Question

On $R$, the set of real numbers, a relation $\rho$ is define as ${}^{\prime }a\rho b^{\prime }$ if an only if $1+ab>0$. Then

• A. $\rho$ is an equivalence relation
• B. $\rho$ is reflexive and transitive but not symmetric
• C. $\rho$ is reflexive and symmetric but not transitive
• D. $\rho$ is only symmetric

Answer 1

Answer: (C)

$\rho$ is reflexive and symmetric but not transitive

$\rho$ is the relation defined on $R$ as $a\rho b$ if and only if $1+ab>0$

(i)

We know that, for any real number $a,a^2>0$

i.e. $aa>0⟹1+aa>1>0⟹1+aa>0⟹a\rho a$

$\therefore \rho$ is reflexive.

(ii)

Let $a\rho b$

i.e. $1+ab>0⟹1+ba>0$

$⟹b\rho a$

$\therefore \rho$ is symmetric.

(iii)

Let $a\rho b$ and $b\rho c$

$⟹1+ab>0$ and $1+bc>0$

we can' conclude $1+ac>0$

For example :-

$1+(-\frac{1}{2})\left(\frac{1}{3}\right)=1-0.1666>0$ and $1+\left(\frac{1}{3}\right)\left(4\right)=2>0$

But $1+(-\frac{1}{3})\left(4\right)=-\frac{1}{3}<0$

Therefore, $\rho$ is transitive.