1.review (odds)

1.review (odds) - GUIDE TO REVIEW QUESTIONS FOR CHAPTER 1...

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Unformatted text preview: GUIDE TO REVIEW QUESTIONS FOR CHAPTER 1 1. 2. a) See p. 3. b) This is not a boring course. a) See pp. 4, 5, 6, and 9. b) Disjunction: r‘l’ll go to the movies tonight or I’ll finish my discrete mathematics homework.” Conjunction: “I’ll go to the movies tonight and I’ll finish my discrete mathematics homework.” Exclusive or: “I’ll go to the movies tonight or I’ll finish my discrete mathematics homework, but not both.” Conditional statement: “If I‘ll go to the movies tonight, then I’ll finish my discrete mathematics homework.” Biconditional: “I’ll go to the movies tonight if and only if I’ll finish my discrete mathematics homework.” a) See p. 6. b) See p. 8. c) Converse: “Ifl go for a walk in the woods tomorrow, then it will be sunny.” Contrapositive: “lfl don’t go for a walk in the woods tomorrow, then it will not be sunny.” . a) See p. 22. b) using truth tables; symbolically, using identities in Tables 678 in Section 1.2; by giving a valid argument about the possible truth values of the propositional variables involved c) Use the fact that 7" —r -uq E or V -q, or use truth tables. . a) Each line of the truth table correSponds to exactly one combination of truth values for the n. atomic propositions involved. we can write down a conjnnction that is true precisely in this case, namer the conjunction of all the atomic propositions that are true and the negations of all the atomic propositions that are false. If we do this for each line of the trnth table for which the value of the compound proposition is to be true, and take the disjunction of the resulting propositions, then we have the desired proposition in its disjunctive normal form. See Exercise 42 in Section 1.2. b) See Exercise 43 in Section 1.2. 0) See Exercises 50 and 52 in Section 1.2. . The negation of Va:P(:J:) is Elm-PRU), and the negation of ElauP(rr) is Veg-Pm). a) In the second, .7; can depend on y. In the first, the same I must “work” for every y. b) See Example 4 in Section 1.4. See pp. 63-«64. This is a. valid argument because it uses the valid rule of inference called modns tollens. This is a. valid argument because it uses the universal modus ponens rule of inference. Therefore if the premises are true, the conclusion must be true. Supplementary Exercises 43 10. a) See pp. 76, 77, and 80. 11. 12. 13. 14. b) For a direct proof, the hypothesis implies that in. : 2% for some is, whence 11+ 4 = 20: + 2), so a + 4 is even. For a proof by contraposition, suppose that 11+ 4 is odd; hence a + 4 = ‘21: + 1 for some is. Then a = 2(k * 2) + 1, so a is odd, hence not even. For a proof by contradiction, assume that n = 2k and n + 4 : 2i + 1 for some I: and l. Subtracting gives 4 : 2(i — + 1, which means that 4 is odd, a contradiction. a) See p. 82. b) Suppose that 3n+ 2 is odd, so that 371+ 2 : 2k+ 1 for some is. Multiply both sides by 3 and subtract 1, obtaining 9n+5 : 6134—2 : 9(3kt+ i). This shows that 9714—5 is even. We prove the converse by contraposition. Suppose that 3n + 2 is not odd, i.e., that it is even. Then 371+ 2 : ‘th for some is. Multiply both sides by 3 and subtract 1 , obtaining 9n + 5 : 6k: 7 l : 2(3k — l) + 1. This shows that 911 + 5 is odd. No ~we could add to dress p2 —> p3 and p1 —> 194, for example. a) Find a counterexample, i.e., an object c such that P(c) is false. b) n = l is a counterexample. Seep. 91. 15. See p. 92. 16. See Example 4 in Section 1.7. SUPPLEMENTARY EXERCISES FOR CHAPTER 1 1. a) q —> p (note that “only if” does not mean “if"‘) b) q A a d) q H 10 (this is another way to say “if and only if" in English words) C) no V np (assuming inclusive use of the English word “or” is intended by the speaker) . We could use truth tables, but we can also argue as follows. a) Since q is false but the conditional statement ,0 ~> q is true, we must conclude that p is also false. b) The disjunction says that either p or q is true. Since p is given to be false, it follows that q must be true. . The inverse of p > q is —up —+ -q. Therefore the converse of the inverse is wq a op. Note that this is the same as the contrapositive of the original statement. The converse of p —r q is q —> 1). Therefore the converse of the converse is p —> q, which was the original statement. The contrapositive of p —i q is -q a -~p. Therefore the converse of the contrapositive is -n}) —> fig, which is the same as the inverse of the original statement. . The straightforward approach is to use disjunctive normal form. There are four cases in which exactly three of the variables are true. The desired proposition is (pAqArA-ns) V (pAqA-urAs) V (pA -uq/\7‘/‘\s) V (mp/\qArAs). . Translating these statements into symbols, using the obvious letters, we have at ~> fig, fly a» fig, 7‘ —r q, and -t /\ 7'. Assume the statements are consistent. The fourth statement tells us that -nt must be true. Therefore by modus ponens with the first statement, we know that mg is true, hence (from the second statement) that fig is true. Also, the fourth statement tells us that r must be true, and so again modus ponens (third statement) makes q true. This is a contraction: q A fig. Thus the statements are inconsistent. 44 11. 13. 15. 17. 19. 21. 23. Chapter 1 The Foundations: Logic and Proofs We are told that exactly one of these people committed the crime, and exactly one (the guilty party) is a knight. We look at the three cases to determine who the knight is. If Amy were the knight, then her protestations of innocence would be true, but that cannot be, since we know that the knight is guilty. If Claire were the knight, then her statement that Brenda is not a normal is true; and since Brenda cannot be the knight in this situation, Brenda must be a knave. That means that Brenda is lying when she says that Amy was telling the truth; therefore Am)r is lying. This means that Amy is guilty, but that cannot be, since Amy isn’t the knight. So Brenda must be the knight. Amy is an innocent normal who is telling the truth when she says she is innocent; Brenda is telling the truth when she says that Amy is telling the truth; and Claire is a normal who is telling the truth when she says that Brenda is not a normal. So Brenda committed the crime. The definition of valid argument is an argument in which the truth of all the premises forces the truth of conclusion. In this example, the two premises can never be true simultaneously, because they are contradictory, irrespective of the true status of the tooth fairy. Therefore it is (vacuously) true that whenever both of the premises are true, the conclusion is also true (irrespective of your luck at finding gold at the end of the rainbow). Because the premises are not both true, we cannot conclude that the conclusion is true. b) T, since 2 divides 4 d) T, since 1 divides every positive integer a) F, since 4 does not divide 5 c) F, by the counterexample in part (a) e) F , since no number is a multiple of all positive integers (No matter what positive integer 1.1 one chooses, if we take m = n. + 1, then P(m, n.) is false, since 71. + 1 does not divide n.) f) T, since 1 divides every positive integer The given statement tells us that there are exactly two elements in the domain. Therefore if we let the domain be anything with size other than 2 the statement will be false. For each person we want to assert the existence of two different people who are that person’s parents. The mast elegant way to do so is Va:3y§lz(y % z AVw(F(w, <—5 (-w = 3; Vin : Note that we are saying that w is a parent of :i: if and only if u: is one of the two people whose existence we asserted. To express the statement that exactly n members of the domain satisfy P, we need to use 71 existential quantifiers, express the fact that these 21 variables all satisfy P and are all different, and express the fact that every other member of the domain that satisfies P must be one of these. a) This is a special case, however. To say that there are no values of a: that make P true we can simply write nElmPUc) or anPm). b) This is the same as Exercise 53 in Section 1.4, because 31:1:P(:L') is the same as EliccP(:i:). write 3:0(P(:c) A Vy(P(y) fir y 2 c) Following the discussion above, we write ElrrlElang(P($1) A P(:z:2)A:r;1 7t :rgAVy(P(y) —> = .121ng = 3:2»). Thus we can d) We expand the previous answer to one more variable: 3x13m23$3(P(:cl)A P(3;2) A P(:1:3) A m1 # 1:2 A 1:1 # 2:3 A $2 is 3:3 A WWW} e (y 2 331V'y :12 V y = :63)»- Suppose that Elzr(P(:r,) —> Q(:r)) is true. Then for some 3:, either Q(.’c) is true or P(.’L‘) is false. If (2(a) is true for some a, then the conditional statement Vth($) —> Ela:Q(:r) is true (having true conclusion). If P(:c) is false for some a, then again the conditional statement VIP(IE) —> 3.1362(3)) is true (having false hypothesis). Conversely, suppose that El$(P(.1:) —> (2(a)) is false. That means that for every :13, the conditional statement P(;c) —> Q(:r) is false, or, in other words, P(:i:) is true and (3(3) is false. The latter statement implies that ElmQ(:t) is false. Thus V3:P(m) ~> 31rQ(:c) has a true hypothesis and a false conclusion and is therefore false. Wri ting Projects 45 25. No. For each a: there may be just. one y making P(:t,y) true, so that the second proposition will not be true. For example, let P(:r,y) be 3: + y = 0, where the domain (universe of discourse) is the integers. Then the first proposition is true, since for each x there exists a y, namely —a‘, such that P(:L', 1;) holds. On the other hand, there is no one as such that (I: + y = O for every y. 27. Let T(s,c,d) be the statement that student .9 has taken class c in department (1. Then, with the domains (universes of discourse) being the students in this class, the courses at this university, and the departments in the school of mathematical sciences, the given statement is VsVdi‘cT(s, c,(l). 29. Let T(.r,y) mean that student :2: has taken class 3;, where the domain is all students in this class. We want to say that there exists exactly one student for whom there exists exactly one class that this student has taken. So we can write simply 315533! T(a:,y). To do this without quantifiers, we need to expand the uniqueness quantifier using Exercise 52 in Section 1.4. Doing so, we have 3:5Vz((§iwa(T(z, w) ¢—> w 2 y)) H z = a"). 31. By universal instantiation we have P(a) 9 (2(0) and (2(a) —> 113(0). By modus tollens we then conclude -|Q(a), and again by modus tollens we conclude fiP(a). 33. We give a proof by contraposition that if is rational, then :2: is rational, assuming throughout that 1L- 2 0. Suppose that : p/q is rational, (1 3A 0. Then a; = : pal/q2 is also rational ((12 is again nOnzero). 35. we can give a constructive proof by letting m : 10500 + 1. Then m2 : (10500 + 1)2 > (10500)2 : 101000. 37. The first three positive cubes are l, 8, and 27. If we want to find a number that. cannot be written as the sum of eight cubes, we would look for a number that is 7 more than a small multiple of 8. Indeed, 23 will do. We can use two 8’s but then would have to use seven 1’s to reach 23, a. total of nine numbers. Clearly no smaller collection will do. This counterexample disproves the statement. 39. The first three positive fifth powers are 1, 32, and 243. If we want to find a number that cannot be written as the sum of 36 fifth powers, we would look for a number that is 31 more than a small multiple of 32. Indeed, 7-32 7 l = 223 will do. \Ve can use six 32’s but then would have to use 31 1’5 to reach 223, a total of 37 numbers. Clearly no smaller collection will do. This counterexample disproves the statement. WRITING PROJECTS FOR CHAPTER 1 Books and articles indicated by bracketed symbols below are listed. near the end of this manual. You should also read. the general comments and advice you will find there about researching and writing these essays. 1. An excellent website for this is http://www.wordsmit]1.demon.co.uk/paradoxes. It includes a bibliogra— phy. 2. Search your library‘s on-line catalog for a book with the word jazzy in the title. You might find [BaGo], [DuPr], [Ka], [K03], or [McFr], for example. 3. Even if you can‘t find a set, you may find some articles about it in materials for high school students and teachers, such as old issues of Mathematics Teacher, published by the National Council of Teachers of Math— ematics. This journal, and possibly even copies of the game, may exist in the education library at your school (if there is one). The company that currently produces it has a website: http: //www . wff—n—proof . com. See also http: //thinkers .law.umich.edu/fileS/WPGames/WFFNPRUF.lltm, which includes the rules. 46 Chapter 1 The Foundations: Logic and Proofs 4. Martin Gardner and others have written some books that annotate Carroll’s writings quite extensively. Lewis U! 10. 11. Carroll has become a cult figure in certain circles. See also [Cal]. [C32], and [Ca3], for original material. A textbook on logic programming and/or the language PROLOG, such as [H02] or [Sal], would be a logical place to start Many bookstores have huge computer science sections these days, so that source should not be ignored. A course on computational logic at Stanford in 200572006 had a Web page with class notes: http://logic.stanford. edu/classeS/cs157/2005fall/c3157.html. Emlerton"s book on logic [En] would be a possible choice for background information. There are books on this subject, such as [Du]. . A place to start might be a recent article on this topic in Science [Re]. As always, a Web search will also turn up more infornmtion. The Web has an encyclopedia made up of articles by contributors. Ren'iarkably. it is usually quite good, with accurate information and useful links and crosserefercnces. See their article on Chomp: http : //en . wikipedia. org/wiki/Chomp. The references given in the text are the obvious place to start. The mathematics education field has bought into Polya’s ideas, especially as they relate to problem-solving. See what the National Council of Teachers of Mathematics (http://www.nctm.org} has to say about it. The classic work in this field is [GrSh]. ...
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1.review (odds) - GUIDE TO REVIEW QUESTIONS FOR CHAPTER 1...

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