Q.Let X={1,2,3,4,5,6,7,8,9}. Let R1 be a relation in X given by R1={(x,y):x−y is divisible by 3} and R2 be another relation on X given by R2={(x,y):{x,y}⊂{1,4,7} or {x,y}⊂{2,5,8} or {x,y}⊂{3,6,9}}. Show that R1=R2.
Concept understanding — Equivalence Relation Proof
Proving a Relation is an Equivalence Relation
A relation R on a set A is an equivalence relation when it satisfies exactly three properties: it is reflexive, symmetric, and transitive. To prove a given relation is an equivalence relation, you check these three — in this order — one at a time.
Antisymmetry plays no role here; that property belongs to partial orders. For an equivalence relation you need only reflexive, symmetric, transitive.
The three checks
- Reflexive — show (a,a)∈R for every a∈A.
- Symmetric — assume (a,b)∈R and deduce (b,a)∈R.
- Transitive — assume (a,b)∈R and (b,c)∈R, and deduce (a,c)∈R.
If all three hold, R is an equivalence relation. If even one fails, produce a single counterexample and you are done.
A worked template
Let R be defined on Z by aRb⟺a−b is divisible by 5.
Reflexive: a−a=0, and 0 is divisible by 5, so aRa for every integer a. ✓
Symmetric: if aRb, then a−b=5k for some integer k. Then b−a=−5k=5(−k), also a multiple of 5, so bRa. ✓
Transitive: if aRb and bRc, then a−b=5k and b−c=5m. Adding, a−c=5(k+m), a multiple of 5, so aRc. ✓
All three hold, so R is an equivalence relation.
Once a relation is proved to be an equivalence relation, it splits A into disjoint equivalence classes — here, the five classes of integers grouped by their remainder on division by 5.
Reflexivity must hold for every element, not just some. A relation that pairs many elements correctly but misses even one self-pair is not reflexive, and so not an equivalence relation.
Searches like "how to prove a relation is an equivalence relation" and "equivalence relation class 12 maths examples" are common around exam time, reflecting how central this proof technique is to the NCERT/CBSE Class 12 Relations and Functions chapter. Mastering this three-step check (reflexive, symmetric, transitive) also pays off directly in JEE Main and state CET set-theory questions.
Concept: Equivalence Relation Proof — Two relations defined differently on the same set are equal if they contain exactly the same ordered pairs.
Step 1: Understand R1.
x−y divisible by 3 means x≡y(mod3). On X={1,…,9}, the residues modulo 3 partition the set into three classes:
{1,4,7}, {2,5,8}, {3,6,9}.
So (x,y)∈R1 iff x and y belong to the same residue class.
Step 2: Understand R2.
R2 explicitly pairs elements only if they lie together in one of those same three subsets: {1,4,7}, {2,5,8}, or {3,6,9}.
Step 3: Compare.
Both relations require x and y to be in the same one of these three groups. Hence every pair in R1 is in R2, and every pair in R2 is in R1.
R1=R2 because both relations consist of all ordered pairs whose elements belong to the same residue class modulo 3.
Both relations partition X into the same three residue classes modulo 3, so they are exactly the same set of ordered pairs. We prove R1=R2 by showing each is a subset of the other.
The key insight is that both R1 and R2 are describing the same thing: two numbers are related if and only if they leave the same remainder when divided by 3. R1 says this directly using divisibility. R2 says it by listing the three groups {1,4,7}, {2,5,8}, {3,6,9} — which are precisely the numbers that are congruent to 1, 2, and 0 modulo 3 respectively.
To prove two sets are equal, we show R1⊆R2 and R2⊆R1.
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Show R1⊆R2.
Take any (x,y)∈R1. By definition, x−y is divisible by 3, meaning x≡y(mod3).
Now look at the residues modulo 3 of the numbers in X:
- Numbers congruent to 0 mod 3: {3,6,9}
- Numbers congruent to 1 mod 3: {1,4,7}
- Numbers congruent to 2 mod 3: {2,5,8} Since x and y have the same residue, they must belong to the same one of these three sets. Therefore {x,y} is a subset of one of the three listed sets, which is exactly the condition for (x,y)∈R2. So R1⊆R2.
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Show R2⊆R1.
Take any (x,y)∈R2. Then {x,y} is contained in one of the three sets {1,4,7}, {2,5,8}, or {3,6,9}.
Within each of these sets, all numbers are congruent modulo 3:
- In {1,4,7}, each number ≡1(mod3)
- In {2,5,8}, each number ≡2(mod3)
- In {3,6,9}, each number ≡0(mod3) Hence x≡y(mod3), so x−y is divisible by 3. Thus (x,y)∈R1, and R2⊆R1.
Since R1⊆R2 and R2⊆R1, we have R1=R2.
A common mistake is to think R2 only relates pairs within the same listed set, but forgets that the condition {x,y}⊂{1,4,7} includes the case x=y (since a set with one element is still a subset). Both relations are reflexive, symmetric, and transitive — they are equivalence relations.
You can also see this by noting that R1 partitions X into three equivalence classes: [1]={1,4,7}, [2]={2,5,8}, [3]={3,6,9}. R2 explicitly defines the same partition. Two relations that generate the same partition are identical.
We have shown that R1=R2 by proving mutual inclusion, since both relations pair numbers with the same remainder modulo 3.
Method: Proving Two Relations Are Equal by Mutual Inclusion
When asked to show R1=R2 for two relations described in different ways, treat them as sets of ordered pairs and prove each is contained in the other. This double-inclusion technique works for any set-equality claim, not just relations.
Steps
Step 1: Restate both relations in one common language.
Here both descriptions are really about congruence modulo 3: "x−y divisible by 3" and "x,y lie in the same listed group" both say x≡y(mod3). Finding the shared underlying idea is what makes the two inclusions routine.
Step 2: Show R1⊆R2.
Take an arbitrary pair (x,y)∈R1, use its defining property, and deduce it satisfies R2's condition. (Same residue mod 3 ⇒ both lie in one of the listed classes.)
Step 3: Show R2⊆R1.
Take an arbitrary (x,y)∈R2 and deduce it satisfies R1's condition. (Sharing a listed class ⇒ same residue ⇒ difference divisible by 3.)
Step 4: Conclude equality.
Two inclusions in opposite directions give R1=R2. A quicker equivalent finish: show both relations induce the same partition into equivalence classes — identical partitions mean identical relations.
Common Mistakes
Mistake 1: Proving only one inclusion.
Why it's wrong: showing R1⊆R2 alone does not prove equality; the two relations could differ if R2 has extra pairs. Correct approach: prove both R1⊆R2 and R2⊆R1 (or show they induce the identical partition).
Mistake 2: Miscomputing the residue classes modulo 3.
Why it's wrong: students sometimes group {3,6,9} as residue 3; on division by 3 these leave remainder 0, not 3. A wrong grouping breaks the correspondence between R1 and R2. Correct approach: compute remainders carefully — {1,4,7}→1, {2,5,8}→2, {3,6,9}→0.
Mistake 3: Forgetting that {x,y}⊂{1,4,7} includes the case x=y.
Why it's wrong: a single-element subset like {1} still satisfies the subset condition, so the reflexive pairs (x,x) are genuinely in R2. Overlooking this makes R2 look smaller than R1. Correct approach: treat equal-element pairs as included in both relations.
Showing the 12 most recent of 13 on this concept.
- CBSE 2026Set ANNUAL1 markQ.Write the answer in one word/sentence: Define equivalence relation.
›Reveal solutionSolution
Equivalence relation = reflexive + symmetric + transitive.
A relation R on a set A is an equivalence relation if it is reflexive (aRa for all a), symmetric (aRb⇒bRa), and transitive (aRb,bRc⇒aRc).
✓Final answerA relation which is reflexive, symmetric and transitive.
- CBSE 2025Set ANNUAL1 markQ.Define an equivalence relation. OR If R={(1,−1),(2,−2),(3,−1)} is a relation, then find the domain and the range of R.
›Reveal solutionSolution
State the three defining properties of an equivalence relation.
A relation R on a non-empty set A is called an equivalence relation if it satisfies all three of the following:
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Reflexive: (a,a)∈R for every a∈A.
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Symmetric: if (a,b)∈R then (b,a)∈R.
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Transitive: if (a,b)∈R and (b,c)∈R then (a,c)∈R.
✓Final answerR is an equivalence relation iff it is reflexive, symmetric and transitive.
Alternative (Or):
Read off first components (domain) and second components (range) of the ordered pairs.
For R={(1,−1),(2,−2),(3,−1)}:
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The domain is the set of all first coordinates: {1,2,3}.
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The range is the set of all second coordinates: {−1,−2,−1}={−1,−2}.
✓Final answerDomain ={1,2,3}, Range ={−1,−2}
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- CBSE 2024Set A1 markQ.A Relation R in a set A is said to be ______ relation if R is reflexive, symmetric, and transitive.
›Reveal solutionSolution
A relation that is reflexive, symmetric and transitive is called an equivalence relation.
By definition (NCERT), a relation R in a set A is called an equivalence relation if it is reflexive (every a∈A satisfies (a,a)∈R), symmetric ((a,b)∈R⇒(b,a)∈R), and transitive ((a,b)∈R,(b,c)∈R⇒(a,c)∈R). Such a relation partitions A into disjoint equivalence classes.
✓Final answerequivalence relation.
- CBSE 2024Set ANNUAL1 markMCQQ.Let R be the relation in the set Z of all integers defined as, R = {(x, y) : x – y is an integer}, then R is(a) Reflexive(b) Symmetric(c) Transitive(d) Equivalence relation
›Reveal solutionSolution
R is reflexive, symmetric, and transitive, hence an equivalence relation.
For any x∈Z, x−x=0, which is an integer, so (x,x)∈R for every x. Hence R is reflexive.
If (x,y)∈R, then x−y is an integer. Then y−x=−(x−y) is also an integer, so (y,x)∈R. Hence R is symmetric.
If (x,y)∈R and (y,z)∈R, then x−y and y−z are both integers. Then x−z=(x−y)+(y−z) is a sum of two integers, hence an integer. So (x,z)∈R. Hence R is transitive.
Since R is reflexive, symmetric and transitive, R is an equivalence relation.
✓Final answer(d) Equivalence relation
- CBSE 2023Set ANNUAL1 markQ.If R is an equivalence relation on A, then link the domain of R and the range of R.
›Reveal solutionSolution
For an equivalence relation R on A, both the domain and the range of R equal A itself.
Since R is an equivalence relation on A, it is reflexive: for every a∈A, (a,a)∈R. This means every element of A appears as a first coordinate (so it is in the domain) and also as a second coordinate (so it is in the range) of some pair in R. Hence Domain(R)=A and Range(R)=A, i.e. the domain and range of R coincide and both equal A.
✓Final answerDomain(R) = Range(R) = A.
- CBSE 2023Set ANNUAL1 markQ.Define an equivalence relation. OR A relation R in the set N of natural numbers is defined as R={(x,y):y=x+5 and x<4}. Find the range of R.
›Reveal solutionSolution
An equivalence relation is one that is simultaneously reflexive, symmetric and transitive; the alternative just lists the images of the allowed x-values.
A relation R on a set A is called an equivalence relation when it satisfies all three of the following properties:
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Reflexive: (a,a)∈R for every a∈A.
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Symmetric: if (a,b)∈R then (b,a)∈R.
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Transitive: if (a,b)∈R and (b,c)∈R then (a,c)∈R.
✓Final answerA relation R on a set is an equivalence relation if it is reflexive, symmetric and transitive.
Alternative (Or):
List the natural numbers x<4, apply y=x+5, and collect the y-values as the range.
Here R={(x,y):y=x+5 and x<4} with x∈N.
The natural numbers satisfying x<4 are x=1,2,3. Applying y=x+5:
x=1⇒y=6,x=2⇒y=7,x=3⇒y=8.
So R={(1,6),(2,7),(3,8)}, and the range is the set of second coordinates.
✓Final answerRange of R={6, 7, 8}.
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- CBSE 2023Set ANNUAL1 markQ.Define an equivalence relation.
›Reveal solutionSolution
An equivalence relation is a relation satisfying all three properties: reflexivity, symmetry, and transitivity.
A relation R defined on a set A is called an equivalence relation if it satisfies all of the following three conditions:
- Reflexive: (a,a)∈R for every a∈A.
- Symmetric: If (a,b)∈R, then (b,a)∈R, for all a,b∈A.
- Transitive: If (a,b)∈R and (b,c)∈R, then (a,c)∈R, for all a,b,c∈A.
An equivalence relation partitions the set A into disjoint equivalence classes.
✓Final answerA relation R on a set A that is reflexive, symmetric, and transitive is called an equivalence relation.
- CBSE 2023Set ANNUAL1 markMCQQ.Case study based question: An organization conducted a bike race under two different categories- boys and girls. Totally there were 250 participants, out of which three from category 1 and two from category 2 were selected for the final race. John forms two sets B and G with these participants for his college project. Let B={b1,b2,b3} and G={g1,g2} where B and G represents the set of boys and girls respectively, who were selected for the final race. Answer the following using the above information. John wishes to form all the relations possible from B to G. How many such relations are possible?(a) 26(b) 25(c) 0(d) 23
›Reveal solutionSolution
The number of relations from a set of size m to a set of size n is 2mn; here m=3,n=2.
B={b1,b2,b3} has ∣B∣=3 elements; G={g1,g2} has ∣G∣=2 elements.
A relation from B to G is any subset of B×G. The number of elements in B×G is ∣B∣×∣G∣=3×2=6.
The number of subsets of a set with 6 elements is 26.
So the number of relations possible from B to G is 26.
✓Final answer(a) 26
- CBSE 2022Set ANNUAL1 markMCQQ.Which of the following relations on A = {1, 2, 3} is an equivalence relation?(a) {(1, 1), (2, 2), (3, 3)}(b) {(1, 1), (2, 2), (3, 3), (1, 2)}(c) {(1, 1), (3, 3), (1, 3), (3, 1)}(d) None of these
›Reveal solutionSolution
Only option (a) is reflexive, symmetric, and transitive at once - the other two options fail reflexivity or symmetry.
An equivalence relation on A={1,2,3} must be reflexive (contain (1,1),(2,2),(3,3)), symmetric (if (x,y) is in it so is (y,x)), and transitive.
(a) {(1,1),(2,2),(3,3)} - contains all three diagonal pairs (reflexive), has no off-diagonal pair to break symmetry, and is trivially transitive. This IS an equivalence relation (it is the identity relation on A).
(b) {(1,1),(2,2),(3,3),(1,2)} - reflexive, but (1,2) is present while (2,1) is not, so it fails symmetry.
(c) {(1,1),(3,3),(1,3),(3,1)} - (2,2) is missing, so it fails reflexivity (2 is not related to itself).
So among the given options, only (a) satisfies all three properties.
✓Final answer(a) {(1,1),(2,2),(3,3)} is the equivalence relation.
- CBSE 2022Set ANNUAL1 markMCQQ.Case study based question: Students of class-XII planned to plant saplings along straight lines, parallel to each other to one side of the playground ensuring that they had enough play area. Let us assume that they planted one of the rows of the saplings along the line y=x−4. Let L be the set of all lines which are parallel on the ground and R be a relation on L. Answer the following using the above information. Let relation R be defined by R={(L1,L2):L1∥L2 where L1,L2∈L}, then R is _______ relation.(a) Equivalence(b) Only reflexive(c) Not reflexive(d) symmetric but not transitive
›Reveal solutionSolution
The relation 'is parallel to' on a set of lines is reflexive, symmetric and transitive, hence an equivalence relation.
R={(L1,L2):L1∥L2} on the set L of all lines.
Reflexive: every line is parallel to itself, so (L,L)∈R for all L. (True)
Symmetric: if L1∥L2 then L2∥L1, so (L1,L2)∈R⇒(L2,L1)∈R. (True)
Transitive: if L1∥L2 and L2∥L3, then L1∥L3, so (L1,L2),(L2,L3)∈R⇒(L1,L3)∈R. (True)
Since R is reflexive, symmetric, and transitive, it is an equivalence relation.
✓Final answer(a) Equivalence
- CBSE 2020Set ANNUAL1 markQ.What is meant by an equivalence relation?
›Reveal solutionSolution
definition recall
A relation R on a set A is an equivalence relation if it is reflexive ((a,a)∈R for all a), symmetric ((a,b)∈R⇒(b,a)∈R), and transitive ((a,b),(b,c)∈R⇒(a,c)∈R).
✓Final answerA relation that is simultaneously reflexive, symmetric and transitive.
- CBSE 2019Set ANNUAL1 markMCQQ.Write the correct option from the following if R{(a,b):a and b both are either even or odd} in the set A{1,2,3,4,5,6,7}:(a) No relation(b) Trivial(c) Equivalence relation(d) Not symmetric
›Reveal solutionSolution
“Same parity” partitions A into evens and odds, so R is reflexive, symmetric and transitive — an equivalence relation.
R = {(a, b) : a and b are both even or both odd} on A = {1,2,3,4,5,6,7}.
Step 1 (Reflexive): Any a has the same parity as itself, so (a, a) ∈ R. ✓
Step 2 (Symmetric): If a and b have the same parity, so do b and a; (a,b) ∈ R ⇒ (b,a) ∈ R. ✓
Step 3 (Transitive): If a, b same parity and b, c same parity, then a, c same parity; (a,b),(b,c) ∈ R ⇒ (a,c) ∈ R. ✓
All three properties hold, so R is an equivalence relation.
✓Final answer(c) Equivalence relation.
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