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(B) Let the relation $R$ be defined on the set of real numbers $\mathbb{R}$ as $xRy \iff x^2 = xy$.$1$. Reflexivity: For any $x \in \mathbb{R}$,we che

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Reflexive

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(B) Let the relation $R$ be defined on the set of real numbers $\mathbb{R}$ as $xRy \iff x^2 = xy$.$1$. Reflexivity: For any $x \in \mathbb{R}$,we check if $xRx$ holds. $xRx \iff x^2 = x \cdot x$,which is $x^2 = x^2$. This is true for all $x \in \mathbb{R}$. Thus,the relation is reflexive.$2$. Symmetry: For $x, y \in \mathbb{R}$,if $xRy$,then $x^2 = xy$. Does this imply $yRx$,i.e.,$y^2 = yx$? If $x=1, y=0$,then $1^2 = 1 \cdot 0$ is $1=0$,which is false. If $x=2, y=4$,then $2^2 = 2 \cdot 4$ is $4=8$,which is false. Let us test $x=2, y=2$,then $4=4$ (True). But for $x=0, y=1$,$0^2 = 0 \cdot 1$ is $0=0$ (True),but $y^2 = yx$ gives $1^2 = 1 \cdot 0$,which is $1=0$ (False). Thus,it is not symmetric.$3$. Transitivity: If $xRy$ and $yRz$,then $x^2 = xy$ and $y^2 = yz$. From $x^2 = xy$,if $x 
eq 0$,then $x=y$. From $y^2 = yz$,if $y 
eq 0$,then $y=z$. Thus $x=z$,so $x^2 = xz$. If $x=0$,then $0=0$,which is always true. However,consider $x=2, y=2, z=4$. $2^2 = 2 \cdot 2$ (True),$2^2 = 2 \cdot 4$ (False). The relation is not transitive.Therefore,the relation is reflexive.

$x^2 = xy$ is a relation which is

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