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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