Prob. Let A,B,C be any three sets, then prove that-
AX(B∩C) = (AXB) ∩ (AXC)
Solution:
(x,y) ∈ Ax(B∩C)
x∈A and (y∈(B∩C))
x∈A and (y∈B and y∈C)
(x∈A and y∈B) and (x∈A and y∈C)
(x,y) ∈ (A x B) and (x,y) ∈(A x C) // by Cartesian Product.
(x,y) ∈ (AxB)∩(AxC)
Prob. Prove that-
A∩(B∪C) = (A∩B) ∪ (A∩C)
Solution:
Let x ∈ A ∩ (B U C).
Then x ∈ A and x ∈ (B U C).
(x ∈ A and x ∈ B) or (x ∈ A and x ∈ c).
x ∈ (A and B) or x ∈ ( A and c).
x ∈ (A ∩ B) U ( A ∩ C).
Prob. If A, B, C, D are any four sets then prove that –
(A∩B)X(C∩D) = (AXC)∩(BXD)
Solution:
Consider(x,y)
(x,y)∈(A∩B)×(C∩D)
x∈(A∩B) ∧ y∈(C∩D)
(x∈A and x∈B) ∧ (y∈C and y∈D)
(x∈A ∧ y∈C) and (x∈B ∧ y∈D)
(x,y)∈(A∧C) and (x,y)∈(B∧D)
(x,y)∈((A∧C) and (B∧D))
(x,y)∈((A×C) ∩ (B×D))
(A×C) ∩ (B×D)
Prob. Show that-
(P∩Q)X(R∩S) = (PXR)∩(QXS)
For some arbitrary sets P, Q, R and S
Solution:
Consider(x,y)
(x,y)∈(P∩Q)X(R∩S)
x∈(P∩Q) ∧ y∈(R∩S)
(x∈P and x∈Q) ∧ (y∈R and y∈S)
(x∈P ∧ y∈R) and (x∈Q ∧ y∈S)
(x,y)∈(P∧R) and (x,y)∈Q∧S)
(x,y)∈((P∧R) and (Q∧S))
(x,y)∈((P×R) ∩ (Q×S))
(PXR)∩(QXS)