In this chapter we will learn some math operations performed on two or more sets.
You can understand set operation just like operation on two or more numbers by addition/subtraction/division or multiplication.
But as we are doing operation on sets we have to follow some different rules and techniques.
In order to understand the chapter you should have basic understanding of sets and its properties which have been already discussed in previous chapters.
Operations on Sets in Math
Given below are some basic operations used in two or more sets:
(a) Union of sets
(b) Intersection of sets
(c) Difference of sets
(d) Complement of set
Let’s discuss all the above operations in brief.
Union of Sets
Let A & B are the two given sets. The union of set A & B will contain:
⟹ Unique elements of set A
⟹ Unique elements of set B
⟹ Common elements of set A & B
The notation of union of set is ” ∪ “.
Hence, the union of set A & B is expressed as ” A ∪ B “
Note:
Do not get confused Union symbol “∪” with the Universal set symbol U.
While both looks the same, they have completely different meaning, while the Union is a set operation and the Universal set is a collection of elements.
Examples of Union operation
Example 01
Given below is two sets A & B.
A = { 1 , 2, 3 }
B = { 3, 7, 8, 9 }
Find union of set A & B (i.e. A ∪ B )
Solution
We know that union of two sets results in combining all unique elements of given set and also the elements which is common.
A ∪ B = { 1, 2, 3, 7, 8, 9 }
Example 02
Given below are two sets P & Q
P = { x : x=2n+1 & x 𝜖 N }
Q = { x : x=2n & x 𝜖 N }
Find the union of sets P & Q (i.e. P ∪ Q )
Solution
P = { x : x=2n+1 & x 𝜖 N }
Given above is the set name P.
Set P contains element x such that:
⟹ x = 2n+1
⟹ x belongs to natural number N
After Putting values we will get;
P = {3, 5, 7, 9, 11 . . . }
Similarly after putting value we will get set Q;
Q = { 2, 4, 6, 8, 10 . . . }
The union of set P & Q is given as:
P ∪ Q = {2, 3, 4, 5, 6, 7 . . . . }
Intersection of sets
If A & B are two sets then intersection of set A & B will results in common element present in both set A & B.
Notation of set intersection is ” \mathtt{\cap } “
Hence, the intersection of set A & B is expressed as \mathtt{A\ \cap \ B} .
Examples of Intersection operation
Example 01
Given below are two sets A & B;
A = { 3, 16, 18, 23 }
B = { 16, 29, 23, 45 }
Find the intersection of set A & B (i.e. A ∩ B)
Solution
We know that in intersection of sets we take the common element present between the given set.
Observe that out of given elements, digits 16 & 23 is common between the sets A & B.
Hence, A ∩ B = { 16, 23 }
Example 02
Given below are the sets P & Q;
P = { 17, 19, 25 }
Q = { 13, 21, 36 }
Find the intersection of set P & Q (i.e. P ∩ Q)
Solution
Observe that there is no common element between set P & Q.
Hence, the intersection of set P & Q results in empty set.
(P ∩ Q) = 𝜙
Difference of Sets
If A & B are two given sets then;
A – B ⟹ results in removal of all common elements from set A.
B – A ⟹ results in removal of all common elements from set B
Let us understand the concept with example;
Examples of Difference of Sets
Example 01
Given below are two sets A & B.
A = { 31, 45, 63, 72, 89, 100 }
B = ( 39, 47, 63, 77, 89, 100, 110 }
Find set A – B and B – A
Solution
A – B can be found by removing common elements from set A
A – B = { 31, 45, 72}
B – A can be found by removing the common elements from B
B – A = { 39, 47, 77, 110 }
Example 02
Given below are two sets P & Q
P = { 3, 6, 9 }
Q = { 4, 6, 8 }
Find P – Q and Q – P
Solution
(i) P – Q can be found by removing common elements of P & Q from P
P – Q = { 3, 9 }
(ii) Q – P can be found by removing the common elements from set Q.
Q – P = { 4, 8 }
Complement of Set
If A is the given set and U is the universal set, then complement of set is calculated by U – A.
Hence Complement of Set A is all the elements which are not in set A.
The complement of set A is represented as A’ (read as ” A dash ” ) or \mathtt{A^{C}}
Let us see some examples for further understanding.
Examples of Complement of Set
Example 01
Given below is Universal set U and set A.
U = { 2, 5, 7, 9, 12, 15, 17 }
A = { 2, 7, 17 }
Find the complement of set A
Solution
Complement of A is the set with elements not present in A.
Its calculated using below expression.
A’ = U – A
A’ = { 2, 5, 7, 9, 12, 15, 17 } – { 2, 7, 17 }
A’ = { 5, 9, 12, 15 }
Example 02
Given below are universal set U and set Q.
U = { 5, 10, 15, 20, 25, 30}
Q = { 5, 15, 25, 30 }
Find the complement of set Q (or Q’)
Solution
Complement of Q can be found by removing elements of Q from universal set U
Q’ = U – Q
Q’ = { 5, 10, 15, 20, 25, 30} – { 5, 15, 25, 30 }
Q’ = { 10, 20 }