Solution Manual For Analysis with an Introduction to Proof, 5th Edition

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’SSOLUTIONSMANUALANALYSIS WITH ANINTRODUCTION TOPROOFFIFTHEDITIONSteven Lay

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2This manual is intended to accompany the 5th edition ofAnalysis with an Introduction to Proofby Steven R. Lay (Pearson, 2013).It contains solutions to nearly every exercise in the text.Thoseexercises that have hints (or answers) in the back of the book are numbered inboldprint, and the hintsare included here for reference. While many of the proofs have been given in full detail, some of themore routine proofs are only outlines.For some of the problems, other approaches may be equallyacceptable. This is particularly true for those problems requesting a counterexample. I have not triedto be exhaustive in discussing each exercise, but rather to be suggestive.Let me remind you that the starred exercises are not necessarily the more difficult ones.Theyare the exercises that are used in some way in subsequent sections.There is a table on page 3 thatindicates where starred exercises are used later.The following notations are used throughout thismanual:= the set of natural numbers {1, 2, 3, 4, …}= the set of rational numbers= the set of real numbers= “for every”= “there exists”= “such that”I have tried to be accurate in the preparation of this manual.Undoubtedly, however, somemistakes will inadvertently slip by.I would appreciate having any errors in this manual or the textbrought to my attention.Steven R. Layslay@leeuniversity.edu

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3Table of Starred ExercisesNote: The prefix P indicates a practice problem, the prefix E indicates an example, the prefix T refers to atheorem or corollary, and the absence of a prefix before a number indicates an exercise.StarredStarredExerciseLater UseExerciseLater use2.1.26T3.4.112.2.102.4.262.3.322.5.33.1.3E7.1.73.1.47.1.73.1.6E8.1.13.1.74.3.10, 4.3.15, E8.1.7, T9.2.93.1.8P8.1.33.1.244.1.7f, E5.3.73.1.273.3.143.1.30b3.3.11, E4.1.11, 4.3.143.2.6a4.1.9a, T4.2.1, 6.2.23, 7.2.16, T9.2.93.2.6bT6.3.83.2.6cT4.1.143.2.7T8.2.53.3.7T7.2.4, 7.2.33.3.127.1.14, T7.2.43.4.153.5.12, T4.3.123.4.213.5.73.5.89.2.153.6.125.5.94.1.6bE4.2.24.1.7fT4.2.7, 4.3.10, E8.1.74.1.9a5.2.10, 9.2.174.1.11E4.3.44.1.125.1.154.1.135.1.134.1.15b4.4.11, 4.4.18, 5.3.124.1.165.1.154.2.17E6.4.34.2.185.1.14, T9.1.104.3.144.4.54.4.108.2.144.4.168.3.94.4.17T8.3.35.1.146.2.85.1.16T6.2.95.1.185.2.14, 5.3.155.1.195.2.175.2.10T7.2.85.2.117.2.9b5.2.13T5.3.5, T6.1.7, 7.1.135.2.169.2.155.3.13bT6.2.8, T6.2.106.1.66. 2.14, 6.2.196.1.87.3.136.1.17b6.4.96.2.8T7.2.16.3.13d9.3.167.1.12P7.2.57.1.137.2.57.1.167.2.177.2.9aP7.3.77.2.11T8.2.137.2.157.3.207.2.20E7.3.98.1.7E8.2.68.1.88.2.138.1.13a9.3.88.2.129.2.7, 9.2.88.2.14T8.3.49.1.15a9.2.9

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Section 1.1xLogical Connectives4Analysiswith an Introduction to Proof5th Editionby Steven R. Layslay@leeuniversity.eduChapter 1 – Logic and ProofSolutions to ExercisesSection 1.1 – Logical Connectives1. (a)False: A statement may be false.(b)False: A statement cannot be both true and false.(c)True: See the comment after Practice 1.1.4.(d)False: See the comment before Example 1.1.3.(e)False: If the statement is false, then its negation is true.2. (a)False:pis the antecedent.(b)True: Practice 1.1.6(a).(c)False: See the paragraph before Practice 1.1.5.(d)False: “pwheneverq” is “ifq, thenp”.(e)False: The negation ofpºqisp~q.3.Answers in Book:(a) The 3 × 3 identity matrix is not singular.(b)The functionf(x) = sinxis not bounded on.(c)The functionfis not linear or the functiongis not linear.(d)Six is not prime and seven is not odd.(e)xis inDandf(x)t5.

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Section 1.1xLogical Connectives5(f)(an) is monotone and bounded, but (an) is not convergent.(g)fis injective, andSis not finite and not denumerable.4. (a)The functionf(x) =x2– 9 is not continuous atx= 3.(b)The relationRis not reflexive and not symmetric.(c)Four and nine are not relatively prime.(d)xis not inAandxis inB.(e)x < 7 andf(x) is in C.(f ) (an) is convergent, but (an) is not monotone or not bounded.(g)fis continuous andAis open, butf– 1(A) is not open.5.Answers in book:(a) Antecedent:Mis singular; consequent:Mhas a zero eigenvalue.(b) Antecedent: linearity; consequent: continuity.(c) Antecedent: a sequence is Cauchy; consequent: it is bounded.(d) Antecedent:y> 5; consequent:x< 3.6. (a) Antecedent: it is Cauchy; consequent: a sequence is convergent.(b) Antecedent: boundedness; consequent: convergence.(c) Antecedent: orthogonality; consequent: invertability.(d) Antecedent: K is closed and bounded; consequent: K is compact.7and 8 are routine.9. Answers in book:(a)TTisT.(b)FTisT.(c)FFisF.(d)TºTisT.(e)FºFisT.(f)TºFisF.(g)(TF)ºTisT.(h)(TF)ºFisF.(i)(TF)ºFisT.(j)~(FT)isF.10. (a)TFisF.(b)FFisF.(c)FTisT.(d)TºFisF.(e)FºFisT.(f)FºTisT.(g)(FT)ºFisF.(h)(TºF)ºTisT.(i)(TT)ºFisF.(j)~(FT)isT.11. Answers in book:(a)p~q;(b) (pq)~ (pq);(c) ~qºp;(d) ~pºq;(e)p~q.12. (a)n~m;(b) ~m~nor ~ (mn);(c)nºm;(d)mº~n;(e)~ (mn).13. (a) and (b) are routine.(c)pq.14. These truth tables are all straightforward. Note that the tables for (c) through (f) have 8 rows because there are 3letters and therefore 23= 8 possible combinations of T and F.Section 1.2 - Quantifiers1. (a)True: See the comment before Example 1.2.1.(b)False: The negation of a universal statement is an existential statement.(c)True: See the comment before Example 1.2.1.2. (a)False: It means there exists at least one.(b)True: Example 1.2.1.(c)True: See the comment after Practice 1.2.4.3.(a)No pencils are red.(b)Some chair does not have four legs.(c)Someone on the basketball team is over 6 feet 4 inches tall.(d)x> 2,f(x)z7.

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Section 1.2xQuantifiers6(e)xinAy> 2,f(y)d0 orf(y)tf(x).(f)xx> 3 andH> 0,x29 +H.4. (a)Someone does not like Robert.(b)No students work part-time.(c)Some square matrices are triangular.(d)xinB,f(x)k.(e)xx> 5 and 3df(x)d7.(f)xinAyinB,f(y)f(x).5.Hints in book:The True/False part of the answers.(a) True. Letx= 3.(b) True. 4 is less than 7 and anything smaller than 4 will also be less than 7.(c) True. Letx=5.(d) False. Choosexz5rsuch asx= 2.(e) True. Letx= 1, or any other real number.(f) True. The square of a real number cannot be negative.(g) True. Letx= 1, or any real number other than 0.(h) False. Letx= 0.6. (a) True. Letx= 5.(b) False. Letx= 3.(c) True. Choosexz3rsuch asx= 2.(d) False. Letx=3.(e) False. The square of a real number cannot be negative.(f) False. Letx= 1, or any other real number.(g) True. Letx= 1, or any other real number.(h) True.x–x=x+ (–x) and a number plus its additive inverse is zero.7.Answers in book: (a)You can use (ii) to prove (a) is true.(b)You can use (i) to prove (b) is true.Additional answers: (c) You can use (ii) to prove (c) is false.(d) You can use (i) to prove (d) is false.8. The best answer is (c).9. Hints in book:The True/False part of the answers.(a)False. For example, letx= 2 andy= 1. Thenx>y.(b)True. For example, letx= 2 andy= 3. Thenxy.(c)True. Given anyx, lety=x+ 1. Thenxy.(d)False. Given anyy, letx=y+ 1. Thenx>y.10. (a)True. Given anyx, lety= 0.(b)False. Letx= 0. Then for allywe havexy= 0z1.(c)False. Lety= 0. Then for allxwe havexy= 0z1.(d)True. Given anyx, lety= 1. Thenxy=x.11.Hints in book:The True/False part of the answers.(a)True. Letx= 0. Then given anyy, letz=y. (A similar argument works for anyx.)(b)False. Given anyxand anyy, letz=x+y+ 1.(c)True. Letz=y–x.(d)False. Letx= 0 andy= 1. (It is a true statement forxz0.)(e)True. Letxd0.(f)True.Takezdy.This makes “z!y” false so that the implication is true. Or, choosez!x+y.12. (a)True. Givenxandy, letz=x+y.(b)False. Letx= 0. Then given anyy, letz=y+ 1.(c)True. Letx= 1. Then given anyy, letz=y. (Anyxz0 will work.)(d)False. Letx= 1 andy= 0. (Anyxz0 will work.)(e)False. Letx= 2. Given anyy, letz=y+ 1. Then “z!y” is true, but “z!x+y” is false.(f)True. Given anyxandy, either choosez!x+yorzdy.

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Section 1.2xQuantifiers713.Answer in book:(a)x,f(x) =f(x);(b)xf(x)zf(x).14. (a)k!0x,f(x+k) =f(x).(b)k!0,xf(x+k)zf(x).15.Answer in book:(a)xandy,xdyºf(x)df(y).(b)xandyxdyandf(x) >f(y).16. (a)xandy,xyºf(x)!f(y).(b)xandyxyandf(x)df(y).17.Answer in book:(a)xandy,f(x) =f(y)ºx=y.(b)x andyf(x) =f(y) andxzy.18. (a)yinBxinAf(x) =y.(b)yinBxinA,f(x)zy.19.Answer in book:(a)H!0,G!0xD, |xc|Gº|f(x)f(c)|H.(b)H!0G!0,xD|xc| <Gand |f(x)f(c)|tH.20. (a)H!0G!0xandyinS, |x–y|Gº|f(x) –f(y)|F.(b)F!0E!0,xandyinS|x–y|Eand|f(x) –f(y)|tF.21.Answer in book:(a)H!0,G!0xD,0|xc|Gº|f(x)L|H.(b)H!0G!0,xD0|xc|Gand |f(x)L|tH.22. Answers will vary.Section 1.3 – Techniques of Proof: I1. (a)False:pis the hypothesis.(b)False: The contrapositive is ~qº~p.(c)False: The inverse is ~pº~q.(d)False:p(n) must be true for alln.(e)True: Example 1.3.1.2. (a)True: See the comment after Practice 1.3.4.(b)False: It’s called a contradiction.(c)True: See the comment after Practice 1.3.8.(d)True: See the end of Example 1.3.1.(e)False: Must showp(n) is true for alln.3. Answers in book: (a)If all violets are not blue, then some roses are not red.(b)IfAis invertible, then there is no nontrivial solution toAx=0.(c)Iff(C) is not connected, then eitherfis not continuous orCis not connected.4. (a)If some violets are blue, then all roses are red.(b)IfAis not invertible, then there exists a nontrivial solution toAx=0.(c)Iff(C) is connected, thenfis continuous andCis connected.5. (a)If some roses are not red, then no violets are blue.(b)IfAx=0has no nontrivial solutions, thenAis invertible.(c)Iffis not continuous orCis not connected, thenf(C) is not connected.6. For some of these, other answers are possible.(a)Letx= –4.(b)Letn=2.(c)If 0x1, thenx3x2. In particular, (1/2)3= 1/8 < 1/4 = (1/2)2.(d)An equilateral triangle.(e)n= 40 orn= 41.(f)2 is prime, but not odd.(g)101, 103, etc.(h)35+ 2 = 245 is not prime.(i)Letn= 5 or any odd greater than 3.(j)Letx= 2 andy= 18.(k) Letx= 0.(l)The reciprocal of 1 is not less than 1.(m) Letx= 0.(n)Letx=1.

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Section 1.3xTechniques of Proof: I87. (a)Supposep= 2k+ 1 andq= 2r+ 1 for integerskandr. Thenp+q= (2k+ 1) + (2r+ 1) = 2(k+r+ 1), sop+qis even.(b)Supposep= 2k+ 1 andq= 2r+ 1 for integerskandr. Thenpq= (2k+ 1)(2r+ 1) = 4kr+ 2k+ 2r+ 1 =2(2kr+k+r) + 1, sopqis odd.(c)Here are two approaches. The first mimics part (a) and the second uses parts (a) and (b).Proof 1: Supposep= 2k+ 1 andq= 2r+ 1 for integerskandr. Thenp+ 3q= (2k+ 1) + 3(2r+ 1) =2(k+ 3r+ 2), sop+ 3qis even.Proof 2: Supposepandqare both odd.By part (b), 3qis odd since 3 is odd. So by part (a),p+ 3qis even.(d)Supposep= 2k+ 1 andq= 2rfor integerskandr. Thenp+q= (2k+ 1) + 2r= 2(k+r) + 1, sop+qis odd.(e)Supposep= 2kandq= 2rfor integerskandr. Thenp+q= 2k+ 2r= 2(k+r), sop+qis even.(f)Supposep= 2k, thenpq= 2(kq), sopqis even. A similar argument applies whenqis even.(g)This is the contrapositive of part (f).(h)Hint in book:look at the contrapositive.Proof: To prove the contrapositive, supposep= 2k+ 1. Thenp2= (2k+ 1)2= 4k2+ 4k+ 1 =2(2k2+ 2k) + 1, sop2is odd.(i)To prove the contrapositive, supposep= 2k. Thenp2= (2k)2= 4k2= 2(2k2), sop2is even.8. Supposef(x1) =f(x2). That is, 4x1+ 7 = 4x2+ 7. Then 4x1= 4x2, sox1=x2.9. Answers in book:(a)rº~shypothesis~sº~tcontrapositive of hypothesis: 1.3.12(c)rº~tby 1.3.12(l)(b)~tº(~r~s)contrapositive of hypothesis: 1.3.12(c)~r~sby 1.3.12(h)~sby 1.3.12(j)(c)r~rby 1.3.12(d)~sº~vcontrapositive of hypothesis 4 [1.3.12(c)]rº~vhypothesis 1 and 1.3.12(l)~rºuhypotheses 2 and 3 and 1.3.12(l)~vuby 1.3.12(o)10. (a)~rhypothesis~rº(r~s)contrapositive of hypothesis: 1.3.12(c)r~sby 1.3.12(h)~sby lines 1 and 3, and 1.3.12(j)(b)~thypothesis~tº(~r~s)contrapositive of hypothesis: 1.3.12(c)~r~sby lines 1 and 2, and 1.3.12(h)~sby line 3 and 1.3.12(k)(c)sºrcontrapositive of hypothesis: 1.3.12(c)tºuhypothesissthypothesisruby 1.3.12(o)

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Section 1.3xTechniques of Proof: I911. Letp:The basketball center is healthy.q: The point guard is hot.r:The team will win.s:The fans are happy.t:The coach is a millionaire.u:The college will balance the budget.The hypotheses are (pq)º(rs) and (st)ºu. The conclusion ispºu.Proof:pº(pq)from the contrapositive of 1.3.12(k)(pq)º(rs)hypothesis(rs)ºsby 1.3.12(k)sº(st)from the contrapositive of 1.3.12(k)(st)ºuhypothesispºuby 1.3.12(m)Section 1.4 – Techniques of Proof: II1. (a)True: See the comment before Example 1.4.1.(b)False: Indirect proofs avoid this.(c)False: Only the “relevant” steps need to be included.2. (a)True: See the comment before Practice 1.4.2.(b)False: The left side of the tautology should be [(p~q)ºc].(c)True: See the comment after Practice 1.4.8.3. Given anyH> 0, letG=H/3. ThenGis also positive and whenever 2 –G<x< 2 +G, we have2233xHHsothat 6 –H< 3x< 6 +Hand 11 –H< 3x+ 5 < 11 +H, as required.4. Given anyH> 0, letG=H/5. ThenGis also positive and whenever 1 –G<x< 1 +G, we have1155xHHsothat 5 –H< 5x< 5 +Hand –2 –H< 5x– 7 < –2 +H. Now multiply by –1 and reverse the inequalities: 2 +H>7 – 5x> 2 –H. This is equivalent to 2 –H< 7 – 5x< 2 +H.5. Letx= 1. Then for any real numbery, we havexy=y.6. Letx= 0. Then for any real numbery, we havexy=xbecausexy= 0.7. Given any integern, we haven2+n3=n2(1 +n). Ifnis even, thenn2is even. Ifnis odd, then 1 +nis even. Ineither case, their product is even. [This uses Exercise 1.3.7(f).]8. Ifnis odd, thenn= 2m+ 1 for some integerm, son2= (2m+ 1)2= 4m2+ 4m+ 1 = 4m(m+ 1) + 1. Ifmis even,thenm= 2pfor some integerp. But thenn2= 4(2p)(m+ 1) + 1 = 8[p(m+ 1)] + 1. Son2= 8k+ 1, wherekis theintegerp(m+ 1). On the other hand, ifmis odd, thenm+ 1 is even andm+ 1 = 2qfor some integerq. But then,n2= 4m(2q) + 1 = 8(mq) + 1. In this case,n2= 8k+ 1, wherekis the integermq. In either case,n2= 8k+ 1 forsome integerk.9.Answer in book:Let2.nThen2334( 2)431,22nn§·¨¸©¹as required. The integerisunique.10. Existence follows from Exercise 3. It isnotunique.x=2 orx=1/2.11.Answer in book:Letxbe a real number greater than 5 and lety= 3x/(5x). Then 5x0 and 2x!0, soy0.Furthermore,35515155,3333(5)1535xyxxxxxyxxx§·¨¸©¹§· ¨¸©¹as required.

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Section 1.4xTechniques of Proof: II1012. Solve the quadraticy2– 6xy+ 9 = 0 to obtain2331.yxxrCall one solutionyand the otherz. The proof isorganized like Exercise 11.13.Answer in book:Suppose thatx2x6t0 andx!3. It follows that (x2)(x3)t0 and, sincex3!0, itmust be thatx2t0. That is,xt2.14. Suppose32xxdandx!2. Thenx– 2 > 0, so we can multiply both sides of the first inequality byx– 2 toobtainxd3(x– 2). Thusxd3x– 6. That is,xt3. Ifx= 2, then2xxis not defined.15.Hint in book:Suppose log27is rational and find a contradiction.Proof: Suppose log27 =a/b, whereaandbare integers. We may assume thata> 0 andb> 0. We have 2a/b= 7,which implies 2a= 7b. But the number 2ais even and the number 7bis odd, a contradiction. Thus log27 must beirrational.16. Suppose |x+ 1|d3 and consider the two cases:xt–1 andx–1. Ifx–1, thenx+ 10 so that |x+ 1| =x+ 1.This impliesx+ 13 andx2. So in this case we have –1x2.On the other hand, ifx< –1, thenx+ 1 < 0 and |x+ 1| = –(x+ 1). This leads to –(x+ 1)3,x+ 1– 3, andx–4. In this case we have –4x< –1.Combining the two cases, we get –4x2.17. (a) This proves the converse, which is not a valid proof of the original implication.(b)This is a valid proof using the contrapositive.18. (a) This is a valid proof in the form of Example 1.3.12 (p): [pº(qr)][(p~q)ºr].(b)This proves the converse, which is not a valid proof of the original implication.19. (a) Ifpqxand,mnythen,ppnqmmqnqnxysox+yis rational.(b) Ifpqxand,mnythen,pmqnxysoxyis rational.(c)Hint in book:Use a proof by contradiction.Proof: Suppose,pqxyis irrational, and suppose.rsxyThen(),prqpsqqsrsyxyxso thatyis rational, a contradiction. Thusx+ymust be irrational.20. (a)False. For example, letx=2andy=2.Thenx+y= 0.(b)True. This is the contrapositive of Exercise 7(a).(c)False. For example, letx=y=2 .Thenxy= 2.(d)True. This is the contrapositive of Exercise 13(b).21. (a) Letx= 0 and2 .yThenxy= 0.(b)xyxyzwhenx= 0.(c) Ifxz0, then the conclusion is true.

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Section 1.4xTechniques of Proof: II1122. It is false as stated, and a counterexample isx=2 .Sincexis negative, it’s square root is not real, and hencenot irrational. (Every irrational number is a real number.) Ifxt0, then the result is true, and can be established bylooking at the contrapositive. (In order to use the contrapositive, you have to know that the negation of “xisirrational” is “xis rational.” This is only true if you know thatxis a real number.)23. Hint in book:Find a counterexample.Solution:62+ 82= 102or (2)2+ 02= 22.24. Ifa,b,care consecutive odd integers, thena= 2k+ 1,b= 2k+ 3, andc= 2k+ 5 for some integerk. Supposea2+b2=c2. Then (2k+ 1)2+ (2k+ 3)2= (2k+ 5)2. Whence 4k2– 4k– 15 = 0 andk=5/2ork=3/2. Thiscontradictskbeing an integer.25. It is true. We may label the numbers asn,n+ 1,n+ 2,n+ 3, andn+ 4. We have the sumS=n+ (n+ 1) + (n+ 2) + (n+ 3) + (n+ 4) = 5n+ 10 = 5(n+ 2),so thatSis divisible by 5.26. It is true. We may label the numbers asn,n+ 1,n+ 2, andn+ 3. We have the sumS=n+ (n+ 1) + (n+ 2) + (n+ 3) = 4n+ 6 = 2(2n+ 3).Since 2n+ 3 is odd,Sis not divisible by 4.27.Hint in book:Consider two cases depending on whethernis odd or even.Proof: Ifnis odd, thenn2and 3nare both odd, son23nis even. Thusn23n8 is even. Ifnis even, then so aren2, 3n, andn23n8. (This uses Exercise 1.3.7.)28. False. Letn= 1. Thenn24n8 = 13. Any other oddnwill also work.29. Hint in book:Letz=22.Ifzis rational, we are done. Ifzis irrational, look at2.zSolution: We have2(22)2222222.z§·¨¸©¹30. Counterexample: letx= 1/3 andy= –8. Then (1/3)– 8= 38is a positive integer and (–8)1/3is a negative integer.31. Supposex!0. Then (x+ 1)2=x2+ 2x+ 1!x2+ 1, so the first inequality holds. For the second inequality,consider the difference 2(x2+1)(x+1)2. We have2(x2+1)(x+1)2= 2x2+ 2x22x1 =x22x+ 1 = (x1)2t0,for allx, so the second inequality also holds.

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Section 2.1xBasic Set Operations12Analysiswith an Introduction to Proof5th Editionby Steven R. Layslay@leeuniversity.eduChapter 2 – Sets and FunctionsSolutions to ExercisesThe following notations are used throughout this document:= the set of natural numbers {1, 2, 3, 4, …}= the set of rational numbers= the set of real numbers= “for every”= “there exists”= “such that”Section 2.1 – Basic Set Operations1. (a)True: Definition 2.1.3.(b)False:is the set of positive integers.(c)True: Example 2.1.5.(d)True: Theorem 2.1.7.This work is protected by United States copyright laws and is provided solely forthe use of instructors in teaching their courses and assessing student learning.Dissemination or sale of any part of this work (including on the World Wide Web)will destroy the integrity of the work and is not permitted. The work and materialsfrom it should never be made available to students except by instructors usingthe accompanying text in their classes. All recipients of this work are expected toabide by these restrictions and to honor the intended pedagogical purposes andthe needs of other instructors who rely on these materials.

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Section 2.1xBasic Set Operations132. (a)False:AB=meansAandBare disjoint.(b)True: Definition 2.1.8.(c)False:xA\BmeansxAandxB.(d)False: This is OK to use sinceSbeing nonempty is the only nontrivial case.3.Answer in book:(a) True;(b) False;(c) True;(d) True;(e) False;(f) False;(g) False;(h) True.4. (a){6, 8};(b) {2, 4, 6, 8, 10};(c) {2, 4};(d) {6, 8};(e) {10};(f) {5, 7, 10};(g);(h) {5, 7}.5.The answer to (a) is in the book.Part (b) is similar.6. (a)U;(b);(c)AB;(d)AB;(e)A;(f)A.7. (a) The diagram is the same as (AB)\(AB);(b);(c)A;(d)U\A.8. (a) True.is a subset of every set.(b) True.is an element ofS.(c) True.is an element ofS, so {}S.(d) True. {} is an element ofS.9, 10, and 11 are routine.12. Beginning sentence: “LetxA” or “SupposexA” or “IfxA…” You would have to show thatxB.13. Beginning sentence: “LetxA” or “SupposexA” or “IfxA…” You would have to show thatxB.14. a, b, d.15. a, c.16. b, d.17. a, b, c, d.18. This is routine.19.Hint in book:Suppose thatU=ABandAB=. To showAU\B, letxA. ThenxB.(Why?) Sincexis not inB, where is it?On the other hand, to showU\BA, supposexU\B. Expand what it means forxto be inU\B, andcombine this with one of the original hypotheses to conclude thatxA.20. This is similar to Exercise 19.21. True. Both are equal toAB. Here is one of the proofs: IfxAB, thenxAandxB. ThusxA\B,soxA\(A\B). Conversely, ifxA\(A\B), thenxAandxA\B. IfxB, then sincexA,xA\B,a contradiction. ThusxBand soxAB.22. False. The left side isAand the right side isB.23 and 24 are routine.25.(a) Answer in book:[1, 2],{1};BBBB*BB(b)*B= (1,2),B=;(c)*B= [2,f),B= {2};(d)*B= [0,5),B= [2,3].26 is routine.Section 2.2 - Relations1. (a)True: Theorem 2.2.2.(b)False: “ordered” subset is meaningless.

Solution Manual For Analysis with an Introduction to Proof, 5th Edition - Page 15 preview image

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Section 2.2xRelations14(c)True: Theorem 2.2.15.(d)False: True for an equivalence relation, but not in general.2. (a)False: It should be (a,b).(b)True: Definition 2.2.9.(c)False: {Ex:xS} determines the partition.(d)True: Definition 2.2.12.3. (a,a) = {{a}, {a,a}}, but {a,a} = {a}, so (a,a) = {{a}, {a}} = {{a}}.4. {a}u{a} = {(x,y) :x{a} andy{a}} = {(a,a)}. But this is equal to {{{a}}} by Exercise 3.5. (2,3)(3,2) = {{2,3}}.6.AuB=.7., {(a,1)}, {(a,2)}, {(a,3)}, {(a,1), (a,2)}, {(a,1), (a,3)}, {(a,2), (a,3)}, {(a,1), (a,2), (a,3)}8. (a) 4;(b)24= 16;(c)29= 512.9. The answers are in the back of the book.10. (a)False. LetA= {1} andB= {2}. ThenAuB= {(1,2)} andBuA= {(2,1)}.(b)True;(c) True;(d) False. See Practice 2.2.6 for a counterexample.11.(a)Answer in book:reflexive, transitive(b)reflexive and transitive(c)reflexive (if everyone likes themselves)(d)reflexive and transitive(e)Answer in book:all three(f)symmetric(g)symmetric and transitive(h)reflexive and symmetric12. Examples with the setS= {1, 2, 3}:(a)reflexive only: {(1,1), (2,2), (3,3), (1,2), (2,3)}(b)symmetric only: {(1,2), (2,1)}(c)transitive only: {(1,2), (2,3), (1,3)}(d)all but transitive: {(1,1), (2,2), (3,3), (1,2), (2,3), (2,1), (3,2)}(e)all but symmetric: {(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)}(f)all but reflexive: {(1,1), (2,2)} orExamples from other sets:(a)See Exercise 5(c).(b)Perpendicular lines in the plane; numbers whose gcd is 1;xzy; |x–y|!1.(c)xy(d)|x–y|1(e)xdy;xdividesy;AB.(f)LetS= {2} and definexRyiffxy; letS=andR= {(1,1), (2,2), (3,3)}.13.Answer in book:(, )a bEis a vertical line through the point (a,b).

Solution Manual For Analysis with an Introduction to Proof, 5th Edition - Page 16 preview image

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Section 2.2xRelations1514. Rewriting the equation asa–b=c–dmakes it easier to verify the equivalence relation properties.E(7,3)is the linex–y= 4.15. (a) The partitionPis a family of parallel lines of the formx+ 2y=k.(b)E(5,3)is the linex+ 2y= 11.16. (a) The partitionPis a family of parallel lines of the formy= 3x+k.(b)E(2,5)is the liney= 3x– 1.17.Ris an equivalence relation.P={{1}, {2,3}}.18.Ris not symmetric.19.Answer in book:R= {(a,a), (b,b), (c,c), (d,d), (b,c), (c,b)}.20.R= {(a,a), (b,b), (c,c), (d,d), (b,c), (c,b), (b,d), (d,b), (c,d), (d,c)}.21.P= {{a,c}, {b}, {d}, {e}}.22.P= {{a,b,d}, {c}, {e}}.23. The verification thatRis an equivalence relation is straightforward.E5is the set of odd numbers. There are twodistinct classes.24. The verification thatRis an equivalence relation is straightforward.E5is the set of integers of the form 3k+ 2 forsomek'. There are 3 distinct classes.25. It is an equivalence relation.26. It is not reflexive and not transitive.27. (a) is routine;(b) {(9,2), (2,9), (6,3), (3,6), (1,18), (18,1)};(c){(1,3), (3,1)}.(d){(1,4), (4,1), (2,2)}(e) {(1,8), (8,1), (2,4), (4,2).(f)E(9, 2)is the hyperbolaxy= 18.28. (a) is routine;(b)E(9,2)= {(3,4), (9,2), (81,1)};(c){(5,2), (25,1)};(d){(2,6), (4,3), (8,2), (64,1)}.29. (a)ab=ba, soRis reflexive. Ifay=bx, thenxb=ya, soRis symmetric. For the transitive property, it is easier toreplaceay=bxwith the equivalent conditiona/b=x/y. Clearly, ifa/b=x/y, andx/y=c/d, thena/b=c/d, soRis transitive.(b) Each equivalence class consists of ordered pairs of numbers where the ratio of the first number to the secondnumber is the same for all pairs in the class.30. (a)R= {(1,1), (2,2), (3,3)}. There are three equivalence classes andP= {{1}, {2}, {3}}.(b)R= {(1,1), (2,2), (3,3), (1,2), (2,1)}. There are two equivalence classes andP={{1,2}, {3}}.(c)R= {(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2), (1,3), (3,1)}. There is one equivalence class andP= {{1,2,3}}.31. (a)True: IfRandSare reflexive, thenxA, (x,x)Rand (x,x)S. Hence (x,x)RSandRSisreflexive.(b)True: If eitherRorSis reflexive, thenRSis reflexive. For example, ifRis reflexive, thenxA,(x,x)R. But then (x,x)RS.(c)Answer in book:If (x,y)RS, then (x,y)Rand (x,y)S. SinceRandSare symmetric, this implies(y,x)Rand (y,x)S. Thus (y,x)RS.

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