De Morgan’s law states that ‘The complement of the union of two sets A and B is equal to the intersection of the complement of the sets A’ and B’. Also, according to De Morgan’s law, the complement of the intersection of two sets A and B is equal to the union of the complement of the sets A and B i.e.,
(A∪B)’ = A’ ∩ B’
And (A ∩ B)’ = A’ ∪ B’
What is the complement of a Set?
The complement of a set is defined as
A’ = {x : x ∈ U and x ∉ A}
where A’ denotes the complement.
Complement of Sets Properties
i) Complement Laws: The union of a set A and its complement A’ gives the universal set U of which, A and A’ are a subset.
A ∪ A’ = U
Also, the intersection of a set A and its complement A’ gives the empty set ∅.
A ∩ A’ = ∅
For Example: If U = {1 , 2 , 3 , 4 , 5 } and A = {1 , 2 , 3 } then A’ = {4 , 5}. From this it can be seen that
A ∪ A’ = U = { 1 , 2 , 3 , 4 , 5}
Also
A ∩ A’ = ∅
ii) Law of Double Complementation: According to this law if we take the complement of the complemented set A’ then, we get the set A itself.
(A’)’ = A
In the previous example we can see that, if U = {1 , 2 , 3 , 4 , 5} and A = {1 , 2 ,3} then A’ ={4 , 5}. Now if we take the complement of set ‘A’ we get,
(A’)’ = {1 , 2 , 3} = A
This gives back the set A itself.
iii) Law of empty set and universal set:
According to this law the complement of the universal set gives us the empty set and vice-versa i.e.,
∅’ = U And U’ = ∅
This law is self-explanatory.
To understand this law better let us consider the following example:
Solved Problem
Example:
A universal set U which consists of all the natural numbers which are multiples of 3, less than or equal to 20. Let A be a subset of U which consists of all the even numbers and the set B is also a subset of U consisting of all the prime numbers. Verify De Morgan Law.
Solution: We have to verify (A ∪ B)’ = A’ ∩ B’ and (A ∩ B)’ = A’∪B’. Given that,
U = {3 , 6 , 9 , 12 , 15 , 18}
A = {6 , 12 , 18}
B = {3}
The union of both A and B can be given as,
A ∪ B = {3 , 6 , 12 , 18}
The complement of this union is given by,
(A ∪ B)’={9 , 15}
Also, the intersection and its complement are given by:
A ∩ B = ∅
(A ∩ B)’ = {3 , 6 , 9 , 12 , 15 ,18}
Now, the complement of the sets A and B can be given as:
A’ = {3 , 9 , 15}
B’ = {6 , 9 , 12 , 15 , 18}
Taking the union of both these sets, we get,
A’∪B’ = {3 , 6 , 9 , 12 , 15 ,18}
And the intersection of the complemented sets is given as,
A’ ∩ B’ = {9 , 15}
We can see that:
(A ∪ B)’ = A’ ∩ B’ = {9 , 15}
And also,
(A ∩ B)’ = A’ ∪ B’ = {3 , 6 , 9 , 12 , 15 ,18}
Hence, the above result is true in general and is known as De Morgan Law.
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