Let the relation R_{1} be defined on R as a R | vedclass

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(D) Reflexivity: For any $a \in R$,we have $1 + a \cdot a = 1 + a^{2}$. Since $a^{2} \ge 0$,$1 + a^{2} \ge 1 > 0$. Thus,$(a, a) \in R_{1}$ for all $a

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$R_{1}$ is reflexive and symmetric but not transitive.

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(D) Reflexivity: For any $a \in R$,we have $1 + a \cdot a = 1 + a^{2}$. Since $a^{2} \ge 0$,$1 + a^{2} \ge 1 > 0$. Thus,$(a, a) \in R_{1}$ for all $a \in R$. So,$R_{1}$ is reflexive.Symmetry: If $(a, b) \in R_{1}$,then $1 + ab > 0$. Since $ab = ba$,we have $1 + ba > 0$,which implies $(b, a) \in R_{1}$. So,$R_{1}$ is symmetric.Transitivity: Consider $a = 1$,$b = 1/2$,and $c = -1$. We have $1 + (1)(1/2) = 1.5 > 0$,so $(1, 1/2) \in R_{1}$. Also,$1 + (1/2)(-1) = 0.5 > 0$,so $(1/2, -1) \in R_{1}$. However,$1 + (1)(-1) = 0$,which is not $> 0$. Thus,$(1, -1) 
otin R_{1}$. Therefore,$R_{1}$ is not transitive.

Let the relation $R_{1}$ be defined on $R$ as $a R_{1} b$ if $1+ab > 0$. Then

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