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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
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Misc. 8 Important You are here
Example 42 Important
Example 41
To prove relation reflexive, transitive, symmetric and equivalent
Last updated at Jan. 30, 2020 by Teachoo
Misc 8 (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 β 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 = { π, {1} , {2} , {3}, {1, 2} , {2, 3} , {1, 3}, {1, 2, 3} } Since {1} β {1, 2} β΄ {1} R {1, 2} If A β B, all elements of A are in B Misc 8 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 β B. Is R an equivalence relation on P(X)? Justify you answer: ARB means A β B Here, relation is R = {(A, B): A & B are sets, A β B} Check reflexive Since every set is a subset of itself, A β A β΄ (A, A) β R. β΄R is reflexive. Check symmetric To check whether symmetric or not, If (A, B) β R, then (B, A) β R If (A, B) β R, A β B. But, B β A is not true Example: Let A = {1} and B = {1, 2}, As all elements of A are in B, A β B But all elements of B are not in A (as 2 is not in A), So B β A is not true β΄ R is not symmetric. If A β B, all elements of A are in B Checking transitive Since (A, B) β R & (B, C) β R If, A β B and B β C. then A β C β (A, C) β R So, If (A, B) β R & (B, C) β R , then (A, C) β R β΄ R is transitive. Hence, R is reflexive and transitive but not symmetric. Hence, R is not an equivalence relation since it is not symmetric.