Let P, Q, R be subsets of a Universal set U.
Prove that (i) Px(QR) =(PxQ)(PxR)
(ii) (P-Q)xR=(PxR)-(QxR)
Answers
Step-by-step explanation:
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Concept
Cartesian product of two sets A and B is defined as the set of ordered pairs (a,b) such that a∈ A and b ∈ B. In set-builder notation,
A × B = {(a, b) | a ∈ A ∧ b ∈ B}.
Difference of two sets is defined as
A - B = {x | x ∈ A ∧ x ∉ B}.
Union of two sets A and B is defined as
A∪B = {x | x ∈ A or x ∈ B}.
Prove
- P × (Q ∪ R) = (P × Q) ∪ (P × R)
- (P - Q) × R = (P × R) - (Q × R)
Proof
1. Let (p, x) ∈ P × (Q ∪ R)
⇒ p ∈ P and x ∈ (Q ∪ R) (Using the definition of cartesian product)
⇒ p ∈ P and (x ∈ Q or x ∈ R) (Using the definition of union)
⇒ (p ∈ P and x ∈ Q) or (p ∈ P and x ∈ R) (and is distributive over or)
⇒ (p, x) ∈ P × Q or (p, x) ∈ P × R (Using the definition of cartesian product)
⇒ (p, x) ∈ (P × Q) ∪ (P × R) (Using the definition of union)
⇒ P × (Q ∪ R) ⊆ (P × Q) ∪ (P × R).
Let (p, x) ∈ (P × Q) ∪ (P × R)
⇒ (p, x) ∈ P × Q or (p, x) ∈ P × R
⇒ (p ∈ P and x ∈ Q) or (p ∈ P and x ∈ R)
⇒ p ∈ P and (x ∈ Q or x ∈ R)
⇒ p ∈ P and x ∈ (Q ∪ R)
⇒ (p, x) ∈ P × (Q ∪ R)
⇒ (P × Q) ∪ (P × R) ⊆ P × (Q ∪ R).
If two sets are subsets of each other then the two sets are equal. So,
P × (Q ∪ R) = (P × Q) ∪ (P × R).
Hence, proved.
2. Let (x, r) ∈ (P - Q) × R
⇔ x ∈ (P - Q) and r ∈ R
⇔ (x ∈ P and x ∉ Q) and r ∈ R
⇔ (x ∈ P and x ∈ Q') and r ∈ R (x ∈ U and x ∉ A ⇒ x ∈ U - A = A')
The above statement is same as
(x ∈ P and r ∈ R) and (x ∈ Q' and r ∈ R)
⇔ (x ∈ P and r ∈ R) and (x ∉ Q and r ∈ R)
⇔ x ∈ (P × R) and x ∉ (Q × R)
⇔ x ∈ (P × R) - (Q × R) (Using the definition of set difference)
⇒ (P - Q) × R ⊆ (P × R) - (Q × R) and (P × R) - (Q × R) ⊆ (P - Q) × R
⇒ (P - Q) × R = (P × R) - (Q × R).
Hence, proved.
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