De Morgans law : The complement of the union of two sets is the intersection of their complements and the complement of the intersection of two sets is the union of their complements.These are called De Morgan’s laws.
These are named after the mathematician De Morgan.
1) Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}. Show that (A ∪B)^{ '} = A^{ '} ∩ B^{'}.
Solution :
U = {1, 2, 3, 4, 5, 6}
A = {2, 3}
B = {3, 4, 5}
A ∪ B = {2, 3} ∪ {3, 4, 5}
= {2, 3, 4, 5}
∴ (A ∪ B)^{ '} = {1, 6}
Also A^{ '} = {1, 4, 5, 6}
B ^{ '} = {1, 2, 6}
∴ A^{'} ∩ B^{'} = {1, 4, 5, 6} ∩ {1, 2, 6}
= {1, 6}
Hence (A ∪ B)^{'} = A^{ '} ∩ B^{'}
________________________________________________________________
2) If ξ = {a,b,c,d,e}, A = { a,b,d} and B = {b,d,e}. Prove De Morgan's law of intersection.
Solution :
ξ = {a,b,c,d,e}
A = { a,b,d}
B = {b,d,e}
(A ∩ B) = { a,b,d} ∩ {b,d,e}
(A ∩ B) = {b,d}
∴ (A ∩ B)' = {a, c,e} ----->(1)
A' = {c,e} and B' = {a,c}
∴ A' ∪ B' = {c,e} ∪ {a,c}
A' ∪ B'= { a, c,e} ----->(2)
From (1) and (2)
(A ∩ B)' = A' ∪ B' (which is a De Morgan's law of intersection). Set Theory