Table Of ContentSTUDY GUIDE
for
APPLIED FINITE MATHEMATICS
Third Edition
by
NICHOLAS A. MACRI
Temple University
ACADEMIC PRESS
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CHAPTER 1
SET THEORY
Exercise set 1.1
Exercise 5 (a) {A,R,D,V,K}. (b) {M,I,S,P}. (c) {T,A,B,L,E}. Note that
although the letters A and R appear in the word AARDVARK more than once, we
do not include them in the set more than once. Similarly, the letters I, S
and P appear more than once in the word MISSISSIPPI, but they are included
only once in the set.
2
Exercise 1ft Since A = {x|x is an integer and x < 25}, it follows that
A = {-5,-4,-3,-2,-1,0,1,2,3,4,5}.
(a) {x|x is a positive integer and x < 16} c A is true, since {x|x is a
positive integer and x < 16} = {1,2,3,4}, and clearly {1,2,3,4} c A.
(b) {2,-2,6} c A is false, since 6 i A.
(c) A c {1,2,3,4,5} is false, since it is not true that every element of A
is an element of {1,2,3,4,5}. For example, -3 e A, but -3 i {1,2,3,4,5}.
(d) {x|x is an integer and x < 5} c A is false because {x|x is an integer
and x < 5} contains at least one member that is not in A. For example,
-6 is in {x|x is an integer and x < 5} but -6 is not in A.
Exercise set 1.2
Results to remember:
1. The intersection of two sets A and B, denoted An B, is that set whose
elements belong to both A and B.
2. The union of two sets A and B, denoted A u B, is the set whose elements
belong to either A or B and possibly to both A and B.
3. Two sets A and B are said to be disjoint if they have no elements in
common; that is, A n B = <J>.
1
1: Set Theory
4. The first distributive law: If A, B, and C are any three sets, then
C n (A u B) = (C n A) u (C n B).
5. The second distributive law: If A, B, and C as?e any three sets, then
C u (A n B) = (C u A) n (C u B).
Exercise 5
In this problem, the distinct regions in the figure have been marked 1, 2,
3, 4, 5, 6, and 7. We will answer questions (a) through (d) by first writing
down the set of numbers that constitute the given set.
Note that set S is made up of regions 1, 2, and 3; set C is made up of
regions 2, 3, 4, 6, and 7, while set T is made up of regions 3, 4, and 5.
We therefore will write
S = {1,2,3}, C = {2,3,4,6,7}, and T = {3,4,5}.
(a) C u (S n T) = {2,3,4,6,7} u ({1,2,3} n {3,4,5})
= {2,3,4,6,7} u {3} = {2,3,4,6,7} = C.
(b) C n (S u T) = {2,3,4,6,7} n ({1,2,3} u {3,4,5})
= {2,3,4,6,7} n {1,2,3,4,5} = {2,3,4}.
2
Exercise Set 1.2
(c) (C n S) u T = ({2,3,4,6,7} n {1,2,3}) u {3,4,5}
= {2,3} u {3,4,5} = {2,3,4,5}.
(d) (C u S) n (C n T) = ({2,3,4,6,7} u {1,2,3}) n ({2,3,4,6,7} n {3,4,5})
= {1,2,3,4,6,7} n {3,4} = {3,4}.
Exercise 15 (a) Show that A c A u B and B c A u B.
A u B is made up of regions labeled 1,2, and 3. A is made up of regions
1 and 2. A, then, is clearly a subset of A u B. B is made up of regions
2 and 3. Here we can see that B is a subset of A u B.
(b) Show that A n B c A and A n B c B. A n B is the region labeled 2.
A is made up of regions 1 and 2. B is made up of regions 2 and 3.
Clearly, then, we see that A n B is a subset of both A and B.
3
1: Set Theory
Exercise 17 M = the set of all male employees.
F = the set of all female employees.
A = the set of all administrative employees.
T = the set of all technical employees.
S = the set of all employees working for the company for at
least 5 years.
We will represent these sets graphically as discs labeled with the letters
M, F, A, T, or S, as the situation requires.
(a)
The graphic representation of M n A (shaded region) shows that M n A
represents the set of male administrative employees.
(b)
M u F is the set of employees who are either male or female; in other
words, the set of all employees.
(c)
A n T n F is the set of administrative technical female employees.
4
Exercise Set 1.3
(d)
A u T u F is the set of employees who are administrative, technical, or
female.
(e)
M n A n S is the set of male administrative employees who have been
with the company for at least 5 years.
Exercise Set 1.3
Results to remember:
1. A universal set for a problem is some fixed set U from which all the
elements involved in the problem come.
2. If U is a universal set and A is a subset of U, the complement of A,
denoted Af, is the set of all elements in U that are not in A. Think
of the complement of a set as those objects outside of the set but in
the universal set.
3. For any two sets A and B, the DeMorgan laws are as follows:
First DeMorgan law: (A u B)f = A1 n Bf. Stated in words, this
law says that the complement of a union is the intersection of the
complements.
5
1: Set Theory
Second DeMorgan law: (A n B)' = A1 u B'. This law says that the
complement or an intersection is the union of the complements.
4. If U is the universal set and A is a subset of U, then
a) (A1)1 = A.
b) A u A' = U.
c) A n A1 = <j>.
5. An ordered pair (a,b) is a listing of two objects a and b in definite
order. The element a is called the first entry in the ordered pair,
and the element b is called the second entry in the ordered pair.
6. If A and B are two sets, then the set of all ordered pairs (a,b), where
a e A and b £ B, is called the Cartesian product of A and B; it is
denoted by the symbol A x B. We can write A x B in set-builder notation
as A x B = {(a,b)|a £ A and b e B}.
Exercise 3 U = the set of all stocks traded on the New York Stock Exchange.
A = the set of stocks traded on the New York Stock Exchange that have paid
a dividend for the past 10 years without any interruption. B = the set of
stocks traded on the New York Stock Exchange that have a price-to-earnings
ratio of no more than 12.
(a)
A' is the set of stocks that have not paid a dividend for the past 10
years without any interruption.
00
B' is the set of stocks that have a price-to-earnings ratio of more
than 12.
(c)
6
Exercise Set 1.3
Since (A u B)f = A' n B' we see that (A u B)f is the set of those
stocks that have not paid a dividend for the past 10 years without any
interruption and have a price-to-earnings ratio of more than 12.
(d)
Since (A n B)' = A' u B', we see that (A n B) ' is the set of those
stocks that either have not_ paid a dividend for the past 10 years with
out any interruption or have a price-to-earnings ratio of more than 12.
Exercise 5 A = {x|x is an integer satisfying x > 0}. If we let U = set of
all real numbers, then A1 is the set of all real numbers that are ncrt posi
tive integers. In set-builder notation we write
A' = {x|x is a real number and not a positive integer}.
If we make U = set of all integers, and we write U = {...,-4,-3,-2,-1,0,
1,2,3,4,...}, then we can write A1 = {...,-4,-3,-2,-1,0}. Take note that
the complement of a set is dependent on the choice of the universal set.
Exercise 9 Note that the universal set is made up of the regions labeled
1,2,...,8.
(a) Af = {1,2,5,6}, B' = {1,2,3,8}, and C = {1,6,7,8}. So A' n Bf = {1,2}
and (A1 n Bf) n C = {1}. Another way of determining the set
Af n Bf n C is to note that A' n B' n Cf is that region outside of A
and outside of B and outside of C.
7
1: Set Theory
(b) Note that A u B u C = {2,3,4,5,6,7,8}, so (A u B u C)' = {1}. We may
also solve the problem by using the first DeMorgan law as follows:
(A u B u C)f = ((A u B) u C)f = (A u B) ' n Cf = Af n Bf n Cf
= {1}, from part (a). The region for this has been shaded alone.
(c) Note that A n B n C = {4}, so (A n B n C)' = {1,2,3,5,6,7,8}.
mmmm
(d) A u B = {3,4,5,6,7,8} and C = {2,3,4,5}. Therefore (A u B) n C = {3,4,5}
and [(A u B) n C]f = {1,2,6,7,8}.
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