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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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