How do I check whether a relation is an equivalence relation in Class 12 Relations and Functions?
GradeHunt_TeamGradeHunt Team · · Haryana Board 12th
HBSE 12th Maths, first chapter Relations and Functions. I know the definitions of reflexive, symmetric and transitive, but when a relation is given as a set of ordered pairs or as a rule like a - b divisible by 3, I don't know how to actually prove each property. Please explain with an example.
equivalence relationrelations and functionsclass 12 mathshbse
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A relation R on a set A is an equivalence relation if it is reflexive, symmetric and transitive. You must prove all three. If even one fails, it is not an equivalence relation.
What to show for each:
1. Reflexive: (a, a) is in R for every a in A.
2. Symmetric: if (a, b) is in R, then (b, a) is in R.
3. Transitive: if (a, b) and (b, c) are in R, then (a, c) is in R.
Example 1 (a rule): R = {(a, b) : a - b is divisible by 3} on the set of integers Z.
Reflexive: a - a = 0, and 0 is divisible by 3. So (a, a) is in R for every integer a.
Symmetric: let (a, b) be in R, so a - b = 3k for some integer k. Then b - a = -3k = 3(-k), which is divisible by 3. So (b, a) is in R.
Transitive: let (a, b) and (b, c) be in R, so a - b = 3k and b - c = 3m. Adding, a - c = 3(k + m), which is divisible by 3. So (a, c) is in R.
All three hold, so R is an equivalence relation.
Example 2 (a set of pairs): A = {1, 2, 3}, R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}.
Reflexive: yes, (1,1), (2,2) and (3,3) are all there.
Symmetric: no. (1, 2) is in R but (2, 1) is not.
Transitive: no. (1, 2) and (2, 3) are in R but (1, 3) is not.
So R is reflexive only, and not an equivalence relation.
Tips:
1. To disprove a property, one counter-example is enough. Give the exact pair.
2. To prove a property for a rule, use general letters (a, b, c), not one numerical case.
3. Two out of three is not enough. 'a is less than or equal to b' on real numbers is reflexive and transitive, but not symmetric (2 is less than or equal to 5, but 5 is not less than or equal to 2), so it is not an equivalence relation.