**Complement of a Set**

Let U be the universal set which consists of all prime numbers and A be the subset of U which consists of all those prime numbers that are not divisors of 42. Thus, A = {x : x ∈ U and x is not a divisor of 42 }. We see that 2 ∈ U but 2 ∉ A, because 2 is divisor of 42. Similarly, 3 ∈ U but 3 ∉ A, and 7 ∈ U but 7 ∉ A. Now 2, 3 and 7 are the only elements of U which do not belong to A. The set of these three prime numbers, i.e., the set {2, 3, 7} is called the **Complement** of A with respect to U, and is denoted by A’. So we have A’ = {2, 3, 7}. Thus, we see that

A’ = {x : x ∈ U and x ∉ A }. This leads to the following definition.

**Definition 8** 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 ∈ U and x ∉ A }. Obviously A′ = U – A

We note that the complement of a set A can be looked upon, alternatively, as the difference between a universal set U and the set A.

**Example 20** Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {1, 3, 5, 7, 9}. Find A′.

**Solution** We note that 2, 4, 6, 8, 10 are the only elements of U which do not belong to A.

Hence A′ = { 2, 4, 6, 8,10 }.

**Example 21** Let U be universal set of all the students of Class XI of a coeducational school and A be the set of all girls in Class XI. Find A′.

**Solution** Since A is the set of all girls, A’ is clearly the set of all boys in the class.

**Note:-** If A is a subset of the universal set U, then its complement A′ is also a subset of U.

Again in Example 20 above, we have A′ = { 2, 4, 6, 8, 10 }

Hence (A′ )′ = {x : x ∈ U and x ∉ A′ } = {1, 3, 5, 7, 9} = A

It is clear from the definition of the complement that for any subset of the universal set U,

we have ( A′ )′ = A

Now, we want to find the results for ( A ∪ B )′ and A′ ∩ B′ in the following example.

**Example 22** Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}. Find A′, B′ , A′ ∩ B′, A ∪ B and hence show that ( A ∪ B )′ = A′ ∩ B′.

**Solution** Clearly A′ = {1, 4, 5, 6}, B′ = { 1, 2, 6 }. Hence A′ ∩ B′= { 1, 6 }

Also A ∪ B = { 2, 3, 4, 5 }, so that (A ∪ B )′ = { 1, 6 }

(A ∪ B)′ = { 1, 6 } = A′ ∩ B′

It can be shown that the above result is true in general. If A and B are any two subsets of the universal set U, then

( A ∪ B)′ = A′ ∩ B′. Similarly, ( A ∩ B )′ = A′ ∪ B′. These two results are stated in words as follows :

**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.

The complement A′ of a set A can be represented by a Venn diagram as shown in Fig 1.10.

The shaded portion represents the complement of the set A.

**Some Properties of Complement Sets**

1. Complement laws: (i) A ∪ A′ = U (ii) A ∩ A′ = φ

2. De Morgan’s law: (i) (A ∪ B)´ = A′ ∩ B′ (ii) (A ∩ B )′ = A′ ∪ B′

3. Law of double complementation : (A′ )′ = A

4. Laws of empty set and universal set φ′ = U and U′ = φ.

These laws can be verified by using Venn diagrams.