Abstract Algebra - Lovet (Solu�ons Manual All Chapters, 100% Ori...

Abstract Algebra Structures and Applica�ons, 1e Stephen Lovet (Solu�ons Manual All Chapters, 100% Original Verified, A+ Grade) 1 / 4

1jSet Theory 1.1 { Sets and Functions

Exercise:1 Section 1.1

Question:LetU=fn2Njn10gand consider the subsetsA=f1;3;5;7;9g,B=f1;2;3;4;5g, and

C=f1;2;5;7;8g. Calculate the following operations.a)A\B b)B[C)A c)(A\B)\(A\C)\(B\C)

d) AB)C)\(A(BC))

Solution:We apply the denitions of set operations: a)A\B=f1;3;5g b)B[C)A=f1;2;3;4;5;7;8g f1;3;5;7;9g=f2;4;8g c)A\B\A\C\B\C=f1;3;5g\f1;5;7g\f1;2;5g=f2;4;6;7;8;9;10g\f2;3;4;6;8;9;10g\f4;6;8;9;10g

d) AB)C)\(A(BC)) = (f7;9g C)\(A f3;4g) =f9g \ f1;5;7;9g=f9g

Exercise:2 Section 1.1

Question:LetU=fa; b; c; d; e; f; ggand consider the subsetsA=fa; b; dg,B=fb; c; eg, andC=fc; d; fg.Calculate the following operations.a)C\(A[B) b)A[C)B c)A[B[C)(A\B\C) d)AB)[(BC) Solution:We apply the denitions of set operations: a)C\(A[B) =C\ fa; b; c; d; eg=fc; dg b)A[C)B=fa; b; c; d; fg B=fa; d; fg c)A[B[C)(A\B\C) =fa; b; c; d; e; fg ;=fa; b; c; d; e; fg d)AB)[(BC) =fa; dg [ fb; eg=fa; b; d; eg

Exercise:3 Section 1.1

Question:As subsets of the reals, describe the dierences between the setsf3;5g, [3;5] and (3;5).Solution:The setf3;5gcontains the integers 3 and 5. The closed interval [3;5] contains all real numbers between 3 and 5 including 3 and 5, while the open interval (3;5) contains all real numbers between 3 and 5 not including 3 and 5.

Exercise:4 Section 1.1

Question:Prove that the following are true for all setsAandB.

a)A\BA.b)AA[B.Solution:We use the denitions of subsets and the intersection and union of sets.

a) x2A\B:Thenx2Aandx2B=)x2A, soA\BA.

• x2A. We know thatA[B=fyjy2A or y2Bg, sox2A=)x2A[B. HenceAA[B.

Exercise:5 Section 1.1

Question:LetAandBbe subsets of a setS.

a) ABif and only ifP(A)P(B)

b) P(A\B) =P(A)\P(B).

c) P(A[B) =P(A)[P(B) if and only ifABorBA.

Solution:

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2CHAPTER 1. SET THEORY

• )):SupposeAB. Then,8a2A,a2B. SinceP(B) contains all the possible subsets ofB, all the

possible subsets ofAmust be inP(B) becauseAB. Therefore,P(A)P(B).((=):SupposeP(A)P(B). Then8fag 2P(A),fag 2P(B). Therefore, there must exist a subsetC ofP(B) that contains everyfagfromP(A). The subsetCleads to the conclusion that everya2A must also be inB. Therefore,AB.b)P(A\B) =fft1; t2; :::; tng jti2A; ti2Bg. This impliesftig 2P(A) andftig 2P(B).Therefore, by denition of intersection,P(A\B) =P(A)\P(B).

c) )):Suppose there are two setsAandBsuch that neitherABnorBA. Leta2ABand

b2BA. Then the setfa; bgis inP(A[B) but not inP(A) or inP(B). Therefore by the contrapositive,P(A[B) =P(A)[P(B) ifABorBA.

((=):SupposeAB. Then,A[B=BsoP(A[B) =P(B). Now supposeBA. ThenA[B=Aso

P(A[B) =P(A). Either way,P(A[B) =P(A)[P(B).

Exercise:6 Section 1.1

Question:Give the list description ofP(f1;2;3;4g).

Solution:Using the denition of a power set,

P(f1;2;3;4g) =f;;f1g;f2g;f3g;f4g;f1;2g;f1;3g;f1;4g;f2;3g;f2;4g;f3;4g;

f1;2;3g;f1;2;4g;f1;3;4g;f2;3;4g;f1;2;3;4gg:

Exercise:7 Section 1.1

Question:Give the list description offfa1; a2; : : : ; akg 2P(f1;2;3;4;5g)

a1+a2+ +ak= 8g.Solution:We need to nd all the subsets off1;2;3;4;5gwhose elements add to a total of 8. Recall that no subset has repeated elements sof4;4;gdoes not make sense. The set is

ff1;2;5g;f1;3;4g;f3;5gg:

Exercise:8 Section 1.1

Question:LetA,B, andCbe subsets of a setS.

a) AB)C= (AC)(BC).

b)A(BC).

Solution:

a)

SABCSABC

In the rst Venn diagram, the lighter shade represents (AB), and the darker shade, which overlaps some of (AB), represents (AB)C. In the second Venn diagram, the lighter shade represents (AC), while the darker shade represents (AC)(BC). We observe from the diagrams that the darker regions are equal. 3 / 4

1.1. SETS AND FUNCTIONS3

b)

SABCSABC

In the Venn diagram above, the lighter shade representsBC, and the darker shade representsA(BC).In the second diagram, the lighter region representsAB, and the darker region representsAC, which overlaps some ofAB. Thus, (AB)[(AC) =A(BC).

Exercise:9 Section 1.1

Question:LetA,B, andCbe subsets of a setS.

a) A4B=;if and only ifA=B.

b) A\(B4C) = (A\B)4(A\C).

Solution:LetA,B, andCbe subsets of a setS.

a)A4B=;. Then by denition of the symmetric dierence

(AB)[(BA) =;:

If the union of two sets is the empty set, then each of the two sets must be empty. Hence we deduce that AB=;andBA=;. Now for and two setsUandT, the identityUT=;is equivalent toUT.Hence we deduce thatABandBA. Consequently,A=B.The argument of the opposite direction is identical. Suppose thatA=B. ThenABandBA. Thus AB=;andBA=;. We deduce thatA4B= (AB)[(BA) =;.b)A\(B4C) = (A\B)4(A\C). We could use a well designed Venn diagram. We could also use a membership table which lists all possibilities of an element whether it is in or not in one of the given three sets. Here is a membership table for both side of the equality.In this table, we put an p in a column to indicate membership and nothing to indicate non-membership.Hence if there is a p in theAandCcolumn and nothing in theBcolumn, that refers to the situations of an element inA, not inBand inC.

ABC(B4C)A\(B4C)(A\B)(A\C)(A\B)4(A\C)

pppp p pp p p pp p p p pp p p pp p p p p Since theA\(B4C) and column and the (A\B)4(A\C) of this membership table are the same, then the sets are equal.

Exercise:10 Section 1.1

Question:LetSbe a set and letfAigi2Ibe a collection of subsets ofS. Prove the following.a) [ i2I Ai= \ i2I Ai.b) \ i2I Ai= [ i2I Ai.Solution:LetSbe a set and letfAigi2Ibe a collection of subsets ofS.

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