Let S be the set of all real numbers. Then on | vedclass

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(A) The relation is $R = \{ (a, b) : 1 + ab > 0 \}$.$1$. Reflexive: For any $a \in S$,$(a, a) \in R$ if $1 + a^2 > 0$. Since $a^2 \ge 0$ for all real

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Reflexive and symmetric but not transitive

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(A) The relation is $R = \{ (a, b) : 1 + ab > 0 \}$.$1$. Reflexive: For any $a \in S$,$(a, a) \in R$ if $1 + a^2 > 0$. Since $a^2 \ge 0$ for all real $a$,$1 + a^2 \ge 1 > 0$. Thus,$R$ is reflexive.$2$. Symmetric: If $(a, b) \in R$,then $1 + ab > 0$. Since $ab = ba$,$1 + ba > 0$,which implies $(b, a) \in R$. Thus,$R$ is symmetric.$3$. Transitive: For $R$ to be transitive,if $(a, b) \in R$ and $(b, c) \in R$,then $(a, c) \in R$. Let $a = -8, b = -2, c = 0.25$. $1 + ab = 1 + (-8)(-2) = 17 > 0$.$1 + bc = 1 + (-2)(0.25) = 1 - 0.5 = 0.5 > 0$.However,$1 + ac = 1 + (-8)(0.25) = 1 - 2 = -1$,which is not $> 0$. Thus,$R$ is not transitive.Conclusion: The relation is reflexive and symmetric,but not transitive.

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

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