Key Concepts and Formulas

• Relation Properties:
  • Reflexive: For all $x \in A$, $x R x$.
  • Symmetric: For all $x, y \in A$, if $x R y$, then $y R x$.
  • Transitive: For all $x, y, z \in A$, if $x R y$ and $y R z$, then $x R z$.
• Modular Arithmetic: $a \equiv b \pmod{m}$ means $a-b$ is divisible by $m$.
• Divisibility: An integer $k$ is divisible by an integer $n$ if $k = mn$ for some integer $m$.

Step-by-Step Solution

We are given a relation $R$ on the set $Z \times Z$ defined by $(a, b) R (c, d)$ if and only if $ad - bc$ is divisible by 5. This can be expressed using modular arithmetic as $ad - bc \equiv 0 \pmod{5}$. We need to determine if $R$ is reflexive, symmetric, and transitive.

Step 1: Checking for Reflexivity

• Objective: To determine if for every element $(a, b) \in Z \times Z$, the relation $(a, b) R (a, b)$ holds.
• Procedure: We substitute $(a, b)$ for both $(a, b)$ and $(c, d)$ in the definition of the relation and check if the condition is met.
• Working: For $(a, b) R (a, b)$ to hold, we must have $a \cdot b - b \cdot a$ divisible by 5. $a \cdot b - b \cdot a = ab - ba = 0$ Since 0 is divisible by any non-zero integer, 0 is divisible by 5 ($0 = 5 \cdot 0$). Thus, $ab - ba \equiv 0 \pmod{5}$ is always true.
• Conclusion: The relation $R$ is reflexive.

Step 2: Checking for Symmetry

• Objective: To determine if for any two elements $(a, b), (c, d) \in Z \times Z$, if $(a, b) R (c, d)$, then $(c, d) R (a, b)$ must also hold.
• Procedure: We assume the premise $(a, b) R (c, d)$ is true and check if the conclusion $(c, d) R (a, b)$ follows from it.
• Working: Assume $(a, b) R (c, d)$. This means $ad - bc \equiv 0 \pmod{5}$. We need to check if $(c, d) R (a, b)$ holds, which means $c \cdot b - d \cdot a$ is divisible by 5. $c \cdot b - d \cdot a = cb - da$ We can rewrite this expression as: $cb - da = -(da - cb) = -(ad - bc)$ Since we assumed $ad - bc \equiv 0 \pmod{5}$, it implies $ad - bc = 5k$ for some integer $k$. Then, $cb - da = -(5k) = 5(-k)$. Since $-k$ is also an integer, $cb - da$ is divisible by 5. Thus, $cb - da \equiv 0 \pmod{5}$ holds.
• Conclusion: The relation $R$ is symmetric.

Step 3: Checking for Transitivity

• Objective: To determine if for any three elements $(a, b), (c, d), (e, f) \in Z \times Z$, if $(a, b) R (c, d)$ and $(c, d) R (e, f)$, then $(a, b) R (e, f)$ must also hold.

• Procedure: We assume the premises $(a, b) R (c, d)$ and $(c, d) R (e, f)$ are true and try to deduce if $(a, b) R (e, f)$ follows. If we can find a counterexample where the premises are true but the conclusion is false, then the relation is not transitive.

• Working: Assume $(a, b) R (c, d)$ and $(c, d) R (e, f)$. This implies:

  1. $ad - bc \equiv 0 \pmod{5}$
  2. $cf - de \equiv 0 \pmod{5}$ We need to check if $af - be \equiv 0 \pmod{5}$.

  Let's try to find a counterexample. Consider the following pairs:

  • Let $(a, b) = (1, 0)$.
  • Let $(c, d) = (0, 5)$.
  • Let $(e, f) = (1, 1)$.

  Check if $(a, b) R (c, d)$: $ad - bc = (1)(5) - (0)(0) = 5 - 0 = 5$. Since $5$ is divisible by 5, $(1, 0) R (0, 5)$ holds.

  Check if $(c, d) R (e, f)$: $cf - de = (0)(1) - (5)(1) = 0 - 5 = -5$. Since $-5$ is divisible by 5, $(0, 5) R (1, 1)$ holds.

  Now, check if $(a, b) R (e, f)$: $af - be = (1)(1) - (0)(1) = 1 - 0 = 1$. Since $1$ is not divisible by 5, $(1, 0) \not R (1, 1)$.

  We have found a case where $(a, b) R (c, d)$ and $(c, d) R (e, f)$ are true, but $(a, b) R (e, f)$ is false.

• Conclusion: The relation $R$ is not transitive.

Common Mistakes & Tips

• Algebraic Manipulation in Transitivity: When checking transitivity, direct algebraic manipulation of the modular equations can be tricky. Sometimes it's easier to work with the definition of divisibility ($ad-bc=5k$) or look for a counterexample if the direct proof isn't obvious.
• Counterexample Strategy: For transitivity, if you suspect the relation is not transitive, try to construct a simple counterexample using small integers. This is often faster than a lengthy algebraic proof.
• Modular Arithmetic Properties: Remember that if $a \equiv 0 \pmod{m}$, then $-a \equiv 0 \pmod{m}$, and $ka \equiv 0 \pmod{m}$ for any integer $k$. This is crucial for symmetry.

Summary

We analyzed the given relation $R$ on $Z \times Z$ by checking its reflexivity, symmetry, and transitivity. We found that for any element $(a, b)$, $(a, b) R (a, b)$ holds because $ab - ba = 0$, which is divisible by 5, proving reflexivity. We also showed that if $(a, b) R (c, d)$, then $ad - bc \equiv 0 \pmod{5}$, which implies $cb - da = -(ad - bc) \equiv 0 \pmod{5}$, proving symmetry. However, by constructing a counterexample with $(1, 0) R (0, 5)$ and $(0, 5) R (1, 1)$, but $(1, 0) \not R (1, 1)$, we demonstrated that the relation is not transitive. Therefore, $R$ is reflexive and symmetric but not transitive.

The final answer is \boxed{A}.