Let S be the set of all real numbers. Then, the relation $R=\{(a, b):1+ ab > 0\}$ on S is
Answer & explanation
Correct answer: option 1
We observe the following properties:
Reflexivity: Let a be an arbitrary element of R. Then,
$a ∈ R$
$⇒1+a.a=1+a%2 >0$ [$∵ a^2>0$ for all $a ∈ R$]
$⇒ (a, a) ∈ R_1$ [By def. of $R_1$]
Thus, $(a, a) ∈ R_1$ for all $a ∈ R$. So, $R_1$ is reflexive on R.
Symmetry: Let $(a, b) ∈ R$. Then,
$(a, b) ∈ R_1$
$⇒1 + ab > 0$
$⇒1 + ba > 0$ [∵ ab = ba for all a, b ∈ R]
$⇒(b, a) ∈ R_1$ [By def. of $R_1$]
Thus, $(a, b) ∈ R_1⇒(b, a) ∈ R_1$ for all $a, b ∈ R$.
So, $R_1$ is symmetric on R.
Transitivity: We observe that $(1, 1/2) ∈ R_1$ and $(1/2, -1) ∈ R_1$ but $(1, -1) ∉ R_1$ because $1+1 × (-1)=0≯ 0$.
So, $R_1$ is not transitive on R.