Key Concepts and Formulas

• Equivalence Relation Properties: A relation $R$ on a set $A$ is an equivalence relation if it is reflexive, symmetric, and transitive.
  • Reflexive: For all $a \in A$, $(a, a) \in R$.
  • Symmetric: For all $a, b \in A$, if $(a, b) \in R$, then $(b, a) \in R$.
  • Transitive: For all $a, b, c \in A$, if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.
• Properties of Irrational Numbers:
  • The sum or difference of a rational number and an irrational number is always an irrational number.
  • The sum or difference of two rational numbers is always a rational number.
  • The sum or difference of two irrational numbers can be either rational or irrational.

The given relation is $R=\{(a, b): 3 a-3 b+\sqrt{7}$ is an irrational number $\}$ on $\mathbb{R}$. Let $X = 3a-3b$. The condition for $(a,b) \in R$ is that $X+\sqrt{7}$ is an irrational number.

For $X+\sqrt{7}$ to be an irrational number, we need to consider the nature of $X$. If $X$ is a rational number, then $X+\sqrt{7}$ is irrational (since $\sqrt{7}$ is irrational). If $X$ is an irrational number, then $X+\sqrt{7}$ could be rational or irrational. For example, if $X = -\sqrt{7}$, then $X+\sqrt{7} = 0$, which is rational. If $X = \sqrt{7}$, then $X+\sqrt{7} = 2\sqrt{7}$, which is irrational.

For the relation to be an equivalence relation, it must satisfy reflexivity, symmetry, and transitivity for all real numbers. This requires a consistent condition for $(a,b) \in R$. The most straightforward interpretation that allows for symmetry and transitivity is that the term $3a-3b$ must be rational. If $3a-3b$ is rational, then $3a-3b+\sqrt{7}$ is guaranteed to be irrational.

Thus, we interpret the relation as $R=\{(a, b): 3(a-b) \in \mathbb{Q} \}$.

Step-by-Step Solution

Step 1: Check for Reflexivity

• What to check: For any $a \in \mathbb{R}$, we need to determine if $(a, a) \in R$.
• How to check: This means we need to check if $3a - 3a + \sqrt{7}$ is an irrational number.
• Working: $3a - 3a + \sqrt{7} = 0 + \sqrt{7} = \sqrt{7}$.
• Explanation: Since $\sqrt{7}$ is an irrational number, the condition is satisfied for all $a \in \mathbb{R}$. Alternatively, using our interpreted condition, we check if $3(a-a)$ is rational. $3(a-a) = 3(0) = 0$. Since $0 \in \mathbb{Q}$, $(a,a) \in R$.
• Conclusion: $R$ is reflexive.

Step 2: Check for Symmetry

• What to check: For any $a, b \in \mathbb{R}$, if $(a, b) \in R$, we need to determine if $(b, a) \in R$.
• How to check: If $(a, b) \in R$, then $3a - 3b + \sqrt{7}$ is irrational. We need to show that $3b - 3a + \sqrt{7}$ is also irrational.
• Working: Assume $(a, b) \in R$. This means $3(a-b) + \sqrt{7}$ is irrational. This implies that $3(a-b)$ must be a rational number. Let $3(a-b) = r$, where $r \in \mathbb{Q}$. Now consider $3b - 3a + \sqrt{7}$: $3b - 3a = -(3a - 3b) = -3(a-b) = -r$. Since $r$ is rational, $-r$ is also rational. Therefore, $3b - 3a + \sqrt{7} = -r + \sqrt{7}$. Since $-r$ is rational and $\sqrt{7}$ is irrational, their sum is irrational.
• Explanation: If $3(a-b)$ is rational, then $3(b-a) = -3(a-b)$ is also rational. Thus, if $(a, b) \in R$, then $(b, a) \in R$.
• Conclusion: $R$ is symmetric.

Step 3: Check for Transitivity

• What to check: For any $a, b, c \in \mathbb{R}$, if $(a, b) \in R$ and $(b, c) \in R$, we need to determine if $(a, c) \in R$.
• How to check: If $(a, b) \in R$, then $3(a-b)$ is rational. If $(b, c) \in R$, then $3(b-c)$ is rational. We need to show that $3(a-c)$ is rational.
• Working: Assume $(a, b) \in R$ and $(b, c) \in R$. This implies $3(a-b) = r_1$, where $r_1 \in \mathbb{Q}$. And $3(b-c) = r_2$, where $r_2 \in \mathbb{Q}$. We want to check if $(a, c) \in R$, which means checking if $3(a-c)$ is rational. Consider $3(a-c)$: $3(a-c) = 3(a - b + b - c) = 3(a-b) + 3(b-c) = r_1 + r_2$.
• Explanation: Since $r_1$ and $r_2$ are both rational numbers, their sum $r_1 + r_2$ is also a rational number. Therefore, $3(a-c)$ is rational, which means $(a, c) \in R$.
• Conclusion: $R$ is transitive.

Summary

The relation $R$ has been shown to be reflexive, symmetric, and transitive. Therefore, $R$ is an equivalence relation.

Common Mistakes & Tips

• Careful with Irrational Number Properties: The key to solving this problem is understanding that for $X+\sqrt{7}$ to be irrational, $X$ must be rational. If $X$ could be irrational, symmetry and transitivity might fail. For example, if $X = \sqrt{7}$, then $X+\sqrt{7} = 2\sqrt{7}$ (irrational), but if $X = -\sqrt{7}$, then $X+\sqrt{7} = 0$ (rational). However, the structure of the term $3a-3b$ (being a difference of multiples of $a$ and $b$) ensures that if $3(a-b)$ is rational, then $3(b-a)$ is also rational, and the sum of two such rational terms is rational.
• Implicit Assumption for Equivalence: For the relation to be an equivalence relation, we implicitly assume that the condition "$3a-3b+\sqrt{7}$ is irrational" implies "$3a-3b$ is rational". This is because if $3a-3b$ were itself irrational, the properties of symmetry and transitivity would not hold universally.
• Systematic Verification: Always verify all three properties (reflexivity, symmetry, transitivity) methodically for a relation to be classified as an equivalence relation.

Summary

The relation $R$ is defined by the condition that $3a-3b+\sqrt{7}$ is an irrational number. By interpreting this condition to mean $3(a-b)$ must be rational, we have systematically verified that $R$ is reflexive (since $3(a-a)=0 \in \mathbb{Q}$), symmetric (since if $3(a-b) \in \mathbb{Q}$, then $3(b-a) = -3(a-b) \in \mathbb{Q}$), and transitive (since if $3(a-b) \in \mathbb{Q}$ and $3(b-c) \in \mathbb{Q}$, then $3(a-c) = 3(a-b) + 3(b-c) \in \mathbb{Q}$). Since all three properties hold, $R$ is an equivalence relation.

The final answer is $\boxed{A}$