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Complement of a Set Class 11 Maths: Formula,Rules and Examples

October 13, 2026 11th Mathematics(English), JEE MAINS Maths(English) No Comments
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1 1.Complement of a Set Class 11 Maths: Formula,Rules and Examples
1.1 2.Properties of Complement Sets
1.2 3.Complement of a Set Class 11 Maths Examples
1.2.1 4.Practice Questions of Complement of a Set Class 11 Maths for Students
1.2.2 5.Frequently Asked Questions Related to Complement of a Set Class 11 Maths
1.2.3 Q:1.What is meant by the complementary set of the set?
1.2.4 Q:2.What is the De Morgan Rule about Complementary Set?
1.2.5 Q:3.What is the Complementary Set of a Subset of a Universal set?
1.2.6 5.छात्र-छात्राओं से आज का प्रश्न (Today’s Question to Students):

1.Complement of a Set Class 11 Maths: Formula,Rules and Examples

Learn about the Complement of a Set for Class 11 Maths.Master its definition,universal set relations,Venn diagrams,properties and solved examples step-by-step.

Also Read This Article:- Complement (set theory)

2.Properties of Complement Sets

\text{(1.) Complement Rules} \\\text{(i) } A \cup A' = U \\\text{Proof: For universal set } U \\A \subset U \text{ and } A' \subset U \\\therefore A \subset U \text{ and } A' \subset U \implies A \cup A' \subset U \quad \cdots{ (1)} \\\text{If } x \in U \text{ then} \\x \in U \implies x \in A \text{ or } x \in A' \\\implies x \in A \cup A' \\\therefore U \subset A \cup A' \quad \cdots{ (2)} \\\text{From (1) and (2) we get} \\A \cup A' = U \\\text{(ii) } A \cap A' = \phi \\\text{Proof: Let } x \in A \text{ then } \dots
Let x \in A \cap A' then
\Rightarrow x \in A \wedge x \in A'
\Rightarrow x \in A \wedge x \notin A
\Rightarrow x \in \phi
\therefore A \cap A' \subset \phi \quad \cdots{ (1)}
But \phi \subset A \cap A' \quad [\because \phi \text{ subset of each set}] \quad \cdots{ (2)}
From (1) and (2), we get
A \cap A' = \phi
(2.) De Morgan’s law
(i) (A \cup B)' = A' \cap B'
Proof: x \in (A \cup B)' \Rightarrow x \notin A \cup B
\Rightarrow x \notin A \wedge x \notin B
\Rightarrow x \in A' \wedge x \in B'
\Rightarrow x \in A' \cap B'
(A \cup B)' \subset A' \cap B' \quad \cdots{ (1)}
Conversely:
x \in A' \cap B' \Rightarrow x \in A' \wedge x \in B'
\Rightarrow x \notin A \wedge x \notin B
\Rightarrow x \notin A \cup B
\Rightarrow x \in (A \cup B)'
A' \cap B' \subset (A \cup B)' \quad \cdots{ (2)}
From (1) and (2), we get
A' \cap B' = (A \cup B)'
(ii) (A \cap B)' = A' \cup B'
Proof: x \in (A \cap B)' \Rightarrow x \notin A \cap B
\Rightarrow x \notin A \vee x \notin B
\Rightarrow x \in A' \vee x \in B'
\Rightarrow x \in A' \cup B'
(A \cap B)' \subset A' \cup B' \quad \cdots{ (1)}
Conversely: x \in A' \cup B'
\Rightarrow x \in A' \vee x \in B'
\Rightarrow x \notin A \vee x \notin B
\Rightarrow x \notin A \cap B
\Rightarrow x \in (A \cap B)'
\Rightarrow A' \cup B' \subset (A \cap B)' \quad \cdots{ (2)}
From (1) and (2), we get
A' \cup B' = (A \cap B)'

(3.) Rule of double complementation: (A’)’= A
Proof: Let x \in (A')' then
x \in (A')' \Rightarrow x \notin A'
\Rightarrow x \in A \quad [\because x \in A \vee x \in A']
\Rightarrow (A')' \subset A \quad \cdots{ (1)}
Again if x \in A then
x \in A \Rightarrow x \notin A' \quad [\because A \cap A' = \phi]
\Rightarrow x \in (A')'
\therefore A \subset (A')' \quad \cdots{ (2)}
From (1) and (2), we get
(A’)’ = A
(4.) Rule of empty set and universal set \phi' = U and U' = \phi
Proof: Let x \in \phi' then
x \in \phi' \Rightarrow x \in U \vee x \notin \phi
\Rightarrow x \in U \quad [\because \phi \text{ has not any element}]
\therefore \phi' \subset U \quad \cdots{ (1)}
Again if x \in U then
x \in U \Rightarrow x \notin \phi \quad [\because \phi \text{ has not any element}]
\Rightarrow x \in \phi'
\therefore U \subset \phi' \quad \cdots{ (2)}
From (1) and (2), we get
\phi' = U
(ii) U' = \phi
Proof: Let x \in U' then
x \in U' \Rightarrow x \in U \wedge x \notin U
\Rightarrow x \in \phi
\therefore U' \subset \phi \quad \cdots{ (1)}
But \phi \subset U' \quad [\because \phi \text{ is subset of each set}] \quad \cdots{ (2)}
From (1) and (2), we get
\Rightarrow U' = \phi

Also Read This Article:- Properties of Sets Class 11 with Proof

3.Complement of a Set Class 11 Maths Examples

Example: 1. Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}.
Find (i) A’ (ii) B’ (iii) (A \cup C)’ (iv) (A \cup B)’ (v) (A’)’ (vi) (B – C)’
Solution: (i) A = {1, 2, 3, 4},
\cup = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A’ = {5, 6, 7, 8, 9}
(ii) B = {2, 4, 6, 8}, \cup = {1, 2, 3, 4, 5, 6, 7, 8, 9}
B’ = {1, 3, 5, 7, 9}
(iii) A = {1, 2, 3, 4}, C = {3, 4, 5, 6}
\cup = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A \cup C = {1, 2, 3, 4, 5, 6}
(A \cup C)’ = {7, 8, 9}
(iv) A = {1, 2, 3, 4}, B = {2, 4, 6, 8}
\cup = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A \cup B = {1, 2, 3, 4, 6, 8}
(A \cup B)’ = {5, 7, 9}
(v) A = {1, 2, 3, 4}
\cup = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A’ = {5, 6, 7, 8, 9}
\Rightarrow (A’)’ = {1, 2, 3, 4}
(vi) B = {2, 4, 6, 8}, C = {3, 4, 5, 6}
\cup = {1, 2, 3, 4, 5, 6, 7, 8, 9}
B – C = {2, 8}
(B – C)’ = {1, 3, 4, 5, 6, 7, 9}
Example: 2. If \cup = {a, b, c, d, e, f, g, h}, find the complements of the following sets:
Example: 2 (i). A = {a, b, c}
Solution: A = {a, b, c}, \cup = {a, b, c, d, e, f, g, h}
A’ = {d, e, f, g, h}
Example: 2 (ii). B = {d, e, f, g}
Solution: B = {d, e, f, g}, \cup = {a, b, c, d, e, f, g, h}
B’ = {a, b, c, h}
Example: 2 (iii). C = {a, c, e, g}
Solution: C = {a, c, e, g}
\cup = {a, b, c, d, e, f, g, h}
C’ = {b, d, f, h}
Example: 2 (iv). D = {f, g, h, a}
Solution: D = {f, g, h, a}
\cup = {a, b, c, d, e, f, g, h}
D’ = {b, c, d, e}
Example: 3. Taking the set of natural numbers as the universal set, write down the complements of the following sets:
Example: 3 (i). {x : x is an even natural number}
Solution: Let A = {x : x is an even natural number}
A’ = {x : x is an odd natural number}
Example: 3 (ii). {x : x is an odd natural number}
Solution: Let A = {x : x is an odd natural number}
A’ = {x : x is an even natural number}
Example: 3 (iii). {x : x is a positive multiple of 3}
Solution: Let A = {x : x is a positive multiple of 3}
A’ = {x : x \in N and x is not a multiple of 3}
Example: 3 (iv). {x : x is a prime number}
Solution: Let A = {x : x is a prime number}
A’ = {x : x is a positive composite number or x = 1}
Example: 3 (v). {x : x is a natural number divisible by 3 and 5}
Solution: Let A = {x : x is a natural number divisible by 3 and 5}
A’ = {x : x is a positive integer which is not divisible by 3 or not divisible by 5}
Example: 3 (vi). {x : x is a perfect square}
Solution: Let A = {x : x is a perfect square}
A’ = {x : x \in N and x is not a perfect square}
Example: 3 (vii). {x : x is a perfect cube}
Solution: Let A = {x : x is a}
Example: 3 (viii) {x : x + 5 = 8}
Solution: Let A = {x : x + 5 = 8}
A’ = {x : x \in N and x \neq 3}
Example: 3 (ix) {x : 2x + 5 = 9}
Solution: Let A = {x : 2x + 5 = 9}
A’ = {x : x \in N and x \neq 2}
Example: 3 (x) \{x : x \geq 7 \}
Solution: Let A = \{x : x \geq 7 \}
A’ = {x : x \in N and x < 7}
Example: 3 (xi) {x : x \in N and 2x + 1 > 10}
Solution: Let A = {x : x \in N and 2x + 1 > 10}
2x + 1 > 10 \Rightarrow x > \frac{9}{2}
A’ = \{x : x \in N \text{ and } x \leq \frac{9}{2} \}
Example: 4. If \cup = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}, verify that:
Example: 4 (i) (A \cup B)’ = A’ \cap B’
Solution: (A \cup B)’ = A’ \cap B’
A = {2, 4, 6, 8}, B = {2, 3, 5, 7}
\cup = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A \cup B = {2, 3, 4, 5, 6, 7, 8}
(A \cup B)’ = {1, 9} \cdots (1)
A’ = {1, 3, 5, 7, 9}
B’ = {1, 4, 6, 8, 9}
A’ \cap B’ = {1, 9} \cdots (2)
From (1) and (2), we get
(A \cup B)’ = A’ \cap B’
Example: 4 (ii) (A \cap B)’ = A’ \cup B’
Solution: (A \cap B)’ = A’ \cup B’
A = {2, 4, 6, 8}, B = {2, 3, 5, 7}
\cup = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A \cap B = {2}
(A \cap B)’ = {1, 3, 4, 5, 6, 7, 8, 9} \cdots (1)
A’ = {1, 3, 5, 7, 9}
B’ = {1, 4, 6, 8, 9}
A’ \cup B’ = {1, 3, 4, 5, 6, 7, 8, 9} \cdots (2)
From (1) and (2), we get
(A \cap B)’ = A’ \cup B’
Example: 5. Draw appropriate Venn diagram for each of the following:
Example: 5 (i). (A \cup B)’
Solution: (A \cup B)’

Example: 5 (ii). A’ \cap B’
Solution: A’ \cap B’

Example: 5 (iii). (A \cap B)’
Solution: (A \cap B)’

Example: 5 (iv). A’ \cup B’
Solution: A’ \cup B’

Example: 6. Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60°, what is A’?
Solution: A’ = {Set of all equilateral triangles}
Example: 7. Fill in the blanks to make each of the following a true statement:
Example: 7 (i). A \cup A’ = \dots
Solution: A \cup A’ = U
Example: 7 (ii). \phi‘ \cap A = \dots
Solution: \phi‘ \cap A = A
Example: 7 (iii). A \cap A’ = \dots
Solution: A \cap A’ = \phi
Example: 7 (iv). U’ \cap A = \dots
Solution: U’ \cap A = \phi
Practice Questions of Complement of a Set Class 11 Maths for Students

4.Practice Questions of Complement of a Set Class 11 Maths for Students

Using properties of sets, show that:
(1.) A \cup (A \cap B) = A
(2.) A \cap (A \cup B) = A
By solving the above questions,you can understand the Complement of a Set Class 11 Maths: Formula,Rules and Examples well because the concept is well understood when you solve it practically.

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Also Read This Article:- Subsets in Class 11

5.Frequently Asked Questions Related to Complement of a Set Class 11 Maths

Q:1.What is meant by the complementary set of the set?

Ans:Let U be the universal set and A a subset of U.Then the complement of A is the set of all elements of U which are not the elements of A.Symbolically,we write A’ to denote the complement of A with respect to U, Thus A' = {x : x \in U \text{ and } x \notin A}, obviously A' = U - A 

Q:2.What is the De Morgan Rule about Complementary Set?

Ans:The complement of the union of two sets is the intersection of their complements and the complement of the intersection of two sets is the union of their complements. These are called De Morgan’s laws.These are named after the mathematician De Morgan.

Q:3.What is the Complementary Set of a Subset of a Universal set?

Ans:If A is a subset of the universal set U,then its complement A’ is also a subset of U.
By answering the above questions,you can know about the primary terms of Complement of a Set Class 11 Maths: Formula,Rules and Examples.

5.छात्र-छात्राओं से आज का प्रश्न (Today’s Question to Students):

🎯 विनर्स कॉर्नर:क्या आप इस सवाल का सही जवाब दे सकते हैं?अपने नाम के साथ नीचे कमेंट करें!सही जवाब देने वाले Top Students के नाम हमारी आगे post/is article ke update में Photo ya Special Mention के साथ publish की जाएगी।अपना जवाब अभी दर्ज करें!👇
छात्र-छात्राओं से आज का सवाल (Today’s Question to students)
यदि किसी संख्या के \frac{1}{4} का \frac{2}{5} का \frac{1}{3} का मान 40 है तो वह संख्या क्या है?
(If the value of \frac{1}{3} of \frac{2}{5} of \frac{1}{4} of a number is 40, then what is that number?)
*पिछली प्रश्नोत्तरी का हल*
x^2 - 1 = (x + 1)(x - 1)
\because 2x^3 + 4x^2 + 2ax + b, x^2 - 1 से पूरी तरह भाज्य है, अतः व्यंजक में x = -1 और x = 1 रखने पर
-2 + 4 - 2a + b = 0
और 2 + 4 + 2a + b = 0
या, -2a + b = -2
और 2a + b = -6
\therefore 2b = -8
\Rightarrow b = -4
\therefore -2a = 2 \Rightarrow a = -1
*Previous Quiz Solution*
Since 2x^3 + 4x^2 + 2ax + b fully divisible by x^2 - 1, when x = -1 and x = 1 are placed in the expression:
\because -2 + 4 - 2a + b = 0
and 2 + 4 + 2a + b = 0
or -2a + b = -2
and 2a + b = -6
\therefore 2b = -8 \Rightarrow b = -4
\Rightarrow -2a = 2 \Rightarrow a = -1
*”This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*

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