Let $R_1$ and $R_2$ be two relations defined on $R$ by $aR_1b \Leftrightarrow ab \ge 0$ and $aR_2b \Leftrightarrow a \ge b$. Then,
(A)
$R_1$ is an equivalence relation but not $R_2$
(B)
$R_2$ is an equivalence relation but not $R_1$
(C)
both $R_1$ and $R_2$ are equivalence relations
(D)
neither $R_1$ nor $R_2$ is an equivalence relation
MEDIUM
Correct Answer:D
Explanation
For $R_1$: $aR_1b \Leftrightarrow ab \ge 0$. Reflexive: $(a,a) \in R_1$ since $a^2 \ge 0$ for all $a \in R$. Symmetric: If $(a,b) \in R_1$, then $ab \ge 0$, so $ba \ge 0$, and $(b,a) \in R_1$. Transitive: If $(a,b) \in R_1$ and $(b,c) \in R_1$, then $ab \ge 0$ and $bc \ge 0$. This does not necessarily imply $ac \ge 0$. For example, $a = -5$, $b = 0$, $c = 5$. Then $ab = 0 \ge 0$ and $bc = 0 \ge 0$, but $ac = -25 < 0$. Thus $R_1$ is not transitive, and hence not an equivalence relation. For $R_2$: $aR_2b \Leftrightarrow a \ge b$. Reflexive: $(a,a) \in R_2$ since $a \ge a$ for all $a \in R$. Symmetric: If $(a,b) \in R_2$, then $a \ge b$. This does not necessarily imply $b \ge a$. For example, $a = 2$, $b = 1$. Then $2 \ge 1$, but $1 \ngeq 2$. Thus $R_2$ is not symmetric, and hence not an equivalence relation. Therefore, neither $R_1$ nor $R_2$ is an equivalence relation.