Question

6. Given that U = { a, b, c, d, e, f, g, h }, A = {a , b, f, g} and B= { a,b,c } , verify
De Morgan’s laws of complementation

Answer 1

Hi ,

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De-morgan's Laws :

If A and B are any two sets ,

1 ) ( A U B )' = A' and B'

2 ) ( A and B )' = A' U B'

And

We know that ,

A' = U - A

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Now ,

U = { a ,b , c , d , e , f , g , h }

A = { a , b , f , g }

B = { a , b , c }

i ) A U B = { a , b , f , g } U { a , b , c }

= { a , b , c , f , g }

( A U B )' = U - ( A U B )

= { a,b,c,d,e,f,g,h } - { a,b,c,f,g }

= { d , e ,h } ----( 1 )

A' = U - A

= { a,b,c,d,e,f,g,h } - { a ,b ,f ,g }

= { c , d ,e , h }

B' = U - B

= { a,b,c,d,e,f,g,h } - { a ,b ,c }

= { d,e, f ,g ,h }

A' and B' = { c ,d,e,h } and {d, e,f,g,h }

= {d, e , h } ----( 2 )

From ( 1 ) and ( 2 ), we get

( A U B )' = A' and B'

I hope this helps you.

: )

Answer 2

Union of two sets :
The union of the sets A and B is the set of all the element that belongs to either A or B or both. It is denoted by A U B(“A union B”).
Intersection of two sets :
The intersection of the sets a and b is the set of all the elements which belong to both A and B. It is denoted by A ∩ B (“ A intersection B”).
•If A and B do not have any element in common then A ∩ B= a null set = Ø

•A’ == The complementary set of A
• To find A’ , list all the members of the universal set U which are not members of A.
De-morgan's Laws :
(A U B )’ = A’ ∩ B’
(A ∩ B )’ = A’ U B’

SOLUTION :
GIVEN :
U = { a, b, c, d, e, f, g, h }, A = {a ,b,f, g} and B= { a,b,c }

A ∩ B = {a,b}
Then, (A ∩ B )’ = U - (A ∩ B )
(A ∩ B )’ = {c,d,e,f,g,h}...............(1)

A’ = U - A = {c,d,e,h}
B’ = U - B = {d,e,f,g,h}
A’ U B’ = {c,d,e,f,g,h}................(2)

Hence , from eq 1 & 2
(A ∩ B )’ = A’ U B’

A U B = = {a,b,c,f,g}
Then , (A U B)’= U - A U B ={d,e,h}
(A U B)’= {d,e,h}....................(3)

A’ = U - A = {c,d,e,h}
B’ = U - B = {d,e,f,g,h}
A’ ∩ B’ = {d,e,h}................(4)

Hence , from eq 3 & 4
(A U B )’ = A’ ∩ B’

HOPE THIS WILL HELP YOU….