If U = {2, 3, 5, 7, 9} is the universal set and A = {3, 7}, B = {2, 5, 7, 9}, then prove that:
(i) $(A \cup B)' = A' \cap B'$
(ii) $(A \cap B)' = A' B' .$

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Solution

Given:
U = {2, 3, 5, 7, 9}
A = {3, 7}
B = {2, 5, 7, 9}

To prove :
(i) $(A \cup B)' = A' \cap B'$
(ii) $(A \cap B)' = A' \cup B'$

Proof :

(i) LHS:
$(A \cup B) = {2, 3, 5, 7, 9} (A \cup B)' = ϕ$

RHS:
$A' = {2, 5, 9} B' = {3} A' \cap B' = ϕ LHS = RHS$

∴ $(A \cup B)' = A' \cap B'$

(ii) LHS:
$(A \cap B) = {7} (A \cap B)' = {2, 3, 5, 9}$

RHS:
$A' = {2, 5, 9} B' = {3} A' \cup B' = {2, 3, 5, 9}$

LHS = RHS

∴ $(A \cap B)' = A' \cup B'$

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Similar questions

If U= {2,3,5,7,9} is the universal set and A = {3,7} , B = {2,5,7,9}, then prove that :

(i) $A \cup B)^{'} = A^{'} \cap B^{'}$

(ii) $A \cap B)^{'} = A^{'} \cup B^{'}$

(De Morgan's laws) For any two sets A and B, prove that:
$I . (A \cup B)^{'} = (A^{'} \cap B^{'})$

$I I . (A \cap B)^{'} = (A^{'} \cup B^{'})$

U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8}

B = {2, 3, 5, 7}

(A \cup B)^{'} = A^{'} \cap B^{'}

(A \cap B)^{'} = A^{'} \cup B^{'}

A' \cup B = U \Rightarrow A \subset B

B' \subset A' \Rightarrow A \subset B

If A and B are subsets of the universal set U, then show that,

$A \subset B \Leftrightarrow A \cup B = B$

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