Misc 3 - Chapter 1 Class 12 Relation and Functions
Last updated at April 16, 2024 by Teachoo
To prove relation reflexive, transitive, symmetric and equivalent
Example 4 Important
Ex 1.1, 6
Ex 1.1, 15 (MCQ) Important
Ex 1.1, 7
Ex 1.1, 1 (i)
Ex 1.1, 2
Ex 1.1, 3
Ex 1.1, 4
Ex 1.1, 5 Important
Ex 1.1, 10 (i)
Ex 1.1, 8
Ex 1.1, 9 (i)
Example 5
Example 6 Important
Example 2
Ex 1.1, 12 Important
Ex 1.1, 13
Ex 1.1, 11
Example 3
Ex 1.1, 14
Misc 3 Important You are here
Example 19 Important
Example 18
To prove relation reflexive, transitive, symmetric and equivalent
Last updated at April 16, 2024 by Teachoo
Misc 3 (Introduction) Given a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if A â B. Is R an equivalence relation on P(X)? Justify you answer: Taking an example Let X = {1, 2, 3} P(X) = Power set of X = Set of all subsets of X = { ð, {1} , {2} , {3}, {1, 2} , {2, 3} , {1, 3}, {1, 2, 3} } Since {1} â {1, 2} âī {1} R {1, 2} If A â B, all elements of A are in B Misc 3 Given a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if A â B. Is R an equivalence relation on P(X)? Justify you answer: ARB means A â B Here, relation is R = {(A, B): A & B are sets, A â B} Check reflexive Since every set is a subset of itself, A â A âī (A, A) â R. âīR is reflexive. Check symmetric To check whether symmetric or not, If (A, B) â R, then (B, A) â R If (A, B) â R, A â B. But, B â A is not true Example: Let A = {1} and B = {1, 2}, As all elements of A are in B, A â B But all elements of B are not in A (as 2 is not in A), So B â A is not true âī R is not symmetric. If A â B, all elements of A are in B Checking transitive Since (A, B) â R & (B, C) â R If, A â B and B â C. then A â C â (A, C) â R So, If (A, B) â R & (B, C) â R , then (A, C) â R âī R is transitive. Hence, R is reflexive and transitive but not symmetric. Hence, R is not an equivalence relation since it is not symmetric.