Q7:

Let $A = \{1, 2, 3\}$. Then number of equivalence relations containing $(1, 2)$ is

Step 1: Theoretical Foundation of Equivalence Relations

An equivalence relation $R$ on a set $A$ must satisfy three reflexive, symmetric, and transitive properties [Per the definition of Equivalence Relations]:

• Reflexivity: $(a, a) \in R$ for all $a \in A$. Since $A = \{1, 2, 3\}$, $R$ must contain $\{(1, 1), (2, 2), (3, 3)\}$.
• Symmetry: If $(a, b) \in R$, then $(b, a) \in R$.
• Transitivity: If $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.

Step 2: Identifying Mandatory Elements

Given $A = \{1, 2, 3\}$ and the condition that $(1, 2) \in R$, we must satisfy the properties above:

1. By Reflexivity: $\{(1, 1), (2, 2), (3, 3)\} \subseteq R$.
2. By Symmetry: Since $(1, 2) \in R$, it follows that $(2, 1) \in R$.
3. Current subset $R_0 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$.

We check for Transitivity in $R_0$: All pairs $(a, b)$ and $(b, c)$ lead to $(a, c)$ already within the set. Thus, $R_0$ is an equivalence relation.

Step 3: Determining Additional Possible Equivalence Relations

An equivalence relation partitions a set into disjoint equivalence classes. The number of equivalence relations corresponds to the number of partitions of the set $\{1, 2, 3\}$ where $1$ and $2$ are in the same subset.

Any other partition would require $1$ and $2$ to be in different blocks, which contradicts the condition that $(1, 2) \in R$. For instance, in partition $\{\{1\}, \{2\}, \{3\}\}$, $(1, 2) \notin R$. In partitions like $\{\{1, 3\}, \{2\}\}$, $(1, 2) \notin R$.

Step 4: Visualizing the Partitions

The first case corresponds to the relation $R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$. The second case corresponds to the universal relation $R_2 = A \times A = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)\}$.

Final Solution: 2