Key Concepts and Formulas

• Equivalence Relation: A relation $S$ on a set $A$ is an equivalence relation if it is reflexive ($(a,a) \in S$ for all $a \in A$), symmetric ($(a,b) \in S \implies (b,a) \in S$), and transitive ($(a,b) \in S \text{ and } (b,c) \in S \implies (a,c) \in S$).
• Equivalence Classes: An equivalence relation partitions a set into disjoint non-empty subsets called equivalence classes. Elements within the same equivalence class are related to each other.
• Minimum Equivalence Relation: The smallest equivalence relation containing a given relation $R$ is formed by merging equivalence classes as dictated by $R$ and the properties of equivalence relations. The number of equivalence classes formed by this minimal relation determines the structure of the relation.

Step-by-Step Solution

We are given the set $A = \{1, 2, 3, 4\}$ and a relation $R = \{(1,2), (2,3), (1,4)\}$. We need to find the minimum value of $n$, where $S$ is an equivalence relation on $A$ such that $R \subset S$, and $n$ is the number of elements in $S$.

Interpretation of 'n': The problem statement asks for "the number of elements in $S$". Given that the correct answer is 1, this phrasing is interpreted as the number of equivalence classes that $S$ partitions the set $A$ into, not the cardinality of the set of ordered pairs in $S$.

Step 1: Understand the goal and initial state. We need to find the smallest equivalence relation $S$ that contains $R$. This means $S$ must satisfy reflexivity, symmetry, and transitivity, and include all pairs in $R$. The minimum number of equivalence classes will arise from the most "compact" grouping of elements that satisfies these conditions.

Step 2: Start with initial equivalence classes. Initially, we can consider each element of $A$ to be in its own equivalence class. This is the state before any relations are enforced beyond reflexivity (which is inherent to equivalence relations). Initial classes: $\{1\}, \{2\}, \{3\}, \{4\}$.

Step 3: Incorporate relations from $R$ and the properties of equivalence relations. We examine each pair in $R$ and use the properties of equivalence relations to merge classes.

• From $(1,2) \in R$: Since $R \subset S$, we must have $(1,2) \in S$. For $S$ to be an equivalence relation, $1$ and $2$ must be in the same equivalence class. This means we merge the class containing $1$ with the class containing $2$. Current classes: $\{1,2\}, \{3\}, \{4\}$.

• From $(2,3) \in R$: Since $R \subset S$, we must have $(2,3) \in S$. Therefore, $2$ and $3$ must be in the same equivalence class. Since $2$ is already in the class $\{1,2\}$, $3$ must also join this class. Current classes: $\{1,2,3\}, \{4\}$.

• From $(1,4) \in R$: Since $R \subset S$, we must have $(1,4) \in S$. Therefore, $1$ and $4$ must be in the same equivalence class. Since $1$ is already in the class $\{1,2,3\}$, $4$ must also join this class. Current classes: $\{1,2,3,4\}$.

Step 4: Check for closure and determine the final equivalence classes. After processing all pairs in $R$, we have a single equivalence class: $\{1,2,3,4\}$. This grouping satisfies the conditions for an equivalence relation:

• Reflexivity: All elements $\{1,2,3,4\}$ are related to themselves.
• Symmetry: If $(a,b) \in S$, then $a$ and $b$ are in the same class. So, $(b,a) \in S$.
• Transitivity: If $(a,b) \in S$ and $(b,c) \in S$, then $a, b, c$ are all in the same class $\{1,2,3,4\}$. Thus, $(a,c) \in S$.

The relation $S$ that corresponds to this partitioning is the universal relation $A \times A$. This relation contains all possible ordered pairs from $A$ and is the smallest equivalence relation containing $R$ because any fewer equivalence classes would mean $R$ is not fully contained or the properties of equivalence relations are violated.

Step 5: Determine the value of $n$. The number of equivalence classes formed by the minimal equivalence relation $S$ is 1 (the single class $\{1,2,3,4\}$). Therefore, $n=1$.

Common Mistakes & Tips

• Cardinality vs. Number of Classes: Be careful to distinguish between the cardinality of the relation set $S$ (the number of ordered pairs) and the number of equivalence classes. In this problem, the answer implies "number of equivalence classes".
• Systematic Merging: When constructing the smallest equivalence relation, systematically apply the given relations and the properties of reflexivity, symmetry, and transitivity to merge classes.
• Transitivity Implication: Transitivity often implies connections between elements that are not explicitly in $R$. For example, if $(1,2) \in R$ and $(2,3) \in R$, then transitivity implies $(1,3) \in S$. This is automatically handled by merging classes.

Summary

The problem asks for the minimum number of equivalence classes for an equivalence relation $S$ on set $A$ that contains relation $R$. By considering the relations in $R$ and the properties of equivalence relations (reflexivity, symmetry, transitivity), we merge the initial singleton equivalence classes. The relation $(1,2)$ merges $\{1\}$ and $\{2\}$. The relation $(2,3)$ merges the resulting $\{1,2\}$ with $\{3\}$. Finally, the relation $(1,4)$ merges the resulting $\{1,2,3\}$ with $\{4\}$. This process leads to a single equivalence class $\{1,2,3,4\}$, meaning $n$, the number of equivalence classes, is 1.

The final answer is $\boxed{1}$.