3v +v 12vnv2v +v 1v6nv +v(nv +v1)2
3 2
=v 1v(nv+v1)v +v1v(nv +v1)v +v1v(nv+v1),
3 2 6
aftervsomevelementaryvalgebraicvmanipulation.
1.6v Provev thatv 13v +v23v +v·v·v·v +vn3v =v 14vn4v +v 1v2n3v +v 1v4n2.
Solution.vBasevstep:vWhenvnv =v1,vbothvsidesvequalv1.
Inductivevstep:
[13v +v23v +v·v·v·v +vnv3v]+v (nv +v1)3v =v 1vn4v +v 1vn3v +v 1vn2v +v(nv +v1)3.
4 2 4
Expandingvgivesv 1vn4v +v 3vn3v +v 13vn2v +v3nv +v 1,
4 2 4
whichvis
4 (nv+v1) +v2 (nv+v1) +v 4 (nv +v1) .
1v 4v 1v 3v 1v 2
1.7v Provev thatv 14v +v24v +v·v·v·v +vn4v=v 1vn55v +v 1vn24v +v 1vn33v −vv1v n.30
Solution.v Thevproofvisvby1indu1ction1vonvn1vv≥v1.v Ifvnv−v1,v thenv thevleftvsidevis
1,vwhilevthevrightvsidevis + + − =v1vasvwell.
5 2 3 30
Forvthevinductivevstep,
14v +v24v +v·v·v·v +vn4v +v(nv +v1)4v =v 1v5n5v +v 1vn24v +v 1vn33v −v 1v 30
nv +v(nv +v1)4.
Itvisvnowvstraightforwardvtovcheckvthatvthisvlastvexpressionvisvequalvto
5v(nv+v1) +v12v(nv+v1) +v13v(nv+v1) −v 130(nv+v1).
1 5v 4v 3v v
1.8 Findvavformulavforv1 +3v+5 +·v·v·+v(2nv−1),vandvusevmathematicalvinductionvt
ovprovevthatvyourvformulavisvcorrect.
Solution.vWevprovevbyvinductionvonvnv ≥v1vthatvthevsumvisvn2.
BasevStep.vWhenvnv=v1,vwevinterpretvthevleftvsidevtovmeanv1.vOfvcourse,v
12v =v1,vandvsovthevbasevstepvisvtrue.
InductivevStep.
1v +v3v +v5v +v·v·v·v +v(2nv −v1)v +v(2nv +v1)
=v 1v +v3v +v5v +v··· v +v(2nv −v 1)]+v (2nv +v1)
=vn2v +v2nv +v1
=v(nv+v1)2.
, 4
.n
1.9 Findv avvformulav forv 1v +v jv!vjv,vvandv usev inductionv tov provevvthatvvyour
j=1
formulavisvcorrect.
Solution.vvAvvlistv ofv thevvsumsv forvvnvv1=
,vv2,vv3,vv4,vv5vvisv 2,vv6,vv24,vv120,vv720.vT
hesevarevfactorials;vbetter,vtheyvarev2v!,v3v!,v4!,v5!,v6!v.vWevhavevbeenvledvtovth
evguess
n
S(n)v:v1v+ jv!vjv=v (nv +v1)!.
jv=1
Wevnowvusevinductionvtovprovevthatvthevguessvisvalwaysvtrue.vThevbasevstepv
S(1)vhasvalreadyvbeenvchecked;vitvisvonvthevlist.vForvthevinductivevstep,vwev
mustvprove
n+1
S(nv +v1)v :v1v+ jv!vjv=v (nv +v2)!.
jv=1
Rewritevthevleftvsidevas
n
1v+ jv!vjv +v(nv+v1)!(nv+v1).
jv=1
Byvthevinductivevhypothesis,vthevbracketedvtermvisv(nv +v1)!,vandvsovthevleftvsid
evequals
(nv +v1)!+v(nv+v1)!(nv+v1)v=v(nv+v1)![1v+v(nv+v1)]
=v(nv+v1)!(nv+v2)
=v(nv+v2)!.
Byvinduction,v S(n)visvtruevforvallvnv ≥v1.
1.10 (M.vBarr)vTherevisvavfamousvanecdotevdescribingvavhospitalvvisitvofvG.vH.v
HardyvtovRamanujan.vHardyvmentionedvthatvthevnumberv1729vofvthevtaxivh
evhadvtakenvtovthevhospitalvwasvnotvanvinterestingvnumber.vRamanujanvdisa
greed,vsayingvthatvitvisvthevsmallestvpositivevintegervthatvcanvbevwrittenvasvth
evsumvofvtwovcubesvinvtwovdifferentvways.
(i) ProvevthatvRamanujan’svstatementvisvtrue.
Solution.vFirst,v1729visvthevsumvofvtwovcubesvinvtwovdifferentvwa
ys:
1729v=v13v +v123; 1927v =v93v +v103.
Second,v nov smallerv numberv nv hasv thisv property.v Ifv nv =va3vv+v b3,
3vvv 3
thenvva,v b≤12.vvItv isv nowv avvmatterv ofv checkingv allv pairsvva+ b vvforvs
uchvav andvb.
SOLUTION MANUAL FOR A FIRST COURS v v v v v
E IN ABSTRACT ALGEBRA, WITHAPPLIC
v v v v v
ATIONS8TH EDITION BY JOSEPH v
v
v v
J. ROTMAN LATEST 2024
v v v
Exercises for Chapter 1
v v v
1.1 Truevorvfalsevwithvreasons.
(i) Therevisvavlargestvintegervinveveryvnonemptyvsetvofvnegativevinte-
gers.
v
Solution.vTrue.vIfvCv isvavnonemptyvsetvofvnegativevintegers,vthen
−Cv =v {−nv :v nv ∈v Cv}
isvavnonemptyvsetvofvpositivevintegers.vIfv−avisvthevsmallestvelementv
ofv−Cv,vwhichvexistsvbyvthevLeastvIntegervAxiom,vthenv−av≤v−cvf
orvallvcv ∈vC,vsovthatvav ≥vcvforvallvcv ∈vC.
(ii) Therevisvavsequencevofv13vconsecutivevnaturalvnumbersvcontainingv
exactlyv2vprimes.
Solution.vTrue.vThevintegersv48vthroughv60vformvsuchvavsequence;v
onlyv53vandv59varevprimes.
(iii) Therev arev atv leastv twov primesv inv anyv sequencev ofv 7v consecutivev
naturalvnumbers.
Solution.v False.v Thev integersv 48v throughv 54v arev 7v consecutivev
naturalvnumbers,vandvonlyv53visvprime.
(iv) Ofvallvthevsequencesvofvconsecutivevnaturalv numbersvnotvcontainingv
2vprimes,vtherevisvavsequencevofvshortestvlength.
Solution.vTrue.vThevsetvCv consistingvofvthevlengthsvofvsuchv(finite)v
sequencesvisvavnonemptyvsubsetvofvthevnaturalvnumbers.
(v) 79visvavprime.
√v √v
Solution.vTrue.v 79v< 81v =9,vandv79visvnotvdivisiblevbyv2,v3,
5,vorv7.
(vi) Therev existsv av sequencev ofv statementsv S(1),vS(2),v... v withv S(2n)vt
ruevforvallvnv≥v1vandvwithvS(2nv−v1)vfalsevforveveryvnv≥v1.vSoluti
on.vTrue.vDefinevS(2nv−v1)vtovbevthevstatementvnv/=vn,vandvdefinev S(
2n)vtovbevthevstatementvnv =vn.
(vii) Forv allv n ≥v 0,v wev havev nv ≤v Fnv,v wherev Fnv isv thev nthv Fibonacci
number.
,2
Solution.vTrue.v Wev havev 0v =v F0,v 1v =v F1,v 1v=v F2,v andv 2v =vF
3.v Usevthevsecondvformvofvinductionvwithvbasevstepsvnv =v2vandvnv
=v3v(verifyingvthevinductivevstepvwillvshowvwhyvwevchoosevthesev
numbers).v Byvthevinductivevhypothesis,vnv—v2v≤vFn−2v andvnv—
1v Fn≤1.vHence,v2nv 3v Fnv.−Bu≤
tvnv 2nv 3vforval≤lvnv 3,− ≥
−
asvdesired.
(viii) Ifvmv andvnv arevnaturalvnumbers,vthenv(mn)!= vm!n!.
Solution.v False.v Ifvmv =v2v=vn,vthenv(mn)!= v24vandvm!n!= v4.
1.2 (i) Forvanyvnv ≥v0vandvanyvrv /=v1,vprovevthat
1v+vrv +vrv2v +vrv3v +v·v·v·v +vrnv =v (1v −vrn+1)/(1v −vr).
Solution.v Wevusevinductionvonvnv ≥v 1.v Whenvnv =v 1,v bothvsidesve
qualv1v+vrv.v Forvthev inductivev step,vnotev that
[1v +vrv +vrv2v +vrv3v +v·v·v·v +vrnv]+ v rn+1v =v (1v −vrn+1)/(1v −vr)v +vrn+1
1v −vrn+1v +v(1v −vr)rn+1
=
1v−vr
1v−v rn+2
=v
.
1v−vr
(ii)v Provevthat
1v +v2v +v22v +v·v·v·v +v2nv =v 2n+1v −v1.
Solution.v Thisv isv thev specialv casev ofv thev geometricv seriesv whenvr
v =v2;vhence,vthevsumvisv(1v−v2
n+1)/(1v−v2)v=v2n+1v−v1.vOnevcanval
sovprovevthisvdirectly,vbyvinductionvonvnv ≥v0.
1.3 Show,vforvallvnv≥v1,vthatv10nvleavesvremainderv1vaftervdividingvbyv9.vSolut
ion.v Thisv mayv bev rephrasedv tov sayv thatv therev isv anv integerv qnv withv10nv =v
9qnv+v1.vIfvwevdefinevq1v =v1,vthenv10v=vq1v+v1,vandvsovthevbasevstepvisvtrue.
Forvthevinductivevstep,vtherevisvanvintegervqnvwith
10n+1v =v10v×v10nv =v10(9qnv +v1)
=v90qnv +v10v =v9(10qnv +v1)v+v1.
Definev qn+1v =v10qnv +v1,v whichv isv anv integer.
1.4 Provevthatvifv0v≤vav ≤vb,vthenvanv ≤vbnv forvallvnv ≥v0.
Solution.vBasevstep.va0v =v1v=vb0,vandvsova0v ≤vb0.
Inductivevstep.vThevinductivevhypothesisvis
anv≤vbn.
, 3
Sincevavisvpositive,vTheoremv1.4(i)vgivesvan+1v=vaanv≤vabnv;vsincevbvisvpo
sitive,vTheoremv1.4(i)vnowvgivesvabnv ≤vbbnv =vbn+1.
1.5v Provev thatv 12v +v22v +v·v·v·v +vn2v =v 1v6n(nv +v1)(2nv +v1)v =v 1vn33v +v 1vn22v +v 1vn.6
Solution.vThevproofvis1byvin1vduc1tionvonvnv ≥v1.vWhenvnv =v1,vthevleftvsidevis
1vandvthevrightvsidevis + + =v1.
3 2 6
Forvthevinductivevstep,
[12v +v22v +v··· v +vn2v]+v (nv +v1)2v =v 13vn3v +v 12vnv2v +v 1v6nv +v(nv +v1)2
3 2
=v 1v(nv+v1)v +v1v(nv +v1)v +v1v(nv+v1),
3 2 6
aftervsomevelementaryvalgebraicvmanipulation.
1.6v Provev thatv 13v +v23v +v·v·v·v +vn3v =v 14vn4v +v 1v2n3v +v 1v4n2.
Solution.vBasevstep:vWhenvnv =v1,vbothvsidesvequalv1.
Inductivevstep:
[13v +v23v +v·v·v·v +vnv3v]+v (nv +v1)3v =v 1vn4v +v 1vn3v +v 1vn2v +v(nv +v1)3.
4 2 4
Expandingvgivesv 1vn4v +v 3vn3v +v 13vn2v +v3nv +v 1,
4 2 4
whichvis
4 (nv+v1) +v2 (nv+v1) +v 4 (nv +v1) .
1v 4v 1v 3v 1v 2
1.7v Provev thatv 14v +v24v +v·v·v·v +vn4v=v 1vn55v +v 1vn24v +v 1vn33v −vv1v n.30
Solution.v Thevproofvisvby1indu1ction1vonvn1vv≥v1.v Ifvnv−v1,v thenv thevleftvsidevis
1,vwhilevthevrightvsidevis + + − =v1vasvwell.
5 2 3 30
Forvthevinductivevstep,
14v +v24v +v·v·v·v +vn4v +v(nv +v1)4v =v 1v5n5v +v 1vn24v +v 1vn33v −v 1v 30
nv +v(nv +v1)4.
Itvisvnowvstraightforwardvtovcheckvthatvthisvlastvexpressionvisvequalvto
5v(nv+v1) +v12v(nv+v1) +v13v(nv+v1) −v 130(nv+v1).
1 5v 4v 3v v
1.8 Findvavformulavforv1 +3v+5 +·v·v·+v(2nv−1),vandvusevmathematicalvinductionvt
ovprovevthatvyourvformulavisvcorrect.
Solution.vWevprovevbyvinductionvonvnv ≥v1vthatvthevsumvisvn2.
BasevStep.vWhenvnv=v1,vwevinterpretvthevleftvsidevtovmeanv1.vOfvcourse,v
12v =v1,vandvsovthevbasevstepvisvtrue.
InductivevStep.
1v +v3v +v5v +v·v·v·v +v(2nv −v1)v +v(2nv +v1)
=v 1v +v3v +v5v +v··· v +v(2nv −v 1)]+v (2nv +v1)
=vn2v +v2nv +v1
=v(nv+v1)2.
, 4
.n
1.9 Findv avvformulav forv 1v +v jv!vjv,vvandv usev inductionv tov provevvthatvvyour
j=1
formulavisvcorrect.
Solution.vvAvvlistv ofv thevvsumsv forvvnvv1=
,vv2,vv3,vv4,vv5vvisv 2,vv6,vv24,vv120,vv720.vT
hesevarevfactorials;vbetter,vtheyvarev2v!,v3v!,v4!,v5!,v6!v.vWevhavevbeenvledvtovth
evguess
n
S(n)v:v1v+ jv!vjv=v (nv +v1)!.
jv=1
Wevnowvusevinductionvtovprovevthatvthevguessvisvalwaysvtrue.vThevbasevstepv
S(1)vhasvalreadyvbeenvchecked;vitvisvonvthevlist.vForvthevinductivevstep,vwev
mustvprove
n+1
S(nv +v1)v :v1v+ jv!vjv=v (nv +v2)!.
jv=1
Rewritevthevleftvsidevas
n
1v+ jv!vjv +v(nv+v1)!(nv+v1).
jv=1
Byvthevinductivevhypothesis,vthevbracketedvtermvisv(nv +v1)!,vandvsovthevleftvsid
evequals
(nv +v1)!+v(nv+v1)!(nv+v1)v=v(nv+v1)![1v+v(nv+v1)]
=v(nv+v1)!(nv+v2)
=v(nv+v2)!.
Byvinduction,v S(n)visvtruevforvallvnv ≥v1.
1.10 (M.vBarr)vTherevisvavfamousvanecdotevdescribingvavhospitalvvisitvofvG.vH.v
HardyvtovRamanujan.vHardyvmentionedvthatvthevnumberv1729vofvthevtaxivh
evhadvtakenvtovthevhospitalvwasvnotvanvinterestingvnumber.vRamanujanvdisa
greed,vsayingvthatvitvisvthevsmallestvpositivevintegervthatvcanvbevwrittenvasvth
evsumvofvtwovcubesvinvtwovdifferentvways.
(i) ProvevthatvRamanujan’svstatementvisvtrue.
Solution.vFirst,v1729visvthevsumvofvtwovcubesvinvtwovdifferentvwa
ys:
1729v=v13v +v123; 1927v =v93v +v103.
Second,v nov smallerv numberv nv hasv thisv property.v Ifv nv =va3vv+v b3,
3vvv 3
thenvva,v b≤12.vvItv isv nowv avvmatterv ofv checkingv allv pairsvva+ b vvforvs
uchvav andvb.
