Solution Manual for Discrete Mathematics, 8th Edition

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’SSOLUTIONSMANUALDISCRETEMATHEMATICSEIGHTHEDITIONRichard JohnsonbaughDePaul University, Chicago

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’s Solutions Manual,Discrete Mathematics, 8eiContentsChapter 11Chapter 214Chapter 339Chapter 456Chapter 577Chapter 685Chapter 7115Chapter 8136Chapter 9158Chapter 10182Chapter 11187Chapter 12196Appendices208

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Solutions to Selected ExercisesSection 1.12.{2,4}3.{7,10}5.{2,3,5,6,8,9}6.{1,3,5,7,9,10}8.A9.∅11.B12.{1,4}14.{1}15.{2,3,4,5,6,7,8,9,10}18.{n∈Z+|n≥6}19.{2n−1|n∈Z+}21.{n∈Z+|n≤5orn= 2m, m≥3}22.{2n|n≥3}24.{1,3,5}25.{n∈Z+|n≤5orn= 2m+ 1, m≥3}27.{n∈Z+|n≥6orn= 2 orn= 4}29. 130. 333. We find thatB={2,3}. SinceAandBhave the same elements, they are equal.34. Letx∈A. Thenx= 1,2,3. Ifx= 1, since 1∈Z+and 12<10, thenx∈B. Ifx= 2, since 2∈Z+and22<10, thenx∈B. Ifx= 3, since 3∈Z+and 32<10, thenx∈B. Thus ifx∈A, thenx∈B.Now suppose thatx∈B. Thenx∈Z+andx2<10. Ifx≥4, thenx2>10 and, for these values ofx,x /∈B. Thereforex= 1,2,3. For each of these values,x2<10 andxis indeed inB. Also, for each ofthe valuesx= 1,2,3,x∈A. Thus ifx∈B, thenx∈A. ThereforeA=B.37. Since (−1)3−2(−1)2−(−1) + 2 = 0,−1∈B. Since−1/∈A,A6=B.38. Since 32−1>3, 3/∈B. Since 3∈A,A6=B.41. Equal42. Not equal45. Letx∈A. Thenx= 1,2. Ifx= 1,x3−6x2+ 11x= 13−6·12+ 11·1 = 6.Thusx∈B. Ifx= 2,x3−6x2+ 11x= 23−6·22+ 11·2 = 6.Againx∈B. ThereforeA⊆B.46. Letx∈A. Thenx= (1,1) orx= (1,2). In either case,x∈B. ThereforeA⊆B.49. Since (−1)3−2(−1)2−(−1) + 2 = 0,−1∈A. However,−1/∈B. ThereforeAis not a subset ofB.50. Consider 4, which is inA. If 4∈B, then 4∈Aand 4 +m= 8 for somem∈C. However, the only valueofmfor which 4 +m= 8 ism= 4 and 4/∈C. Therefore 4/∈B.Since 4∈Aand 4/∈B,Ais not asubset ofB.

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2SOLUTIONS53.UAB54.UAB56.UABC57.UABC59.UABC62. 3263. 10565. 5167. Suppose thatnstudents are taking both a mathematics course and a computer science course.Then4nstudents are taking a mathematics course, but not a computer science course, and 7nstudents aretaking a computer science course, but not a mathematics course.The following Venn diagram depictsthe situation:

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SOLUTIONS3&%'$&%'$4nn7nMathCompSciThus, the total number of students is4n+n+ 7n= 12n.The proportion taking a mathematics course is5n12n=512,which is greater than one-third.69.{(a,1),(a,2),(b,1),(b,2),(c,1),(c,2)}70.{(1,1),(1,2),(2,1),(2,2)}73.{(1, a, a),(2, a, a)}74.{(1,1,1),(1,2,1),(2,1,1),(2,2,1),(1,1,2),(1,2,2),(2,1,2),(2,2,2)}77. Vertical lines (parallel) spaced one unit apart extending infinitely to the left and right.79. Consider all points on a horizontal line one unit apart. Now copy these points by moving the horizontallinenunits straight up and straight down for all integersn >0. The set of all points obtained in thisway is the setZ×Z.80. Ordinary 3-space82. Take the lines described in the instructions for this set of exercises and copy them by movingnunits outand back for alln >0. The set of all points obtained in this way is the setR×Z×Z.84.{1,2}{1},{2}85.{a, b, c}{a, b},{c}{a, c},{b}{b, c},{a}{a},{b},{c}88. False89. True91. False92. True94.∅,{a},{b},{c},{d},{a, b},{a, c},{a, d},{b, c},{b, d},{c, d},{a, b, c},{a, b, d},{a, c, d},{b, c, d},{a, b, c, d}. All except{a, b, c, d}are proper subsets.95. 210= 1024; 210−1 = 102398.B⊆A99.A=U102. The symmetric difference of two sets consists of the elements in one or the other but not both.103.A4A=∅,A4A=U,U4A=A,∅4A=A105. The set of primes

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4SOLUTIONSSection 1.22. Is a proposition. Negation: 6 + 96= 15.3. Not a proposition4. Is a proposition. Negation:π6= 3.14.6. Is a proposition. Negation: For every positive integern, 193406=n·17.7. Is a proposition. Negation: Audrey Meadows was not the original “Alice” in the “Honeymooners.”9. Is a proposition. Negation: The line “Play it again, Sam” does not occur in the movieCasablanca.10. Is a proposition. Some even integer greater than 4 is not the sum of two primes.12. Not a proposition. The statement is neither true nor false.13. No heads were obtained.15. No heads or no tails were obtained.18. True19. True21. False22. False24.pq(¬p∨ ¬q)∨pTTTTFTFTTFFT25.pq(p∨q)∧ ¬pTTFTFFFTTFFF27.pq(p∧q)∨(¬p∨q)TTTTFFFTTFFT28.pqr¬(p∧q)∨(r∧ ¬p)TTTFTTFFTFTTTFFTFTTTFTFTFFTTFFFT

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SOLUTIONS530.pqr¬(p∧q)∨(¬q∨r)TTTTTTFFTFTTTFFTFTTTFTFTFFTTFFFT32.¬(p∧q). True.33.p∨ ¬(q∧r). True.35. Lee takes computer science and mathematics.36. Lee takes computer science or mathematics.38. Lee takes computer science but not mathematics.39. Lee takes neither computer science nor mathematics.41. You do not miss the midterm exam and you pass the course.42. You play football or you miss the midterm exam or you pass the course.44. Either you play football and you miss the midterm exam or you do not miss the midterm exam and youpass the course.46. It is not Monday and either it is raining or it is hot.47. It is not the case that today is Monday or it is raining, and it is hot.50. Today is Monday and either it is raining or it is hot, and it is hot or either it is raining or today isMonday.51.p∧q52.p∧ ¬q54.p∨q55. (p∨q)∧ ¬p57.p∧r∧q58. (p∨r)∧q60. (q∨ ¬p)∧ ¬r62.p∧ ¬r63.p∧q∧r65.¬p∧ ¬q∧r66.¬(p∨q∨ ¬r)67.pqpexorqTTFTFTFTTFFF69. Inclusive-or70. Inclusive-or72. Exclusive-or73. Exclusive-or77.”lung disease” -cancer78.”minor league” baseball team illinois -”midwest league”

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6SOLUTIONSSection 1.32. If Rosa has 160 quarter-hours of credits, then she may graduate.3. If Fernando buys a computer, then he obtains $2000.5. If a person gets that job, then that person knows someone who knows the boss.6. If you go to the Super Bowl, then you can afford the ticket.8. If a better car is built, then Buick will build it.9. If the chairperson gives the lecture, then the audience will go to sleep.11. If the switch is not turned properly, then the light will not be on.13. Contrapositive of Exercise 2: If Rosa does not graduate, then she does not have 160 quarter-hours ofcredits.15. False16. False18. False19. True21. True22. True24. Unknown25. Unknown27. True28. Unknown30. Unknown31. Unknown34. True35. True37. True38. False40. True41. False44. (p∧r)→q45.¬((r∧ ¬q)→r)48. (¬p∨ ¬r)→ ¬q49.r→q51.q→(p∨r)52. (q∧p)→ ¬r54. If it is not raining, then it is hot and today is Monday.55. If today is not Monday, then either it is raining or it is hot.57. If today is Monday and either it is raining or it is hot, then either it is hot, it is raining, or today isMonday.58. If today is Monday or (it is not Monday and it is not the case that (it is raining or it is hot)), then eithertoday is Monday or it is not the case that (it is hot or it is raining).60. Letp: 4>6 andq: 9>12. Given statement:p→q; true. Converse:q→p; if 9>12, then 4>6; true.Contrapositive:¬q→ ¬p; if 9≤12, then 4≤6; true.61. Letp:|1|<3 andq:−3<1<3.Given statement:q→p; true.Converse:p→q; if|1|<3, then−3<1<3; true. Contrapositive:¬p→ ¬q; if|1| ≥3, then either−3≥1 or 1≥3; true.64.P6≡Q65.P≡Q67.P6≡Q68.P≡Q70.P6≡Q71.P6≡Q74. Either Dale is not smart or not funny.75. Shirley will not take the bus and not catch a ride to school.78.(a) Ifpandqare both false, (pimp2q)∧(qimp2p) is false, butp↔qis true.(b) Making the suggested change does not alter the last line of theimp2table.79.pq¬(p∧q)¬p∨ ¬qTTFFTFTTFTTTFFTT

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SOLUTIONS7Section 1.42. Invalidp→q¬r→ ¬q...r3. Validp↔rr... p5. Validp→(q∨r)¬q∧ ¬r...¬p7. Valid(p∨q)→(r∨s)p∧ ¬r... s8. Invalidp→rq→s¬(q∧p)¬p... s11. If 4 megabytes of memory is better than no memory at all, then either we will buy a new computer or wewill buy more memory. If we will buy a new computer, then we will not buy more memory. Therefore if4 megabytes of memory is better than no memory at all, then we will buy a new computer. Invalid.12. If 4 megabytes of memory is better than no memory at all, then we will buy a new computer. If we willbuy a new computer, then we will buy more memory. Therefore, we will buy more memory. Invalid.14. If 4 megabytes of memory is better than no memory at all, then we will buy a new computer. If we willbuy a new computer, then we will buy more memory. 4 megabytes of memory is better than no memoryat all. Therefore we will buy more memory. Valid.16. If the hardware is unreliable or the output is correct, then the while loop is not faulty. If the output iscorrect, then the while loop is faulty. Either the for loop is faulty or the output is correct. Therefore thehardware is unreliable. Invalid.17. If, if the for loop is faulty, then the hardware is unreliable, then the while loop is faulty. If, if the whileloop is faulty, then the output is correct, then the for loop is faulty. The hardware is unreliable and theoutput is correct. Either the for loop is faulty or the while loop is faulty. Therefore the for loop is faultyand the while loop is faulty. Invalid.19. If the for loop is faulty, then the while loop is faulty or the hardware is unreliable. If the while loop isfaulty, then the for loop is faulty or the output is correct. Either the for loop is faulty or the while loopis not faulty. The output is not correct.Therefore the for loop is faulty or the hardware is unreliable.Invalid.

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8SOLUTIONS21. Valid22. Valid24. Valid25. Suppose thatp1, p2, . . . , pnare all true.Since the argumentp1, p2/... pis valid,pis true.Sincep, p3, . . . , pnare all true and the argumentp, p3, . . . , pn/... cis valid,cis true. Therefore the argumentp1, p2, . . . , pn/... cis valid.28. Modus ponens29. Disjunctive syllogism31. Letpdenote the proposition “there is gas in the car,” letqdenote the proposition “I go to the store,”letrdenote the proposition “I get a soda,” and letsdenote the proposition “the car transmission isdefective.” Then the hypotheses are:p→q,q→r,¬r.Fromp→qandq→r, we may use the hypothetical syllogism to concludep→r.Fromp→rand¬r, we may use modus tollens to conclude¬p. From¬p, we may use addition to conclude¬p∨s. Since¬p∨srepresents the proposition “there is not gas in the car or the car transmission is defective,” weconclude that the conclusion does follow from the hypotheses.32. Letpdenote the proposition “Jill can sing,” letqdenote the proposition “Dweezle can play,” letrdenotethe proposition “I’ll buy the compact disk,” and letsdenote the proposition “I’ll buy the compact diskplayer.” Then the hypotheses are:(p∨q)→r,p,s.Fromp, we may use addition to concludep∨q. Fromp∨qand (p∨q)→r, we may use modus ponensto concluder.Fromrands, we may use conjunction to concluder∧s.Sincer∧srepresents theproposition “I’ll buy the compact disk and the compact disk player,” we conclude that the conclusiondoes follow from the hypotheses.34. The truth tablepqp∨qTTTTFTFTTFFFshows that wheneverpis true,p∨qis also true. Therefore addition is a valid argument.35. The truth tablepqp∧qTTTTFFFTFFFFshows that wheneverp∧qis true,pis also true. Therefore simplification is a valid argument.

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SOLUTIONS937. The truth tablepqrp→qq→rp→rTTTTTTTTFTFFTFTFTTTFFFTFFTTTTTFTFTFTFFTTTTFFFTTTshows that wheneverp→qandq→rare true,p→ris also true. Therefore hypothetical syllogism is avalid argument.38. The truth tablepqp∨q¬pTTTFTFTFFTTTFFFTshows that wheneverp∨qand¬pare true,qis also true.Therefore disjunctive syllogism is a validargument.Section 1.52. The statement is a command, not a propositional function.3. The statement is a command, not a propositional function.5. The statement is not a propositional function since it has no variables.6. The statement is a propositional function. The domain of discourse is the set of real numbers.8. 1 divides 77. True.9. 3 divides 77. False.11. For somen,ndivides 77. True.12. For everyn,ndoes not divide 77. False.14. It is false that for everyn,ndivides 77. True.15. It is false that for somen,ndivides 77. False.17. False18. True20. True21. True23. False24. True26.¬P(1)∧ ¬P(2)∧ ¬P(3)∧ ¬P(4)27.¬(P(1)∧P(2)∧P(3)∧P(4))29.¬P(1)∨ ¬P(2)∨ ¬P(3)∨ ¬P(4)30.¬(P(1)∨P(2)∨P(3)∨P(4))33. Some student is taking a math course.34. Every student is not taking a math course.36. It is not the case that every student is taking a math course.

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10SOLUTIONS37. It is not the case that some student is taking a math course.40. There is some person such that if the person is a professional athlete, then the person plays soccer. True.41. Every soccer player is a professional athlete. False.43. Every person is either a professional athlete or a soccer player. False.44. Someone is either a professional athlete or a soccer player. True.46. Someone is a professional athlete and a soccer player. True.49.∃x(P(x)∧Q(x))50.∀x(Q(x)→P(x))54. True55. True57. False58. True60. No. The suggested replacement returns false if¬P(d1) is true, and true if¬P(d1) is false.62. Literal meaning: Every old thing does not covet a twenty-something. Intended meaning: Some old thingdoes not covet a twenty-something. LetP(x) denote the statement “xis an old thing” andQ(x) denotethe statement “xcovets a twenty-something.” The intended statement is∃x(P(x)∧ ¬Q(x)).63. Literal meaning: Every hospital did not report every month. (Domain of discourse: the 74 hospitals.)Intended meaning (most likely): Some hospital did not report every month. LetP(x) denote the state-ment “xis a hospital” andQ(x) denote the statement “xreports every month.” The intended statementis∃x(P(x)∧ ¬Q(x)).65. Literal meaning:Everyone does not have a degree.(Domain of discourse:People in Door County.)Intended meaning: Someone does not have a degree. LetP(x) denote the statement “xhas a degree.”The intended statement is∃x¬P(x).66. Literal meaning: No lampshade can be cleaned. Intended meaning: Some lampshade cannot be cleaned.LetP(x) denote the statement “xis a lampshade” andQ(x) denote the statement “xcan be cleaned.”The intended statement is∃x(P(x)∧ ¬Q(x)).68. Literal meaning: No person can afford a home. Intended meaning: Some person cannot afford a home.LetP(x) denote the statement “xis a person” andQ(x) denote the statement “xcan afford a home.”The intended statement is∃x(P(x)∧ ¬Q(x)).69. The literal meaning is as Mr. Bush spoke. He probably meant: Someone in this country doesn’t agreewith the decisions I’ve made. LetP(x) denote the statement “xagrees with the decisions I’ve made.”Symbolically, the clarified statement is∃x¬P(x).71. Literal meaning: Every move does not work out. Intended meaning: Some move does not work out. LetP(x) denote the statement “xis a move” andQ(x) denote the statement “xworks out .” The intendedstatement is∃x(P(x)∧ ¬Q(x)).74. Letp(x) :xis good.q(x) :xis too long.r(x) :xis short enough.The domain of discourse is the set of movies. The assertions are

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SOLUTIONS11∀x(p(x)→ ¬q(x))∀x(¬p(x)→ ¬r(x))p(Love Actually)q(Love Actually).By universal instantiation,p(Love Actually)→ ¬q(Love Actually).Sincep(Love Actually) is true, then¬q(Love Actually) is also true. But this contradicts,q(Love Actually).77. LetP(x) denote the propositional function “xis a member of the Titans,” letQ(x) denote the propo-sitional function “xcan hit the ball a long way,” and letR(x) denote the propositional function “xcanmake a lot of money.” The hypotheses areP(Ken),Q(Ken),∀x Q(x)→R(x).By universal instantiation, we haveQ(Ken)→R(Ken). FromQ(Ken) andQ(Ken)→R(Ken), we mayuse modus ponens to concludeR(Ken). FromP(Ken) andR(Ken), we may use conjunction to concludeP(Ken)∧R(Ken).By existential generalization, we have∃x P(x)∧R(x) or, in words, someone is amember of the Titans and can make a lot of money. We conclude that the conclusion does follow fromthe hypotheses.78. LetP(x) denote the propositional function “xis in the discrete mathematics class,” letQ(x) denotethe propositional function “xloves proofs,” and letR(x) denote the propositional function “xhas takencalculus.” The hypotheses are∀x P(x)→Q(x),∃x P(x)∧ ¬R(x).By existential instantiation, we haveP(d)∧ ¬R(d) for somedin the domain of discourse. FromP(d)∧¬R(d), we may use simplification to concludeP(d) and¬R(d).By universal instantiation, we haveP(d)→Q(d). FromP(d)→Q(d) andP(d), we may use modus ponens to concludeQ(d). FromQ(d)and¬R(d), we may use conjunction to concludeQ(d)∧ ¬R(d).By existential generalization, we have∃Q(x)∧ ¬R(x) or, in words, someone who loves proofs has never taken calculus. We conclude that theconclusion does follow from the hypotheses.80. By definition, the proposition∃x∈D P(x) is true whenP(x) is true for somexin the domain ofdiscourse. Takingxequal to ad∈Dfor whichP(d) is true, we find thatP(d) is true for somed∈D.81. By definition, the proposition∃x∈D P(x) is true whenP(x) is true for somexin the domain ofdiscourse. SinceP(d) is true for somed∈D,∃x∈D P(x) is true.Section 1.62. Everyone is taller than someone.3. Someone is taller than everyone.The solutions for 7–20 are for Exercise 2.7. False8. False10. False11. False13. False14. False16. False17. True19. True20. False

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12SOLUTIONS23. Everyone is taller than or the same height as someone.24. Someone is taller than or the same height as everyone.29. For every person, there is a person such that if the persons are distinct, the first is taller than the second.30. There is a person such that, for every person, if the persons are distinct, the first is taller than the second.35.∀x∀yL(x, y). False.36.∃x∃yL(x, y). True.40.∀x¬A(x,Profesor Sandwich)41.∀x∃yE(x)→A(x, y)44. True45. False49. True50. False52. False53. False55. False56. False58. True59. True61. True62. False64. True65. True67. fori= 1 tonif (foralldj(i))return truereturn falseforalldj(i){forj= 1 tonif (¬P(di, dj))return falsereturn true}68. fori= 1 tonforj= 1 tonif (P(di, dj))return truereturn false70. Since the first two quantifiers are universal and the last quantifier is existential, Farley choosesxandy,after which, you choosez. If Farley chooses values that makex≥y, sayx=y= 0, whatever value youchoose forz,(z > x)∧(z < y)is false. Since Farley can always win the game, the quantified propositional function is false.71. Since the first two quantifiers are universal and the last quantifier is existential, Farley choosesxandy, after which, you choosez. Whatever values Farley chooses, you can choosezto be one less than theminimum ofxandy; thus making(z < x)∧(z < y)true. Since you can always win the game, the quantified propositional function is true.73. Since the first two quantifiers are universal and the last quantifier is existential, Farley choosesxandy,after which, you choosez. If Farley chooses values such thatx≥y, the proposition(x < y)→((z > x)∧(z < y))is true by default (i.e., it is true regardless of what value you choose forz). If Farley chooses values suchthatx < y, you can choosez= (x+y)/2 and again the proposition(x < y)→((z > x)∧(z < y))is true. Since you can always win the game, the quantified propositional function is true.

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SOLUTIONS1375. The proposition must be true.P(x, y) is true for allxandy; therefore, no matter which value forxwechoose, the proposition∀yP(x, y) is true.76. The proposition must be true. SinceP(x, y) is true for allxandy, we may chooseanyvalues forxandyto makeP(x, y) true.78. The proposition can be false. LetNdenote the set of persons James James, Terry James, and Lee James;let the domain of discourse beN×N; and letP(x, y) be the statement “x’s first name is the same asy’s last name.” Then∃x∀yP(x, y) is true, but∀x∃yP(x, y) is false.79. The proposition must be true. Since∃x∀yP(x, y) is true, there is some value forxfor which∀yP(x, y)is true. Choosing any value forywhatsoever makesP(x, y) true. Therefore∃x∃yP(x, y) is true.81. The proposition can be false.LetP(x, y) be the statementx > yand let the domain of discourse beZ+×Z+. Then∃x∃yP(x, y) is true, but∀x∃yP(x, y) is false.82. The proposition can be false.LetP(x, y) be the statementx > yand let the domain of discourse beZ+×Z+. Then∃x∃yP(x, y) is true, but∃x∀yP(x, y) is false.84. The proposition can be true.LetP(x, y) be the statementx≤yand let the domain of discourse beZ+×Z+. Then∀x∀yP(x, y) is false, but∃x∀yP(x, y) is true.85. The proposition can be true.LetP(x, y) be the statementx≤yand let the domain of discourse beZ+×Z+. Then∀x∀yP(x, y) is false, but∃x∃yP(x, y) is true.87. The proposition can be true. LetNdenote the set of persons James James, Terry James, and Lee James;let the domain of discourse beN×N; and letP(x, y) be the statement “x’s first name is different fromy’s last name.” Then∀x∃yP(x, y) is false, but∃x∀yP(x, y) is true.88. The proposition can be true.LetP(x, y) be the statementx > yand let the domain of discourse beZ+×Z+. Then∀x∃yP(x, y) is false, but∃x∃yP(x, y) is true.90. The proposition can be true.LetP(x, y) be the statementx < yand let the domain of discourse beZ+×Z+. Then∃x∀yP(x, y) is false, but∀x∃yP(x, y) is true.91. The proposition can be true.LetP(x, y) be the statementx≤yand let the domain of discourse beZ×Z. Then∃x∀yP(x, y) is false, but∃x∃yP(x, y) is true.93.∀x∃y P(x, y) must be false.Since∃x∃y P(x, y) is false, for everyxand for everyy,P(x, y) is false.Choosex=x′in the domain of discourse.For this choice ofx,P(x, y) is false for everyy.Therefore∀x∃y P(x, y) is false.94.∃x∀y P(x, y) must be false.Since∃x∃y P(x, y) is false, for everyxand for everyy,P(x, y) is false.Choosey=y′in the domain of discourse. Now, for any choice ofx,P(x, y) is false fory=y′. Therefore∃x∀y P(x, y) is false.96. Not equivalent. LetP(x, y) be the statementx > yand let the domain of discourse beZ+×Z+. Then¬(∀x∃yP(x, y)) is true, but∀x¬(∃yP(x, y)) is false.97. Equivalent by De Morgan’s law100.∃ε >0∀δ >0∃x((0<|x−a|< δ)∧(|f(x)−L| ≥ε))101.∀L∃ε >0∀δ >0∃x((0<|x−a|< δ)∧(|f(x)−L| ≥ε))102. Literal meaning: No school may be right for every child. Intended meaning: Some school may not beright for some child.LetP(x, y) denote the statement “schoolxis right for childy.”The intendedstatement is∃x∃y¬P(x, y).

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