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

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Reflexive and symmetric but not transitive

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(A) $1$. Reflexivity: For any $a \in S$,we have $1 + a \cdot a = 1 + a^2$. Since $a^2 \ge 0$,$1 + a^2 \ge 1 > 0$. Thus,$(a, a) \in R$ for all $a \in S$. So,$R$ is reflexive.$2$. Symmetry: If $(a, b) \in R$,then $1 + ab > 0$. Since $ab = ba$,it follows that $1 + ba > 0$,which means $(b, a) \in R$. So,$R$ is symmetric.$3$. Transitivity: Consider $a = 1$,$b = -1/2$,and $c = -1$. For $(a, b) = (1, -1/2)$,$1 + (1)(-1/2) = 1 - 0.5 = 0.5 > 0$,so $(1, -1/2) \in R$.For $(b, c) = (-1/2, -1)$,$1 + (-1/2)(-1) = 1 + 0.5 = 1.5 > 0$,so $(-1/2, -1) \in R$.However,for $(a, c) = (1, -1)$,$1 + (1)(-1) = 1 - 1 = 0$,which is not $> 0$. Thus,$(1, -1) 
otin R$. Therefore,$R$ is not transitive.

Let $S$ be the set of all real numbers. Then the relation $R = \{(a, b) : 1 + ab > 0\}$ on $S$ is

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