Chapter#01: Sets
Teacher Name: Ms. Hafsa Bukhari
Campus: Korangi
Class: VIII
Subject: Mathematics
Student Learning Outcomes
S.L.O Numbers
Objectives
1.1 Sets
1.1.1
Recall Sets and its kinds
1.1.2
Comprehend the concept of proper and improper subset
1.1.3
Find the Power set of a given set
1.1.4
Recall the concept of Union, Intersection, Complement and Difference of two
or more sets
1.2 Laws on Set
1.2.1
Verify Commutative, Associative, Distributive properties and illustrate through
Venn Diagrams
1.2.2
Verify De-Morgan’s Law when two sets are subsets of universal set.
S.L.O: 1.1.1 Recall Sets and its kinds
What is Set?
Definition: A set is a collection of well-defined distinct objects. The objects that
make up a set (also known as the set's elements or members) can be anything:
numbers, people, letters of the alphabet, other sets, and so on.
Examples
Types of Sets
From book page#9
S.L.O 1.1.2 Comprehend the concept of proper and improper subset
An Improper subset is a subset containing every
element of the original set.
A proper subset contains some but not all of the
elements of the original set.
For example; consider a set {1,2,3,4,5}.
{1,2,4} and {1} are the two proper subsets
while
{1,2,3,4,5} is an improper subset.
S.L.O: 1.1.3 Find the Power Set of the given set.
Power Set: A set which contains every possible subset of a Set is
called the power set of a given set.
If S is a finite set with |S| = n elements, then the number of subsets
of S is
|P(S)| = 2n
If S is the set {x, y, z}, then the subsets of S are
{}
{x}
{y}
{z}
{x, y}
{x, z}
{y, z}
{x, y, z}
and hence the power set of S is {{}, {x}, {y}, {z}, {x, y}, {x, z}, {y, z}, {x, y, z}}
Exercise
(i) N={a, e, i, o, u}
(ii) A = { 6,7,8}
(iii) S = { p,q,r,s}
(iv) R = { set of whole numbers less than 4}
(v) A = { x: x ∈ 𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟𝑠, 10 < 𝑥 < 18}
Exercise of Power Set
Solution (i) N={a, e, i, o, u}
There will be total of 2𝑛 = 25 = 2 × 2 × 2 × 2 × 2 = 32 subsets in the power set
of N
P(N) = {{ }, {a}, {e}, {i}, {o}, {u},
{a,e},{a,i},{a,o},{a,u},{e,i},{e,o},{e,u},{i,o},{i,u},{o,u},
{a,e,i,},{a,e,o},{a, e, u}, {a, i, o}, {a, i, u}, {a, o, u}, {e, i, o}, {e, i, u}, {e, o, u},
{i, o, u},
{a, e, i, o}, {a, e, i, u}, {a, e, o, u}, {a, i, o, u}, {e, i, o, u},
{a, e, i, o, u}}
Solution (ii) A = { 6,7,8}
P(A) = { { }, {6}, {7}, {8}, {6,7}, {6,8}, {7,8}, {6, 7, 8} }
Solution (iii) S = { p,q,r,s}
P(S) = { { }, {p}, {q}, {r}, {s}, {p,q}, {p,r}, {p,s}, {q,r}, {q,s}, {r,s}, {p,q,r},
{p,r,s}, {q,r,s}, {p,q,s}, {p,q,r,s} }
Solution (iv) R = { set of whole numbers less than 4}
R = {0,1,2,3}
P(R) = { { }, {0}, {1}, {2}, {3}, {0,1}, {0,2}, {0,3}, {1,2}, {1,3}, {2,3}, {0,1,2},
{0,1,3}, {0,2,3}, {1,2,3}, {0,1,2,3} }
Solution (v)
A = { x: x ∈ 𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟𝑠, 10 < 𝑥 < 18}
The Set builder form of A is { 12, 14, 16}
There will be 2𝑛 = 23 = 2 × 2 × 2 = 8 subsets in the power set of set A.
P(A) = {{ }, {12}, {14}, {16}, {12, 14}, {12, 16}, {14, 16}, {12, 14, 16}}
S.L.O 1.1.4: Union, Intersection, Complement and Difference of two or
more sets.
• The union (∪) of two sets contains all the elements contained in either set (or
both sets).
• The intersection (⋂) of two sets contains only the elements that are in both
sets.
• The complement (A’or Ac) of a set A contains everything that is not in the
set A.
Questions: Union, Intersection, Complement and Difference of two or
more sets.
Let A, B and C be three sets such that:
Set A = {2, 4, 6, 8, 10, 12}, set B = {3, 6, 9, 12, 15} and set
C = {1, 4, 7, 10, 13, 16}.
(i) A ∪ B
A ∪ B = {2, 4, 6, 8, 10, 12}∪{3, 6, 9, 12, 15}
A ∪ B = {2, 3, 4, 6, 8, 9, 10, 12, 15}
A
B
2, 4 A
8, 10,
12
6
12
3, 9
15
Let A, B and C be three sets such that:
Set A = {2, 4, 6, 8, 10, 12}, set B = {3, 6, 9, 12, 15} and set
C = {1, 4, 7, 10, 13, 16}.
Find:
(ii) A ∩ B
A ∩ B = {2, 4, 6, 8, 10, 12}∩{3, 6, 9, 12, 15}
A ∩ B = { 6, 12 }
A
B
6
12
Let A, B and C be three sets such that:
Set A = {2, 4, 6, 8, 10, 12}, set B = {3, 6, 9, 12, 15} and set
C = {1, 4, 7, 10, 13, 16}.
(iii) B ∩ A
B ∩ A = {3, 6, 9, 12, 15} ∩ {2, 4, 6, 8, 10, 12},
B ∩ A = {6, 12}
B
A
6
12
Let A, B and C be three sets such that:
Set A = {2, 4, 6, 8, 10, 12}, set B = {3, 6, 9, 12, 15} and set
C = {1, 4, 7, 10, 13, 16}.
(iv) B ∪ A
B ∪ A = {3, 6, 9, 12, 15} ∪ {2, 4, 6, 8, 10, 12}
B ∪ A = {2, 3, 4, 6, 8, 9, 10, 12, 15}
Let A, B and C be three sets such that:
Set A = {2, 4, 6, 8, 10, 12}, set B = {3, 6, 9, 12, 15} and set
C = {1, 4, 7, 10, 13, 16}.
(v) B ∪ C
B ∪ C = {3, 6, 9, 12, 15} ∪ {1, 4, 7, 10, 13, 16}.
B ∪ C = {1, 3, 4, 6, 7, 9, 10, 12, 13, 15, 16}
B
3, 6, 9
12, 15
C
1, 4, 7,
10, 13,
16
Let A, B and C be three sets such that:
Set A = {2, 4, 6, 8, 10, 12}, set B = {3, 6, 9, 12, 15} and set
C = {1, 4, 7, 10, 13, 16}.
(vi) Is A ∪ B = B ∪ A?
Yes
Let A, B and C be three sets such that:
Set A = {2, 4, 6, 8, 10, 12}, set B = {3, 6, 9, 12, 15} and
C = {1, 4, 7, 10, 13, 16}.
(vii) Is B ∩ C = B ∪ C?
B ∩ C = {3, 6, 9, 12, 15} ∩ {1, 4, 7, 10, 13, 16}.
B∩C={ }
B ∪ C = {1, 3, 4, 6, 7, 9, 10, 12, 13, 15, 16}
B∩C≠B∪C
Set A = {2, 4, 6, 8, 10, 12}, set B = {3, 6, 9, 12, 15} and set C = {1, 4, 7, 10, 13, 16}
(viii) (A ∪ B) ∩ (A ∪ C)
Solution:
(A ∪ B)
A ∪ B = {2, 4, 6, 8, 10, 12} ∪ {3, 6, 9, 12, 15}
A ∪ B = { 2, 3, 4, 6, 8, 10, 12, 15}
A
B
(A ∪ C)
A ∪ C = {2, 4, 6, 8, 10, 12} ∪ {1, 4, 7, 10, 13, 16}
2, 8
6,
12
A ∪ C = {1, 2, 4, 6, 7, 8, 10, 12, 13, 16}
4, 10
(A ∪ B) ∩ (A ∪ C)
1, 7, 13,
16
(A ∪ B) ∩ (A ∪ C) = { 2, 3, 4, 6, 8, 10, 12, 15} ∩ {1, 2, 4, 6, 7, 8, 10, 12, 13, 16}
C
(A ∪ B) ∩ (A ∪ C) = { 2, 4, 6, 8, 10, 12}
3, 9,
15
Exercise of Complement and Difference
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4} and B = {2, 4, 6, 8}.
(i) Find A’
A’ = U-A
A’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4}
A’ = { 5,6,7,8,9 }
U
A
5
8
6
7
1,2,3,4
9
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4} and B = {2, 4, 6, 8}.
(ii) Find B’
B’ = U-B
B’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 4, 6, 8}
B’ = {1,3,5,7,9}
U
B
5
7
3
1
2,4,6,8
9
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4} and B = {2, 4, 6, 8}.
(iii) (A ∩ B)’
(A ∩ B)
A ∩ B = {1, 2, 3, 4} ∩ {2, 4, 6, 8}
A ∩ B = {2, 4}
(A ∩ B)’ = U - (A ∩ B)
(A ∩ B)’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 4}
(A ∩ B)’ = {1, 3, 5, 6, 7, 8, 9}
A
5
1,
3
7
U
B
4
g
2
6,
8
9
•
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4} and B = {2, 4, 6, 8}.
•
(iv) A' ∪ B’
A’ = U-A
A’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4}
A’ = { 5,6,7,8,9 }
B’ = U-B
B’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 4, 6, 8}
B’ = {1,3,5,7,9}
A' ∪ B’ = { 5,6,7,8,9 } ∪ {1,3,5,7,9}
A' ∪ B’ = {1, 3, 5, 6, 7, 8, 9}
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4} and B = {2, 4, 6, 8}.
(v) (A ∪ B)’
(A ∪ B)
A ∪ B = {1, 2, 3, 4} ∪ {2, 4, 6, 8}
A ∪ B = {1,2,3,4,6,8}
(A ∪ B)’ = U - (A ∪ B)
(A ∪ B)’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4, 6, 8}
(A ∪ B)’ = {5, 7, 9}
A
5
1,
3
7
U
B
4
g
2
6,
8
9
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4} and B = {2, 4, 6, 8}.
(vi) A' ∩ B’
A’ = U-A
A’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4}
A’ = { 5,6,7,8,9 }
B’ = U-B
B’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 4, 6, 8}
B’ = {1,3,5,7,9}
A' ∩ B’ = { 5,6,7,8,9 } ∩ {1,3,5,7,9}
A' ∩ B’ = {5, 7, 9}
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4} and B = {2, 4, 6, 8}.
(vii) A-B
A-B = {1, 2, 3, 4}- {2, 4, 6, 8}.
A-B = { 1, 3 }
A
1,
3
B
4
2
6,
8
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4} and B = {2, 4, 6, 8}.
(viii) B-A
B-A = {2, 4, 6, 8} - {1, 2, 3, 4}
B-A = { 6, 8 }
A
1,
3
B
4
2
6,
8
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4} and B = {2, 4, 6, 8}.
(ix) (A-B)’
A-B = {1, 2, 3, 4}- {2, 4, 6, 8}.
A-B = { 1, 3 }
(A-B)’ = U - (A-B)
(A-B)’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - { 1, 3 }
A
5
1,
3
(A-B)’ = {2, 4, 5, 6, 7, 8, 9}
7
U
B
4
g
2
6,
8
9
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4} and B = {2, 4, 6, 8}.
(x) (B-A)’
B-A = {2, 4, 6, 8} - {1, 2, 3, 4}
B-A = { 6, 8 }
(B-A)’ = U- (B-A)
A
5
1,
(B-A)’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - { 6, 8 }
3
7
(B-A)’ = {1, 2, 3, 4, 5, 7, 9}
1,
3
U
B
4
g
2
6,
8
9
1.2 Laws on Sets
 Commutative law
 Associative law
 Distributive law
 De Morgan’s law
Commutative Law:
This Property states that for Union and Intersection both, the order of sets
in which we do operation does not change the result.
General Properties: A ∪ B = B ∪ A
A ∩ B = B ∩ A.
Solution:
1) Let A = {x : x is a whole number between 4 and 8} and B =
{x : x is an even natural number less than 10}. Verify Commutative
Property with respect to Union and Intersection. Also draw Venn
diagram.
A = { 5, 6, 7 } and B = { 2, 4, 6, 8 }
Commutative Law with respect to union
A∪B = {5, 6, 7} ∪ {2, 4, 6, 8}
A∪B = {2, 4, 5, 6, 7, 8}
and
Commutative law with respect to intersection
A∩B = {5, 6, 7}∩{2, 4, 6, 8}
and
A∩B = {6}
A
B
g
B∪A = {2, 4, 6, 8} ∪ {5, 6, 7}
B∪A = {2, 4, 5, 6, 7, 8}
B ∩ A = {2, 4, 6, 8} ∩ {5, 6, 7}
B ∩ A = {6}
U
A
B
g
U
Associative Property:
This Property states that for Union and Intersection both, that how the sets are
grouped does not change the result.
General Property: (A ∪ B) ∪ C = A ∪ (B ∪ C)
and
(A ∩ B) ∩ C = A ∩ (B ∩ C)
Solution#1 Let A = {a, n, t}, B = {t, a, p}, and C = {s, a, p}.
Associative Property of Union: (A ∪ B) ∪ C = A ∪ (B ∪ C)
(A ∪ B) ∪ C = ({a, n, t} ∪ {t, a, p}) ∪ {s, a, p}.
(A ∪ B) ∪ C = {a, n, p, t} ∪ {s, a, p}.
(A ∪ B) ∪ C = {a, n, p, s, t}
Left Hand Side
A ∪ (B ∪ C) = {a, n, t} ∪ ({t, a, p} ∪ {s, a, p}).
A ∪ (B ∪ C) = {a, n, t} ∪ {a, p, s, t}.
A ∪ (B ∪ C) = {a, n, p, s, t}
Right Hand Side
LHS = RHS
Associative Property of Intersection: (A∩B) ∩ C = A ∩ (B ∩ C)
(A ∩ B) ∩ C = ({a, n, t} ∩ {t, a, p}) ∩ {s, a, p}.
(A ∩ B) ∩ C = {a, t} ∩ {s, a, p}.
(A ∩ B) ∩ C = {a}
Left Hand Side
A ∩ (B ∩ C) = {a, n, t} ∩ ({t, a, p} ∩ {s, a, p}).
A ∩ (B ∩ C) = {a, n, t} ∩ {a, p}.
A ∩ (B ∩ C) = {a}
Right Hand Side
LHS = RHS
Solution#2 Let X = {1, 2, 3, 4, 5, 6}, Y = { 4, 5, 6, 7}, and
Z = { 2, 3, 6, 7}
Associative Property of Union: (A ∪ B) ∪ C = A ∪ (B ∪ C)
(A ∪ B) ∪ C = ({1, 2, 3, 4, 5, 6} ∪{ 4, 5, 6, 7}) ∪ { 2, 3, 6, 7}
(A ∪ B) ∪ C = {1, 2, 3, 4, 5, 6, 7} ∪ { 2, 3, 6, 7}
(A ∪ B) ∪ C = {1, 2, 3, 4, 5, 6, 7}
Left Hand Side
A ∪ (B ∪ C) = {1, 2, 3, 4, 5, 6} ∪ ({ 4, 5, 6, 7} ∪ { 2, 3, 6, 7})
A ∪ (B ∪ C) = {1, 2, 3, 4, 5, 6, 7} ∪ { 2, 3, 4, 5, 6, 7}
A ∪ (B ∪ C) = {1, 2, 3, 4, 5, 6, 7}
Right Hand Side
Solution#2 Let X = {1, 2, 3, 4, 5, 6}, Y = { 4, 5, 6, 7}, and
Z = { 2, 3, 6, 7}
Associative Property of Union: (A∩B) ∩ C = A ∩ (B ∩ C)
(X ∩ Y) ∩ Z = ({1, 2, 3, 4, 5, 6} ∩{ 4, 5, 6, 7}) ∩ { 2, 3, 6, 7}
(X ∩ Y) ∩ Z = {4, 5, 6} ∩ { 2, 3, 6, 7}
(X ∩ Y) ∩ Z = {6}
Left Hand Side
X ∩ (Y ∩ Z) = {1, 2, 3, 4, 5, 6} ∩ ({ 4, 5, 6, 7} ∩ { 2, 3, 6, 7})
X ∩ (Y ∩ Z) = {1, 2, 3, 4, 5, 6} ∩ {6, 7}
X ∩ (Y ∩ Z) = {6}
Right Hand Side
Distributive Property
The distributive property of union over intersection and the distributive
property of intersection over union show two ways of finding results for
certain problems mixing the set operations of union and intersection.
General Property: A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) and
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
Solution#1 Let A = {a, n, t}, B = {t, a, p}, and C = {s, a, p}.
Distributive Property of Union over intersection:
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
A ∪ (B ∩ C) = {a, n, t} ∪ ({t, a, p} ∩ {s, a, p}).
A ∪ (B ∩ C) = {a, n, t} ∪ {a, p}.
A ∪ (B ∩ C) = {a, n, p, t}
Left Hand Side
(A ∪ B) ∩ (A ∪ C) = ({a, n, t} ∪ {t, a, p}) ∩ ({a, n, t} ∪{s, a, p}).
(A ∪ B) ∩ (A ∪ C) = {a, n, p, t} ∩ {a, n, p, s, t}.
(A ∪ B) ∩ (A ∪ C) = {a, n, p, t}
Right Hand Side
LHS = RHS
Distributive Property of Intersection over union: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩C)
A ∩ (B ∪ C) = {a, n, t} ∩ ({t, a, p} ∪ {s, a, p}).
A ∩ (B ∪ C) = {a, n, t} ∩ {a, p, s, t}
A ∩ (B ∪ C) = { a, t }
Left Hand Side
(A ∩ B) ∪ (A ∩C) = ({a, n, t} ∩ {t, a, p}) ∪ ({a, n, t} ∩{s, a, p}).
(A ∩ B) ∪ (A ∩C) = {a, t} ∪ {a}
(A ∩ B) ∪ (A ∩C) = {a, t}
Right Hand Side
LHS = RHS
Solution#2 Let X = {1, 2, 3, 4, 5, 6}, Y = { 4, 5, 6, 7}, and
Z = { 2, 3, 6, 7}
Distributive Property of Intersection over union: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩C)
X ∩ (Y ∪ Z) = {1, 2, 3, 4, 5, 6} ∩ ({ 4, 5, 6, 7} ∪ { 2, 3, 6, 7})
X ∩ (Y ∪ Z) = {1, 2, 3, 4, 5, 6} ∩ {2, 3, 4, 5, 6, 7}
X ∩ (Y ∪ Z) = {2, 3, 4, 5, 6}
Left Hand Side
(X ∩ Y) ∪ (A ∩Z)=({1, 2, 3, 4, 5, 6}∩{ 4, 5, 6, 7}) ∪ ({1, 2, 3, 4, 5, 6}∩{ 2, 3, 6, 7})
(X ∩ Y) ∪ (A ∩Z) = {4, 5, 6} ∪ { 2, 3, 6}
(X ∩ Y) ∪ (A ∩Z) = {2, 3, 4, 5, 6}
Right Hand Side
Solution#2 Let A = {1, 2, 3, 4, 5, 6}, B = { 4, 5, 6, 7}, and
C = { 2, 3, 6, 7}
Distributive Property of Union over intersection: A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪C)
A ∪ (B ∩ C) = {1, 2, 3, 4, 5, 6}∪ ({ 4, 5, 6, 7} ∩ { 2, 3, 6, 7})
A ∪ (B ∩ C) = {1, 2, 3, 4, 5, 6}∪{6, 7}
A ∪ (B ∩ C) = {1, 2, 3, 4, 5, 6, 7}
Left Hand Side
(A ∪ B) ∩ (A ∪C) = ({1, 2, 3, 4, 5, 6}∪ { 4, 5, 6, 7}) ∩ ({1, 2, 3, 4, 5, 6} ∪ { 2, 3, 6, 7})
(A ∪ B) ∩ (A ∪ C) = {1, 2, 3, 4, 5, 6, 7} ∩ {1, 2, 3, 4, 5, 6, 7}
(A ∪ B) ∩ (A ∪ C) = {1, 2, 3, 4, 5, 6, 7}
Right Hand Side
S.L.O 1.2.2 Verify De-Morgan’s Law when two sets are
subsets of universal set.
De Morgan’s Laws
The complement of the union of two sets is equal to the intersection of their
complements and
the complement of the intersection of two sets is equal to the union of their
complements. These are called De Morgan's laws.
For any two finite sets A and B;
• (i) (A U B)' = A' ∩ B' (which is a De Morgan's law of union).
• (ii) (A ∩ B)' = A' U B' (which is a De Morgan's law of intersection).
De Morgan’s Law
Q#1) If U = {j, k, l, m, n}, X = {j, k, m} and Y = {k, m, n}.
Proof De Morgan's law: (X ∩ Y)' = X' U Y’.
Solution
X ∩ Y= {j, k, m} ∩ {k, m, n}.
X ∩ Y= { k,m }
(X ∩ Y)’ = U- (X ∩ Y)
= {j, k, l, m, n}-{ k,m }
= {j, l, n}
LHS
X’ = U-X
= {j, k, l, m, n}-{j,k,m}
= {l,n}
Y’= {j, k, l, m, n}- {k,m,n}
={j,l}
X' U Y’= {l,n} U {j,l}
= { j,l,n}
RHS
2) Let U = {1, 2, 3, 4, 5, 6, 7, 8}, P = {4, 5, 6} and Q = {5, 6, 8}.
Show that (P ∪ Q)' = P' ∩ Q'.
Solution
P ∪ Q = {4, 5, 6} ∪ {5, 6, 8}.
P ∪ Q = {4, 5,6, 8}
(P ∪ Q)’ = U - (P ∪ Q)
(P ∪ Q)’ = {1, 2, 3, 4, 5, 6, 7, 8} - {4, 5,6, 8}
(P ∪ Q)’ = { 1, 2, 3, 7}
P’ = U-P
P’ = {1, 2, 3, 4, 5, 6, 7, 8}- {4, 5, 6}
P’ = {1, 2, 3, 7, 8}
Q’ = U-Q
Q’ = {1, 2, 3, 4, 5, 6, 7, 8} – {5, 6, 8}
Q’ = {1, 2, 3, 4, 7}
P' ∩ Q’ = {1, 2, 3, 7, 8} ∩ {1, 2, 3, 4, 7}
P' ∩ Q’ = {1, 2, 3, 7}
LHS
RHS