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

Misc 10 You are here

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 10 Show that A ∩ B = A ∩ C need not imply B = C. We have to prove false, so we take a example It is given that A ∩ B = A ∩ C i.e. Common element in set A & B = Common element in set A & C Let A = {0, 1}, B = {0, 2, 3}, and C = {0, 4, 5} A ∩ B = {0} and A ∩ C = {0} Here, A ∩ B = A ∩ C = {0} But B ≠ C as 2 is in set B, but not in A Hence proved Misc 11 Let A and B be sets. If A ∩ X = B ∩ X = ∅ and A ∪ X = B ∪ X for some set X, show that A = B. (Hints: A = A ∩ (A ∪ X), B = B ∩ (B ∪ X) and use distributive law) Given: Let A and B be two sets such that A ∩ X = B ∩ X = ∅ and A ∪ X = B ∪ X for some set X. To prove: A = B Proof: Let A = A ∩ (A ∪ X) A = A ∩ (B ∪ X) (Given A ∪ X = B ∪ X) Let A = A ∩ (A ∪ X) Given A ∪ X = B ∪ X A = A ∩ (B ∪ X) Using distributive law : A ∩ (B ∪ C)= (A ∩ B) ∪ (A ∩ C) = (A ∩ B) ∪ (A ∩ X) As A ∩ X = ∅ given = (A ∩ B) ∪ ∅ A = A ∩ B Let B = B ∩ (B ∪ X) Given A ∪ X = B ∪ X B = B ∩ (A ∪ X) Using distributive law: A ∪ (B ∩ C)= (A ∩ B) ∪ (A ∩ C) = (B ∩ A) ∪ (B ∩ X) As B ∩ X = Φ = (B ∩ A) ∪ Φ B = B ∩ A B = A ∩ B From (1) and (2), A = A ∩ B & B = A ∩ B ∴ A = B Hence proved