h h
Abstract Algebra: An Inte
h h h
ractiveApproach, 2nd Edi
h h h
tion
by
William Paulsen h
**ImmediateDownload
h h
**AllChaptersincluded
h h h
,Answers to Even- h h
Numbered Problems h
Sectionh 0.1
2)h qh =h15,h rh=h12
4)h qh =h −21,h rh =h 17
6)h qh =h87,h rh=h67
8)h qh =h−1,h rh =h215
10)h1h+hnh<h1h+h(n− 1)2h =hn2h+h2(1− n)h <h n2
12)hIfh(n − 1)2h +h3(n − 1)h+h4h=h2k,h thenh n2h+h3nh+h4h=h2(kh+hnh+h1).
14)hIfh4n−1 − 1h=h3k,hthenh4n − 1h=h3(4kh+h1).
16)h(1h+hx)nh=h(1h+hx)(1h+hx)n−1 ≥ (1h+hx)(1h+(n − 1)x)h=h1h+hnxh+hx2(n− 1) ≥
1h+hnx
18)h(nh−h1)2h+h(2nh−h1)h=hn2.
20)h(nh−h1)2((nh−h1)h+h1)2/4h+hn3h =hn2(nh+h1)2/4.
22)h(nh−h1)/((nh−h1)h+h1)h+h1/(n(nh+h1))h=hn/(nh+h1).
24)h 4h·h100h+h(−11)h·h36h=h4.
26)h(−6)h·h464h+h5h·h560h=h16.
28)h(−2)h·h465h+h9h·h105h=h15.
30)h (−54)h·h(487)h+h(−221)h·h(−119)h=h1.
32)hLethch=hgcd(a,hb).h ThenhchishthehsmallesthpositivehelementhofhthehsethAh
=hallhintegershofhthehformhauh+hbv.h IfhwehmultiplyhallhelementhofhAhbyhd,hweh
geththeh seth ofh allh integersh ofh theh formh dauh+hdbv,h andh theh smallesth positiveh ele
menthofhthishsethwouldhbehdc.h Thus,hgcd(da,hdb)h=hdc.
34)h Sinceh bothh x/gcd(x,hy)h andh y/gcd(x,hy)h areh bothh integers,h weh seeh tha
th(x·h y)/gcd(x,hy)h ish ah multipleh ofh bothh xh andh y.h Ifh lcm(x,hy)h =h axh =h byh
ishsmallerhthenh(x · h y)/gcd(x,hy),hthenh(xh ·y)/lcm(x,hy)hwouldhbehgreaterhthanhg
·
cd(x,hy).h Yeth (xh y)/lcm(x, hy)h=hy/ah=hx/bh wouldh beh ah divisorh ofh bothh xh andh
y.
36)h 2h·h3h·h23h·h29.
38)h 7h·h29h·h31.
40)h 3h·h132h ·h101.
42)h uh=h−222222223,h vh=h1777777788.
44)h 34h ·h372h ·h 3336672.
Sectionh 0.2
2)hIfha/bh=hc/d,hsohthathadh=hbc,hthenhab(c2h+d2)h=habc2h+abd2h=ha2cd+b2cdh=
cd(a2h+hb2).h Thus,hab/(a2h+hb2)h=hcd/(c2h+hd2).
4)ha)h One-to-one,h 3xh+h5h=h3yh+h5h⇒hxh=hy.h b)h Onto,hfh((yh−h5)/3)h=hy.
6)ha)hOne-to-one,hx/3h−h2/5h=hy/3h−h2/5h⇒hxh=hy.hb)hOnto,hfh(3yh+h6/5)h=hy.
1
,2 Answersh toh Even-Numberedh Problems
8)h a)h One-to-one,h ifh xh >h 0,h yh <h 0h thenh yh =h 3xh >h 0.h b)h Onto,h ifh y≥ 0,
fh(y/3)h=hy.h Ifh yh <h0,h fh(y)h=hy.
10)ha)hNothone-to-onehfh(1)h=hfh(2)h=h1.h b)hOnto,hfh(2yh−h1)h=hy.
12)h a)h One-to-one,h ifh xh even,h yh odd,h thenh yh =h 2xh+h2h ish even.h b)h Noth onto,
fh(x) ̸=h 3.
14)h a)h Noth one-to-oneh fh(5)h=hfh(8)h=h24.h b)h Noth onto,h fh(x) ̸=h1.
16)hSupposehfhwerehone-to-
one,handhlethB̃h =hfh(A),hsohthathf˜h:hAh→hB̃h wouldhbehahbijection.h Byhlemma
h0.5,h|A|h=h| B̃ | , hbut h| B ˜ | h≤h|B|h<h|A|.
18)hSupposehfhwerehnothone-to-
one.hThenhtherehishahcasehwherehfh(a —1)h{h=
h hhf}h(a2),handhwehcanhconsider →hthehseth
A˜h=hAh a1h ,handhthe |h |
hfunction hf ˜ h:hA˜h Bhwouldhstill hbehonto.h Buth A˜h <h Bh
sohbyhProblemh17hf˜hcannot |h h hbehonto.h Hence,hfh is
one-to-one. |
20)hx4h+h2x2.
22)hx3h−h3x{+h2.
3xh+h14h h ifh xh ish even
24)hfh(x)h= ,
26)hIfhfh(g(x))6xh=hhf+
h(g(y)),
h2 ifhhthen sincehfh ishone-to-
xh ish hodd.
one,hg(x)h=hg(y).h Sincehghishonto,hxh=hy.
28)h Thereh ish someh ch∈hCh suchh thath fh(y) ̸=hch forh allh yh ∈hB.h Thenh fh(g(x)) ̸=hc
sinceh g(x)h∈hB.
30)hIfhxhevenhandhyhodd,hfh(x)h=hfh(y)hmeanshyh=hxh+h8hisheven.h Ontohishprov
{
xh+h3h ifhxhisheven,
enhbyhfindinghthehinverse:h fh−1(x)h=
−hh5=hhhxifh+
∗xhhz) h xh ish odd. h
hyh+hzh
32)h Associative,h (xh∗hy)h∗hzh=hxh∗h(y 2.
−
34)h Noth associative,h (xh∗hy)h∗hzh =hxh−hyh−hz,h xh∗h(yh∗hz)h=hxh−hyh+hz.
36)hYes.
38)hYes.
40)hYes.
42)hfh(x)hishbothhone-to-onehandhonto.
Sectionh 0.3
2)h 55
4)h 25
6)h 36
8)h 7
10)h 10
12)h 91
14)h 43
16)h 223
18)h 73
20)h 1498
22)h 3617
24)h 3875
26)h Firsth findh 0h ≤h qh ≤h uh·hvh suchh thath qh ≡h x(modh u)h andh qh ≡h y(modh v).h Th
enhfindh kh soh thath kh ≡h q(modh uh·hv)h andh kh ≡h z(modh w).
28)h 12
, Answersh toh Even-Numberedh Problems 3
30)h 4
32)h 35
34)h 17
36)h 30
38)h 51
40)h 3684623194282304903214
42)h 21827156424272739145155343596495185185220332
44)h 1334817563332517248
Sectionh 0.4
2)hSinceh1h+2⌊an⌋hishanhinteger,h1h+2⌊an⌋−anhwillhhavehthehsamehdenominato
rhashan.h Thus,hthehnumeratorhan+1hishthehdenominatorhofhan.h Notehthaththeh
fractionshwillhalreadyhbehinhlowesthterms.
4)hSincehthehsequencehbeginshb0h=h0,hb1h=h1,hb2h=h1,hb3h=h2,.h.h.hwehseeht
haththehequationsharehtruehforhnh=h1.h Assumehbothhequationsharehtruehfor
htheh previoush n,h thath is,h b2n−2h =h bn−1h andhb2n−1h =h bn−1h+hbn.h Thenh b
yhthehrecursionhformula,hb2nh=hbn−1h+h(bn−1h+hb— n)h 2(bn−1hmodh(bn−1h+hbn)
).hButh(bn−1hmodh(bn−1h+hbn))h=hbn−1,hsincehbn−1h+hbnh>hbn−1.h Sohb2nh=hbn.
hThenhwehcanhcompute hb2n+1h= hbn−1h+hbnh+hbnh−h2(bn−1h+hbnhmodhbn).h But
h(bn−1h+hbnhmodhbn)h=h(bn−1hmodhbn),handhbn+1h=hbn−1h+hbnh−h2(bn−1h modhbn).
hThus,hb2n+1h=hbnh+hbn+1.
6)h a2n+1h =hb2n+1/b2n+2h =h(bnh+hbn+1)/bn+1h =h(bn/bn+1)h+h1h=hanh+h1.
8)hIfhaih=hajhforhih>hj,hthenhbecausehan+1hishdeterminedhsolelyhonhan,ha2i−jh
=hai.h Inhfact,hthehsequencehwillhrepeathforever,hsohtherehwouldhbehonlyhahfinitehofh
rationalhnumbershinhthehsequence.h Buththishcontradictshthatheveryhrationalhi
shinhthehsequence,hwhichhishanhinfinitehset.
10)hInhcomputinghthehlonghdivisionhofhp/q,hthehremaindershatheachhstagehish giv
enhbyhthehsequencehinhProblemh9.h Sincehthishsequenceheventuallyhrepeats,hth
ehdigitshproducedhbyhthehlonghdivisionhalgorithmhwillheventuallyhrepeat.
12)hIfhp3/q3h=h2hwithhphandhqh coprime,hthenh2|p,hbuthreplacinghph=h2rhshow
sh2|qhtoo.
14)hIfhp2/q2h=h5hwithhphandhqh coprime,hthenh5|p,hbuthreplacinghph=h5rhshow
sh5|qhtoo.
16)hIfhp3/q3h=h3hwithhphandhqh coprime,hthenh3|p,hbuthreplacinghph=h3rhshow
sh3|qhtoo.
18)hIfh1/ah wereh rational,h thenh ah=h1/a−1h wouldh beh rational.
20)hGivenhxhandhy,hchoosehanyhirrationalhz,handhfindhahrationalhqhbetweenh— xh z
andhy — z.h Thenh qh+hzh ish irrationalh byh Problemh 19.
√ √
22)hx2h=h5h+h2 6,handh 6hishirrational,hsohx2h ishtoo.h Ifhxhwerehrational,hthen
x2h wouldh beh rational.
√ √
24)h 2 — 2h andh 2h areh bothh irrational,h buth theh sumh ish 2.
√ √
26)h a6h =h 2h+h4,h a102h =h 2h+h8.
Abstract Algebra: An Inte
h h h
ractiveApproach, 2nd Edi
h h h
tion
by
William Paulsen h
**ImmediateDownload
h h
**AllChaptersincluded
h h h
,Answers to Even- h h
Numbered Problems h
Sectionh 0.1
2)h qh =h15,h rh=h12
4)h qh =h −21,h rh =h 17
6)h qh =h87,h rh=h67
8)h qh =h−1,h rh =h215
10)h1h+hnh<h1h+h(n− 1)2h =hn2h+h2(1− n)h <h n2
12)hIfh(n − 1)2h +h3(n − 1)h+h4h=h2k,h thenh n2h+h3nh+h4h=h2(kh+hnh+h1).
14)hIfh4n−1 − 1h=h3k,hthenh4n − 1h=h3(4kh+h1).
16)h(1h+hx)nh=h(1h+hx)(1h+hx)n−1 ≥ (1h+hx)(1h+(n − 1)x)h=h1h+hnxh+hx2(n− 1) ≥
1h+hnx
18)h(nh−h1)2h+h(2nh−h1)h=hn2.
20)h(nh−h1)2((nh−h1)h+h1)2/4h+hn3h =hn2(nh+h1)2/4.
22)h(nh−h1)/((nh−h1)h+h1)h+h1/(n(nh+h1))h=hn/(nh+h1).
24)h 4h·h100h+h(−11)h·h36h=h4.
26)h(−6)h·h464h+h5h·h560h=h16.
28)h(−2)h·h465h+h9h·h105h=h15.
30)h (−54)h·h(487)h+h(−221)h·h(−119)h=h1.
32)hLethch=hgcd(a,hb).h ThenhchishthehsmallesthpositivehelementhofhthehsethAh
=hallhintegershofhthehformhauh+hbv.h IfhwehmultiplyhallhelementhofhAhbyhd,hweh
geththeh seth ofh allh integersh ofh theh formh dauh+hdbv,h andh theh smallesth positiveh ele
menthofhthishsethwouldhbehdc.h Thus,hgcd(da,hdb)h=hdc.
34)h Sinceh bothh x/gcd(x,hy)h andh y/gcd(x,hy)h areh bothh integers,h weh seeh tha
th(x·h y)/gcd(x,hy)h ish ah multipleh ofh bothh xh andh y.h Ifh lcm(x,hy)h =h axh =h byh
ishsmallerhthenh(x · h y)/gcd(x,hy),hthenh(xh ·y)/lcm(x,hy)hwouldhbehgreaterhthanhg
·
cd(x,hy).h Yeth (xh y)/lcm(x, hy)h=hy/ah=hx/bh wouldh beh ah divisorh ofh bothh xh andh
y.
36)h 2h·h3h·h23h·h29.
38)h 7h·h29h·h31.
40)h 3h·h132h ·h101.
42)h uh=h−222222223,h vh=h1777777788.
44)h 34h ·h372h ·h 3336672.
Sectionh 0.2
2)hIfha/bh=hc/d,hsohthathadh=hbc,hthenhab(c2h+d2)h=habc2h+abd2h=ha2cd+b2cdh=
cd(a2h+hb2).h Thus,hab/(a2h+hb2)h=hcd/(c2h+hd2).
4)ha)h One-to-one,h 3xh+h5h=h3yh+h5h⇒hxh=hy.h b)h Onto,hfh((yh−h5)/3)h=hy.
6)ha)hOne-to-one,hx/3h−h2/5h=hy/3h−h2/5h⇒hxh=hy.hb)hOnto,hfh(3yh+h6/5)h=hy.
1
,2 Answersh toh Even-Numberedh Problems
8)h a)h One-to-one,h ifh xh >h 0,h yh <h 0h thenh yh =h 3xh >h 0.h b)h Onto,h ifh y≥ 0,
fh(y/3)h=hy.h Ifh yh <h0,h fh(y)h=hy.
10)ha)hNothone-to-onehfh(1)h=hfh(2)h=h1.h b)hOnto,hfh(2yh−h1)h=hy.
12)h a)h One-to-one,h ifh xh even,h yh odd,h thenh yh =h 2xh+h2h ish even.h b)h Noth onto,
fh(x) ̸=h 3.
14)h a)h Noth one-to-oneh fh(5)h=hfh(8)h=h24.h b)h Noth onto,h fh(x) ̸=h1.
16)hSupposehfhwerehone-to-
one,handhlethB̃h =hfh(A),hsohthathf˜h:hAh→hB̃h wouldhbehahbijection.h Byhlemma
h0.5,h|A|h=h| B̃ | , hbut h| B ˜ | h≤h|B|h<h|A|.
18)hSupposehfhwerehnothone-to-
one.hThenhtherehishahcasehwherehfh(a —1)h{h=
h hhf}h(a2),handhwehcanhconsider →hthehseth
A˜h=hAh a1h ,handhthe |h |
hfunction hf ˜ h:hA˜h Bhwouldhstill hbehonto.h Buth A˜h <h Bh
sohbyhProblemh17hf˜hcannot |h h hbehonto.h Hence,hfh is
one-to-one. |
20)hx4h+h2x2.
22)hx3h−h3x{+h2.
3xh+h14h h ifh xh ish even
24)hfh(x)h= ,
26)hIfhfh(g(x))6xh=hhf+
h(g(y)),
h2 ifhhthen sincehfh ishone-to-
xh ish hodd.
one,hg(x)h=hg(y).h Sincehghishonto,hxh=hy.
28)h Thereh ish someh ch∈hCh suchh thath fh(y) ̸=hch forh allh yh ∈hB.h Thenh fh(g(x)) ̸=hc
sinceh g(x)h∈hB.
30)hIfhxhevenhandhyhodd,hfh(x)h=hfh(y)hmeanshyh=hxh+h8hisheven.h Ontohishprov
{
xh+h3h ifhxhisheven,
enhbyhfindinghthehinverse:h fh−1(x)h=
−hh5=hhhxifh+
∗xhhz) h xh ish odd. h
hyh+hzh
32)h Associative,h (xh∗hy)h∗hzh=hxh∗h(y 2.
−
34)h Noth associative,h (xh∗hy)h∗hzh =hxh−hyh−hz,h xh∗h(yh∗hz)h=hxh−hyh+hz.
36)hYes.
38)hYes.
40)hYes.
42)hfh(x)hishbothhone-to-onehandhonto.
Sectionh 0.3
2)h 55
4)h 25
6)h 36
8)h 7
10)h 10
12)h 91
14)h 43
16)h 223
18)h 73
20)h 1498
22)h 3617
24)h 3875
26)h Firsth findh 0h ≤h qh ≤h uh·hvh suchh thath qh ≡h x(modh u)h andh qh ≡h y(modh v).h Th
enhfindh kh soh thath kh ≡h q(modh uh·hv)h andh kh ≡h z(modh w).
28)h 12
, Answersh toh Even-Numberedh Problems 3
30)h 4
32)h 35
34)h 17
36)h 30
38)h 51
40)h 3684623194282304903214
42)h 21827156424272739145155343596495185185220332
44)h 1334817563332517248
Sectionh 0.4
2)hSinceh1h+2⌊an⌋hishanhinteger,h1h+2⌊an⌋−anhwillhhavehthehsamehdenominato
rhashan.h Thus,hthehnumeratorhan+1hishthehdenominatorhofhan.h Notehthaththeh
fractionshwillhalreadyhbehinhlowesthterms.
4)hSincehthehsequencehbeginshb0h=h0,hb1h=h1,hb2h=h1,hb3h=h2,.h.h.hwehseeht
haththehequationsharehtruehforhnh=h1.h Assumehbothhequationsharehtruehfor
htheh previoush n,h thath is,h b2n−2h =h bn−1h andhb2n−1h =h bn−1h+hbn.h Thenh b
yhthehrecursionhformula,hb2nh=hbn−1h+h(bn−1h+hb— n)h 2(bn−1hmodh(bn−1h+hbn)
).hButh(bn−1hmodh(bn−1h+hbn))h=hbn−1,hsincehbn−1h+hbnh>hbn−1.h Sohb2nh=hbn.
hThenhwehcanhcompute hb2n+1h= hbn−1h+hbnh+hbnh−h2(bn−1h+hbnhmodhbn).h But
h(bn−1h+hbnhmodhbn)h=h(bn−1hmodhbn),handhbn+1h=hbn−1h+hbnh−h2(bn−1h modhbn).
hThus,hb2n+1h=hbnh+hbn+1.
6)h a2n+1h =hb2n+1/b2n+2h =h(bnh+hbn+1)/bn+1h =h(bn/bn+1)h+h1h=hanh+h1.
8)hIfhaih=hajhforhih>hj,hthenhbecausehan+1hishdeterminedhsolelyhonhan,ha2i−jh
=hai.h Inhfact,hthehsequencehwillhrepeathforever,hsohtherehwouldhbehonlyhahfinitehofh
rationalhnumbershinhthehsequence.h Buththishcontradictshthatheveryhrationalhi
shinhthehsequence,hwhichhishanhinfinitehset.
10)hInhcomputinghthehlonghdivisionhofhp/q,hthehremaindershatheachhstagehish giv
enhbyhthehsequencehinhProblemh9.h Sincehthishsequenceheventuallyhrepeats,hth
ehdigitshproducedhbyhthehlonghdivisionhalgorithmhwillheventuallyhrepeat.
12)hIfhp3/q3h=h2hwithhphandhqh coprime,hthenh2|p,hbuthreplacinghph=h2rhshow
sh2|qhtoo.
14)hIfhp2/q2h=h5hwithhphandhqh coprime,hthenh5|p,hbuthreplacinghph=h5rhshow
sh5|qhtoo.
16)hIfhp3/q3h=h3hwithhphandhqh coprime,hthenh3|p,hbuthreplacinghph=h3rhshow
sh3|qhtoo.
18)hIfh1/ah wereh rational,h thenh ah=h1/a−1h wouldh beh rational.
20)hGivenhxhandhy,hchoosehanyhirrationalhz,handhfindhahrationalhqhbetweenh— xh z
andhy — z.h Thenh qh+hzh ish irrationalh byh Problemh 19.
√ √
22)hx2h=h5h+h2 6,handh 6hishirrational,hsohx2h ishtoo.h Ifhxhwerehrational,hthen
x2h wouldh beh rational.
√ √
24)h 2 — 2h andh 2h areh bothh irrational,h buth theh sumh ish 2.
√ √
26)h a6h =h 2h+h4,h a102h =h 2h+h8.
