MathsCBSE Class 12 · Relations and Functions

Equivalence Relations and Equivalence Classes (Class 12)

Updated 15 October 2026 · Fact-checked

A relation is an equivalence relation when it is reflexive, symmetric and transitive at the same time. It splits the set into equivalence classes: disjoint groups whose members are all related to each other. The board's 5-mark question asks you to prove all three properties and then list one class.

All three properties together split the set into separate classes: here [0], [1] and [2] share no element and together cover every number.

Illustration drawn for QuestionBuddy.

Understand Equivalence Relations and Equivalence Classes

An equivalence relation on a set $A$ is a relation $R$ that is reflexive, symmetric and transitive together. It captures the idea of 'the same in some respect': same remainder on division by 5, same number of sides, same absolute value, equal fractions. Whenever a rule says ' $a$ and $b$ have the same ...', expect an equivalence relation.

Proving it. A full proof has three parts, each argued for arbitrary elements:

Reflexive: show $(a, a) \in R$ for every $a$ .
Symmetric: assume $(a, b) \in R$ and deduce $(b, a) \in R$ .
Transitive: assume $(a, b) \in R$ and $(b, c) \in R$ and deduce $(a, c) \in R$ .

For divisibility rules, write the condition as an equation with an integer, such as $a - b = 5 k$ , and add or rearrange the equations. For rules such as $a d = b c$ , multiply the two given equations and cancel a non-zero factor.

Equivalence classes. For $a \in A$ , the equivalence class of $a$ is $[a] = {x \in A : (x, a) \in R}$ , the set of everything related to $a$ . Three facts follow from the three properties:
1. Every element is in its own class, since $a R a$ .
2. Two classes are either identical or disjoint. If $[a]$ and $[b]$ share even one element, they are the same set.
3. The classes together cover all of $A$ .
So an equivalence relation partitions $A$ into non-overlapping blocks. For example, 'same remainder on division by 3' splits $Z$ into $[0]$ , $[1]$ and $[2]$ . Conversely, any partition of $A$ gives an equivalence relation: relate two elements when they lie in the same block.

Counting with classes. On a finite set, if the classes have sizes $n_{1}, n_{2}, \dots$ , the relation has $n_{1}^{2} + n_{2}^{2} + \dots$ ordered pairs, because within a block every element is related to every element. The number of equivalence relations on a set equals the number of ways to partition it: 1 way for one element, 2 for two, 5 for three and 15 for four.

Minimal equivalence relations. To build the smallest equivalence relation containing some pairs, add the diagonal, add the reverse of every pair, then keep adding shortcuts until the blocks close up. If the given pairs link every element, the result is the universal relation.

In the 2026-27 CBSE syllabus this is part of Unit I. It is NCERT Section 1.2 (Chapter 1), and it is the most frequently asked part of this chapter in recent board papers, usually as a 5-mark long answer or a 4-mark case study.

Equivalence Relations and Equivalence Classes: formulas to remember

Equivalence relation: $R is reflexive + symmetric + transitive$; All three are needed; prove each one separately.
Equivalence class of a: $[a] = {x \in A : (x, a) \in R}$; Always contains $a$ itself.
Classes partition the set: $[a] = [b] or [a] \cap [b] = ϕ; ⋃ [a] = A$; Two classes never partly overlap.
Congruence modulo n on the integers: $a R b ⟺ n ∣ (a - b) ⟺ a - b = nk, k \in Z$; n classes: $[0], [1], \dots, [n - 1]$ .
Number of pairs from class sizes: $∣ R ∣ = n_{1}^{2} + n_{2}^{2} + \dots + n_{k}^{2}$; Every element of a block is related to every element of the same block.
Number of equivalence relations: $∣ A ∣ = 1, 2, 3, 4 \Rightarrow 1, 2, 5, 15$; Equal to the number of partitions of the set.

How to solve Equivalence Relations and Equivalence Classes questions

This is the layout the marking scheme rewards for 'Show that R is an equivalence relation. Also find [a].'

1Restate the rule as an equation: for divisibility write $a - b = nk$ with $k \in Z$ ; for 'same property' write $f (a) = f (b)$ .
2Reflexive: substitute $b = a$ and show the condition holds for every $a$ in the set.
3Symmetric: start from the equation for $(a, b)$ and rearrange or multiply by $- 1$ to get the equation for $(b, a)$ . Say what kind of number the new multiplier is (for example, $- k \in Z$ ).
4Transitive: write both equations, for $(a, b)$ and $(b, c)$ , then add, subtract or multiply them so that $b$ disappears, leaving the condition for $(a, c)$ .
5Conclude: 'R is reflexive, symmetric and transitive, hence an equivalence relation.'
6For the class $[a]$ , solve the condition with $a$ fixed and list every element of the given set that satisfies it, in increasing order.
7Check the class: every listed element must satisfy the rule with $a$ , and $a$ itself must be in the list.

Equivalence Relations and Equivalence Classes 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 classes as groups

Suppose a school of 30 students (roll numbers 1 to 30) assigns sports houses by the remainder when the roll number is divided by 4. Two students are related if they are in the same house.

Show that 'being in the same house' is an equivalence relation on the roll numbers. List the house of roll number 22, find the size of each house, and find the number of ordered pairs in the relation.

Try it first, then show the solutionSolution

$a R b ⟺ a$ and $b$ leave the same remainder on division by 4, that is, $4 ∣ (a - b)$ .
Reflexive: $a - a = 0 = 4 \times 0$ . Symmetric: $a - b = 4 k \Rightarrow b - a = 4 (- k)$ . Transitive: $a - b = 4 k$ , $b - c = 4 m \Rightarrow a - c = 4 (k + m)$ . So R is an equivalence relation.
$22 = 4 \times 5 + 2$ , so roll 22 is in the remainder-2 house: ${2, 6, 10, 14, 18, 22, 26, 30}$ .
Remainder 1: $1, 5, \dots, 29$ (8 students). Remainder 2: 8 students. Remainder 3: $3, 7, \dots, 27$ (7 students). Remainder 0: $4, 8, \dots, 28$ (7 students).
Ordered pairs $= 8^{2} + 8^{2} + 7^{2} + 7^{2} = 64 + 64 + 49 + 49 = 226$ .

Answer: R is an equivalence relation. Roll 22's house is ${2, 6, 10, 14, 18, 22, 26, 30}$ ; the houses have 8, 8, 7 and 7 students; R has 226 ordered pairs.

Memory hook: An equivalence relation is a house system: everyone is in exactly one house, and housemates are related.

Congruence classes

Imagine a 24-hour digital clock and an old 12-hour wall clock. Hours $a$ and $b$ (from 0 to 23) are related if they look the same on the wall clock, that is, if $a - b$ is divisible by 12.

Show that this is an equivalence relation on $H = {0, 1, \dots, 23}$ . Find the class of 15, the number of classes, and the number of ordered pairs.

Try it first, then show the solutionSolution

Reflexive: $a - a = 0 = 12 \times 0$ . Symmetric: $a - b = 12 k \Rightarrow b - a = 12 (- k)$ . Transitive: $a - b = 12 k$ and $b - c = 12 m$ give $a - c = 12 (k + m)$ .
$[15] = {x \in H : 12 ∣ (x - 15)} = {3, 15}$ (3 pm shows as 3 on the wall clock).
Each wall-clock position $0, 1, \dots, 11$ gives one class with two hours, so there are 12 classes.
Ordered pairs $= 12 \times 2^{2} = 48$ .

Answer: It is an equivalence relation; $[15] = {3, 15}$ ; 12 classes; 48 ordered pairs.

Memory hook: The wall clock face is a set of equivalence classes: 3 and 15 are one class.

Tricks to remember Equivalence Relations and Equivalence Classes

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

RST in order
Equivalence = R + S + T, always written in that order: Reflexive, Symmetric, Transitive. Missing any one and the relation is not an equivalence.
Use it when: Structuring every 5-mark proof.
Same-something test
If the rule can be read as 'a and b have the same ___' (remainder, length, absolute value, value of a fraction), it is almost surely an equivalence relation.
Use it when: Guessing the answer to an MCQ before checking it.
Add the equations, b vanishes
For transitivity with differences, $(a - b) + (b - c) = a - c$ : the middle element cancels. With products like $a d = b c$ , multiply the two equations and cancel the middle terms.
Use it when: The transitive step of divisibility and fraction-type relations.
Classes never overlap
Think of rooms in a house: each element sleeps in exactly one room. Two rooms with a common occupant are the same room.
Use it when: Assertion-reason items on equivalence classes.

Equivalence Relations and Equivalence Classes: solved examples

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

Example 1: Same remainder on a finite set · 3 marks · Short answerBoard level

Let $A = {1, 2, 3, \dots, 20}$ and $R = {(a, b) : a$ and $b$ leave the same remainder when divided by 6 $}$ . Show that R is an equivalence relation. Find the equivalence class $[5]$ and the number of equivalence classes.

Try it first, then show the solutionSolution

Same remainder means $6 ∣ (a - b)$ . Reflexive: $a - a = 0 = 6 \times 0$ , so $(a, a) \in R$ . (½ mark)
Symmetric: $a - b = 6 k \Rightarrow b - a = 6 (- k)$ , with $- k \in Z$ , so $(b, a) \in R$ . (½ mark)
Transitive: $a - b = 6 k$ and $b - c = 6 m$ give $a - c = 6 (k + m)$ , so $(a, c) \in R$ . Hence R is an equivalence relation. (1 mark)
$[5]$ : numbers in A of the form $5 + 6 k$ : $5, 11, 17$ (the next, 23, is outside A). (½ mark)
The possible remainders are $0, 1, 2, 3, 4, 5$ , and each occurs in A, so there are 6 classes. (½ mark)

Answer: R is an equivalence relation; $[5] = {5, 11, 17}$ ; 6 equivalence classes.

Example 2: Absolute value relation on the reals · 3 marks · Short answerBoard level

Show that the relation R on $R$ defined by $a R b ⟺ ∣ a ∣ = ∣ b ∣$ is an equivalence relation, and describe the equivalence class of $- 3$ and of $0$ .

Try it first, then show the solutionSolution

Reflexive: $∣ a ∣ = ∣ a ∣$ for every real $a$ . (½ mark)
Symmetric: $∣ a ∣ = ∣ b ∣ \Rightarrow ∣ b ∣ = ∣ a ∣$ . (½ mark)
Transitive: $∣ a ∣ = ∣ b ∣$ and $∣ b ∣ = ∣ c ∣ \Rightarrow ∣ a ∣ = ∣ c ∣$ . So R is an equivalence relation. (1 mark)
$[- 3] = {x : ∣ x ∣ = 3} = {- 3, 3}$ . (½ mark)
$[0] = {x : ∣ x ∣ = 0} = {0}$ , the only class with a single element. (½ mark)

Answer: R is an equivalence relation; $[- 3] = {- 3, 3}$ and $[0] = {0}$ .

Example 3: A disguised congruence · 5 marks · Long answerHarder board level

Show that the relation R on $Z$ given by $a R b ⟺ 7$ divides $3 a + 4 b$ is an equivalence relation. Hence show that $a R b$ exactly when $7$ divides $a - b$ , and state the number of equivalence classes.

Try it first, then show the solutionSolution

Reflexive: $3 a + 4 a = 7 a$ , which is divisible by 7, so $(a, a) \in R$ . (1 mark)
Symmetric: let $3 a + 4 b = 7 k$ . Then $3 b + 4 a = 7 a + 7 b - (3 a + 4 b) = 7 (a + b - k)$ , divisible by 7, so $(b, a) \in R$ . (1 mark)
Transitive: let $3 a + 4 b = 7 k$ and $3 b + 4 c = 7 m$ . Adding, $3 a + 7 b + 4 c = 7 (k + m)$ , so $3 a + 4 c = 7 (k + m - b)$ , divisible by 7. So $(a, c) \in R$ and R is an equivalence relation. (1½ marks)
Since $3 a + 4 b = 7 b + 3 (a - b)$ , 7 divides $3 a + 4 b$ exactly when 7 divides $3 (a - b)$ ; as 7 and 3 have no common factor, this happens exactly when 7 divides $a - b$ . (1 mark)
So the classes are the 7 remainder classes $[0], [1], \dots, [6]$ . (½ mark)

Answer: R is an equivalence relation; $a R b ⟺ 7 ∣ (a - b)$ ; there are 7 equivalence classes.

Example 4: Equal fractions on pairs of natural numbers · 5 marks · Long answerHarder board level

Let $A = {1, 2, 3, 4, 5, 6}$ and define R on $A \times A$ by $(a, b) R (c, d) ⟺ a d = b c$ . Show that R is an equivalence relation and find the equivalence class of $(2, 3)$ .

Try it first, then show the solutionSolution

Reflexive: $ab = ba$ , so $(a, b) R (a, b)$ . (1 mark)
Symmetric: $a d = b c \Rightarrow c b = d a$ , which is the condition for $(c, d) R (a, b)$ . (1 mark)
Transitive: let $a d = b c$ and $c f = d e$ . Multiplying, $a d c f = b c d e$ . Since $c, d \neq = 0$ (they are natural numbers), cancel $c d$ : $a f = b e$ , so $(a, b) R (e, f)$ . (1½ marks)
Hence R is an equivalence relation: $(a, b)$ is related to $(c, d)$ exactly when $\frac{a}{b} = \frac{c}{d}$ . (½ mark)
$[(2, 3)]$ : pairs in $A \times A$ with $3 x = 2 y$ , that is $\frac{x}{y} = \frac{2}{3}$ : $(2, 3)$ and $(4, 6)$ ; $(6, 9)$ is not allowed as $9 \in / A$ . (1 mark)

Answer: R is an equivalence relation; $[(2, 3)] = {(2, 3), (4, 6)}$ .

Example 5: Case study: shelving library books · 4 marks · Case-basedHOTS / competency

Suppose a class library numbers its books from 1 to 30 and shelves together all books whose numbers end in the same digit. Let $B = {1, 2, \dots, 30}$ and $a R b$ if $a$ and $b$ have the same last digit.
(i) Show that R is an equivalence relation. (2 marks)
(ii) Write the equivalence class of book 7. (1 mark)
(iii) How many shelves (equivalence classes) are needed, and how many ordered pairs are in R? (1 mark)

Try it first, then show the solutionSolution

(i) Same last digit means $10 ∣ (a - b)$ . Reflexive: $a - a = 0$ . Symmetric: $a - b = 10 k \Rightarrow b - a = 10 (- k)$ . Transitive: $a - b = 10 k$ , $b - c = 10 m \Rightarrow a - c = 10 (k + m)$ . (2 marks)
(ii) $[7] = {7, 17, 27}$ . (1 mark)
(iii) Last digits $0, 1, \dots, 9$ all occur, so 10 shelves, each with 3 books (for example ${10, 20, 30}$ ). Ordered pairs $= 10 \times 3^{2} = 90$ . (1 mark)

Answer: (i) R is reflexive, symmetric and transitive. (ii) ${7, 17, 27}$ . (iii) 10 classes; 90 ordered pairs.

Example 6: Counting equivalence relations through a given pair · 2 marks · Very short answerHOTS / competency

Find the number of equivalence relations on $A = {1, 2, 3, 4}$ that contain the pair $(1, 2)$ .

Try it first, then show the solutionSolution

An equivalence relation is the same as a partition of A into classes. Containing $(1, 2)$ means 1 and 2 are in the same class. (½ mark)
Glue 1 and 2 into one block $X = {1, 2}$ ; we now partition the three objects $X, 3, 4$ . (½ mark)
Partitions of three objects: all separate (1 way); one pair together (3 ways: $X 3$ , $X 4$ , $34$ ); all together (1 way). Total $1 + 3 + 1 = 5$ . (1 mark)

Answer: 5 equivalence relations.

Where students lose marks in Equivalence Relations and Equivalence Classes

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

Mistake: Not saying why the new multiplier is an integer
In divisibility proofs the conclusion needs ' $a - c = 5 (k + m)$ where $k + m \in Z$ '.
Wrong working
$a - b = 5 k$ , $b - c = 5 m$ , so $a - c = 5 k + 5 m$ . Hence transitive.
Right working
$a - c = (a - b) + (b - c) = 5 (k + m)$ , and $k + m \in Z$ , so $5 ∣ (a - c)$ and $(a, c) \in R$ .
Do this: Always factor the result and name the integer multiplier.
Usually costs: Half a mark in each of the symmetric and transitive steps.
Mistake: Cancelling without saying the factor is non-zero
In $(a, b) R (c, d) ⟺ a d = b c$ , cancelling $c d$ is valid only because $c, d \neq = 0$ .
Wrong working
$a d c f = b c d e \Rightarrow a f = b e$ .
Right working
$a d c f = b c d e$ ; since $c, d \in N$ , $c d \neq = 0$ , so cancelling gives $a f = b e$ .
Do this: Write one phrase justifying every cancellation.
Usually costs: Half to one mark in a 5-mark proof.
Mistake: Listing the class with elements outside the given set
The class is a subset of A, not of $Z$ .
Wrong working
$A = {1, \dots, 20}$ , same remainder mod 6: $[5] = {\dots, - 1, 5, 11, 17, 23, \dots}$ .
Right working
$[5] = {5, 11, 17}$ , since only these lie in A.
Do this: Write the class condition, then filter by the set given in the question.
Usually costs: The class mark (usually 1).
Mistake: Proving only two of the three properties
Students prove reflexive and symmetric and write 'similarly transitive'.
Wrong working
R is reflexive and symmetric, and similarly transitive, so R is an equivalence relation.
Right working
Write the transitive step in full: assume $(a, b), (b, c) \in R$ , combine the conditions and show $(a, c) \in R$ .
Do this: Treat 'transitive' as the main step; it usually carries the most marks.
Usually costs: 1 to 2 marks.

Exam tips

A 5-mark equivalence-relation proof appeared in the 2023 and 2026 board papers and in the 2027 sample paper; give each property its own heading and keep the class calculation separate at the end.
For relations on pairs, write $(a, b) R (c, d)$ explicitly in every line; mixing the order of letters is the most common slip.
If asked for 'the set of all elements related to $a$ ', that is the equivalence class $[a]$ ; list its elements in increasing order.
In case studies, the last part often asks for a count. Use class sizes: pairs $= n_{1}^{2} + n_{2}^{2} + \dots$ .
If a relation fails one property, say so with a counter-example and conclude that it is not an equivalence relation; do not stop after finding the failure without stating the conclusion.

Practice questions on Equivalence Relations and Equivalence Classes

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
Which of the following relations on $A = {1, 2, 3}$ is an equivalence relation?
(a) ${(1, 1), (2, 2), (1, 2), (2, 1)}$
(b) ${(1, 1), (2, 2), (3, 3), (1, 2)}$
(c) ${(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)}$
(d) ${(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}$
Hint
Check reflexive first, then reverse pairs, then chains.
Answer
(c). (a) misses $(3, 3)$ ; (b) is not symmetric; (d) has $(1, 2), (2, 3)$ without $(1, 3)$ , so it is not transitive. (c) has classes ${1}$ and ${2, 3}$ .
Q2 · 1 mark · MCQBoard level
The number of equivalence relations on a set with 3 elements is:
(a) 3
(b) 4
(c) 5
(d) 8
Hint
Count the ways to split ${a, b, c}$ into non-empty groups.
Answer
(c) 5: ${a} {b} {c}$ ; ${a, b} {c}$ ; ${a, c} {b}$ ; ${b, c} {a}$ ; ${a, b, c}$ .
Q3 · 2 marks · Very short answerBoard level
Let P be the set of all polygons in a plane, and let $p R q$ if $p$ and $q$ have the same number of sides. Show that R is an equivalence relation and describe the class of a given square.
Hint
Write $s (p)$ for the number of sides; then the rule is $s (p) = s (q)$ .
Answer
Reflexive: $s (p) = s (p)$ . Symmetric: $s (p) = s (q) \Rightarrow s (q) = s (p)$ . Transitive: $s (p) = s (q)$ , $s (q) = s (r) \Rightarrow s (p) = s (r)$ . The class of a square is the set of all quadrilaterals in the plane.
Q4 · 3 marks · Short answerHarder board level
Show that the relation R on $Z$ defined by $a R b ⟺ 3$ divides $a^{2} - b^{2}$ is an equivalence relation. How many equivalence classes are there? Describe them.
Hint
Squares of integers leave remainder 0 or 1 on division by 3.
Answer
Reflexive: $a^{2} - a^{2} = 0 = 3 \times 0$ . Symmetric: $a^{2} - b^{2} = 3 k \Rightarrow b^{2} - a^{2} = 3 (- k)$ . Transitive: $(a^{2} - b^{2}) + (b^{2} - c^{2}) = 3 (k + m) = a^{2} - c^{2}$ . A square leaves remainder 0 (when 3 divides $a$ ) or 1 (otherwise). So there are 2 classes: the multiples of 3, and the integers not divisible by 3.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If R is an equivalence relation on A and $[a] \cap [b] \neq = ϕ$ , then $[a] = [b]$ .
Reason (R): Every element of A belongs to exactly one equivalence class.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
If $x$ is in both classes, which class does $x$ belong to?
Answer
(a). If $x \in [a] \cap [b]$ and $x$ lies in exactly one class, then $[a]$ and $[b]$ are that same class, so A follows directly from R.
Q6 · 5 marks · Long answerHarder board level
Let $A = {0, 1, 2, 3, 4, 5}$ and define R on $A \times A$ by $(a, b) R (c, d) ⟺ a^{2} + b^{2} = c^{2} + d^{2}$ . Prove that R is an equivalence relation and find the equivalence class of $(3, 4)$ .
Hint
Let $f (a, b) = a^{2} + b^{2}$ ; the rule is $f (a, b) = f (c, d)$ . For the class, find all pairs in $A \times A$ with $x^{2} + y^{2} = 25$ .
Answer
Reflexive, symmetric and transitive follow because the rule is an equality of the numbers $a^{2} + b^{2}$ and $c^{2} + d^{2}$ : $f = f$ ; $f_{1} = f_{2} \Rightarrow f_{2} = f_{1}$ ; $f_{1} = f_{2}$ , $f_{2} = f_{3} \Rightarrow f_{1} = f_{3}$ . $[(3, 4)] = {(x, y) \in A \times A : x^{2} + y^{2} = 25} = {(0, 5), (3, 4), (4, 3), (5, 0)}$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a teacher pairs 24-hour clock readings with the positions on a 12-hour wall clock. On $H = {0, 1, 2, \dots, 23}$ , define $a R b$ if $a - b$ is divisible by 12.
(i) Show that R is an equivalence relation. (2 marks)
(ii) Find $[15]$ . (1 mark)
(iii) How many ordered pairs does R contain? (1 mark)
Hint
Each wall-clock position corresponds to exactly two hours of the day.
Answer
(i) $a - a = 0$ ; $a - b = 12 k \Rightarrow b - a = 12 (- k)$ ; $a - b = 12 k$ , $b - c = 12 m \Rightarrow a - c = 12 (k + m)$ . (ii) $[15] = {3, 15}$ . (iii) 12 classes of 2 hours each: $12 \times 2^{2} = 48$ ordered pairs.

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.What three properties make a relation an equivalence relation?Show answer
Reflexive, symmetric and transitive.
2.Define the equivalence class $[a]$ .Show answer
$[a] = {x \in A : (x, a) \in R}$ , the set of all elements related to $a$ .
3.Can two different equivalence classes share an element?Show answer
No. Two classes are either identical or disjoint.
4.How many equivalence classes does 'same remainder on division by 5' give on $Z$ ?Show answer
Five: $[0], [1], [2], [3], [4]$ .
5.How many equivalence relations are there on a 4-element set?Show answer
15, the number of partitions of a 4-element set.
6.If an equivalence relation on a finite set has classes of sizes 3, 2 and 1, how many ordered pairs does it have?Show answer
$9 + 4 + 1 = 14$ .

Equivalence Relations and Equivalence Classes: frequently asked questions

How do I prove a relation is an equivalence relation?

Prove three things separately, using general elements: reflexive (every element is related to itself), symmetric (a related to b gives b related to a) and transitive (a related to b and b related to c give a related to c). Then write the conclusion in one sentence.

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

The equivalence relation is the whole set of related pairs. An equivalence class is one group of elements, all related to each other, that the relation creates. The classes split the set into non-overlapping parts.

Is the universal relation an equivalence relation?

Yes. The universal relation A × A contains every pair, so it is reflexive, symmetric and transitive. It has just one equivalence class, the whole set A.

Which NCERT section covers equivalence relations?

Section 1.2 (Types of Relations) of Chapter 1, Relations and Functions, in NCERT Mathematics Part I for Class 12, which also introduces equivalence classes.

How many marks are equivalence relation questions worth?

They are most often a 5-mark long answer (prove the relation is an equivalence and find a class) or part of a 4-mark case study, and occasionally a 1-mark MCQ or assertion-reason item.