R is a relation over the set of real numbers | vedclass

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(B) Given,$m R n$ if and only if $mn \ge 0$.Reflexivity:For any real number $m$,we know that $m^2 \ge 0$.Since $mm = m^2$,we have $mm \ge 0$,which imp

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Reflexive and symmetric

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(B) Given,$m R n$ if and only if $mn \ge 0$.Reflexivity:For any real number $m$,we know that $m^2 \ge 0$.Since $mm = m^2$,we have $mm \ge 0$,which implies $m R m$.Thus,$R$ is reflexive.Symmetry:Let $m R n$ hold,which means $mn \ge 0$.Since the product of real numbers is commutative,$nm = mn \ge 0$.This implies $n R m$.Thus,$R$ is symmetric.Transitivity:Let $m R n$ and $n R p$ hold,which means $mn \ge 0$ and $np \ge 0$.If $n = 0$,then $m R 0$ and $0 R p$ are always true $(0 \ge 0)$,but $m R p$ $(mp \ge 0)$ is not necessarily true (e.g.,$m=1, n=0, p=-1$).Wait,let us re-evaluate: $1 R 0$ $(1 	imes 0 = 0 \ge 0)$ and $0 R -1$ $(0 	imes -1 = 0 \ge 0)$,but $1 R -1$ $(1 	imes -1 = -1 Therefore,$R$ is $NOT$ transitive.Conclusion:Since $R$ is reflexive and symmetric but not transitive,the correct option is $B$.

$R$ is a relation over the set of real numbers and it is given by $nm \ge 0$. Then $R$ is

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