One possible answer is: “The set of whole numbers less than 16 that are multiples of 3.” 32.. Neither, both have an infinite amount of num-bers, but you cannot set up a one-to-one cor-re
Trang 1Chapter 1 Sets
Exercises 1.1: Set Concepts
Note: Answers to Exercises1throught7will vary
but should approximate the answers given below
1 A set is a well-defined, unordered collection of
objects having no duplicate members
2 Two sets A and B are said to be equal, written
A = B, if their membership is entirely iden-tical except for the fact that the order of the elements in the sets may be different
3 Two sets A and B are said to be equivalent,
written A ∼ B, if they share the same cardi-nality
4 The cardinality of a set S refers to the
quan-tity of members belonging to the set, and is represented by |S| or n(S)
5 The empty set is a set that has no members
Its cardinality is zero The symbol for the empty set may be either ∅or { }
6 A finite set is a set that has a particular
quan-tity of elements
7 Ellipsis are a succession of three “dots,” which
indicate that a demonstrated pattern of bers continues, either forever or until a num-ber following the ellipsis is reached
8 Well defined; a person is either a paid
em-ployee of the US Government or is not
9 Not well defined; unclear as to meaning of
ef-ficient
10 Well defined; there are 50 odd integers less
than 100, specifically {1, 3, 5, 7, , 99}
11 Not well defined; unclear as to meaning of
well-spoken
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 be-tween 6 and 7
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 trillion, 1,000,000,000,000, has 13 digits
17 Infinite; the decimal expansion of pi never ends
18 Infinite; in Example 1.3.iv, we showed that the numbers between 0 and 2 were infinite This same reasoning would apply to the set of numbers between 4 and 10
19 {M, i, s, p}, 4
20 {1, 2, 3, 4, , 47, 48, 49}, 49
21 {Nebraska, Nevada, New Jersey, New Mexico, New York, North Carolina, North Dakota}, 7
22 {−5}, 1
23 {}, 0
24 {Alaska, Washington, Idaho, Montana, North Dakota, Minnesota, Michigan, New York, Vermont, New Hampshire, Maine, Texas, New Mexico, Arizona, California}, 15
25 At the time this book was written: {Carter, GHW Bush, Clinton, GW Bush, Obama}, 5
Trang 210 Section 1.2 Subsets
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 ÷ 2 ∈ W}
28 {x|xis 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 2 and 6,
includ-ing 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
num-bers between 1 and 19, inclusive.”
36 5
37 4
38 4
39 Equivalent but not equal, because they both have a cardinality of 4, but do not have the same elements
40 Both, because they both have a cardinality of
4 and they have the same elements
41 Both, because |B| = 3 and |C| = 3 and they both have the same elements
42 Equivalent but not equal, because they both have a cardinality of 50, but do not have the same elements
43 Neither, both have an infinite amount of num-bers, but you cannot set up a one-to-one cor-respondence between the elements of A, the natural numbers larger than 2, and the ele-ments in B, all (real) numbers greater than 2
Exercises 1.2: Subsets
Note: Answers to Exercises1throught5will vary
but should approximate the answers given below
1 Set A is a subset of set B if every element of
A is also an element of B
2 A is a proper subset of B if A is a subset of
B and B contains elements that are not in A
3 If A is a subset of B, then B is a superset of
A
4 A power set of a particular set has as its
mem-bers all possible subsets of that particular set
5 Every set has at least one subset, the empty set, so it is not possible for a set to have no subsets
6 A ⊂ B because all the elements of A are the vowels of the English alphabet
7 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
8 A = B because both of these set-builder no-tations describe the same set of numbers
Trang 39 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 ⊃ Bwhere 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= 1
15 27= 128
16 23= 8 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 × 7 = 604,800 seconds
in a week, which is much less than 240
21 True If A and B have the same cardinality, then they both have the same number of ele-ments, 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− 1 This is the same as asking how many proper subsets can be made using the 18 in-gredients 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 5 allow the measure to pass
26 8 There are 11 (4 + 3 + 2 + 2) possible votes and it will take 6 votes for an application to pass We represent each of the voters as P for President, V for Vice-President, A for Archi-tectural Committee Leader, and F for Fence Height Overseer The possible winning coali-tions (and its votes) are: P+V (4 + 3), P+A (4 + 2), P+F (4 + 2), P+A+F (4 + 2 + 2), P+V+A (4 + 3 + 2), P+V+F (4 + 3 + 2), P+V+A+F (4 + 3 + 2 + 2), and V+A+F (3 + 2 + 2) Thus, there are 8 possible winning coalitions
Trang 412 Section 1.3 Venn Diagrams
Exercises 1.3: Venn Diagrams
1
U
2
U
3
U
4 (A ∩ B)C ∪ B = U, so everything is sheded
U
5
U
6 AC∩ A =∅, so nothing is sheded
U
7 AC ∩ A = ∅ and ∅C = U, so everything is shaded
U
8 True, as shown by the following two figures
U
(A B)C
U
A C B C
9 False, as shown by the following two figures
Trang 5A B U
(A B)C
U
A C B C
10 False, as shown by the following two figures
U
U
11 False, as shown by the following two figures
In the first figure, AC∪ U = U and Uc =∅,
so nothing is shaded
U
(A C U)C
U
12 True, the union of the shaded portion of the following two figures is B
U
U
A B C
( )C
13 True, the union of the shaded portions of the following two figures is A
U
U
A B C
14 True, the shaded portion of the first figure is
Trang 614 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
sec-ond figure
U
U
A
15 False, as shown by the following two figures
U
( )C
U
A C B C
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 for-mula 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
1 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
44 35 49
Off campus
2 In the Venn diagram below T represents those who own trucks and C those who own cars
Trang 7Since 21 do not own a vehicle, then 125−21 =
104people 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 num-ber who owned only a truck were 59−43 = 16 and 88 − 43 = 45 only owned a car
No vehicle 21
3 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, 1 cat had all three colors, 6 had none of the col-ors, and 11 had exactly two of the colors
Black Orange
Grey
9 0 5
6 1
6
4 This diagram differs from ones you have seen
earlier since it does not use circles The uni-verse of 63 drivers is divided in two parts verti-cally, male and female, and two parts horizon-tally, teenage drivers and past teenage drivers
From the drawing you can see that there were
13 male drivers who were past their teenage years
Teens
13 6
Past teenage
5 Using the given information we can develop the diagram below
CO res
Non-CO res
3 or more passengers
25 10
5 13
12 21
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 non-Colorado drivers, the 37 must be an error 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
6 From the Venn diagram below we see that 6 majored in engineering only, 8 in biology only, and 15 in chemistry only
0 Eng Bio
Chem
6 8
8 7 15 11
7 There are 447 farmers who grow at least one
of the crops; 20 grew all three; 0 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 farm-ers that are growing only beets and radishes
Trang 816 Section 1.5 Infinite Sets
T 100
BRT BR
BT RT
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
T 100
20 30
25 17
Adding all of the amounts in the Venn dia-gram we get a total of 447 who grow at least one of the crops
8 The children who played only hockey and baseball are shown by the shaded section of the Venn diagram below A total of 35 chil-dren 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 2 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
S
10
3
17 6
48
53
Exercises 1.5: Infinite Sets
Note: Answers to Exercises 1 throught 3 should
approximate the answers given below
1 An infinite sets has unlimited membership
2 A one-to-one correspondence is a relationship
between two sets that associates to each
mem-ber in one set a unique element in the other
set, and vice-versa
3 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
4 The natural numbers are a proper subset of the given numbers, so one possible one-to-one correspondence is (1, 1), (−1, 2), (2, 3), (−2, 4), (3, 5), (−3, 6),
5 One proper subset of the given set is
A = {200, 300, 400, 500, } and a one-to-one correspondence between the given set and this proper subset is (100, 200), (200, 300), (300, 400), (400, 500), (500, 600), , (n, 100+ n)
Trang 9Another 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)
6 One proper subset of the given set is
C = {4, 6, 8, 10, 12, } A one-to-one 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)
7 One proper subset of the given set is C =
{1/2,1/3,1/4,1/5,1/6, } A one-to-one cor-respondence 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)) }
8 Use the one-to-one correspondence {(5, 1),
(10, 2), (15, 3), (20, 4), , (5n, n) }
9 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) }
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 sec-onds 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
0 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 cardinal-ity 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 sec-onds 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 2
Chapter 1 Summary Exercises
1 The set of congresspeople in the United States
is well-defined
2 The set of cats in this room is not well-defined
since we do not know the room being referred to
3 The set of all cats in the world is well-defined
4 The set of odd integers between 20 and 40
in-clusive is {21, 223, 25, 27, 29, 31, 33, 35, 37, 39}
This set is finite and has cardinality 10
5 The set of digits in the full decimal expansion
of all real numbers is finite and has cardinality 10
6 The set of letters in the word Oklahoma is {O,
k, l, a, h, o, m} and its cardinality is 7 No-tice that O and o are considered to be different letters
7 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
8 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 4
Trang 1018 Section 1.6 Chapter Summary Exercises
9 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 4 between 1 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
Is-land, Connecticut, Vermont, New Hampshire,
Maine} is 6
17 The cardinality of {x|x ∈ W, 2 < x < 3} is 0
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 = 16subsets
20 There are 8 winning “coalitions” of votes All
winning coalitions must have the chairman in
them, and any number of the other members
21 True (A ∪ Bc)c= Ac∩ B
22 True Ac∪ Bc= (A ∩ B)c
23 False Ac∩ U 6= A ∩ U
24 True, since both sides are equal to U Ac∪U =
Bc∪ U
25 True, since Ucis empty A ∪ Uc= A
26 True (Ac∩ B) ∪ (A ∩ B) = B
27 False (A ∩ B) ∪ (A ∪ B) 6= A ∩ Bc
28 Ac∩ B
29 (A ∪ B)c
30 A ∩ (B ∪ C)c
31 The left-hand circle in the Venn diagram be-low 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 1 + 2 contain 127 couples, that 2 + 3 contain 227 couples and that region 2 contains 95 cou-ples 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
1
Since the universe contains 283 couples and regions 1 + 2 + 3 contain 259 couples, we de-duce 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 3 must have a total of
732 − 37 = 695 residents We know that re-gions 1 and 2 have a total of 645, so region
3 must have 50 in it We know that regions
2 and 3 have a total of 534, so region 1 must have 695 − 534 = 161 residents Thus region
2 must have 695 − (50 + 161) = 484 residents
So 484 residents had both a computer and a