Miscellaneous

Misc 1

Misc 2 (i)

Misc 2 (ii) Important

Misc 2 (iii) Important

Misc 2 (iv)

Misc 2 (v)

Misc 2 (vi) Important

Misc 3

Misc 4 Important

Misc 5

Misc 6

Misc 7 Important

Misc 8 Important

Misc 9 Important You are here

Misc 10

Misc 11

Misc 12 Important

Misc 13 Important

Misc 14

Misc 15 Important

Misc 16 Important

Last updated at Jan. 27, 2020 by Teachoo

Misc 9 Using properties of sets show that A ∪ (A ∩ B) = A In order to prove A ∪ (A ∩ B) = A, we should prove A ∪ (A ∩ B) is a subset of A i.e. A ∪ (A ∩ B) ⊂ A & A is a subset of A ∪ (A ∩ B) i.e. A ⊂ A ∪ (A ∩ B) As set is a subset of itself, A ⊂ A Also, A is a subset of A ∩ B , i.e. A ⊂ A ∩ B as all elements of set A are in A ∩ B ∪ Union - Combination of two sets ∩ Intersection – Common of two sets Now, A ∪ (A ∩ B) Using distributive law :A ∪ (B ∩ C)= (A ∪ B) ∩ (A ∪ C) = (A) ∩ (A U B) = A = R.H.S Thus, A ∪ (A ∩ B) = A Hence proved Misc 9 Using properties of sets show that A ∪ (A ∩ B) = A Solving L.H.S A ∪ (A ∩ B) Using distributive law :A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) = = (A) ∩ (A U B) = A = R.H.S Thus, A ∪ (A ∩ B) = A Hence proved ∪ Union - Combination of two sets ∩ Intersection – Common of two sets As 2 × (3 + 4) = 2 × 3 + 2 × 4 Misc 9 Using properties of sets show that (ii) A ∩ (A ∪ B) = A. Taking L.H.S A ∩ (A ∪ B) Using distributive law :A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) = = (A) U (A ∩ B) = A = R.H.S Thus A ∩ (A ∪ B) = A. Hence proved ∪ Union - Combination of two sets ∩ Intersection – Common of two sets (From part (i) , s A ∪ (A ∩ B)= A)