First slide

Operations on Sets

Question

Which of the following equality is not true.

Moderate

A

$A \cap (B ~ C) = A \cap B ~ (A \cap C)$
B

$A ~ (A \cap B) = A ~ B$
C

$A ~ (B ~ C) = (A ~ B) \cup (A \cap C)$
D

$A ~ (B ∆ C) = (A ~ B) ∆ (A ~ C)$

Solution

For equality (a),
$A \cap (B ~ C) = A \cap (B \cap C')$
$= ϕ \cup (A \cap B \cap C')$
$= (A \cap B \cap A') \cup (A \cap B \cap C')$
$= A \cap B \cap (A' \cup C')$
$= A \cap B ~ (A \cap C)$
For equality (b),
$A ~ (A \cap B) = A \cap (A' \cup B')$
$= (A \cap A') \cup (A \cap B')$
$= \emptyset \cup (A \cap B') = A \cap B' = A ~ B$
For equality (c)
$A ~ (B ~ C) = A ~ (B \cap C') = A \cap (B' \cup C)$
$= (A \cap B') \cup (A \cap C)$
= (A ~ B) » (A « C)
For (d) Let $A = {1, 2, 3, 4, 5}, B = {3, 4, 5}, C = {1, 2, 3}$
So, $B ∆ C = {4, 5} \cup {1, 2} = {1, 2, 4, 5}$
Thus $A (B ∆ C) = {3}$
$A ~ B = {1, 2}; A ~ C = {4, 5}$
Therefore $(A ~ B) ∆ (A ~ C) = ({1, 2} ~ {4, 5})$
$\cup ({4, 5} ~ {1, 2})$
$= {1, 2} \cup {4, 5} = {1, 2, 4, 5) .$
Hence $A ~ (B ∆ C) \neq (A ~ B) ∆ (A ~ C)$

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