Here we will learn how to proof of De Morgan’s law of union and intersection.
Definition of De Morgan’s law:
The complement of the union of two sets is equal to the intersection of their complements and the complement of the intersection of two sets is equal to the union of their complements. These are called De Morgan’s laws.
For any two finite sets A and B;
(i) (A U B)' = A' ∩ B' (which is a De Morgan's law of union).
(ii) (A ∩ B)' = A' U B' (which is a De Morgan's law of intersection).
Proof of De Morgan’s law: (A U B)' = A' ∩ B'
Let P = (A U B)'
and Q = A' ∩ B'
Let x be an arbitrary element of P then x ∈ P ⇒ x ∈ (A U B)'
⇒ x ∉ (A U B)
⇒ x ∉ A and x ∉ B
⇒ x ∈ A' and x ∈ B'
⇒ x ∈ A' ∩ B'
⇒ x ∈ Q
Therefore, P ⊂ Q …………….. (i)
Again, let y be an arbitrary element of Q then y ∈ Q ⇒ y ∈ A' ∩ B'
⇒ y ∈ A' and y ∈ B'
⇒ y ∉ A and y ∉ B
⇒ y ∉ (A U B)
⇒ y ∈ (A U B)'
⇒ y ∈ P
Therefore, Q ⊂ P …………….. (ii)
Now combine (i) and (ii) we get; P = Q i.e. (A U B)' = A' ∩ B'
Proof of De Morgan’s law: (A ∩ B)' = A' U B'
Let M = (A ∩ B)' and N = A' U B'
Let x be an arbitrary element of M then x ∈ M ⇒ x ∈ (A ∩ B)'
⇒ x ∉ (A ∩ B)
⇒ x ∉ A or x ∉ B
⇒ x ∈ A' or x ∈ B'
⇒ x ∈ A' U B'
⇒ x ∈ N
Therefore, M ⊂ N …………….. (i)
Again, let y be an arbitrary element of N then y ∈ N ⇒ y ∈ A' U B'
⇒ y ∈ A' or y ∈ B'
⇒ y ∉ A or y ∉ B
⇒ y ∉ (A ∩ B)
⇒ y ∈ (A ∩ B)'
⇒ y ∈ M
Therefore, N ⊂ M …………….. (ii)
Now combine (i) and (ii) we get; M = N i.e. (A ∩ B)' = A' U B'
Examples on De Morgan’s law:
1. If U = {j, k, l, m, n}, X = {j, k, m} and Y = {k, m, n}.
Proof of De Morgan's law: (X ∩ Y)' = X' U Y'.
Solution:
We know, U = {j, k, l, m, n}
X = {j, k, m}
Y = {k, m, n}
(X ∩ Y) = {j, k, m} ∩ {k, m, n}
= {k, m}
Therefore, (X ∩ Y)' = {j, l, n} ……………….. (i)
Again, X = {j, k, m} so, X' = {l, n}
and Y = {k, m, n} so, Y' = {j, l}
X' ∪ Y' = {l, n} ∪ {j, l}
Therefore, X' ∪ Y' = {j, l, n} ……………….. (ii)
Combining (i)and (ii) we get;
(X ∩ Y)' = X' U Y'. Proved
2. Let U = {1, 2, 3, 4, 5, 6, 7, 8}, P = {4, 5, 6} and Q = {5, 6, 8}.
Show that (P ∪ Q)' = P' ∩ Q'.
Solution:
We know, U = {1, 2, 3, 4, 5, 6, 7, 8}
P = {4, 5, 6}
Q = {5, 6, 8}
P ∪ Q = {4, 5, 6} ∪ {5, 6, 8}
= {4, 5, 6, 8}
Therefore, (P ∪ Q)' = {1, 2, 3, 7} ……………….. (i)
Now P = {4, 5, 6} so, P' = {1, 2, 3, 7, 8}
and Q = {5, 6, 8} so, Q' = {1, 2, 3, 4, 7}
P' ∩ Q' = {1, 2, 3, 7, 8} ∩ {1, 2, 3, 4, 7}
Therefore, P' ∩ Q' = {1, 2, 3, 7} ……………….. (ii)
Combining (i)and (ii) we get;
(P ∪ Q)' = P' ∩ Q'. Proved
● Set Theory
● Sets
● Subset
● Practice Test on Sets and Subsets
● Problems on Operation on Sets
● Practice Test on Operations on Sets
● Venn Diagrams in Different Situations
● Relationship in Sets using Venn Diagram
● Practice Test on Venn Diagrams
8th Grade Math Practice
From Proof of De Morgan’s Law to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Jul 16, 24 09:27 AM
Jul 16, 24 02:20 AM
Jul 16, 24 01:36 AM
Jul 16, 24 12:17 AM
Jul 15, 24 03:22 PM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.