SOLUTION MANUAL
First Course in Abstract Algebra A
8th Edition by John B. Fraleigh
All Chapters Full Complete
1. Sets and Relations
√ √
1. { 3, − 3} 2. The set is empty.
3. {1, −1, 2,−2, 3, −3,4,−4, 5, −5, 6, −6,10, −10,
...
SOLUTION MANUAL
First Course in Abstract Algebra A
8th Edition by John B. Fraleigh
All Chapters Full Complete
1. Sets and Relations
√ √
1. { 3, − 3} 2. The set is empty.
3. {1, −1, 2,−2, 3, −3,4,−4, 5, −5, 6, −6,10, −10,12, −12,15, −15,20, −20,30, −30,
60, −60}
4. {−10,−9,−8,−7,−6,−5, −4,−3, −2,−1,0, 1, 2,3, 4,5, 6,7, 8, 9,10,11}
5. It is not a well-defined set. (Some may argue that no element of Z
+
is large, because every element
exceeds only a finite number of other elements but is exceeded by an infinite number of other elements.
Such people might claim the answer should be ∅.)
6. ∅ 7. The set is ∅ because 3
3
= 27 and 4
3
= 64.
8. It is not a well-defined set. 9. Q
10. The set containing all numbers that are (positive, negative, or zero) integer multiples of 1, 1/2, or
1/3.
11. {(a, 1), (a, 2), (a, c), (b, 1), (b, 2), (b, c), (c, 1), (c, 2), (c, c)}
12. a. It is a function. It is not one-to-one since there are two pairs with second member 4. It is not onto
B because there is no pair with second member 2.
b. (Same answer as Part(a).)
c. It is not a function because there are two pairs with first member 1.
d. It is a function. It is one-to-one. It is onto B because every element of B appears as second
member ofsome pair.
e. It is a function. It is not one-to-one because there are two pairs with second member 6. It is not
onto B because there is no pair with second member 2.
f. It is not a function because there are two pairs with first member 2.
13. Draw the line through P and x, and let y be its point of intersection with the line segment CD.
14. a. φ : [0, 1] → [0, 2] where φ(x) = 2x b. φ : [1, 3] → [5, 25] where φ(x) = 5 + 10(x − 1)
c. φ : [a, b] → [c, d] where φ(x) = c +
d−
c
(x − a)
b a
15. Let φ : S → R be defined by φ(x) = tan(π(x − )
2
).
16. a. ∅; cardinality 1 b. ∅, {a}; cardinality 2 c. ∅,{a},{b},{a, b}; cardinality 4
d. ∅,{a},{b},{c},{a, b},{a, c},{b, c},{a, b, c}; cardinality 8
17. Conjecture: |P(A)| = 2
s
= 2|
A|
.
Proof The number of subsets of a set A depends only on the cardinality of A, not on what the
elements of A actually are. Suppose B = {1, 2, 3, · · · ,s − 1} and A = {1, 2, 3, ,s}. Then A has all
the elements of B plus the one additional element s. All subsets of B are also subsets of A; these
are precisely the subsets of A that do not contain s, so the number of subsets of A not containing
s is |P(B)|. Any other subset of A must contain s, and removal of the s would produce a subset of
B. Thus the number of subsets of A containing s is also |P(B)|. Because every subset of A either
contains s or does not contain s (but not both), we see that the number of subsets of A is 2|P(B)|.
1
2 0. Sets and Relations
We Qrhave Qrshown Qrthat Qrif QrA Qrhas Qrone Qrmore Qrelement Qrthat QrB, Qrthen Qr|P(A)| Qr= Qr2|P(B)|.
QrNowQr|P(∅)| Qr= Qr1, Qrso Q r if Qr|A| Qr= Qrs, Qrthen Qr|P(A)| Qr= Qr2 Q r .
s
18. We Qrdefine Q ra Q rone-to-one Q rmap Q rφ Q rof Q rB
A Q ronto Q rP(A). Q rLet Q r f Q r∈ Q rB
A
, Q r and Q r let Q rφ(f Q r)
Q r= Q r {x Q r∈ Q rA Q r| Q r f Q r(x) Q r= Q r 1}. Q r Suppose Qrφ(f Qr) Qr= Q rφ(g). Qr QrThen Q r f Qr(x) Q r= Q r1 Q r if Q rand Q ronly
Q r if Q rg(x) Q r= Q r1. Qr QrBecause Q r the Q ronly Q rpossible Q rvalues Q r for Q r f Qr(x) Qrand Qrg(x) Qrare Qr0 Qrand Qr1,
Qrwe Qrsee Qrthat Qrf Qr(x) Qr= Qr0 Qrif Qrand Qronly Qrif Qrg(x) Qr= Qr0. Q r Consequently Qrf Qr(x) Qr= Qrg(x) Qrfor Q r all
Qrx Qr∈ QrA Qrso Qrf Q r= Qrg Qrand Qrφ Qris Qrone Qrto Qrone. Q r To Qrshow Qrthat Qrφ Qris Qronto QrP(A), Qrlet QrS Qr⊆ QrA,
Qrand Qrlet Qrh Qr: QrA Qr→ Qr{0, Qr1} Q r be Qrdefined Qrby Qrh(x) Qr= Qr1 Qrif Qrx Qr∈ QrS Qrand Qrh(x) Qr= Qr0
Qrotherwise. Q r Clearly Qrφ(h) Qr= QrS, Qrshowing Qrthat Qrφ Qris Qrindeed Q r onto Qr Q rP(A).
19. Picking Qrup Qrfrom Qrthe Q rhint, Q rlet Q rZ Q r= Qr{x Qr∈ QrA Qr| Qrx Qr∈/ Q rφ(x)}. Q r We Q rclaim Q rthat Q rfor Q rany
Q ra Qr∈ QrA, Qrφ(a) Qr/= Q rZ. Q r Either Qra Q r∈ Q rφ(a), Q rin Qrwhich Q r case Q ra Q r∈/ QrZ, Q ror Q ra Q r∈/ Qrφ(a), Q r in
Q rwhich Q r case Q ra Q r∈ Q rZ. Q rThus Q rZ Q r and Q rφ(a) Q rare Q r certainly QrdifferentQrsubsetsQrofQrA;QroneQrof
QrthemQrcontainsQraQrandQrtheQrotherQroneQrdoesQrnot.
Based Qron Qrwhat Qrwe Qrjust Qrshowed, Qrwe Qrfeel Qrthat Qrthe Qrpower Qrset Qrof QrA Qrhas Qrcardinality
Qrgreater Qrthan Qr|A|. Q r Proceeding Qrnaively, Q rwe Qrcan Qrstart Qrwith Qrthe Qrinfinite Qrset Q rZ, Q r form Qrits
Qrpower Qrset, Q r then Qrform Qrthe Qrpower Qrset Q r of Qrthat, Qrand Qrcontinue Q rthis Q rprocess Q r indefinitely.
Q r If Qrthere Qrwere Q ronly Qra Q rfinite Q rnumber Qrof Qrinfinite Q rcardinal Q r numbers, Qrthis Qrprocess Qrwould
Qrhave Qrto Qrterminate Qrafter Qra Qrfixed Qrfinite Q rnumber Qrof Qrsteps. Q r Since Qrit Qrdoesn’t, Q r it Qrappears
Qrthat Qrthere Qrmust Qrbe Qran Qrinfinite Qrnumber Qrof QrdifferentQrinfinite Qrcardinal Qrnumbers.
The Q rset Q r of Q r everything Q r is Q r not Q r logically Q r acceptable, Q r because Q r the Q rset Q r of Q r all
Q rsubsets Q r of Q r the Q rset Q r of Q r everything Qrwould Qrbe Qrlarger Qrthan Qrthe Qrset Qrof Qreverything, Qrwhich Qris
Qra Qrfallacy.
20. a. Q r The Qrset Qrcontaining Qrprecisely Qrthe Qrtwo Qrelements Qrof QrA Qrand Qrthe Qrthree Qr(different)
Qrelements Qrof QrB Qris
C Qr= Qr{1,Qr2,Qr3,Qr4,Qr5} Qrwhich Qrhas Qr5 Qrelements.
i) Q r Let Q rA Q r = Q r {−2,Qr−1,Qr0} Q r and Q r B Q r = Q r {1,Qr2,Qr3,Qr·Qr·Qr·} Q r = Q r Z
+
. Qr Qr QrThen Q r |A| Q r =
Q r 3 Q r and Q r |B| Q r = Q r ℵ0, Q r and Q r A
and QrB Q rhave Qrno Qrelements Qrin Qrcommon. Q r The Qrset QrC Q rcontaining Qrall Qrelements Qrin Qreither
QrA Qror QrB Qris QrC Q r=
{−2, Qr−1, Qr0, Qr1, Qr2, Qr3, Qr· Qr· Qr·}. Q r The Qrmap Qrφ Qr: QrC Q r→ QrB Q rdefined Q rby Q rφ(x) Qr= Qrx Qr+ Qr3 Q ris Q rone Q rto
Q rone Q rand Q ronto Q rB, Q rso
|C|Qr= Qr|B| Qr= Qrℵ0. Q rThus Qrwe Qrconsider Qr3Qr+Qrℵ0 Q r= Qrℵ0.
ii) Qr QrLet Q rA Q r = Q r {1,Qr2,Qr3,Qr·Qr·Qr·} Q r and Q r B Qr Qr= Q r {1/2,Qr3/2,Qr5/2,Qr·Qr·Qr·}. Then Q r |A| Qr Qr=
Qr Qr|B| Qr Qr= Qr Qrℵ0 Qr Qr Qrand Q r A Q r and
B Q rhave Q rno Q relements Q r in Q rcommon. Qr Q r The Q rset Q rC Q r containing Q rall Q relements Q r in
Q reither Q rA Q rof Q rB Q r is Q rC Q r =
{1/2,Qr1,Qr3/2,Qr2,Qr5/2,Qr3,Qr·Qr·Qr·}. Qr Q r The Q r map Q r φ Q r : Q r C Qr Qr→Qr QrA Q r defined Q r by Q r φ(x) Q r = Q r 2x Q r is
Q r one Q r to Q r one Q r and Q r onto Q r A, Q rso Q r |C| Qr= Qr|A| Qr= Qrℵ0. Q r Thus Q r we Q r consider Q r ℵ0 Q r + Qrℵ0
Qr Q r = Qrℵ0.
b. Q rWe Qrleave Qrthe Qrplotting Qrof Qrthe Qrpoints Qrin QrAQr×QrB Qrto Qryou. Q rFigure Qr0.14 Qrin Qrthe Qrtext, Qrwhere
Qrthere Qrare Qrℵ0
rowsQreachQrhavingQrℵ0 Q rentries,QrillustratesQrthat Qrwe QrwouldQrconsider QrthatQrℵ0 Qr·Qrℵ0 Qr=Qrℵ0.
21. There Qrare Qr102 Qr
= Qr100 Qrnumbers Qr(.00 Qrthrough Qr.99) Qrof Qrthe Qrform Qr.##, Qrand Qr105 Q r
= Qr100, Qr000
Qrnumbers Qr(.00000 Q r through Qr.99999) Qrof Qrthe Qrform Qr.#####. QrThus Qrfor Qr.##### Qr· Qr· Qr·, Qrwe
Qrexpect Qr10ℵ0 Q rsequences Qrrepresenting Q r all Qrnumbers Qrx Qr∈ QrR Qrsuch Q rthat Qr0 Qr≤ Qrx Qr≤ Qr1,
0
(2 )
)
Qrbut Qra Qrsequence Qrtrailing Qroff Qrin Qr0’s Qrmay Qrrepresent Qrthe Qrsame Q r x Qr∈ QrR Qras Qra Qrsequence
Qrtrailing Qrof Qrin Qr9’s. Q rAt Qrany Qrrate, Qrwe Qrshould Qrhave Qr10ℵ0 Q r≥Qr|[0,Qr1]|Qr= Qr|R|; Qrsee QrExercise
15. Q r On Qrthe Qrother Qrhand, Qrwe Q rcan Qrrepresent Qrnumbers Qrin Q rR Qrusing Qrany Qrinteger Qrbase
Q rn Q r> Q r1, Qrand Qrthese Q r same Qr10ℵ0 Qrsequences Qrusing Qrdigits Qrfrom Qr0 Qrto Qr9 Qrin Qrbase Qrn Qr= Qr12
Qrwould Qrnot Qrrepresent Qrall Qrx Qr∈ Qr[0, Qr1], Qrso Qrwe Q r have Qr10ℵ0 Q r≤ Qr|R|. Q rThus Qrwe Qrconsider Qrthe
Qrvalue Qrof Qr10ℵ0 Qrto Qrbe Qr|R|. Q rWe Qrcould Qrmake Qrthe Qrsame Qrargument Q r using Q rany Q rother
Q rinteger Q rbase Q rn Q r> Q r1, Q r and Q r thus Q rconsider Q rnℵ0 Qr= Q r|R| Q r for Q rn Q r∈ Q rZ
+
, Qrn Q r> Q r1. QrIn
Q rparticular, Q r 12ℵ0 Qr Q r= Q r2ℵ0 Qr Q r= Q r|R|.
[Show More]
