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"Show that R is an equivalence relation" is one of the most predictable questions in Class 12 mathematics, and one of the easiest to lose marks on, usually by checking a few numbers instead of proving a general statement. This post gives you a proof template, works through a full proof, a relation that fails, equivalence classes, and the counting questions NCERT and JEE ask.
Where is this in the syllabus?
The CBSE 2026-27 mathematics curriculum lists, in Class XII Unit I, "Types of relations: reflexive, symmetric, transitive and equivalence relations." NCERT's Class 12 Chapter 1, Relations and Functions (2026-27 reprint) defines each property and introduces equivalence classes. The JEE (Main) 2026 syllabus lists "Relations, type of relations, equivalence relations" in Unit 1, and the JEE (Advanced) 2026 brochure lists "equivalence relation". Check the current documents for your exam year.
For the Class 11 basics, including testing a relation given as a list of pairs, see our post on sets and relations.
How should a proof of each property be written?
NCERT's definitions, for a relation R on a set A:
Reflexive: (a, a) ∈ R for every a ∈ A.
Symmetric: (a₁, a₂) ∈ R implies (a₂, a₁) ∈ R, for all a₁, a₂ ∈ A.
Transitive: (a₁, a₂) ∈ R and (a₂, a₃) ∈ R imply (a₁, a₃) ∈ R.
To prove a property, start with arbitrary elements ("Let a ∈ A") and show the condition holds using only the rule that defines R. To disprove one, give one specific counterexample with numbers. Checking examples never proves a property; one counterexample always disproves it.
Worked example 1: a full proof
On N × N, define (a, b) R (c, d) if a + d = b + c. Show that R is an equivalence relation.
Reflexive. For any (a, b), a + b = b + a, so (a, b) R (a, b).
Symmetric. If (a, b) R (c, d), then a + d = b + c. Rewriting, c + b = d + a, so (c, d) R (a, b).
Transitive. If (a, b) R (c, d) and (c, d) R (e, f), then a + d = b + c and c + f = d + e. Adding, a + d + c + f = b + c + d + e. Cancelling c + d gives a + f = b + e, so (a, b) R (e, f).
So R is an equivalence relation. The equivalence class of (2, 5) is every (c, d) with 2 + d = 5 + c, that is, d − c = 3: (1, 4), (2, 5), (3, 6) and so on. Each class collects the pairs with the same difference.
Worked example 2: a relation that fails
On the real numbers, define a R b if |a − b| ≤ 1.
Reflexive: |a − a| = 0 ≤ 1.
Symmetric: |b − a| = |a − b|.
Not transitive: 0 R 1 and 1 R 2, but |0 − 2| = 2 > 1, so 0 is not related to 2.
One counterexample is a complete answer for the failing property.
What are equivalence classes?
NCERT shows that an equivalence relation divides its set into disjoint subsets, the equivalence classes, whose union is the whole set, and that any such division defines an equivalence relation in return.
Example. On A = {1, 2, …, 10}, let a R b if a − b is divisible by 4. The classes are {1, 5, 9}, {2, 6, 10}, {3, 7} and {4, 8}. Their sizes, 3 + 3 + 2 + 2, add to 10, and no number is in two classes. A quick check like this catches missing or repeated elements.
How do you count equivalence relations on a small set?
Because equivalence relations match divisions into classes, counting one counts the other. On {1, 2, 3} there are 5 divisions: {1}{2}{3}, {1, 2}{3}, {1, 3}{2}, {2, 3}{1} and {1, 2, 3}. So there are 5 equivalence relations.
NCERT's Example 24 and the chapter's miscellaneous exercise ask how many contain (1, 2). The class containing 1 must also contain 2, which leaves {1, 2}{3} and {1, 2, 3}: 2 relations. The same exercise also asks how many relations containing (1, 2) and (1, 3) are reflexive and symmetric but not transitive. The answer is 1: all three (a, a) pairs, plus (1, 2), (2, 1), (1, 3), (3, 1). It is not transitive because (2, 1) and (1, 3) are present but (2, 3) is not.
Which mistakes cost the most marks?
Proving with numbers. "2 R 2 holds" does not show reflexivity.
Missing the a = c case in transitivity. (1, 2) and (2, 1) in R require (1, 1).
Forgetting that reflexive means every element. {(1, 1), (2, 2)} on {1, 2, 3} is not reflexive.
Assuming symmetric and transitive imply reflexive. The empty relation on a non-empty set is symmetric and transitive but not reflexive.
A practice routine
Saying why each line follows is what makes a proof stick. In Chi, Bassok, Lewis, Reimann and Glaser (1989), students who learned well from worked examples explained the steps to themselves and linked them to principles, while weaker students relied on copying the examples.
Annotate one NCERT proof per day, writing why each line follows.
Write three proofs a week from scratch: one equivalence relation, one that fails, one counting question.
List the classes for every equivalence relation you prove, and check that their sizes add up.
If a proof keeps falling apart at the same property, Fermi's AI tutor can work out what you already know and build a plan for the session, so you practise the step you keep missing.
Today, prove or disprove that "a R b if ab > 0" on the non-zero integers is an equivalence relation, and list its classes if it is one.
