Solution Manual For Discrete Mathematics, 7th Edition

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Chapter 1Solutions 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. 119. 322. We find thatB={2,3}. SinceAandBhave the same elements, they are equal.23. Letx∈A. Thenx= 1,2,3. Ifx= 1, since 1∈Z+and 12<10, thenx∈B. Ifx= 2, since2∈Z+and 22<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 thesevalues ofx,x /∈B. Thereforex= 1,2,3. For each of these values,x2<10 andxis indeed inB. Also, for each of the valuesx= 1,2,3,x∈A. Thus ifx∈B, thenx∈A. ThereforeA=B.26. Since (−1)3−2(−1)2−(−1) + 2 = 0,−1∈B. Since−1/∈A,A=B.27. Since 32−1>3, 3/∈B. Since 3∈A,A=B.30. Equal31. Not equal34. 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.35. Letx∈A. Thenx= (1,1) orx= (1,2). In either case,x∈B. ThereforeA⊆B.38. Since (−1)3−2(−1)2−(−1) + 2 = 0,−1∈A. However,−1/∈B. ThereforeAis not a subsetofB.39. Consider 4, which is inA. If 4∈B, then 4∈Aand 4 +m= 8 for somem∈C. However, theonly value ofmfor which 4 +m= 8 ism= 4 and 4/∈C. Therefore 4/∈B. Since 4∈Aand4/∈B,Ais not a subset ofB.1

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2CHAPTER 1SOLUTIONS42.UAB43.UAB45.UABC46.UABC48.UABC51. 3252. 10554. 5156. Suppose thatnstudents are taking both a mathematics course and a computer science courseThen 4nstudents are taking a mathematics course, but not a computer science course, and 7nstudents are taking a computer science course, but not a mathematics course.The followingVenn diagram depicts the situation:

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CHAPTER 1SOLUTIONS34nn7nMathCompSciThus, the total number of students is4n+n+ 7n= 12n.The proportion taking a mathematics course is5n12n=512,which is greater than one-third.58.{(a,1),(a,2),(b,1),(b,2),(c,1),(c,2)}59.{(1,1),(1,2),(2,1),(2,2)}62.{(1, a, a),(2, a, a)}63.{(1,1,1),(1,2,1),(2,1,1),(2,2,1),(1,1,2),(1,2,2),(2,1,2),(2,2,2)}66. Vertical lines (parallel) spaced one unit apart extending infinitely to the left and right.67. Horizontal lines (parallel) spaced one unit apart extending infinitely up and down.69. Consider all points on a horizonal line one unit apart. Now copy these points by moving thehorizonal linenunits straight up and straight down for all integern >0. The set of all pointsobtained in this way is the setZ×Z.70. Ordinary 3-space72. Take the lines described to the solution to Exercise 67 and copy them by movingnunits outand back for alln >0. The set of all points obtained in this way is the setR×Z×Z.74.{1,2}{1},{2}75.{a, b, c}{a, b},{c}{a, c},{b}{b, c},{a}{a},{b},{c}78. False79. True81. False82. True84.∅,{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.85. 210= 1024; 210−1 = 102388.B⊆A89.A=U92. The symmetric difference of two sets consists of the elements in one or the other but not both.93.AA=∅,AA=U,UA=A,∅ A=A95. The set of primes

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4CHAPTER 1SOLUTIONSSection 1.22. Is a proposition. Negation: 6 + 9= 15.3. Not a proposition5. Is a proposition. Negation: For every positive integern, 19340=n·17.6. Is a proposition. Negation: Audrey Meadows was not the original “Alice” in the “Honeymoon-ers.”8. Is a proposition.Negation:The line “Play it again, Sam” does not occur in the movieCasablanca.9. Is a proposition. Some even integer greater than 4 is not the sum of two primes.11. Not a proposition. The statement is neither true nor false.13. No heads were obtained.14. No heads or no tails were obtained.17. True18. True20. False21. False23.pq(¬p∨ ¬q)∨pTTTTFTFTTFFT24.pq(p∨q)∧ ¬pTTFTFFFTTFFF26.pq(p∧q)∨(¬p∨q)TTTTFFFTTFFT27.pqr¬(p∧q)∨(r∧ ¬p)TTTFTTFFTFTTTFFTFTTTFTFTFFTTFFFT

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CHAPTER 1SOLUTIONS529.pqr¬(p∧q)∨(¬q∨r)TTTTTTFFTFTTTFFTFTTTFTFTFFTTFFFT31.¬(p∧q). True.32.p∨ ¬(q∧r). True.34. Lee takes computer science and mathematics.35. Lee takes computer science or mathematics.37. Lee takes computer science but not mathematics.38. Lee takes neither computer science nor mathematics.40. You do not miss the midterm exam and you pass the course.41. You play football or you miss the midterm exam or you pass the course.43. Either you play football and you miss the midterm exam or you do not miss the midterm examand you pass the course.45. It is not Monday and either it is raining or it is hot.46. It is not the case that today is Monday or it is raining, and it is hot.48. Today is Monday and either it is raining or it is hot, and it is hot or either it is raining or todayis Monday.50.p∧q51.p∧ ¬q53.p∨q54. (p∨q)∧ ¬p56.p∧r∧q57. (p∨r)∧q59. (q∨ ¬p)∧ ¬r61.p∧ ¬r62.p∧q∧r64.¬p∧ ¬q∧r65.¬(p∨q∨ ¬r)66.pqpexorqTTFTFTFTTFFF68. Inclusive-or69. Inclusive-or71. Exclusive-or72. Exclusive-or76.”lung disease” -cancer77.”minor league” baseball team illinois -”midwest league”

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6CHAPTER 1SOLUTIONSSection 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.12. Contrapositive of Exercise 2: If Rosa does not graduate, then she does not have 160 quarter-hours of credits.14. False15. False17. False18. True20. True22. Unknown23. Unknown25. True26. Unknown28. Unknown29. Unknown32. True33. True35. True36. False38. True39. False41. (p∧r)→q42.¬((r∧ ¬q)→r)45. (¬p∨ ¬r)→ ¬q46.r→q48.q→(p∨r)49. (q∧p)→ ¬r51. If it is not raining, then it is hot and today is Monday.52. If today is not Monday, then either it is raining or it is hot.54. If today is Monday and either it is raining or it is hot, then either it is hot, it is raining, ortoday is Monday.55. If today is Monday or (it is not Monday and it is not the case that (it is raining or it is hot)),then either today is Monday or it is not the case that (it is hot or it is raining).57. Letp: 4>6 andq: 9>12. Given statement:p→q; true. Converse:q→p; if 9>12, then4>6; true. Contrapositive:¬q→ ¬p; if 9≤12, then 4≤6; true.58. 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.61.P≡Q62.P≡Q64.P≡Q65.P≡Q67.P≡Q68.P≡Q71. Either Dale is not smart or not funny.72. Shirley will not take the bus and not catch a ride to school.75.(a) Ifpandqare both false, (pimp2q)∧(qimp2p) is false, butp↔qis true.(b) Making the suggested change does not alter the last line of theimp2table.

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CHAPTER 1SOLUTIONS776.pq¬(p∧q)¬p∨ ¬qTTFFTFTTFTTTFFTTSection 1.42. Invalidp→q¬r→ ¬q... r3. Validp↔rr... p5. Validp→(q∨r)¬q∧ ¬r...¬p7. If 4 megabytes of memory is better than no memory at all, then either we will buy a newcomputer or we will buy more memory. If we will buy a new computer, then we will not buymore memory. Therefore if 4 megabytes of memory is better than no memory at all, then wewill buy a new computer. Invalid.8. If 4 megabytes of memory is better than no memory at all, then we will buy a new computer.If we will buy a new computer, then we will buy more memory. Therefore, we will buy morememory. Invalid.10. If 4 megabytes of memory is better than no memory at all, then we will buy a new computer. Ifwe will buy a new computer, then we will buy more memory. 4 megabytes of memory is betterthan no memory at all. Therefore we will buy more memory. Valid.12. Valid13. Valid15. Valid16. 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/... c

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8CHAPTER 1SOLUTIONSis valid.19. Modus ponens20. Disjunctive syllogism22. Letpdenote the proposition “there is gas in the car,” letqdenote the proposition “I go to thestore,” letrdenote the proposition “I get a soda,” and letsdenote the proposition “the cartransmission is defective.” 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 toconclude¬p∨s. Since¬p∨srepresents the proposition “there is not gas in the car or the cartransmission is defective,” we conclude that the conclusion does follow from the hypotheses.23. Letpdenote the proposition “Jill can sing,” letqdenote the proposition “Dweezle can play,”letrdenote the proposition “I’ll buy the compact disk,” and letsdenote the proposition “I’llbuy the compact disk player.” Then the hypotheses are:(p∨q)→r,p,s.Fromp, we may use addition to concludep∨q.Fromp∨qand (p∨q)→r, we may usemodus ponens to concluder. Fromrands, we may use conjunction to concluder∧s. Sincer∧srepresents the proposition “I’ll buy the compact disk and the compact disk player,” weconclude that the conclusion does follow from the hypotheses.25. The truth tablepqp∨qTTTTFTFTTFFFshows that wheneverpis true,p∨qis also true. Therefore addition is a valid argument.26. The truth tablepqp∧qTTTTFFFTFFFFshows that wheneverp∧qis true,pis also true. Therefore simplification is a valid argument.

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CHAPTER 1SOLUTIONS928. The truth tablepqrp→qq→rp→rTTTTTTTTFTFFTFTFTTTFFFTFFTTTTTFTFTFTFFTTTTFFFTTTshows that wheneverp→qandq→rare true,p→ris also true.Therefore hypotheticalsyllogism is a valid argument.29. The truth tablepqp∨q¬pTTTFTFTFFTTTFFFTshows that wheneverp∨qand¬pare true,qis also true. Therefore disjunctive syllogism is avalid argument.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.13. False14. True16. True17. True19. False20. True22.¬P(1)∧ ¬P(2)∧ ¬P(3)∧ ¬P(4)23.¬(P(1)∧P(2)∧P(3)∧P(4))25.¬P(1)∨ ¬P(2)∨ ¬P(3)∨ ¬P(4)26.¬(P(1)∨P(2)∨P(3)∨P(4))29. Some student is taking a math course.30. Every student is not taking a math course.32. It is not the case that every student is taking a math course.

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10CHAPTER 1SOLUTIONS33. It is not the case that some student is taking a math course.36. There is some person such that if the person is a professional athlete, then the person playssoccer. True.37. Every soccer player is a professional athlete. False.39. Every person is either a professional athlete or a soccer player. False.40. Someone is either a professional athlete or a soccer player. True.42. Someone is a professional athlete and a soccer player. True.45.∃x(P(x)∧Q(x))46.∀x(Q(x)→P(x))50. True51. True53. False54. True56. No. The suggested replacement returns false if¬P(d1) is true, and true if¬P(d1) is false.58. Literal meaning: Every old thing does not covet a twenty-something. Intended meaning: Someold thing does not covet a twenty-something.LetP(x) denote the statement “xis an oldthing” andQ(x) denote the statement “xcovets a twenty-something.” The intended statementis∃x(P(x)∧ ¬Q(x)).59. Literal meaning:Every hospital did not report every month.(Domain of discourse:the 74hospitals.)Intended meaning (most likely): Some hospital did not report every month.LetP(x) denote the statement “xis a hospital” andQ(x) denote the statement “xreports everymonth.” The intended statement is∃x(P(x)∧ ¬Q(x)).61. Literal meaning:Everyone does not have a degree.(Domain of discourse:People in DoorCounty.) Intended meaning: Someone does not have a degree. LetP(x) denote the statement“xhas a degree.” The intended statement is∃x¬P(x).62. Literal meaning: No lampshade can be cleaned.Intended meaning: Some lampshade cannotbe 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)).64. Literal meaning: No person can afford a home. Intended meaning: Some person cannot afforda home. LetP(x) denote the statement “xis a person” andQ(x) denote the statement “xcanafford a home.” The intended statement is∃x(P(x)∧ ¬Q(x)).65. The literal meaning is as Mr. Bush spoke. He probably meant: Someone in this country doesn’tagree with the decisions I’ve made. LetP(x) denote the statement “xagrees with the decisionsI’ve made.” Symbolically, the clarified statement is∃x¬P(x).69. 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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CHAPTER 1SOLUTIONS11∀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).72. LetP(x) denote the propositional function “xis a member of the Titans,” letQ(x) denote thepropositional function “xcan hit the ball a long way,” and letR(x) denote the propositionalfunction “xcan make 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 may use modus ponens to concludeR(Ken).FromP(Ken) andR(Ken), we may useconjunction to concludeP(Ken)∧R(Ken). By existential generalization, we have∃x P(x)∧R(x)or, in words, someone is a member of the Titans and can make a lot of money. We concludethat the conclusion does follow from the hypotheses.73. LetP(x) denote the propositional function “xis in the discrete mathematics class,” letQ(x) de-note the propositional function “xloves proofs,” and letR(x) denote the propositional function“xhas taken calculus.” 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 universalinstantiation, we haveP(d)→Q(d). FromP(d)→Q(d) andP(d), we may use modus ponensto 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 proofshas never taken calculus. We conclude that the conclusion does follow from the hypotheses.75. By definition, the proposition∃x∈D P(x) is true whenP(x) is true for somexin the domainof discourse. Takingxequal to ad∈Dfor whichP(d) is true, we find thatP(d) is true forsomed∈D.76. By definition, the proposition∃x∈D P(x) is true whenP(x) is true for somexin the domainof discourse. SinceP(d) is true for somed∈D,∃x∈D P(x) is true.

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12CHAPTER 1SOLUTIONSSection 1.62. Everyone is taller than someone.3. Someone is taller than everyone.7.∀x∀yT1(x, y), False;∀x∃yT1(x, y), False;∃x∀yT1(x, y), False;∃x∃yT1(x, y), True.8.∀x∀yT1(x, y), False;∀x∃yT1(x, y), False;∃x∀yT1(x, y), False;∃x∃yT1(x, y), False.11. Everyone is taller than or the same height as someone.12. Someone is taller than or the same height as everyone.16.∀x∀yT2(x, y), False;∀x∃yT2(x, y), True;∃x∀yT2(x, y), True;∃x∃yT2(x, y), True.17.∀x∀yT1(x, y), True;∀x∃yT1(x, y), True;∃x∀yT1(x, y), True;∃x∃yT1(x, y), True.20. For every person, there is a person such that if the persons are distinct, the first is taller thanthe second.21. There is a person such that, for every person, if the persons are distinct, the first is taller thanthe second.25.∀x∀yT3(x, y), False;∀x∃yT3(x, y), True;∃x∀yT3(x, y), False;∃x∃yT3(x, y), True.26.∀x∀yT3(x, y), True;∀x∃yT3(x, y), True;∃x∀yT3(x, y), True;∃x∃yT3(x, y), True.29.∀x∀yL(x, y). False.30.∃x∃yL(x, y). True.34.∀x¬A(x,Profesor Sandwich)35.∀x∃yE(x)→A(x, y)38. True39. False43. True44. False46. False47. False49. False50. False52. True53. True55. True56. False58. True59. True61. fori= 1 tonif (forall dj(i))return truereturn falseforall dj(i){forj= 1 tonif (¬P(di, dj))return falsereturn true}62. fori= 1 tonforj= 1 tonif (P(di, dj))return truereturn false

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CHAPTER 1SOLUTIONS1364. 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 you choose forz,(z > x)∧(z < y)is false. Since Farley can always win the game, the quantified propositional function is false.65. 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 beone less than the minimum ofxandy; thus making(z < x)∧(z < y)true. Since you can always win the game, the quantified propositional function is true.67. 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 choosesvalues such thatx < 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.69. The proposition must be true.P(x, y) is true for allxandy; therefore, no matter which valueforxwe choose, the proposition∀yP(x, y) is true.70. The proposition must be true. SinceP(x, y) is true for allxandy, we may chooseanyvaluesforxandyto makeP(x, y) true.72. The proposition can be false. LetNdenote the set of persons James James, Terry James, andLee James; let the domain of discourse beN×N; and letP(x, y) be the statement “x’s firstname is the same asy’s last name.” Then∃x∀yP(x, y) is true, but∀x∃yP(x, y) is false.73. 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.75. The proposition can be false. LetP(x, y) be the statementx > yand let the domain of discoursebeZ+×Z+. Then∃x∃yP(x, y) is true, but∀x∃yP(x, y) is false.76. The proposition can be false. LetP(x, y) be the statementx > yand let the domain of discoursebeZ+×Z+. Then∃x∃yP(x, y) is true, but∃x∀yP(x, y) is false.78. The proposition can be true. LetP(x, y) be the statementx≤yand let the domain of discoursebeZ+×Z+. Then∀x∀yP(x, y) is false, but∃x∀yP(x, y) is true.79. The proposition can be true. LetP(x, y) be the statementx≤yand let the domain of discoursebeZ+×Z+. Then∀x∀yP(x, y) is false, but∃x∃yP(x, y) is true.

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14CHAPTER 1SOLUTIONS81. The proposition can be true. LetNdenote the set of persons James James, Terry James, andLee James; let the domain of discourse beN×N; and letP(x, y) be the statement “x’s firstname is different fromy’s last name.” Then∀x∃yP(x, y) is false, but∃x∀yP(x, y) is true.82. The proposition can be true. LetP(x, y) be the statementx > yand let the domain of discoursebeZ+×Z+. Then∀x∃yP(x, y) is false, but∃x∃yP(x, y) is true.84. The proposition can be true. LetP(x, y) be the statementx < yand let the domain of discoursebeZ+×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 discoursebeZ×Z. Then∃x∀yP(x, y) is false, but∃x∃yP(x, y) is true.87.∀x∃y P(x, y) must be false. Since∃x∃y P(x, y) is false, for everyxand for everyy,P(x, y) isfalse. 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.88.∃x∀y P(x, y) must be false. Since∃x∃y P(x, y) is false, for everyxand for everyy,P(x, y) isfalse. 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.90. 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.91. Equivalent by De Morgan’s law94.∃ε >0∀δ >0∃x((0<|x−a|< δ)∧(|f(x)−L| ≥ε))95.∀L∃ε >0∀δ >0∃x((0<|x−a|< δ)∧(|f(x)−L| ≥ε))96. Literal meaning: No school may be right for every child. Intended meaning: Some school maynot be right for some child. LetP(x, y) denote the statement “schoolxis right for childy.”The intended statement is∃x∃y¬P(x, y).Problem-Solving Corner: Quantifiers1. The statement of Example 1.6.6 is∀x∃y(x+y= 0).As was pointed out in Example 1.6.6, this statement is true. Now∀x∀y(x+y= 0)is false; a counterexample isx=y= 1. Also∃x∀y(x+y= 0)is false since, given anyx, ify= 1−x, thenx+y= 0.2. Yes; the statement∀m∃n(m < n) with domain of discourseZ×Zof Example 1.6.1 also solvesproblems (a) and (b).

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Chapter 2Solutions to Selected ExercisesSection 2.12. For allx, for ally,x+y=y+x.3. Anisosceles trapezoidis a trapezoid with equal legs.5. The medians of any triangle intersect at a single point.6. If 0< x <1 andε >0, there exists a positive integernsatisfyingxn< ε.8. Letmandnbe odd integers. Then there existk1andk2such thatm= 2k1+1 andn= 2k2+1.Nowm+n= (2k1+ 1) + (2k2+ 1) = 2(k1+k2+ 1).Therefore,m+nis even.9. Letmandnbe even integers.Then there existk1andk2such thatm= 2k1andn= 2k2.Nowmn= (2k1)(2k2) = 2(2k1k2).Therefore,mnis even.11. Letmbe an odd integer andnbe an even integer.Then there existk1andk2such thatm= 2k1+ 1 andn= 2k2. Nowmn= (2k1+ 1)(2k2) = 2(2k1k2+k2).Therefore,mnis even.12. Letmandnbe integers such thatmandm+nare even. Then there existk1andk2such thatm= 2k1andm+n= 2k2. Nown= (m+n)−m= 2k2−2k1= 2(k2−k1).Therefore,nis even.14. Letxandybe rational numbers. Then there exist integersm1, n1, m2, n2such thatx=m1/n1andy=m2/n2. Nowxy= (m1m2)/(n1n2). Thereforexyis rational.15

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