Key Concepts and Formulas

1. Reflexivity: A relation $R$ on a set $A$ is reflexive if $(a,a) \in R$ for all $a \in A$.
2. Symmetry: A relation $R$ on a set $A$ is symmetric if $(a,b) \in R$ implies $(b,a) \in R$ for all $a,b \in A$.
3. Transitivity: A relation $R$ on a set $A$ is transitive if $(a,b) \in R$ and $(b,c) \in R$ implies $(a,c) \in R$ for all $a,b,c \in A$.
4. Equivalence Relation: A relation that is reflexive, symmetric, and transitive.

Step-by-Step Solution

We are given the set $A = \{3,6,9,12\}$ and the relation $R = \{(3,3),(6,6),(9,9),(12,12),(6,12),(3,9),(3,12),(3,6)\}$. We need to determine if $R$ is reflexive, symmetric, and/or transitive.

Step 1: Check for Reflexivity To check for reflexivity, we need to verify if every element in set $A$ is related to itself. This means checking if the pairs $(a,a)$ for all $a \in A$ are present in $R$. The elements of $A$ are $3, 6, 9, 12$. We must check for the presence of $(3,3), (6,6), (9,9), (12,12)$ in $R$.

• $(3,3) \in R$ (given)
• $(6,6) \in R$ (given)
• $(9,9) \in R$ (given)
• $(12,12) \in R$ (given) Since all pairs $(a,a)$ for $a \in A$ are in $R$, the relation $R$ is reflexive.

Step 2: Check for Symmetry To check for symmetry, we need to verify if for every pair $(a,b) \in R$, the corresponding pair $(b,a)$ is also in $R$. Let's examine the non-diagonal elements of $R$:

• Consider $(6,12) \in R$. For symmetry, $(12,6)$ must also be in $R$.
• Checking the given relation $R$, we see that $(12,6) \notin R$. Since we found a pair $(6,12) \in R$ but its reverse $(12,6) \notin R$, the relation $R$ is not symmetric.

Step 3: Check for Transitivity To check for transitivity, we need to verify if for every $(a,b) \in R$ and $(b,c) \in R$, the pair $(a,c)$ is also in $R$. We will systematically check all possible combinations.

Let's consider pairs where the first element is 3:

• $(3,3) \in R$. If $b=3$, we look for $(3,c) \in R$. These are $(3,3), (3,6), (3,9), (3,12)$.
  • For $(3,3)$ and $(3,3)$: Check $(3,3) \in R$. Yes.
  • For $(3,3)$ and $(3,6)$: Check $(3,6) \in R$. Yes.
  • For $(3,3)$ and $(3,9)$: Check $(3,9) \in R$. Yes.
  • For $(3,3)$ and $(3,12)$: Check $(3,12) \in R$. Yes.
• $(3,6) \in R$. If $b=6$, we look for $(6,c) \in R$. These are $(6,6), (6,12)$.
  • For $(3,6)$ and $(6,6)$: Check $(3,6) \in R$. Yes.
  • For $(3,6)$ and $(6,12)$: Check $(3,12) \in R$. Yes.
• $(3,9) \in R$. If $b=9$, we look for $(9,c) \in R$. This is $(9,9)$.
  • For $(3,9)$ and $(9,9)$: Check $(3,9) \in R$. Yes.
• $(3,12) \in R$. If $b=12$, we look for $(12,c) \in R$. This is $(12,12)$.
  • For $(3,12)$ and $(12,12)$: Check $(3,12) \in R$. Yes.

Now let's consider pairs where the first element is 6:

• $(6,6) \in R$. If $b=6$, we look for $(6,c) \in R$. These are $(6,6), (6,12)$.
  • For $(6,6)$ and $(6,6)$: Check $(6,6) \in R$. Yes.
  • For $(6,6)$ and $(6,12)$: Check $(6,12) \in R$. Yes.
• $(6,12) \in R$. If $b=12$, we look for $(12,c) \in R$. This is $(12,12)$.
  • For $(6,12)$ and $(12,12)$: Check $(6,12) \in R$. Yes.

Now let's consider pairs where the first element is 9:

• $(9,9) \in R$. If $b=9$, we look for $(9,c) \in R$. This is $(9,9)$.
  • For $(9,9)$ and $(9,9)$: Check $(9,9) \in R$. Yes.

Now let's consider pairs where the first element is 12:

• $(12,12) \in R$. If $b=12$, we look for $(12,c) \in R$. This is $(12,12)$.
  • For $(12,12)$ and $(12,12)$: Check $(12,12) \in R$. Yes.

We have checked all non-trivial combinations and found that for every instance of $(a,b) \in R$ and $(b,c) \in R$, the pair $(a,c)$ is also in $R$. Thus, the relation $R$ is transitive.

Common Mistakes & Tips

• Reflexivity: Ensure all elements of the set $A$ have their corresponding $(a,a)$ pair in $R$. Missing even one makes it not reflexive.
• Symmetry: To prove a relation is not symmetric, finding just one pair $(a,b) \in R$ such that $(b,a) \notin R$ is sufficient.
• Transitivity: This is the most tedious check. Be systematic. If you find a combination $(a,b) \in R$ and $(b,c) \in R$, but $(a,c) \notin R$, the relation is not transitive. If you check all combinations and find no such counterexample, it is transitive.

Summary

We have analyzed the given relation $R$ on the set $A$.

1. The relation $R$ is reflexive because all elements of $A$ are related to themselves.
2. The relation $R$ is not symmetric because the pair $(6,12) \in R$ but $(12,6) \notin R$.
3. The relation $R$ is transitive as all conditions for transitivity are met.

Therefore, the relation is reflexive and transitive only.

The final answer is $\boxed{D}$ which corresponds to option (D) reflexive and transitive only.