Class-8 Set Operation
Some Facts About Cardinal Number of Set
Introduction to Set Operation
We have already learnt about the basic concept of sets, representing a set by roster or tabular form or by using set builder form, equal set, finite set, infinite sets, and empty set, cardinal number of a finite set, universal set , equivalent sets and sub sets in previous classes. Now we will study about operation on set.
- Union and intersection of sets
- Difference of two sets
- Complement of a set
- Overlapping and disjoint sets
- Basic result about set's cardinal number
Union of Sets
The union of two sets P and Q is the set consisting of all those elements which belong to either P or Q or both. It is written as (P ∪ Q).
Thus, P ∪ Q = {x|x ∈ P or x ∈ Q}
Example. P = {1, 2, 3, 4, 5}
Q = {2, 3, 8, 9}
Then, (P ∪ Q) = {1, 2, 3, 4, 5, 8, 9}
Intersection of Sets
The set consisting of all those elements which belong to both set P and Q and it is called intersection of two sets P and Q. It is written as P ∩ Q. It is read as P intersection Q.
Hence, P ∩ Q = { x| x ∈ P and x ∈ Q}
Example. If P = {a, e, f, g} and Q = {e, f, h, i)
Then, P ∩ Q = {e, f}
Difference of Two Sets
Difference of two set consist of all those elements which belong to one set but do not belong to other. It is written as P − Q.
Hence, P − Q = {x|x ∈ P and x ∉ Q}
Similarly, for Q − P = {x|x ∈ Q and x ∉ P}
So, P − Q is the set consisting of element of P only and Q − P is the set of consisting of elements of Q only.
Example. If P = {0, 4, 5, 6, 7} and Q = {4, 5, 9, 10, 11}
P − Q = {0, 6, 7}, here 0, 6, 7 are elements present in set P only and not present in Q.
Q − p = {9, 10, 11}, here 9, 10, 11 are elements present only in set Q and not present in set P
Complement of Set
If P is any set and U is the universal set then the complement of P is the set consisting of all the elements of universal set and do not belong to set P. It is written as P'.
Thus P' = {x|x ∈ ∪ and x ∉ P}
Example. If P = {1, 2, 3, 6, 9} and U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
P' = {4, 5, 7, 8, 10}
Overlapping Set
Two set said to be overlapping set, if they have at least one element is common. In other way, two sets are overlapping set if P ∩ Q ≠ ∮
Example. Set P = {2, 4, 5, 8, 9} and Set Q = {1, 4, 5, 8, 10, 11}
Here, set P and set Q are overlapping sets because they have elements 4, 5, and 8 are common. So, here P ∩ Q = {4, 5, 8} ≠ ∮
Disjoint Set
If two sets have no common element, then it is said to be disjoint set. Two set P and Q are disjoint if P ∩ Q = ∮
Example. The set P = {February, April, may, June} and Set Q = {Sunday, Monday, Tuesday, Wednesday}
Here, these two sets are disjoint set because they have no common element.
Properties of Set Operation
- Identity Property
- Idempotent Property
- Complement Property
- Commutative Property
- Associative property
- Distributive property
Identity Property
If P is any set, then
P ∪ ∅ = P
P ∩ U = P
Idempotent Property
If P is any set, then
P ∩ P = P
P ∪ P = P
Complement Property
P ∪ P' = U
P ∩ P' = ∅
Commutative Property
Intersection and union of sets follow the commutative property.
A ∩ B = B ∩ A
A ∪ B = B ∪ A
Associative Property
Intersection and union of sets follow the associative property.
(A ∩ B) ∩ C = A ∩ (B ∩ C)
(A ∪ B) ∪ C = A ∪ (B ∪ C)
Distributive Property
Intersection and union of sets follow the distributive property.
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
Let's see some solved examples to understand the properties better.
Example 1.If P = {4, 6, 8, 11, 13, 16, 20 ,22, 25}
Q= { 7, 9 , 11,16, 18 , 22,25}
R = { 0, 5, 9, 18,20, 25}
Find i) P ∪ Q
ii) P ∪ R
iii) P ∩ Q
iv) Q ∩ R
v) Q ∪ R
VI) P ∩ R
Solution. i) P ∪ Q ={4, 6, 8, 11, 13, 16, 20, 22, 25, 7, 9, 18, 20}
ii) P ∪ R = {4, 6, 8, 11, 13, 16, 20, 22, 25, 0, 5, 9, 18)
iii) P ∩ Q = {11, 16, 22, 25}
iv) Q ∩ R = {9, 18, 25}
v) Q ∪ R = {7, 9, 11, 16, 18, 22, 25, 0, 5, 20}
vi) P ∩ R = {20, 25}
Example 2. If universal set U = {10, 20, 30, 40, 50, 60, 70, 80, 90} and set A = { 40, 60, 80, 90}, then find A'.
Solution. Complement of A = A' ={10, 20, 30, 50, 70}
Example 3. If Set A = {BHUBANESWAR} and Set B = {AHMEDABAD},
Find i) A ∪ B ii) A ∩ B iii) A − B IV) B − A
Solution Roster form of set A = {B, H, U, A, N, E, S, W, R}
Roster form of set B = {A, H, M, E, D, B}
i) A ∪ B = {B, H, U, A, N, E, S, W, R, M, D}
ii) A ∩ B = {B, H, A, E}
iii) A − B = {U, N, S, W, R}
iv) B − A = {M, D}
Some Facts About Cardinal Number of Set
If P and Q are two finite sets, then
- n(P ∪ Q) = n(P) + n(Q) − n(P ∩ Q)
- n(P − Q ) = n(P ∪ Q) − n(Q) = n(P) − n(P ∩ Q)
- n(Q − P ) = n(P ∪ Q) − n(P)= n(Q) − n(P ∩ Q)
- n(P ∪ Q) = n(P − Q) + n(Q − P) + n( P ∩ Q)
Example 1. If n(P) = 18, n(Q) = 14, and n(P ∪ Q) = 20 ,Then find n(P ∩ Q)
Solution. n(P ∪ Q) = n(P) + n(P ) − n(P ∩ Q)
=> 20 = 18 + 14 − n(P ∩ Q)
=> n(P ∩ Q) = 32 − 20 = 12
Example 2. n(U) = 30, n(P') =12, n(Q) =15, n(P ∩ Q) = 10, Find
i) n(P')
ii) n(P U Q)
Solution.
i) n(Q') = n(U) − n(Q)
= 30 − 15 = 15
ii) n(P ∪ Q) = n(P) + n(Q) − n( P ∩ Q)
here, we first find out n(P) by using formulae
n(P') = n(U) − n(P)
=> 12 = 30 − n(P)
=> n(P) = 18
Then, we put the value of n(P) below equation
n(P ∪ Q) = 18 + 15 − 10=23
Example 3. If n(U)= 50, n(P)= 30, n(Q)= 10 and n{(P ∪ Q)'} = 18, then find
i) n(P ∪ Q )
ii) n( P ∩ Q)
iii) n(P − Q)
Solution.
i) We know n(P ∪ Q) + n{(P ∪ Q)'} = n(U)
=> n(P ∪ Q) + 18 = 50
=> n(P ∪ Q) = 50 − 18 = 32
ii) n(P ∪ Q) = n(P) + n(Q) − n(P ∩ Q)
=> 32 = 30 + 10 − n(P ∩ Q)
=> n(P ∩ Q) = 40 − 32 = 8
iii) n(P − Q ) = n(P) − n(P ∩ Q)
=> n(P − Q ) = 30 &minus 8 = 22
Class-8 Set Operation Worksheets
Answer Sheet
Set-Operation-AnswerDownload the pdf
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