Key Concepts and Formulas

• A relation $R$ on a set $S$ is:
  • Reflexive if $x R x$ for all $x \in S$.
  • Symmetric if $x R y$ implies $y R x$ for all $x, y \in S$.
  • Transitive if $x R y$ and $y R z$ implies $x R z$ for all $x, y, z \in S$.
• The parity of an integer can be checked using modular arithmetic modulo 2. An integer $n$ is even if $n \equiv 0 \pmod{2}$ and odd if $n \equiv 1 \pmod{2}$.
• The product of two integers $x \cdot y$ is odd if and only if both $x$ and $y$ are odd. Otherwise, the product is even.

Step-by-Step Solution

Step 1: Understand the Set and the Relation's Condition The relation $R$ is defined on the set $A \times B$, where $A=\{2,3,6,8,9,11\}$ and $B=\{1,4,5,10,15\}$. An element of $A \times B$ is an ordered pair $(a, b)$ where $a \in A$ and $b \in B$. The relation $R$ is defined by $(a, b) R (c, d)$ if and only if $3ad - 7bc$ is an even integer.

We can simplify the condition $3ad - 7bc$ being even using modulo 2 arithmetic. $3 \equiv 1 \pmod{2}$ and $7 \equiv 1 \pmod{2}$. So, $3ad - 7bc \equiv 1 \cdot ad - 1 \cdot bc \pmod{2}$. The condition for $R$ becomes $ad - bc \equiv 0 \pmod{2}$, which is equivalent to $ad \equiv bc \pmod{2}$. This means that the products $ad$ and $bc$ must have the same parity (both even or both odd) for $(a, b) R (c, d)$ to be true.

Let's identify the parity of elements in $A$ and $B$: $A_{even} = \{2, 6, 8\}$ $A_{odd} = \{3, 9, 11\}$ $B_{even} = \{4, 10\}$ $B_{odd} = \{1, 5, 15\}$

Step 2: Check for Reflexivity For $R$ to be reflexive, every element $(a, b) \in A \times B$ must satisfy $(a, b) R (a, b)$. According to the relation's condition, this means $a \cdot b \equiv b \cdot a \pmod{2}$. Since multiplication of integers is commutative, $ab = ba$. Thus, $ab \equiv ab \pmod{2}$ is always true for any integers $a$ and $b$. Alternatively, using the original definition: $3a b - 7b a = (3-7)ab = -4ab$. Since $-4$ is an even integer, $-4ab$ is always an even integer for any $a \in A$ and $b \in B$. Therefore, the relation $R$ is reflexive.

Step 3: Check for Symmetry For $R$ to be symmetric, if $(a, b) R (c, d)$, then $(c, d) R (a, b)$ must hold for all $(a, b), (c, d) \in A \times B$. The condition for $(a, b) R (c, d)$ is $ad \equiv bc \pmod{2}$. The condition for $(c, d) R (a, b)$ is $ca \equiv db \pmod{2}$.

Let's try to find a counterexample. We need to find $(a, b)$ and $(c, d)$ such that $ad \equiv bc \pmod{2}$ is true, but $ca \not\equiv db \pmod{2}$.

Consider $(a, b) = (2, 1)$ and $(c, d) = (6, 1)$. Here, $a=2 \in A_{even}$, $b=1 \in B_{odd}$, $c=6 \in A_{even}$, $d=1 \in B_{odd}$.

Check if $(2, 1) R (6, 1)$: $ad = 2 \cdot 1 = 2$ (even) $bc = 1 \cdot 6 = 6$ (even) Since $ad$ and $bc$ are both even, $ad \equiv bc \pmod{2}$. Thus, $(2, 1) R (6, 1)$ holds.

Check if $(6, 1) R (2, 1)$: $ca = 6 \cdot 2 = 12$ (even) $db = 1 \cdot 1 = 1$ (odd) Since $ca$ is even and $db$ is odd, $ca \not\equiv db \pmod{2}$. Thus, $(6, 1) \not R (2, 1)$.

Since we found a case where $(a, b) R (c, d)$ but $(c, d) \not R (a, b)$, the relation $R$ is not symmetric.

Step 4: Check for Transitivity For $R$ to be transitive, if $(a, b) R (c, d)$ and $(c, d) R (e, f)$, then $(a, b) R (e, f)$ must hold for all $(a, b), (c, d), (e, f) \in A \times B$. This means if $ad \equiv bc \pmod{2}$ and $cf \equiv de \pmod{2}$, then we must have $af \equiv be \pmod{2}$.

Let's try to find a counterexample. We need $(a, b) R (c, d)$ and $(c, d) R (e, f)$ to be true, but $(a, b) R (e, f)$ to be false.

Consider $(a, b) = (2, 1)$, $(c, d) = (6, 4)$, and $(e, f) = (3, 1)$. Here, $a=2 \in A_{even}$, $b=1 \in B_{odd}$, $c=6 \in A_{even}$, $d=4 \in B_{even}$, $e=3 \in A_{odd}$, $f=1 \in B_{odd}$.

Check if $(2, 1) R (6, 4)$: $ad = 2 \cdot 4 = 8$ (even) $bc = 1 \cdot 6 = 6$ (even) Since $ad$ and $bc$ are both even, $ad \equiv bc \pmod{2}$. Thus, $(2, 1) R (6, 4)$ holds.

Check if $(6, 4) R (3, 1)$: $cf = 6 \cdot 1 = 6$ (even) $de = 4 \cdot 3 = 12$ (even) Since $cf$ and $de$ are both even, $cf \equiv de \pmod{2}$. Thus, $(6, 4) R (3, 1)$ holds.

Check if $(2, 1) R (3, 1)$: $af = 2 \cdot 1 = 2$ (even) $be = 1 \cdot 3 = 3$ (odd) Since $af$ is even and $be$ is odd, $af \not\equiv be \pmod{2}$. Thus, $(2, 1) \not R (3, 1)$.

Since we found a case where $(a, b) R (c, d)$ and $(c, d) R (e, f)$ but $(a, b) \not R (e, f)$, the relation $R$ is not transitive.

Common Mistakes & Tips

• Misinterpreting Parity Conditions: Always simplify the condition $3ad - 7bc$ being even to $ad \equiv bc \pmod{2}$. This significantly reduces the complexity of checking the relation's properties.
• Errors in Finding Counterexamples: When checking for symmetry or transitivity, be systematic in choosing elements. Mix even and odd numbers from sets $A$ and $B$ to increase the chances of finding a counterexample.
• Confusing Relations on Sets vs. Relations on Cartesian Products: The relation is on $A \times B$, meaning its elements are ordered pairs. Ensure that when checking properties, you are using the definitions for relations on a set of ordered pairs.

Summary We analyzed the given relation $R$ defined on $A \times B$ by simplifying its condition to $ad \equiv bc \pmod{2}$. We found that the relation is reflexive because $ab \equiv ba \pmod{2}$ is always true. However, we demonstrated that $R$ is not symmetric by providing a counterexample where $(2, 1) R (6, 1)$ but $(6, 1) \not R (2, 1)$. Similarly, we showed that $R$ is not transitive with the counterexample involving $(2, 1)$, $(6, 4)$, and $(3, 1)$. Therefore, the relation $R$ is reflexive but not symmetric.

The final answer is \boxed{A}.