Q.Show that the relation R defined by R={(a,b):a−b is an integer} in the set Z of all integers is an equivalence relation.
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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. …
To qualify as an equivalence relation the relation must pass all three tests, reflexivity, symmetry and transitivity, each following from the fact that sums and negatives of integers are again integers. …
Verify the three defining properties — reflexivity, symmetry, transitivity — directly from the definition of R.
R={(a,b):a−b is an integer} on the set Z of all integers.
Reflexive: For any a∈Z, a−a=0, which is an integer. So (a,a)∈R for all a. Hence R is reflexive.
Symmetric: Let (a,b)∈R, i.e. a−b=k for some integer k. Then b−a=−k, which is also an integer. So (b,a)∈R. Hence R is symmetric.
Transitive: Let (a,b)∈R and (b,c)∈R, i.e. a−b=k1 and b−c=k2 for integers k1,k2. Then
a−c=(a−b)+(b−c)=k1+k2 …
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.
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- 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:
- Reflexive: (a,a)∈R for every a∈A.
- Symmetric: if (a,b)∈R then (b,a)∈R.
- Transitive: if (a,b)∈R and (b,c)∈R then (a,c)∈R. …
- 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.
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- 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.
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- 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. H …
- 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:
- Reflexive: (a,a)∈R for every a∈A.
- Symmetric: if (a,b)∈R then (b,a)∈R.
- Transitive: if (a,b)∈R and (b,c)∈R then (a,c)∈R. …
- 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. …
- 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.
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- 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).
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- 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)
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- CBSE 2020Set ANNUAL1 markQ.What is meant by an equivalence relation?
›Reveal solutionSolution
definition recall
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- 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. ✓
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