Distributive laws For any three sets A- B- C prove that-I- A -B - C-A - B -A - C-Distirbutive law of union over intersection-II- A -B - C-A - B -A - C-Distributive law of intersection over union-

The correct option is I. Let x be an arbitary element of A∪ (B∩ C). Then, xϵ A∪ (B∩ C)⇒ xϵ A or xϵ (B∩ C) ⇒ xϵ A or (xϵ B) and xϵ C ⇒ ( xϵ A or xϵ B ...

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Byju's Answer

Standard XI

Mathematics

Definition of a Hyperbola

Distributive ...

Question

(Distributive laws) For any three sets A, B, C prove that:

I. $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$
[Distirbutive law of union over intersection]

II. $A \cap [(B \cup C)] = (A \cap B) \cup (A \cap C)$
[Distributive law of intersection over union]

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Solution

I. Let x be an arbitary element of $A \cup (B \cap C)$ . Then,

$x ϵ A \cup (B \cap C)$ $\Rightarrow x ϵ A o r x ϵ (B \cap C) \Rightarrow x ϵ A o r (x ϵ B) a n d x ϵ C \Rightarrow (x ϵ A o r x ϵ B) a n d (x ϵ A o r x ϵ C)$

[ $∵$ 'or' distributes 'and']

$\Rightarrow x ε (A \cup B) a n d x ϵ (A \cup C) . \Rightarrow x ϵ (A \cup B) \cap (A \cup C)$
$∴ A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C) \dots (i)$

Again. let y be an arbitary element of $(A \cup B) \cap (A \cup C)$ . Then,

$y ϵ (A \cup B) \cap (A \cup C)$ $\Rightarrow y ϵ (A \cup B) a n d y ϵ (A \cup C) \Rightarrow (y ϵ A o r y ϵ B) a n d (y ϵ A o r y ϵ C) \Rightarrow y ϵ A o r (y ϵ B a n d y ϵ C)$

[ $∵$ 'or; distributes 'and']

$\Rightarrow y ϵ A o r y ϵ (B \cap C) \Rightarrow y ϵ A \cup (B \cap C) ∴ (A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C) \dots (i i)$ .

From (i) and (ii), we get $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$

II. Let x be an arbitrary element of $A \cap (B \cup C)$ . Then,

$x ϵ A \cap (B \cup C)$ $\Rightarrow x ϵ A a n d x ϵ (B \cup C) \Rightarrow x ϵ A a n d (x ϵ B o r x ϵ C) \Rightarrow (x ϵ A a n d x ϵ B) o r (x ϵ A a n d x ϵ C)$

[ $∵$ 'and' distributes 'or']

$\Rightarrow x ϵ (A \cap B) o r x ϵ (A \cap C) \Rightarrow x ϵ (A \cap B) \cup (A \cap C)$

$∴ A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C) \dots (i i i)$

Again, let y be an arbitrary element of $(A \cap B) \cup (A \cap C)$ . Then,

$y ϵ (A \cap B) \cup (A \cap C)$ $\Rightarrow y ϵ (A \cap B) o r y ϵ (A \cap C) \Rightarrow (y ϵ A a n d y ϵ B) o r (y ϵ A a n d y ϵ C) \Rightarrow y ϵ A a n d (y ϵ B o r y ϵ C)$

[ $∵$ and' distributes 'or']

$\Rightarrow y ϵ A a n d y ϵ (B \cup C) \Rightarrow y ϵ A \cap (B \cup C)$ .

$∴ (A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C) \dots (i v)$

From (iii) and (iv), we get $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ .

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(Distributive laws) For any three sets A, B, C prove that: I. A∪(B∩C)=(A∪B)∩(A∪C) [Distirbutive law of union over intersection] II. A∩[(B∪C)]=(A∩B)∪(A∩C) [Distributive law of intersection over union]

(Distributive laws) For any three sets A, B, C prove that:

I. $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$
[Distirbutive law of union over intersection]

II. $A \cap [(B \cup C)] = (A \cap B) \cup (A \cap C)$
[Distributive law of intersection over union]