Class JEE Mathematics Sets, Relations, and Functions Q #1140
COMPETENCY BASED
APPLY
4 Marks 2024 JEE Main 2024 (Online) 31st January Morning Shift NUMERICAL
Let $A=\{1,2,3,4\}$ and $R=\{(1,2),(2,3),(1,4)\}$ be a relation on $\mathrm{A}$. Let $\mathrm{S}$ be the equivalence relation on $\mathrm{A}$ such that $R \subset S$ and the number of elements in $\mathrm{S}$ is $\mathrm{n}$. Then, the minimum value of $n$ is __________.
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Let $A = \{1, 2, 3, 4\}$ and $R = \{(1, 2), (2, 3), (1, 4)\}$ be a relation on $A$. We want to find the smallest equivalence relation $S$ on $A$ such that $R \subset S$.

Since $S$ is an equivalence relation, it must be reflexive, symmetric, and transitive.

1. Reflexive: $S$ must contain $\{(1, 1), (2, 2), (3, 3), (4, 4)\}$.

2. $R \subset S$, so $S$ must contain $\{(1, 2), (2, 3), (1, 4)\}$.

3. Symmetric: Since $(1, 2) \in S$, $(2, 1) \in S$. Since $(2, 3) \in S$, $(3, 2) \in S$. Since $(1, 4) \in S$, $(4, 1) \in S$.

4. Transitive: Since $(1, 2) \in S$ and $(2, 3) \in S$, $(1, 3) \in S$. Since $(2, 1) \in S$ and $(1, 4) \in S$, $(2, 4) \in S$. Since $(1, 4) \in S$ and $(4, 1) \in S$, $(1, 1) \in S$.

5. Symmetric (again): Since $(1, 3) \in S$, $(3, 1) \in S$. Since $(2, 4) \in S$, $(4, 2) \in S$.

6. Transitive (again): Since $(2, 3) \in S$ and $(3, 1) \in S$, $(2, 1) \in S$. Since $(1, 2) \in S$ and $(2, 4) \in S$, $(1, 4) \in S$. Since $(2, 1) \in S$ and $(1, 3) \in S$, $(2, 3) \in S$.

So, $S = \{(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 3), (1, 4), (2, 1), (3, 2), (4, 1), (1, 3), (3, 1), (2, 4), (4, 2)\}$.

Now, we need to check if this relation is an equivalence relation.

The equivalence classes are $\{1, 2, 3, 4\}$. We can see that $1, 2, 3$ are related, and $1, 4$ are related. So, $1, 2, 3, 4$ must all be related. Thus, the equivalence classes are $\{1, 2, 3, 4\}$.

Therefore, $S = \{(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)\}$.

The number of elements in $S$ is $n = 16$.

However, we can also consider the transitive closure of $R$. Since $(1, 2)$ and $(2, 3)$ are in $R$, $(1, 3)$ must be in $S$. Also, $(1, 4)$ is in $R$. So, the elements $1, 2, 3$ are related, and $1, 4$ is related. This means that all elements $1, 2, 3, 4$ are related to each other. Therefore, the equivalence relation $S$ must relate every element to every other element.

Thus, $S = A \times A = \{(a, b) \mid a \in A, b \in A\}$. The number of elements in $S$ is $|A \times A| = |A|^2 = 4^2 = 16$.

Therefore, the minimum value of $n$ is 16.

Correct Answer: 16

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply the definitions and properties of relations, equivalence relations, and the concept of minimal sets to construct the required equivalence relation.
Knowledge Dimension: CONCEPTUAL
Justification: The question requires understanding of the concepts of relations, equivalence relations, and how to construct the smallest equivalence relation containing a given relation.
Syllabus Audit: In the context of JEE, this is classified as COMPETENCY. The question assesses the student's ability to apply the theoretical knowledge of relations and equivalence relations to solve a specific problem, rather than directly recalling definitions or theorems.

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