Solution Manual for Mathematics for Information Technology 1st Edition by Basta Full file at https://TestbankDirect.eu/ Chapter Sets Exercises 1.1: Set Concepts Note: Answers to Exercises throught will vary but should approximate the answers given below A set is a well-defined, unordered collection of objects having no duplicate members Two sets A and B are said to be equal, written A = B, if their membership is entirely identical except for the fact that the order of the elements in the sets may be different Two sets A and B are said to be equivalent, written A ∼ B, if they share the same cardinality The cardinality of a set S refers to the quantity of members belonging to the set, and is represented by |S| or n(S) The empty set is a set that has no members Its cardinality is zero The symbol for the empty set may be either ∅ or { } A finite set is a set that has a particular quantity of elements Ellipsis are a succession of three “dots,” which indicate that a demonstrated pattern of numbers continues, either forever or until a number following the ellipsis is reached Well defined; a person is either a paid employee of the US Government or is not Not well defined; unclear as to meaning of efficient 12 Well defined; the astronauts who have piloted the Space Shuttle Atlantis are listed on the NASA Web page 13 Well defined; there are no even integers between and 14 Finite; there are 11 even integers between 10 and 30, inclusive They are: 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30 15 Finite; at the present time there are 50 states in the United States 16 Finite; the number one 1,000,000,000,000, has 13 digits trillion, 17 Infinite; the decimal expansion of pi never ends 18 Infinite; in Example 1.3.iv, we showed that the numbers between and were infinite This same reasoning would apply to the set of numbers between and 10 19 {M, i, s, p}, 20 {1, 2, 3, 4, , 47, 48, 49}, 49 21 {Nebraska, Nevada, New Jersey, New Mexico, New York, North Carolina, North Dakota}, 22 {−5}, 23 {}, 10 Well defined; there are 50 odd integers less than 100, specifically {1, 3, 5, 7, , 99} 24 {Alaska, Washington, Idaho, Montana, North Dakota, Minnesota, Michigan, New York, Vermont, New Hampshire, Maine, Texas, New Mexico, Arizona, California}, 15 11 Not well defined; unclear as to meaning of well-spoken 25 At the time this book was written: {Carter, GHW Bush, Clinton, GW Bush, Obama}, Full file at https://TestbankDirect.eu/ Solution Manual for Mathematics for Information Technology 1st Edition by Basta Full file at https://TestbankDirect.eu/ 10 Section 1.2 26 {x ∈ N|1 ≤ x ≤ 3} or {x ∈ W|1 ≤ x ≤ 3} or {x|x is 1, 2, or 3} 27 {x|x is a non-negative even integer} or {x ∈ W|x ÷ ∈ W} 28 {x|x is a prime number less than 14} 29 No month has exactly 20 days, so D = ∅ 30 E = {x|x is an odd number less than 1000} or E = {x|x = 1, 3, 5, 7, , 975, 977, 999} 31 Answers will vary One possible answer is: “The set of whole numbers less than 16 that are multiples of 3.” 32 Answers will vary One possible answer is: “The set of the four main characters of The Flintstones.” 33 Answers will vary One possible answer is: “The whole numbers between and 6, including 6.” Another is “The numbers 3, 4, 5, and 6.” 34 One possible answer is “The set of the last two states admitted to the United States.” 35 One possible answer is “The set of odd numbers between and 19, inclusive.” Subsets 36 37 38 39 Equivalent but not equal, because they both have a cardinality of 4, but not have the same elements 40 Both, because they both have a cardinality of and they have the same elements 41 Both, because |B| = and |C| = and they both have the same elements 42 Equivalent but not equal, because they both have a cardinality of 50, but not have the same elements 43 Neither, both have an infinite amount of numbers, but you cannot set up a one-to-one correspondence between the elements of A, the natural numbers larger than 2, and the elements in B, all (real) numbers greater than Exercises 1.2: Subsets Note: Answers to Exercises throught will vary but should approximate the answers given below Set A is a subset of set B if every element of A is also an element of B A is a proper subset of B if A is a subset of B and B contains elements that are not in A Every set has at least one subset, the empty set, so it is not possible for a set to have no subsets A ⊂ B because all the elements of A are the vowels of the English alphabet If A is a subset of B, then B is a superset of A A ⊂ B because the elements of A are {2, 4, 6, 8, 10, 12, 14, 16, 18} and every one of these is in B A power set of a particular set has as its members all possible subsets of that particular set A = B because both of these set-builder notations describe the same set of numbers ©2013 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password- protected website for classroom use Full file at https://TestbankDirect.eu/ Solution Manual for Mathematics for Information Technology 1st Edition by Basta Full file at https://TestbankDirect.eu/ Chapter A ⊃ B or B ⊂ A because, while each element of B is a retired professional basketball player there are other retired basketball players in A For example, Shaquille O’Neal is in A but not in B 10 None of these relationships exist since none of the players in set A are in set B and none of the players in set B are in set A 11 A ⊂ B, since A is the empty set and the empty set is a subset of every set 12 A ⊃ B where A = the set of letters in the word Mississippi, so A = {M, i, s, p} and B = set of letters in the word or B = {s, i, p} Thus, every element of B is in A, but the letter M is in A but not in B 13 25 = 32 14 20 = 15 27 = 128 16 23 = since A = {3, 4, 5} 17 210 = 1024 Note: Answers to Exercises 18 throught 24 will vary but should approximate the answers given below 18 False The statement A ⊂ A means that the right-hand set has at least one element that is not in the left-had set But, these two sets are identical, so one of them cannot have more elements than the other 19 False The number of subsets of a set A is 2n where n is the number of elements in A But, there is no whole number n where 2n = 20 20 False Set A has 240 ≈ 1.1 × 1012 subsets There are 60 × 60 × 24 × = 604,800 seconds in a week, which is much less than 240 Sets 11 21 True If A and B have the same cardinality, then they both have the same number of elements, and hence the same number of subsets 22 False The statement ∅ ⊂ ∅ means that the right-hand set has at least one element that is not in the left-had set But, the right-hand set has no elements, so it cannot have more elements than the other 23 True Since every elements of A is in B and every element of B, including all of those in A, are in C, then all the elements of A are in C 24 218 − This is the same as asking how many proper subsets can be made using the 18 ingredients This is 218 If we assume that your Panini sandwich has at least one ingredient, then the empty set is not included, so the number of possible Panini sandwichs is 218 −1 25 If we label the voters A, B, C, D and record their votes in alphabetical order, then there are 16 possible combinations: NNNN, NNNY, NNYN, NYNN, YNNN, NNYY, NYNY, YNNY, YNYN, NYYN, YYNN, YYYN, YYNY, YNYY, NYYY, and YYYY Of these, only the last allow the measure to pass 26 There are 11 (4 + + + 2) possible votes and it will take votes for an application to pass We represent each of the voters as P for President, V for Vice-President, A for Architectural Committee Leader, and F for Fence Height Overseer The possible winning coalitions (and its votes) are: P+V (4 + 3), P+A (4 + 2), P+F (4 + 2), P+A+F (4 + + 2), P+V+A (4 + + 2), P+V+F (4 + + 2), P+V+A+F (4 + + + 2), and V+A+F (3 + + 2) Thus, there are possible winning coalitions ©2013 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password- protected website for classroom use Full file at https://TestbankDirect.eu/ Solution Manual for Mathematics for Information Technology 1st Edition by Basta Full file at https://TestbankDirect.eu/ 12 Section 1.3 Venn Diagrams Exercises 1.3: Venn Diagrams U U A B A B B A U AC ∩ A = ∅ and ∅C = U , so everything is shaded U B A U A B True, as shown by the following two figures U B A (A ∩ B)C ∪ B = U , so everything is sheded U A B (A B)C U B A U A B AC AC ∩ A = ∅, so nothing is sheded BC False, as shown by the following two figures ©2013 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password- protected website for classroom use Full file at https://TestbankDirect.eu/ Solution Manual for Mathematics for Information Technology 1st Edition by Basta Full file at https://TestbankDirect.eu/ Chapter Sets 13 U U (A B)C U B A B A B A 12 True, the union of the shaded portion of the following two figures is B U B A AC BC 10 False, as shown by the following two figures U A U AC (A B U B A B A B A B BC C ) 13 True, the union of the shaded portions of the following two figures is A U B A AC B 11 False, as shown by the following two figures In the first figure, AC ∪ U = U and U c = ∅, so nothing is shaded U A B U B A B A A (AC U )C BC 14 True, the shaded portion of the first figure is ©2013 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password- protected website for classroom use Full file at https://TestbankDirect.eu/ Solution Manual for Mathematics for Information Technology 1st Edition by Basta Full file at https://TestbankDirect.eu/ 14 Section 1.4 Applications of Sets A ∪ B and when that is intersected with A, the result is A, which is the same as the second figure U B A U B A AC (A B) A U B A A 15 False, as shown by the following two figures U B A (A C B) BC 16 Think of the calculation for n(A ∪ B ∪ C) as finding the number of elements in two sets where one of the sets is A ∪ B and the other set is C Then n[(A ∪ B) ∪ C] = n(A ∪ B) + n(C) − n[(A ∪ B) ∩ C] Since n(A ∪ B) = n(A) + n(B) − n(A ∩ B) the formula becomes n[(A∪B)∪C] = n(A)+n(B)− n(A ∩ B) + n(C) − n[(A ∪ B) ∩ C] Rewrite n[(A∪B)∩C] as n[(A∩C)∪(B ∩C)] Using the formula for the union of two sets we have n[(A ∩ C) ∪ (B ∩ C)] = n(A ∩ C) + n(B ∩ C) − n[(A ∩ C) ∩ (B ∩ C)] We can simplify the last term as n(A ∩ B ∩ C) Thus, n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A∩B)−n(A∩C)−n(B ∩C)+n(A∩B ∩C) 17 Solving the formula in Exercise 16 for n(A ∩ B ∩ C), we get n(A ∩ B ∩ C) = n(A ∪ B ∪ C) − n(A) − n(B) − n(C) + n(A ∩ B) + n(A ∩ C) + n(B ∩ C) Exercises 1.4: Applications of Sets In the Venn diagram below A represents those who live in apartments and C those who live on campus From the diagram we see that there are 49 students who live in apartments off campus and 35 who lived on campus but NOT in an apartment We can compute that 47 students lived neither on campus nor in an apartment Off campus A 49 C 44 35 In the Venn diagram below T represents those who own trucks and C those who own cars ©2013 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password- protected website for classroom use Full file at https://TestbankDirect.eu/ Solution Manual for Mathematics for Information Technology 1st Edition by Basta Full file at https://TestbankDirect.eu/ Chapter Sets Since 21 not own a vehicle, then 125−21 = 104 people own a vehicle, n(T )+n(C)−n(T ∩ C) = 104 From the given information, n(T ∩ C) = n(T ) + n(C) − 104 = 59 + 88 − 104 = 43 people own both a car and a truck The number who owned only a truck were 59 − 43 = 16 and 88 − 43 = 45 only owned a car No vehicle T C 15 Using the given information we can develop the diagram below Male Female CO res 12 21 Non-CO res 11 21 25 13 10 3 or more passengers The Venn diagram below shows the number of cats that had the color schemes described From this we see that 24 cats had some black Also, 44 cats had some of the three colors, cat had all three colors, had none of the colors, and 11 had exactly two of the colors Orange Black 12 11 Grey Another error is that the State Patrol officer indicated that 45 cars were driven by men, yet the diagram has a total of 50 mle drivers From the Venn diagram below we see that majored in engineering only, in biology only, and 15 in chemistry only Bio Eng 11 This diagram differs from ones you have seen earlier since it does not use circles The universe of 63 drivers is divided in two parts vertically, male and female, and two parts horizontally, teenage drivers and past teenage drivers From the drawing you can see that there were 13 male drivers who were past their teenage years Past teenage 31 13 13 Teens Female Of the 95 cars sampled, 63 were driven by Colorado residents and so 95 − 63 = 32 had to be driven by non-Colorado residents If you add the four figures in the “non-CO res” boxes the total is 37 Since there were only 32 nonColorado drivers, the 37 must be an error Male 15 Chem There are 447 farmers who grow at least one of the crops; 20 grew all three; grew none of the crops; and 72 grow exactly two of the crops In the Venn diagram below we have put the numbers for the ones that are growing just one crop: 135 only beets, 120 only radishes, and 100 only turnips The other four parts of the circles are labeled to indicate what those farmers are growing Thus, BR are the farmers that are growing only beets and radishes ©2013 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password- protected website for classroom use Full file at https://TestbankDirect.eu/ Solution Manual for Mathematics for Information Technology 1st Edition by Basta Full file at https://TestbankDirect.eu/ 16 Section 1.5 B 135 Infinite Sets R BR BRT BT 120 RT 100 T The number who are growing beets is 210 and this includes the 135 who are growing only beets plus those who are growing both beets and radishes [n(BR) + n(BRT ) = 50] and those who are growing both beets and turnips [n(BT ) + n(BRT ) = 45] Thus, 210 − 135 = 75 = n(BR) + n(BRT ) + n(BT ) But, n(BR) + n(BRT ) + n(BT ) + n(BRT ) = 50 + 45 = 95 From these two formulas, we see the n(BRT ) = 20 and we conclude the n(BR) = 30 and n(BT ) = 25 Since those who grow radishes and turnips, 37, includes those who grow all, three, those who grow just radishes and turnips is 37−20 = 17 B 135 Adding all of the amounts in the Venn diagram we get a total of 447 who grow at least one of the crops The children who played only hockey and baseball are shown by the shaded section of the Venn diagram below A total of 35 children played hockey From the Venn diagram below we see the number who played only hockey, 17, those who played only soccer and hockey, 6, and those who played all three sports, 10, totals 33 That leaves children who played hockey and baseball only A total of 30 children played soccer From the Venn diagram below we see the number who played soccer only, 3, those who played all three sports, 10, and those who played only soccer and hockey, 6, total 19 Subtracting, 30 − 19 = 11, we see that 11 children play only baseball and soccer R 30 20 25 100 T 48 10 120 17 B H 17 53 S Exercises 1.5: Infinite Sets Note: Answers to Exercises throught should approximate the answers given below An infinite sets has unlimited membership A one-to-one correspondence is a relationship between two sets that associates to each member in one set a unique element in the other set, and vice-versa A countable set is a set that is finite, or that can be placed in one-to-one correspondence with the set of natural numbers The natural numbers are a proper subset of the given numbers, so one possible one-toone correspondence is (1, 1), (−1, 2), (2, 3), (−2, 4), (3, 5), (−3, 6), One proper subset of the given set is A = {200, 300, 400, 500, } and a one-toone correspondence between the given set and this proper subset is (100, 200), (200, 300), (300, 400), (400, 500), (500, 600), , (n, 100+ n) ©2013 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password- protected website for classroom use Full file at https://TestbankDirect.eu/ Solution Manual for Mathematics for Information Technology 1st Edition by Basta Full file at https://TestbankDirect.eu/ Chapter Another possible subset in B = {1000, 2000, 3000, 4000, } and a one-to-one correspondence between B and the given set is (100, 1000), (200, 2000), (300, 3000), (400, 4000), (500, 5000), (n, 10n) One proper subset of the given set is C = {4, 6, 8, 10, 12, } A one-toone correspondence between the given set and C and the given set is (2, 4), (4, 6), (6, 8), (8, 10), (10, 12), (n, n + 2) One proper subset of the given set is C = { 1/2, 1/3, 1/4, 1/5, 1/6, } A one-to-one correspondence between the given set and D is {(1, 1/2), ( 1/2, 1/3), ( 1/3, 1/4), ( 1/4, 1/5), ( 1/5, 1/6), ( 1/n, 1/(n+1)) } Use the one-to-one correspondence {(5, 1), (10, 2), (15, 3), (20, 4), , (5n, n) } One possible answer is {(−1, 1), (−2, 2), (−3, 3), , (n, |n|) } 10 One answer is {(10, 1), (14, 2), (18, 3), , (n, (n − 6)/4) } Sets 17 11 One answer is {(3, 1), (6, 2), (9, 3), (12, 4), , (n, n/3), } Another possible answer is (n, |n/3|) 12 Finite There are 60 × 60 × 24 = 86,400 seconds in a day and approximately 86,400 × 365.25 = 31,557,600 seconds in a year If this in multiplied by the number of years since AD, we get a large but finite number 13 Finite The set of digits in the full decimal representation of e is finite and has cardinality 10 14 Infinite There are an infinite number of points on a number line and that is just one part of the Cartesian plane 15 Finite There are 60 × 60 × 24 = 86,400 seconds in a day 16 Finite There is only one even prime number, 2, and it is the only prime number that can be divided by Chapter Summary Exercises The set of congresspeople in the United States is well-defined The set of cats in this room is not well-defined since we not know the room being referred to The set of all cats in the world is well-defined The set of odd integers between 20 and 40 inclusive is {21, 223, 25, 27, 29, 31, 33, 35, 37, 39} This set is finite and has cardinality 10 The set of digits in the full decimal expansion of all real numbers is finite and has cardinality 10 The set of letters in the word Oklahoma is {O, k, l, a, h, o, m} and its cardinality is Notice that O and o are considered to be different letters The set of all whole numbers less than 30 is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29} and its cardinality is 30 A = {2, 3, 4, 5} in set-builder notation is A = {x ∈ N|1 < x < 6} or A = {x ∈ N|2 ≤ x ≤ 5} The cardinality is ©2013 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password- protected website for classroom use Full file at https://TestbankDirect.eu/ Solution Manual for Mathematics for Information Technology 1st Edition by Basta Full file at https://TestbankDirect.eu/ 18 Section 1.6 B is the set of all rational numbers less than 100 In set-builder notation, this is B = {x ∈ Q|x < 100} Its cardinality is infinite 10 {4, 8, 12, 16, 20} is the set of numbers divisible by between and 20 inclusive 11 B = {10, 11, 12, 13, 14, 15} and C = {a, b, c, d, e, f } Sets B and C are equivalent 12 S = {5, 4, 3, 2}, T = {2, 4, 3, 5} Sets S and T are equal 13 A ⊂ B 14 A ⊃ B 15 A ⊂ B since A does not include 21 16 The cardinality of {Massachusetts, Rhode Island, Connecticut, Vermont, New Hampshire, Maine} is Chapter Summary Exercises 27 False (A ∩ B) ∪ (A ∪ B) = A ∩ B c 28 Ac ∩ B 29 (A ∪ B)c 30 A ∩ (B ∪ C)c 31 The left-hand circle in the Venn diagram below represents the number of couples who lived with one of the couple’s parents, the right-hand circle represent those that lived in apartments Thus we know that regions + contain 127 couples, that + contain 227 couples and that region contains 95 couples Thus region 1, the number of couples who lived with one of the couple’s parents but not in an apartment, must contain 32 couples, and region 3, those that live in a apartment but not with one of the couple’s parents must contain 132 couples 17 The cardinality of {x|x ∈ W, < x < 3} is 18 There exists a set A such that A ⊃ A This statement is false It implies that A has a member that A does not have 19 There exists a set A having exactly 16 subsets This statement is true The set {1, 2, 3, 4} has 24 = 16 subsets 20 There are winning “coalitions” of votes All winning coalitions must have the chairman in them, and any number of the other members 21 True (A ∪ B c )c = Ac ∩ B 22 True Ac ∪ B c = (A ∩ B)c 23 False Ac ∩ U = A ∩ U 24 True, since both sides are equal to U Ac ∪U = Bc ∪ U 25 True, since U c is empty A ∪ U c = A 26 True (Ac ∩ B) ∪ (A ∩ B) = B Since the universe contains 283 couples and regions + + contain 259 couples, we deduce that 24 couples live neither with one of the couple’s parents nor in an apartment 32 Use the same Venn diagram as in Exercise 31 The left-hand circle represents those residents with computers, 645 of them The right-hand circle represents the 534 that had swimming pools The universe has 732 residents and the area outside the two circles has 37 residents Thus regions 1, 2, and must have a total of 732 − 37 = 695 residents We know that regions and have a total of 645, so region must have 50 in it We know that regions and have a total of 534, so region must have 695 − 534 = 161 residents Thus region must have 695 − (50 + 161) = 484 residents So 484 residents had both a computer and a ©2013 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password- protected website for classroom use Full file at https://TestbankDirect.eu/ Solution Manual for Mathematics for Information Technology 1st Edition by Basta Full file at https://TestbankDirect.eu/ Chapter Sets swimming pool, 161 had only a computer, and 50 had only a swimming pool 33 In the Venn diagram below, let the circle containing region be the wavy-coated dogs, the circle containing region be the brown dogs, and the circle containing region be the female dogs The universe, the sum of regions 0–7, contains 342 dogs Regions 1+2+3+4 = 158 dogs, regions + + + = 37 dogs, and regions + + + = 162 dogs Region contains dogs, regions + contain 78 dogs, and regions + contain 160 dogs Curly females would be given by regions 6+7 Regions 3+4 must contain 158−78 = 80 dogs, thus regions 6+7 must contain 162−80 = 82 dogs, so there are 82 curly females Brown wavy males would be given by region 2, and there is insufficient data to determine how many there are women, the circle containing region be those that have declared a major, and the circle containing region be the freshmen The universe has 512 students, regions + + + = 263, regions + + + = 412 students, regions + + + = 368 Region contains 130 students Regions + = students Regions + = 182 Thus regions + = 368 − 182 = 186 students, and that is the number of freshmen women There is insufficient data to determine region 6, the number of freshmen men who have declared a major 36 Let S be the set of things that get sad sometimes, H be the set of humans, and P be the set of professors The Euler diagram below shows that the syllogism is valid S H P 19 34 Use the same Venn diagram as in Exercise 33 Let the circle containing region be the books, the circle containing region be the “new” items, and the circle containing region be the checked-out items Then regions + + + = 15,832 while regions 0+5+6+7 = 169 items Regions 2+3+5+6 = 453, regions 3+4+6+7 = 1,832 items Region has 32 items Regions + = 302 items Region has items Regions + = 1,700 items Thus region contains 1,700 items Thus region + + has 1832 − 32 = 1800 items Thus region has 1800 − 1700 = 100 items So 100 “new” books were checked out, and 1,700 old books were checked out 35 Use the same Venn diagram as in Exercise 33 Let the circle containing region be the 37 Let B be the set of things that bark, D be the set of dogs, and E be the set of things that eat too much As the Euler diagram shows this syllogism is not valid, as there can be dogs that bark but don’t eat too much B D E 38 Let I be the set of independent things, C the set of cats, and P the set of people Then the Euler diagram demonstrates that the syllogism is invalid ©2013 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password- protected website for classroom use Full file at https://TestbankDirect.eu/ Solution Manual for Mathematics for Information Technology 1st Edition by Basta Full file at https://TestbankDirect.eu/ 20 Section 1.6 Chapter Summary Exercises 42 {1, 2, 4, 8, 16, 32, } is infinite as shown by the mapping {(1, 2), (2, 4), (4, 8), , (i, 2i), }, which does not include in its range I P C 43 {1, 1/3, 1/5, 1/7, 1/9, } is infinite as shown by the mapping {(1, 1/3), ( 1/3, 1/5), , ( 1/i, 1/(i+2)), }, 39 Let P be the set of things that are printed, B be the set of books, and W be the set of things that are wrong Then the Euler diagram shows that this syllogism is invalid whose range does not include 44 The correspondence between {3, 8, 13, 18, } and N is {(3, 1), (8, 2), (13, 3), , (5i − 2, i), } P B W 45 The correspondence between {7, 8, 9, 10, 11, } and N is {(7, 1), (8, 2), (9, 3), , (6 + i, i), } 40 Let G be the set of things that give off light, L be the set of lamps, B be the set of bright things, H be the set of things that are hard to look at, and Z be the set of bronze things Then, as the Euler diagram shows, this syllogism is invalid 46 The correspondence between {9, 12, 15, 18, } and N is {(9, 1), (12, 2), (15, 3), , (6 + 3i, i), } 47 The correspondence between {1, 1/4, 1/9, 1/16, 1/25, } and N is H G {(1, 1), ( 1/4, 2), ( 1/9, 3), , ( 1/i2 , i), } L B 48 The set of all meals eaten by all human beings since there were human beings is finite, large but finite There have only been so many (a finite number) human beings and each of them had only so many (another finite number) meals Z ! 41 {1, 3, 5, 7, } is infinite as shown by the mapping {(1, 3), (3, 5), (5, 7), , (i, i + 2), }, which does not include in its range 49 The set of√digits in the full decimal representation of is finite and has cardinality 10 50 The set of prime numbers divisible by 15 is finite and empty ©2013 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password- protected website for classroom use Full file at https://TestbankDirect.eu/ .. .Solution Manual for Mathematics for Information Technology 1st Edition by Basta Full file at https://TestbankDirect.eu/ 10 Section 1.2... classroom use Full file at https://TestbankDirect.eu/ Solution Manual for Mathematics for Information Technology 1st Edition by Basta Full file at https://TestbankDirect.eu/ Chapter A ⊃ B or... classroom use Full file at https://TestbankDirect.eu/ Solution Manual for Mathematics for Information Technology 1st Edition by Basta Full file at https://TestbankDirect.eu/ 12 Section 1.3 Venn