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Types of Relations: Reflexive, Symmetric, Transitive and Equivalence

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Student-friendly explanation

How to write this in exams

1
Start with the exact idea
A relation R on a set A is a subset of A × A. It is reflexive if every a in A satisfies (a, a) in R, symmetric if (a, b) in R implies (b, a) in R, and transitive if (a, b) in R and (b, c) in R imply (a, c) in R. A relation that is reflexive, symmetric, and transitive is an equivalence relation.
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Then show how to use it
1. Identify the set A and list the ordered pairs of R. 2. Check reflexivity by verifying every diagonal pair. 3. Check symmetry by reversing each non-diagonal pair. 4. Check transitivity by testing chains of the form (a,b) and (b,c). 5. State the final classification clearly.
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Add one concrete example
On A = {1, 2, 3}, R = {(1,1), (2,2), (3,3), (1,2), (2,1)} is reflexive because all diagonal pairs are present. It is symmetric because (1,2) and (2,1) occur together. It is not transitive if a required pair produced by two connected pairs is missing; here the listed pairs do not create a missing transitive requirement, so transitivity must be checked pair by pair.
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Avoid this incomplete answer
Marking a relation as transitive without checking chains such as (a,b) and (b,c), especially when the required pair (a,c) is absent.

Definition

Example

On A = {1, 2, 3}, R = {(1,1), (2,2), (3,3), (1,2), (2,1)} is reflexive because all diagonal pairs are present. It is symmetric because (1,2) and (2,1) occur together. It is not transitive if a required pair produced by two connected pairs is missing; here the listed pairs do not create a missing transitive requirement, so transitivity must be checked pair by pair.

Rule to remember

Reflexive: for every a in A, (a,a) in R. Symmetric: (a,b) in R implies (b,a) in R. Transitive: (a,b) in R and (b,c) in R imply (a,c) in R. Equivalence relation: all three properties hold together. Empty relation on non-empty A is not reflexive; universal relation on A is reflexive, symmetric, and transitive.

Memory hook

Reflexive checks self, symmetric checks reverse, transitive checks chain.

Examples and method

Worked example

Let A = {1,2,3} and R = {(1,1),(2,2),(3,3),(1,2),(2,1)}. Reflexive: (1,1),(2,2),(3,3) are present. Symmetric: (1,2) has (2,1), and all diagonal pairs reverse to themselves. Transitive: check non-trivial chains: (1,2) and (2,1) give (1,1), present; (2,1) and (1,2) give (2,2), present. Other chains involve diagonal pairs and remain present. Hence R is reflexive, symmetric, and transitive, so it is an equivalence relation.

Method to apply

1. Identify the set A and list the ordered pairs of R. 2. Check reflexivity by verifying every diagonal pair. 3. Check symmetry by reversing each non-diagonal pair. 4. Check transitivity by testing chains of the form (a,b) and (b,c). 5. State the final classification clearly.

Diagram support

A relation on a finite set can be shown using directed arrows, but a diagram is not necessary for standard proof questions. The essential support is the ordered-pair checklist.

How CBSE asks it

Usually asked as: determine whether a given relation is reflexive, symmetric, transitive, or equivalence; justify with reasons. Assertion-reason questions often test whether one property implies another, which is generally false.

Avoid common mistakes

Common confusion

Students often call a relation reflexive after seeing one pair like (1,1). Reflexive means every element of the set must have its diagonal pair.

Common wrong answer

Marking a relation as transitive without checking chains such as (a,b) and (b,c), especially when the required pair (a,c) is absent.

Exam tip

Write the property test in symbols before applying it. For a relation on A, the diagonal checklist (a, a) for all a in A is the fastest way to test reflexivity.

Quick check

Let A = {1,2} and R = {(1,1),(2,2),(1,2)}. Is R symmetric?

No. Since (1,2) is in R but (2,1) is not in R, the relation is not symmetric.

Answer writing and exam use

1-mark answer

2-mark answer

A relation R on a set A is a subset of A × A. It is reflexive if every a in A satisfies (a, a) in R, symmetric if (a, b) in R implies (b, a) in R, and transitive if (a, b) in R and (b, c) in R imply (a, c) in R. A relation that is reflexive, symmetric, and transitive is an equivalence relation. Reflexive: for every a in A, (a,a) in R. Symmetric: (a,b) in R implies (b,a) in R. Transitive: (a,b) in R and (b,c) in R imply (a,c) in R. Equivalence relation: all three properties hold together. Empty relation on non-empty A is not reflexive; universal relation on A is reflexive, symmetric, and transitive. On A = {1, 2, 3}, R = {(1,1), (2,2), (3,3), (1,2), (2,1)} is reflexive because all diagonal pairs are present. It is symmetric because (1,2) and (2,1) occur together. It is not transitive if a required pair produced by two connected pairs is missing; here the listed pairs do not create a missing transitive requirement, so transitivity must be checked pair by pair.

3-mark answer

To classify a relation, always test it against the whole set A. A single example can disprove a property, but a general argument is needed to prove it. Empty and universal relations are also special cases: the empty relation has no ordered pairs, while the universal relation is A × A. Reflexive: for every a in A, (a,a) in R. Symmetric: (a,b) in R implies (b,a) in R. Transitive: (a,b) in R and (b,c) in R imply (a,c) in R. Equivalence relation: all three properties hold together. Empty relation on non-empty A is not reflexive; universal relation on A is reflexive, symmetric, and transitive. Let A = {1,2,3} and R = {(1,1),(2,2),(3,3),(1,2),(2,1)}. Reflexive: (1,1),(2,2),(3,3) are present. Symmetric: (1,2) has (2,1), and all diagonal pairs reverse to themselves. Transitive: check non-trivial chains: (1,2) and (2,1) give (1,1), present; (2,1) and (1,2) give (2,2), present. Other chains involve diagonal pairs and remain present. Hence R is reflexive, symmetric, and transitive, so it is an equivalence relation. Usually asked as: determine whether a given relation is reflexive, symmetric, transitive, or equivalence; justify with reasons. Assertion-reason questions often test whether one property implies another, which is generally false. Marking a relation as transitive without checking chains such as (a,b) and (b,c), especially when the required pair (a,c) is absent.

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