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Question: Answered & Verified by Expert
Let the relation $R_{1}$ be defined on $R$ as $a R_{1}b$ if $1+a b > 0 .$ Then
MathematicsSets and RelationsJEE Main
Options:
  • A $R_{1}$ is reflexive only.
  • B $R_{1}$ is equivalence relation.
  • C $R_{1}$ is reflexive and transitive but not symmetric
  • D $R_{1}$ is reflexive and symmetric but not transitive.
Solution:
2499 Upvotes Verified Answer
The correct answer is: $R_{1}$ is reflexive and symmetric but not transitive.
We observe the following properties:
Reflexivity : Let a be an arbitrary element of $R$, Then, $a \in R$ $\Rightarrow 1+a \cdot a=1+a^{2}>0 \quad\left[\because a^{2}>0\right.$ for all $\left.a \in R\right]$
$\Rightarrow \quad(a, a) \in R_{1}$
$\left[\right.$ By def. of $\left.R_{1}\right]$
Thus, $(a, a) \in R_{1}$ for all $a \in R$. So, $R_{1}$ is reflexive on $R$.
Symmetry : Let $(a, b) \in R$ Then, $(a, b) \in R_{1}$
$\Rightarrow \quad 1+a b>0$
$[\because a b=b a$ for all $]$
$\Rightarrow \quad 1+b a>0 \Rightarrow(b, a) \in R_{1}$
Thus, $(a, b) \in R_{1} \neq(b, a) \in R,$ for all $a, b \in R$
So, $R_{1}$ is symmetric on $R$.
Transitivity : we observe that $\left(1, \frac{1}{2}\right) \in R_{1}$ and $\left(\frac{1}{2},-1\right) \in R_{1}$ but (1,-1)$\notin R_{1}$ because
$1+1 \times(-1)=0 \gg 0$
So, $R_{1}$ is not transitive on $R$.

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