Key Concepts and Formulas

• Reflexive Relation: A relation $R$ on a set $A$ is reflexive if for every element $a \in A$, the pair $(a,a)$ is in $R$.
• Transitive Relation: A relation $R$ on a set $A$ is transitive if for all $a,b,c \in A$, whenever $(a,b) \in R$ and $(b,c) \in R$, it must follow that $(a,c) \in R$.
• Symmetric Relation: A relation $R$ on a set $A$ is symmetric if for all $a,b \in A$, whenever $(a,b) \in R$, it must follow that $(b,a) \in R$.
• Not Symmetric Relation: A relation $R$ on a set $A$ is not symmetric if there exists at least one pair $(a,b) \in R$ such that $(b,a) \notin R$.

Step-by-Step Solution

Step 1: Identify the base elements required for the relation. The set is $A=\{1,2,3\}$. For a relation to be reflexive, it must contain all pairs of the form $(a,a)$ for every $a \in A$. So, $R$ must contain $\{(1,1), (2,2), (3,3)\}$. The problem also states that the relation must contain the element $(1,2)$. Therefore, the minimal set of elements that $R$ must contain is $R_{min} = \{(1,1), (2,2), (3,3), (1,2)\}$. The number of elements in $R_{min}$ is 4.

Step 2: Analyze the constraints on the size of the relation and the "not symmetric" condition. The relation $R$ must contain at most 6 elements, so $|R| \le 6$. Since $R_{min}$ already has 4 elements, we can add at most $6 - 4 = 2$ more elements to $R_{min}$. The relation must be "not symmetric". The definition of a not symmetric relation is that there exists at least one pair $(a,b) \in R$ such that $(b,a) \notin R$. Since $(1,2) \in R$, for $R$ to be not symmetric, it is sufficient that $(2,1) \notin R$. If $(2,1) \in R$, then $R$ would be symmetric with respect to the pair $(1,2)$.

Step 3: Consider the implications of $(2,1) \notin R$ for the "not symmetric" condition. If $(2,1) \notin R$, then the condition "not symmetric" is satisfied by the presence of $(1,2)$ and the absence of $(2,1)$. This implies that the relation is symmetric for all other distinct pairs of elements. That is, if $(x,y) \in R$ where $x \ne y$ and $\{x,y\} \ne \{1,2\}$, then $(y,x)$ must also be in $R$.

Step 4: Enumerate possible relations by adding elements to $R_{min}$ while satisfying all conditions. We start with $R_{min} = \{(1,1), (2,2), (3,3), (1,2)\}$, and $|R_{min}|=4$. We can add 0, 1, or 2 more elements.

• Case 1: Add 0 elements. Let $R_1 = \{(1,1), (2,2), (3,3), (1,2)\}$.

  • Reflexive: Yes, it contains $(1,1), (2,2), (3,3)$.
  • Contains $(1,2)$: Yes.
  • Not Symmetric: Yes, because $(1,2) \in R_1$ and $(2,1) \notin R_1$.
  • Transitive: To check transitivity, we look for pairs $(a,b) \in R_1$ and $(b,c) \in R_1$. The only non-reflexive pair is $(1,2)$. There are no other pairs $(b,c)$ that start with $2$. Thus, there are no implications for transitivity that require adding new elements. So, $R_1$ is transitive.
  • Size: $|R_1|=4 \le 6$. Yes. This is a valid relation.

• Case 2: Add 1 element. If we add only one element, say $(1,3)$, then $R = \{(1,1), (2,2), (3,3), (1,2), (1,3)\}$. According to Step 3, if $(1,3) \in R$ and $1 \ne 3$, then $(3,1)$ must also be in $R$ for the relation to be symmetric for pairs other than $(1,2)$. So, adding a single non-reflexive element is not allowed under the interpretation derived in Step 3. Thus, we cannot add exactly one element.

• Case 3: Add 2 elements. The two elements must be a symmetric pair of non-reflexive elements, as per Step 3. The possible pairs of distinct elements from $A$ are $(1,2), (1,3), (2,3)$. Since $(1,2)$ is already included and $(2,1)$ must be excluded, we can only consider adding $\{(1,3), (3,1)\}$ or $\{(2,3), (3,2)\}$.

  • Subcase 3a: Add $\{(1,3), (3,1)\}$. Let $R_2 = \{(1,1), (2,2), (3,3), (1,2), (1,3), (3,1)\}$.

    • Reflexive: Yes.
    • Contains $(1,2)$: Yes.
    • Not Symmetric: Yes, $(1,2) \in R_2$ and $(2,1) \notin R_2$.
    • Size: $|R_2|=6 \le 6$. Yes.
    • Transitive: We check for implications. We have $(3,1) \in R_2$ and $(1,2) \in R_2$. For transitivity, $(3,2)$ must be in $R_2$. However, $(3,2) \notin R_2$. Therefore, $R_2$ is not transitive. This relation is invalid.

  • Subcase 3b: Add $\{(2,3), (3,2)\}$. Let $R_3 = \{(1,1), (2,2), (3,3), (1,2), (2,3), (3,2)\}$.

    • Reflexive: Yes.
    • Contains $(1,2)$: Yes.
    • Not Symmetric: Yes, $(1,2) \in R_3$ and $(2,1) \notin R_3$.
    • Size: $|R_3|=6 \le 6$. Yes.
    • Transitive: We check for implications. We have $(1,2) \in R_3$ and $(2,3) \in R_3$. For transitivity, $(1,3)$ must be in $R_3$. However, $(1,3) \notin R_3$. Therefore, $R_3$ is not transitive. This relation is invalid.

Step 5: Conclude the number of valid relations. From the analysis in Step 4, only one relation satisfies all the given conditions: $R_1 = \{(1,1), (2,2), (3,3), (1,2)\}$.

Common Mistakes & Tips

• Misinterpreting "Not Symmetric": The standard definition of "not symmetric" is existential. However, in questions designed to have a unique small answer, it often implies a specific asymmetry (like only one pair breaking symmetry). Always check if the problem context implies a stricter interpretation.
• Forgetting Transitivity: After ensuring reflexivity, the presence of $(1,2)$, and the "not symmetric" condition, it's crucial to rigorously check transitivity. Transitivity can impose requirements for additional elements.
• Counting Elements: Keep track of the number of elements at each step to ensure the size constraint $|R| \le 6$ is met.

Summary We are looking for relations on $A=\{1,2,3\}$ that are reflexive, contain $(1,2)$, are transitive, not symmetric, and have at most 6 elements. Reflexivity requires $\{(1,1), (2,2), (3,3)\}$. The condition of containing $(1,2)$ means $R$ must include at least $\{(1,1), (2,2), (3,3), (1,2)\}$. For the relation to be "not symmetric" given $(1,2) \in R$, it must be that $(2,1) \notin R$. This implies that for any other non-reflexive pair $(x,y)$ in $R$, $(y,x)$ must also be in $R$. Considering relations with 4, 5, or 6 elements, we find that only the relation $R = \{(1,1), (2,2), (3,3), (1,2)\}$ satisfies all conditions. Adding any other elements, while maintaining the implied symmetry for pairs other than $(1,2)$ and $(2,1)$, violates transitivity or exceeds the size limit.

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