Question

Among the relations 
$S=\left\{(a, b): a, b \in R -\{0\}, 2+\frac{a}{b}>\right\}$ 
and $T=\left\{(a, b): a, b \in R , a^2-b^2 \in Z\right\}$,

• A. $S$ is transitive but $T$ is not
• B. both $S$ and $T$ are symmetric
• C. neither $S$ nor $T$ is transitive
• D. $T$ is symmetric but $S$ is not

Hint:

When analyzing relations for symmetry, check if \( (a, b) \) implies \( (b, a) \) for the relation to be symmetric.

Answer 1

The Correct Option is D

For relation $T=a^2-b^2=-I$
Then, (b, a) on relation $R$
$\Rightarrow b^2-a^2=-I$
$\therefore T\text{is symmetric}$
$S=\left\{(a,b):a,b\in R-\{0\},2+\frac{a}{b}>0\right\}$
$2+\frac{a}{b}>0\Rightarrow \frac{a}{b}>-2,\Rightarrow \frac{b}{a}<\frac{-1}{2}$
If $(b,a)\in S$ then
$2+\frac{b}{a}$ not necessarily positive
$\therefore S$ is not symmetric
So, the correct option is (D) : $T$ is symmetric but $S$ is not

Answer 2

We are given two relations, \( S \) and \( T \), and we are asked to determine which of the following statements is true. 
For relation \( T \): We know \( T = \{(a, b): a, b \in \mathbb{R}, a^2 - b^2 \in \mathbb{Z}\} \). From this, we deduce that \[ b^2 - a^2 = -1 \quad \Rightarrow \quad b = -a \quad \text{(relation \( T \) is symmetric)}. \] Thus, \( T \) is symmetric. 
For relation \( S \): We know \( S = \{(a, b): a, b \in \mathbb{R} \setminus \{ 0 \}, a^2 + b^2 > 0\} \). For \( S \), we see that \[ \frac{a}{b} = \frac{a}{-b} \quad \text{(the relation is not necessarily symmetric)}. \] So, \( S \) is not symmetric. Thus, the correct answer is \( T \) is symmetric but \( S \) is not symmetric.