Let N be the set of natural numbers. A relation R on $N \times N$ is defined by "(a, b) R (c, d) if and only if ad = bc, for all a, b, c, d $\in$ N". Show that R is an equivalence relation.

> [!TLDR] > To show that a relation is an equivalence relation, we must prove it is reflexive, symmetric, and transitive. For the given relation $R$ on $N \times N$ defined by $(a, b) R (c, d)$ if $ad = bc$, we demonstrate all three properties hold, thus proving $R$ is an equivalence relation. An equivalence relation is a fundamental concept in mathematics that partitions a set into disjoint subsets, called equivalence classes. Each element within an equivalence class is "equivalent" to every other element in that same class according to the relation. To prove a relation $R$ on a set $A$ is an equivalence relation, we must demonstrate that it satisfies three specific properties: 1. **Reflexivity**: Every element is related to itself. For all $x \in A$, $x R x$. 2. **Symmetry**: If $x$ is related to $y$, then $y$ is related to $x$. For all $x, y \in A$, if $x R y$, then $y R x$. 3. **Transitivity**: If $x$ is related to $y$ and $y$ is related to $z$, then $x$ is related to $z$. For all $x, y, z \in A$, if $x R y$ and $y R z$, then $x R z$. The given relation $R$ is defined on the set $N \times N$, where $N$ is the set of natural numbers. An element in $N \times N$ is an ordered pair $(a, b)$ where $a, b \in N$. The relation is defined as $(a, b) R (c, d)$ if and only if $ad = bc$. This definition is reminiscent of how equivalent fractions are defined (e.g., $\frac{a}{b} = \frac{c}{d}$ implies $ad=bc$), which intuitively suggests it might be an equivalence relation. We will now formally prove this. Let's verify each property: ### 1. Reflexivity A relation $R$ is reflexive if for every element $(a, b) \in N \times N$, we have $(a, b) R (a, b)$. According to the definition of $R$, $(a, b) R (a, b)$ means that $a \cdot b = b \cdot a$. This statement is true for all natural numbers $a$ and $b$ due to the commutative property of multiplication. Since $ab = ba$ is always true for any $a, b \in N$, the condition $ad=bc$ holds when $(c,d) = (a,b)$. Therefore, $R$ is

Let N be the set of natural numbers. A relation R on N × N is defined by "(a, b) R (c, d) if and only if ad =

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N \times N

CBSECBSE Class XII Board 2020Subjective· 4mImportance★★★★★

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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	$\forall a,\ a\,R\,a$

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To show that a relation is an equivalence relation, we must prove it is reflexive, symmetric, and transitive. For the given relation $R$ on $N \times N$ defined by $(a, b) R (c, d)$ if $ad = bc$ , we demonstrate all three properties hold, thus proving $R$ is an equivalence relation.

An equivalence relation is a fundamental concept in mathematics that partitions a set into disjoint subsets, called equivalence classes. Each element within an equivalence class is "equivalent" to every other element in that same class according to the relation. To prove a relation $R$ on a set $A$ is an equivalence relation, we must demonstrate that it satisfies three specific properties:

Reflexivity: Every element is related to itself. For all $x \in A$ , $x R x$ .
Symmetry: If $x$ is related to $y$ , then $y$ is related to $x$ . For all $x, y \in A$ , if $x R y$ , then $y R x$ .
Transitivity: If $x$ is related to $y$ and $y$ is related to $z$ , then $x$ is related to $z$ . For all $x, y, z \in A$ , if $x R y$ and $y R z$ , then $x R z$ .

The given relation $R$ is defined on the set $N \times N$ , where $N$ is the set of natural numbers. An element in $N \times N$ is an ordered pair $(a, b)$ where $a, b \in N$ . The relation is defined as $(a, b) R (c, d)$ if and only if $ad = bc$ . This definition is reminiscent of how equivalent fractions are defined (e.g., $\frac{a}{b} = \frac{c}{d}$ implies $ad=bc$ ), which intuitively suggests it might be an equivalence relation. We will now formally prove this.

Let's verify each property:

1. Reflexivity

A relation $R$ is reflexive if for every element $(a, b) \in N \times N$ , we have $(a, b) R (a, b)$ .

According to the definition of $R$ , $(a, b) R (a, b)$ means that $a \cdot b = b \cdot a$ .

This statement is true for all natural numbers $a$ and $b$ due to the commutative property of multiplication.

Since $ab = ba$ is always true for any $a, b \in N$ , the condition $ad=bc$ holds when $(c,d) = (a,b)$ .

Therefore, $R$ is reflexive.

2. Symmetry

A relation $R$ is symmetric if for any two elements $(a, b), (c, d) \in N \times N$ , whenever $(a, b) R (c, d)$ , it implies $(c, d) R (a, b)$ .

Assume $(a, b) R (c, d)$ . By the definition of $R$ , this means $ad = bc$ .

We need to show that $(c, d) R (a, b)$ , which by definition means $cb = da$ .

Given $ad = bc$ .

We know that multiplication of natural numbers is commutative, so $ad = da$ and $bc = cb$ .

Thus, $ad = bc$ can be rewritten as $da = cb$ , or $cb = da$ .

This is exactly the condition for $(c, d) R (a, b)$ .

Therefore, $R$ is symmetric.

3. Transitivity

A relation $R$ is transitive if for any three elements $(a, b), (c, d), (e, f) \in N \times N$ , whenever $(a, b) R (c, d)$ and $(c, d) R (e, f)$ , it implies $(a, b) R (e, f)$ .

Assume $(a, b) R (c, d)$ and $(c, d) R (e, f)$ .

From $(a, b) R (c, d)$ , we have $ad = bc$ (Equation 1).

From $(c, d) R (e, f)$ , we have $cf = de$ (Equation 2).

We need to show that $(a, b) R (e, f)$ , which means $af = be$ .

Let's manipulate the given equations to arrive at the desired result.

From Equation 1, $ad = bc$ . …

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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 41 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 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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CBSE 2025Set ANNUAL1 markMCQ
$Q. The relation R = {(a, a), (b, b), (c, c)} on the set A = {a, b, c} is ...........$
(a) Identity
(b) Reflexive only
(c) Symmetric only
(d) Equivalence
›Reveal solutionSolution

R = {(a,a), (b,b), (c,c)} is reflexive, symmetric, and transitive on A = {a, b, c}, so by definition it is an equivalence relation (it is in fact the identity relation, which is always the smallest possible equivalence relation on a set).

Check reflexive: For every element $x \in A$ , $(x,x) \in R$ . Indeed $(a,a), (b,b), (c,c)$ are all present. So R is reflexive.

Check symmetric: If $(x,y) \in R$ then $(y,x) \in R$ . Since every pair here has $x = y$ , this holds trivially.

Check transitive: If $(x,y) \in R$ and $(y,z) \in R$ then $(x,z) \in R$ . Again since all pairs are of the form $(x,x)$ , this holds trivially.

…

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Q.Let N be the set of natural numbers. A relation R on N×NN \times NN×N is defined by "(a, b) R (c, d) if and only if ad = bc, for all a, b, c, d ∈\in∈ N". Show that R is an equivalence relation.

1. Reflexivity

2. Symmetry

3. Transitivity

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Q.Let N be the set of natural numbers. A relation R on $N \times N$ is defined by "(a, b) R (c, d) if and only if ad = bc, for all a, b, c, d $\in$ N". Show that R is an equivalence relation.

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