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INSTRUCTOR SOLUTIONS MANUAL for Discrete Mathematics with Applications, 5th Edition by Susanna Epp Instructor’s Manual Section 1.1

Chapter 1: Speaking Mathematically Many college and university students have difficulty using and interpreting language involving if-then statements and quantification. Section 1.1 is a gentle introduction to the relation between informal and formal ways of expressing such statements. The exercises are intended to start the process of helping students improve their ability to interpret mathematical statements and evaluate their truth or falsity. Sections 1.2 - 1.4 are a brief introduction to the language of sets, relations, functions, and graphs. Including Sections 1.2 and 1.3 at the beginning of the course can help students relate discrete mathematics to the pre-calculus or calculus they have studied previously while enlarging their perspective to include a greater proportion of discrete examples. Section 1.4 is designed to broaden students’ understanding of the way the word graph is used in mathematics and to show them how graph models can be used to solve some significant problems. Proofs of set properties, such as the distributive laws, and proofs of properties of relations and functions, such as transitivity and surjectivity, are considerably more complex than those used in Chapter 4 to give students their first practice in constructing mathematical proofs. For this reason set theory as a theory is left to Chapter 6, properties of functions to Chapter 7, and properties of relations to Chapter 8. By making slight changes about exercise choices, instructors could cover Section 1.2 just before starting Chapter 6 and Section 1.3 just before starting Chapter 7. The material in Section 1.4 lays the groundwork for the discussion of the handshake theorem and its applications in Section 4.9. Instructors who wish to offer a self-contained treatment of graph theory can combine both sections with the material in Chapter 10. College and university mathematics instructors may be surprised by the way students understand the meaning of the term “real number.” When asked to evaluate the truth or falsity of a statement about real numbers, it is not unusual for students to think only of integers. Thus an informal description of the relationship between real numbers and points on a number line is given in Section 1.2 to illustrate that there are many real numbers between any pair of consecutive integers, Examples 3.3.5 and 3.3.6 show that while there is a smallest positive integer there is no smallest positive real number, and the discussion in Chapter 7, which precedes the proof of the uncountability of the real numbers between 0 and 1, describes a procedure for approximating the (possibly infinite) decimal expansion for an arbitrarily chosen point on a number line.

Section 1.1 1. a. x2 = −1 (Or : the square of x is −1)

b. a real number x

2. a. a remainder of 2 when it is divided by 5 and a remainder of 3 when it is divided by 6 b. an integer n; n is divided by 6 the remainder is 3 3. a. between a and b

b. distinct real numbers a and b; there is a real number c

4. a. a real number; greater than r b. real number r; there is a real number s 5. a. r is positive b. positive; the reciprocal of r is positive (Or : positive; 1/r is positive) c. is positive; 1/r is positive (Or : is positive; the reciprocal of r is positive) √ 6. a. s is negative b. negative; the cube root of s is negative (Or : 3 s is negative) √ c. is negative; 3 s is negative (Or : the cube root of s is negative)

Instructor’s Manual: Chapter 1 7. a. There are real numbers whose sum is less than their difference. True. For example, 1 + (−1) = 0, 1 − (−1) = 1 + 1 = 2, and 0 < 2. b. There is a real number whose square is less than itself. True. For example, (1/2)2 = 1/4 < 1/2 . c. The square of each positive integer is greater than or equal to the integer. True. If n is any positive integer, then n ≥ 1. Multiplying both sides by the positive number n does not change the direction of the inequality (see Appendix A, T20), and so n2 ≥ n. d. The absolute value of the sum of any two numbers is less than or equal to the sum of their absolute values. True. This is known as the triangle inequality. It is discussed in Section 4.4. 8. a. have four sides

b. has four sides

c. has four sides d. is a square; has four sides

e. J has four sides 9. a. have at most two real solutions b. has at most two real solutions c. has at most two real solutions d. is a quadratic equation; has at most two real solutions e. E has at most two real solutions 10. a. have reciprocals

b. a reciprocal

11. a. have positive square roots

c. s is a reciprocal for r

b. a positive square root

c. r is a square root for e

12. a. real number; product with every number leaves the number unchanged b. a positive square root

c. rs = s

13. a. real number; product with every real number equals zero b. with every number leaves the number unchanged

c. ab = 0

Section 1.2 1. A = C and B = D 2. a. The set of all positive real numbers x such that 0 is less than x and x is less than 1 b. The set of all real numbers x such that x is less than or equal to zero or x is greater than or equal to 1 c. The set of all integers n such that n is a factor of 6 d. The set of all positive integers n such that n is a factor of 6 3. a. No, 4 { is } a set with one element, namely 4, whereas 4 is just a symbol that represents the number 4 b. Three: the elements of the set are 3, 4, and 5. c. Three: the elements are the symbol 1, the set {1}, and the set {1, {1}} 4.

a. Yes: { 2} is the set whose only element is 2. b. One: 2 is the only element in this set c. Two: The two elements are 0 and {0} d. Yes: { 0} is one of the elements listed in the set. e. No: The only elements listed in the set are 0{and } 1 , {and } 0 is not equal to either of these.

Instructor’s Manual Section 1.4

5. The only sets that are equal to each other are A and D. A contains the integers 0, 1, and 2 and nothing else. B contains all the real numbers that are greater than or equal to −1 and less than 3. C contains all the real numbers that are greater than —1 and less than 3. Thus − 1 is in B but not in C. D contains all the integers greater than −1 and less than 3. Thus D contains the integers 0, 1, and 2 and nothing else, and so D = {0, 1, 2} = A. E contains all the positive integers greater than −1 and less than 3. Hence E contains the integers 1 and 2 and nothing else, that is, E = {1, 2}. 6. T2 and T−3 each have two elements, and T0 and T1 each have one element. JustiFIcation: T2 = {2, 22} = {2, 4}, T−3 = {−3, (−3)2} = {−3, 9}, T1 = {1, 12} = {1, 1} = {1}, and T0 = {0, 02} = {0, 0} = {0}. 7. a. {1, −1} b. T = {m ∈ Z | m = 1+(−1)k for some integer k} = {0, 2}. Exercises in Chapter 4 explore the fact that (−1)k = −1 when k is odd and (−1)k = 1 when k is even. So 1+(−1)k = 1+(−1) = 0 when k is odd, and 1 + (−1)k = 1 + 1 = 2 when k is even. c. the set has no elements d. Z (every integer is in the set) e. There are no elements in W because there are no integers that are both greater than 1 and less than −3. f. X = Z because every integer u satisfies at least one of the conditions u ≤ 4 or u ≥ 1. 8. a. No, B ¢ A because j ∈ B and j ∈ /A b. Yes, because every element in C is in A. c. Yes, because every element in C is in C. c.. Yes, because it is true that every element in C is in C. d. Yes, C is a proper subset of A. Both elements of C are in A, but A contains elements (namely c and f ) that are not in C. 9. a. Yes b. No, the number 1 is not a set and so it cannot be a subset. c. No: The only elements in {1, 2} are 1 and 2, and {2} is not equal to either of these. d. Yes: {3} is one of the elements listed in {1, {2}, {3}}. e. Yes: {1} is the set whose only element is 1. f. No, the only element in {2} is the number 2 and the number 2 is not one of the three elements in {1, {2}, {3}}. g. Yes: The only element in {1} is 1, and 1 is an element in {1, 2}. h. No: The only elements in {{1}, 2} are {1} and 2, and 1 is not equal to either of these. i. Yes, the only element in {1} is the number 1, which is an element in {1, {2}}. j. Yes: The only element in {1} is 1, which is is an element in {1}. So every element in {1} is in {1}. 10. a. No. Observe that (−2)2 = (−2)(−2) = 4, whereas −22 = −(22) = −4. So ((−2)2, −22) = (4, −4), whereas (−22, (−2)2) = (−4, 4). And (4, −4) ̸= (−4, 4) because −4 ̸= 4. b. No: For two ordered pairs to be equal, the elements in each pair must occur in the same order. In this case the first element of the first pair is 5, whereas the first element of the second

Instructor’s Manual: Chapter 1 pair is —5, and the second element of the first pair is − 5 whereas the second element of the second pair is 5. √ √ c. Yes. Note that 8 − 9 = −1 and 3 −1 = −1, and so (8 − 9, 3 −1) = (−1, −1). d. Yes The first elements of both pairs equal 1 2, and the second elements of both pairs equal −8. 11. a. {(w, a), (w, b), (x, a), (x, b), (y, a), (y, b), (z, a), (z, b)} A × B has 4 · 2 = 8 elements. b. {(a, w), (b, w), (a, x), (b, x), (a, y), (b, y), (a, z), (b, z)} B × A has 4 · 2 = 8 elements. c. {(w, w), (w, x), (w, y), (w, z), (x, w), (x, x), (x, y), (x, z), (y, w), (y, x), (y, y), (y, z), (z, w), (z, x), (z, y), (z, z)} A × A has 4 · 4 = 16 elements. d. {(a, a), (a, b), (b, a), (b, b)} B × B has 2 · 2 = 4 elements. 12. All four sets have nine elements. a. S × T = {(2, 1), (2, 3), (2, 5), (4, 1), (4, 3), (4, 5), (6, 1), (6, 3), (6, 5)} b. T × S = {(1, 2), (3, 2), (5, 2), (1, 4), (3, 4), (5, 4), (1, 6), (3, 6), (5, 6)} c. S × S = {(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)} d. T × T = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)} 13. a. A × (B × C) = {(1, (u, m)), (1, (u, n)), (2, (u, m)), (2, (u, n)), (3, (u, m)), (3, (u, n))} b. (A × B) × C = {((1, u), m), ((1, u), n), ((2, u), m), ((2, u), n), ((3, u), m), ((3, u), n)} c. A × B × C = {(1, u, m), (1, u, n), (2, u, m), (2, u, n), (3, u, m), (3, u, n)} 14. a.R × (S × T ) = {(a, (x, p)), (a, (x, q)), (a, (x, r)), (a, (y, p)), (a, (y, q)), (a, (y, r))} b. (R × S) × T = {((a, x), p), ((a, x), q), ((a, x), r), ((a, y), p), ((a, y), q), ((a, y), r)} c. R × S × T = {(a, x, p), (a, x, q), (a, x, r), (a, y, p), (a, y, q), (a, y, r)} 15. 0000, 0001, 0010, 0100, 1000 16. yxxxx, xyxxx, xxyxx, xxxyx, xxxxy

Section 1.3 1. a. No. Yes. No. Yes. b. R = {(2, 6), (2, 8), (2, 10), (3, 6), (4, 8)} c. Domain of R = A = {2, 3, 4}, co-domain of R = B = {6, 8, 10} d. R 2

2. a. 2 S 2 because 1 − 1 = 0, which is an integer. 2

2 1

— 1 S −1 because −1 − −1 = 0, which is an integer. 3 S 3 because 13 − 1 3= 0, which is an integer. 3 S/ − 3 because 1 − 1 = 2 , which is not an integer. 3

−3

Instructor’s Manual Section 1.4 b. S = {(−3, −3), (−2, −2), (−1, −1), (1, 1), (2, 2), (3, 3), (1, −1), (−1, 1), (2, −2), (−2, 2)} c. The domain and co-domain of S are both {−3, −2, −1, 1, 2, 3}. d.

S –3

–3

–2

–1

3. a. 3 T 0 because 3−0 = 3 = 1, which is an integer. 3

1 ̸ T (−1) because 1−(−1) = 2 , which is not an integer. 3

(2, −1) ∈ T because 2−(−1) = 3 3= 1, which is an integer. 3 (3, −2) ∈ / T because 3−(−2) = 5 , which is not an integer. 3

b. T = {(1, −2), (2, −1), (3, 0)} c. Domain of T = E = {1, 2, 3}, co-domain of T = F = {−2, −1, 0} d.

T 1

–2

–1

4. a. 2 V 6 because 2−6 = −44 = −1, which is an integer. 4 (−2)−8 = −64, which is not an integer. 4 (−2)−8 (−2) ̸ V 8 because = −64, which is not an integer. 4 0 ̸ V 6 because 0−6 = −6 , which is not an integer. 4 4 2 ̸ V 4 because 2−4 = −2 , which is not an integer. 4 4

1 ̸ V (−1) because

b. V = {(−2, 6), (0, 4), (0, 8), (2, 6)} c. Domain of V = {−2, 0, 2}, co-domain of V = {4, 6, 8} d. G

–2

5. a. (2, 1) ∈ S because 2 ≥ 1. (2, 2) ∈ S because 2 ≥ 2. 2 ̸S 3 because 2 § 3. (−1)S(−2) because −1 ≥ −2.

Instructor’s Manual: Chapter 1 b.

graph of S

x $ y in shaded region

x 1

6. a. (2, 4) ∈ R because 4 = 22. (4, 2) ∈ / R because 2 ̸= 42 . (−3, 9) ∈ R because 9 = (−3)2. (9, −3) ∈ / R because −3 ̸= 92 . b.

y 9 8 7 6

y = x2

5 4 3 2 1 -6

-5

-4

-3

-2

-1

-1 -2

7. a.

R 4

A 4 5 6

5 6 7

4 5 6

B 5 6 7

b. R is not a function because it satisfies neither property (1) nor property (2) of the definition. It fails property (1) because (4, y) ∈/ R, for any y in B. It fails property (2) because (6, 5) ∈ R and (6, 6) ∈ R and 5 = ̸ 6. S is not a function because (5, 5) ∈ S and (5, 7) ∈ S and 5 = ̸ 7. So S does not satisfy property (2) of the definition of function. T is not a function both because (5, x) ∈ / T for any x in B and because (6, 5) ∈ T and (6, 7) ∈ T and 5 = ̸ 7. So T does not satisfy either property (1) or property (2) of the definition of function.

Instructor’s Manual Section 1.4 8. a.

U A

V B 1

B 1

A 2 4

B 1 3 5

b. None of U , V , or W are functions. U is not a function because (4, y) is not in U for any y in B, and so U does not satisfy property (1) of the definition of function. V is not a function because (2, y) is not in V for any y in B, and so V does not satisfy property (1) of the definition of function. W is not a function because both (2, 3) and (2, 5) are in W and 3 ≠ 5, and so W does not satisfy property (2) of the definition of function. 9. a. There is only one: {(0, 1), (1, 1)} b. {(0, 1)}, {(1, 1)} 10. The following sets are relations from {a, b} to {x, y} that are not functions: {(a, x)}, {(a, y)}, {(b, x)}, {(b, y)}, {(a, x), (a, y)}, {(b, x), (b, y)}, {(a, x), (a, y), (b, x)}, {(a, x), (a, y), (b, y)}, {(b, x), (b, y), (a, x)}, {(b, x), (b, y), (a, y)}, {(a, x), (a, y), (b, x), (b, y)}. 11. L(0201) = 4, L(12) = 2 12. C(x) = yx, C(yyxyx) = yyyxyx 13. a. Domain = A = {−1, 0, 1}, co-domain = B = {t, u, v, w} b. F (−1) = u, F (0) = w, F (1) = u 14. a. Domain of G = {1, 2, 3, 4}, co-domain of G = {a, b, c, d} b. G(1) = G(2) = G(3) = G(4) = c 15. a. This diagram does not determine a function because 2 is related to both 2 and 6. b. This diagram does not determine a function because 5 is in the domain but it is not related to any element in the co-domain. c. This diagram does not determine a function because 4 is related to both 1 and 2, which violates property (2) of the definition of function. d. This diagram defines a function; both properties (1) and (2) are satisfied. e. This diagram does not determine a function because 2 is in the domain but it is not related to any element in the co-domain. ( ) ( )2 16. f (−1) = (−1)2 = 1, f (0) = 02 = 0, f 1 = 1 = 1 . 2

17. g(−1000) = −999, g(0) = 1, g(999) = 1000 18. h(− 12 ) = h( 0 ) = h( 9 ) = 2 5

2x3 + 2x 2x(x2 + 1) = = 2x = f (x). Therefore, by definition of 2 x +1 x2 + 1 equality of functions, f = g.

19. For each x ∈ R, g(x) =

Instructor’s Manual: Chapter 1 20. For all x ∈ R, K(x) = (x − 1)(x − 3) + 1 = (x2 − 4x + 3) + 1 = x2 + 4x + 4 = (x − 2)2 = H(x). Therefore, by definition of equality of functions, H = K.

Section 1.4 1. V (G) = {v1, v2, v3, v4}, E(G) = {e1, e2, e3} Edge-endpoint function: Edge e1 e2 e3

Endpoints {v1, v2} {v1, v3} {v3}

2. V (G) = {v1, v2, v3, v4}, E(G) = {e1, e2, e3, e4, e5} Edge-endpoint function: Edge e1 e2 e3 e4 e5

Endpoints {v1, v2} {v2, v3} {v2, v3} {v2, v4} {v4}

e1 1 2

e3 3

4. e2

e3 v1 e1

e4 v5

5. Imagine that the edges are strings and the vertices are knots. You can pick up the left-hand figure and lay it down again to form the right-hand figure as shown below. 5

e6 e7

e2 1

e4 3

Instructor’s Manual Section 1.4 6.

e v1

7. v

v5 e9

e5 v2

v3 e

e e

8. (i) e1, e2, e7 are incident on v1. (ii) v 1, v 2, and v 3 are adjacent to v 3. (iii) e2, e8, e9, and e3 are adjacent to e1. (iv) Loops are e6 and e7. (v) e8 and e9 are parallel; e4 and e5 are parallel. (vi) v 6 is an isolated vertex. (vii) degree of v3 = 5 9.

(i) e1, e2, e7 are incident on v1. (ii) v1 and v2 are adjacent to v3. (iii) e2 and e7 are adjacent to e1. (iv) e1 and e3 are loops. (v) e4 and e5 are parallel. (vi) v4 is an isolated vertex. (vii) degree of v3 = 2 10. a.Yes. According to the graph, Sports Illustrated is an instance of a sports magazine, a sports magazine is a periodical, and a periodical contains printed writing. b. Yes. According to the graph, Poetry Magazine is an instance of a Literary journal which is a Scholarly journal and, therefore, contains Long words. 11. (vvccB/) → (vc/Bvc) → (vvcB/c) → (c/Bvvc) → (vcB/vc) → (/Bvvcc) (vvccB/) → (vv/Bcc) → (vvcB/c) → (c/Bvvc) → (vcB/vc) → (/Bvvcc) (vvccB/) → (vv/Bcc) → (vvcB/c) → (c/Bvvc) → (ccB/vv) → (/Bvvcc)

Instructor’s Manual: Chapter 1

12. To solve this puzzle using a graph, introduce a notation in which, for example, wc/fg means that the wolf and the cabbage are on the left bank of the river and the ferryman and the goat are on the right bank. Then draw those arrangements of wolf, cabbage, goat, and ferryman that can be reached from the initial arrangement (wgcf /) and that are not arrangements to be avoided (such as (wg /fc)). At each stage ask yourself, “Where can I go from here?” and draw lines or arrows pointing to those arrangements. This method gives the graph shown below.

Examining the diagram reveals the solutions (wgcf /) → (wc/gf ) → (wcf /g ) → (w /gcf ) → (wgf /c) → (g /wcf ) → (gf /wc) → (/wgcf ) and (wgcf /) → (wc/gf ) → (wcf /g ) → (c/wgf ) → (gcf /w ) → (g /wcf ) → (gf /wc) → (/wgcf ) 13. vvvc/Bcc vvvccB/c vvvcccB/

vvcc/Bvc

vvv/Bccc

vvvcB/cc

vc/Bvvcc

vvvcc/Bc vcB/vvcc vvccB/vc

cccB/vvv

/Bvvvccc

c/Bvvvcc ccB/vvvc

ccB/vvvc

The diagram shows several solutions. Among them is (vvvcccB/) → (vvcc/Bvc) → (vvvccB/c) → (vvv/Bccc) → (vvvcB/cc) → (vc/Bvvcc) → (vvccB/vc) → (cc/Bvvvc) → (cccB/vvv) → (c/Bvvvcc) → (vcB/vvcc) → (/Bvvvccc), or one can end with (c/Bvvvcc) → (ccB/vvvc) → (/Bvvvccc), or one can start with (vvvcccB/) → (vvvc/Bcc) → (vvvccB/c). 14. Represent possible amounts of water in jugs A and B by ordered pairs with, say, the ordered pair (1,3) indicating that there is one quart of water in jug A and three quarts in jug B. Starting with (0,0), draw an edge from one ordered pair to another if it is possible to go from the situation represented by the one pair to that represented by the other and back by either

Instructor’s Manual Section 1.4

filling a jug from the tap, emptying a jug into the drain, or transferring water from one jug to another. Except for (0,0), only draw edges from states that have edges incident on them (since these are the only states that can be reached). The resulting graph is shown as follows: (0,5)

(1,5)

(2,5)

(0,4)

(1,4)

(2,4)

(3,5)

(3,4)

(0,3)

(1,3)

(2,3) (3,3)

(0,2)

(1,2)

(2,2)

(0,1)

(1,1)

(2,1)

(0,0)

(1,0)

(2,0)

(3,2)

(3,1)

(3,0)

It is clear from the graph that one solution is (0, 0) → (3, 0) → (0, 3) → (3, 3) → (1, 5) → (1, 0) and another solution is (0, 0) → (0, 5) → (3, 2) → (0, 2) → (2, 0) → (2, 5) → (3, 4) → (0, 4) → (3, 1) → (0, 1). Note that it would be possible to add arrows to the above graph from each reachable state to each other state that could be obtained from it either by filling one of the jugs to the top or by emptying the entire contents of one of the jugs. For instance, one could draw an arrow from (0,3) to (0,5) or from (0,3) to (0,0). Because the graph is connected, all such arrows would point to states already reachable by other means, so that it is not necessary to add such additional arrows to find solutions to the problem (and it makes the diagram look more complicated). However, if the problem were to find all possible solutions, the arrows would have to be added. 15.

3 b

1 a

c 3

1 e

Vertex e has maximal degree, so color it with color #1. Vertex a does not share an edge with e, and so color #1 may also be used for it. From the remaining uncolored vertices, all of d, g, and f have maximal degree. Choose any one of them—say, d—and use color #2 for it. Observe that vertices c and f do not share an edge with d, but they do share an edge with each other, which means that color #2 may be used for one but not the other. Choose to color f with color #2 because the degree of f is greater than the degree of c. The remaining uncolored vertices, b, c, and g, are unconnected, and so color #3 may be used for all three.

Instructor’s Manual: Chapter 1

16. Represent each committee name by a vertex, labeled with the first letter of the name of the committee, and join vertices if, and only if, the corresponding committees have a member in common. Figure (a) shows one way to color the graphs. Vertex H has maximum degree, so use color #1 for it. Vertex L does not share an edge with H, and so color #1 may also be used for it. From the remaining uncolored vertices, each of P , U , and G is adjacent to three other vertices. Choose any one of them, say P , and use color #2 for it. Vertices U and C do not share an edge with P or with each other, and so color #2 may also be used for them. Color #3 can then be used for the remaining vertex, G, at which point all vertices will be colored. Note that the same result is obtained if U is chosen instead of P in step 2. However, if G is chosen instead of P in step 2, the result is the coloring indicated in Figure (b). 2

3 C

2 P

2 C

3 P

L 1

(a)

(b)

To use the results of Figure (a) to schedule the meetings, let color n correspond to meeting time n. Then Time 1: hiring, library Time 2: personnel, undergraduate education, colloquium Time 3: graduate education Using the results of Figure (b) to schedule the meetings produces this result: Time 1: hiring, library Time 2: graduate education, colloquium Time 3: personnel, undergraduate education 17. In the following graph each course number is represented as a vertex. Vertices are joined if, and only if, the corresponding courses have a student in common. 3

101

102

2 100

110 4 135

120 130

Vertex 135 has maximum degree, so use color #1 for it. All vertices share edges with vertex 135, and so color #1 cannot be used on any other vertex. From the remaining uncolored vertices, only vertex 120 has maximum degree. So use color #2 for it. Because vertex 100 does not share an edge with vertex 120, color #2 may also be used for it. From the remaining uncolored vertices, all of 101, 102, 110, and 130 have maximum degree. Choose any one of them, say vertex 101, and use color #3 for it. Neither vertex 102 nor

Instructor’s Manual Section 1.4

vertex 110 shares an edge with vertex 101, but they do share an edge with each other. So color #3 may be used for only one of them. If color #3 is used for vertex 110, then, since the remaining vertices 130 and 102 are connected, two additional colors would be needed for them to have different colors. On the other hand, if color #3 is used for vertex 102, then, since the remaining vertices, 110 and 130, are not connected to each other, color 4 may be used for both. Therefore, to minimize the number of colors, color #3 should be used for vertex 102 and color #4 for vertices 110 and 130. The result is indicated in the annotations on the graph. To use the results for scheduling exams, let color n correspond to exam time n. Then Time 1: MCS135 Time 2: MCS 100 and MCS120 Time 3: MCS101 and MCS102 Time 4: MCS110 and MCS130 Note that because, for example, MSC135, MSC102, MSC110, and MSC 120 are all connected to each other, they must all be given different colors, and so the schedule for the seven exams must use at least four time periods.

Instructor’s Manual: Chapter 2

Chapter 2: The Logic of Compound Statements The ability to reason using the principles of logic is essential for solving problems in abstract mathematics and computer science and for understanding the reasoning used in mathematical proof and disproof. Because a significant number of students who come to college have had limited opportunity to develop this ability, a primary aim of Chapters 2 and 3 is to help students develop an inner voice that speaks with logical precision. Consequently, the various rules used in logical reasoning are developed both symbolically and in the context of their somewhat limited but very important use in everyday language. Exercise sets for Sections 2.1–2.3 and 3.1–3.4 contain sentences for students to negate, write the contrapositive for, and so forth. Virtually all students benefit from doing these exercises. Another aim of Chapters 2 and 3 is to teach students the rudiments of symbolic logic as a foundation for a variety of upper-division courses. Symbolic logic is used in, among others, the study of digital logic circuits, relational databases, artificial intelligence, and program verification.

Suggestions 1. In Section 2.1 a surprising number of students apply De Morgan’s law to write the negation of, for example, “1 < x≤3” as “1 ≥ x > 3.” You may find that it takes some effort to teach them to avoid making this mistake. 2. In Sections 2.1 and 2.4, students have more difficulty than you might expect simplifying statement forms and circuits. Only through trial and error can you learn the extent to which this is the case at your institution. If it is, you might either assign only the easier exercises or build in extra time to teach students how to do the more complicated ones. Discussion of simplification techniques occurs again in Chapter 6 in the context of set theory. At this later point in the course most students are able to deal with it successfully. 3. In ordinary English, the phrase “only if” is often used as a synonym for “if and only if.” But it is possible to find informal sentences for which the intuitive interpretation is the same as the logical definition. It is helpful to give examples of such statements when you introduce the logical definition. For instance, it is not hard to get students to agree that “The team will win the championship only if it wins the semifinal game” means the same as “If the team does not win the semifinal game then it will not win the championship.” Once students see this, you can suggest that they remember this example when they encounter more abstract “only if”statements. Through guided discussion, students also come to agree that the statement “Winning the semifinal game is a necessary condition for winning the championship” translates to “If the team does not win the semifinal game then it will not win the championship.” They can be encouraged to use this (or a similar statement) as a reference to help develop intuition for general statements of the form “A is a necessary condition for B.” With students who have weaker backgrounds, you may find yourself tying up excessive amounts of class time discussing “only if” and “necessary and sufficient conditions.” You might just assign the easier exercises, or you might assign exercises on these topics to be done for extra credit (putting corresponding extra credit problems on exams) and use the results to help distinguish A from B students. It is probably best not to omit these topics altogether, though, because the language of “only if” and “necessary and sufficient conditions” is a standard part of the technical vocabulary of textbooks used in upper-division courses, as well as occurring regularly in non-mathematical writing. 4. In Section 2.3, many students mistakenly conclude that an argument is valid if, when they compute the truth table, they find a single row in which both the premises and the conclusion are true. The source of students’ difficulty appears to be their tendency to ignore quantification and to misinterpret if-then statements as “and” statements. Since the definition of validity includes both a universal quantifier and if-then, it is helpful to go back over the definition and the procedures for testing for validity and invalidity after discussing the general topic of universal conditional statements

Solutions for Exercises: The Logic of Compound Statements

in Section 3.1. As a practical measure to help students assess validity and invalidity correctly, the first example in Section 2.3 is of an invalid argument whose truth table has eight rows, several of which have true premises and a true conclusion. To further focus students’ attention on the situations where all the premises are true, the truth values for the conclusions of arguments are omitted when at least one premise is false. 5. In Section 2.3, you might suggest that students just familiarize themselves with, but not memorize, the various forms of valid arguments covered in Section 2.3. It is wise, however, to have them learn the terms modus ponens and modus tollens (because these are used in some upper-division computer science courses) and converse and inverse errors (because these errors are so common).

Section 2.1 If p then q. p q

1. Common form: Therefore,

(a + 2b)(a2 − b) can be written in prefix notation. All algebraic expressions can be written in prefix notation. 2. Common form:

If p then q. ∼q ∼p

Therefore,

All prime numbers are odd. 2 is odd 3. Common form:

p∨q ∼p q

Therefore,

My mind is shot. Logic is confusing. 4. Common form:

If p then q. If q then r. If p then r.

Therefore,

Has 4 vertices and 6 edges. Is complete; Any two of its vertices can be joined by a path 5. a. It is a statement because it is a true sentence. 1,024 is a perfect square because 1,024 = 322, and the next smaller perfect square is 312 = 961, which has fewer than four digits. b. The truth or falsity of this sentence depends on the reference for the pronoun “she.” Considered on its own, the sentence cannot be said to be either true or false, and so it is not a statement. c. This sentence is false; hence it is a statement. d. This is not a statement because its truth or falsity depends on the value of x. 6. a. s ∧ i

b. ∼s ∧ ∼i

7. m ∧ ∼ c 8. a. (h ∧ w) ∧ ∼s

b. ∼ w ∧ (h ∧ s)

d. (∼w ∧ ∼s) ∧ h

c. ∼ w∧ ∼ h ∧ ∼ s

e. w∧ ∼ (h ∧ s) (w ∧ (∼ h ∨ ∼ s) is also acceptable)

9. a. p ∨ q

b. r ∧ p

c. r ∧ ( p ∨ q)

10. a. p ∧ q ∧ r

b. p ∧ ∼ q

c. p ∧ (∼q ∨ ∼r)

d. (∼ p ∧ q) ∧ ∼ r

e. ∼ p ∨ (q ∧ r)

Instructor’s Manual: Section 2.1

11. Inclusive or. For instance, a team could win the playoff by winning games 1, 3, and 4 and losing game 2. Such an outcome would satisfy both conditions. 12.

13.

p T T F F

q T F T F

∼p F F T T

∼p ∧ q F F T F

p∧q

p∨q

∼ (p ∧ q)

∼ (p ∧ q) ∨ (p ∨ q)

T T F F

T F T F

T F F F

T T T F

F T T T

T T T T

14.

15.

p q

q∧r

p ∧ (q ∧ r)

T T T T F F F F

T T F F T T F F

T F T F T F T F

T F F F T F F F

T F F F F F F F

∼q

∼q∨r

p ∧ (∼ q ∨ r)

T T T T F F F F

T T F F T T F F

T F T F T F T F

F F T T F F T T

T F T T T F T T

T F T T F F F F

p∧q

p ∨ (p ∧ q )

T T F F

T F T F

T F F F

T T F F

T T F F x

16.

˛¸

same truth values

The truth table shows that p ∨ (p ∧ q) and p always have the same truth values. Thus they are logically equivalent. (This proves one of the absorption laws.)

Solutions for Exercises: The Logic of Compound Statements

4 17.

p T T F F

q T F T F

p∧q T F F F

∼p F F T T

∼q F T F T

∼ (p ∧ q) F T T T `

∼ p∧ ∼ q F F F T ˛¸

← ← x

different truth values in rows 2 and 3

The truth table shows that ∼ (p ∧ q) and ∼ p ∧ ∼ q do not always have the same truth values. Therefore they are not logically equivalent. 18. p

p∨t

T F

T T `

T T

˛¸

same truth values

The truth table shows that p ∨ t and t always have the same truth values. Thus they are logically equivalent. (This proves one of the universal bound laws.) 19.

p ∧t

T F

T T

T F

˛¸

same truth values

The truth table shows that p ∧ t and p always have the same truth values. Thus they are logically equivalent. This proves the identity law for ∧. 20.

p T F

c F F

p ∧c F F `

p ∨c T F ˛¸

← x

different truth values in row 1

The truth table shows that p ∧ c and p ∨c do not always have the same truth values. Thus they are not logically equivalent. 21. p

p q r

p∧q

q ∧r

(p ∧ q ) ∧ r

p ∧ (q ∧ r )

T T T T F F F F

T T F F T T F F

T F T F T F T F

T T F F F F F F

T F F F T F F F

T F F F F F F F

˛¸

same truth values

The truth table shows that (p ∧ q) ∧ r and p ∧ (q ∧ r) always have the same truth values. Thus they are logically equivalent. (This proves the associative law for ∧.)

Instructor’s Manual: Section 2.1 22.

q∨r

p∧q

p∧r

p ∧ (q ∨ r)

(p ∧ q) ∨ (p ∧ r)

T T T T F F F F

T T F F T T F F

T F T F T F T F

T T T F T T T F

T T F F F F F F

T F T F F F F F

T T T F F F F F

˛¸

same truth values

The truth table shows that p ∧ (q ∨ r) and (p ∧ q) ∨ (p ∧ r) always have the same truth values. Therefore they are logically equivalent. This proves the distributive law for ∧ over ∨. 23. p T T T T F F F F

q T T F F T T F F

r T F T F T F T F

p∧q T T F F F F F F

q ∨r T T T F T T T F

(p ∧ q ) ∨ r T T T F T F T F `

p ∧ (q ∨ r ) T T T F F F F F ˛¸

← ← x

different truth values in rows 5 and 7

The truth table shows that (p ∧ q) ∨ r and p ∧ (q ∨ r) have different truth values in rows 5 and 7. Thus they are not logically equivalent. (This proves that parentheses are needed with ∧ and ∨.) 24.

p T T T T F F F F

q T T F F T T F F

r T F T F T F T F

p∨q T T T T T T F F

p∧r T F T F F F F F

(p ∨ q) ∨ (p ∧ r) T T T T T T F F `

(p ∨ q) ∧ r T F T F T F F F ˛¸

← ← ← x

different truth values in rows 2, 3, and 6

The truth table shows that (p ∨ q) ∨ (p ∧ r) and (p ∨ q) ∧ r have different truth values in rows 2, 3, and 6. Hence they are not logically equivalent. 25. Hal is not a math major or Hal’s sister is not a computer science major. 26. Sam is not an orange belt or Kate is not a red belt. 27. The connector is not loose and the machine is not unplugged. 28. The train is not late and my watch is not fast.

Solutions for Exercises: The Logic of Compound Statements

29. This computer program does not have a logical error in the first ten lines and it is not being run with an incomplete data set. 30. The dollar is not at an all-time high or the stock market is not at a record low. 31. a. 01, 02, 11, 12

b. 21, 22

c. 11, 10, 21, 20

32. −2 ≥ x or x ≥ 7 33. −10 ≥ x or x ≥ 2 34. 2 ≤ x ≤ 5 35. x > −1 and x ≤ 1 36. 1 ≤ x or x < −3 37. 0 ≤ x or x < −7 38. This statement’s logical form is (p ∧ q) ∨ r, so its negation has the form ∼((p ∧ q) ∨ r) ≡ ∼(p ∧ q) ∧ ∼r ≡ (∼p ∨ ∼q) ∧ ∼r. Thus a negation for the statement is (num orders ≤ 100 or num instock > 500) and num instock ≥ 200. 39. The statement’s logical form is (p ∧ q) ∨ ((r ∧ s) ∧ t), so its negation has the form ∼ ((p ∧ q) ∨ ((r ∧ s) ∧ t))

≡ ≡ ≡

∼ (p ∧ q) ∧ ∼ ((r ∧ s) ∧ t)) (∼ p ∨ ∼ q) ∧ (∼ (r ∧ s)∨ ∼ t)) (∼ p ∨ ∼ q) ∧ ((∼ r ∨ ∼ s) ∨ ∼ t)).

Thus a negation is (num orders ≥ 50 or num instock ≤ 300) and ((50 > num orders or num orders ≥ 75) or num instock ≤ 500). 40. q

∼p

∼q

p∧q

p∧∼q

∼ p ∨ (p∧ ∼ q )

(p ∧ q ) ∨ (∼ p ∨ (p ∧ ∼ q ))

T T T F F T F F

F F T T

F T F T

T F F F

F T F F

F T T T

T T T T

Since all the truth values of (p ∧ q) ∨ (∼p ∨ (p ∧ ∼q)) are T, (p ∧ q) ∨ (∼p ∨ (p ∧ ∼q)) is a tautology. 41. q

∼p

∼q

p∧∼q

∼p∨q

(p ∧ ∼ q )(p ∨ q )

T T T F F T F F

F F T T

F T F T

F T F F

T F T T

F F F F

Since all the truth values of (p ∧ ∼q) ∧ (∼p ∨ q) are F, (p ∧ ∼q) ∧ (∼p ∨ q) is a contradiction.

Instructor’s Manual: Section 2.1 22.

∼p

∼q

∼p∧q

q∧r

((∼ p ∧ q) ∧ (q ∧ r))

((∼ p ∧ q) ∧ (q ∧ r))∧ ∼ q

T T T T F F F F

T T F F T T F F

T F T F T F T F

F F F F T T T T

F F T T F F T T

F F F F T T F F

T F F F T F F F

F F F F T F F F

F F F F F F F F ˛¸ all F ′ s

Since all the truth values of (( ∼ p ∧ q) ∧ (q ∧ r)) ∧ ∼ q are F , (( ∼ p ∧ q) ∧ (q ∧ r)) ∧ ∼ q is a contradiction. 43.

∼p

∼q

∼p∨q

p∧ ∼ q

(∼ p ∨ q) ∨ (p ∧ ∼ q)

T T F F

T F T F

F F T T

F T F T

T F T T

F T F F

T T T T ˛¸ all T ′ s

44. a. No real numbers satisfy this inequality b. No real numbers satisfy this inequality. 45. Let b be “Bob is a double math and computer science major,” m be “Ann is a math major,” and a be “Ann is a double math and computer science major.” Then the two statements can be symbolized as follows: a. (b∧ m)∧ ∼ a and b. ∼ (b ∧ a) ∧ (m ∧ b). Note: The entries in the truth table assume that a person who is a double math and computer science major is also a math major and a computer science major. b T T T T F F F F

m T T F F T T F F

a T F T F T F T F

∼a F T F T F T F T

b∧m T F T F F F F F

m∧b T T F F F F F F

b∧a T F T F F F F F

∼ (b ∧ a) F T F T T T T T

(b ∧ m)∧ ∼ a ∼ (b ∧ a) ∧ (m ∧ b) F F T T F F F F F F F F F F F F ` ˛¸ x same truth values

The truth table shows that (b ∧ m) ∧ ∼ a and ∼ (b ∧ a) ∧ (m ∧ b) always have the same truth values. Hence they are logically equivalent. 46. a. Solution 1: Construct a truth table for p ⊕ p using the truth values for exclusive or. p T F

p ⊕p F F

because an exclusive or statement is false when both components are true and when both components are false, and the two components in p ⊕ p are both p. Since all its truth values are false, p ⊕ p ≡ c, a contradiction.

Solutions for Exercises: The Logic of Compound Statements

Solution 2: Replace q by p in the logical equivalence p ⊕ q ≡ (p ∨ q) ∧ ∼ (p ∧ q), and simplify the result. p ⊕ p ≡ (p ∨ p) ∧ ∼(p ∧ p) by definition of ⊕ ≡ p ∧ ∼p by the identity laws ≡c by the negation law for ∧ b. Yes.

p⊕q

q⊕r

(p ⊕ q) ⊕ r

p ⊕ (q ⊕ r)

T T T T F F F F

T T F F T T F F

T F T F T F T F

F F T T T T F F

F T T F F T T F

T F F T F T T F

˛¸ same truth values

The truth table shows that (p ⊕ q) ⊕ r and p ⊕ (q ⊕ r) always have the same truth values. So they are logically equivalent. c. Yes.

p⊕q

p∧r

q∧r

(p ⊕ q) ∧ r

(p ∧ r) ⊕ (q ∧ r)

T T T T F F F F

T T F F T T F F

T F T F T F T F

F F T T T T F F

T F T F F F F F

T F F F T F F F

F F T F T F F F

˛¸ same truth values

The truth table shows that (p ⊕q) ∧ r and (p ∧ r) ⊕(q ∧ r) always have the same truth values. So they are logically equivalent. 47. There is a famous story about a philosopher who once gave a talk in which he observed that whereas in English and many other languages a double negative is equivalent to a positive, there is no language in which a double positive is equivalent to a negative. To this, another philosopher, Sidney Morgenbesser, responded sarcastically, “Yeah, yeah.” [Strictly speaking, sarcasm functions like negation. When spoken sarcastically, the words “Yeah, yeah” are not a true double positive; they just mean “no.”] 48.

a. the distributive law c. the negation law for ∨

49.

a. the commutative law for ∨ b. the distributive law c. the negation law for ∧ d. the identity law for ∨ (p ∧ ∼q) ∨ p ≡ p ∨ (p ∧ ∼q) by the commutative law for ∨ ≡p by the absorption law (with ∼q in place of q )

50.

51. Solution 1 : p ∧ (∼ q ∨ p)

b. the commutative law for ∨ d. the identity law for ∧

≡ ≡

p ∧ (p ∨ ∼ q) p

commutative law for ∨ absorption law

Instructor’s Manual: Section 2.2 Solution 2 : p ∧ (∼ q ∨ p)

≡ ≡ ≡

(p ∧ ∼ q) ∨ (p ∧ p) (p ∧ ∼ q) ∨ p p

52.

∼ (p ∨ ∼ q) ∨ (∼ p ∧ ∼ q)

≡ ≡ ≡ ≡ ≡

(∼ p ∧ ∼ (∼ q)) ∨ (∼ p ∧ ∼ q) (∼ p ∧ q) ∨ (∼ p ∧ ∼ q) ∼ p ∧ (q∨ ∼ q) ∼ p∧ t ∼p

53.

∼((∼p ∧ q) ∨ (∼p ∧ ∼q)) ∨ (p ∧ q)

54.

(p ∧ (∼ (∼ p ∨ q))) ∨ (p ∧ q)

≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡

distributive law identity law for ∧ by exercise 50. De Morgan’s law double negative law distributive law negation law for ∨ identity law for ∧

≡ ∼[∼p ∧ (q ∨ ∼q)] ∨ (p ∧ q) by the distributive law ≡ ∼(∼p ∧ t) ∨ (p ∧ q) by the negation law for ∨ ≡ ∼(∼p) ∨ (p ∧ q) by the identity law for ∧ ≡ p ∨ (p ∧ q) by the double negative law ≡ p by the absorption law

(p ∧ (∼ (∼ p)∧ ∼ q)) ∨ (p ∧ q) (p ∧ (p∧ ∼ q)) ∨ (p ∧ q) ((p ∧ p)∧ ∼ q)) ∨ (p ∧ q) (p ∧ ∼ q)) ∨ (p ∧ q) p ∧ (∼ q ∨ q) p ∧ (q∨ ∼ q) p∧ t p

De Morgan’s law double negative law associative law for ∧ idempotent law for ∧ distributive law commutative law for ∨ negation law for ∨ identity law for ∧

Section 2.2 1. If this loop does not contain a stop or a go to, then it will repeat exactly N times. 2. If I catch the 8:05 bus, then I am on time for work. 3. If you do not freeze, then I’ll shoot. 4. If you don’t fix my ceiling, then I won’t pay my rent. 5.

p T T F F

q T F T F

∼p F F T T

∼q F T F T

∼p

∼p∧q

p∨q

(p ∨ q) ∨ (∼ p ∧ q)

(p ∨ q) ∨ (∼ p ∧ q) → q

T T F F

T F T F

F F T T

F F T F

T T T F

T F T T

p T T T T F F F F

q T T F F T T F F

r T F T F T F T F

∼q F F T T F F T T

∼p∨q T F T T

p ∧∼q F F T T F F F F

∼ p ∨ q →∼ q F T F T

p ∧∼q →r T T T F T T T T

Solutions for Exercises: The Logic of Compound Statements

10 8.

∼p

∼p∨q

∼p∨q →r

T T T T F F F F

T T F F T T F F

T F T F T F T F

F F F F T T T T

T T F F T T T T

T F T T T F T F

p T T T T F F F F

q T T F F T T F F

r T F T F T F T F

∼r F T F T F T F T

p ∧∼r F T F T F F F F

p→r

q→r

(p → r) ↔ (q → r)

T T T T F F F F

T T F F T T F F

T F T F T F T F

T F T F T T T T

T F T T T F T T

T T T F T F T T

q→r

p → (q → r)

p∧q

p∧q →r

(p → (q → r)) ↔ (p ∧ q → r)

T T T T F F F F

T T F F T T F F

T F T F T F T F

T F T T T F T T

T F T T T T T T

T T F F F F F F

T F T T T T T T

T T T T T T T T

10.

11.

q ∧r T T T F T T T F

p ∧∼r ↔q ∨r F T F F F F F T

12. If x > 2 then x2 > 4, and if x < −2 then x2 > 4. 13. a.

∼p

p→q

∼p ∧q

T T F F

T F T F

F T F T

T F T T

˛¸

same truth values

Instructor’s Manual: Section 2.2

The truth table shows that p → q and ∼p ∨ q always have the same truth values. Hence they are logically equivalent. b.

∼q

p→q

∼ (p → q)

p∧∼q

T T F F

T F T F

F T F T

T F T T

F T F F

˛¸

same truth values

The truth table shows that ∼ (p → q) and p ∧ ∼ q always have the same truth values. Hence they are logically equivalent. 14. a. p T T T T F F F F

q T T F F T T F F

r T F T F T F T F

∼q F F T T F F T T

∼r F T F T F T F T

q∨r T T T F T T T F

p∧ ∼ q F F T T F F F F

p∧ ∼ r F T F T F F F F

p →q∨r T T T F T T T T `

p ∧∼q→r p ∧∼r→q T T T T T T F F T T T T T T T T ˛¸ x same truth values

The truth table shows that the three statement forms p → q ∨ r, p ∧ ∼ q → r, and p ∧ ∼ r → q always have the same truth values. Thus they are all logically equivalent. b. If n is prime and n is not odd, then n is 2. And: If n is prime and n is not 2, then n is odd. 15.

p T T T T F F F F

q T T F F T T F F

r T F T F T F T F

q→r T F T T T F T T

p→q T T F F T T T T

p → (q → r) (p → q) → r T T F F T T T T T T T F T F T F ` ˛¸ x

← ← ←

different t r uth values

The truth table shows that p →(q →r) and (p → q) → r do not always have the same truth values. (They differ for the combinations of truth values for p, q, and r shown in rows 6, 7, and 8.) Therefore they are not logically equivalent. 16. Let p represent “You paid full price” and q represent “You didn’t buy it at Crown Books.” Thus, “If you paid full price, you didn’t buy it at Crown Books” has the form p → q. And “You didn’t buy it at Crown Books or you paid full price” has the form q ∨ p.

Solutions for Exercises: The Logic of Compound Statements p q T T T F F T F F

p→q T F T T `

q ∨p T T T F ˛¸

← ← x

different truth values

These two statements are not logically equivalent because their forms have different truth values in rows 2 and 4. (An alternative representation for the forms of the two statements is p → ∼ q and ∼q ∨ p. In this case, the truth values differ in rows 1 and 3.) 17. Let p represent “2 is a factor of n,” q represent “3 is a factor of n,” and r represent “6 is a factor of n.” The statement “If 2 is a factor of n and 3 is a factor of n, then 6 is a factor of n” has the form p ∧ q → r. And the statement “2 is not a factor of n or 3 is a not a factor of n or 6 is a factor of n” has the form ∼ p ∨ ∼ q ∨ r. p

∼p

∼q

p∧q

p∧q →r

∼ p∨ ∼ q ∨ r

T T T T F F F F

T T F F T T F F

T F T F T F T F

F F F F T T T T

T T F F T T F F

T T F F F F F F

T F T T T T T T

˛¸

same truth values

The truth table shows that p ∧ q → r and ∼ p ∨ ∼ q ∨ r always have the same truth values. Therefore they are logically equivalent. 18. Part 1 : Let p represent “It walks like a duck,” q represent “It talks like a duck,” and r represent “It is a duck.” The statement “If it walks like a duck and it talks like a duck, then it is a duck” has the form p ∧ q → r. And the statement “Either it does not walk like a duck or it does not talk like a duck or it is a duck” has the form ∼ p ∨ ∼ q ∨ r. p

∼p

∼q

p ∧q

∼ p∨ ∼ q

p ∧q →r

(∼ p ∨ ∼ q) ∨ r

T T T T F F F F

T T F F T T F F

T F T F T F T F

F F F F T T T T

F F T T F F T T

T T F F F F F F

F F T T T T T T

T F T T T T T T

˛¸

same truth values

The truth table shows that p ∧q → r and ( ∼ p ∨ ∼ q)∨ r always have the same truth values. Thus the following statements are logically equivalent:“If it walks like a duck and it talks like a duck, then it is a duck” and “Either it does not walk like a duck or it does not talk like a duck or it is a duck.”

Instructor’s Manual: Section 2.2

Part 2 : The statement “If it does not walk like a duck and it does not talk like a duck then it is not a duck” has the form ∼ p ∧ ∼ q →∼ r. p T T T T F F F F

q T T F F T T F F

r T F T F T F T F

∼p F F F F T T T T

∼q F F T T F F T T

∼r F T F T F T F T

p ∧q T T F F F F F F

∼ p∧ ∼ q F F F F F F T T

p ∧ q → r (∼ p ∧ ∼ q) →∼ r T T F T T T T T T T T T T F T T ` ˛ ¸ x different t r uth values

←

The truth table shows that p ∧ q →r and ( ∼p ∧ ∼ q) →∼ r do not always have the same truth values. (They differ for the combinations of truth values of p, q, and r shown in rows 2 and 7.) Thus they are not logically equivalent, and so the statement “If it walks like a duck and it talks like a duck, then it is a duck” is not logically equivalent to the statement “If it does not walk like a duck and it does not talk like a duck then it is not a duck.” In addition, because of the logical equivalence shown in Part 1, we can also conclude that the following two statements are not logically equivalent: “Either it does not walk like a duck or it does not talk like a duck or it is a duck” and “If it does not walk like a duck and it does not talk like a duck then it is not a duck.” 19. False. The negation of an if-then statement is not an if-then statement. It is an and statement. 20. a. Negation: P is a square and P is not a rectangle. b. Negation: Today is New Year’s Eve and tomorrow is not January. c. Negation: The decimal expansion of r is terminating and r is not rational. d. Negation: n is prime and both n is not odd and n is not 2. Or: n is prime and n is neither odd nor 2. e. Negation: x is nonnegative and x is not positive and x is not 0. Or: x is nonnegative but x is not positive and x is not 0. Or: x is nonnegative and x is neither positive nor 0. f. Negation: Tom is Ann’s father and either Jim is not her uncle or Sue is not her aunt. g. Negation: n is divisible by 6 and either n is not divisible by 2 or n is not divisible by 3. 21. By assumption, p → q is false. By definition of a conditional statement, the only way this can happen is for the hypothesis, p, to be true and the conclusion, q, to be false. a. The only way ∼ p → q can be false is for ∼ p to be true and q to be false. But since p is true, ∼ p is false. Hence ∼ p → q is not false and so it is true. b. Since p is true, then p ∨ q is true because if one component of an and statement is true, then the statement as a whole is true. c The only way q → p can be false is for q to be true and p to be false. Thus, since q is false, q → p is not false and so it is true. 22. a. Contrapositive: If P is not a rectangle, then P is not a square. b. Contrapositive: If tomorrow is not January, then today is not New Year’s Eve. c. Contrapositive: If r is not rational, then the decimal expansion of r is not terminating. d. Contrapositive: If n is not odd and n is not 2, then n is not prime.

Solutions for Exercises: The Logic of Compound Statements e. Contrapositive: If x is not positive and x is not 0, then x is not nonnegative. Or: If x is neither positive nor 0, then x is negative. f. Contrapositive: If either Jim is not Ann’s uncle or Sue is not her aunt, then Tom is not her father. g. Contrapositive: If n is not divisible by 2 or n is not divisible by 3, then n is not divisible by 6.

23. a. Converse: If P is a rectangle, then P is a square. Inverse: If P is not a square, then P is not a rectangle. b. Converse: If tomorrow is January, then today is New Year’s Eve. Inverse: If today is not New Year’s Eve, then tomorrow is not January. c. Converse: If r is rational then the decimal expansion of r is terminating. Inverse: If the decimal expansion of r is not terminating, then r is not rational. d. Converse: If n is odd or n is 2, then n is prime. Inverse: If n is not prime, then n is not odd and n is not 2. e. Converse: If x is positive or x is 0, then x is nonnegative. Inverse: If x is not nonnegative, then both x is not positive and x is not 0. Or: If x is negative, then x is neither positive nor 0. f. Converse: If Jim is Ann’s uncle and Sue is her aunt, then Tom is her father. Inverse: If Tom is not Ann’s father, then Jim is not her uncle or Sue is not her aunt g. Converse: If n is divisible by 2 and n is divisible by 3, then n is divisible by 6 Inverse: If n is not divisible by 6, then n is not divisible by 2 or n is not divisible by 3. 24.

p T T F F

q T F T F

p→q T F T T `

q→p T T F T ˛¸

← ← x

different truth values

The truth table shows that p → q and q → p have different truth values in the second and third rows. Hence they are not logically equivalent. 25.

p T T F F

q T F T F

∼p F F T T

∼q F T F T

p→q T F T T `

∼ p →∼ q T T F T ˛¸ x

← ←

different truth values

The truth table shows that p → q and ∼ p →∼ q have different truth values in rows 2 and 3, so they are not logically equivalent. Thus a conditional statement is not logically equivalent to its inverse.

Instructor’s Manual: Section 2.2 26.

∼p

∼q

∼ q →∼ p

p→q

T T F F

T F T F

F F T T

F T F T

T F T T

˛¸

same truth values

The truth table shows that ∼q → ∼p and p → q always have the same truth values and thus are logically equivalent. It follows that a conditional statement and its contrapositive are logically equivalent to each other.

27.

∼p

∼q

q→p

∼ p →∼ q

T T F F

T F T F

F F T T

F T F T

T T F T

˛¸

same truth values

The truth table shows that q → p and ∼ p →∼ q always have the same truth values and thus are logically equivalent. It follows that the converse and inverse of a conditional statement are logically equivalent to each other.

28. The if-then form of “I say what I mean” is “If I mean something, then I say it.” The if-then form of “I mean what I say” is “If I say something, then I mean it.” Thus “I mean what I say” is the converse of “I say what I mean,” and so the two statements are not logically equivalent.

29. The corresponding tautology is (p → (q ∨ r)) ↔ ((p ∧ ∼q) → r) p

∼q

q ∨r

p∧∼q

p → (q ∨ r )

p∧∼q →r

T T T T F F F F

T T F F T T F F

T F T F T F T F

F F T T F F T T

T T T F T T T F

F F T T F F F F

T T T F T T T T

(p → (q ∨ r )) ↔ ((p ∧ ∼ q ) → r ) T T T T T T T T

The truth table shows that (p → (q ∨ r)) ↔ ((p ∧ ∼q) → r) is a tautology because all of its truth values are T.

30. The corresponding tautology is p ∧ (q ∨ r) ↔ (p ∧ q) ∨ (p ∧ r)

Solutions for Exercises: The Logic of Compound Statements

q∨r

p ∧q

p ∧r

p ∧ (q ∨ r)

(p ∧ q) ∨ (p ∧ r)

T T T T F F F F

T T F F T T F F

T F T F T F T F

T T T F T T T F

T T F F F F F F

T F T F F F F F

T T T F F F F F

p ∧ (q ∨ r) ↔ (p ∧ q) ∨ (p ∧ r) T T T T T T T T ` ˛¸ x all T ’s

The truth table shows that p ∧(q ∨ r) ↔ (p ∧ q) ∨(p ∧ r) is always true. Hence it is a tautology. 31. The corresponding tautology is (p → (q → r)) ↔ ((p ∧ q) → r). p

q→r

p ∧q

p → (q → r)

(p ∧ q) → r)

p → (q → r) ↔ (p ∧ q) → r

T T T T F F F F

T T F F T T F F

T F T F T F T F

T F T T T F T T

T T F F F F F F

T F T T T T T T

T T T T T T T T ˛¸ all T ’s

The truth table shows that (p → (q → r)) ↔ ((p ∧ q) → r) is always true. Hence it is a tautology. 32. If this quadratic equation has two distinct real roots, then its discriminant is greater than zero, and if the discriminant of this quadratic equation is greater than zero, then the equation has two real roots. 33. If this integer is even, then it equals twice some integer, and if this integer equals twice some integer, then it is even. 34. If the Cubs do not win tomorrow’s game, then they will not win the pennant. If the Cubs win the pennant, then they will have won tomorrow’s game. 35. If Sam is not an expert sailor, then he will not be allowed on Signe’s racing boat. If Sam is allowed on Signe’s racing boat, then he is an expert sailor. 36. The Personnel Director did not lie. By using the phrase “only if,” the Personnel Director set forth conditions that were necessary but not sufficient for being hired: if you did not satisfy those conditions then you would not be hired. The Personnel Director’s statement said nothing about what would happen if you did satisfy those conditions. 37. If a new hearing is not granted, payment will be made on the fifth. 38. If it doesn’t rain, then Ann will go. 39. If a security code is not entered, then the door will not open. 40. If I catch the 8:05 bus, then I am on time for work. 41. If this triangle has two 45◦ angles, then it is a right triangle.

Instructor’s Manual: Section 2.2

42. If this number is not divisible by 3, then it is not divisible by 9. If this number is divisible by 9, then it is divisible by 3. 43. If Jim does not do his homework regularly, then Jim will not pass the course. If Jim passes the course, then he will have done his homework regularly. 44. If Jon’s team wins the rest of its games, then it will win the championship. 45. If this computer program produces error messages during translation, then it is not correct. If this computer program is correct, then it does not produce error messages during translation. 46. a. This statement is the converse of the given statement, and so it is not necessarily true. For instance, if the actual boiling point of compound X were 200◦C, then the given statement would be true but this statement would be false. b. This statement must be true. It is the contrapositive of the given statement. c. must be true

d. not necessarily true e. must be true

f. not necessarily true

Note: To solve this problem, it may be helpful to imagine a compound whose boiling point is greater than 150◦ C. For concreteness, suppose it is 200◦ C. Then the given statement would be true for this compound, but statements a, d, and f would be false. 47. a. p ∧ ∼q → r ≡ ∼(p ∧ ∼q) ∨ r b.

p ∧ ∼q → r

≡

∼(p ∧ ∼q) ∨ r

≡ ≡

∼[∼(∼(p ∧ ∼q)) ∧ ∼r] ∼[(p ∧ ∼q) ∧ ∼r]

by the identity for → shown in the directions [an acceptable answer] by De Morgan’s law [another acceptable answer] by the double negative law [another acceptable answer]

Any of the expressions in part (b) would also be acceptable answers for part (a). 48. a.

p∨ ∼ q → r ∨ q

≡

∼ (p ∨ ∼ q) ∨ (r ∨ q)

≡

(∼ p ∧ ∼ (∼ q)) ∨ (r ∨ q)

≡

(∼ p ∧ q) ∨ (r ∨ q)

by the identity for → shown in the directions [an acceptable answer] by De Morgan’s law [another acceptable answer] by the double negative law [another acceptable answer]

≡ (∼ p ∧ q) ∨ (r ∨ q) by part (a) ≡ ∼ (∼ (∼ p ∧ q)∧ ∼ (r ∨ q)) by De Morgan’s law ≡ ∼ (∼ (∼ p ∧ q) ∧ (∼ r ∧ ∼ q)) by De Morgan’s law Any of the expressions in part (b) would also be acceptable answers for part (a). 49. a.

p∨ ∼ q → r ∨ q

(p → r) ↔ (q → r)

≡ ≡

(∼p ∨ r) ↔ (∼q ∨ r) [∼(∼p ∨ r) ∨ (∼q ∨ r)] ∧ [∼(∼q ∨ r) ∨ (∼p ∨ r)] by the identity for ↔ shown in the

≡

[(p ∧ ∼r) ∨ (∼q ∨ r)] ∧ [(q ∧ ∼r) ∨ (∼p ∨ r)]

directions [an acceptable answer] by De Morgan’s law [another acceptable answer]

(∼p ∨ r) ↔ (∼q ∨ r)

≡

∼[∼(p ∧ ∼r) ∧ ∼(∼q ∨ r)] ∧ ∼[∼(q ∧ ∼r) ∧ ∼(∼p ∨ r)]

≡

∼[∼(p ∧ ∼r) ∧ (q ∧ ∼r)] ∧ ∼[∼(q ∧ ∼r) ∧ (p ∧ ∼r)]

by De Morgan’s law by De Morgan’s law

Any of the expressions in part (b) would also be acceptable answers for part (a).

Solutions for Exercises: The Logic of Compound Statements

18 50. a.

(p → (q → r)) ↔ ((p ∧ q) → r)

≡ ≡ ≡

[∼ p ∨ (q → r)] ↔ [∼ (p ∧ q) ∨ r] [∼ p ∨ (∼ q ∨ r)] ↔ [∼ (p ∧ q) ∨ r] ∼ [∼ p ∨ (∼ q ∨ r)] ∨ [∼ (p ∧ q) ∨ r] ∧∼ [∼ (p ∧ q) ∨ r] ∨ [∼ p ∨ (∼ q ∨ r)]

b. By part (a), De Morgan’s law, and the double negative law, (p → (q → r)) ↔ ((p ∧ q) → r)

≡

∼ [∼ p ∨ (∼ q ∨ r)] ∨ [∼ (p ∧ q) ∨ r] ∧∼ [∼ (p ∧ q) ∨ r] ∨ [∼ p ∨ (∼ q ∨ r)] ≡ ∼ [∼ p ∨ (∼ q ∨ r)]∧ ∼ [∼ (p ∧ q) ∨ r] ∧ ∼ ∼ [(p ∧ q)∧ ∼ r]∧ ∼ [∼ p ∨ (∼ q ∨ r)] ≡ ∼ ∼ [p ∧ ∼ (∼ q ∨ r)] ∧ [(p ∧ q)∧ ∼ r] ∧ ∼ ∼ [(p ∧ q)∧ ∼ r] ∧ [p ∧ ∼ (∼ q ∨ r)] ≡ ∼ ∼ [p ∧ (q ∧ ∼ r)] ∧ [(p ∧ q)∧ ∼ r] ∧ ∼ ∼ [(p ∧ q)∧ ∼ r] ∧ [p ∧ (q∧ ∼ r)]. Any of the expressions in the right-hand column would also be acceptable answers for part (a). 51. Yes. As in exercises 47-50, the following logical equivalences can be used to rewrite any statement form in a logically equivalent way using only ∼ and ∧: p → q ≡∼ p ∨ q p ∨ q ≡∼ (∼ p ∧ ∼ q)

p ↔ q ≡ (∼ p ∨ q) ∧ (∼ q ∨ p) ∼ (∼ p) ≡ p

The logical equivalence p ∧ q ≡ ∼ (∼ p ∨ ∼ q) can then be used to rewrite any statement form in a logically equivalent way using only ∼ and ∨.

Section 2.3 1.

√ 2 is not rational.

2. 1 − 0.99999... is less than every positive real number. 3. Logic is not easy. 4. This graph cannot be colored with two colors. 5. They did not telephone. 6.

conclusion

premises

p T

q T

p ¸

q p x` →

q ˛

p∨q T (

critical row

→

T T critical row T F F T Rows 2 and 4T of the truth table are the critical rows in which all the premises are true, but F T F row 4 shows that it is possible for (argument of this form to have true premises and a false Fan F T T T conclusion. Thus this argument form is invalid.

Instructor’s Manual: Section 2.2 7.

conclusion

premises

p T

q T

∼q F

r T

∼ ∨ T

→

r T

critical row

T T T F F T T F T F T T T F T T F F T T F T F T T F F T T F T F F F T F This row describes the only situation in which all the premises are true. Because the conclusion F Tform is valid. T isFalsoFtrue Fhere, T the argument 8.

conclusion

premises

p T

q T

∼q F

r T

∨ ¸T

→ T ˛

x` →∼

critical row

F T T F F T F F critical row T ( critical row T F T T T T T T F F T T T F T ( F T T F T T T F ( F T F F T T T This row shows that it is possible for an argument of this form to have true premises and a F conclusion. F T TThus this F argument T form isTinvalid. false F F F T F T T conclusion

premises

p T

q T

r T

∼q F

∼r F

p∧q T

p ¸

q ∧ →∼ F

p q x` ∨∼

q ∼

p ˛

→ T

∼r

( critical row critical row critical row

T T T F F T T T T T T ( F ( T F T T F F T T T T F F T T F T T T T F T T F F F T F T F T F F T F T F T F F T T F F T T F Rows 2, 3, and 4 of the truth table are the critical rows in which all the premises are true, but F F F T T F T T F

row 3 shows that it is possible for an argument of this form to have true premises and a false conclusion. Hence the argument form is invalid. 10. premise

p T T T T F F F F

q T T F F T T F F

r T F T F T F T F

∼p F F F F T T T T

∼q F F T T F F T T

∼r F T F T F T F T

∼ p∧ ∼ q F F F F F F T T

p ∨q T T T T T T F F

r T F T F T F T F

p ∨q →r T F T F T F T T

conclusion

∼ r → ∼ p∧ ∼ q T ( critical row F T ( critical row F T ( critical row F T( T(

critical row critical row

Solutions for Exercises: The Logic of Compound Statements

This form of argument has just one premise. Rows 1, 3, 5, 7, and 8 of the truth table represent all the situations in which the premise is true, and in each of these rows the conclusion is also true. Therefore, the argument form is valid. 11.

conclusion

premises

∼p

∼q

∼r

q∨r

q x`

→

∨

r ˛

∼ ∨∼ F

T T

∼ p∨ ∼ r

(

T ( F

critical row critical row

critical row critical row critical row

T F F F T T F F T F T T T F F T T F ( F T2, 3,F6, 7,Tand 8Fof the T truth Ttable represent T T Rows the situations in which all the premises are

( true, forTan argument F but F row T 3 TshowsTthat it F is possible T T of this form to have true premises ( and a false conclusion. Hence the argument form is invalid. F

12. a.

q T

¸ x` p→q

conclusion

premises

p T

p( T

˛ q

critical row critical row

T T T F F F Rows 1 and 3 F T T of the T truth table F ( represent the situations in which all the premises are true, but F row T 3 shows T that F it is possible for an argument of this form to have true premises and a false conclusion. Hence the argument form is invalid. b.

conclusion

premises

p T

q T

¸ x` ˛ p→q ∼p T F F T

∼q T (

critical row

critical row F T T F F T T T F ( Rows 2 and 4 of the truth table represent the situations in which all the premises are true, but row 4 shows that it is possible for an argument of this form to have true premises and a false conclusion. Hence the argument form is invalid.

13.

conclusion

premises

→

∼p

∼

critical row T( T F Row T 4Fof theFtruth table T represents the only situation in which all the premises are true, and inFthisTrow the is also true. Therefore, the argument form (modus tollens) is valid. T conclusion F F F T T

Instructor’s Manual: Section 2.2 14. p T T F F

q T F T F

premise

conclusion

p T F T F

p∨q T (

critical row

T (

critical row

The truth table shows that in the two situations (represented by rows 1 and 3) in which the premise is true, the conclusion is also true. Therefore, Generalization, version (a), is valid.

15. p T T F F

q T F T F

premise

conclusion

q T F T F

p∨q T (

critical row

T (

critical row

The truth table shows that in the two situations (represented by rows 1 and 3) in which the premise is true, the conclusion is also true. Therefore, Generalization, version (b), is valid.

16.

premise

p ∧q

T T F F

T F T F

T F F F

conclusion

critical row

The truth table shows that in the only situation (represented by row 1) in which both premises are true, the conclusion is also true. Therefore, Specialization, version (sa), is valid.

17.

premise

p ∧q

T T F F

T F T F

T F F F

conclusion

critical row

The truth table shows that in the only situation (represented by row 1) in which both premises are true, the conclusion is also true. Therefore, Specialization, version (b), is valid.

Solutions for Exercises: The Logic of Compound Statements

18.

conclusion

premises

p T

q T

∨ ¸ T x`

˛ ∼

critical row

F T( T F T T Row 2 represents the only situation in which both premises are true. Because the conclusion F T T F isFalsoTtrue here the argument form is valid. F T 19.

conclusion

premises

¸ x` ˛ p∨q ∼p

T T F F

T F T F

T T T F

F F T T

T (

critical row

The truth table shows that in the only situation (represented by row 3) in which both premises are true, the conclusion is also true. Therefore, Elimination, version (b), is valid. 20.

conclusion

premises

p→r ¸ x` ˛ T ( critical row T T T p→q q→r T T T T F T F T F T F T T F F F T critical row F T T T T T ( F T F T F critical row F F T T T T ( critical row F F F T T T ( The truth table shows that in the four situations (represented by rows 1, 5, 7, and 8) in which both premises are true, the conclusion is also true. Therefore, Transitivity is valid. p

21.

conclusion

premises

q ∨

r x` →

r ˛

r (

critical row

→ critical row

T T T critical row T T F T F F ( T F T T T T T T F F T F T F T T T T T T ( F T F T T F The truth table shows that in the three situations (represented by rows 1, 3, 5) in which all F F T F T T three premises are true, the conclusion is also true. Therefore, proof by division into cases is F F F F T T valid.

Instructor’s Manual: Section 2.2

22. Let p represent “Tom is on team A” and q represent “Hua is on team B.” Then the argument has the form ∼p → q ∼q → p ∴ ∼p ∨ ∼q premises

∼p

∼q

conclusion

∼p∨∼q ( F

critical row T ¸ T x` ˛ critical row ∼ critical row T F T ( F T T T F T T ( T F T T Rows 1, 2, and F F T 3 of T the truth F table are Fthe critical rows in which all the premises are true, but row 1 shows that it is possible for an argument of this form to have true premises and a false conclusion. Thus this argument form is invalid.

23.

form: . ..

p ∨q p→r q∨ ∼ r conclusion

premises

∼r

x` ∨

q∨ ∼ r T (

critical row

→ critical row

F T T critical row T T F T T F ( critical row T F T F T T F T F F T T F F T T F T T T ( ( F T1, 3,F5, andT 6 represent T Rows theT situationsTin which both premises are true, but in row 3 the F F Tis false. F Hence, F it is Tpossible for an argument of this form to have true premises conclusion F F F T F so theT given argument is invalid. and a false conclusion, and 24.

form: p → q q ... p invalid: converse error

25.

form: p ∨ q ∼p . .. q

valid: elimination

form: p → q q→r . .. p→r

valid: transitivity

form: p → q ∼p . .. ∼q

invalid: inverse error

26.

27.

Solutions for Exercises: The Logic of Compound Statements p→q q p

invalid, converse error

p→q ∼p ∼q

invalid, inverse error

invalid, converse error

...

p→q q p

31.

form: ...

p ∧q q

valid, generalization

32.

form: . ..

p→r q→r

28.

form: ...

29.

form: . ..

30.

form:

valid, proof by division into cases

p∨q →r

33. A valid argument with a false conclusion must have at least one false premise. In the following example, the second premise is false. (The first premise is true because its hypothesis is false.) If the square of every real number is positive, then no real number is negative. The square of every real number is positive. Therefore, no real number is negative. 34. An invalid argument with a true conclusion can have premises that are either true or false. In the following example the first premise is true for either one of following two reasons: its hypothesis is false and its conclusion is true. If the square of every real number is positive, then some real numbers are positive. Some real numbers are positive. Therefore, the square of every real number is positive. 35. A correct answer should indicate that for a valid argument, any argument of the same form that has true premises has a true conclusion, whereas for an invalid argument, it is possible to find an argument of the same form that has true premises and a false conclusion. The validity of an argument does not depend on whether the conclusion is true or not. The validity of an argument only depends on the formal relationship between its premises and its conclusion. 36. The program contains an undeclared variable. One explanation: 1. There is not a missing semicolon and there is not a misspelled variable name. (by (c) and (d) and definition of ∧) 2. It is not the case that there is a missing semicolon or a misspelled variable name. (by (1) and De Morgan’s laws) 3. There is not a syntax error in the first five lines. (by (b) and (2) and modus tollens) 4. There is an undeclared variable. (by (a) and (3) and elimination) 37. The treasure is buried under the flagpole. One explanation: 1. The treasure is not in the kitchen. (by (c) and (a) and modus ponens) 2. The tree in the front yard is not an elm. (by (b) and (1) and modus tollens) 3. The treasure is buried under the flagpole. (by (d) and (2) and elimination)

Instructor’s Manual: Section 2.2 38. a. A is a knave and B is a knight. One explanation: 1. Suppose A is a knight. 2. ∴ What A says is true. (by definition of knight) 3. ∴ B is a knight also. (That’s what A said.) 4. ∴ What B says is true. (by definition of knight) 5. ∴ A is a knave. (That’s what B said.) 6. ∴ We have a contradiction: A is a knight and a knave. (by (1) and (5)) 7. ∴ The supposition that A is a knight is false. (by the contradiction rule) 8. ∴ A is a knave. (negation of supposition) 9. ∴ What B says is true. (B said A was a knave, which we now know to be true.) 10. ∴ B is a knight. (by definition of knight) b. C is a knave and D is a knight One explanation: 1. Suppose C is a knight. 2. ... C is a knave (because what C said was true). 3. ... C is both a knight and a knave (by (1) and (2)), which is a contradiction. 4. ... C is not a knight (because by the contradiction rule the supposition is false). 5. ... What C says is false (because since C is not a knight he is a knave and knaves always speak falsely). 6. ... At least one of C or D is a knight (by De Morgan’s law). 7.D is a knight (by (4) and (6) and elimination). 8. ... C is a knave and D is a knight (by (4) and (7)). To check that the problem situation is not inherently contradictory, note that if C is a knave and D is a knight, then each could have spoken as reported. c. One is a knight and the other is a knave. One explanation: There is one knave. E and F cannot both be knights because then both would also be knaves (since each would have spoken the truth), which is a contradiction. Nor can E and F both be knaves because then both would be telling the truth which is impossible for knaves. Hence, the only possible answer is that one is a knight and the other is a knave. But in this case both E and F could have spoken as reported, without contradiction. d. U , Z, X, and V are knaves and W and Y are knights. One explanation: 1. The statement made by U must be false because if it were true then U would not be a knight (since none would be a knight), but since he spoke the truth he would be a knight and this would be a contradiction. 2. ... there is at least one knight, and U is a knave (since his statement that there are no knights is false). 3. Suppose Z spoke the truth. Then so did W (since if there is exactly one knight then it is also true that there are at most three knights). But this implies that there are at least two knights, which contradicts Z′s statement. Hence Z cannot have spoken the truth. 4. ... there are at least two knights, and Z is a knave (since his statement that there is exactly one knight is false). Also X′s statement is false because since both U and Z are knaves it is impossible for there to be exactly five knights. Hence X also is a knave. 5. ... there are at least three knaves (U , Z, and X), and so there are at most three knights. 6. ... W ′s statement is true, and so W is a knight.

Solutions for Exercises: The Logic of Compound Statements 7. Suppose V spoke the truth. Then V , W , and Y are all knights (otherwise there would not be at least three knights because U , Z, and X are known to be knaves). It follows that Y spoke the truth. But Y said that exactly two were knights. This contradicts the result that V , W , and Y are all knights. 8. ... V cannot have spoken the truth, and so V is a knave. 9. ... U , Z, X, and V are all knaves, and so there are at most two knights. 10. Suppose that Y is a knave. Then the only knight is W , which means that Z spoke the truth. But we have already seen that this is impossible. Hence Y is a knight. 11. By 6, 9, and 10, the only possible solution is that U , Z, X, and V are knaves and W and Y are knights. Examination of the statements shows that this solution is consistent: in this case, the statements of U , Z, X, and V are false and those of W and Y are true.

39. The chauffeur killed Lord Hazelton. One explanation: 1. Suppose the cook was in the kitchen at the time of the murder. 2. ∴ The butler killed Lord Hazelton with strychnine. (by (c) and (1) and modus ponens) 3. ∴ We have a contradiction: Lord Hazelton was killed by strychnine and a blow on the head. (by (2) and (a)) 4. ∴ The supposition that the cook was in the kitchen is false. (by the contradiction rule) 5. ∴ The cook was not in the kitchen at the time of the murder. (negation of supposition) 6. ∴ Sara was not in the dining room when the murder was committed. (by (e) and (5) and modus ponens) 7. ∴ Lady Hazelton was in the dining room when the murder was committed. (by (b) and (6) and elimination) 8. ∴ The chauffeur killed Lord Hazelton. (by (d) and (7) and modus ponens) 40. One solution: Suppose Socko is telling the truth. Then Fats is also telling the truth because if Lefty killed Sharky then Muscles didn’t kill Sharky. Consequently, two of the men were telling the truth, which contradicts the fact that all were lying except one. Therefore, Socko is not telling the truth: Lefty did not kill Sharky. Hence Muscles is telling the truth and all the others are lying. It follows that Fats is lying, and so Muscles killed Sharky. Another solution: The statements of Socko and Muscles contradict each other, which implies that one is lying and the other is telling the truth. If Socko is telling the truth, then Fats is also telling the truth, which contradicts the fact that only one person told the truth. So Muscles is the only one who told the truth. Hence Muscles is telling the truth and all the others are lying. It follows that Fats is lying, and so Muscles killed Sharky. 41.

(1) p → t by premise (d) ∼p by premise (c) ∴ ∼p by modus tollens (2) ∼p ∴ ∼p ∨ q

by (1) by generalization

∼p ∨ q → r ∼p ∨ q ∴ r

(3)

∼p r ∴ ∼p ∧ r

(4)

by premise (a) by (2) by modus ponens

by (1) by (3) by conjunction

∼p ∧ r → ∼s by premise (e) ∼p ∧ r by (4) ∴ ∼s by modus ponens

(5)

Instructor’s Manual: Section 2.2 s ∨ ∼q ∼s ∴ ∼q

by premise (b) by (5) by elimination

(1) . ..

q→r ∼r

by premise (b) by premise (d)

∼q

by modus tollens

(2) . ..

p ∨q ∼q

by premise (a) by (1)

by elimination

(3) . ..

∼q →u∧s ∼q

by premise (e) by (1)

u∧s

by modus ponens

(4) ...

u∧s s

by (3) by specialization

(5)

p s p ∧s

by (2) by (4) by conjunction

(6)

42.

... (6) . .. 43.

p ∧s→t p ∧s

by premise (c) by (5)

by modus ponens

(1) ∼w u∨w ∴ u

by premise (d) by premise (e) by elimination

u → ∼ p by premise (c) u by (1) ∴ ∼p by modus ponens

(2)

∼p → r ∧ ∼s by premise (a) ∼p by (2) ∴ r ∧ ∼s by modus ponens

(3)

(4) r ∧ ∼s by (3) ∴ ∼s by specialization (5) ∼t → s by premise (b) ∼s by (4) ∴ ∼t by modus tollens 44.

(1) . ..

∼q∨s ∼s

by premise (d) by premise (e)

∼q

by elimination

(2) . ..

p→q ∼q

by premise (a) by (1)

∼p

by modus tollens

(3) . ..

r∨s ∼s

by premise (b) by premise (e)

by elimination

(4)

∼p r ∼p ∧r

...

by (2) by (3) by conjunction

Solutions for Exercises: The Logic of Compound Statements (5) . ..

∼p ∧r →u ∼p ∧r

by premise (f) by (4)

by modus ponens

(6) . ..

∼ s →∼ t ∼s

by premise (c) by premise (e)

∼t

by modus ponens

(7) . ..

w∨t ∼t

by premise (g) by (6)

by elimination

(8)

u w u∧w

by (5) by (7) by conjunction

...

Section 2.4 1. R = 1 2. R = 1 3. S = 1 4. S = 1 5. The input/output table is as follows: Input

Output

P Q 1 1 1 0 0 1 0 0

R 1 1 0 1

6. The input/output table is as follows: Input

Output

P Q R 1 1 1 1 1 0 1 0 1 1 0 0 0 1 1 0 1 0 0 0 1 0 0 0

S 1 0 1 1 1 0 1 0

7. The input/output table is as follows: Input P 1 1 0 0

Q 1 0 1 0

Output R 0 1 0 0

Instructor’s Manual: Section 2.2 8. The input/output table is as follows: Input P 1 1 1 1 0 0 0 0

Q 1 1 0 0 1 1 0 0

Output R 1 0 1 0 1 0 1 0

S 1 1 1 1 1 1 1 1

9. P ∨ ∼Q 10. (P ∨ Q)∧ ∼ Q 11. (P ∧ ∼Q) ∨ R 12. (P ∨ Q)∨ ∼ (Q ∧ R) 13. NOT

14. P

NOT

15. P OR NOT

NOT

16.

P Q R

AND OR NOT

AND

Solutions for Exercises: The Logic of Compound Statements

17. P

AND

NOT

NOT AND

18. a. (P ∧ Q ∧ ∼R) ∨ (∼P ∧ Q ∧ R) b. P Q

AND NOT

NOT AND

19. a. (P ∧ Q∧ ∼ R) ∨ (P ∧ ∼ Q∧ ∼ R) ∨ (∼ P ∧ Q ∧ ∼ R) b. One circuit (among many) having the given input/output table is the following: P Q

AND

NOT

AND

NOT

AND

NOT

20. a. (P ∧ Q ∧ R) ∨ (P ∧ ∼Q ∧ R) ∨ (∼P ∧ ∼Q ∧ ∼R)

Instructor’s Manual: Section 2.2 b.

P AND

Q R

OR NOT

AND

NOT AND NOT

NOT

21. a. (P ∧ Q ∧ ∼ R) ∨ (∼ P ∧ Q ∧ R) ∨ (∼ P ∧ Q ∧ ∼ R) b. One circuit (among many) having the given input/output table is the following: P Q R

AND NOT

NOT

AND

NOT AND NOT

22. The input/output table is Input

Output

P Q R 1 1 1 1 1 0 1 0 1 1 0 0 0 1 1 0 1 0 0 0 1 0 0 0

S 0 1 0 0 0 0 1 0

One circuit (among many) having this input/output table is shown below.

Solutions for Exercises: The Logic of Compound Statements P AND

Q NOT

OR NOT AND NOT

23. The input/output table is as follows: Input P 1 1 1 1 0 0 0 0

Q 1 1 0 0 1 1 0 0

Output R 1 0 1 0 1 0 1 0

S 1 0 0 0 0 0 0 1

One circuit (among many) having this input/output table is the following: P Q R

AND

NOT

AND

NOT

24. Let P and Q represent the positions of the switches in the classroom, with 0 being “down” and 1 being “up.” Let R represent the condition of the light, with 0 being “off” and 1 being “on.” Initially, P = Q = 0 and R = 0. If either P or Q (but not both) is changed to 1, the light turns on. So when P = 1 and Q = 0, then R = 1, and when P = 0 and Q = 1, then R = 1. Thus when one switch is up and the other is down the light is on, and hence moving the switch that is down to the up position turns the light off. So when P = 1 and Q = 1, then R = 0. It follows that the input/output table has the following appearance: Input

Output

P Q 1 1 1 0 0 1 0 0

R 0 1 1 0

Instructor’s Manual: Section 2.2

One circuit (among many) having this input/output table is the following: P Q

NOT

AND

NOT AND

25. Let P , Q, and R indicate the positions of the switches, with 1 indicating that the switch is in the on position. Let an output of 1 indicate that the security system is enabled. The complete input/output table is as follows: Input P 1 1 1 1 0 0 0 0

Q 1 1 0 0 1 1 0 0

Output R 1 0 1 0 1 0 1 0

S 1 1 1 0 1 0 0 0

One circuit (among many) having this input/output table is the following: P Q R

AND

AND NOT OR

NOT

AND

Note: One alternative answer interchanges the 1’s and 0’s. 26. The Boolean expression for (a) is (P ∧ Q) ∨ Q, and for (b) it is (P ∨ Q) ∧ Q. We must show that if these expressions are regarded as statement forms, then they are logically equivalent.

Solutions for Exercises: The Logic of Compound Statements Now (P ∧ Q) ∨ Q

≡ Q ∨ (P ∧ Q) ≡ (Q ∨ P ) ∧ (Q ∨ Q) ≡ (Q ∨ P ) ∧ Q ≡ (P ∨ Q) ∧ Q

by the commutative law for ∨ by the distributive law by the idempotent law by the commutative law for ∧

Alternatively, by the absorption laws, both statement forms are logically equivalent to Q. 27. The Boolean expression for circuit (a) ∼ is P∧ (∼ (∼ P ∧Q)) and for circuit (b) it is∼ (P ∨Q). We must show that if these expressions are regarded as statement forms, then they are logically equivalent. Now ∼ P ∧ (∼ (∼ P ∧ Q))

≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡

∼ P ∧ (∼ (∼ P )∨ ∼ Q) ∼ P ∧ (P ∨ ∼ Q) (∼ P ∧ P ) ∨ (∼ P ∧ ∼ Q) (P ∧ ∼ P ) ∨ (∼ P ∧ ∼ Q) c ∨(∼ P ∧ ∼ Q) (∼ P ∧ ∼ Q)∨c ∼P∧∼Q ∼ (P ∨ Q)

by De Morgan’s law by the double negative law by the distributive law by the commutative law for ∧ by the negation law for ∧ by the commutative law for ∨ by the identity law for ∨ by De Morgan’s law.

28. The Boolean expression for circuit (a) is (P ∧ Q) ∨ (P ∧ ∼ Q) ∨ (∼ P ∧ ∼ Q) and for circuit (b) it is P ∨ ∼ Q. We must show that if these expressions are regarded as statement forms, then they are logically equivalent. Now (P ∧ Q) ∨ (P ∧ ∼ Q) ∨ (∼ P ∧ ∼ Q) ≡

((P ∧ Q) ∨ (P ∧ ∼ Q)) ∨ (∼ P ∧ ∼ Q)

≡ ≡ ≡ ≡ ≡ ≡ ≡

(P ∧ (Q∨ ∼ Q)) ∨ (∼ P ∧ ∼ Q) (P ∧ t) ∨ (∼ P ∧ ∼ Q) P ∨ (∼ P ∧ ∼ Q) (P ∨ ∼ P ) ∧ (P ∨ ∼ Q) t ∧ (P ∨ ∼ Q) (P ∨ ∼ Q) ∧ t P∨ ∼ Q

by inserting parentheses (which is legal by the associative law for ∨) by the distributive law by the negation law for ∨ by the identity law for ∧ by the distributive law by the negation law for ∨ by the commutative law for ∧ by the identity law for ∧.

29. The Boolean expression for circuit (a) is (P ∧ Q) ∨ (∼ P ∧ Q) ∨ (P ∧ ∼ Q) and for circuit (b) it is P ∨ Q. We must show that if these expressions are regarded as statement forms, then they are logically equivalent. Now (P ∧ Q) ∨ (∼ P ∧ Q) ∨ (P ∧ ∼ Q) ≡

((P ∧ Q) ∨ (∼ P ∧ Q)) ∨ (P ∧ ∼ Q)

≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡

((Q ∧ P ) ∨ (Q∧ ∼ P )) ∨ (P ∧ ∼ Q) (Q ∧ (P ∨ ∼ P )) ∨ (P ∧ ∼ Q) (Q∧ t) ∨(P ∧ ∼ Q) Q ∨ (P ∧ ∼ Q) (Q ∨ P ) ∧ (Q ∨ ∼ Q) (Q ∨ P )∧ t Q∨P P ∨Q

by inserting parentheses (which is legal by the associative law for ∨) by the commutative law for ∧ by the distributive law by the negation law for ∨ by the identity law for ∧ by the distributive law by the negation law for ∨ by the identity law for ∧ by the commutative law for ∨.

Instructor’s Manual: Section 2.2 30. (P ∧ Q) ∨ (∼P ∧ Q) ∨ (∼P ∧ ∼Q) ≡ (P ∧ Q) ∨ ((∼P ∧ Q) ∨ (∼P ∧ ∼Q) by inserting parentheses (which is legal by the associative law) ≡ (P ∧ Q) ∨ (∼P ∧ (Q ∨ ∼Q)) by the distributive law ≡ (P ∧ Q) ∨ (∼P ∧ t) by the negation law for ∨ ≡ (P ∧ Q) ∨ ∼P by the identity law for ∧ ≡ ∼P ∨ (P ∧ Q) by the commutative law for ∨ ≡ (∼P ∨ P ) ∧ (∼P ∨ Q) by the distributive law ≡ (P ∨ ∼P ) ∧ (∼P ∨ Q) by the commutative law for ∨ ≡ t ∧ (∼P ∨ Q) by the negation law for ∨ ≡ (∼P ∨ Q) ∧ t by the commutative law for ∧ ≡ ∼P ∨ Q by the identity law for ∧

31. (∼ P ∧ ∼ Q) ∨ (∼ P ∧ Q) ∨ (P ∧ ∼ Q) ≡ ((∼ P ∧ ∼ Q) ∨ (∼ P ∧ Q)) ∨ (P ∧ ∼ Q) ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡

(∼ P ∧ (∼ Q ∨ Q)) ∨ (P ∧ ∼ Q) (∼ P ∧ (Q∨ ∼ Q)) ∨ (P ∧ ∼ Q) (∼ P ∧ t) ∨(P ∧ ∼ Q) ∼ P ∨ (P ∧ ∼ Q) (∼ P ∨ P ) ∧ (∼ P ∨ ∼ Q) (P ∨ ∼ P ) ∧ (∼ P ∨ ∼ Q) t ∧(∼ P ∨ ∼ Q) (∼ P ∨ ∼ Q)∧ t ∼P∨∼Q ∼ (P ∧ Q)

by inserting parentheses (which is legal by the associative law) by the distributive law by the commutative law for ∨ by the negation law for ∨ by the identity law for ∧ by the distributive law by the commutative law for ∨ by the negation law for ∨ by the commutative law for ∧ by the identity law for ∧ by De Morgan’s law.

32. (P ∧ Q ∧ R) ∨ ((P ∧ ∼ Q ∧ R) ∨ (P ∧ ∼ Q∧ ∼ R) ≡

≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡

((P ∧ (Q ∧ R)) ∨ (P ∧ (∼ Q ∧ R))) ∨ (P ∧ (∼ Q∧ ∼ R)) ∨ (P ∧ ∼ Q) by inserting parentheses (which is legal by the associative law) (P ∧ [(Q ∧ R) ∨ (∼ Q ∧ R)] ∨ (P ∧ [∼ Q∧ ∼ R]) by the distributive law P ∧ ([(Q ∧ R) ∨ (∼ Q ∧ R)] ∨ [∼ Q∧ ∼ R]) by the distributive law P ∧ ([(R ∧ Q) ∨ (R∧ ∼ Q)] ∨ [∼ Q∧ ∼ R]) by the commutative law for ∧ P ∧ ([(R ∧ (Q∨ ∼ Q)] ∨ [∼ Q∧ ∼ R]) by the distributive law P ∧ ([(R∧ t] ∨[∼ Q∧ ∼ R]) by the negation law for ∨ P ∧ (R ∨ [∼ Q∧ ∼ R]) by the identity law for ∧ P ∧ ((R ∨ ∼ Q) ∧ (R∨ ∼ R)) by the distributive law P ∧ ((R ∨ ∼ Q) ∧ t]) by the negation law for ∨ P ∧ (R ∨ ∼ Q) by the identity law for ∧.

33. a. (P | Q) | (P | Q)

≡ ≡ ≡ ≡

∼ [(P | Q) ∧ (P | Q)] ∼ (P | Q) ∼ [∼ (P ∧ Q)] P ∧Q

by definition of | by the idempotent law for ∧ by definition of | by the double negative law.

b. P ∧ (∼ Q ∨ R) ≡ (P | (∼ Q ∨ R)) | (P | (∼ Q ∨ R)) by part (a)

≡ (P | [(∼ Q |∼ Q) | (R | R)]) | (P | [(∼ Q |∼ Q) | (R | R)]) by Example 2.4.7(b)

≡ (P | [((Q | Q) | (Q | Q)) | (R | R)]) | (P | [((Q | Q) | (Q | Q)) | (R | R)]) by Example 2.4.7(a)

Solutions for Exercises: The Logic of Compound Statements

34. a. (P ↓ Q) ↓ (P ↓ Q) ≡ ∼(P ↓ Q) by part (a) ≡ ∼[∼(P ∨ Q)] by definition of ↓ ≡P ∨Q by the double negative law b. P ∨Q

≡ ∼ (∼ (P ∨ Q)) by the double negative law ≡ ∼ (P ↓ Q) by definition of ↓ ≡ (P ↓ Q) ↓ (P ↓ Q) by part (a).

c. P ∧Q

d. P →Q

e. P ↔Q

≡ ≡ ≡

∼ (∼ P ∨ ∼ Q) by De Morgan’s law and the double negative law ∼ P ↓∼ Q by definition of ↓ (P ↓ P ) ↓ (Q ↓ Q) by part (a).

≡ ≡ ≡

∼P ∨Q by Exercise 13(a) of Section 2.2 (∼ P ↓ Q) ↓ (∼ P ↓ Q) by part (b) ((P ↓ P ) ↓ Q) ↓ ((P ↓ P ) ↓ Q) by part (a).

≡

(P → Q) ∧ (Q → P )

≡

([(P ↓ P ) ↓ Q] ↓ [(P ↓ P ) ↓ Q)]) ∧ ([(Q ↓ Q) ↓ P ] ↓ [(Q ↓ Q) ↓ P )])

≡

(([(P ↓ P ) ↓ Q] ↓ [(P ↓ P ) ↓ Q)]) ↓ ([(P ↓ P ) ↓ Q] ↓ [(P ↓ P ) ↓ Q)])) ↓ (([(Q ↓ Q) ↓ P ] ↓ [(Q ↓ Q) ↓ P )]) ↓ ([(Q ↓ Q) ↓ P ] ↓ [(Q ↓ Q) ↓ P )]))

by the truth table on page 46 of the text

by part (d)

by part (c)

Section 2.5 1. 1910 = 16 + 2 + 1 = 100112 2. 55 = 32 + 16 + 4 + 2 + 1 = 1101112 3. 287 = 256 + 16 + 8 + 4 + 2 + 1 = 1000111112 4. 45810 = 256 + 128 + 64 + 8 + 2 = 1110010102 5. 1609 = 1024 + 512 + +64 + 8 + 1 = 110010010012 6. 1424 = 1024 + 256 + 128 + 16 = 101100100002 7. 11102 = 8 + 4 + 2 = 1410 8. 101112 = 16 + 4 + 2 + 1 = 2310 9. 1101102 = 32 + 16 + 4 + 2 = 5410 10. 11001012 = 64 + 32 + 4 + 1 = 10110 11. 10001112 = 64 + 4 + 2 + 1 = 7110 12. 10110112 = 64 + 16 + 8 + 2 + 1 = 9110 13.

+ 1

1 0 1 12 1 0 12 0 0 0 02

Instructor’s Manual: Section 2.2 14.

1 0 0 12 + 1 0 1 12 1 0 1 0 02 15.

+ 1

16.

1 0 1 1 0 12 1 1 1 0 12 0 0 1 0 1 02 1

1 1 0 1 1 1 0 1 12 1 0 0 1 0 1 1 0 1 02 1 0 0 0 0 0 1 0 1 0 12

17.

1 − 18.

0 1

1 1 1

0 0 1

02 12 12

10 0

1 −

1 1 1

0 1 1

1 02 0 12 0 12

1 −

0 1 1 1 0 0 1 1 0

0 1 1

1 −

1 1 1

0 0 1

1 1 1

0 1 0

02 12 12

19.

20.

21. a. S = 0, T = 1

12 12 02

b. S = 0, T = 1

c. S = 0, T = 0

22. Note that +

111111112 12 1000000002

and 1000000002 = (28)10. Because 12 = 110, we have that −

111111112 + 12 = 12 111111112 =

23. |−23|10 = 2310 = (16 + 4 + 2 + 1)10 = 000101112 11101001. So the answer is 11101001.

−

(28)10 110 (28 − 1)10

flip the bits

−→

d→ d 1 11101000 −

Solutions for Exercises: The Logic of Compound Statements

24. |−67|10 = 6710 = (64 + 2 + 1)10 = 010000112

flip the bits

−→

add 1

10111100 −→ 10111101.

So the 8-bit two’s complement is 10111101. 25. |−4|10 = 410 = 000001002

flip the bits

−→

d→ 1 11111100. So the answer is 11111100. 11111011 −

26. |−115|10 = 11510 = (64 + 32 + 16 + 2 + 1)10 = 011100112

flip the bits

−→

add 1

10001100 −→ 10001101

So the 8-bit two’s complement is 10001101. 27. Because the leading bit is 1, this is the 8-bit two’s complement of a negative integer. flip the bits

add 1

11010011 −→ 00101100 −→ 001011012 = (32 + 8 + 4 + 1)10 = |−45|10. So the answer is −45. 28. Because the leading bit is 1, this is the 8-bit two’s complement of a negative integer. 10011001 a d→ d 1 01100111 = (64 + 32 + 4 + 2 + 1) 01100110 − = |−103| . So the answer is −103. 10 2

flip the bits

−→

29. Because the leading bit is 1, this is the 8-bit two’s complement of a negative integer. flip the bits

add 1

11110010 −→ 00001101 −→ 000011102 = (8 + 4 + 2)10 = |−14|10. So the answer is −14. 30. Because the leading bit is 1, this is the 8-bit two’s complement of a negative integer. 10111010 a d→ d 1 01000110 = (64 + 4 + 2) 01000101 − = |−70| . So the answer is −70. 10 2

flip the bits

31. 5710 = (32 + 16 + 8 + 1)10 = 1110012 → 00111001 |−118|10 = (64 + 32 + 16 + 4 + 2)10 = 011101102

flip the bits

−→

add 1

10001001 −→ 10001010.

So the 8-bit two’s complements of 57 and −118 are 00111001 and 10001010. Adding the 8-bit two’s complements in binary notation gives

00111001 10001010 11000011

Since the leading bit of this number is a 1, the answer is negative. Converting back to decimal flip the bits a d→ 1 001111012 = (32 + 16 + 8 + 4 + 1)10 = |61|10. 00111100 − form gives 11000011 −→ So the answer is −61. 32. 6210 = (32 + 16 + 8 + 4 + 2)10 = 1111102 → 00111110 |−18|10 = (16 + 2)10 = 00010010

flip the bits

−→

add 1

11101101 −→ 11101110

Thus the 8-bit two’s complements of 62 and −18 are 00111110 and 10110111. Adding the 8-bit two’s complements in binary notation gives

00111110 11101110 00101100

Truncating the 1 in the 28th position gives 00101100. Since the leading bit of this number is a 0, the answer is positive. Converting back to decimal form gives 00101100 → 1011002 = (32 + 8 + 4)10 = 4410. So the answer is 44.

−→

Instructor’s Manual: Section 2.2 33. |−6|10 = (4 + 2)10 = 1102

flip the bits

d→ 1 11111010 00000110 → 11111001 −

−→

|−73|10 = (64 + 8 + 1)10 = 01001001

flip the bits

−→1 10110111 10110110 add

−→

Thus the 8-bit two’s complements of —6 and − 73 are 11111010 and 10110111. Adding the 8-bit two’s complements in binary notation gives 11111010 10110111 10110001

Truncating the 1 in the 28th position gives 10110001. Since the leading bit of this number is a 1, the answer is negative. Converting back to decimal form gives 10110001

flip the bits

−→

add 1

01001110 −→ 010011112 = (64 + 8 + 4 + 2 + 1) = 7910 = |−79|10

So the answer is −79. 34. 8910 = (64 + 16 + 8 + 1)10 = 010110012 |−55|10 = (32 + 16 + 4 + 2 + 1)10 = 001101112

flip the bits

−→

add 1

11001000 −→ 11001001

So the 8-bit two’s complements of 89 and − 55 are 01001111 and 11010101. Adding the 8-bit two’s complements in binary notation gives 01011001 11001001 100100010

Truncating the 1 in the 28th position gives 00100010. Since the leading bit of this number is a 0, the answer is positive. Converting back to decimal form gives 001000102 = (32 + 2)10 = 3410. So the answer is 34. 35. |−15|10 = (8 + 4 + 2 + 1)10 = 000011112

flip the bits

|−46|10 = (32 + 8 + 4 + 2)10 = 001011102

−→

flip the bits

−→

add 1

11110000 −→ 11110001 add 1

11010001 −→ 11010010

So the 8-bit two’s complements of —15 and − 46 are 11110001 and 10100010. Adding the 8-bit two’s complements in binary notation gives +

11110001 11010010 111000011

Truncating the 1 in the 28th position gives 11000011. Since the leading bit of this number is a 1, the answer is negative. Converting back to decimal form gives add 1 11000011 flip the bits −→ 00111100 −→ 001111012 = −(32 + 16 + 8 + 4 + 1)10 = |−61|10 .

So the answer is −61.

Solutions for Exercises: The Logic of Compound Statements

36. 12310 = (64 + 32 + 16 + 8 + 2 + 1)10 = 011110112 |−94|10 = (64 + 16 + 8 + 4 + 2)10 = 010111102

flip the bits

−→

add 1

10100001 −→ 10100010

So the 8-bit two’s complements of 123 and − 94 are 01111011 and 10100010. Adding the 8-bit two’s complements in binary notation gives +

01111011 10100010 100011101

Truncating the 1 in the 28th position gives 00011101. Since the leading bit of this number is a 0, the answer is positive. Converting back to decimal form gives 000111012 = (16 + 8 + 4 + 1)10 = 2910. So the answer is 29. 37. a. The 8-bit two’s complement of −128 is computed as follows: |−128|10 = 12810 = (27)10 = 100000002

flip the bits

−→

add 1

01111111 −→ 10000000.

So the 8-bit two’s complement of −128 is 10000000. If the two’s complement procedure is flip the bits a 01111111 dd−→1 10000000. applied to this result, the following is obtained 10000000 −→ So the 8-bit two’s complement of the 8-bit two’s complement of −128 is 10000000, which is the 8-bit two’s complement of −128. b. Suppose a, b, and a + b are integers in the range from 1 through 128. Then 1 ≤ a ≤ 128 = 27

1 ≤ b ≤ 128 = 27

1 ≤ a + b ≤ 128 = 27.

Multiplying all parts of all three inequalities by −1 gives −1 ≥ −a ≥ −2 7

− 1 ≥ −b ≥ −2 7

− 1 ≥ −(a + b) ≥ −2 7.

Thus 28 − (a + b) ≥ 28 − 27 ≥ 27(2 − 1) = 27, and (28 − a) + (28 − b) = 28 + (28 − (a + b)) ≥ 28 + 27. Therefore, the 8-bit two’s complement of (28 — a) + (28 − b) has 1’s in both the 28th and the 27th positions. The 1 in the 28th position is truncated, and the 1 in the 27th position shows that the sum is negative. 38. A2BC16 = 10·163 + 2·162 + 11·16 + 12 = 41,66010 39. E0D16 = 14·162 + 0 + 13 = 359710 40. 39EB16 = 3·163 + 9·162 + 14·16 + 11 = 14, 82710 41. 0001 1100 0000 1010 1011 11102 42. B53DF 816 = 1011 0101 0011 1101 1111 10002 43. 4ADF 8316 = 0100 1010 1101 1111 1000 00112 44. 2E16 45. 1011 0111 1100 01012 = B7C516

Instructor’s Manual: Section 2.2

46. 1011 0111 1100 01012 = B7C516 47. a. 6·84 + 1·83 + 5·82 + 0·8 + 2·1 = 25,41010 b. 207638 = 2·84 + 0·83 + 7·82 + 6·8 + 3 = 8, 69110 c. To convert an integer from octal to binary notation: (i) Write each octal digit of the integer in fixed 3-bit binary notation (and include leading zeros as needed). Note that octal digit 3-bit binary equivalent

0 000

1 001

2 010

3 011

4 100

5 101

6 110

7 111

(ii) Juxtapose the results. As an example, consider converting 615028 to binary notation: 68 = 1102 18 = 0012 58 = 1012 08 = 0002 28 = 0102. So in binary notation the integer should be 110 001 101 000 0102. This result can be checked by writing the integer in decimal notation and comparing it to the answer obtained in part (a): 110 001 101 000 0102 = (1 · 214 + 1 · 213 + 1 · 29 + 1 · 28 + 1 · 26 + 1 · 2)10 = 2541010. This is the same as the answer obtained in part (a). So the two methods give the same result. To convert an integer from binary to octal notation: (i) Group the digits of the binary number into sets of three, starting from the right and adding leading zeros as needed: (ii) Convert the binary numbers in each set of three into octal digits; (iii) Juxtapose those octal digits. As an example consider converting 11010111012 to octal notation. Grouping the binary digits in sets of three and adding two leading zeros gives 001 101 011 101. To convert each group into an octal digit, note that 0012 = 18 1012 = 58 0112 = 38 1012 = 58. So the octal version of the integer should be 15358. To check this result, observe that 1 101 011 1012 = (1 · 29 + 1 · 28 + 1 · 26 + 1 · 24 + 1 · 23 + 1 · 22 + 1)10 = 86110 and 15358 = (1 · 83 + 5 · 82 + 3 · 8 + 5)10 = 86110 also. So the two methods give the same result.

Instructor’s Manual: Chapter 3

Chapter 3: The Logic of Quantified Statements Ability to use the logic of quantified statements correctly is necessary for doing mathematics because mathematics is, in a very broad sense, about quantity. The main purpose of this chapter is to familiarize students with the language of universal and existential statements. The various facts about quantified statements developed in this chapter are used extensively in Chapter 4 and are referred to throughout the rest of the book. Experience with the formalism of quantification is especially useful to students planning to study LISP or Prolog, program verification, or relational databases. ∀ In this edition of Discrete Mathematics with Applications a -statement with a single variable ∀ is translated using the words “for every” rather than “for all.” The reason is that when the “for all” translation is given, a significant number of students use a plural verb rather than a singular one when translating a ∀-statement into English. For example they write “For all primes p, p are positive” rather than “For all primes p, p is positive.” The problem is that an important step in learning to prove mathematical statements is coming to think of the variable as singular but capable of representing any element in a set. Many students come to college with inconsistent interpretations of quantified statements. In tests made at DePaul University, over 60% of students chose the statement “No fire trucks are red” as the negation of “All fire trucks are red.” Yet, through guided discussion, these same students came fairly quickly to accept that “Some fire trucks are not red” conveys the negation more accurately, and most learned to take negations of general statements of the form “∀x, if P (x) then Q(x),” “∀x, ∃y such that P (x, y),” and so forth with reliable accuracy. One thing to keep in mind is the tolerance for potential ambiguity in ordinary language, which is typically resolved through context or inflection. For instance, as indicated on page 124 of the text, the sentence “All mathematicians do not wear glasses” can be a way to phrase a negation to “All mathematicians wear glasses.” (To see this, say it out loud, stressing the word “not.”) Some grammarians ask us to avoid such phrasing because of its potentially ambiguity, but the usage is widespread even in formal writing in high-level publications (“All juvenile offenders are not alike,” Anthony Lewis, The New York Times, 19 May 1997, Op-Ed page) and in literary works (“All that glisters is not gold,” William Shakespeare, The Merchant of Venice, Act 2, Scene 7, 1596-1597). Even rather complex sentences can be negated in this way. For instance, when asked to write a negation for “The sum of any two irrational numbers is irrational,” a student wrote “The sum of any two irrational numbers is not irrational,” which is an acceptable informal negation (again, say it out loud, stressing the word “not”). To avoid such responses, it may be necessary to specify to students that simply inserting the word “not” is not an acceptable answer to a problem that asks for a negation. The modified formal language of the text includes the words “such that” in statements containing an existential quantifier because when students write multiply-quantified statements “formally,” they often insert the words “such that” in the wrong place. That is, they insert it in a place that changes the meaning of the statement they were given. If they were not required to include the words “such that,” an opportunity to correct their misunderstanding would be missed. It can also be helpful to write out the words “if-then” in addition to using an arrow to denote the conditional. This encourages students to take the word “if” more seriously than they may otherwise be inclined to do1). The discussion about the geometry of the real numbers begun in Section 1.2 is continued in this chapter. Examples 3.3.5 and 3.3.6 show that while there is a smallest positive integer there is no smallest positive real number. The topic is further explored in Section 7.4, where a proof of the uncountability of the real numbers between 0 and 1 is given. 1Many students (and other people) mistakenly interpret if-then statements as and statements. For instance, when students are asked to state what it means for a relation R to be symmetric, a significant fraction write “aRb and bRa.”

Solutions for Exercises: The Logic of Quantified Statements

Suggestions 1. The exercises in Sections 3.1–3.3 are designed to try to imprint new language patterns on students’ minds. Because it takes time to develop new habits, it is helpful to continue assigning exercises from these sections for several days after covering them in class. To prepare for Chapter 4, universal conditional statements should especially be emphasized. As in Section 2.2, care may need to be taken not to spend excessive class time going over the more difficult exercises, such as those on the meaning of “necessary” and “sufficient” conditions. 2. The most important idea of Section 3.4 is also the simplest: the rule of universal instantiation. Yet this inference rule drives an enormous amount of mathematical reasoning. If you wish to move rapidly through Chapter 3, you could focus on this rule and its immediate consequences in Section 3.4 and omit the discussion of how to use diagrams to check validity of arguments.

Section 3.1 1. a. False

b. True

c. False

d. True

e. False

f. True

2. a. The statement is true. The integers correspond to certain of the points on a number line, and the real numbers correspond to all the points on the number line. b. The statement is false; 0 is neither positive nor negative. c. The statement is false. For instance, let r = −2. Then −r = −(−2) = 2, which is positive. d. The statement is false. For instance, the number 21 is a real number, but it is not an integer. 3. a. When m = 25 and n = 10, the statement “m is a factor of n2” is true because n2 = 100 and 100 = 4 · 25. But the statement “m is a factor of n” is false because 10 is not a product of 25 times any integer. Thus the hypothesis of R(m, n) is true and the conclusion is false, so the statement as a whole is false. b. Sample answer :R(m, n) is false when m = 8 and n = 4 because 8 is a factor of 42 = 16, but 8 is not a factor of 4. c. When m = 5 and n = 10, both statements “m is a factor of n2” and “m is a factor of n” are true because n = 10 = 5· 20 = m · 20. Thus both the hypothesis and conclusion of R(m, n) are true, and so the statement as a whole is true. d. Two sample answers: (1) Let m = 2 and n = 6. Then both statements “m is a factor of n2” and “m is a factor of n” are true because n = 6 = 2· 3 = m· 3 and n2 = 36 = 2· 18 = m ·18. Thus both the hypothesis and conclusion of R(m, n) are true, and so the statement as a whole is true. (2) Let m = 6 and n = 2. Then both statements “m is a factor of n2” and “m is a factor of n” are false because n = 2 ̸ = 6· k, for any integer k, and n2 = 4 ̸ = 6· j, for any integer j. Thus both the hypothesis and conclusion of R(m, n) are false, and so the statement as a whole is true. 4. a. Q(−2, 1) is the statement “If −2 < 1 then (−2)2 < 12.” The hypothesis of this statement is −2 < 1, which is true. The conclusion is (−2)2 < 12, which is false because (−2)2 = 4 and 12 = 1 and 4 ≮ 1. Thus Q(−2, 1) is a conditional statement with a true hypothesis and a false conclusion. So Q(−2, 1) is false. b. Let x = — 1 and y = 0. Then x < y because −1 < 0 but x2 ≮ y2 because (− 1)2 = 1 ≮ 02 = 0. Thus the hypothesis x < y is true and the conclusion x2 < y2 is false, so the statement as a whole is false. c. Q(3, 8) is the statement “If 3 < 8 then 32 < 82.” The hypothesis of this statement is 3 < 8, which is true. The conclusion is 32 < 82, which is also true because 32 = 9 and 82 = 64 and

Instructor’s Manual: Section 3.2

9 < 64. Thus Q(3, 8) is a conditional statement with a true hypothesis and a true conclusion. So Q(3, 8) is true. d. Three sample answers: (1) Let x = 2 and y = 3. Then x < y because 2 < 3 and x2 < y2 because 22 = 4 < 32 = 9. Thus both the hypothesis and the conclusion are true, so the statement as a whole is true. (2) Let x = 3 and y = 2. Then x ≮ y because 3 ≮ 2 and x2 ≮ y2 because 32 = 9 ≮ 22 = 4. Thus both the hypothesis and the conclusion are false, so the statement as a whole is true. (3) Let x = 2 and y = —3. Then x ≮ y because 2 ≮ − 3 and x2 < y2 because 22 = 4 < (− 3)2 = 9. Thus the hypothesis is false and the conclusion is true, so the statement as a whole is true. 5.

a. The truth set is the set of all integers d such that 6/d is an integer, so the truth set is {−6, −3, −2, −1, 1, 2, 3, 6}. b. Truth set = {1, 2, 3, 6} c. The truth set is the set of all real numbers x with the property that 1 ≤ x2 ≤ 4, so the truth set is {x ∈ R | −2 ≤ x ≤ −1 or 1 ≤ x ≤ 2}. In other words, the truth set is the set of all real numbers between −2 and −1 inclusive together with those between 1 and 2 inclusive. d. Truth set = {−2, −1, 1, 2} 6. a. {−9, −8, −7, −6, −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9} b. Truth set = {1, 2, 3, 4, 5, 6, 7, 8.9} c. Truth set = {−8, −6, −4, −2, 0, 2, 4, 6, 8} 7. baa, bab, bac, bba, bbb, bbc, bca, bcb 8. 110, 001, 010, 011 9. Counterexample: Let x = 1/2. Then

1 x

1 (1/2)

= 2, and 1/2 ≯ 2. (This is one counterexample

among many.) 10. Counterexample 1: Let a = 1, and note that (a − 1)/a = (1 − 1)/1 = 0 is an integer. Counterexample 2: Let a = −1, and note that (a − 1)/a = (−1 − 1)/(−1) = 2 is an integer. 11. Counterexample: Let m = 1 and n = 1. Then m · n = 1 · 1 = 1 and m + n = 1 + 1 = 2. But 1 ≯ 2, and so m · n ≯ m + n. (This is one counterexample among many.) 12. Counterexample: Let x = 1 and y = 1, and note that √ √ √ x+y = 1+1= 2 whereas

√ √ √ √ x + y = 1 + 1 = 1 + 1 = 2,

and 2

√

(This is one counterexample among many. Any real numbers x and y with xy = ̸ 0 will produce a counterexample.) 13. (a), (e), (f) 14. (b), (c), (e), (f)

Solutions for Exercises: The Logic of Quantified Statements 15. a. Some acceptable answers: All rectangles are quadrilaterals. If a figure is a rectangle then that figure is a quadrilateral. Every rectangle is a quadrilateral. All figures that are rectangles are quadrilaterals. Any figure that is a rectangle is a quadrilateral. b. Some acceptable answers: There is a set with sixteen subsets. Some set has sixteen subsets. Some sets have sixteen subsets. At least one set has 16 subsets. There is at least one set that has sixteen subsets. 16. a. ∀ dinosaur x, x is extinct. b. ∀ real number x, x is positive, negative, or zero. c. ∀ irrational number x, x is not an integer. d. ∀ logician x, x is not lazy. e. ∀ integer x, x2 does not equal 2, 147, 581, 953. f. ∀ real number x, x2 ̸= −1.

17.

a. ∃ an exercise x such that x has an answer. b. ∃ a real number x such that x is rational. 18. a. ∃s ∈ D such that E(s) and M (s). (Or: ∃s ∈ D such that E(s) ∧ M (s).) b. ∀s ∈ D, if C(s) then E(s). (Or: ∀s ∈ D, C(s) → E(s).) c. ∀s, if C(s) then ∼ E(s). d. ∃x such that C(s) ∧ M (s). e. (∃s ∈ D such that C(s) ∧ E(s)) ∧ (∃s ∈ D such that C(s) ∧ ∼E(s)) 19. (b), (d), (e) 20. Some acceptable answers: If a real number is positive, then its square root is positive. All positive real numbers have positive square roots. Every real number that is positive has a positive square root. Each real number that is positive has a positive square root. Given any positive real number, that number has a positive square root. The square root of a positive real number is positive. 21. a. The total degree of G is even, for any graph G. b. The base angles of T are equal, for any isosceles triangle T . c. p is even, for some prime number p. d. f is not differentiable, for some continuous function f . 22. a. ∀x, if x is a Java program, then x has at least 5 lines. b. ∀x, if x is a valid argument with true premises, then x has a true conclusion. Or : ∀ argument x, if x is valid and x has true premises then x has a true conclusion. Or : ∀ valid argument x, if x has true premises then x has a true conclusion. 23. a. ∀x if x is an equilateral triangle, then x is isosceles. ∀ equilateral triangles x, x is isosceles. b. ∀x, if x is a computer science student then x needs to take data structures. ∀ computer science student x, x needs to take data structures. 24. a. ∃ a hatter x such that x is mad. ∃x such that x is a hatter and x is mad. b. ∃ a question x such that x is easy. ∃x such that x is a question and x is easy.