Solution Manual for Mathematical Reasoning for Elementary Teachers, 7th Edition

Chapter 2 Sets and Whole Numbers COOPERATIVE INVESTIGATION Counting Cars and Trains 1. 8 trains: P, (LG)W, W(LG), RR, RWW, WRW, WWR, WWWW 2. 16 trains: Y, PW, WP, (LG)R, R(LG), (LG)WW, W(LG)W, WW(LG), RRW, RWR, WRR, RWWW, WRWW, WWRW, WWWR, WWWWW 3. The number of trains of lengths 3, 4, and 5 are respectively 4, 8, and 16, revealing a doubling pattern. Continuation of the pattern shows there will be 32, 64, and 128 trains of respective lengths 6, 7, and 8. 4. 2n−1 of length n can be formed. It must be shown that there are twice as many trains of length n as there are of length n – 1. There are two ways to lengthen a train of length n – 1 by one unit: (1) add a white caboose, giving us 2n− 2 trains of length n, all of which have white cabooses; and (2) lengthen the caboose by one unit, giving us another 2n− 2 trains of length n, all of which have nonwhite cabooses. Altogether, the two methods give us 2n − 2 + 2n − 2 = 2(2n − 2 ) = 2n−1 different trains of length n. This counts all of the trains of length n, since any train of length n corresponds to a unique train of length n – 1 by either removing a white caboose or shortening a nonwhite caboose by one unit. Since the number of trains of length one is 20, there are n – 1 doublings to give us 2n−1 trains of length n. Alternatively, think of starting with a train of n white cars. Those could be glued together in all possible ways to form all possible trains of length n. To do this, at each of the n – 1 junctions between two white cars one must make one of two choices—to glue or not to glue. Hence, there are 2n−1 possibilities in all, as above. 5. (i) 5 trains: WWWW, WWR, WRW, RWW, RR (ii) 8 trains: WWWWW, WWWR, WWRW, WRWW, RWWW, WRR, RWR, RRW (iii) The number of RW-trains of length eight is F9 = 34 ; 10 of these trains have five cars (WWRRR and its rearrangements). (iv) The number of RW-trains of length 1, 2, 3, 4, 5, … is 1, 2, 3, 5, 8, …, the Fibonacci numbers beginning with F2. Recall that F1 = F2 = 1 and Fn + 2 = Fn +1 + Fn . Note that adding a white caboose to the trains of length n + 1 and adding a red caboose to trains of length n yields all of the RWtrains of length n + 2. Thus, the number of trains of length n is Fn +1 . 6. (i) 2 trains: P, RR (ii) 3 trains: Y, (LG)R, R(LG) (iii) 13 trains: Br, (DG)R, R(DG), Y(LG), (LG)Y, PP, PRR, RPR, RRP, (LG)(LG)R, (LG)R(LG), R(LG)(LG), RRRR. None of these trains contain five cars. (iv) The number of W-trains of length 1, 2, 3, 4, 5, … is 0, 1, 1, 2, 3, …, again the Fibonacci numbers where we agree to set F0 = 0. Note that adding cabooses of length n – 1, n – 2, n – 3, …, 2 to the W-trains of length 2, 3, 4, .., n – 1 gives all the W-trains of length n + 1, except for the one additional train that consists of a single rod. This corresponds to the formula 1 + F1 + F2 + + Fn − 2 = Fn . Thus the number of W-trains of length n is Fn −1 . 7. Answers will vary. Section 2.1 Sets and Operations on Sets Problem Set 2.1 1. (a) {Arizona, California, Idaho, Oregon, Utah} (b) {Maine, Maryland, Massachusetts, Michigan, Minnesota, Mississippi, Missouri, Montana} (c) {Arizona} 24 Copyright © 2015 Pearson Education, Inc. (c) True. The sets contain the same elements. Order doesn’t matter. 2. (a) {l, i, s, t, h, e, m, n, a, o, y, c} (b) {a, e, m, t} 8. (a) True. 7 is in {6, 7, 23}, but the two sets are not the same. (c) ∅ 3. (a) {7, 8, 9, 10, 11, 12,13} (b) False. The sets are equal. (b) {9, 11, 13} (c) False. 6 is not a member of {7}. (c) Since 2 = 2 · 1, 4 = 2 · 2, 6 = 2 · 3, 8 = 2 · 4, 10 = 2 · 5, etc., the set is {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}. 9. 4. (a) {4, 8, 12, 16, 20} (b) Since 3 = 2 · 1 + 1, 5 = 2 · 2 + 1, 7 = 2 · 3 + 1, 9 = 2 · 4 + 1, 11 = 2 · 5 + 1, 13 = 2 · 6 + 1, 15 = 2 · 7 + 1, 17 = 2 · 8 + 1, and 19 = 2 · 9 + 1, the set is {3, 5, 7, 9, 11, 13, 15, 17, 19}. (c) Since 1 = 12, 4 = 22, 9 = 32, and 16 = 42, the set is {1, 4, 9, 16}. 5. Answers will vary. (a) { x ∈U 11 ≤ x ≤ 14} or { x ∈U 10 < x 12} or { x ∈ N x = 2n for n ∈ N and n > 6} (c) B ∩ C = {a, b} (d) A ∪ B = {a, b, c, d, e} (e) A = {f, g, h} (f) A ∩ C = {a, b} (g) A ∪ ( B ∩ C ) = {a, b, c, d, e} 10. (a) M = {45, 90, 135, 180, 225, 270, 315, …} (b) L ∩ M = {90, 180, 270, …} This can be described as the set of natural numbers that are divisible by both 6 and 45, or as the set of natural numbers that are divisible by 90. {x ∈ N x = n for n ∈ N with n odd 11. (a) D = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144} (b) G ∩ D = {1, 2, 3, 6, 9, 18} 2 and n ≥ 5} (c) (b) A ∩ B = {a, b, c} (c) 90 6. Answers will vary. (a) (a) B ∪ C = {a, b, c, h} { x ∈ N x is divisible by 3} (c) 18 12. (a) A ∩ B ∩ C is the set of elements common to all three sets. 7. (a) True. The sets contain the same elements. (b) True. Every element in {6}—namely, 6— is in {6, 7, 8}. 25 Copyright © 2015 Pearson Education, Inc. 26 Chapter 2 Sets and Whole Numbers (b) A ∪ ( B ∩ C ) is the set of elements in B but not in C, together with the elements of A. (c) ( A ∩ B ) ∪ C is the set of elements in C or in both A and B. (d) A ∪ ( B ∩ C ) is the set of all elements that are not in A, together with additional elements that are in both B and C. (c) Answers may vary. 14. No. It is possible that there are elements of A that are also elements of B but not C, or C but not B. For example, let A = {1, 2}, B = {2, 3}, C = {3}. A ∪ B = A ∪ C = {1, 2, 3}, but B ≠ C. 15. No, it is not necessarily true that F = G. For example, if D is any subset of the whole numbers, then F = {0, 1} and G = {7, 9} is a counterexample. In other words, suppose that D = {2, 4, 6, 8} and F and G are defined as before. D ∩ F = ∅ = D ∩ G , but F ≠ G. 16. (a) Since A ∩ B = {6, 12, 18} , (e) A ∪ B ∪ C is the set of all elements in A, B, and/or C. A ∩ B = {1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 19, 20} Since A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} and B = {1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17,19, 20}, (f) A ∩ B ∩ C is the set of elements that are in both B and C, but not in A. A ∪ B = {1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 19, 20} Since A ∪ B = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20} A ∪ B = {1, 5, 7, 11, 13, 17, 19} . Since A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} and 13. (a) A and B must be disjoint sets contained in C. B = {1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20}, A ∩ B = {1, 5, 7, 11, 13, 17, 19} . (b) Answers may vary. (b) The results of part (a) suggest that A ∩ B = A ∪ B and A ∪ B = A ∩ B 17. (a) R ∩ C is the set of red circles. Draw two red circles—one large and one small. Copyright © 2015 Pearson Education, Inc. Solutions to Problem Set 2.1 (b) L ∩ H is the set of large hexagons. Draw two large hexagons—one red and one blue. (c) T ∪ H is the set of shapes that are either triangles or hexagons. Draw two large triangles, two large hexagons, two small triangles, and two small hexagons—a red and a blue of each. (d) L ∩ T is the set of large triangles. Draw two large triangles—one red and one blue. (e) B ∩ C is the set of blue shapes that are not circles. That is, the set of four elements consisting of the large and small blue hexagons and the large and small blue triangles. (f) H ∩ S ∩ R is the set of hexagons that are both small and red. Draw one small red hexagon. 18. (a) L ∩ T (b) B ∩ (T ∪ H ) or B ∩ C (c) S ∪T (d) ( B ∩ C ) ∪ R 19. (a) Answers will vary. One possibility is B = set of students taking piano lessons, C = set of students learning a musical instrument. of shapes used. The number of cards is probably 3 × 3 × 2 × 3 = 54 . We cannot be sure of this result without knowing for sure that the deck does not contain any different shapes, colors, etc. (b) Answers will vary. (c) Answers will vary. 21. Answers will vary. 22. Children often interpret “or” as the “exclusive or” in which either one of two possibilities holds, but not both. Have the child consider the statement “Tomorrow, we will have music class or we will go to the gym”, and point out that this remains true even if both music class and gym class take place. 23. Answers will vary. Sample answer: The natural numbers (also called the counting numbers) are a subset of the whole numbers. Zero is a whole number but not a natural number, but all natural numbers are a subset of the whole numbers. 24. Joe always seems to find two sets which are disjoint. His teacher can use any pair of sets such as the set of all boys in his class and the other set of all students in the class of those who have brown eyes. If the two sets have an overlap, Joe won’t get the addition that he expected. 25. (a) 8 regions (b) Answers will vary. (c) Answers will vary. 20. (a) There appear to be 3 choices for shape, 3 choices for color, 2 choices for shading (shaded or not), and 3 choices for number 27 (b) By counting, there are 14 regions. Copyright © 2015 Pearson Education, Inc. 28 Chapter 2 Sets and Whole Numbers (c) Verify that A ∩ B ∩ C ∩ D has no region by observing that the region for A ∩ D is entirely contained in B ∪ C . Likewise, the region for B ∩ C is entirely contained in A ∪ D. The other missing region is A∩ B ∩C ∩ D . (d) Yes. Each loop contains 8 different regions, and there are 16 regions all together. 30. Note that in this diagram, O refers to the region outside both circles. 31. Answers will vary. (The loop shown on the bottom could have represented negative instead of positive, reversing all signs.) Note that in this diagram, O– refers to the region outside all circles. 26. These Venn diagrams show that A∩ B = A∪ B : A∩ B 32. B. 4 students like all three foods, so there must be a region common to all three loops in the Venn diagram. A∪ B 27. These Venn diagrams show that A∪ B = A∩ B : A∪ B 33. A ∩ B = {6, 12, 18} These are the elements that are divisible by both 2 and 3. A∩ B 28. (a) There are 6 choices for shape, 2 choices for size, and 2 choices for color, so the number of pieces is 6 × 2 × 2 = 24 . (b) 6 × 2 × 2 × 2 = 48 29. (a) There are eight subsets: ∅, {P}, {N}, {D}, {P, N}, {P, D}, {N, D}, {P, N, D}. (b) There are 16 subsets: ∅, {P}, {N}, {D}, {P, N}, {P, D}, {N, D}, {P, N, D}, {Q}, {P, Q}, {N, Q}, {D, Q}, {P, N, Q}, {P, D, Q}, {N, D, Q}, {P, N, D, Q}. (c) Half of the subsets of {P, N, D, Q} contain Q. (d) The number of subsets doubles with each additional element, so a set with n elements has 2n subsets. 34. Choice H (42) only. It is both divisible by 7 (as in set V) and divisible by 3 (as in set W). 35. The number must be a multiple of both 10 and 12. The question asks for the least number with that property, so the answer is LCM(10, 12) = 3 ⋅ 2 ⋅ 2 ⋅ 5 = 60, choice D. 36. There are 28 cello students, of whom 12 also play violin Therefore, there are 28 − 12 = 16 who only play cello. There are 23 violin students, of whom 12 also play cello. Therefore, there are 23 − 12 = 11 who only play violin. Thus, there are 16 + 12 + 11 = 39 students who play either the cello or violin. There are 54 students, so there are 54 − 39 = 15 students who do not play either cello or violin. Copyright © 2015 Pearson Education, Inc. Solutions to Problem Set 2.2 Section 2.2 Sets, Counting, and the Whole Numbers Problem Set 2.2 8. (a) The solutions for ( x − 1)( x − 9) = 0 are x = 1 and x = 9, so C = {1, 9} and n(C) = 2. (b) n(D) = 2 because D = {40, 80}. 1. (a) 13(th): ordinal first: ordinal (b) fourth: ordinal; second: ordinal 93: cardinal. In common usage, it represents the number, 93, correct out of 100. 9. (a) The correspondence 0 ↔ 1, 1 ↔ 2, 2 ↔ 3,…, w ↔ w + 1,… shows W ⬃ N. (b) The correspondence 1 ↔ 2, 3 ↔ 4,… , n ↔ n + 1,… shows D ⬃ E. (c) The correspondence 2. (a) fourth: ordinal 13: cardinal (b) 2213762789: nominal; 23rd: ordinal 18: cardinal. 3. (a) Equivalent, since there are five letters in the set {A, B, M, N, P} (b) Not equivalent since the sets have different numbers of elements. (c) Equivalent, say by the correspondence o ↔ t,n ↔ w,e ↔ o (d) Not equivalent, since the set {0} has one element and ∅ has no elements. 1 ↔ 10, 2 ↔ 100 = 102 ,… , n ↔ 10n ,… shows that the sets are equivalent. 10. (a) Finite. The number of grains of sand is large, but finite. (b) Infinite (c) Infinite 11. (a) Answers will vary. For example, Q1 ↔ Q2 , Q3 ↔ Q4 , and so on. (Note that P need not be the center—it can be any fixed point inside the small circle.) 4. (a) Equivalent, since there are four elements in each set. (b) Equivalent, since there are 24 elements in each set. (c) Not equivalent, since {0, 1} has two elements and the other set is infinite. 5. (a) n(A) is 0, 1, 2, 3, or 4, since it has strictly fewer elements than of set B. (b) Answers will vary. For example, Q1 ↔ Q2 , Q3 ↔ Q4 , and so on. (b) n(C) is 5, 6, 7, …, any whole number greater than or equal to 5. 6. Yes, A = {1, 8, 27, 64}, B = {California, Arizona, New Mexico, Texas}. Since both sets have the same number of elements, A ⬃ B. We could have the correspondence 1 ↔ California, 8 ↔ Arizona, 27 ↔ New Mexico, 64 ↔ Texas. 12. (a) Answers will vary. For example, Q1 ↔ Q2 , etc. 7. (a) n(A) = 7 because A = {21, 22, 23, 24, 25, 26, 27}. (b) n(B) = 0 because B = ∅ . There is no natural number x such that x + 1 = x. Copyright © 2015 Pearson Education, Inc. 29 30 Chapter 2 Sets and Whole Numbers that A ∩ B = A ∪ B . Thus A ∪ B has no additional elements besides those in A ∩ B, and so neither A nor B has any additional elements. Since A = A ∩ B and B = A ∩ B , we conclude that A = B. (Caution: This reasoning would not be valid if infinite sets were allowed.) (b) Answers will vary. For example, Q1 ↔ Q2 , etc. 13. (a) True. A set B cannot have fewer elements than its subset A. (b) False. Possible counterexample: A = {1}, B = {2, 3}. Then n(A) = 1 and n(B) = 2, so n(A) < n(B), but A ⊄ B . (c) True. The union A ∪ B does not include any more elements than just A, so the elements of B must already be elements of A. (d) True. If an element of A was not also an element of B, then this element would be missing in A ∩ B but included in A. 14. (a) True. 16. (a) 1000 ÷ 6 = 166.666…, so the largest element of S is 166 · 6 = 996. Therefore, n(S) = 166. (b) The elements in F ∩ S are the multiples of 30. Since 1000 ÷ 30 = 33.33…, the largest element of F ∩ S is 33 · 30 = 990. Therefore n ( F ∩ S ) = 33. (c) 17. Three of the four regions in the 2-loop diagram account for 500 of the households. Thus there are 200 households in the overlapped region representing the number of households having both a TV and a computer. (b) True. (c) False. For example, if B = { x | x is any positive integer except 3} , then B = {3} . (d) False. This is true only if A and B are disjoint. 15. (a) n( A ∩ B) ≤ n( A) The set A ∩ B contains only the elements of A that are also elements of B. That is, A ∩ B ⊆ A . Thus, A ∩ B cannot have more elements than A. (b) n( A) ≤ n( A ∪ B ) 18. Use the fact that the eight regions pictured below are mutually disjoint sets and apply the strategy of working backwards, that is, start with n( A ∩ B ∩ C ) = 7 . Next use n( A ∩ B) , n( B ∩ C ) , and n( A ∩ C ) to find the values 10, 5, and 8, respectively. Then use n(A), n(B), and n(C) to find the values 15, 28, and 10, respectively, and finally n(U) to find the value 17. The set A ∪ B contains all of the elements of the set A and any additional elements of B that are not already included. Then A ⊆ A ∪ B . Therefore, A ∪ B must have at least as many elements as A. (c) Since n( A ∩ B) = n( A ∪ B) and A ∩ B ⊆ A ⊆ A ∪ B , we can conclude Copyright © 2015 Pearson Education, Inc. Solutions to Problem Set 2.2 19. The 3-loop Venn diagram can be filled in the manner of problem 18. We now see there are 20 + 5 + 25 = 50 percent of the students who like just one sport and 5 percent do not like any of the three sports. 20. Let A, H, and P be the sets of students taking anthropology, history, and psychology, respectively. Then n(A) = 40, n(H) = 11, n(P) = 12, n( A ∩ H ∩ P) = 3, n( A ∩ H ) = 6, and n( A ∩ P) = 6. Note: If one chooses to solve this problem by using a Venn diagram, there are several possibilities, since the number of students taking both history and psychology cannot be determined. However, the answers to parts (a), (b), and (c) are uniquely determined. 31 23. Answers will vary. One possible answer uses a cup with 5 marbles. Ask “how many marbles are in the cup?” to get the response “five.” Now remove a marble, and again ask how many marbles are in the cup, eliciting the response “four.” Continue to remove one marble at a time, until no marbles remain in the cup. Explain that the number of marbles in the empty cup is “zero.” 24. Answers will vary, but should include the concept that if one strip is shorter than a second strip, and the second strip is shorter than the third, then the first must be shorter than the third. 25. 400 is a square number since 400 is the product of two squares, 4 and 100. Also, 400 = 202 , as illustrated below. (a) n( A − ( A ∩ H ) ∪ ( A ∩ P)) = n( A) − n( A ∩ H ) − n( A ∩ P ) + n( A ∩ H ∩ P ) = 40 − 6 − 6 + 3 = 31 Note: One might argue that the answer is zero, since all of the students in this situation are also taking mathematics. (b) n( A ∪ H ) = n( A) + n( H ) − n( A ∩ H ) = 40 + 11 − 6 = 45 (c) n(( H ∩ A) − P ) = n( H ∩ A) − n( H ∩ A ∩ P) = 6−3=3 26. Annabelle recognized that “third” and “sixth” are nominal. There isn’t a cardinal number in the conversation, so there is no addition in this setting. Annabelle laughed with the math professor, as did her parents. 21. (a) 27. (a) The numbers that have been inserted are bold. Explanations to Zack will vary. (b) 22. Answers will vary. Both “number” and “color” are abstract concepts that can be taught by giving concrete examples. (b) Answers will vary Copyright © 2015 Pearson Education, Inc. 32 Chapter 2 Sets and Whole Numbers 28. (a) Richard is multiplying by 2, rather than squaring. He must think that the exponent 2 is just a multiplier. (b) Drawing pictures of squares of size 1 × 1, 2 × 2, 3 × 3, 4 × 4, and 5 × 5 would show that areas don’t double, they grow much faster than that. 31. (a) Row 0 1 Row 1 1 1 Row 2 1 2 1 Row 3 1 3 3 1 Row 4 1 4 6 4 1 (b) The table is the same as Pascal’s triangle each entry in the table is the sum of the numbers in the row just above that are directly above and one column to the left. Using this pattern, we get Row 5: 1 5 10 10 5 1 and Row 6: 1 6 15 20 15 6 1. 29. (a) Jeff is multiplying the base by the exponent, which is the wrong procedure. In this case, he is multiplying by three, not taking numbers to the third power. 32. Evelyn’s assumptions give the following diagram. (b) Use cubes whose sides are of length 1 cm, 2 cm, 3 cm, 4 cm, and 5 cm. 30. (a) Six ways: yrg, rgy, gyr, ygr, gry, ryg (b) Six. The numbers 1, 2, 3 can be viewed as the bottom, middle, top positions for balls in the can. Therefore, each of the six ways to place the balls in the can are a distinct one-to-one correspondence. (c) There are 12 ways to place the first egg, and 11 empty spaces in which to place the second egg. Therefore, the first two eggs can be put into the carton in 12 11 ways. This leaves 10 empty choices for the third egg, so there are 12 11 10 ways to place the first three eggs. Continuing in this way shows there are 12 11 10 9 8 7 6 5 4 3 2 1 = 479,001,600 ways to place the dozen eggs. (d) If {1, 2, 3, …, 12} and {a, b, c, d, e, f, g, h, i, j, k, l} represent the set of eggs and the set of spaces in the carton from part (c), the number of one-to-one correspondences is the same as the number of ways to put the eggs into the carton, namely 12 11 10 9 8 7 6 5 4 3 2 1 = 479,001,600. The zero values are obtained from Evelyn’s assumption that anyone with a VCR also has a TV. Evelyn should question the survey because only 94 of the 100 households are represented. (Evelyn’s assumption may be wrong—for example, there may be people who use a monitor instead of a TV to view videotapes—but these are very unlikely to account for 6% of a given sample.) Also, the stereo and VCR category is missing 33. Consider the 35 students taking Arabic: 20 take only Arabic, 7 take both Arabic and Bulgarian (and possibly Chinese), telling us that 8 students are taking Arabic and Chinese and not Bulgarian. Similarly, there are 5 students taking Bulgarian and Chinese but not Arabic. The rest of the values in the Venn diagram are now easy to fill in. In particular, 3 students take all three languages, and 26 are not taking any of the three languages. 34. Since we are given that k < l and l < m, we can choose sets K, L, and M satisfying K ⊂ L ⊂ M and n(K) = k, n(L) = l, and n(M) = m. By the transitive property of set inclusion (see Section 2.1, or just look at a Venn diagram) we know that K ⊂ M and so k 19) (f) Closed (since the sum of two multiples of 3 is a multiple of 3: 3a + 3b = 3(a + b)) 9. (a) Commutative property of addition (b) Closure property (c) Additive-identity property of zero (b) (d) Associative and commutative properties (e) Associative and commutative properties 10. (a) (1 + 20) + (2 + 19) + (3 + 18) + + (10 + 11) = 21 + 21 + 21 + + 21 (for 10 terms) = (10)(21) = 210 (b) Associative and commutative properties 11. (a) (b) 15. (a) comparison 12. (a) (b) measurement (number-line) 16. (1) missing addend (b) (b) take-away 17. Answers will vary. For example, 13. (a) 5 + 7 = 12 7 + 5 = 12 12 – 7 = 5 12 – 5 = 7 (b) 4 + 8 = 12 8 + 4 = 12 12 – 8 = 4 12 – 4 = 8 (a) Take-away: Maritza bought a booklet of 20 tickets for the amusement park rides. She used 6 tickets for the roller coaster. How many tickets does she have left? Copyright © 2015 Pearson Education, Inc. 36 Chapter 2 Sets and Whole Numbers (b) Missing Addend: Nancy’s school has 23 proof-of-purchase coupons for graphing calculators. Her school will be entitled to a free overhead projection calculator with 40 proof-of-purchase coupons. How many more coupons do they need to get the free calculator? (c) Comparison: Oak Ridge Elementary School has an enrollment of 482 students and Crest Hill School has an enrollment of 393 students. How many more students does Oak Ridge have than Crest Hill? (d) Measurement: A fireman climbed up 11 rungs on a ladder, but the smoke was too thick and he came down 3 rungs. How many rungs up the ladder is the fireman? 18. Jeff has read through page 240. Therefore, the number of pages is 257 – 240 = 17 pages or 257 – 241 + 1 = 17 pages. 19. Use the guess and check method. Some answers will vary. 22. Write all possible combinations of 3 different numbers whose sum is the indicated number in the triangle. Then, place numbers which occur in more than one sum at the vertices. (a) 1+3+6 2+3+5 4+5+1 (b) 1+4+6 2+3+6 5+4+2 (c) 1+5+6 2+4+6 3+4+5 (a) ((8 – 5) – 2) – 1 = 0 23. Blake has one more marble than before, and Andrea has one less than before, so now Blake has two more marbles than Andrea. Use a small number of marbles, say five each, to demonstrate what happened. After giving Blake one marble, he has 6 and Andrea has 4, and thus clearly Blake has two more marbles than Andrea. (b) 8 + (5 – 2) + 1 = 12 24. Answers will vary. (c) (8 + 5) – (2 + 1) = 10 25. Answers will vary. (a) (8 – 5) – (2 – 1) = 2 (b) 8 – (5 – 2) – 1 = 4 20. Use the guess and check method. Some answers will vary. 21. (a) First fill in the squares by noting that 3 – 1 = 2, 4 – 2 = 2, and 7 – 2 = 5. Then complete the circles. 26. (a) 0 is the cardinal number of the empty set, n{∅} = 0. If the set A has a elements in it (i.e., n(A) = a), then n( A) + n(∅) = n( A ∪ ∅). But, A ∪ ∅ = A, since there are no elements in ∅ to add to the set A. Thus, n( A ∪ ∅) = n( A) = a. (b) Use the guess and check method. (b) Answers will vary, but the major idea is to give an example of starting with a group of children in the room, playing the role of A, and adding all others in the room who are over 80 years old. As there is no addition to A, the number of elements in it doesn’t change, and so a + 0 = a. Copyright © 2015 Pearson Education, Inc. Solutions to Problem Set 2.3 37 27. Answers will vary but should include that given two of the minuend, subtrahend, and answer, the third can be found. Furthermore, students often struggle when problems are presented in “non-conventional” ways. (This is worth the struggle as it is the beginning of algebra.) Many elementary school students have difficulty when there is a blank in the front of the number sentence. 28. Answers will vary. 29. First find the top row and the left column. For example, the fourth entry in the left column must be 6 – 2 = 4, and then the third entry in the top row is 5 – 4 = 1, and so on. The completed table is shown below. + 5 4 1 6 9 2 0 8 7 3 3 8 7 4 9 12 5 3 11 10 6 9 14 13 10 15 18 11 9 17 16 12 6 11 10 7 12 15 8 6 14 13 9 4 9 8 5 10 13 6 4 12 11 7 0 5 4 1 6 9 2 0 8 7 3 7 12 11 8 13 16 9 7 15 14 10 5 10 9 6 11 14 7 5 13 12 8 2 7 6 3 8 11 4 2 10 9 5 1 6 5 2 7 10 3 1 9 8 4 8 13 12 9 14 17 10 8 16 15 11 30. They must both love some of the same mice. In fact, there are five mice that Angel and Jane both love since 8 = 7 + 6 − n (where there are n mice they both love). 33. (a) Yes, Carmen’s method is correct using the associativity property and a good understanding of position value. (b) 31. (a) This student is “subtracting up”, which is a common mistake. (b) Answers will vary. For example: After reminding her about place value, ask her to do a two-digit subtraction such as 73 − 35 and reflect on the similarities between how to subtract two-digit numbers and how to subtract three-digit numbers. That could be followed with 736 − 327 = ? as a next level problem that is easier than the one in the problem. 32. Phillip does two things incorrectly. First, he sees that he can’t subtract the 9 from the 6, so he goes to the next column to do an exchange. He changes the 8 to a 7, but then, instead of having 16 ones, he adds 1 to 6 to give 7. He has misunderstood that the 1 he got from the exchange is actually 10, not 1 unit. Phillip then proceeds to subtract up anyway (he mistakenly writes 7 − 9 as 2). 60 + 11 = 71 − (30 + 8) = −38 30 + 3 = 33 34. Note that elements of A ∩ B are counted twice when calculating n(A) + n(B). We compensate by subtracting n( A ∩ B) , giving n( A ∪ B). 35. Let T = {n ∈ N n is a multiple of 12 and n ≤ 200} and F = {n ∈ N n is a multiple of 5 and n ≤ 200} . The number we need to find is n(T ∪ F ) = n(T ) + n( F ) − n(T ∩ F ) = 16 + 40 − 3 = 53 Copyright © 2015 Pearson Education, Inc. 38 Chapter 2 Sets and Whole Numbers 22 = 21 + 1 23 = 10 + 10 + 3 24 = 21 + 3 25 = 15 + 10 36. (c) For example, choose 73 and 74. 73 = 45 + 28, and 74 = 45 + 28 + 1. Three is the maximum number of triangular numbers needed. 37. (d) Any whole number may be written as the sum of at most three triangular numbers. 38. {0}, since 0 – 0 = 0. 39. (a) {2, 3, 4, 5, 6, 7, 8, 9, …} since 2 + 2 = 4, 2 + 3 = 5, 2 + 4 = 6, etc. (b) {0, 1} (c) No, because the set of all whole numbers, {0, 1, 2, …}, fits the given description. (d) Whole numbers that must be in C are all even numbers ≥ 2. Zero and odd numbers may or may not be in C. 40. (a) Use the formula tn = n(n + 1) . 2 n 1 2 3 4 5 6 tn 1 3 6 10 15 21 n 7 8 9 10 11 12 tn 28 36 45 55 66 78 n 13 14 15 tn 91 105 120 (b) 11 = 10 + 1 12 = 6 + 6 13 = 10 + 3 14 = 10 + 3 + 1 15 = 15 16 = 15 + 1 17 = 15 + 1 + 1 18 = 15 + 3 19 = 10 + 6 + 3 20 = 10 + 10 21 = 21 41. It can be shown that Sameer is correct. The easiest way to find a sum is to use the largest Fibonacci number possible as the next summand, a procedure sometimes called a “greedy algorithm.” For example, to express 100 as a Fibonacci sum, first write down 89 since it is the largest Fibonacci number less than 100. This leaves 11, so next write down 8. The 3 that still remains to be accounted for is a Fibonacci number, so 100 = 89 + 8 + 3. 42. (a) Corey is correct. The sum of two even numbers can be thought of as the sum of two rows having m columns with two rows having n columns. The result is two rows having m + n columns. Since the result can always be represented using two rows each having the same number of columns, the result is always an even number. (b) Maya is mistaken. This can be shown by a counterexample. For example, the sum of the odd numbers 3 and 5 is the even number 8. 43. Statement B does not correspond to the number sentence 15 – 8 = 䊐. The other problems illustrate take-away (A), comparison (C), and missing addend (D). 44. C. Jordan started with his allowance, x, and subtracted $4 from it because he spent that amount. He is left with x − 4 which is the $16 he has left. Thus x − 4 = 16. 45. D. The number must be more than 500. 46. A. 45,749. Just subtracting the last two digits in your head gives the only possible correct answer. 47. C. 1975 + 16 + 4 = 1995 48. B. 54 + 17 = 71 Copyright © 2015 Pearson Education, Inc. Solutions to Problem Set 2.4 39 Section 2.4 Multiplication and Division of Whole Numbers Problem Set 2.4 1. (a) 3 5 = 15, set model (repeated addition) (b) 6 3 = 18, number-line model (c) 5 3 = 15, set model (d) 3 6 = 18, number-line model (e) 8 4 = 32, rectangular area model (f) 3 2 = 6, multiplication tree model 2. (a) A set of dominoes, containing 11 stacks of 5 dominoes is modeled by a rectangular array. 11 × 5 = 55 The set contains 55 dominoes. (b) The outfits that can be made from 3 skirts and 6 blouses is modeled by a Cartesian product, shown below. 3 × 6 = 18, so she has 18 outfits she can wear. b1 b2 b3 b4 b5 b6 s1 (s1 , b1 ) (s1 , b 2 ) (s1 , b3 ) (s1 , b 4 ) (s1 , b5 ) (s1 , b6 ) s2 (s 2 , b1 ) (s 2 , b2 ) (s 2 , b3 ) (s 2 , b4 ) (s 2 , b5 ) (s 2 , b6 ) s3 (s3 , b1 ) (s3 , b 2 ) (s3 , b3 ) (s3 , b 4 ) (s3 , b5 ) (s3 , b6 ) (c) A hike of 10 miles each day for 5 days can be modeled by a number line. 5 · 10 = 50 He hiked 50 miles. (d) The production of 35 widgits a day in a 5-day workweek can be modeled by a set. 5 · 35 = 175 The company makes 175 widgets in a 5-day workweek. Copyright © 2015 Pearson Education, Inc. 40 Chapter 2 Sets and Whole Numbers (e) The sunroom can be modeled by a rectangular area. (iv) Using the commutative property of multiplication, 56,108 6 = 6 56,108 = 56,108 + 56,108 + 56,108 + 56,108 + 56,108 + 56,108 = 336,648 4. (a) Each of the a lines from set A intersects each of the b lines from set B. Since the lines for A are parallel and the lines for B are parallel, the intersection points will all be distinct. (b) 9 × 18 = 162 She needs 162 1-square-foot tiles. (f) The outcomes that result from rolling a die and flipping a coin can be modeled with a multiplication tree. 5. (a) Not closed. For example, 2 × 2 = 4 , which is not in the set. (b) Closed. 0 × 0 = 0, 0 × 1 = 0, 1 × 1 = 1, 1 × 0 = 0 . All products are in the set. (c) Not closed. For example, 2 × 4 = 8 , which is not in the set. (d) Closed. The product of any two even whole numbers is always another even whole number. (e) Closed. The product of any two odd whole numbers is always another odd whole number. (f) Not closed. For example, 2 × 23 = 24 , which is not in the set. 6 × 2 = 12 12 outcomes are possible. 3. (a) Answers will vary. (b) (i) 4 · 9 = 9 + 9 + 9 + 9 = 36 (ii) 7 × 536 = 536 + 536 + 536 + 536 + 536 + 536 + 536 = 3752 (iii) 6 47,819 = 47,819 + 47,819 + 47,819 + 47,819 + 47,819 + 47,819 = 286,914 (g) Closed. 2m × 2n = 2m + n for any whole numbers m and n. (h) Closed. 7 a × 7b = 7 a + b for any whole numbers a and b. 6. (a) Closed. The product of any two whole numbers in the set W – {5} is also in the set, and 5 is not a product of any numbers in the set (5 is prime). (b) Not closed, since 2 × 3 = 6 . (c) Closed. The product of any two whole numbers in the set W – {2, 3} is also in the set, and 2 and 3 are not products of numbers in the set (2 and 3 are prime). Copyright © 2015 Pearson Education, Inc. Solutions to Problem Set 2.4 41 7. (a) Commutative property of multiplication (b) Distributive property of multiplication over addition (c) Multiplication-by-zero property 8. (a) Distributive property of multiplication over addition (b) Associative property of multiplication (c) Multiplicative identity property of 1 9. Commutative property: 5 × 3 = 3 × 5 10. Commutative property: 6 × 2 = 2 × 6 11. (a) (b) (c) 12. If we assume the numbers are in feet, the swimming pool is 2 ft by (5 + 4) = 9 ft and thus the pool is 18 ft2. The other side of the equation means that for the two pools, one is 2 ft by 5 ft (10 ft2) and the other is 2 ft by 4 ft (8 ft2). The larger pool and these two pools next to each other show the Distributive property. 13. The rectangle is a + b by c + d, so its area is (a + b) (c + d). The rectangle labeled F is a by c so its area is ac. Similarly, the areas of the rectangles O, I, and L are given by the respective products ad, bc, and bd. Summing the areas of the four rectangles labeled F, O, I, and L gives the area of the large rectangle, so (a + b) (c + d) = ac + ad + bc + bd. 14. a + b + c a·d a·e a·f b·d b·e b·f c·d c·e c·f d + e + f (a + b + c ) · (d + e + f) =a·d+a·e+a·f+b·d+b·e+b·f +c·d+c·e+c·f 15. (a) Distributive property of multiplication over addition: 7 · 19 + 3 · 19 = (7 + 3) · 19 = 10 · 19 = 190 (b) Distributive property of multiplication over addition: 24 · 17 + 24 · 3 = 24 · (17 + 3) = 24 · 20 = 480 Copyright © 2015 Pearson Education, Inc. 42 Chapter 2 Sets and Whole Numbers (c) Distributive property, associative property, and/or multiplication property of zero: 36 · 15 – 12 · 45 = (12 · 3) · 15 – 12 · (3 · 15) = 12 · (3 · 15) – 12 · (3 · 15) = (12 – 12) · (3 · 15) = 0 · (3 · 15) = 0 16. (a) Commutative and Distributive properties: 7 ⋅ 16 + 7 ⋅ 24 = 7 ⋅ 40 = 280 (b) Associative and multiplicative property of zero: 23 ⋅ ( 25 ⋅ 35 ⋅ 24 ⋅ 0) = ( 23 ⋅ 25 ⋅ 35 ⋅ 24) ⋅ 0 =0 (c) Distributive property, commutative, and associative properties: 36 ⋅ 15 + 12 ⋅ 55 = 3 ⋅ 12 ⋅ 15 + 12 ⋅ 55 = 12 ⋅ 3 ⋅ 15 + 12 ⋅ 55 = 12 ⋅ 45 + 12 ⋅ 55 = 12 ⋅ 100 = 1200 17. 22. (a) 19 − 5 = 14, which is greater than 5, so we must subtract 5 again. 14 − 5 = 9, which is still greater than 5. Subtracting 5 again, 9 − 5 = 4, which is less than 4, so we are done. Because we subtracted 5 three times, the quotient is 3. The remainder is 4. This is represented as −5 −5 −5 a = 19 → 19 − 5 = 14 → 14 − 5 = 9 → 9 − 5 = 4 (done) (b) 18 − 9 = 9. Since 9 ≤ 9, subtract 9 again to get 9 − 9 = 0. The remainder is 0. We subtracted twice, so the quotient is 2. −9 −9 a = 18 → 18 − 9 = 9 → 9 − 9 = 0 (done) (c) 25 − 8 = 17, which is greater than 8, so we must subtract 8 again. 17 − 8 = 9, which is still greater than 8. Subtracting 8 again gives 9 − 8 = 1, which is less than 8, so we are done. Because we subtracted 8 three times, the quotient is 3. The remainder is 1. −8 −8 −8 a = 25 → 25 − 8 = 17 → 17 − 8 = 9 → 9 − 8 = 1 (done) (d) Answered in each part separately. 23. (a) 14 − 7 = 7. Since 7 ≤ 7, subtract 7 again to get 7 − 7 = 0. The remainder is 0. We subtracted twice, so the quotient is 2. −7 −7 a = 14 → 14 − 7 = 7 → 7 − 7 = 0 (done) 18. (a) 18 ÷ 6 = 3 (b) 18 ÷ 3 = 6 19. (a) 4 8 = 32, 8 4 = 32, 32 ÷ 8 = 4, 32 ÷ 4 = 8 (b) 6 5 = 30, 5 6 = 30, 30 ÷ 5 = 6, 30 ÷ 6 = 5 20. (a) 7 9 = 63, 9 7 = 63, 63 ÷ 7 = 9, 63 ÷ 9 = 7 (b) 11 5 = 55, 5 11 = 55, 55 ÷ 5 = 11, 55 ÷ 11 = 5 21. (a) Repeated subtraction (b) Partition (c) Missing factor or repeated subtraction (b) 7 is already less than 14, so there is no subtraction. The remainder is 7 and, since there were no subtractions, the quotient is 0. (c) Answered in each part separately. 24. (a) Since 78 – 13 = = = = = = gives 0, 78 ÷ 13 = 6. (b) Since 832 – 52 = = = = = = = = = = = = = = = = gives 0, 832 ÷ 52 = 16. (c) Since 96 – 14 = = = = = = gives 12, 96 ÷ 14 = 6 R 12. (Note: Stop pressing = when the result is less than the divisor, 14.) (d) Since 548,245 – 45,687 = = = = = = = = = = = = gives 1, 548,245 ÷ 45,687 = 12 R 1. Copyright © 2015 Pearson Education, Inc. Solutions to Problem Set 2.4 25. (a) y = (5 · 5) + 4 = 25 + 4 = 29 (b) 3x + 2 = 20, so x = 6. 26. (a) z = 3 ⋅ 15 + 4 = 45 + 4 = 49 (b) 20 = 2 x (3) + 2 ⇒ 20 = 6 x + 2 ⇒ 18 = 6 x ⇒ x = 3 (c) w + 2 = 6 ( 4) + 5 ⇒ w + 2 = 24 + 5 ⇒ w + 2 = 29 ⇒ w = 27 27. (a) 320 ⋅ 315 = 320 +15 = 335 Next have a second group come forward, and verify that 2 4 = 8. And so on. (b) Have the children form rectangles of various sizes, and then count off row by row to obtain the total number of children in the array. (c) Choose how many teams to be formed (the divisor) and see how large the teams will be, with an equal number on each team. (d) For a given number of children on a team, add teams until the total number of children reaches a certain total. Then count how many teams were needed. (b) (32 )5 = 32⋅5 = 310 (c) 43 y 3 ⋅ z 3 = ( y ⋅ z )3 or ( yz )3 28. (a) 48 ⋅ 78 = (4 ⋅ 7)8 = 288 (b) x 7 ⋅ x9 = x7 + 9 = x16 (c) (t 3 )4 = t 3⋅4 = t12 29. (a) 8 = 2 ⋅ 2 ⋅ 2 = 23 (b) 4 ⋅ 8 = (2 ⋅ 2) ⋅ (2 ⋅ 2 ⋅ 2) = 25 (c) 1024 = 210 (d) 84 = (23 )4 = 23⋅4 = 212 30. Use the guess and check method. (a) m = 4; 3 ⋅ 3 ⋅ 3 ⋅ 3 = 81 (b) n = 12 (c) p = 10 (d) q = 20 31. Answers will vary, but should include the fact that, if we are dividing a by b, then if a b, the either a − b = 0 or a − b > 0. The remainder cannot be negative in either case because the process stops before that can happen. 33. Answers will vary. These are sample problems: Peter has a board 14 feet long, and each box he makes requires 3 feet of board. How many completed boxes can be made? Answer = 4. Tina has a collection of 14 antique dolls. A display box can hold at most three dolls. How many boxes does Tina require to display her entire collection? Answer = 5. Andrea has 14 one-by-one foot paving stones to place on her 3-foot-wide walkway. How many feet of walk can she pave, and how many stones will she have left over for a future project? Answer = 4 feet of walk paved, with 2 stones remaining. 34. Answers will vary, but should make the point that the division a b is approached by forming a train of length a, and then seeing how many cars of length b are needed to form an equally long train. 35. As listed in the theorem, starting with the second one, • It doesn’t matter in which order you multiply two numbers. • It doesn’t matter which way you group the terms when multiplying three numbers. • Multiplying by 1 never changes the number (and 1 is the only number for which that happens.) • Zero times any number is zero. 32. Answers will vary. The following are sample answers. (a) Have children hold hands in, say, groups of four. Have one group come to the front of the class, and verify that 1 4 = 4. Copyright © 2015 Pearson Education, Inc. 44 36. Chapter 2 Sets and Whole Numbers 0 = 2 × (3 − 3) 1 = 2(3− 3) 2 = 2 + (3 − 3) 3 = 3(3 − 2) 4 = 3 + (3 − 2) 5 = 23 − 3 6 = 32 − 3 7 = 3× 3− 2 8 = 3+ 3+ 2 9 = 3 × 3 10 = ? 11 = 23 + 3 or (3 × 3) + 2 12 = 32 + 3 13 = ? 14 = ? 15 = 3 × (3 + 2) 16 = ? 17 = ? 18 = 2 × 3 × 3 37. This student struggles to see the relationship between multiplication and division exercises. In addition, students typically have more difficulty when the blank or box is at the beginning of the equation. 38. Answers will vary, but you could first ask for her solution. She may say that since the total cost of one nut and one bolt together is $1.00, and she is buying 18 of them, the total cost is $18.00. You could then ask her to justify her response: 18 ⋅ 86 + 18 ⋅ 14 = 18 ⋅ (86 + 14) = 18 ⋅ 100 = 1800 cents or $18.00. You could then ask her a question that would lead to her saying it is the distributive property. 39. The student subtracted the number of cupcakes Nelson baked in each pan instead of dividing to find the number of pans needed to bake all of the cupcakes. This is common because students understand they need to do something with the numbers, but they aren’t sure what. In this case, the student is not thinking of grouping the cupcakes into pans, which should be a signal that he should divide rather than subtract, which suggests removal. 40. Draw a rectangular pool of 30 + 2 feet in length and 20 + 3 feet in width. Then identify the partial products in the picture. 32 × 23 = (30 + 2)( 20 + 3) = 30 ( 20) + 2 (20) + 3 (30) + 3 ( 2) = 736 41. The equation is x 2 − 3 = 33 ⇒ x 2 = 36 ⇒ x = 6 or x = −6. Since we are dealing only with whole numbers, Jing’s answer is x = 6. 42. Andrea is clever to realize that there can’t be ft2 added to ft; that is, area with a length measurement together in one equation can’t be correct. She needed to notice that 2(5 + 1) = 2ft × 5 ft + 2ft × 1ft so there really is a ft2 throughout the equation. 43. The operation is closed, commutative, and associative. The circle is the identity, since if it is either of the “factors” the outcome of the operation is the other factor. Copyright © 2015 Pearson Education, Inc. Solutions to Chapter 2 Review Exercises 1, 0. Thus the number we seek is 59, which we note has a remainder of 1 when divided by 2. 44. (a) The magic multiplication constant is 4096 = 2 . 12 (b) The exponents form a magic addition square with the magic addition constant 12. It is clear how the multiplication square has been formed. For example, the product of the upper row is 23 × 28 × 21 = 23 +8 +1 = 212. That is, the 12 product is always 2 . (c) 23 28 21 22 24 26 27 20 25 3 8 1 2 4 6 7 0 5 45 48. D 49. B 50. B 51. A 52. C 53. C Chapter 2 Review Exercises 1. (a) S = {4, 9, 16, 25} P = {2, 3, 5, 7, 11, 13, 17, 19, 23} T = {2, 4, 8, 16} (b) P = {4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25} S ∩ T = {4, 16} S ∪ T = {2, 4, 8, 9, 16, 25} S ∩ T = {9, 25} 2. 27 = 33 6561 = 38 3 = 31 9 = 32 81 = 34 729 = 36 2187 = 37 1 = 30 243 = 35 3. (a) (b) If A ⊆ B and A is not equal to B, then A ⊂ B. 45. (a) Since 2 × 185 = 370,500 − 370 = 130, and 130 ÷ 2 = 65, the question is “How many tickets must still be sold?’ (b) 67 ÷ 12 = 5 R 7. Thus 5 answers “How many cartons will be filled?” and 7 answers “How many eggs are in the partially filled carton?” 47. When divided by 5 the remainder is 4, so the number is in the list 4, 9, 14, 19, …, 94, 99. When divided by 4 the remainder is 3, the list of remainders are 0, 1, 2, 3, 0, 1, 2, 3, …, 0, 1, 2, 3, so the number we seek is one of 19, 39, 59, 79, or 99. When divided by 3 the remainders in this list of 5 numbers is 1, 0, 2, A ∩ (B ∪ C ) = ( A ∩ B) ∪ ( A ∩ C ) (c) (d) A ∪ ∅ = A 4. n( S ) = 3, n(T ) = 6, n( S ∪ T ) = n({s, e, t, h, o, r, y}) = 7, n( S ∩ T ) = n({e, t}) = 2 , 46. Note that 318 ÷ 14 = 22 R 10. (a) 22 is the answer to “How many full rows of chairs are set up in the auditorium?” (b) 10 is the answer to “How many chairs are in the incomplete row at the rear of the auditorium?” A ⊆ A∪ B n( S ∩ T ) = n({s}) = 1, n(T ∩ S ) = n({h, o, r, y)} = 4 5. 1 ↔ a 36 ↔ f 4 ↔ b 49 ↔ g 9 ↔ c 64 ↔ h 16 ↔ d 81 ↔ i 25 ↔ e 100 ↔ j Copyright © 2015 Pearson Education, Inc. Chapter 2 Sets and Whole Numbers 46 6. There is a one-to-one correspondence between the set of cubes and a proper subset. For example, 1 ↔ 1 8 ↔ 2 27 ↔ 3 64 ↔ 4 125 ↔ 5 (c) (d) ↔ K 3 ↔ K (e) 7. 12. (a) A B = {(p, x), (p, y), (q, x), (q, y), (r, x), (r, y), (s, x), (s, y)} 8. (a) Suppose A = {a, b, c, d, e} and B = { , } . Then n(A) = 5, n(B) = 2, ■★ A ∩ B = ∅ and n( A ∪ B) = n({a, b, c, d, e, ■, ★}) = 7 . (b) Since n(A) = 4, n(B) = 2, and n(A B) = 8, the Cartesian product models 4 2 = 8. 13. Answers will vary. Since 36 = 3 × 3 × 4 , one possibility is 6′′ × 6′′ × 8′′ . (b) 14. Since 92 ÷ 12 = 7 R 8, there are eight rows (7 full rows, and a partial row of 8 soldiers in the back). 9. (a) Commutative property of addition: 7+3=3+7 15. (a) (b) Additive-identity property of zero: 7+0=7 10. (a) (b) (b) 11. (a) (c) (b) Copyright © 2015 Pearson Education, Inc.