Key Concepts and Formulas

1. Symmetric Relation: A relation $R$ on a set $A$ is symmetric if for every pair $(x, y) \in R$, the pair $(y, x)$ is also in $R$. Mathematically, $\forall x, y \in A, (x, y) \in R \implies (y, x) \in R$.

2. Transitive Relation: A relation $R$ on a set $A$ is transitive if for every pair of pairs $(x, y) \in R$ and $(y, z) \in R$, the pair $(x, z)$ is also in $R$. Mathematically, $\forall x, y, z \in A, ((x, y) \in R \land (y, z) \in R) \implies (x, z) \in R$.

Step-by-Step Solution

The set is $A = \{a, b, c\}$. We need to analyze the given relations $R_1$ and $R_2$ for symmetry and transitivity.

Analysis of Relation $R_1$

$R_1 = \{(c, a), (b, b), (a, c), (c, c), (b, c), (a, a)\}$

Step 1: Check for Symmetry of $R_1$ To determine if $R_1$ is symmetric, we examine each ordered pair $(x, y) \in R_1$ and check if its reverse $(y, x)$ is also present in $R_1$.

• $(c, a) \in R_1$. We check for $(a, c)$. Indeed, $(a, c) \in R_1$.
• $(b, b) \in R_1$. We check for $(b, b)$. It is present. (Self-loops are always symmetric with themselves).
• $(a, c) \in R_1$. We check for $(c, a)$. Indeed, $(c, a) \in R_1$.
• $(c, c) \in R_1$. We check for $(c, c)$. It is present.
• $(b, c) \in R_1$. We check for $(c, b)$. Upon inspecting $R_1$, we find that $(c, b) \notin R_1$. Since we found a pair $(b, c) \in R_1$ for which $(c, b) \notin R_1$, the relation $R_1$ is not symmetric.

Step 2: Check for Transitivity of $R_1$ To determine if $R_1$ is transitive, we look for pairs $(x, y) \in R_1$ and $(y, z) \in R_1$ and check if $(x, z) \in R_1$.

• Consider $(b, c) \in R_1$ and $(c, a) \in R_1$. For transitivity, we must have $(b, a) \in R_1$. However, $(b, a) \notin R_1$. Since we found a case where $(x, y) \in R_1$ and $(y, z) \in R_1$ but $(x, z) \notin R_1$, the relation $R_1$ is not transitive.

Conclusion for $R_1$: $R_1$ is neither symmetric nor transitive.

Analysis of Relation $R_2$

$R_2 = \{(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c)\}$

Step 3: Check for Symmetry of $R_2$ We examine each ordered pair $(x, y) \in R_2$ and check if its reverse $(y, x)$ is also present in $R_2$.

• $(a, b) \in R_2$. We check for $(b, a)$. Indeed, $(b, a) \in R_2$.
• $(b, a) \in R_2$. We check for $(a, b)$. Indeed, $(a, b) \in R_2$.
• $(c, c) \in R_2$. We check for $(c, c)$. It is present.
• $(c, a) \in R_2$. We check for $(a, c)$. Indeed, $(a, c) \in R_2$.
• $(a, a) \in R_2$. We check for $(a, a)$. It is present.
• $(b, b) \in R_2$. We check for $(b, b)$. It is present.
• $(a, c) \in R_2$. We check for $(c, a)$. Indeed, $(c, a) \in R_2$. For every pair $(x, y) \in R_2$, its reverse $(y, x)$ is also in $R_2$. Therefore, $R_2$ is symmetric.

Step 4: Check for Transitivity of $R_2$ We look for pairs $(x, y) \in R_2$ and $(y, z) \in R_2$ and check if $(x, z) \in R_2$.

• Consider $(c, a) \in R_2$ and $(a, b) \in R_2$. For transitivity, we must have $(c, b) \in R_2$. However, $(c, b) \notin R_2$. Since we found a case where $(x, y) \in R_2$ and $(y, z) \in R_2$ but $(x, z) \notin R_2$, the relation $R_2$ is not transitive.

Conclusion for $R_2$: $R_2$ is symmetric but not transitive.

Summary of Findings

• $R_1$ is not symmetric and not transitive.
• $R_2$ is symmetric and not transitive.

Now, let's evaluate the given options:

• (A) both $R_1$ and $R_2$ are not symmetric. This is incorrect because $R_2$ is symmetric.
• (B) $R_1$ is not symmetric but it is transitive. This is incorrect because $R_1$ is not transitive.
• (C) $R_2$ is symmetric but it is not transitive. This statement aligns with our findings.
• (D) both $R_1$ and $R_2$ are transitive. This is incorrect because neither $R_1$ nor $R_2$ is transitive.

Common Mistakes & Tips

• Carefully check all pairs: For symmetry, ensure you check the reverse of every pair in the relation. For transitivity, systematically identify all $(x, y)$ and $(y, z)$ combinations and verify the existence of $(x, z)$.
• Counterexamples are sufficient: To prove a relation is not symmetric or transitive, you only need to find one instance that violates the definition.
• Self-loops: Pairs like $(a, a)$ do not violate symmetry or transitivity in themselves. They satisfy these properties trivially.

Summary We systematically analyzed both relations $R_1$ and $R_2$ by checking the definitions of symmetry and transitivity. We found that $R_1$ fails both conditions, while $R_2$ satisfies symmetry but fails transitivity due to a specific counterexample. Comparing these findings with the given options, we conclude that option (C) accurately describes the properties of $R_2$.

The final answer is \boxed{C}