A class teacher is keen to assess the learning of the concept of 'relations' taught to her students. She writes the following five relations defined on each set $A = \{1, 2, 3\}$: $R_1 = \{(2, 3), (3, 2)\}$ $R_2 = \{(1, 2), (1, 3), (3, 2)\}$ $R_3 = \{(1, 2), (2, 1), (1, 1)\}$ $R_4 = \{(1, 1), (1, 2), (3, 3), (2, 2)\}$ $R_5 = \{(1, 1), (1, 2), (3, 3), (2, 2), (2, 1), (2, 3), (3, 2)\}$ Students are asked to answer the following questions for the above relations: (i) Find the relation which is reflexive and transitive but not symmetric. (ii) Find the relation which is reflexive and symmetric but not transitive. (iii) (A) Find the relation which is symmetric but neither reflexive nor transitive. OR (iii) (B) To make relation $R_2$ an equivalence relation, write the pairs that need to be added.

> [!TLDR] > (i) $R_4$ is reflexive and transitive but not symmetric; (ii) $R_5$ is reflexive and symmetric but not transitive; (iii)(A) $R_1$ is symmetric but neither reflexive nor transitive; (iii)(B) to make $R_2$ an equivalence relation add $(1,1),(2,2),(3,3),(2,1),(2,3),(3,1)$. On $A=\{1,2,3\}$ recall: **reflexive** requires all three self-pairs $(1,1),(2,2),(3,3)$; **symmetric** requires $(b,a)$ whenever $(a,b)$ is present; **transitive** requires $(a,c)$ whenever $(a,b)$ and $(b,c)$ are present. ## Part (a) **(i) Reflexive and transitive but not symmetric.** $R_4=\{(1,1),(1,2),(3,3),(2,2)\}$ contains all three self-pairs, so it is reflexive. It has $(1,2)$ but not $(2,1)$, so it is not symmetric. The only non-self pair is $(1,2)$; the only chain it could start, $(1,2)$ with $(2,2)$, yields $(1,2)$ which is present — no required pair is missing, so it is transitive. Hence $R_4$ satisfies (i). **(ii) Reflexive and symmetric but not transitive.** $R_5=\{(1,1),(1,2),(3,3),(2,2),(2,1),(2,3),(3,2)\}$ has all self-pairs (reflexive); every off-diagonal pair has its reverse — $(1,2)/(2,1)$ and $(2,3)/(3,2)$ — so it is symmetric. But $(1,2)$ and $(2,3)$ require $(1,3)$, which is absent, so it is not transitive. Hence $R_5$ satisfies (ii). **(iii)(A) Symmetric but neither reflexive nor transitive.** $R_1=\{(2,3),(3,2)\}$ is symmetric. It has no self-pairs, so it is not reflexive. And $(2,3)$ with $(3,2)$ requires $(2,2)$, which is missing, so it is not transitive. Hence $R_1$ satisfies (iii)(A). …

A class teacher is keen to assess the learning of the concept of 'relations' taught to her students. She write

Question

A = \{1, 2, 3\}

(i) Find the relation which is reflexive and transitive but not symmetric.

(ii) Find the relation which is reflexive and symmetric but not transitive.

(iii)
(A) Find the relation which is symmetric but neither reflexive nor transitive.

(OR)

(iii)
(B) To make relation

R_2

an equivalence relation, write the pairs that need to be added.

CBSECBSE Class XII Board 2025Subjective· 4mImportance★★★★★

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Part (a)Concept understanding — Relation Properties

Properties of a Relation

A relation $R$ on a set $A$ pairs elements of $A$ with one another. Some relations behave in regular, predictable ways, and we name these behaviours properties. Three matter most for CBSE Class 12 — reflexive, symmetric, transitive (together they build an equivalence relation); a fourth, antisymmetric, is worth knowing for order relations.

Reflexive — everything relates to itself

$R$ is reflexive if $a\,R\,a$ for every $a\in A$ . "Has the same age as" is reflexive; "is taller than" is not. If even one element misses its self-pair, reflexivity fails: on $\{1,2,3\}$ , $\{(1,1),(2,2)\}$ is not reflexive because $(3,3)$ is absent.

Symmetric — the relation runs both ways

$R$ is symmetric if $a\,R\,b \implies b\,R\,a$ . "Is married to" is symmetric; "is taller than" is not. Symmetry does not demand that every pair be related — only that any pair which appears also appears reversed. So $\{(1,2),(2,1),(3,3)\}$ is symmetric, but $\{(1,2),(2,1),(1,3)\}$ is not, since $(3,1)$ is missing.

Transitive — relations chain

$R$ is transitive if $a\,R\,b$ and $b\,R\,c$ together force $a\,R\,c$ . "Is an ancestor of" is transitive; "is a friend of" is not. A single broken chain breaks the property: $\{(1,2),(2,3)\}$ is not transitive because $(1,3)$ is missing.

Antisymmetric — two-way ties force equality

$R$ is antisymmetric if $a\,R\,b$ and $b\,R\,a$ together force $a=b$ . The order relation $\le$ is antisymmetric: $a\le b$ and $b\le a$ give $a=b$ . It does not ban self-pairs like $(1,1)$ ; it only rules out distinct elements related both ways.

Tip

Test the properties in order of ease — reflexivity first, then symmetry, transitivity. A single counterexample is enough to disprove any of them.

Property	Condition
Reflexive	$\forall a,\ a\,R\,a$

Part (b)Concept understanding — Equivalence Relation Proof

Proving a Relation is an Equivalence Relation

A relation $R$ on a set $A$ is an equivalence relation when it satisfies exactly three properties: it is reflexive, symmetric, and transitive. To prove a given relation is an equivalence relation, you check these three — in this order — one at a time.

Note

Antisymmetry plays no role here; that property belongs to partial orders. For an equivalence relation you need only reflexive, symmetric, transitive.

The three checks

Reflexive — show $(a, a) \in R$ for every $a \in A$ .
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 $(a, c) \in R$ .

If all three hold, $R$ is an equivalence relation. If even one fails, produce a single counterexample and you are done.

A worked template

Let $R$ be defined on $\mathbb{Z}$ by $a\,R\,b \iff a - b$ is divisible by $5$ .

Reflexive: $a - a = 0$ , and $0$ is divisible by $5$ , so $a\,R\,a$ for every integer $a$ . ✓

Symmetric: if $a\,R\,b$ , then $a - b = 5k$ for some integer $k$ . Then $b - a = -5k = 5(-k)$ , also a multiple of $5$ , so $b\,R\,a$ . ✓

Transitive: if $a\,R\,b$ and $b\,R\,c$ , then $a - b = 5k$ and $b - c = 5m$ . Adding, $a - c = 5(k + m)$ , a multiple of $5$ , so $a\,R\,c$ . ✓

All three hold, so $R$ is an equivalence relation. …

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(i) $R_4$ is reflexive and transitive but not symmetric; (ii) $R_5$ is reflexive and symmetric but not transitive; (iii)(A) $R_1$ is symmetric but neither reflexive nor transitive; (iii)(B) to make $R_2$ an equivalence relation add $(1,1),(2,2),(3,3),(2,1),(2,3),(3,1)$ .

On $A=\{1,2,3\}$ recall: reflexive requires all three self-pairs $(1,1),(2,2),(3,3)$ ; symmetric requires $(b,a)$ whenever $(a,b)$ is present; transitive requires $(a,c)$ whenever $(a,b)$ and $(b,c)$ are present.

Part (a)

Reflexive and transitive but not symmetric. $R_4=\{(1,1),(1,2),(3,3),(2,2)\}$ contains all three self-pairs, so it is reflexive. It has $(1,2)$ but not $(2,1)$ , so it is not symmetric. The only non-self pair is $(1,2)$ ; the only chain it could start, $(1,2)$ with $(2,2)$ , yields $(1,2)$ which is present — no required pair is missing, so it is transitive. Hence $R_4$ satisfies (i).
Reflexive and symmetric but not transitive. $R_5=\{(1,1),(1,2),(3,3),(2,2),(2,1),(2,3),(3,2)\}$ has all self-pairs (reflexive); every off-diagonal pair has its reverse — $(1,2)/(2,1)$ and $(2,3)/(3,2)$ — so it is symmetric. But $(1,2)$ and $(2,3)$ require $(1,3)$ , which is absent, so it is not transitive. Hence $R_5$ satisfies (ii). (iii)(A) Symmetric but neither reflexive nor transitive. $R_1=\{(2,3),(3,2)\}$ is symmetric. It has no self-pairs, so it is not reflexive. And $(2,3)$ with $(3,2)$ requires $(2,2)$ , which is missing, so it is not transitive. Hence $R_1$ satisfies (iii)(A). …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 54 on this concept.

CBSE 2026Set V11 markMCQ
$R$
(a) symmetric but not transitive
(b) transitive but not symmetric
(c) symmetric and transitive
(d) neither symmetric nor transitive
›Reveal solutionSolution

$R=\{(1,1),(2,2)\}$ is both symmetric and transitive, so the answer is (c).

Symmetry: whenever $(a,b)\in R$ we need $(b,a)\in R$ . Here the only pairs are $(1,1)$ and $(2,2)$ , and each is its own reverse, so symmetry holds. …
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CBSE 2026Set A1 markMCQ
$Q. What type of a relation is "less than" in the set of real numbers?$
(a) Only symmetric
(b) Only transitive
(c) Only reflexive
(d) Equivalence
›Reveal solutionSolution
" $<$ " on $\mathbb{R}$ is transitive only.

Consider the relation " $a < b$ " on $\mathbb{R}$ :

Reflexive? $a < a$ is false for every $a$ , so NOT reflexive.

Symmetric? $a < b$ does not imply $b < a$ , so NOT symmetric. …
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CBSE 2026Set ANNUAL1 markMCQ
$R$
(a) Reflexive but not symmetric
(b) Reflexive but not transitive
(c) Symmetric and transitive
(d) Neither symmetric nor transitive
›Reveal solutionSolution

$R$ is reflexive (it contains every $(x,x)$ ) but not symmetric, since $(1,2)\in R$ while $(2,1)\notin R$ .

Given: $R=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}$ on $A=\{1,2,3\}$ .

Reflexivity: $R$ is reflexive if $(x,x)\in R$ for every $x\in A$ . Here $(1,1),(2,2),(3,3)$ are all present, so $R$ is reflexive.

Symmetry: $R$ is symmetric if $(x,y)\in R\Rightarrow(y,x)\in R$ . Here $(1,2)\in R$ but $(2,1)\notin R$ . So $R$ is not symmetric.

…

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CBSE 2026Set ANNUAL1 markMCQ
$R$
(a) $(2, 4) \in R$
(b) $(3, 8) \in R$
(c) $(6, 8) \in R$
(d) $(8, 6) \in R$
›Reveal solutionSolution
Check each pair against both conditions $a=b-2$ and $b>6$ .

$R=\{(a,b): a=b-2,\ b>6\}$

$(2,4)$ : $a=b-2 \Rightarrow 2=2$ ✓, but $b>6 \Rightarrow 4>6$ ✗. Not in $R$ .

$(3,8)$ : $a=b-2 \Rightarrow 3=6$ ✗. Not in $R$ . …
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CBSE 2026Set ANNUAL1 markMCQ
$R$
(a) $R$ is reflexive and symmetric but not transitive
(b) $R$ is reflexive and transitive but not symmetric
(c) $R$ is symmetric and transitive but not reflexive
(d) $R$ is an equivalence relation
›Reveal solutionSolution
Test $R$ against the three defining properties (reflexive, symmetric, transitive) one at a time, directly from its listed ordered pairs.

$R=\{(1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3,2)\}$ on $\{1,2,3,4\}$ .

Reflexive? Need $(a,a)\in R$ for every $a\in\{1,2,3,4\}$ : $(1,1)$ ✓, $(2,2)$ ✓, $(3,3)$ ✓, $(4,4)$ ✓. All present — $R$ is reflexive.

Symmetric? Need: whenever $(a,b)\in R$ , also $(b,a)\in R$ . Take $(1,2)\in R$ : is $(2,1)\in R$ ? It is not in the list. So $R$ is not symmetric.

Transitive? Need: whenever $(a,b)\in R$ and $(b,c)\in R$ , also $(a,c)\in R$ . Checking every chain:

$(1,1),(1,2)\Rightarrow(1,2)$ ✓

$(1,1),(1,3)\Rightarrow(1,3)$ ✓

$(1,2),(2,2)\Rightarrow(1,2)$ ✓

$(1,3),(3,3)\Rightarrow(1,3)$ ✓

$(1,3),(3,2)\Rightarrow(1,2)$ ✓ …
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CBSE 2026Set ANNUAL1 markMCQ
$R$
(a) $(2, 4) \in R$
(b) $(3, 8) \in R$
(c) $(6, 8) \in R$
(d) $(8, 7) \in R$
›Reveal solutionSolution

Only $(6,8)$ satisfies $a=b-2$ with $b>6$ .

…

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CBSE 2026Set ANNUAL1 mark
$Q. Write the answer in one word/sentence: Define equivalence relation.$
›Reveal solutionSolution

Equivalence relation = reflexive + symmetric + transitive.

…

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CBSE 2025Set X11 markMCQ
$R$
(a) $(a, a) \in R$ for all $a \in A$
(b) $(a, a) \in R$ for atleast one $a \in A$
(c) $(a, b) \in R$ implies $(b, a) \in R$
(d) $(a, b) \in R$ and $(b, c) \in R$ implies $(a, c) \in R$
›Reveal solutionSolution

Tests the definition of a reflexive relation — correct option is (a).

A relation $R$ on a set $A$ is reflexive when every element is related to itself. Formally, $(a,a)\in R$ must hold for all $a\in A$ — not just for at least one element. Option (c) states symmetry ( $(a,b)\in R\Rightarrow(b,a)\in R$ ) and option (d) states transitivit …
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CBSE 2025Set IX1 markMCQ
$R = \{(a, b) : a = b - 1,\ b \geq 3\}$
(a) $(2, 4) \in R$
(b) $(4, 5) \in R$
(c) $(4, 6) \in R$
(d) $(1, 3) \in R$
›Reveal solutionSolution
Only $(4,5)$ satisfies $a=b-1$ with $b\ge 3$ ; option (b).

Concept. A pair $(a,b)$ belongs to $R$ only if it satisfies both conditions: $a=b-1$ and $b\ge 3$ .

$(2,4)$ : $b-1=3\ne 2$ . ✗ …
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CBSE 2025Set ANNUAL1 markMCQ
$A=\{1,2,3,4\}$
(i) Reflexive
(ii) Symmetric
(iii) Transitive
(iv) Equivalent
›Reveal solutionSolution

Check each property directly: $R$ turns out to be symmetric only.

$A=\{1,2,3,4\}$ , $R=\{(a,b): a+b \text{ is odd}\}$ .

Reflexive? $(a,a)$ needs $a+a=2a$ to be odd — but $2a$ is always even. So $R$ is not reflexive.

Symmetric? If $a+b$ is odd, then $b+a=a+b$ is the same sum, also odd. So $(a,b)\in R \Rightarrow (b,a)\in R$ . $R$ is symmetric.

…

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CBSE 2025Set ANNUAL1 markMCQ
$A = \{a, b, c\}$
(a) reflexive and symmetric but not transitive
(b) reflexive and transitive but not symmetric
(c) symmetric and transitive but not reflexive
(d) an equivalence relation
›Reveal solutionSolution

R is symmetric (the only cross-pair (a,b)/(b,a) both appear) and not transitive ((b,a) & (a,b) would force (b,b), which is missing) — matching option (a) once we note a reflexivity caveat below.

Step 1 — Reflexive? Full reflexivity on A = {a, b, c} needs (a,a), (b,b), (c,c) all in R. Only (a,a) is present; (b,b) and (c,c) are not. So strictly, R is not fully reflexive on {a,b,c}.

Step 2 — Symmetric? The only pair with a 'partner' is (a,b), and its reverse (b,a) is also in R. There is no pair in R whose reverse is missing. So R is symmetric.

Step 3 — Transitive? Take (b,a) ∈ R and (a,b) ∈ R: transitivity would require (b,b) ∈ R. It is not. So R is not transitive.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Relation R = {(x, y) : x < y² where x, y \in R} is:$
(a) Reflexive but not symmetric.
(b) Symmetric and transitive but not Reflexive.
(c) Reflexive and Symmetric.
(d) Neither reflexive nor symmetric nor transitive.
›Reveal solutionSolution

Test each property of $R=\{(x,y): x<y^2\}$ with concrete counterexamples — all three fail.

Reflexive? Need $x<x^2$ for every $x\in\mathbb R$ . Take $x=\dfrac12$ : is $\dfrac12 < \dfrac14$ ? No. So $R$ is not reflexive.

Symmetric? Need $x<y^2 \Rightarrow y<x^2$ . Take $x=0,\,y=1$ : $0<1^2$ is true, so $(0,1)\in R$ . But is $1<0^2=0$ ? No. So $(1,0)\notin R$ , and $R$ is not symmetric.

…

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Part (a)

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