On the set R of real numbers,we define x P y | vedclass

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(D) For every real number $x$,$x^2 \geq 0$. $	herefore (x, x) \in P$. Hence,$P$ is reflexive. Now,let $(x, y) \in P$. $\Rightarrow xy \geq 0$. $\Righ

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reflexive and symmetric but not transitive

Vedclass · Answer

(D) For every real number $x$,$x^2 \geq 0$. $	herefore (x, x) \in P$. Hence,$P$ is reflexive. Now,let $(x, y) \in P$. $\Rightarrow xy \geq 0$. $\Rightarrow yx \geq 0$. $	herefore (y, x) \in P$. Hence,$P$ is symmetric. Again,consider $(-1, 0) \in P$ because $(-1)(0) = 0 \geq 0$,and $(0, 2) \in P$ because $(0)(2) = 0 \geq 0$. However,$(-1, 2) 
otin P$ because $(-1)(2) = -2 Therefore,$P$ is not transitive. Thus,the relation $P$ is reflexive and symmetric but not transitive.

On the set $R$ of real numbers,we define $x P y$ if and only if $x y \geq 0$. Then,the relation $P$ is

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