Using properties of sets, show that $\begin{matrix} (i) A \cup (A \cap B) = A (i i) A \cap (A \cup B) = A \end{matrix}$

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Solution

$(i) A \cup (A \cap B) = A l e t B i s a s u b s e t o f A (B \subset A) ∴ A \cap B = B [c o m m o n e l e m e n t s o f A & B] N o w, A \cup (A \cap B) = A \cup B = A [A l l E l e m e n t s b e l o n g s t o A & B] (i i) A \cap (A \cup B) = A l e t A \subset B (A i s a s u b s e t o f A) ∴ A \cup B = B [A l l E l e m e n t s b e l o n g s t o A & B] A \cap (A \cup B) = A \cap B = A [c o m m o n e l e m e n t s o f A & B]$

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(A \cup B) \cap (A \cap B') = A

Using properties of sets, show that

(i) $A \cup (A \cap B) = A$ (ii) $A \cap (A \cup B) = A .$

(i) A \cup (A \cap B) = A

(i i) A \cap (A \cup B) = A

A = (A \cap B) \cap (A - B)

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