Show that- (P∩Q)X(R∩S) = (PXR)∩(QXS)
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 … Read more
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 … Read more
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 … Read more
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 ∈ … Read more
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) … Read more
Mathematical induction is a unique and special way to prove the things, in only two steps. Step 1. Show that it is true for n = … Read more