MATHS PRATICE QUESTION PAPERS RELEASED BY APOLLO IAS ACADEMY

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POINTS TO REMEMBER

 De Morgan’s Laws


Augustus De Morgan (1806 – 1871) was a British mathematician. 


He was born on 27th June 1806 in Madurai, Tamilnadu, India. His
father was posted in India by the East India Company. When he was
seven months old, his family moved back to England. De Morgan was
educated at Trinity College, Cambridge, London. He formulated laws for
set difference and complementation. These are called De Morgan’s laws.

De Morgan’s Laws for Set Difference

These laws relate the set activities association, crossing point, and set distinction. 


Let us consider three sets A, B and C as A = −{ } 5 2 , , − 1 3, , B = −{ } 3,−203 ,,,5 and
C = −{ } 2 1 , , − 045 , , .
Now, B CÈ = { , --- 321 , , , 0345 ,,}
 A B − ∪ ( ) C = { } −5 1, ... (1)
Then, A B - = { , -5 1} and A C− = { } −5 1, , 3
 ( ) A B − ∪( ) A C− = { } −5 1, , 3 ... (2)
 ( ) A B − ∩( ) A C− = { } −5 1, ... (3)
From (1) and (2), we see that
 A B − ∪ ( ) C ≠ − ( ) A B ∪ − ( ) A C
But note that from (1) and (3), we see that

 A B − ∪ C = A B − ∩ A C

Now, B CÇ = { } −2 0, ,5
 A B − ∩ ( ) C ={ } −5 1, , 3 ... (4)
From (3) and (4) we see that
 A B − ∩ ( ) C ≠ − ( ) A B ∩ − ( ) A C
But note that from (2) and (4), we get A B − ∩ ( ) C A =−∪ ( ) B A( ) −C


De Morgan's laws for set contrast: For any three sets A, B, and C.


(i) A B − ∪ ( ) C A =−∩ ( ) B A( ) −C (ii) A B − ∩ ( ) C A =−∪ ( ) B A( ) −C
( ) A B − ∪( ) A C− ∪( ) A B ∩ =____

 De Morgan’s Laws for Complementation


These laws relate the set operations on union, intersection, and complementation.
Let us consider universal set U={0,1,2,3,4,5,6}, A={1,3,5} and B={0,3,4,5}.
Check whether
A B − = A B ∩ ′
Now, A B È = { , 0 1, , 345, } Thinking Corner
Then, ( ) A B ∪ ′
= { , 2 6} .....(1)
Next, A¢
= { , 0 246 ,,} and B¢
 = { ,1 2, }6
Then, A B ′ ∩ ′
= { , 2 6} .....(2)
From (1) and (2), we get ( ) A B ∪ ′ = A B ′ ∩ ′

A B − ∪B A − ′
=____
Also, A B ∩ = { } 3 5, , Thinking Corner
 ( ) A B ∩ , , , , ′ = { } 0 1 246 .....(3)
 A′ = { } 0, , 246, andB′ = { } 1 2, ,6
 A B ′ ∪ ′ = { } 0 1, ,246 , , .....(4)
From (3) and (4), we get ( ) A B ∩ ′ = A B ′ ∪ ′

De Morgan's laws for complementation: Let 'U' be the general set containing 

finite sets A and B. Then (i) ( ) A B ∪ ′ = A B ′ ∩ ′ (ii) ( ) A B ∩ ′ = A B ′ ∪ ′

Example 1.26 If U = ∈ { } x x: ,  − ≤2 1 x ≤ 0 ,
A x = = { } : , x p2 1 + ∈p p , − ≤1 4 ≤ , B x = = { } : , x q3 1 + ∈q q  , − ≤1 4 < ,
verify De Morgan’s laws for complementation.
A A ( ) B =____
Thinking Corner Solution
Given U = −{ , 2 1 − , , 0 1, , 234, , , 567,,, 8 9, } 10 ,

A = −{ ,1 1, , 357,,}9 and B = −{ , 2 1,,, 4 7 10}

Law (i) ( ) A B ∪ ′ =A B ′ ∩ ′
Now, A B È = { , - -2 1, ,1 345 , , , , 7 9, } 10

 ( ) A B ∪ ′ = { , 0 2, , 6 8} ..... (1)
( ) A B ∪ ′ ∪( ) A B ′ ∩ =___
Thinking Corner
Then, A¢ = { , -2 0, , 2468 , , , } 10 and B¢ = { , -1 0, , 2 3 5, ,689 ,,}
 A B ′ ∩ ′ = { , 0 2, , 6 8} ..... (2)
 From (1) and (2), it is verified that ( ) A B ∪ ′ = A B ′ ∩ ′
Law (ii) ( ) A B ∩ ′ = A B ′ ∪ ′
Now, A B Ç = { ,1 7}

 ( ) A B ∩ ′ = { , - -2 1, , 0 2345 , , ,,, 689, ,10} ..... (3)

Then, A B ′ ∪ ′ = { , - -2 1, , 0 2345 , , ,,, 689, ,10} ..... (4)

From (3) and (4), it is verified that ( ) A B ∩ ′ = A B ′ ∪ ′

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