MathsCBSE Class 12 · Relations and Functions

Reflexive, Symmetric and Transitive Relations (Class 12)

Updated 15 October 2026 · Fact-checked

A relation R on a set A is reflexive if every element is related to itself, symmetric if a R b always gives b R a, and transitive if a R b and b R c always give a R c. To prove a property, argue for arbitrary elements; to disprove it, one specific counter-example is enough.

Reflexive needs a loop on every element, symmetric needs arrows both ways, transitive needs a shortcut for every two-step chain.

Illustration drawn for QuestionBuddy.

Understand Reflexive, Symmetric and Transitive Relations

A relation $R$ on a set $A$ is simply a subset of $A \times A$ . When $(a, b) \in R$ we write $a R b$ and say ' $a$ is related to $b$ '. Two relations sit at the extremes: the empty relation $ϕ$ (no pair at all) and the universal relation $A \times A$ (every pair). Most board questions are about three properties a relation may or may not have.

Reflexive: $(a, a) \in R$ for every $a \in A$ . Think of the diagonal of $A \times A$ : all of it must be inside $R$ . Missing even one $(a, a)$ breaks it.
Symmetric: whenever $(a, b) \in R$ , the reverse pair $(b, a)$ is also in $R$ . It is a two-way street.
Transitive: whenever $(a, b) \in R$ and $(b, c) \in R$ , the pair $(a, c)$ is also in $R$ . Every two-step chain must have its one-step shortcut.

The three properties are independent: a relation can have any combination of them. On a finite set given as a list of pairs, you check each one by scanning the list. On an infinite set such as $N$ , $Z$ or $R$ , the relation is given by a rule, and the logic matters more than the arithmetic.

Proving versus disproving. To show a property holds, take arbitrary elements (say $a, b, c$ ) that satisfy the hypothesis and deduce the conclusion in general; checking a few numbers is not a proof. To show a property fails, give one specific counter-example with actual numbers, and show both what holds and what fails. For instance, $a R b ⟺ a \leq b$ on $R$ is not symmetric because $1 \leq 2$ but $2 \neq \leq 1$ .

Watch the 'if-then' structure. Symmetry and transitivity are conditional statements. If no pair ever triggers the 'if', the property holds automatically (it is true vacuously). That is why the empty relation on a non-empty set is symmetric and transitive but not reflexive. Similarly, a relation such as ${(1, 1)}$ on ${1, 2}$ is symmetric and transitive but not reflexive, because $(2, 2)$ is missing.

Counting relations on a finite set. If $A$ has $n$ elements, $A \times A$ has $n^{2}$ pairs and each pair is either in or out, so there are $2^{n^{2}}$ relations. A reflexive relation must contain all $n$ diagonal pairs and is free on the other $n^{2} - n$ pairs, giving $2^{n^{2} - n}$ . For a symmetric relation you choose freely only the diagonal pairs and one pair from each mirror couple ${(a, b), (b, a)}$ , which is $\frac{n ( n + 1 )}{2}$ free choices, giving $2^{n (n + 1) /2}$ .

This topic is the base for equivalence relations, which are relations that are reflexive, symmetric and transitive all at once. In NCERT Class 12 it is Section 1.2 of Chapter 1.

Reflexive, Symmetric and Transitive Relations: formulas to remember

Reflexive: $(a, a) \in R for every a \in A$; Every diagonal pair must be present. One missing $(a, a)$ makes R not reflexive.
Symmetric: $(a, b) \in R \Rightarrow (b, a) \in R$; Check each pair's reverse. Pairs of the form $(a, a)$ are their own reverse.
Transitive: $(a, b) \in R and (b, c) \in R \Rightarrow (a, c) \in R$; Every chain needs its shortcut, including chains such as $(1, 2), (2, 1)$ , which need $(1, 1)$ .
Number of relations on a set with n elements: $2^{n^{2}}$; Each of the $n^{2}$ ordered pairs is either in R or not.
Number of reflexive relations: $2^{n^{2} - n}$; The n diagonal pairs are forced in; the rest are free.
Number of symmetric relations: $2^{n (n + 1) /2}$; Free choices: n diagonal pairs plus $\frac{n ( n - 1 )}{2}$ mirror couples.
Empty and universal relations: $ϕ : symmetric, transitive; A \times A : reflexive, symmetric, transitive$; The empty relation on a non-empty set is not reflexive.

How to solve Reflexive, Symmetric and Transitive Relations questions

Use this routine for every 'check whether R is reflexive, symmetric or transitive' question, whether R is a list of pairs or a rule on an infinite set.

1Write the set and the rule clearly: $R = {(a, b) : \dots}$ on $A$ . Note any restriction such as $a, b \in N$ (positive) or $Z$ (includes 0 and negatives).
2Reflexive: test the rule with $b = a$ . If it holds for every $a$ , prove it in one line. If it fails, name one element, for example 'since $(\frac{1}{2}, \frac{1}{2}) \in / R$ '.
3Symmetric: assume $(a, b) \in R$ , write what the rule gives, and try to deduce the rule for $(b, a)$ . If the rule is lopsided (like $\leq$ , 'divides', $a = 2 b$ ), look for a counter-example.
4Transitive: assume $(a, b) \in R$ and $(b, c) \in R$ , write both conditions, and try to combine them into the condition for $(a, c)$ . If they do not combine, search for a counter-example; try 0, 1, negatives and equal values first.
5For a finite list of pairs, scan systematically: list the diagonal pairs, the reverse of every pair, and every chain $(a, b), (b, c)$ including chains that loop back like $(1, 2), (2, 1)$ .
6State each counter-example in full: what is in R, what is not, and which property therefore fails.
7End with a summary sentence: 'Hence R is reflexive and transitive but not symmetric.'

Reflexive, Symmetric and Transitive 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.

Symmetric and transitive relations

Suppose three friends, Asha, Ben and Chetan, use a messaging app. Let $x R y$ mean ' $x$ has saved $y$ 's number'. Asha has saved Ben's number, Ben has saved Asha's and Chetan's numbers, and nobody else has saved anyone.

Write R as a set of ordered pairs on {Asha, Ben, Chetan}. Decide whether R is reflexive, symmetric and transitive. What is the least number of pairs that must be added to make R symmetric?

Try it first, then show the solutionSolution

Using initials, $R = {(A, B), (B, A), (B, C)}$ .
Reflexive: $(A, A) \in / R$ (Asha has not saved her own number), so R is not reflexive.
Symmetric: $(B, C) \in R$ but $(C, B) \in / R$ , so R is not symmetric.
Transitive: $(A, B) \in R$ and $(B, A) \in R$ but $(A, A) \in / R$ , so R is not transitive.
To make it symmetric, every pair needs its reverse. $(A, B)$ and $(B, A)$ already pair up; only $(C, B)$ is missing. Adding it gives ${(A, B), (B, A), (B, C), (C, B)}$ , which is symmetric.

Answer: $R = {(A, B), (B, A), (B, C)}$ is neither reflexive, nor symmetric, nor transitive. Adding one pair, $(C, B)$ , makes it symmetric.

Memory hook: Saving a number is one-way: symmetry needs the other person to save you back.

Transitivity as shortcuts

Imagine four junctions P, Q, R and S joined by one-way roads P to Q, Q to R and R to S. Let $x T y$ mean 'there is a single road from $x$ to $y$ '.

Is T transitive? Find the smallest relation containing T that is transitive, and say in words what it describes.

Try it first, then show the solutionSolution

$T = {(P, Q), (Q, R), (R, S)}$ .
$(P, Q), (Q, R) \in T$ but $(P, R) \in / T$ , so T is not transitive.
Add the shortcut of every two-step chain: $(P, R)$ from P to Q to R, and $(Q, S)$ from Q to R to S.
Now $(P, R)$ and $(R, S)$ form a chain, so add $(P, S)$ as well.
Check the new set ${(P, Q), (Q, R), (R, S), (P, R), (Q, S), (P, S)}$ : every chain now has its shortcut, so it is transitive.

Answer: T is not transitive. Adding the 3 pairs $(P, R)$ , $(Q, S)$ and $(P, S)$ gives the smallest transitive relation containing T. It describes 'you can drive from $x$ to $y$ ', by one or more roads.

Memory hook: Transitive means every two-step route has a direct shortcut in the list.

Tricks to remember Reflexive, Symmetric and Transitive Relations

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

Mirror, two-way street, relay
Reflexive is a mirror: everyone sees themselves. Symmetric is a two-way street: if you can go, you can come back. Transitive is a relay race: if a passes to b and b passes to c, then a reaches c.
Use it when: Recalling the three definitions quickly in MCQs and assertion-reason items.
Prove with letters, disprove with numbers
A property that holds needs a general argument with $a, b, c$ . A property that fails needs only one numerical counter-example.
Use it when: Deciding how much to write for each property in a 3- or 5-mark answer.
Try 0, 1, a negative, and equals
When hunting a counter-example, try 0, 1, a negative number, a fraction like $\frac{1}{2}$ , and $a = b$ first. Most rules break at one of these.
Use it when: Rules like $ab \geq 0$ (fails transitivity through 0) or $a \leq b^{2}$ (fails reflexivity at $\frac{1}{2}$ ).
Counting: n squared, drop the diagonal, fold the square
All relations: $2^{n^{2}}$ . Reflexive: the diagonal is fixed, so drop it: $2^{n^{2} - n}$ . Symmetric: fold the square along the diagonal and choose on one half plus the diagonal: $2^{n (n + 1) /2}$ .
Use it when: One-mark counting MCQs.

Reflexive, Symmetric and Transitive Relations: solved examples

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

Example 1: Reading properties from a list of pairs · 2 marks · Very short answerBoard level

Let $A = {1, 2, 3}$ and $R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}$ . Determine whether R is reflexive, symmetric and transitive.

Try it first, then show the solutionSolution

Reflexive: $(1, 1), (2, 2), (3, 3)$ are all in R, so R is reflexive. (½ mark)
Symmetric: $(1, 2) \in R$ but $(2, 1) \in / R$ , so R is not symmetric. (½ mark)
Transitive: $(1, 2) \in R$ and $(2, 3) \in R$ , but $(1, 3) \in / R$ , so R is not transitive. (1 mark)

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

Example 2: A rule on the integers · 3 marks · Short answerBoard level

A relation R on $Z$ is defined by $a R b$ if and only if $a - b \leq 2$ . Check whether R is reflexive, symmetric and transitive.

Try it first, then show the solutionSolution

Reflexive: for any $a \in Z$ , $a - a = 0 \leq 2$ , so $(a, a) \in R$ . R is reflexive. (1 mark)
Symmetric: $(0, 5) \in R$ since $0 - 5 = - 5 \leq 2$ , but $(5, 0) \in / R$ since $5 - 0 = 5 > 2$ . R is not symmetric. (1 mark)
Transitive: $(5, 3) \in R$ since $5 - 3 = 2 \leq 2$ , and $(3, 1) \in R$ since $3 - 1 = 2 \leq 2$ , but $(5, 1) \in / R$ since $5 - 1 = 4 > 2$ . R is not transitive. (1 mark)

Answer: R is reflexive but neither symmetric nor transitive.

Example 3: Common factor relation on the natural numbers · 3 marks · Short answerHarder board level

On $N$ , define $a R b$ if $a$ and $b$ have a common factor greater than 1. Show that R is symmetric but neither reflexive nor transitive.

Try it first, then show the solutionSolution

Reflexive: the only factor of 1 is 1, so 1 and 1 have no common factor greater than 1. Thus $(1, 1) \in / R$ and R is not reflexive. (1 mark)
Symmetric: if $a$ and $b$ share a factor $d > 1$ , then $b$ and $a$ share the same factor $d$ . So $(a, b) \in R \Rightarrow (b, a) \in R$ ; R is symmetric. (1 mark)
Transitive: $(2, 6) \in R$ (common factor 2) and $(6, 3) \in R$ (common factor 3), but 2 and 3 have no common factor greater than 1, so $(2, 3) \in / R$ . R is not transitive. (1 mark)

Answer: R is symmetric only; $(1, 1) \in / R$ breaks reflexivity and the chain $2, 6, 3$ breaks transitivity.

Example 4: The zero trap · 3 marks · Short answerHarder board level

A relation R on $Z$ is defined by $a R b ⟺ ab \geq 0$ . Prove that R is reflexive and symmetric. Is R transitive? Justify.

Try it first, then show the solutionSolution

Reflexive: for every integer $a$ , $a \cdot a = a^{2} \geq 0$ , so $(a, a) \in R$ . (1 mark)
Symmetric: if $ab \geq 0$ then $ba = ab \geq 0$ , so $(a, b) \in R \Rightarrow (b, a) \in R$ . (1 mark)
Transitive: take $a = - 1$ , $b = 0$ , $c = 1$ . Then $ab = 0 \geq 0$ and $b c = 0 \geq 0$ , so $(- 1, 0), (0, 1) \in R$ . But $a c = - 1 < 0$ , so $(- 1, 1) \in / R$ . R is not transitive. (1 mark)

Answer: R is reflexive and symmetric but not transitive; zero links a negative and a positive number.

Example 5: Fewest pairs to add · 2 marks · Very short answerHarder board level

Let $A = {1, 2, 3, 4}$ and $R = {(1, 2), (2, 3)}$ . Find the least number of ordered pairs that must be added to R so that the new relation is reflexive and transitive. Is the new relation symmetric?

Try it first, then show the solutionSolution

Reflexive needs all four diagonal pairs: add $(1, 1), (2, 2), (3, 3), (4, 4)$ . (½ mark)
Transitive: the chain $(1, 2), (2, 3)$ needs $(1, 3)$ , so add it. (½ mark)
Check $R^{'} = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 3), (1, 3)}$ : chains through a diagonal pair give back a pair already present, and $(1, 3)$ starts no new chain, so $R^{'}$ is transitive. (½ mark)
So 5 pairs are needed. $R^{'}$ is not symmetric, since $(1, 2) \in R^{'}$ but $(2, 1) \in / R^{'}$ . (½ mark)

Answer: 5 pairs: $(1, 1), (2, 2), (3, 3), (4, 4), (1, 3)$ . The new relation is not symmetric.

Example 6: Case study: stations on a metro line · 4 marks · Case-basedHOTS / competency

Suppose five metro stations on a straight line are numbered 1 to 5 in order. A travel card allows a free ride between stations $a$ and $b$ if they are the same station or next to each other. Define R on $S = {1, 2, 3, 4, 5}$ by $a R b ⟺ ∣ a - b ∣ \leq 1$ .
(i) Is R reflexive? (1 mark)
(ii) Is R symmetric? (1 mark)
(iii) Show that R is not transitive and explain what this means for a passenger. (1 mark)
(iv) How many ordered pairs does R contain? (1 mark)

Try it first, then show the solutionSolution

(i) $∣ a - a ∣ = 0 \leq 1$ for each station, so R is reflexive. (1 mark)
(ii) $∣ a - b ∣ = ∣ b - a ∣$ , so $a R b \Rightarrow b R a$ ; R is symmetric. (1 mark)
(iii) $(1, 2) \in R$ and $(2, 3) \in R$ but $∣1 - 3∣ = 2 > 1$ , so $(1, 3) \in / R$ . A passenger can ride free from 1 to 2 and from 2 to 3, but not from 1 to 3 in one free ride. (1 mark)
(iv) Diagonal pairs: 5. Neighbouring stations: 4 adjacent couples, each giving 2 ordered pairs, so 8. Total $5 + 8 = 13$ . (1 mark)

Answer: (i) Yes. (ii) Yes. (iii) No: $(1, 2), (2, 3) \in R$ but $(1, 3) \in / R$ . (iv) 13 ordered pairs.

Example 7: Counting reflexive and symmetric relations · 3 marks · Short answerHOTS / competency

Let $A = {a, b, c}$ . Find the number of relations on A that are (i) reflexive, (ii) symmetric, (iii) both reflexive and symmetric.

Try it first, then show the solutionSolution

$A \times A$ has $3^{2} = 9$ ordered pairs: 3 diagonal pairs and 6 off-diagonal pairs forming 3 mirror couples. (½ mark)
(i) Reflexive: the 3 diagonal pairs are forced in; each of the 6 others is free. Number $= 2^{6} = 64$ . (1 mark)
(ii) Symmetric: choose each diagonal pair freely (3 choices) and each mirror couple as a whole (3 choices): $2^{3 + 3} = 64$ . (1 mark)
(iii) Both: diagonal pairs forced in, mirror couples free: $2^{3} = 8$ . (½ mark)

Answer: (i) 64 (ii) 64 (iii) 8.

Where students lose marks in Reflexive, Symmetric and Transitive Relations

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

Mistake: Proving a property with one numerical example
One example can disprove a property but never proves it for all elements of an infinite set.
Wrong working
Reflexive: $(3, 3) \in R$ because $3 - 3 = 0$ . So R is reflexive.
Right working
Reflexive: for every $a \in Z$ , $a - a = 0$ , which satisfies the rule, so $(a, a) \in R$ for all $a$ .
Do this: Use a general element $a$ (and $b, c$ ) when a property holds; save numbers for counter-examples.
Usually costs: The mark for that property in a 3- or 5-mark proof.
Mistake: Missing the loop-back chain when checking transitivity
A chain like $(1, 2), (2, 1)$ requires $(1, 1)$ ; students only look at chains with three different elements.
Wrong working
$R = {(1, 2), (2, 1)}$ : no chain $a, b, c$ with different elements, so R is transitive.
Right working
$(1, 2), (2, 1) \in R$ but $(1, 1) \in / R$ , so R is not transitive.
Do this: When scanning chains, allow $c = a$ : every symmetric couple $(a, b), (b, a)$ needs both $(a, a)$ and $(b, b)$ .
Usually costs: 1 mark, and the MCQ when options test this.
Mistake: Writing a counter-example without saying what fails
Examiners need both halves: what is in R and what is not.
Wrong working
Not symmetric, e.g. 2 and 4.
Right working
Not symmetric: $(2, 4) \in R$ since 2 divides 4, but $(4, 2) \in / R$ since 4 does not divide 2.
Do this: Write the counter-example as a full sentence with both membership statements and the reason for each.
Usually costs: Half to one mark per property.
Mistake: Ignoring the set the relation is defined on
Whether 0, negatives or fractions are allowed decides which counter-examples exist.
Wrong working
$a R b ⟺ ab > 0$ on $N$ : not reflexive since $0 \cdot 0 = 0$ .
Right working
On $N$ , 0 is not an element, and $a \cdot a > 0$ for every natural number, so R is reflexive on $N$ .
Do this: Underline the set ( $N$ , $Z$ , $R$ or a finite set) and choose counter-examples only from it.
Usually costs: The whole property, sometimes the final conclusion too.

Exam tips

Write the three properties as three headed paragraphs, 'Reflexive:', 'Symmetric:', 'Transitive:', so each mark is easy for the examiner to award.
When a property holds, start with 'Let $a, b \in A$ such that $(a, b) \in R$ ' and end with 'Hence $(b, a) \in R$ '. This structure is what the marking scheme rewards.
For finite sets, a missing diagonal pair or a missing reverse pair is the fastest counter-example; name it exactly.
Counting questions on reflexive or symmetric relations appear as 1-mark MCQs (one was asked in the 2023 board paper); learn $2^{n^{2} - n}$ and $2^{n (n + 1) /2}$ .
Before concluding 'not transitive', double-check your counter-example really has $(a, b)$ and $(b, c)$ in R with the same middle element.

Practice questions on Reflexive, Symmetric and Transitive 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
Let $A = {1, 2, 3}$ and $R = {(1, 2), (2, 1)}$ . Then R is:
(a) reflexive and symmetric
(b) symmetric and transitive
(c) symmetric only
(d) transitive only
Hint
Check the chain $(1, 2), (2, 1)$ : what shortcut does it need?
Answer
(c). The diagonal pairs are missing, so R is not reflexive. Each pair's reverse is present, so R is symmetric. The chain $(1, 2), (2, 1)$ needs $(1, 1)$ , which is absent, so R is not transitive.
Q2 · 1 mark · MCQBoard level
The number of reflexive relations on a set with 4 elements is:
(a) $2^{4}$
(b) $2^{12}$
(c) $2^{16}$
(d) $2^{10}$
Hint
Remove the 4 diagonal pairs from the 16 pairs of $A \times A$ .
Answer
(b) $2^{12} = 4096$ , since the 4 diagonal pairs are fixed and the other $16 - 4 = 12$ pairs are free.
Q3 · 3 marks · Short answerBoard level
Show that the relation R on $N$ defined by $a R b$ if $a$ divides $b$ is reflexive and transitive but not symmetric.
Hint
For transitivity, write $b = k a$ and $c = mb$ .
Answer
Reflexive: $a = 1 \cdot a$ , so $a$ divides $a$ . Symmetric fails: 2 divides 4 but 4 does not divide 2. Transitive: if $b = k a$ and $c = mb$ for natural numbers $k, m$ , then $c = (mk) a$ , so $a$ divides $c$ . Hence R is reflexive and transitive but not symmetric.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The empty relation on $A = {1, 2}$ is symmetric and transitive.
Reason (R): A statement of the form 'if $(a, b) \in R$ then ...' cannot fail when R has no pairs.
(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
Can you find a pair in the empty relation that violates symmetry or transitivity?
Answer
(a). Symmetry and transitivity are 'if-then' conditions, and with no pairs there is no case where they fail, so both hold. (The empty relation is not reflexive, because $(1, 1)$ is missing, but that is not claimed.)
Q5 · 3 marks · Short answerHarder board level
Let R be the relation on $R$ defined by $a R b ⟺ ab > 0$ . Show that R is symmetric and transitive but not reflexive.
Hint
Transitivity: $ab > 0$ means $a$ and $b$ have the same sign and neither is zero.
Answer
Not reflexive: $0 \cdot 0 = 0$ , so $(0, 0) \in / R$ . Symmetric: $ab > 0 \Rightarrow ba > 0$ . Transitive: $ab > 0$ means $a, b$ are non-zero with the same sign; $b c > 0$ means $b, c$ have the same sign; so $a$ and $c$ have the same sign as $b$ and $a c > 0$ . (Alternatively, $ab \cdot b c = a b^{2} c > 0$ with $b^{2} > 0$ gives $a c > 0$ .)
Q6 · 2 marks · Very short answerHarder board level
Give an example of a relation on $A = {1, 2, 3}$ that is reflexive and symmetric but not transitive. Justify each property.
Hint
Start with the diagonal, then add two mirror couples that share one element.
Answer
$R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}$ . All diagonal pairs are present (reflexive); every pair's reverse is present (symmetric); $(1, 2), (2, 3) \in R$ but $(1, 3) \in / R$ (not transitive).
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose five students A, B, C, D and E score 72, 85, 85, 60 and 91 marks in a test. Define R on the set of these students by $x R y$ if $x$ scored at least as many marks as $y$ .
(i) Is R reflexive? (1 mark)
(ii) Is R symmetric? Use the pairs (B, C) and (E, A). (1 mark)
(iii) Is R transitive? (1 mark)
(iv) Find the number of ordered pairs in R. (1 mark)
Hint
For (iv), count for each student how many students (including themselves) scored at most as much.
Answer
(i) Yes: each student's score is at least their own score. (ii) No: $(B, C)$ and $(C, B)$ are both in R since $85 \geq 85$ , but $(E, A) \in R$ ( $91 \geq 72$ ) while $(A, E) \in / R$ , so R is not symmetric. (iii) Yes: if $s_{x} \geq s_{y}$ and $s_{y} \geq s_{z}$ then $s_{x} \geq s_{z}$ . (iv) A: 2 (A, D); B: 4; C: 4; D: 1; E: 5. Total $= 16$ .

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 condition for a relation R on A to be reflexive.Show answer
$(a, a) \in R$ for every $a \in A$ .
2.How many relations are there on a set with $n$ elements?Show answer
$2^{n^{2}}$ .
3.Is the empty relation on ${1, 2}$ transitive?Show answer
Yes, vacuously: there are no pairs, so no chain can fail. It is also symmetric, but not reflexive.
4.What does $(1, 2), (2, 1) \in R$ force if R is transitive?Show answer
Both $(1, 1)$ and $(2, 2)$ must be in R.
5.How many symmetric relations are there on a set with 3 elements?Show answer
$2^{3 \cdot 4/2} = 2^{6} = 64$ .
6.Give a one-line reason why ' $a \leq b$ ' on $R$ is not symmetric.Show answer
$1 \leq 2$ but $2 \neq \leq 1$ , so $(1, 2) \in R$ while $(2, 1) \in / R$ .

Reflexive, Symmetric and Transitive Relations: frequently asked questions

Can a relation be symmetric and transitive but not reflexive?

Yes. A symmetric, transitive relation is reflexive only on the elements that appear in some pair. If an element is never related to anything, such as 3 in the relation {(1, 1), (1, 2), (2, 1), (2, 2)} on {1, 2, 3}, its diagonal pair is missing and the relation is not reflexive.

Is it enough to show one example to prove a relation is reflexive?

No. A property that holds must be shown for every element, using a general element a. A single numerical example is only enough to show that a property fails.

Which NCERT section covers types of relations?

Section 1.2, Types of Relations, in Chapter 1 (Relations and Functions) of NCERT Mathematics Part I for Class 12. Its exercise has questions on checking these properties for finite and infinite sets.

Are antisymmetric relations or partial orders in the 2026-27 syllabus?

No. The 2026-27 CBSE syllabus lists reflexive, symmetric, transitive and equivalence relations only. Antisymmetry is not asked.

How many marks do relation-property questions carry in the board exam?

They appear as 1-mark MCQs, 2- or 3-mark short answers and as part of 5-mark equivalence relation proofs or 4-mark case studies, so they are worth practising in full sentences.