Proof of De Morgan’s Law

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

Representation of a Set

Types of Sets

Pairs of Sets

Subset

Practice Test on Sets and Subsets

Complement of a Set

Problems on Operation on Sets

Operations on Sets

Practice Test on Operations on Sets

Word Problems on Sets

Venn Diagrams

Venn Diagrams in Different Situations

Relationship in Sets using Venn Diagram

Examples on Venn Diagram

Practice Test on Venn Diagrams

Cardinal Properties of Sets



7th Grade Math Problems

8th Grade Math Practice

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