On the set R of real numbers, the relation rh | vedclass

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(D) Given the relation $\rho$ on the set $R$ of real numbers defined by $x \rho y \iff x > |y|$.$1$. For reflexive property:Check if $x \rho x$ holds

Vedclass · Accepted Answer

If $x > |y|$, then $\rho$ is transitive but neither reflexive nor symmetric.

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(D) Given the relation $\rho$ on the set $R$ of real numbers defined by $x \rho y \iff x > |y|$.$1$. For reflexive property:Check if $x \rho x$ holds for all $x \in R$.$x \rho x \iff x > |x|$.This is false for all $x \leq 0$ (e.g.,if $x = -1$, $-1 > |-1| = 1$ is false).Thus, $\rho$ is not reflexive.$2$. For symmetric property:Check if $x \rho y \implies y \rho x$.$x \rho y \implies x > |y|$.$y \rho x \implies y > |x|$.If we take $x = 2$ and $y = 1$, $2 > |1|$ is true, but $1 > |2|$ is false.Thus, $\rho$ is not symmetric.$3$. For transitive property:Check if $x \rho y$ and $y \rho z \implies x \rho z$.$x \rho y \implies x > |y|$.$y \rho z \implies y > |z|$.Since $y > |z|$, we have $|y| \geq y > |z|$, so $|y| > |z|$.Since $x > |y|$ and $|y| > |z|$, by transitivity of inequality, $x > |z|$.Therefore, $x \rho z$ holds.Thus, $\rho$ is transitive.Conclusion: The relation $\rho$ defined by $x > |y|$ is transitive but neither reflexive nor symmetric.

On the set $R$ of real numbers, the relation $\rho$ is defined by $x \rho y$ if $x > |y|$. Which of the following statements is true regarding the properties of $\rho$?

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