MathsISC Class 12 · Relations and Functions

Equivalence Relations: Reflexive, Symmetric and Transitive (ISC Class 12)

Q: What is the difference between an equivalence relation and an equivalence class?

An equivalence relation is the rule that relates pairs of elements. An equivalence class is one of the groups it creates: all the elements related to a given element. The classes never overlap and together make up the whole set.

Q: Can a relation be symmetric and transitive but not reflexive?

Yes. On A = {1, 2, 3}, the relation {(1, 1), (1, 2), (2, 1), (2, 2)} is symmetric and transitive but not reflexive, because 3 is not related to itself.

Q: How do I show that a relation is not transitive?

Find three elements a, b and c with (a, b) and (b, c) in R but (a, c) not in R, and state all three facts clearly. One such example is a complete answer.

Q: Is every identity relation an equivalence relation?

Yes. The identity relation on A is reflexive, symmetric and transitive, and it is the smallest equivalence relation on A. Its equivalence classes are the single elements.

Q: Are relations in the ISC Class 12 Maths syllabus for 2027?

Yes. The CISCE 2027 syllabus lists types of relations (reflexive, symmetric, transitive and equivalence) along with the identity, empty and universal relations, under Unit 1, Relations and Functions.

Updated 15 October 2026 · Fact-checked

A relation R on a set A is an equivalence relation if it is reflexive (every a is related to itself), symmetric (a R b gives b R a) and transitive (a R b and b R c give a R c). It splits A into disjoint equivalence classes of mutually related elements.

A relation with all three properties splits the set into separate classes, and every element belongs to exactly one class.

Illustration drawn for QuestionBuddy.

Understand Equivalence Relations

A relation $R$ on a set $A$ is any subset of $A \times A$ . We write $a R b$ when $(a, b) \in R$ . Three special relations come up in one-mark questions: the empty relation $ϕ$ (no pair is related), the universal relation $A \times A$ (every pair is related) and the identity relation $I_{A} = {(a, a) : a \in A}$ .

A relation can have three properties, and each has a precise test.

Reflexive: $(a, a) \in R$ for every $a \in A$ . One missing pair $(a, a)$ is enough to fail.
Symmetric: whenever $(a, b) \in R$ , also $(b, a) \in R$ .
Transitive: whenever $(a, b) \in R$ and $(b, c) \in R$ , also $(a, c) \in R$ .

A relation with all three properties is an equivalence relation. Typical examples: "has the same remainder on division by 5" on $Z$ , "is similar to" on triangles, "is parallel to or coincident with" on lines. "Is perpendicular to" on lines is symmetric but neither reflexive nor transitive, since two lines perpendicular to a third are parallel to each other.

Notice the logic. Reflexivity is a statement about every element, so it is proved for a general $a$ . Symmetry and transitivity are "if ... then" statements, so they are true automatically when the "if" part never happens: the empty relation on a non-empty set is symmetric and transitive but not reflexive. Every identity relation is reflexive, but a reflexive relation can contain extra pairs, so it need not be the identity relation.

Equivalence classes. If $R$ is an equivalence relation on $A$ , the class of $a$ is $[a] = {x \in A : x R a}$ . Two classes are either identical or disjoint, and together they cover $A$ . So an equivalence relation is the same thing as a way of sorting $A$ into non-overlapping groups. Congruence modulo 3 on $Z$ sorts the integers into $[0], [1], [2]$ : multiples of 3, numbers leaving remainder 1, and numbers leaving remainder 2.

Smallest relations. The smallest equivalence relation on $A$ is the identity relation $I_{A}$ ; the largest is $A \times A$ . To build the smallest equivalence relation containing some given pairs, start with $I_{A}$ , add the given pairs, add their reverses, and keep adding pairs forced by transitivity until nothing new is needed.

In proofs on infinite sets, write each property as a general statement: start from the hypothesis, use the rule that defines $R$ , and reach the conclusion. A disproof needs just one concrete counter-example.

Equivalence Relations: formulas to remember

Reflexive: $(a, a) \in R \forall a \in A$; Fails if even one element is not related to itself.
Symmetric: $(a, b) \in R \Rightarrow (b, a) \in R$; Check every pair in a listed relation, or argue generally.
Transitive: $(a, b), (b, c) \in R \Rightarrow (a, c) \in R$; A single chain that breaks is a complete disproof.
Equivalence class: $[a] = {x \in A : x R a}$; Classes are equal or disjoint, and their union is A.
Identity relation: $I_{A} = {(a, a) : a \in A}$; The smallest equivalence relation on A.
Number of relations: $∣ A ∣ = n \Rightarrow 2^{n^{2}} relations, 2^{n^{2} - n} reflexive$; Each of the n² ordered pairs is either in or out; reflexive ones must contain the n diagonal pairs.

How to solve Equivalence Relations questions

Use this order for the ISC question "Show that R is an equivalence relation" and for "Check whether R is reflexive, symmetric or transitive".

1Write the rule of $R$ in symbols, for example $a R b ⟺ 5 ∣ (a - b)$ .
2Reflexive: take any $a \in A$ and show the rule holds for $(a, a)$ . Example: $a - a = 0 = 5 \times 0$ .
3Symmetric: assume $a R b$ , write what that means (such as $a - b = 5 k$ ), and rewrite it to show $b R a$ (such as $b - a = 5 (- k)$ ).
4Transitive: assume $a R b$ and $b R c$ , write both equations with different letters ( $5 k$ and $5 m$ ), add or combine them, and show $a R c$ .
5If a property seems to fail, look for the smallest numerical counter-example and state it clearly, such as "(1, 2) ∈ R but (2, 1) ∉ R".
6Conclude in a sentence: "R is reflexive, symmetric and transitive, hence an equivalence relation."
7If equivalence classes are asked, describe each class in words and set notation, and check that they do not overlap and cover the whole set.

Equivalence Relations in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Equivalence relation and equivalence classes

Suppose a hospital records every event by the number of hours $h$ after midnight on 1 January, while the ward clock shows only 12-hour time. Two events are related when they show the same reading on the ward clock, that is, $h_{1} R h_{2} ⟺ 12 ∣ (h_{1} - h_{2})$ , for whole numbers $h_{1}, h_{2}$ .

Show that $R$ is an equivalence relation. Which reading does the clock show for events at $h = 39$ and $h = 100$ ? Describe the equivalence class of $h = 0$ .

Try it first, then show the solutionSolution

Reflexive: $h - h = 0 = 12 \times 0$ .
Symmetric: if $h_{1} - h_{2} = 12 k$ , then $h_{2} - h_{1} = 12 (- k)$ .
Transitive: if $h_{1} - h_{2} = 12 k$ and $h_{2} - h_{3} = 12 m$ , then adding, $h_{1} - h_{3} = 12 (k + m)$ . So $R$ is an equivalence relation.
$39 = 12 \times 3 + 3$ , so the clock shows 3 o'clock. $100 = 12 \times 8 + 4$ , so it shows 4 o'clock.
$[0] = {0, 12, 24, 36, \dots}$ : every event that falls exactly on 12 o'clock.

Answer: R is an equivalence relation with 12 classes, one for each clock reading; $h = 39$ shows 3 o'clock, $h = 100$ shows 4 o'clock, and $[0]$ is the set of multiples of 12.

Memory hook: A 12-hour clock is an equivalence relation: hours that look the same on the dial fall in one class.

Reflexive and symmetric but not transitive

Imagine three friends living along a straight road: Asha at the 0 km mark, Bilal at 0.8 km and Chitra at 1.6 km. Call two people "neighbours" if they live within 1 km of each other.

Is "is a neighbour of" reflexive, symmetric and transitive on these three people? Is it an equivalence relation?

Try it first, then show the solutionSolution

Reflexive: each person is 0 km from themselves, which is within 1 km.
Symmetric: distance from A to B equals distance from B to A, so the relation reverses.
Transitive: Asha and Bilal are 0.8 km apart and Bilal and Chitra are 0.8 km apart, so both pairs are neighbours, but Asha and Chitra are 1.6 km apart.
So (Asha, Bilal) and (Bilal, Chitra) are in R but (Asha, Chitra) is not: R is not transitive.

Answer: Reflexive and symmetric but not transitive, so not an equivalence relation.

Memory hook: "Close to" leaks: small gaps add up to a big one, so nearness is never transitive.

Tricks to remember Equivalence Relations

Short memory aids. Say each one aloud once, then use it on a question the same day.

RST
Equivalence = R + S + T: Reflexive (self), Symmetric (swap), Transitive (chain). Test them in that order.
Use it when: Structuring a 2- or 3-mark proof.
Mirror, swap, bridge
Reflexive: every element sees itself in a mirror. Symmetric: every arrow can be swapped. Transitive: every two-step bridge has a direct road.
Use it when: Checking listed relations quickly in MCQs.
Perpendicular is the trap
"Perpendicular to" is symmetric only. "Parallel to" (counting a line as parallel to itself) is all three.
Use it when: The common geometry MCQ.
Classes are boxes
An equivalence relation sorts the set into boxes: every element in exactly one box, and two boxes never share an element.
Use it when: Writing equivalence classes and checking you have them all.

Equivalence Relations: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Congruence modulo 5 · 2 marks · Short answerBoard level

Show that the relation $R$ on $Z$ defined by $a R b ⟺ 5 ∣ (a - b)$ is an equivalence relation.

Try it first, then show the solutionSolution

Reflexive: $a - a = 0 = 5 \times 0$ , so $a R a$ for every integer $a$ . (½ mark)
Symmetric: $a R b \Rightarrow a - b = 5 k$ for some $k \in Z \Rightarrow b - a = 5 (- k) \Rightarrow b R a$ . (½ mark)
Transitive: $a - b = 5 k$ and $b - c = 5 m$ give $a - c = 5 (k + m)$ , so $a R c$ . (1 mark)

Answer: $R$ is reflexive, symmetric and transitive, so it is an equivalence relation on $Z$ .

Example 2: Relations on triangles · 1 mark · MCQBoard level

Let $T$ be the set of all triangles in a plane and $R = {(T_{1}, T_{2}) : T_{1} is similar to T_{2}}$ . Then $R$ is:
(a) reflexive only
(b) symmetric and transitive only
(c) an equivalence relation
(d) neither reflexive nor symmetric

Try it first, then show the solutionSolution

Every triangle is similar to itself; similarity reverses; and two triangles similar to a third are similar to each other (equal corresponding angles).

Answer: (c) an equivalence relation.

Example 3: A relation on ordered pairs · 3 marks · Short answerHarder board level

Show that the relation $R$ on $N \times N$ defined by $(a, b) R (c, d) ⟺ a + d = b + c$ is an equivalence relation. Write the equivalence class of $(2, 5)$ when $R$ is restricted to $A \times A$ , where $A = {1, 2, 3, 4, 5, 6}$ .

Try it first, then show the solutionSolution

Reflexive: $a + b = b + a$ , so $(a, b) R (a, b)$ . (½ mark)
Symmetric: $(a, b) R (c, d) \Rightarrow a + d = b + c \Rightarrow c + b = d + a \Rightarrow (c, d) R (a, b)$ . (½ mark)
Transitive: $a + d = b + c$ and $c + f = d + e$ . Adding, $a + d + c + f = b + c + d + e$ , so $a + f = b + e$ , which means $(a, b) R (e, f)$ . (1 mark)
Class of $(2, 5)$ : $(c, d)$ with $2 + d = 5 + c$ , that is $d - c = 3$ . In $A \times A$ these are $(1, 4), (2, 5), (3, 6)$ . (1 mark)

Answer: $R$ is an equivalence relation; $[(2, 5)] = {(1, 4), (2, 5), (3, 6)}$ , the pairs whose second entry exceeds the first by 3.

Example 4: Equivalence classes on a finite set · 3 marks · Short answerHarder board level

Let $A = {1, 2, 3, 4, 5, 6}$ and $R = {(a, b) : ∣ a - b ∣ is even}$ . Show that $R$ is an equivalence relation, list its equivalence classes and show that no element of ${1, 3, 5}$ is related to any element of ${2, 4, 6}$ .

Try it first, then show the solutionSolution

Reflexive: $∣ a - a ∣ = 0$ , which is even. Symmetric: $∣ a - b ∣ = ∣ b - a ∣$ . (1 mark)
Transitive: if $a - b$ and $b - c$ are even, then $a - c = (a - b) + (b - c)$ is even. (1 mark)
$∣ a - b ∣$ is even exactly when $a$ and $b$ have the same parity. So $[1] = {1, 3, 5}$ and $[2] = {2, 4, 6}$ . (½ mark)
An odd number minus an even number is odd, so no element of ${1, 3, 5}$ is related to an element of ${2, 4, 6}$ . (½ mark)

Answer: Equivalence relation with two classes, ${1, 3, 5}$ and ${2, 4, 6}$ .

Example 5: Disproving all three properties · 3 marks · Short answerHarder board level

Check whether the relation $R$ on $R$ defined by $a R b ⟺ a \leq b^{2}$ is reflexive, symmetric or transitive.

Try it first, then show the solutionSolution

Reflexive: take $a = \frac{1}{2}$ . Is $\frac{1}{2} \leq \frac{1}{4}$ ? No. So $R$ is not reflexive. (1 mark)
Symmetric: $1 \leq 2^{2}$ , so $(1, 2) \in R$ ; but $2 \leq 1^{2}$ is false, so $(2, 1) \in / R$ . Not symmetric. (1 mark)
Transitive: $(3, 2) \in R$ since $3 \leq 4$ , and $(2, 1.5) \in R$ since $2 \leq 2.25$ ; but $3 \leq 2.25$ is false, so $(3, 1.5) \in / R$ . Not transitive. (1 mark)

Answer: $R$ is neither reflexive, nor symmetric, nor transitive.

Example 6: Building equivalence relations on {1, 2, 3} · 5 marks · Long answerHOTS / competency

Let $A = {1, 2, 3}$ .
(i) Write the smallest equivalence relation on $A$ containing $(1, 2)$ .
(ii) How many equivalence relations on $A$ contain $(1, 2)$ ?
(iii) Give a relation on $A$ that contains $(1, 2)$ and $(2, 3)$ , is reflexive and symmetric, but is not transitive.

Try it first, then show the solutionSolution

(i) Start with $I_{A} = {(1, 1), (2, 2), (3, 3)}$ , add $(1, 2)$ and, for symmetry, $(2, 1)$ . Transitivity needs nothing more, since the chains $(1, 2), (2, 1)$ and $(2, 1), (1, 2)$ give $(1, 1)$ and $(2, 2)$ , already present. (2 marks)
$R_{1} = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}$ .
(ii) An equivalence relation containing $(1, 2)$ puts 1 and 2 in the same class. Either 3 is alone, giving $R_{1}$ , or 3 joins them, giving the universal relation $A \times A$ . So there are 2. (2 marks)
(iii) $R_{2} = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}$ . It is reflexive and symmetric, but $(1, 2), (2, 3) \in R_{2}$ while $(1, 3) \in / R_{2}$ . (1 mark)

Answer: (i) ${(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}$ ; (ii) 2; (iii) ${(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}$ .

Where students lose marks in Equivalence Relations

The slips examiners see most often, and what to write instead.

Mistake: Proving reflexivity with one example
Reflexive is a "for every a" statement; one example shows nothing.
Wrong working
$(2, 2) \in R$ since $5 ∣ 0$ . Hence $R$ is reflexive.
Right working
For any $a \in Z$ , $a - a = 0 = 5 \times 0$ , so $(a, a) \in R$ . Hence $R$ is reflexive.
Do this: Use a general letter for reflexive, symmetric and transitive proofs; use numbers only for counter-examples.
Usually costs: The reflexive mark.
Mistake: Confusing symmetric with transitive
The ISC 2024 examiners reported this confusion on the "perpendicular lines" MCQ.
Do this: Symmetric uses one pair and swaps it; transitive uses two pairs that share a middle element.
Usually costs: 1 mark in Section A.
Mistake: Treating "reflexive" as "identity"
A reflexive relation must contain every $(a, a)$ but may contain more pairs.
Do this: Remember: identity ⇒ reflexive, but reflexive does not imply identity.
Usually costs: 1 mark on a two-statement MCQ.
Mistake: Using the same constant twice in the transitive step
Writing $a - b = 5 k$ and $b - c = 5 k$ assumes the two multiples are equal, which is not given.
Wrong working
$a - b = 5 k$ , $b - c = 5 k \Rightarrow a - c = 10 k$ .
Right working
$a - b = 5 k$ , $b - c = 5 m \Rightarrow a - c = 5 (k + m)$ .
Do this: Use a new letter for each new hypothesis.
Usually costs: Half to one mark for an incorrect argument.

Exam tips

Write the three headings "Reflexive", "Symmetric", "Transitive" and finish with "Hence R is an equivalence relation". Each heading usually earns its own part-mark.
For a listed relation on a small set, check reflexivity first by looking for every $(a, a)$ ; it is the quickest property to fail.
When asked for "the smallest equivalence relation", start from the identity relation and add only the pairs that symmetry and transitivity force.
For geometry relations (parallel, perpendicular, similar, congruent), state the geometric fact you use in one line.
If a question asks for equivalence classes, name every class and confirm that they cover the whole set.

Practice questions on Equivalence Relations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The number of ordered pairs in the smallest equivalence relation on $A = {1, 2, 3, 4}$ is:
(a) 0
(b) 4
(c) 8
(d) 16
Hint
Which pairs must every reflexive relation contain?
Answer
(b) 4. The smallest equivalence relation is the identity relation ${(1, 1), (2, 2), (3, 3), (4, 4)}$ .
Q2 · 2 marks · Short answerBoard level
Show that the relation $R$ on $Z$ given by $a R b ⟺ a + b$ is even, is an equivalence relation.
Hint
For transitivity, add $a + b$ and $b + c$ .
Answer
Reflexive: $a + a = 2 a$ is even. Symmetric: $a + b = b + a$ . Transitive: if $a + b$ and $b + c$ are even, then $a + c = (a + b) + (b + c) - 2 b$ is even. Hence an equivalence relation.
Q3 · 2 marks · Short answerBoard level
Let $B$ be the set of books in a library and $R = {(x, y) : x and y have the same number of pages}$ . Show that $R$ is an equivalence relation and describe the equivalence class of a book with 200 pages.
Hint
Equality of numbers is reflexive, symmetric and transitive.
Answer
Each property follows from the same property of "=" for page counts. The class of a 200-page book is the set of all books in the library with exactly 200 pages.
Q4 · 3 marks · Short answerHarder board level
Show that the relation $R$ on $R$ defined by $a R b ⟺ 1 + ab > 0$ is reflexive and symmetric but not transitive.
Hint
Try $a = 1$ , $b = - \frac{1}{2}$ , $c = - 3$ .
Answer
Reflexive: $1 + a^{2} > 0$ . Symmetric: $1 + ab = 1 + ba$ . Not transitive: $1 + (1) (- \frac{1}{2}) = \frac{1}{2} > 0$ and $1 + (- \frac{1}{2}) (- 3) = \frac{5}{2} > 0$ , but $1 + (1) (- 3) = - 2 < 0$ .
Q5 · 1 mark · MCQHarder board level
If $A$ has 3 elements, the number of reflexive relations on $A$ is:
(a) 8
(b) 64
(c) 512
(d) 27
Hint
The 3 diagonal pairs are fixed; the other 6 pairs are free.
Answer
(b) 64, since $2^{9 - 3} = 2^{6}$ .
Q6 · 3 marks · Short answerHOTS / competency
For the relation "congruence modulo 3" on $Z$ , write the three equivalence classes and show that every integer lies in exactly one of them.
Hint
Use the division algorithm $a = 3 q + r$ , $0 \leq r < 3$ .
Answer
$[0] = {\dots, - 3, 0, 3, \dots}$ , $[1] = {\dots, - 2, 1, 4, \dots}$ , $[2] = {\dots, - 1, 2, 5, \dots}$ . By the division algorithm every integer is $3 q + r$ with a unique $r \in {0, 1, 2}$ , so it lies in exactly one class; the classes are disjoint and their union is $Z$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose every student of a school belongs to exactly one of four houses: Red, Blue, Green and Yellow. On the set $S$ of students define $R = {(a, b) : a and b are in the same house}$ and $T = {(a, b) : a and b are in different houses}$ .
(i) Show that $R$ is an equivalence relation. (2 marks)
(ii) How many equivalence classes does $R$ have? (1 mark)
(iii) Is $T$ transitive? Justify. (1 mark)
Hint
For (iii), take a chain that comes back to the start.
Answer
(i) Each student is in the same house as themselves; "same house" reverses; if $a, b$ share a house and $b, c$ share a house, all three are in that house. (ii) 4, one per house (assuming every house has at least one student). (iii) No: if $a$ is in Red and $b$ in Blue, then $(a, b), (b, a) \in T$ but $(a, a) \in / T$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the three conditions for an equivalence relation.Show answer
Reflexive: $(a, a) \in R$ for all $a$ . Symmetric: $(a, b) \in R \Rightarrow (b, a) \in R$ . Transitive: $(a, b), (b, c) \in R \Rightarrow (a, c) \in R$ .
2.Is the empty relation on $A = {1, 2}$ transitive?Show answer
Yes. There are no pairs, so the condition is never broken. It is also symmetric, but it is not reflexive.
3.Which properties does "is perpendicular to" have on lines in a plane?Show answer
Only symmetric.
4.How many relations are there on a set with 2 elements?Show answer
$2^{2^{2}} = 16$ .
5.Write the equivalence classes of congruence modulo 2 on $Z$ .Show answer
The even integers and the odd integers.
6.What is the largest equivalence relation on a set $A$ ?Show answer
The universal relation $A \times A$ .

Equivalence Relations: frequently asked questions

What is the difference between an equivalence relation and an equivalence class?

An equivalence relation is the rule that relates pairs of elements. An equivalence class is one of the groups it creates: all the elements related to a given element. The classes never overlap and together make up the whole set.

Can a relation be symmetric and transitive but not reflexive?

Yes. On A = {1, 2, 3}, the relation {(1, 1), (1, 2), (2, 1), (2, 2)} is symmetric and transitive but not reflexive, because 3 is not related to itself.

How do I show that a relation is not transitive?

Find three elements a, b and c with (a, b) and (b, c) in R but (a, c) not in R, and state all three facts clearly. One such example is a complete answer.

Is every identity relation an equivalence relation?

Yes. The identity relation on A is reflexive, symmetric and transitive, and it is the smallest equivalence relation on A. Its equivalence classes are the single elements.

Are relations in the ISC Class 12 Maths syllabus for 2027?

Yes. The CISCE 2027 syllabus lists types of relations (reflexive, symmetric, transitive and equivalence) along with the identity, empty and universal relations, under Unit 1, Relations and Functions.