(i) A'∩B' = {1,6,8,10}
(ii) A - (B∪C) = {2,4}
(iii) (A-B)∪(B-C) = {2,4}
(i) We have verified that A∪(B∩C) = (A∪B)∩(A∪C)
(ii) We have verified De-Morgan's law
Sets
From the question, we are to find the given set operations
From the given information,
U = {1,2,3,...10}
A = {2,3,4,5}
B = {3,5,7,9}
C = {1,3,5,7,9}
We are to find
(i) A'∩B'
A' = {1,6,7,8,9,10}
B' = {1,2,4,6,8,10}
∴ A'∩B' = {1,6,8,10}
(ii) A - (B∪C)
A = {2,3,4,5}
Now, find (B∪C)
B∪C = {1,3,5,7,9}
∴ A - (B∪C) = {2,4}
(iii) (A-B)∪(B-C)
A-B = {2,4}
B-C = { }
∴ (A-B)∪(B-C) = {2,4}
We are to verify
(i) A∪(B∩C) = (A∪B)∩(A∪C)
First, determine A∪(B∩C)
A = {2,3,4,5}
B∩C = {3,5,7,9}
∴ A∪(B∩C) = {2,3,4,5,7,9}
Now, determine (A∪B)∩(A∪C)
A∪B = {2,3,4,5,7,9}
A∪C = {1,2,3,4,5,7,9}
∴ (A∪B)∩(A∪C) = {2,3,4,5,7,9}
Thus,
A∪(B∩C) = (A∪B)∩(A∪C)
(ii) De-Morgan's law
De-Morgan's law states that
(A∪B)' = A'∩B'
(A∩B)' = A'∪B'
For (A∪B)' = A'∩B'
A∪B = {2,3,4,5,7,9}
∴ (A∪B)' = {1,6,8,10}
A' = {1,6,7,8,9,10}
B' = {1,2,4,6,8,10}
∴ A'∩B' = {1,6,8,10}
Thus,
(A∪B)' = A'∩B'
Also,
For (A∩B)' = A'∪B'
A∩B = {3,5}
(A∩B)' = {1,2,4,6,7,8,9,10}
A' = {1,6,7,8,9,10}
B' = {1,2,4,6,8,10}
∴ A'∪B' = {1,2,4,6,7,8,9,10}
Thus,
(A∩B)' = A'∪B'
Hence, we have verified that A∪(B∩C) = (A∪B)∩(A∪C), and we have also verified De-Morgan's law
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