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A relation $\rho$ on the set of real number $R$ is defined as $\{x \rho y: x y>0\} .$ Then, which of the following is/are true?
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Verified Answer
The correct answers are:
$\rho$ is symmetric but not reflexive, $\rho$ is symmetric and transitive
We have, $x p y: x y > 0$
(i) Reflexive Suppose $x \rho x \in R$
$\Rightarrow x^{2} > 0$
which is not true when $x=0$. Hence, relation is not reflexive.
(ii) Symmetric $x p y \in R$ $\Rightarrow \quad x y>0$
$\Rightarrow \quad y x>0$
$\Rightarrow \quad y \rho x \in R$
If $(x, y) \in R,$ then, relation is symmetric.
(iii) Transitive $\mathrm{xp} \mathrm{y} \in \mathrm{R}$
$x y>0$
and $\quad y \rho z \in R$ $y z>0$
$\begin{array}{ll}
\Rightarrow x y^{2} z > 0 \\
\Rightarrow x z > 0 \\
\Rightarrow (x, z) \in R \\
\text {If }(x, y) \in R, \text { then }(y, z) \in R \\
\Rightarrow (x, z) \in R
\end{array}$
Hence, relation is transitive.
(i) Reflexive Suppose $x \rho x \in R$
$\Rightarrow x^{2} > 0$
which is not true when $x=0$. Hence, relation is not reflexive.
(ii) Symmetric $x p y \in R$ $\Rightarrow \quad x y>0$
$\Rightarrow \quad y x>0$
$\Rightarrow \quad y \rho x \in R$
If $(x, y) \in R,$ then, relation is symmetric.
(iii) Transitive $\mathrm{xp} \mathrm{y} \in \mathrm{R}$
$x y>0$
and $\quad y \rho z \in R$ $y z>0$
$\begin{array}{ll}
\Rightarrow x y^{2} z > 0 \\
\Rightarrow x z > 0 \\
\Rightarrow (x, z) \in R \\
\text {If }(x, y) \in R, \text { then }(y, z) \in R \\
\Rightarrow (x, z) \in R
\end{array}$
Hence, relation is transitive.
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