Let R_{1} and R_{2} be two relations defined | vedclass

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(D) For $R_{1}$: $a R_{1} b \iff ab \geq 0$.$1$. Reflexive: $a \cdot a = a^{2} \geq 0$ for all $a \in \mathbb{R}$. So,$R_{1}$ is reflexive.$2$. Symmet

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Neither $R_{1}$ nor $R_{2}$ is an equivalence relation

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(D) For $R_{1}$: $a R_{1} b \iff ab \geq 0$.$1$. Reflexive: $a \cdot a = a^{2} \geq 0$ for all $a \in \mathbb{R}$. So,$R_{1}$ is reflexive.$2$. Symmetric: If $ab \geq 0$,then $ba \geq 0$. So,$R_{1}$ is symmetric.$3$. Transitive: Let $a=2, b=0, c=-2$. Then $ab = 2 \cdot 0 = 0 \geq 0$ and $bc = 0 \cdot (-2) = 0 \geq 0$. However,$ac = 2 \cdot (-2) = -4 Therefore,$R_{1}$ is not an equivalence relation.For $R_{2}$: $a R_{2} b \iff a \geq b$.$1$. Reflexive: $a \geq a$ is true for all $a \in \mathbb{R}$. So,$R_{2}$ is reflexive.$2$. Symmetric: If $a \geq b$,it does not imply $b \geq a$ (e.g.,$2 \geq 1$ but $1 
geq 2$). So,$R_{2}$ is not symmetric.$3$. Transitive: If $a \geq b$ and $b \geq c$,then $a \geq c$. So,$R_{2}$ is transitive.Since $R_{2}$ is not symmetric,it is not an equivalence relation.Thus,neither $R_{1}$ nor $R_{2}$ is an equivalence relation.

Let $R_{1}$ and $R_{2}$ be two relations defined on the set of real numbers $\mathbb{R}$ by $a R_{1} b \iff ab \geq 0$ and $a R_{2} b \iff a \geq b$. Then:

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