On R, the set of real numbers, a relation rho | vedclass

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(C) For reflexivity: For any $a \in R$, we have $1 + a^2 > 0$. Thus, $(a, a) \in \rho$. So, $\rho$ is reflexive. For symmetry: If $(a, b) \in \rho$, t

Vedclass · Accepted Answer

$\rho$ is reflexive and symmetric but not transitive

Vedclass · Answer

(C) For reflexivity: For any $a \in R$, we have $1 + a^2 > 0$. Thus, $(a, a) \in ho$. So, $ho$ is reflexive. For symmetry: If $(a, b) \in ho$, then $1 + ab > 0$. Since $ab = ba$, we have $1 + ba > 0$, which implies $(b, a) \in ho$. So, $ho$ is symmetric. For transitivity: Consider $a = 1$, $b = -0.5$, and $c = -9$. Check $(a, b)$: $1 + (1)(-0.5) = 0.5 > 0$, so $(1, -0.5) \in ho$. Check $(b, c)$: $1 + (-0.5)(-9) = 1 + 4.5 = 5.5 > 0$, so $(-0.5, -9) \in ho$. Check $(a, c)$: $1 + (1)(-9) = 1 - 9 = -8 Since $(1, -0.5) \in ho$ and $(-0.5, -9) \in ho$ but $(1, -9) 
otin ho$, the relation is not transitive. Therefore, $ho$ is reflexive and symmetric but not transitive.

On $R$, the set of real numbers, a relation $\rho$ is defined as $a \rho b$ if and only if $1+a b > 0$. Then,

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