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Mathematics

Let U = {3, 6, 9, 12, 15, 18, 21, 24} be the universal set and let A = {6, 12, 18, 24} be its subset.

Verify that:

(i) A ∪ A = A

(ii) A ∩ A = A

(iii) A ∩ A' = Φ

(iv) A ∪ A' = U

(v) (A')' = A

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Answer

Given:

Universal set U = {3, 6, 9, 12, 15, 18, 21, 24}

Subset A = {6, 12, 18, 24}

(i) A ∪ A = A

LHS = A ∪ A = {6, 12, 18, 24} ∪ {6, 12, 18, 24}

LHS = {6, 12, 18, 24} \quad[since repeated elements are written once]

RHS = A = {6, 12, 18, 24}

Since LHS = RHS,

∴ The statement A ∪ A = A is verified.

(ii) A ∩ A = A

LHS = A ∩ A = {6, 12, 18, 24} ∩ {6, 12, 18, 24}

LHS = {6, 12, 18, 24}

RHS = A = {6, 12, 18, 24}

Since LHS = RHS,

∴ The statement A ∩ A = A is verified.

(iii) A ∩ A' = Φ

A' = U - A = {3, 6, 9, 12, 15, 18, 21, 24} - {6, 12, 18, 24} = {3, 9, 15, 21}

LHS = A ∩ A' = {6, 12, 18, 24} ∩ {3, 9, 15, 21}

LHS = Φ

RHS = Φ

Since LHS = RHS,

∴ The statement A ∩ A' = Φ is verified.

(iv) A ∪ A' = U

A' = {3, 9, 15, 21} \quad[From previous step]

LHS = A ∪ A' = {6, 12, 18, 24} ∪ {3, 9, 15, 21}

LHS = {3, 6, 9, 12, 15, 18, 21, 24}

RHS = U = {3, 6, 9, 12, 15, 18, 21, 24}

Since LHS = RHS,

∴ The statement A ∪ A' = U is verified.

(v) (A')' = A

We know A' = {3, 9, 15, 21}

LHS = (A')' = U - A' = {3, 6, 9, 12, 15, 18, 21, 24} - {3, 9, 15, 21}

LHS = {6, 12, 18, 24}

RHS = A = {6, 12, 18, 24}

Since LHS = RHS,

∴ The statement (A')' = A is verified.

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