JEE MainMathematics

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Testing Reflexivity, Symmetry, and Transitivity on a Product Set

Let R be a relation on ℕ × ℕ defined by (a, b)R(c, d) if and only if ad(b - c) = bc(a - d). Then R is

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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 $a b (b - a) \neq b a (a - b)$ 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 $a d (b - c)$ to $c b (d - a) = b c (d - a) = - b c (a - d)$ , and changes $b c (a - d)$ to $d a (c - b) = - a d (b - c)$ , yielding identical equality.

B · symmetric but neither reflexive nor transitive

This is the correct option. Correctly identifying that $(a, b) R (a, b)$ requires $a b (b - a) = b a (a - b) ⟹ 2 a b (b - a) = 0$ , which fails whenever $a \neq b$ in $N$ , 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; $(1, 2) R (c, d)$ and $(c, d) R (e, f)$ does not imply $(1, 2) R (e, f)$ 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: $a b (b - a) = b a (a - b)$ requires $2 a b (b - a) = 0$ , which is false for any pair where $a \neq b$ .

Reviewed route

Solution

StepWorking

01Given

Relation $R$ on $N \times N$ defined by $(a, b) R (c, d) ⟺ a d (b - c) = b c (a - d)$ .

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 $a b c d$ ) or by a counterexample.

04Execute

For reflexivity: We check if $(a, b) R (a, b)$ holds for all $(a, b) \in N \times N$ . Substituting $(c, d) = (a, b)$ gives $a b (b - a) = b a (a - b) = - a b (b - a)$ , which implies $2 a b (b - a) = 0$ . Since $a, b \in N$ , this holds only if $a = b$ . Thus, for $a \neq b$ (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 $a d (b - c) = b c (a - d)$ . We need to test $(c, d) R (a, b)$ , which requires $c b (d - a) = d a (c - b)$ . Rewriting both sides: $c b (d - a) = - b c (a - d)$ and $d a (c - b) = - a d (b - c)$ . Multiplying both sides by $- 1$ gives $b c (a - d) = a d (b - c)$ , 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 $a d (b - c) = b c (a - d)$ by $a b c d$ (since $a, b, c, d > 0$ ) yields: $\frac{b - c}{b c} = \frac{a - d}{a d} ⟹ \frac{1}{c} - \frac{1}{b} = \frac{1}{d} - \frac{1}{a} ⟹ \frac{1}{a} - \frac{1}{b} = \frac{1}{d} - \frac{1}{c} = - (\frac{1}{c} - \frac{1}{d})$ . 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 $f (a, b) = f (e, f) \neq - f (e, f)$ (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 $(5, 30)$ , $f (2, 3) = \frac{1}{6} \neq - f (5, 30) = - \frac{1}{6}$ . 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 B.

Hints that build this answer step by step

Does the reflexivity condition $(a, b) R (a, b)$ hold for every $(a, b) \in N \times N$ ?
No, because $a b (b - a) = b a (a - b)$ simplifies to $2 a b (b - a) = 0$ , which fails when $a \neq b$ .
If $(a, b) R (c, d)$ holds, what equation describes $(c, d) R (a, b)$ ?
$c b (d - a) = d a (c - b)$ , which is algebraically identical to the given condition.
Dividing both sides of $a d (b - c) = b c (a - d)$ by $a b c d$ 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
Academic status: Answer verified against the official NTA key
Source: Previous-year question
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 $a b (b - a) \neq b a (a - b)$ 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 $a d (b - c)$ to $c b (d - a) = b c (d - a) = - b c (a - d)$ , and changes $b c (a - d)$ to $d a (c - b) = - a d (b - c)$ , yielding identical equality.
B · correct: symmetric but neither reflexive nor transitive - This is the correct option. Correctly identifying that $(a, b) R (a, b)$ requires $a b (b - a) = b a (a - b) ⟹ 2 a b (b - a) = 0$ , which fails whenever $a \neq b$ in $N$ , 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; $(1, 2) R (c, d)$ and $(c, d) R (e, f)$ does not imply $(1, 2) R (e, f)$ 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: $a b (b - a) = b a (a - b)$ requires $2 a b (b - a) = 0$ , which is false for any pair where $a \neq b$ .

Step-by-step solution

Given: Relation $R$ on $N \times N$ defined by $(a, b) R (c, d) ⟺ a d (b - c) = b c (a - d)$ .
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 $a b c d$ ) or by a counterexample.
Execute: For reflexivity: We check if $(a, b) R (a, b)$ holds for all $(a, b) \in N \times N$ . Substituting $(c, d) = (a, b)$ gives $a b (b - a) = b a (a - b) = - a b (b - a)$ , which implies $2 a b (b - a) = 0$ . Since $a, b \in N$ , this holds only if $a = b$ . Thus, for $a \neq b$ (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 $a d (b - c) = b c (a - d)$ . We need to test $(c, d) R (a, b)$ , which requires $c b (d - a) = d a (c - b)$ . Rewriting both sides: $c b (d - a) = - b c (a - d)$ and $d a (c - b) = - a d (b - c)$ . Multiplying both sides by $- 1$ gives $b c (a - d) = a d (b - c)$ , 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 $a d (b - c) = b c (a - d)$ by $a b c d$ (since $a, b, c, d > 0$ ) yields: $\frac{b - c}{b c} = \frac{a - d}{a d} ⟹ \frac{1}{c} - \frac{1}{b} = \frac{1}{d} - \frac{1}{a} ⟹ \frac{1}{a} - \frac{1}{b} = \frac{1}{d} - \frac{1}{c} = - (\frac{1}{c} - \frac{1}{d})$ . 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 $f (a, b) = f (e, f) \neq - f (e, f)$ (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 $f (x, y) = \frac{1}{x} - \frac{1}{y}$ 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 $(5, 30)$ , $f (2, 3) = \frac{1}{6} \neq - f (5, 30) = - \frac{1}{6}$ . 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 B.

Testing Reflexivity, Symmetry, and Transitivity on a Product Set

What feels right?

The relation is symmetric, but fails to be reflexive because $a b (b - a) \neq b a (a - b)$ 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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