Key Concepts and Formulas

• Equivalence Relation: A relation $R$ on a set $A$ is an equivalence relation if it is reflexive, symmetric, and transitive.
  • Reflexive: For all $a \in A$, $a \,R\, a$.
  • Symmetric: For all $a, b \in A$, if $a \,R\, b$, then $b \,R\, a$.
  • Transitive: For all $a, b, c \in A$, if $a \,R\, b$ and $b \,R\, c$, then $a \,R\, c$.

Step-by-Step Solution

Step 1: Analyze Relation $R_1$ ($a \,R_{1} \,b \Leftrightarrow a b \geq 0$)

We need to check if $R_1$ is reflexive, symmetric, and transitive on $\mathbb{R}$.

• Reflexivity: For any $a \in \mathbb{R}$, we check if $a \,R_{1} \,a$. This means $a \cdot a \geq 0$, which is $a^2 \geq 0$. Since the square of any real number is always non-negative, $R_1$ is reflexive.

• Symmetry: For any $a, b \in \mathbb{R}$, if $a \,R_{1} \,b$, we check if $b \,R_{1} \,a$. If $a \,R_{1} \,b$, then $a b \geq 0$. Since multiplication of real numbers is commutative ($a b = b a$), we have $b a \geq 0$. Thus, $b \,R_{1} \,a$ is true. $R_1$ is symmetric.

• Transitivity: For any $a, b, c \in \mathbb{R}$, if $a \,R_{1} \,b$ and $b \,R_{1} \,c$, we check if $a \,R_{1} \,c$. This means $a b \geq 0$ and $b c \geq 0$. Consider the case when $b=0$. If $b=0$, then $a \cdot 0 \geq 0$ (which is $0 \geq 0$, true for any $a \in \mathbb{R}$) and $0 \cdot c \geq 0$ (which is $0 \geq 0$, true for any $c \in \mathbb{R}$). However, if we choose $a=1$ and $c=-1$, then $a \,R_{1} \,b$ and $b \,R_{1} \,c$ are satisfied with $b=0$, but $a \,R_{1} \,c$ means $1 \cdot (-1) \geq 0$, which is $-1 \geq 0$, and this is false. Therefore, $R_1$ is not transitive.

Since $R_1$ is not transitive, it is not an equivalence relation.

Step 2: Analyze Relation $R_2$ ($a \,R_{2} \,b \Leftrightarrow a \geq b$)

We need to check if $R_2$ is reflexive, symmetric, and transitive on $\mathbb{R}$.

• Reflexivity: For any $a \in \mathbb{R}$, we check if $a \,R_{2} \,a$. This means $a \geq a$. This is true for all real numbers. $R_2$ is reflexive.

• Symmetry: For any $a, b \in \mathbb{R}$, if $a \,R_{2} \,b$, we check if $b \,R_{2} \,a$. If $a \,R_{2} \,b$, then $a \geq b$. This does not imply $b \geq a$. For example, if $a=2$ and $b=1$, then $2 \geq 1$ is true, but $1 \geq 2$ is false. Therefore, $R_2$ is not symmetric.

• Transitivity: For any $a, b, c \in \mathbb{R}$, if $a \,R_{2} \,b$ and $b \,R_{2} \,c$, we check if $a \,R_{2} \,c$. If $a \geq b$ and $b \geq c$, then by the transitive property of inequalities, $a \geq c$. Therefore, $R_2$ is transitive.

Since $R_2$ is not symmetric, it is not an equivalence relation.

Step 3: Conclusion

From Step 1, $R_1$ is not an equivalence relation because it is not transitive. From Step 2, $R_2$ is not an equivalence relation because it is not symmetric.

Therefore, neither $R_1$ nor $R_2$ is an equivalence relation.

Common Mistakes & Tips

• Thorough Checking: Always verify all three properties (reflexivity, symmetry, transitivity) for a relation to be an equivalence relation. Failing even one property is sufficient to disqualify it.
• Counterexamples: To prove a property is false, a single, clear counterexample is sufficient. For $R_1$'s transitivity, the case with $b=0$ is a critical counterexample.
• Understanding the Domain: The set on which the relation is defined ($\mathbb{R}$ in this case) is crucial. Properties might hold on one set but not another.

Summary

We analyzed both relations $R_1$ and $R_2$ against the three conditions for an equivalence relation: reflexivity, symmetry, and transitivity. Relation $R_1$ was found to be reflexive and symmetric but not transitive. Relation $R_2$ was found to be reflexive and transitive but not symmetric. Since neither relation satisfies all three conditions, neither is an equivalence relation.

The final answer is \boxed{D}.