Chapter 2 Answers. Lesson 2.1

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1 Chapter Answers Lesson.1 1. a. For example, the mother might cut the cake or decide which child gets which piece. b. No. Often one child feels that the other got the better portion. c. For example, legal matters decided by a judge or corporate decisions decided by a chief executive officer. a. For example, a coin flip might be used to decide the distribution of the pieces if there is disagreement. b. No. Often one child feels that the random distribution leaves him or her with the lesser portion. c. For example, a coin flip to decide the opening of a football game or the use of spinners to determine the winner of a lottery. 3. a. For example, the pieces might be divided and weighed on a scale. Adjustments can then be made until the two pieces weigh the same. b. No. For example, although the two pieces might have the same weight, icing might make one piece more desirable than the other. c. For example, dividing a drink of some type between two children. 4. a. She feels that the two pieces are equal. b. He may not feel that they are equal. c. The chooser. The cutter feels that the two pieces are equal, but the chooser might feel that one is better than the other and, of course, gets to choose it. 5. For example, a good model results in each person feeling that he or she gets a fair share. That is, if there are n people, each feels that his or her share is at least 1 n th of the cake. 6. For example, the model can seem unfair if one of the heirs feels the house is priceless, perhaps for sentimental reasons, and thereby feels strongly that it should remain in the family. 7. For example, if one heir places a higher value on the house than the other, that heir could receive the house and be required to compensate the other heir in some way. 8. a. Students often award the house to Mary because she places a higher value on it. b. Sample answers: Mary can be required to pay Juan $60,000, half the value she places on the house. Mary can be required to pay Juan $50,000, which is half the value Juan places on the house. A compromise might be to pay Juan $55,000, which is the average of the previous two amounts. Chapter Answers 1

2 9. For example, one can require that each of n heirs receive a settlement that the heir feels is at least 1 th of the house s value. n 10. For example, if you place a higher value on the house than you feel it really has, you might receive a larger settlement than if you assigned the value truthfully. 11. If there were differences, students should point this out and discuss the aspects of the methods that caused the differences to occur. 1. Based on the percentage of the student body, the sophomores deserve 7.4 seats, the juniors 6.4, and the seniors 6.. Students might assign 7, 7, and 6 to the sophomores, juniors, and seniors respectively, arguing that the larger decimal parts mean the seat should go to either the sophomores or juniors, and choosing the juniors because of seniority. 13. For example, each class should receive seats so that the percentage of seats in the council is as close as possible to the percentage of the class s population in the school. 14. Cake division and division of a house are similar because the people involved can value the object or pieces of it differently. Division of a house and division of seats in a council are similar because the objects of division cannot be split into any number of parts, as can a cake. 15. a. Sample answer: The three problems of this lesson involve sharing a benefit. This problem is about sharing a cost. That is, the participants give something instead of receiving something. b. Sample answer: The cost of sharing something seems most like the problem of dividing a piece of cake because cake and costs can be divided into any number of parts of any size. A difference is the receiving/giving characteristic noted in the answer to part a; another difference is the possibility that the shares are supposed to be equal in the cake division problem, but they might not be equal (by agreement) when costs are shared. Chapter Answers

3 Lesson. 1. For example, that the bids are made independently (there is no collusion among the heirs); that the heirs bid sincerely; that heirs are capable of paying into the estate, if necessary; that the heirs are willing to accept cash instead of items in the estate.. a. Calvin: $5; Hobbes: $.50 b. Calvin receives the wagon, which he thinks is worth $0, an initial cash payment of $5, and $3.75 of the remaining cash. Therefore, Calvin thinks his settlement is worth $8.75. Hobbes gets the sled ($15), initial cash of $7.50, and remaining cash of $3.75 for a total of $ a. $40,000. $35, a. b. $80,000 $40,000 = $40,000 c. $40,000 d. $35,000. Because there is $5,000 left in the estate, each receives $,500. Garfield s share is $80,000 $40,000 + $,500 = $4,500. Marmaduke s is $35,000 + $,500 = $37,500. e. For example, students might argue that the loan is strictly Garfield s affair, and that it should not be considered. f. Marmaduke receives $40,000 instead of $37,500. Garfield receives $40,000 instead of $4,500. Which is fairest is a matter of opinion, but the algorithm of this lesson results in each receiving more than the perceived fair share, rather than just one of them receiving more than the perceived fair share. Amy Bart Carol Fair share 3, , , Items received motorcycle, ticket painting Cash , Share of remaining cash Final settlement 4, , ,7.3 b. For example, an auction could be held to sell one or more of the items awarded to the heir who cannot pay into the estate. The resulting cash could then be included in the estate's cash and the division recalculated. Chapter Answers 3

4 5. Step of the algorithm should be changed to reflect a 0% 40% 40% split. The same is true of the remaining cash in step 5. Amy Bart Carol Fair share,300 4,840 4,50 Items received motorcycle, ticket painting Cash, ,50 Share of remaining cash ,136 +1,136 Final settlement,868 5,976 5, One possibility is a coin flip. Another is negotiating. 7. a.! # # # " 13,000 11, ,000 17, ,000 13,000 0 b. The value to Alan of the items received by Betty. c. The value to Betty of the items received by Betty. $ & & & % 8. For example, one of the heirs might make a high bid on an object he or she does not want in order to increase the amount of cash he or she receives. However, doing so risks being given the unwanted item at too high a price. 9. The division of an island is a continuous problem, but the case of Ellis Island is not so simple. If the division involves buildings and specific tracts of land on which they sit, then the division is discrete. 10. Sample answer: Each friend should divide the rent into two parts so that the larger amount is what that person is willing to pay for the nicer bedroom and the smaller amount is what that person is willing to pay for the lesser bedroom. The nicer bedroom goes to the person who assigns the higher value to it. The difference between the two high bids is split between them. For example, one person bids 750/550, the other 700/600. The difference between 750 and 700 is 50, and half of 50 is 5. The nicer bedroom goes to the 750 bidder, who pays a rent of = 75; the lesser bedroom goes for a rent of = 575. Chapter Answers 4

5 Lesson.3 1. a. For example, students might argue in favor of giving the extra seat to the seniors if they were given only 4 seats in the 0-seat council by the Jefferson model, or to the juniors if they were given only 5 seats in the 0-seat council by the Hamilton model. Reasoning could be based on the decimal parts of the quota and seniority. b c. Sophomore quota: 10.83; junior quota: 5.6; senior quota: 4.57 d. Sophomore seats: 11; junior seats: 6; senior seats: 4 e. Although the size of the council increases and the class sizes don't change, the seniors lose a seat.. a. Sophomore adjusted ratio: = 4.18; junior adjusted ratio: 40 6 = 40; senior adjusted ratio: = 39. b. Decrease the ratio until it drops below 40, but not below 39.. Sophomore seats: 11; junior seats: 6; senior seats: 4. c. No, because no class loses seats from the 0-member council Jefferson apportionment of 11, 5, For example, after the first seat is awarded, repeat steps 5 and 6 to award the next seat. Repeat until all seats are assigned. 4. a When the ideal ratio is, the decimal part of the 100-member class is larger than the decimal part of the 30-member class. The situation is reversed when the ideal ratio drops to 1. b. For a small class: Decreases in a divisor result in larger changes in the decimal part of the quotient when dividing a small number than when dividing a large one. c. At the time, Virginia was the largest state in the Union, and the Jefferson model favors large states. 5. a. Sophomore quota: 10.71; junior quota: 5.69; senior quota: Sophomore seats: 11; junior seats: 6; senior seats: 4. b. Sophomore quota: 10.39; junior quota: 6.19; senior quota: Sophomore seats: 10; junior seats: 6; senior seats: 5. c. The sophomore class increases in size and the senior class decreases, but the senior class takes a seat from the sophomores. d. Slight changes in a small class result in greater changes in the decimal part of the quotient than similar changes in a large class. 6. It is an indiscreet attempt because it appears blatant. Chopping a seat into parts is treating a discrete object as if it were continuous, like a cake. 7. The ideal ratio is 308,745, ,760. Chapter Answers 5

6 a. Texas: ; New York: 7.363; California: 5.61 b. Texas and California have apportionments above their quotas, but New York's is below. c. Some students will say that New York was treated unfairly. d. Sample answer: Louisiana s quota is 6.416, but it gets only 6 seats. Minnesota s quota is 7.488, but it gets 8 seats. The apportionment does not seem to have treated Liousiana less fairly than Minnesota, because Minnesota s quota is closer to 8 than Louisiana s is to a b. To three decimal places, the quotas are.003, 10.31,.967, and.709, respectively. c., 10, 3, and 3 seats, respectively. d. No. The initial Jefferson apportionment is, 10,, and seats, respectively. e., 11, 3, and seats, respectively. An adjusted ratio of, say 95, apportions all 18 seats. f. In this case, the Jefferson model gives a seat to the largest district. Small districts tend to dislike the Jefferson model because it favors large districts. Chapter Answers 6

7 Lesson.4 1. a b. Sophomores: 11; juniors: 6; seniors: 4 c d (Sr.) (Jr.) (So.) Sophomore seats: 11; junior seats: 6; senior seats: 4 e (Jr.) (Sr.) 44.4 (So.) Sophomore seats: 11; junior seats: 5; senior seats: 5 f. The model favored by a given class can be seen in the following table of final apportionment results: Hamilton Jefferson Webster Hill Sophomore Junior Senior Juniors favor any model except Hill; seniors favor Hill only.. a b. Hamilton c. Probably not, since this measure appears to favor the Hamilton model, which can produce nasty paradoxes. 3. a. The representations are 48 people per seat for the juniors and 39. people per seat for the seniors. The difference is 18.3% of the junior representation. b. The representations are 49 people per seat for the seniors and 40 people per seat for the juniors. The difference is 18.4% of the senior representation. c. Taking a seat from the juniors is slightly less unfair than taking a seat from the seniors. Only the Hill model produces the same result in Exercise a = 50 b Chapter Answers 7

8 c d. Increase the ratio until it passes , but stays below 5. Freshman: 1; sophomore: 4; junior: 3; senior:. e. The freshman quota is.1, so the number of seats should be either or a. The ideal ratio is 50, which produces quotas of 11., 4,, and.78 for parties A, B, C, and D, respectively. The adjusted ratios are 46.75, 40, 33.33, and 46.3, respectively, so party A gets the extra seat. b. The adjusted ratio for combined parties C and D is 47.8, so the extra seat now goes to this new party. c. No. The other models award the extra seat to A, so the merger of parties C and D does not result in the movement of a seat. 6. The Hamilton and Jefferson models apportion 1, 3, 3, and 7 seats, respectively. The Webster and Hill models apportion, 3, 3, and 6 seats, respectively. With current populations, districts B and C are unlikely to favor a particular model. District D is likely to favor Hamilton or Jefferson because they award the district 7 seats instead of 6. District A is likely to favor the Webster or Hill models because they award the district seats instead of Cumberland s quota is = (Note that if the board is increased to 9 seats, Cumberland's quota is = 6.0.) This situation is different from weighted voting because a community has several seats, each having one vote, rather than a single seat with several votes. 8. The Hill round-off point approaches the Webster as the quota increases, but the Webster is always larger. Chapter Answers 8

9 Lesson.5 1. a. Ava feels that she has exactly one-third because she cut the cake into pieces that she feels are equal. b. Bert and Carlos each could feel that he receives more than one-third because each has the opportunity to choose.. a. Three portions; there would probably be six pieces b. This does not violate the definition or any of the assumptions. 3. In Step 1, the first assumption is applied. In Step, the second assumption is applied and perhaps the third. In Step 3, the first assumption is applied. In Step 4, the second assumption is applied and perhaps the third. 4. One-sixth or 0.16 because Carlos feels that both of Ava s cuts are even. 5. a. Because Carlos feels that Ava s initial cut is even. Therefore, he values each of Ava s and Bert s pieces as 0.5. b. One-sixth. Carlos chooses the largest of the three pieces, which must be at least a third of half the cake. c. Four-sixths or 0.67 d. One-third or Balavan must be the cutter because the cutter divides the cake into pieces that he or she feels are equal. 7. Although cookies seem discrete, they can be broken into parts, so the problem is continuous. However, values placed on cookies or pieces of cookies can differ if, for example, some have more chocolate chips than others. 8. a. 3 = 6 b. k(k + 1) or k + k c. (k + 1)(k + 5) or k + 6k + 5 d. 3 x 4 = 1 9. a. Yes. No. b. Yes. Yes. c. Probably not. Yes. 10. a. Yes. Yes. b. Yes. Yes. c. Yes. Yes. Chapter Answers 9

10 11. a. For example, Carlos might value the pieces in the initial division as 0.9 and 0.1, and feel that the subsequent divisions are equal. Carlos now feels his share is = 0.33, but Carlos also feels that the person who receives the portion that Carlos feels is 0.9 now has a portion that Carlos values as 0.6. b. Yes. The example in the answer to part a shows that Carlos could be jealous of another s portion. 1. a. The first person to press a key (or say cut ) is agreeing that the piece cut first is at least one-third of the cake. The others feel it is no more than that, or they would have pressed a key sooner. The same is true of each of the remaining key presses. b. Yes. For example, after receiving a piece, a person might feel that the next piece is much bigger than his or hers. 13. The moving knife method does. 14. All can be extended. For the cut-and-choose model, apply the three-person model then have each divide his or her portion in four portions. Have the fourth person select one of each. Both inspection and moving knife can be extended without adjustments. Chapter Answers 10

11 Lesson.6 1. a. k + 1 k 1 b. k + k k + 1 k 1. a. New handshakes: 3. Total handshakes: 6. b. New handshakes: 4. Total handshakes: a. 7, because the new person must shake hands with each of the seven. 4. a. b. k, because the new person must shake hands with each of the k. c. H n = H n 1 + (n 1); H n+1 = H n + n b. c. d. k e. f. 10(10! 1) k(k! 1) k(k! 1) (k + 1)k k(k! 1) = 45 k(k! 1) + k = k! k (k + 1)k + k = k + k (k + 1)k = 5. a. Sample answer: With 10 people there are 45 potential conflicts, and with 0 people there are 190. The number of potential conflicts more than doubles. b. The demand for police and other legal services might grow more rapidly than the population because the number of potential conflicts outpaces the population. 6. a. V k+1 = V k 7. b. V n = n c. V k+1 = k+1 d. V k+1 = ( k ) = 1 ( k ) = k+1 k + 3k + or (k + 1)(k + ) (k + 1)(k + ) Key points of the proof: Assume there are dominoes in a double-k (k + )(k + 3) set and prove there are dominoes in a double-(k + 1) set. Note that in forming a double-(k + 1) set from a double-k set, k + new dominoes are required because a collection of k + 1 spots must be paired with a blank, with each of 1 through k spots, and with itself. Chapter Answers 11

12 8.! ( 9 + 1) ( 9 + ) $ The image of Gauss contains 48 " # % & = 640 dominoes. k + k or k(k + 1) k(k + 1) Key points of the proof: Assume there are pins when there are k rows (k + 1)(k + ) and prove there are pins when there are k + 1 rows. Note that in adding a row to a configuration of k rows, k + 1 new pins are required. 9. k 1 Key points of the proof: Assume there are k 1 grains on the chessboard when the kth square is filled, and prove there are k+1 1 grains on the board when the k + 1st square is filled. Note that there are k 1 grains on the kth square, so the number of grains that are placed on the (k + 1)st square is k. Thus the total number of grains on k + 1 squares is k 1 + k = k 1 = k k + k or k( k + 1) Key points of the proof: Assume it takes k( k + 1) toothpicks to make a k k square, and prove that it takes (k + 1)(k + ) toothpicks to make a (k + 1) (k + 1) square. Note that in moving from a k k square to a (k + 1) (k + 1) square, an additional 4(k + 1) toothpicks are required. 11. k! k or k( k! 1) ( ) Key points of the proof: Assume there are k k! 1 when there are k choices on the ballot, and prove there are ways of selecting items ( k + 1)k ways of selecting items when there are k + 1 choices on the ballot. Note that when an additional choice is added to a ballot with k choices, the number of new pairs that can be selected is k because the new choice can be paired with each of the original k choices. 1. k + k or k( k + 1) Key points of the proof: Assume there are k( k + 1) gifts given on the kth day, and ( k + 1) ( k + ) prove there are gifts given on the k + 1st day. Note that in moving from the kth day to the k + 1st day, the number of new gifts is k + 1. Chapter Answers 1

13 13. k + k or k( k + 1) ( ) Key points of the proof: Assume there are k k + 1 building blocks in a k k set of steps. Note that a set of (k + 1) (k + 1) steps can be created from a k k set by adding a column of k + 1 steps (or a row of k + 1 steps). 14. a. This is a 1 1 square that is covered by single tromino except for one corner. Thus, the theorem is true for the first value of k. b. The following figure shows that the theorem is true for a (4 4) square. c. A 3 3 (8 8) square is composed of four 4 4 squares. If you tile each of these 4 4 squares in the same way shown in the answer to part b, you get something like this: If you rotate the upper left 4 4 square 90 clockwise, the upper right 4 4 square 180, and lower right 4 4 square 90 counterclockwise, the four 1 1 squares that are not tiled come together in the center of the figure: A tromino can now be used to cover three of the four 1 1 square. The 4 4 square with the uncovered 1 1 tile can be rotated so that the uncovered square is back in a corner of the 8 8 square. Chapter Answers 13

14 d. Divide the k+1 k+1 square into four k k squares. Each of these four can be tiled with the exception of one corner. Rotate three of these four so that the 1 1 squares that are not tiled come together in the center. The square that is not tiled can be covered with a tromino except for a single 1 1 square that can be rotated back to a corner of the k+1 k+1 square. 15. a. To show that the formula works for k = 1, note that the first odd integer is 1 and that (1) 1 = 1. Assume that the kth odd integer is k 1. It must be proved that the (k + 1)st odd integer is (k + 1) 1 = k + 1. Since odd integers are apart, the (k + 1)st odd integer can be obtained from the kth by adding : k 1 + = k + 1. b. A formula for the sum of the first k odd integers is k. Note that the formula is true for k = 1 because the sum of all odd integers through the first is 1, and 1 = 1. Assume that the sum of the first k odd integers is k, and prove that the sum of the first k + 1 odd integers is (k + 1). Since the sum of the first k odd integers is k and the (k + 1)st odd integer is k + 1, the sum of the first k + 1 odd integers is k + k + 1 = (k + 1). 16. a. A formula for the kth even integer is k. To show that the formula works for k = 1, note that the first even integer is and that (1) =. Assume that the kth even integer is k. It must be proved that the (k + 1)st even integer is (k + 1) = k +. Since even integers are apart, the (k + 1)st even integer can be obtained from the kth by adding : k +. b. A formula for the sum of the first k even integers is k + k. Note that the formula is true for k = 1 because the sum of all even integers through the first is, and =. Assume that the sum of the first k even integers is k + k and prove that the sum of the first k + 1 even integers is (k + 1) + (k + 1). Since the sum of the first k even integers is k + k and the (k + 1)st even integer is k +, the sum of the first k + 1 even integers is k + k + k + = (k + k + 1) + (k + 1) = (k + 1) + (k + 1). Chapter Answers 14

15 Chapter Review 1. A reasonable summary should include the following points: Fair division problems can be either continuous or discrete. Cake division is an example of the continuous case; legislative apportionment is an example of the discrete case. The division of an estate among heirs usually demonstrates aspects of both types since objects such as a house are discrete, but cash can be considered continuous. A discussion of cake division models, including cut-and-choose, moving knife, and inspection A discussion of apportionment models, including Hamilton, Jefferson, Webster, and Hill; the discussion should include paradoxes that can arise and mention the work of Balinski and Young. A discussion of estate division, including the algorithm that results in each heir getting more than a fair share by his or her own evaluation of the estate. Answers are rounded to the nearest dollar. Joan Henry Sam Fair share $9,13 $8,933 $8,767 Items received lot computer, stereo boat Cash $1,13 $6,333 $067 Final settlement $9,675 $9,395 $9,9 3. Answers are rounded to the nearest dollar. Anne Beth Jay Fair share $3,00 $,967 $3,017 Items received car, computer stereo Cash $5,600 $,967 $,017 Final Settlement $3,405 $3,17 $3, 4. Answers are rounded to the nearest dollar. Lynn Pauline Tim Fair share $4,5 $,150 $,163 Items received car, kayak guitar, watch Cash $,475 $750 $,163 Final settlement $4,756 $,416 $,48 Chapter Answers 15

16 5. a. 10 h , , b. 64.7, 4.7, 10.6 i. 64, 5, 11 c. 65, 5, 10 j. 65, 5, 11 d. 64, 4, 10 k , , e , , l. 64, 5, 11 f. 65, 5, 10 m. 64, 5, 11 g. 65, 5, 11 n. State A gains population and state C loses population, but A loses a seat to C. 6. The Hamilton, Webster, and Hill models apportion, 13, 3, and seats to A, B, C, and D, respectively. However, the Jefferson model gives 14 seats to B and only 1 to A, while leaving C and D the same. Therefore, B would strongly favor the Jefferson model, while A would strongly oppose it. 7. Balinski and Young proved that any apportionment model sometimes produces one of three undesirable results: violation of quota, the loss of a seat when the size of the legislative body increases even if population doesn t decrease, and the loss of a seat by one state whose population increases to another whose population decreases. 8. Arnold and Betty 9. Have each of the original four divide his or her piece into five pieces that he or she considers equal. Have the new person select a piece from each of the others. 10. A table of data: Teams Games A formula: k k or k(k 1). Note that the formula is correct for the base case k = 1 since there clearly are no games if there is just a single team, and the formula gives 0 when k = 1. Assume there are k(k 1) games when there are k teams, and prove there are (k + 1)k games when there are k + 1 teams. When a new team is added to a league of k teams, the number of new games is k since the new team must play each of the existing k teams twice. Thus, the total number of games when there are k + 1 teams is k(k 1) + k = k k + k = k + k = (k + 1)k. Chapter Answers 16

17 11. A table of data: Circles Rings A formula: k! k or k( k 1). Note that the formula is correct for the base case k = 1 since there clearly are no rings if there is just a single circle, and the formula gives 0 when k = 1. Assume there are k( k 1) ( k + 1)k rings when there are k + 1 circles. rings when there are k circles, and prove there are When a new circle is added to a set of k circles, the number of new rings is k since there is a new ring between the new circle and each of the existing circles. Thus, the total number of rings when there are k + 1 circles is k( k 1) + k = k( k 1) + k = k! k + k = k + k ( = k + 1)k. 1. Answers to this exercise should demonstrate an understanding that this is an apportionment problem. Thus, students might recommend the Hamilton, Jefferson, Webster, or Hill model to resolve the issue. However, any recommendation should include mention of problems that might occur, such as paradoxes with the Hamilton model or favoring large departments with the Jefferson model. 13. Answers to this exercise should include a discussion of the necessity of changing the definition of fairness: wording, for example, might change to no more than instead of at least. The answer might also note that some chores could be considered divisible if they are split among two or more children. If an allowance is involved, reduction of the allowance might be considered similar to paying into an estate. If allowances are distributed according to chores done, parents might award chores to the lowest bidder. Chapter Answers 17