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JEE Mains Maths · Relations and Functions

Reflexive, Symmetric, Transitive and Equivalence Relations

Deciding whether a relation is reflexive, symmetric or transitive, proving each property in general or breaking it with one counterexample, and recognising equivalence relations and their classes.

All Relations and Functions notesJEE Mains Maths strategy

Why this matters

Twenty-five PYQs, and most ask the same thing: which of the three properties hold. Proving a property needs a general argument; breaking one needs a single counterexample, and choosing a good one is the skill. Two ideas cover the page.

Concept 1 of 2: Testing reflexive, symmetric and transitive

Reflexive: every element relates to itself. Symmetric: whenever aRbaRb, also bRabRa. Transitive: whenever aRbaRb and bRcbRc, also aRcaRc. To show a property holds, argue for all elements; to show it fails, one counterexample is enough. Good counterexamples use 0, a repeated element, or a short chain of three numbers that crosses a boundary.

Definition

  • Reflexive: aRaaRa for every aa.
  • Symmetric: aRb⇒bRaaRb\Rightarrow bRa.
  • Transitive: aRbaRb and bRc⇒aRcbRc\Rightarrow aRc.
  • A property with no pair to test holds by default: the empty relation is symmetric and transitive.

Transitivity

aRb and bRc ⇒ aRcaRb\ \text{and}\ bRc\ \Rightarrow\ aRc

Worked example

On R\mathbb R, aRbaRb if ∣a−b∣≤2|a-b|\le2. Which properties hold?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 8 Apr 2023 · Q170Moderate

Example 1 · Relations and Functions · Reflexive, Symmetric, Transitive and Equivalence

Let A={1,2,3,4,5,6,7}\mathbf{A = \{ 1,2,3,4,5,6,7\}}. Then the relation R=R = {(x,y)∈A×A:x+y=7}\mathbf{\{(x,y) \in A \times A:x + y = 7\}} is

A chain back to the start

Transitivity applies to aRbaRb and bRabRa too: together they force aRaaRa. A symmetric relation missing (a,a)(a,a) for such an aa is not transitive.
Drill 15 more on testing reflexive, symmetric and transitive

Concept 2 of 2: Equivalence relations and their classes

An equivalence relation has all three properties. Its usual shape is 'aRbaRb when g(a)=g(b)g(a)=g(b)' for some function gg, and every relation of that shape is an equivalence. Rewrite the condition into that shape: '2a+3b2a+3b is a multiple of 5' is the same as 'aa and bb leave the same remainder on division by 5'. The classes split the set into disjoint pieces, one for each value of gg.

Definition

  • Equivalence: reflexive, symmetric and transitive.
  • aRb  ⟺  g(a)=g(b)aRb\iff g(a)=g(b) is always an equivalence (  ⟺  \iff reads 'if and only if').
  • Class of aa: all bb with bRabRa. Classes are disjoint and cover the set.
  • Classes of sizes k1,k2,…k_1,k_2,\dots give a relation with k12+k22+…k_1^2+k_2^2+\dots pairs.

The usual shape

aRb  ⟺  g(a)=g(b)aRb\iff g(a)=g(b)

Worked example

On Z\mathbb Z, aRbaRb if 3 divides a−ba-b. Is it an equivalence, and what are its classes?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 29 January 2023 · Q154Moderate

Example 2 · Relations and Functions · Reflexive, Symmetric, Transitive and Equivalence

Let RR be a relation defined on NN as aRbaRb if 2a+3 b2a + 3\text{ }b is a multiple of 5,a,b∈N5,a,b \in N. Then RR is

Reflexive and symmetric is not enough

Many relations here are reflexive and symmetric but fail transitivity, for example ∣a−b∣≤1|a-b|\le1. Test a chain that crosses the limit before calling it an equivalence.
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Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Watch out for (2)

Test yourself on Relations and Functions

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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