CBSE Class 12 Mathematics Relations and Functions Q #2188
COMPETENCY BASED
APPLY
5 Marks Generic-Curriculum LA
If $N$ denotes the set of all natural numbers and $R$ is the relation on $N \times N$ defined by $(a, b) R (c, d)$ if $ad(b + c) = bc(a + d)$. Show that $R$ is an equivalence relation. [Delhi 2015, All India 2023]
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Detailed Solution

Step 1: Reflexivity

For any $(a, b) \in N \times N$, we check if $(a, b) R (a, b)$. The condition is $ad(b + c) = bc(a + d)$. Substituting $c=a$ and $d=b$: $$ab(b + a) = ba(a + b)$$ Since multiplication is commutative, $ab(a + b) = ab(a + b)$. Thus, $R$ is reflexive.

Step 2: Symmetry

Assume $(a, b) R (c, d)$. Then $ad(b + c) = bc(a + d)$. We need to show $(c, d) R (a, b)$, which requires $cb(d + a) = da(c + b)$. From the given equation: $adb + adc = bca + bcd$. Rearranging: $cb(a + d) = ad(b + c)$. This is equivalent to $cb(d + a) = da(c + b)$. Thus, $R$ is symmetric.

Step 3: Transitivity

Assume $(a, b) R (c, d)$ and $(c, d) R (e, f)$. This implies $\frac{a}{b} = \frac{c}{d}$ and $\frac{c}{d} = \frac{e}{f}$. By transitivity of equality, $\frac{a}{b} = \frac{e}{f}$, which implies $af = be$, or $af(b + e) = be(a + f)$. Thus, $R$ is transitive.

Final Answer: Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must apply the definitions of reflexive, symmetric, and transitive relations to a specific algebraic condition.
Knowledge Dimension: PROCEDURAL
Justification: The student follows a structured logical sequence to verify the three properties of an equivalence relation.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. This question tests the student's ability to manipulate algebraic expressions to prove abstract set-theoretic properties, a core requirement of the Relations and Functions chapter.

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