Equivalence Relations
On the set of all integers, define a relation R by aRb if and only if the difference a-b is divisible by 5; which property fails to hold so that R may or may not be an equivalence relation?
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About This Question
- Subject
- mathematics
- Chapter
- sets, relations and functions
- Topic
- equivalence relations
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
None, R is an equivalence relation
An equivalence relation must be reflexive, symmetric, and transitive, a definition central to partitioning sets in JEE Advanced. Reflexivity: a-a = 0 is divisible by 5, so aRa holds for every integer. Symmetry: if a-b is divisible by 5 then b-a = -(a-b) is also divisible by 5, so bRa holds. Transitivity: if a-b and b-c are each multiples of 5, their sum (a-b)+(b-c)=a-c is a multiple of 5, so aRc holds. All three properties are satisfied, making R the congruence-modulo-5 relation that partitions integers into five residue classes. Option reflexivity-fails is false because zero is divisible by 5. Option symmetry-fails is false since divisibility is preserved under negation. Option transitivity-fails is false because sums of multiples of 5 remain multiples of 5. Hence R is a genuine equivalence relation. Plausibility check: the five classes {...,-5,0,5,...} through {...,-1,4,9,...} are disjoint and cover all integers, exactly as an equivalence relation requires. Equivalence relations and partitions are two faces of the same structure, since every partition of a set induces an equivalence relation and every equivalence relation produces a partition into disjoint classes. Treating congruence modulo a fixed integer as the canonical prototype lets a student verify reflexivity, symmetry and transitivity almost mechanically, which is exactly the efficiency JEE Advanced rewards under time pressure.
This medium difficulty mathematics question is from the chapter sets, relations and functions, covering the topic of equivalence relations. It appeared in the 2025 exam.
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