We check whether each relation is an equivalence relation.

An equivalence relation must be:

1. Reflexive
2. Symmetric
3. Transitive

1. Relation $R_1$

Given $a \, R_1 \, b \iff ab \ge 0$

We test the three properties.

(i) Reflexive

For every $a \in \mathbb{R}$, $a \cdot a = a^2 \ge 0$ So, $a \, R_1 \, a$ for all $a$. Hence, $R_1$ is reflexive.

(ii) Symmetric

If $a \, R_1 \, b$, then $ab \ge 0$ But $ba = ab \ge 0$ So $b \, R_1 \, a$. Hence, $R_1$ is symmetric.

(iii) Transitive

We need to check whether $a \, R_1 \, b \text{ and } b \, R_1 \, c \implies a \, R_1 \, c$ That is, $ab \ge 0 \text{ and } bc \ge 0 \implies ac \ge 0?$

This is not always true. Take $a=1, \quad b=0, \quad c=-1$ Then $ab = 1\cdot 0 = 0 \ge 0$ $bc = 0\cdot (-1)=0 \ge 0$ So, $1 \, R_1 \, 0 \quad \text{and} \quad 0 \, R_1 \, (-1)$ But $ac = 1\cdot (-1) = -1 < 0$ So, $1 \not\, R_1 \, (-1)$ Thus, $R_1$ is not transitive.

Therefore, $R_1$ is not an equivalence relation.

2. Relation $R_2$

Given $a \, R_2 \, b \iff a \ge b$

Again test the three properties.

(i) Reflexive

For every $a \in \mathbb{R}$, $a \ge a$ So $a \, R_2 \, a$. Hence, $R_2$ is reflexive.

(ii) Symmetric

If $a \ge b$, does it imply $b \ge a$? Not always. For example, let $a=2, \quad b=1$ Then $2 \ge 1$ so $2 \, R_2 \, 1$ is true, but $1 \ge 2$ is false. So $1 \, R_2 \, 2$ is false. Hence, $R_2$ is not symmetric.

(iii) Transitive

If $a \ge b \quad \text{and} \quad b \ge c$ then $a \ge c$ So $R_2$ is transitive.

But since symmetry fails, $R_2$ is not an equivalence relation.

3. Final conclusion

• $R_1$ is not an equivalence relation.
• $R_2$ is not an equivalence relation.

Therefore, the correct option is $\boxed{\text{D: neither } R_1 \text{ nor } R_2 \text{ is an equivalence relation}}$

4. Comparison with stored correct answer

Stored correct answer: D

Our derived answer also gives D, so they agree.