Review Exercise
1. Express following sets using the listing method.
a. 𝐴 is the set of positive factors of 18
A = { 1,2,3,6,9,18}
b. 𝐵 is the set of positive even numbers below or equal to 30
B = {2,4,6,8,10,12,14,16,18,20,22,24,26,28,30}
c. 𝐶 = {2𝑛 | 𝑛 = 0, 1, 2, 3, … }
C = { 0,2,4,6,8,10, …..}
d. D = {x I x² = 9}
2. Express following sets using the set-builder method.
a. {2, 4, 6, … . .}
First element 2 = 1 x 2
Second element 4 = 2 x 2
Third element 6 = 2 x 3………………..as we can see 2 is constant so our general equation becomes 2n where n = {1,2,3,4,.....}
={ 2n</span><span> | n = 1, 2, 3, … }
b. {1, 3, 5, … , 99}
1 = 2 - 1 = 2(1) - 1 = 1
3 = 4 - 1 = 2(2) - 1 = 3
5 = 6 - 1 = 2(3) - 1 = 5
………..
99 = 100 - 1 = 2(50) - 1 = 99
So the set-builder expression is = { 2n - 1 | n = 1, 2, 3,....,50 }
c. {1, 4, 9, … , 81}
1 = 1²
4 = 2²
9 = 3²
…..
81 = 9²
So the set-builder expression is = { n² | n = 1, 2, 3,....,9 }
3. Find all the subsets of the following sets.
a. {3, 4, 5}
Number of element = n = 3
So number of subset = 2n = 2³ = 8
The subsets are { } ,{3} ,{4} ,{5} ,{3,4} ,{3,5} ,{4,5} ,{3,4,5}
b. {𝑎, 𝑏}
Number of element = n = 2
So number of subset = 2n = 2² = 4
The subsets are { } ,{a} ,{b} ,{a,b}
4. Find 𝐴 ⋃ 𝐵 and 𝐴 ⋂ 𝐵 of the following.
a. 𝐴 = {2, 3, 5, 7, 11} and 𝐵 = {1, 3, 5, 8, 11}
A = {2, 3, 5, 7, 11} , B = {1, 3, 5, 8, 11}
A ⋃ B = {1,2,3,5,7,8,11}
A ⋂ B = {3,5,11}
b. 𝐴 = {𝑥 | 𝑥 is the factor of 12} and 𝐵 = {𝑥 | 𝑥 is the factor of 18}
A = factor of 12 = {1,2,3,4,6,12} , B = factor of 18 = {1,2,3,6,9,18}
A ⋃ B = {1,2,3,4,6,9,12,18}
A ⋂ B = {1,2,3,6}
c. 𝐴 = {3𝑥 | 𝑥 ∈ ℕ, 𝑥 ≤ 20 } and 𝐵 = {4𝑥 | 𝑥 ∈ ℕ, 𝑥 ≤ 15}
A = 3x where x ≤ 20 = {3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60}
B = 4x where x ≤ 15 ={4,8,12,16,20,24,28,32,36,40,44,48,52,56,60}
A ⋃ B = {3,4,6,8,9,12,15,16,18,20,21,24,27,28,30,32,33,36,39,40,42,44,45,48,51,52,54,56,57,60}
A ⋂ B = {12,24,36,48,60}
5. If 𝐵 ⊆ 𝐴, 𝐴 ∩ 𝐵 ′ = {1, 4, 5}, and 𝐴 ∪ 𝐵 = {1, 2, 3, 4, 5, 6}, then find set 𝐵.
Since B ⊆ A = B - A = ∅
A ∩ B ′ = {1, 4, 5}
A ∪ B = {1, 2, 3, 4, 5, 6}
A ∩ B ′ = A - B
A ∪ B = (A - B) ∪ (B ⊆ A)
But B ⊆ A = B
{1, 2, 3, 4, 5, 6} = {1, 4, 5} ∪ B….. since it is union the value of B is the remaining values that are in A ∪ B but not in A ∩ B′
So B = {2,3,6}
6. Let 𝐴 = {2, 4, 6, 7, 8, 9}, 𝐵 = {1, 3, 5, 6, 10} and 𝐶 = {𝑥 | ∈ ℤ, 3𝑥 + 6 = 0 or 2𝑥 + 6 = 0}.
Find
a) 𝐴 ∪ 𝐵
A = {2,4,5,7,8,9} , B={1,3,5,6,10}, C = {</span><span>𝑥 | ∈ ℤ, 3𝑥 + 6 = 0 or 2𝑥 + 6 = 0}
3x + 6 = 0 or 2x + 6 = 0
3x = -6 or 2x = -6
x = -2 or x = -3
C = { -3,-2}
A ∪ B = {1,2,3,4,5,6,7,8,9,10}
b) Is (𝐴 ∪ 𝐵) ∪ 𝐶 = 𝐴 ∪ (𝐵 ∪ 𝐶)?
A = {2,4,5,7,8,9} , B={1,3,5,6,10}, C = {</span><span>𝑥 | ∈ ℤ, 3𝑥 + 6 = 0 or 2𝑥 + 6 = 0}
3x + 6 = 0 or 2x + 6 = 0
3x = -6 or 2x = -6
x = -2 or x = -3
C = { -3,-2}
(A ∪ B) ∪ C = {-3,-2,1,2,3,4,5,6,7,8,9,10}
A ∪ (B ∪ C) = {-3,-2,1,2,3,4,5,6,7,8,9,10}
As we ca see the result (A ∪ B) ∪ C = A ∪ (B ∪ C) ……..associative property
7. Suppose the universal set 𝑈 be the set of one-digit numbers, and set 𝐴 = {𝑥 | 𝑥 is an even natural number less than or equal to 9 }. Describe each set by complete listing method:
a. 𝐴 ′
U = {1,2,3,4,5,6,7,8,9}
A = {2,4,6,8}
A′ = U - A = {1,3,5,7,9,}
b. 𝐴 ∩ 𝐴′
U = {1,2,3,4,5,6,7,8,9}
A = {2,4,6,8}
A′ ={1,3,5,7,9,}
A ∩ A′ = ∅
c. 𝐴 ∪ 𝐴′
U = {1,2,3,4,5,6,7,8,9}
A = {2,4,6,8}
A′ ={1,3,5,7,9,}
A∪ A′ = { 1,2,3,4,5,6,7,8,9} = U
d. (𝐴′)′
U = {1,2,3,4,5,6,7,8,9}
A = {2,4,6,8}
A′ = {1,3,5,7,9,}
(A′)′ = A = {2,4,6,}
e. 𝜙 \ 𝑈
U = {1,2,3,4,5,6,7,8,9}
𝜙 \ U = ∅
f. 𝜙′
U = {1,2,3,4,5,6,7,8,9}
𝜙′ = U
g. 𝑈′
U = {1,2,3,4,5,6,7,8,9}
U′ = ∅
8. Let 𝑈 = {1, 2, 3, 4, … … . ,10}, 𝐴 = {1, 3, 5, 7}, 𝐵 = {1, 2, 3, 4}.
Evaluate 𝑈\(𝐴∆𝐵).
Given Required
U = {1,2,3,4,...., 10} U\(A∆B) = ?
A ={1,3,5,7}
B = {1,2,3,4}
To determine U\(A∆B) first we have to find (A∆B) = (A - B) ∪ (B - A)
= {5,7} ∪ {2,4}
= {2,4,5,7}
U\(A∆B) = U - (A∆B) = {1,2,3,4,...., 10} - {2,4,5,7}
= {1,3,6,8,9,10}
9. Consider a universal set 𝑈 = {1, 2, 3, . . . , 14}, 𝐴 = {2, 3, 5, 7, 11}, 𝐵 = {2, 4, 8, 9, 10, 11}. Then, which one of the following is true?
A. (𝐴 ∪ 𝐵) ′ = {1, 4, 6, 12, 13, 14}
B. 𝐴 ∩ 𝐵 = 𝐴 ′ ∪ 𝐵 ′
C. 𝐴∆𝐵 = (𝐴 ∩ 𝐵) ′
D. 𝐴\𝐵 = {3, 5, 7}
E. None
Given
U = {1, 2, 3, . . . , 14}
A = {2, 3, 5, 7, 11}
B = {2, 4, 8, 9, 10, 11}
A. (A ∪ B) ′ = {1, 4, 6, 12, 13, 14}
A ∪ B = {2,3,4,5,7,8,9,10,11}
(A ∪ B) ′ = U - (A ∪ B) = {1,6,12,13,14}
Choice A is not true
B. A ∩ B = A ′ ∪ B ′
A ∩ B = {2,11}
A ′ = {1,4,6,8,9,10,12,13,14}
B ′ = {1,3,5,6,7,12,13,14}
A ′ ∪ B ′ = {1,3,4,5,6,7,8,9,10,12,13,14}
Choice B is not true
C. A∆B = (A ∩ B) ′
A∆B = (A - B) ∪ (B - A) = {3,5,7} ∪ {4,8,9,10}
= {3,4,5,7,8,9,10}
A ∩ B ={2,11}
(A ∩ B)′= U - A ∩ B = {1,2,3,....,14} - {2,11}
= { 1,3,4,5,6,7,8,9,10,12,13,14}
Choice C is not true
D. A\B = {3, 5, 7}
Choice D is true
So the answer is D
10. Let 𝐴 = {3, 7, 𝑎</span><span>2</span><span> } and 𝐵 = {2, 4, 𝑎 + 1, 𝑎 + 𝑏} be two sets and all the elements of the two sets are integers. If 𝐴 ∩ 𝐵 = {4, 7}, then find a and b. In addition, find 𝐴 ⋃ 𝐵.
11.In a survey of 200 students in Motta Secondary School, 90 students are members of Nature club, 31 students are members of Mini-media club, 21 students are members of both clubs. Answer the following questions.
a. How many students are members of either of the clubs?
Given Required
n(U) = 200 A ∪ B = ?
n(A) = numbers of nature = 90
n(B) = number of minimedia = 31
n(A ∩ B) = 21
(A ∪ B) = n(A) + n(B) - n(A ∩ B)
= 90 + 31 - 21
= 121 - 21
= 100
b. How many students are not members of either of the clubs?
Given Required
n(U) = 200 (n(A ∪ B))’ = ?
n(A) = numbers of nature = 90
n(B) = number of minimedia = 31
n(A ∩ B) = 21
(n(A ∪ B))’ = n(U) - n(A ∪ B)
= 200 - 100
= 100
c. How many students are only in Nature club?
Given Required
n(U) = 200 only nature club = ?
n(A) = numbers of nature = 90
n(B) = number of minimedia = 31
n(A ∩ B) = 21
only nature club = n(A) - n(A ∩ B)
= 90 - 21
= 69
12.A survey was conducted in a class of 100 children and it was found out that 45 of them like Mathematics whereas only 35 like Science and 10 students like both subjects. How many like neither of the subjects?
A) 70 B) 30 C) 100 D) 40
Given Required
n(U) = 100 (n(M ∪ S))’ = ?
n(M) = 45
n(S) = 35
n(M ∩ S) = 10
n(M ∪ S) = n(M) + n(S) - n(M ∩ S)
= 45 + 35 - 10
= 70
(n(M ∪ S))’ = n(U) - n(M ∪ S)
= 100 - 70
= 30
The answer is 30