DE MORGANS LAW FOR COMPLIMENTS
About "De Morgans law for compliments"
De Morgans Law for Compliments :
Here we are going to see De morgan's law for set complements.
Demorgans law :
De Morgan’s father (a British national) was in the service of East India Company, India. Augustus De Morgan (1806-1871) was born in Madurai, Tamilnadu, India. His family moved to England when he was seven months old. He had his education at Trinity college, Cambridge, England.
De Morgan’s laws relate the three basic set operations union, intersection and complement.
De morgan's law for set complementation :
Let U be the universal set containing sets A and B. Then
(i) (A u B)' = A' n B'
(ii) (A n B)' = A' u B'
Proof by Venn diagram
(A n B)' = A' u B'
From the above Venn diagrams (2) and (5), it is clear that
(A n B)' = A' u B'
Hence, De morgan's law for complementation is verified.
(A U B)' = A' n B'
De morgan's laws for complements - Examples
Example 1 :
Let U = { - 2, -1, 0, 1, 2, 3, 4, 5, .......10 }, A = {- 2, 2, 3, 4, 5 } and B = { 1, 3, 5, 8, 9 }. Verify De Morgan’s laws of complementation
Solution :
First, we shall verify (A u B)' = A' n B'
To do this, we consider
A u B = {- 2, 2, 3, 4, 5 } u { 1, 3, 5, 8, 9 }
A u B = { -2, 1, 2, 3, 4, 5, 8, 9 }
We know that
(A u B)' = U \ { -2, 1, 2, 3, 4, 5, 8, 9 }
(A u B)' = { -1, 0, 6, 7, 10 } ---------(1)
A' = U \ A = U \ { -2, 2, 3, 4, 5 }
= { -1, 0, 1, 6, 7, 8, 9, 10 }
B' = U \ B = U \ { 1, 3, 5, 8, 9 }
= { -2, -1, 0, 2, 4, 6, 7, 10 }
A'nB' = { -1, 0, 1, 6, 7, 8, 9, 10 } n { -2, -1, 0, 2, 4, 6, 7, 10 }
A' n B' = { -1, 0, 6, 7, 10 } ---------(2)
From (1) and (2), it is clear that (A u B)' = A' n B'
First, we shall verify (A n B)' = A'u B'
To do this, we consider
A n B = {- 2, 2, 3, 4, 5 } u { 1, 3, 5, 8, 9 }
A n B = { 3, 5 }
We know that
(A n B)' = U \ { 3, 5 }
(A n B)' = { 2, -1, 0, 1, 2, 4, 6, 7, 8, 9, 10}
A' = U \ A = U \ { -2, 2, 3, 4, 5 }
= { -1, 0, 1, 6, 7, 8, 9, 10 }
B' = U \ B = U \ { 1, 3, 5, 8, 9 }
= { -2, -1, 0, 2, 4, 6, 7, 10 }
A'UB' = { -1, 0, 1, 6, 7, 8, 9, 10 } n { -2, -1, 0, 2, 4, 6, 7, 10 }
A' U B' = { 2, -1, 0, 1, 2, 4, 6, 7, 8, 9, 10} ---------(2)
From (1) and (2), it is clear that (A n B)' = A' U B'
Example 2 :
Let U = {4, 8, 12, 16, 20, 24, 28} , A = {8, 16, 24} and B = {4, 16, 20, 28}. Find (AU B)' and (A n B)'.
Solution :
A u B = {8, 16, 24} U {4, 16, 20, 28}
A u B = { 4, 8, 16, 20, 24, 28 }
We know that
(A u B)' = U \ { 4, 8, 16, 20, 24, 28 }
(A u B)' = { 12 }
(A n B) = {8, 16, 24} n {4, 16, 20, 28}
= {16}
(A n B)' = U \ { 16 }
= {4, 8, 12, 20, 24, 28}
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