medium difficulty10 min

Types of Relations: Reflexive, Symmetric, Transitive and Equivalence

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.

In one line: Reflexive checks self, symmetric checks reverse, transitive checks chain.

Practice This Concept

Understand it

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.

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.

Formula and rule

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.

See it worked out

Follow the example line by line, then use the same method on new questions.

Worked example

Let A = {1,2,3} and R = {(1,1),(2,2),(3,3),(1,2),(2,1)}.

1Reflexive: (1,1),(2,2),(3,3) are present.
2Symmetric: (1,2) has (2,1), and all diagonal pairs reverse to themselves.
3Transitive: check non-trivial chains: (1,2) and (2,1) give (1,1), present; (2,1) and (1,2) give (2,2), present.
4Other chains involve diagonal pairs and remain present.
5Hence R is reflexive, symmetric, and transitive, so it is an equivalence relation.

Method you can reuse

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

Visual tip

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.

Check yourself

Answer the question in your head before opening the answer.

Try it yourself

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

Think first, then show answer

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

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.

Don't mix this up

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.

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.

Write it in the exam

Each point is worth about one mark. Write point 1 for a 1-mark question and keep adding points as the marks go up.

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.

1
Enough for 1 mark
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.
2
Add for 2 marks
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.
3
Add for 3 marks
To classify a relation, always test it against the whole set A.

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