Back to Concepts
MathsMediumClass 11
Relations — Equivalence relation, types
Sets, Relations & Functions
10
JEE Qs
8%
Hard
75
min
Master the precise definitions of reflexive, symmetric, and transitive relations, and practice proving/disproving them using general elements or clear counterexamples.
🧮 Key Formulas
A relation R on a set A is reflexive if (a,a) ∈ R for all a ∈ A.
A relation R on a set A is symmetric if (a,b) ∈ R implies (b,a) ∈ R for all a,b ∈ A.
A relation R on a set A is transitive if (a,b) ∈ R and (b,c) ∈ R implies (a,c) ∈ R for all a,b,c ∈ A.
A relation R on a set A is an equivalence relation if R is reflexive, symmetric, and transitive.
For an equivalence relation R on A, the equivalence class of an element a ∈ A is [a] = {x ∈ A | (x,a) ∈ R}.
✅ Key Points for JEE
- 1To prove a relation is an equivalence relation, explicitly verify all three properties (reflexive, symmetric, transitive) using general elements, not specific examples.
- 2To disprove a relation from being an equivalence relation, find a single counterexample for any of the three properties.
- 3For transitivity, remember the condition is 'if (a,b) ∈ R and (b,c) ∈ R'. If no such pair (a,b) and (b,c) exists for any a,b,c, the relation is vacuously transitive.
- 4Equivalence classes partition the set into disjoint, non-empty subsets. Every element belongs to exactly one equivalence class.
- 5Problems often involve relations defined by 'a divides b', 'a is congruent to b modulo m', or geometric conditions (e.g., 'lines are parallel').
⚠️ Common Mistakes
- ✕Failing to check the 'for all' condition for reflexivity; e.g., only checking a few elements instead of ensuring (a,a) is in R for *every* element in the set.
- ✕Misinterpreting transitivity, especially in cases where the premise (a,b) ∈ R and (b,c) ∈ R is not met, leading to incorrect conclusions about its absence.
- ✕Confusing symmetric property with transitive property, or incorrectly assuming one implies the other.
- ✕Not understanding that equivalence classes are disjoint or that their union forms the entire set, leading to errors in partitioning problems.
NCERT Chapters
- Class 12 Maths Ch 1: Relations and Functions