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Lecture 1.4
Contemporary Mathematics
Instruction: Set Notation
We have already studied roster or list as well as descriptive notation. In this lecture, we
will discuss set builder and set interval notation.
Set builder notation begins with a variable, frequently x, followed by a vertical slash (|) or
a colon (:). The vertical slash (and/or the colon) reads as "such that." A description of the values
(or other objects) that the variable can take follows the "such that" symbol. Accordingly,
J = {a : a > 0, a ∈ } describes the set of positive integers (values of a greater than zero and
elements of the set of integers) while K = {b | b < 0, b ∈
}
describes the negative real numbers.
To further elucidate set builder notation, let's recall , the set of natural numbers, and
consider two subsets of : H = {2} and G = {the natural numbers between one and five} . The
phrase "between one and five" does not include one or five. In roster notation, G = {2, 3, 4} . Set
builder notation describes G as { x | 2 ≤ x ≤ 4, x ∈ } also written { x : 2 ≤ x ≤ 4, x ∈ } , which
reads "x such that two is less than or equal to x and x is less than or equal to four, and x is an
element of the natural numbers." Similarly, set builder notation represents set H as { x | x = 2} ,
which reads "x such that x equals two." Below, set builder notation is used to represent G − H .
G − H = { x : 3 ≤ x ≤ 4, x ∈
}
As we have seen, set builder notation can be used to describe an interval of permitted real
values that a variable can take (as with K = {b | b < 0, b ∈ } ). Similarly, set interval notation
describes an interval of permitted real values that a variable can take using the lower and upper
limits on the value of the variable. The lower and upper limits may or may not be included in
the interval.
Let's consider S = { x : −5 x + 7 > 12, x ∈ } . This set includes the real number solutions to
the inequality −5 x + 7 > 12 . To elucidate interval notation, let's solve the inequality.
− 5 x + 7 > 12
−5 x > 12 − 7
−5 x > 5
−5 x 5
>
−5 −5
x < −1
The solution to the inequality includes the set of real numbers smaller than negative one. The
number line below represents the solution set.
Lecture 1.4
In set interval notation, we write the solution set as ( −∞, −1) . The symbol −∞ reads "negative
infinity." The notation ( −∞, −1) signifies a set of values "from negative infinity to negative one,
negative one excluded." An astute student might wonder how to write set interval notation if the
solution had been x ≤ −1 . In such a case, the proper interval notation is ( −∞, −1] , which
signifies a set of values "from negative infinity to negative one, negative one included."
Let's consider T = { x : 9 ≤ 7 x + 2 < 23, x ∈ } . This set includes the real number solutions
to the compound inequality 9 ≤ 7 x + 2 < 23 . Solving the compound inequality reveals a solution
set that includes the real numbers between one and three, including one but excluding three as
shown below.
9 ≤ 7 x + 2 < 23
9 − 2 ≤ 7 x < 23 − 2
7 7 x 21
≤
<
7 7
7
1≤ x < 3
In set interval notation, we write the solution set as [1,3) .
Before presenting a problem set, this lecture will now review solving linear equations and
inequalities containing absolute value expressions. Consider the equation below.
12 − 3 x − 8 = −2
To solve, we will start by isolating the absolute value expression. For this example, we need to
add eight to both sides.
12 − 3x − 8 + 8 = −2 + 8
12 − 3x = 6
Now, we need to eliminate the absolute value expression. To do this, we will write a compound
statement. One of the statements will look exactly like the equation as is without the absolute
value, but the other statement will change one side of the equation to the exact opposite. We do
this because we know if a = k , then a = k or a = − k .
12 − 3 x = 6
12 − 3x = 6 or 12 − 3 x = −6
Finally, we solve each equation separately.
Lecture 1.4
12 − 3 x = 6
12 − 3 x = −6
−3x = −6
−3 x = −18
or
−6
−18
x=
x=
−3
−3
x=2
x=6
Both two and six are solutions to the equation 12 − 3 x − 8 = −2 . Using set builder notation, we
can represent the solutions as a set: V = { x : 12 − 3 x − 8 = −2} = { x : x = 2, x = 6} .
Before continuing, let's examine properties of absolute value below.
If a > k , then a > k or a < −k .
If a ≥ k , then a ≥ k or a ≤ −k .
If a < k , then a < k and a > −k .
If a ≤ k , then a ≤ k and a ≥ −k .
Note the subtle shift in the conjunction, which switches from "or" in the first two statements to
"and" in the second two statements.
Consider the inequality below. Note that the inequality uses a greater than symbol.
2x +1 − 4 > 5
To solve, we will start by isolating the absolute value expression. For this example, we need to
add four to both sides.
2x +1 − 4 + 4 > 5 + 4
2x +1 > 9
Now, we need to eliminate the absolute value expression. To do this, we will write a compound
inequality using "or." We use "or" because the inequality symbol involves a "greater than"
symbol.
One of the compound inequalities will look exactly like the inequality as is without the
absolute value, but the other inequality will change one side to the exact opposite with the
inequality symbol reversed. We do this because we know if a > k , then a > k or a < − k .
2x +1 > 9
2x +1 > 9
or
2 x + 1 < −9
Lecture 1.4
Finally, we solve each inequality separately.
2x +1 > 9
2x > 9 −1
2 x > 8 or
8
x>
2
x>4
2 x + 1 < −9
2 x < −9 − 1
2 x < −10
−10
x<
2
x < −5
Since the compound inequality uses "or" as the conjunction, any number that satisfies one of the
statements is a solution, i.e., any number greater than four is a solution as well as any number
less than negative five. We can represent the solution as a set. Using set builder notation, we
write S = { x | x < −5 ∪ x > 4, x ∈ } . In set interval notation, we write ( −∞, −5 ) ∪ ( 4, ∞ ) .
Now, let's consider another inequality. Note that the inequality below uses a "less than or
equal to" symbol.
7 2 x + 1 ≤ 35
To solve, we will start by isolating the absolute value expression. For this example, we need to
divide both sides by seven.
7 2x +1
≤
7
2x +1 ≤ 5
35
7
Now, we need to eliminate the absolute value expression. To do this, we will write a compound
inequality using "and." We use "and" because the inequality symbol involves a "less than"
symbol.
One of the compound inequalities will look exactly like the inequality as is without the
absolute value, but the other inequality will change one side to the exact opposite with the
inequality symbol reversed. We do this because we know if a ≤ k , then a ≤ k and a ≥ − k .
2x +1 ≤ 5
2 x + 1 ≤ 5 and
Finally, we solve each inequality separately.
2 x + 1 ≥ −5
Lecture 1.4
2x +1 ≤ 5
2x ≤ 5 −1
2 x ≤ 4 and
4
x≤
2
x≤2
2 x + 1 ≥ −5
2 x ≥ −5 − 1
2 x ≥ −6
−6
x≥
2
x ≥ −3
Since the compound inequality uses "and" as the conjunction, the solution only includes numbers
that satisfy both statements in the compound inequality, i.e., the solution only includes numbers
that are both less than or equal to two and greater than or equal to negative three. In set builder
notation, we write S = { x | −3 ≤ x ≤ 2, x ∈ } . In set interval notation, we write, [ −3, 2] .
Application Exercise 1.4
Problems
Consider the following sets.
A = { x :1 − 5 x > −19, x ∈ `}
B = { NASA's five space shuttle orbiters built for space flight}
D =
{x
: 3x + 1 > 7, x ∈ \ }
E = { x : 2 − x ≤ 4, x ∈ \}
#1
Find n( A) , the cardinal number of set A.
#2
Use roster notation to represent set B.
#3
Use interval notation to represent set D.
#4
Use interval notation to represent set E.
1. n ( A ) = 3
3.
( 2,∞ )
2. {Columbia, Challenger, Endeavour, Atlantis, Discovery}
4. [ −2,6]
Assignment 1.4
Problems
Consider the following sets.
A = { x : 6 x + 12 = 42, x ∈ \}
B = { x : 6 x + 12 = 42, x ∈ \}
D = { x : 6 x + 12 ≥ 42, x ∈ \}
E = { x : 6 x + 12 < 42, x ∈ \}
F = { x : 6 x + 12 < 42, x ∈ `}
#1
Find n( A) , the cardinal number of set A.
#2
Use roster notation to represent set B.
#3
Use interval notation to represent set D.
#4
Use interval notation to represent set E.
#5
Determine the number of subsets of F.
Lecture 1.5
Contemporary Mathematics
Instruction: Addition Rules, Complement Principle, & Multiplication Principle
This lecture is our final section strictly on sets. Soon we will use sets to think about
counting, probability, statistics, and functions. This lecture provides a bridge from set theory to
counting theory.
Let's assume a universal set of the first six whole numbers and a few proper subsets.
U = {0, 1, 2, 3, 4, 5}
A = {0, 1, 2}
B = {3, 4, 5}
C = {2, 3}
Our interest in this lecture lies with the cardinal number of the various proper subsets of U
above. We note that n(A) = 3, n(B) = 3, and n(C) = 2, but what of n ( A ∪ B ) , n ( A ∪ C ) ,
n ( Ac ) and n ( A × B ) or n ( A × C ) ?
To answer this question, observe that A and B have no common elements, a characteristic
of disjointed or mutually exclusive sets as defined below.
Sets are mutually exclusive if they contain no common elements.
Consequently, E and F are mutually exclusive if E ∩ F = {
}.
Returning to our example above, sets A and B are mutually exclusive because they contain no
common elements. We see that the union of the two sets will have as many elements as the sum
of the number of elements in each set as shown below.
A ∪ B = {0, 1, 2} ∪ {3, 4, 5}
A ∪ B = {0, 1, 2, 3, 4, 5}
∴
n( A ∪ B) = 6
This observation is generally true so that we have the Addition Rule for Mutually Exclusive Sets.
If E and F are mutually exclusive, then n ( E ∪ F ) = n ( E ) + n ( F )
Sets A and C are not mutually exclusive because they have a common element. We
found n ( A ∪ B ) , but what of n ( A ∪ C ) , the cardinal number of the union of set A with set C,
two sets not mutually exclusive? We see that the union of two sets not mutually exclusive will
Lecture 1.5
have as many elements as the sum of the number of elements in each set less the number of
common elements as shown below.
A ∪ C = {0, 1, 2} ∪ {2, 3}
A ∪ C = {0, 1, 2, 3}
∴
n( A ∪ C ) = 4
This observation is generally true so that we have the General Addition Rule.
n( E ∪ F ) = n( E) + n(F ) − n( E ∩ F )
The General Addition Rule gives the cardinal number of the union of any two sets whether or not
the two sets are mutually exclusive. In the case that the sets are mutually exclusive the
intersection is an empty set whose cardinal number is zero.
To find n ( AC ) , we turn to the definition of the complement set. The complement of A is
the set of elements in the universal set not in A. Accordingly, n ( AC ) + n ( A ) = n (U ) .
Subtracting n ( A ) from both sides, we have n ( AC ) = n (U ) − n ( A ) , which can be generalized as
the Complement Principle below.
n ( E C ) = n (U ) − n ( E )
n ( E ) = n (U ) − n ( E C )
We have found n ( A ∪ B ) , n ( A ∪ C ) , and n ( AC ) , but what of n ( A × B ) or n ( A × C ) ?
The number of elements in a Cartesian product equals the product of the number of elements in
each factor of the Cartesian product as seen below.
A × B = {0, 1, 2} × {3, 4, 5}
A × B = {( 0,3) , ( 0, 4 ) , ( 0,5 ) , (1,3) , (1, 4 ) , (1,5 ) , ( 2,3) , ( 2, 4 ) , ( 2,5 )}
∴
n( A× B) = 9
This observation is generally true so that we have the Multiplication Principle.
n (E × F ) = n (E)⋅n (F )
The Multiplication Principle can be extended for cross product of the elements of a set of sets
{E1 , E2 , … , E p } :
n ( E1 × E2 ×
× E p ) = n ( E1 ) ⋅ n ( E2 ) ⋅
⋅ n ( Ep )
Application Exercise 1.5
Problems
As of November 2004, 439 individuals have flown on spaceflights including 96 from
Russia/Soviet Union, 277 from America, and 66 from other nations. Twenty-two of these heroes
died while in a spacecraft including eighteen astronauts (Americans) and four cosmonauts
(Russians).
Let the universal set be the 439 individuals that have flown on spaceflights as of November
2004. Let A represent the American members of U. Let R represent the members of U from
Russia/Soviet Union. Let N represent the individuals from other nations. Finally, let D represent
the individuals that have died in a spacecraft.
#1
Find n ( A ∪ D ) .
#2
Find n ( D C ) .
#3
Find n ( N ∩ A ) .
#4
Find n ( N ∪ A ) .
#5
n (( D ∩ A) × ( D ∩ R ) )
1. n ( A ∪ D ) = 281
3. n ( N ∩ A ) = 0
5. n ( ( D ∩ A ) × ( D ∩ R ) ) = 72
2. n ( D C ) = 417
4. n ( N ∪ A ) = 343
Assignment 1.5
Problems
#1
Assume M and P are mutually exclusive, find n ( M ∪ P ) if n ( M ) = 22 and n ( P ) = 18 .
#2
Assume n ( J ) = 9 , n ( J ∪ K ) = 17 , and ( J ∩ K ) = {
#3
Find n ( G ∪ H ) , assuming n ( G ) = 15 , n ( H ) = 12 , and n ( G ∩ H ) = {3} .
#4
Find n ( E ∩ F ) , assuming n ( E ) = 200 , n ( F ) = 250 , and n ( E ∪ F ) = {380} .
#5
Show that sets X and Y are mutually exclusive given that n ( X ) = 44 , n (Y ) = 16 , and
} , find n ( K ) .
n ( X ∪ Y ) = 60 .
#6
What can be said about sets L and M if n ( L ) = 12 , n ( M ) = 22 , and n ( X ∪ Y ) = 31 ?
#7
Find n (U ) , assuming n ( A ) = 52 and n ( AC ) = 48 .
#8
Find n (T ) , assuming n (U ) = 487 and n (T C ) = 44 .
#9
Find n ( C × D ) , given that n ( C ) = 20 and n ( D ) = 7 .
#10
Find n (V ) , assuming n (W ) = 8 and n (V × W ) = 96 .