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Let \mathrm{R_{1}} and \mathrm{R_{2}} be two relations defined on \mathbb{R} by \mathrm{a\; R_{1} b \Leftrightarrow a b \geq 0 } and \mathrm {a \; R_{2} b \Leftrightarrow a \geq b} Then,

Option: 1

\mathrm{R_{1}} is an equivalence relation but not  \mathrm{R_{2}}


Option: 2

\mathrm{R_{2}} is an equivalence relation but not \mathrm{R_{1}}


Option: 3

 both \mathrm{ R_{1}} and \mathrm{ R_{2}} are equivalence relations 


Option: 4

 neither \mathrm{ R_{1}} nor \mathrm{ R_{2}}  is an equivalence relation 


Answers (1)

best_answer

\mathrm{R_{1}=\{x y \geq 0, x, y \in R\}}

\mathrm{for\; reflexive\; x \times x \geq 0 \;which \;is\; true\; for\; Symmetric}

\mathrm{if\; x y \geq 0 \Leftrightarrow y x \geq 0}

\mathrm{if \;x=2, y=0 \;and\; z=-2}

\mathrm{Then x \cdot y \geq 0 \&\;y.z \geq 0 \;but \;x \cdot z \geq 0 \;is\; not\; true}

\mathrm{\Rightarrow not \;transitive\; relation}

\mathrm{\Rightarrow R_{1}\; is\; not\; equivalence}

\mathrm{R_{2}\; if\; a \geq b\; it\; does\; not\; implies\; b \geq a}

\mathrm{\Rightarrow R_{2}\; is\; not\; equivalence \;relation }

Hence correct option is  4

Posted by

Pankaj

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