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Sets, Relations and Functions: Mathematics | JEE Main

Let R be a relation on

\mathbb{N} \times \mathbb{N}

defined by

(a, b)\text{R}(c, d)

if and only if

ad(b - c) = bc(a - d)

. Then R is

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

Atransitive but neither reflexive nor symmetricBsymmetric but neither reflexive nor transitiveCsymmetric and transitive but not reflexiveDreflexive and symmetric but not transitive

Step-by-step solutionView

Correct answer

The relation is symmetric, but fails to be reflexive because $ab(b - a) \neq ba(a - b)$ when $a \neq b$ , and fails to be transitive.

Option analysis

Why each option works or fails

A · transitive but neither reflexive nor symmetric

Believing that interchanging pairs reverses signs symmetrically while preserving an overarching transitive structure. Check symmetry directly: swapping $(a,b)$ and $(c,d)$ changes $ad(b-c)$ to $cb(d-a) = bc(d-a) = -bc(a-d)$ , and changes $bc(a-d)$ to $da(c-b) = -ad(b-c)$ , yielding identical equality.

B · symmetric but neither reflexive nor transitive

None. This is the correct option. Correctly identifying that $(a,b)\text{R}(a,b)$ requires $ab(b-a) = ba(a-b) \implies 2ab(b-a)=0$ , which fails whenever $a \neq b$ in $\mathbb{N}$ , while swapping $(a,b)$ and $(c,d)$ negates both sides identically.

C · symmetric and transitive but not reflexive

Assuming that algebraic relations resembling proportionality or cross-multiplication automatically satisfy transitivity. Test transitivity with explicit counterexamples or solve for common terms; $(1,2)\text{R}(c,d)$ and $(c,d)\text{R}(e,f)$ does not imply $(1,2)\text{R}(e,f)$ because the condition does not decouple into independent coordinate functions.

D · reflexive and symmetric but not transitive

Assuming reflexivity holds by overlooking the fact that $b - a \neq a - b$ for distinct natural numbers. Substitute $(c,d) = (a,b)$ into the definition: $ab(b-a) = ba(a-b)$ requires $2ab(b-a) = 0$ , which is false for any pair where $a \neq b$ .

Reviewed route

Solution

StepWorking

01given

Relation $R$ on $\mathbb{N} \times \mathbb{N}$ defined by $(a, b)R(c, d) \iff ad(b - c) = bc(a - d)$ .

02goal

Determine whether $R$ is reflexive, symmetric, or transitive.

03approach

Test reflexivity by checking $(a, b)R(a, b)$ . Test symmetry by swapping $(a, b)$ and $(c, d)$ . Test transitivity either algebraically by simplifying the condition (dividing by $abcd$ ) or by a counterexample.

04execute

For reflexivity: We check if $(a, b)R(a, b)$ holds for all $(a, b) \in \mathbb{N} \times \mathbb{N}$ . Substituting $(c, d) = (a, b)$ gives $ab(b - a) = ba(a - b) = -ab(b - a)$ , which implies $2ab(b - a) = 0$ . Since $a, b \in \mathbb{N}$ , this holds only if $a = b$ . Thus, for $a \ne b$ (e.g., $(1, 2)$ ), $(a, b)R(a, b)$ fails. Hence, $R$ is not reflexive.

05execute

For symmetry: Assume $(a, b)R(c, d)$ , which means $ad(b - c) = bc(a - d)$ . We need to test $(c, d)R(a, b)$ , which requires $cb(d - a) = da(c - b)$ . Rewriting both sides: $cb(d - a) = -bc(a - d)$ and $da(c - b) = -ad(b - c)$ . Multiplying both sides by $-1$ gives $bc(a - d) = ad(b - c)$ , which is precisely the given condition. Thus, $(c, d)R(a, b)$ always holds whenever $(a, b)R(c, d)$ holds. Hence, $R$ is symmetric.

06execute

For transitivity: Dividing the condition $ad(b - c) = bc(a - d)$ by $abcd$ (since $a,b,c,d > 0$ ) yields: $\frac{b - c}{bc} = \frac{a - d}{ad} \implies \frac{1}{c} - \frac{1}{b} = \frac{1}{d} - \frac{1}{a} \implies \frac{1}{a} - \frac{1}{b} = \frac{1}{d} - \frac{1}{c} = -\left(\frac{1}{c} - \frac{1}{d}\right)$ . Notice the negative sign! If $(a, b)R(c, d)$ , then $f(a, b) = -f(c, d)$ where $f(x, y) = \frac{1}{x} - \frac{1}{y}$ . Then $(c, d)R(e, f)$ implies $f(c, d) = -f(e, f)$ , which means $f(a, b) = f(e, f) \ne -f(e, f)$ (unless $f(a,b) = 0$ ). Explicitly, for $(2, 3)$ , $f(2, 3) = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$ . For $(3, 2)$ , $f(3, 2) = -\frac{1}{6}$ , so $(2, 3)R(3, 2)$ holds. For $(5, 30)$ , $f(5, 30) = \frac{1}{5} - \frac{1}{30} = \frac{1}{6}$ , so $f(3, 2) = -f(5, 30)$ , meaning $(3, 2)R(5, 30)$ holds. However, for $(2, 3)$ and $(5, 30)$ , $f(2, 3) = \frac{1}{6} \ne -f(5, 30) = -\frac{1}{6}$ . Thus $((2, 3), (5, 30)) \notin R$ . Hence, $R$ is not transitive.

✓verify

$R$ is symmetric, but neither reflexive nor transitive. This corresponds to Option (1).

Hints that build this answer step by step

Does the reflexivity condition $(a, b)\text{R}(a, b)$ hold for every $(a, b) \in \mathbb{N} \times \mathbb{N}$ ?
No, because $ab(b - a) = ba(a - b)$ simplifies to $2ab(b - a) = 0$ , which fails when $a \neq b$ .
If $(a, b)\text{R}(c, d)$ holds, what equation describes $(c, d)\text{R}(a, b)$ ?
$cb(d - a) = da(c - b)$ , which is algebraically identical to the given condition.
Dividing both sides of $ad(b - c) = bc(a - d)$ by $abcd$ yields $\frac{1}{c} - \frac{1}{b} = \frac{1}{d} - \frac{1}{a}$ , or $\frac{1}{a} - \frac{1}{b} = \frac{1}{d} - \frac{1}{c}$ . Does this imply transitivity for R?
No, because the intermediate pair $(c,d)$ has its coordinates swapped in the difference ( $\frac{1}{d} - \frac{1}{c}$ instead of $\frac{1}{c} - \frac{1}{d}$ ), preventing the terms from chaining.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why isn't the relation reflexive if it holds when a = b?

A relation is reflexive on a set S only if (x, x) belongs to R for EVERY element x in S. Since it fails whenever a ≠ b, it is not reflexive on N x N.

How did dividing by abcd make transitivity so easy to disprove?

Dividing separates the variables into (a, b) and (c, d) terms: 1/a - 1/b = -(1/c - 1/d). This reveals that R relates pairs whose differences of reciprocals are negatives of each other, making transitivity impossible in general.

Answer

The relation is symmetric, but fails to be reflexive because $ab(b - a) \neq ba(a - b)$ when $a \neq b$ , and fails to be transitive.

Why each option works or fails

A: transitive but neither reflexive nor symmetric - Believing that interchanging pairs reverses signs symmetrically while preserving an overarching transitive structure. Check symmetry directly: swapping $(a,b)$ and $(c,d)$ changes $ad(b-c)$ to $cb(d-a) = bc(d-a) = -bc(a-d)$ , and changes $bc(a-d)$ to $da(c-b) = -ad(b-c)$ , yielding identical equality.
B · correct: symmetric but neither reflexive nor transitive - None. This is the correct option. Correctly identifying that $(a,b)\text{R}(a,b)$ requires $ab(b-a) = ba(a-b) \implies 2ab(b-a)=0$ , which fails whenever $a \neq b$ in $\mathbb{N}$ , while swapping $(a,b)$ and $(c,d)$ negates both sides identically.
C: symmetric and transitive but not reflexive - Assuming that algebraic relations resembling proportionality or cross-multiplication automatically satisfy transitivity. Test transitivity with explicit counterexamples or solve for common terms; $(1,2)\text{R}(c,d)$ and $(c,d)\text{R}(e,f)$ does not imply $(1,2)\text{R}(e,f)$ because the condition does not decouple into independent coordinate functions.
D: reflexive and symmetric but not transitive - Assuming reflexivity holds by overlooking the fact that $b - a \neq a - b$ for distinct natural numbers. Substitute $(c,d) = (a,b)$ into the definition: $ab(b-a) = ba(a-b)$ requires $2ab(b-a) = 0$ , which is false for any pair where $a \neq b$ .

Step-by-step solution

given: Relation $R$ on $\mathbb{N} \times \mathbb{N}$ defined by $(a, b)R(c, d) \iff ad(b - c) = bc(a - d)$ .
goal: Determine whether $R$ is reflexive, symmetric, or transitive.
approach: Test reflexivity by checking $(a, b)R(a, b)$ . Test symmetry by swapping $(a, b)$ and $(c, d)$ . Test transitivity either algebraically by simplifying the condition (dividing by $(1,2)\text{R}(c,d)$ ) or by a counterexample.
execute: For reflexivity: We check if $(a, b)R(a, b)$ holds for all $(a, b) \in \mathbb{N} \times \mathbb{N}$ . Substituting $(c, d) = (a, b)$ gives $ab(b - a) = ba(a - b) = -ab(b - a)$ , which implies $2ab(b - a) = 0$ . Since $a, b \in \mathbb{N}$ , this holds only if $a = b$ . Thus, for $a \ne b$ (e.g., $(1, 2)$ ), $(a, b)R(a, b)$ fails. Hence, $R$ is not reflexive.
execute: For symmetry: Assume $(a, b)R(c, d)$ , which means $ad(b - c) = bc(a - d)$ . We need to test $(c, d)R(a, b)$ , which requires $cb(d - a) = da(c - b)$ . Rewriting both sides: $cb(d - a) = -bc(a - d)$ and $da(c - b) = -ad(b - c)$ . Multiplying both sides by $-1$ gives $bc(a - d) = ad(b - c)$ , which is precisely the given condition. Thus, $(c, d)R(a, b)$ always holds whenever $(a, b)R(c, d)$ holds. Hence, $R$ is symmetric.
execute: For transitivity: Dividing the condition $ad(b - c) = bc(a - d)$ by $abcd$ (since $a,b,c,d > 0$ ) yields: $\frac{b - c}{bc} = \frac{a - d}{ad} \implies \frac{1}{c} - \frac{1}{b} = \frac{1}{d} - \frac{1}{a} \implies \frac{1}{a} - \frac{1}{b} = \frac{1}{d} - \frac{1}{c} = -\left(\frac{1}{c} - \frac{1}{d}\right)$ . Notice the negative sign! If $(a, b)R(c, d)$ , then $f(a, b) = -f(c, d)$ where $f(x, y) = \frac{1}{x} - \frac{1}{y}$ . Then $(c, d)R(e, f)$ implies $f(c, d) = -f(e, f)$ , which means $f(a, b) = f(e, f) \ne -f(e, f)$ (unless $f(a,b) = 0$ ). Explicitly, for $(2, 3)$ , $f(2, 3) = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$ . For $(3, 2)$ , $f(3, 2) = -\frac{1}{6}$ , so $(2, 3)R(3, 2)$ holds. For $(5, 30)$ , $f(5, 30) = \frac{1}{5} - \frac{1}{30} = \frac{1}{6}$ , so $f(3, 2) = -f(5, 30)$ , meaning $(3, 2)R(5, 30)$ holds. However, for $(2, 3)$ and $(5, 30)$ , $f(2, 3) = \frac{1}{6} \ne -f(5, 30) = -\frac{1}{6}$ . Thus $((2, 3), (5, 30)) \notin R$ . Hence, $R$ is not transitive.
verify: $R$ is symmetric, but neither reflexive nor transitive. This corresponds to Option (1).

Sets, Relations and Functions: Mathematics | JEE Main

What feels right?

The relation is symmetric, but fails to be reflexive because $ab(b - a) \neq ba(a - b)$ when $a \neq b$ , and fails to be transitive.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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