Mathematics
Let U = {a, b, c, d, e, f, g} be the universal set and let its subsets be A = {a, b, d, e} and B = {b, e, g}.
Verify that:
(i) (A ∪ B)' = (A' ∩ B')
(ii) (A ∩ B)' = (A' ∪ B')
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Answer
Given:
Universal set U = {a, b, c, d, e, f, g}
Subset A = {a, b, d, e}
Subset B = {b, e, g}
(i) (A ∪ B)' = (A' ∩ B')
First let us find A ∪ B:
A ∪ B = {a, b, d, e} ∪ {b, e, g} = {a, b, d, e, g}
(A ∪ B)' = Elements in U which are not in (A ∪ B).
LHS = (A ∪ B)' = U - (A ∪ B)
LHS = {a, b, c, d, e, f, g} - {a, b, d, e, g}
LHS = {c, f}
Now, find A' and B':
A' = U - A
A' = {a, b, c, d, e, f, g} - {a, b, d, e} = {c, f, g}
B' = U - B
B' = {a, b, c, d, e, f, g} - {b, e, g} = {a, c, d, f}
RHS = (A' ∩ B') = {c, f, g} ∩ {a, c, d, f}
RHS = {c, f}
Since LHS = RHS,
∴ The statement (A ∪ B)' = (A' ∩ B') is verified.
(ii) (A ∩ B)' = (A' ∪ B')
First let us find A ∩ B:
A ∩ B = {a, b, d, e} ∩ {b, e, g} = {b, e}
LHS = (A ∩ B)' = U - (A ∩ B).
LHS = (A ∩ B)' = {a, b, c, d, e, f, g} - {b, e}
LHS = {a, c, d, f, g}
Now, find A' and B':
A' = U - A
A' = {a, b, c, d, e, f, g} - {a, b, d, e} = {c, f, g}
B' = U - B
B' = {a, b, c, d, e, f, g} - {b, e, g} = {a, c, d, f}
RHS = (A' ∪ B') = {c, f, g} ∪ {a, c, d, f}
RHS = {a, c, d, f, g}
Since LHS = RHS,
∴ The statement (A ∩ B)' = (A' ∪ B') is verified.
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