Sets, Relations & Functions
Symmetry of relations S and T
nta_pyq_2023_jan
Grade None
Question:
Among the relations \( S = \left\{(a, b) : a, b \in \mathbb{R} - \{0\},\ 2 + \frac{a}{b} > 0\right\} \) and \( T = \{(a, b) : a, b \in \mathbb{R},\ a^{2} - b^{2} \in \mathbb{Z}\} \),
S is transitive but T is not
T is symmetric but S is not
Neither S nor T is transitive
Both S and T are symmetric
Step-by-Step Solution
Key Concept: For T: if a²−b²$\in$Z then b²−a²=−(a²−b²)$\in$Z, so T is symmetric. For S: 2+a/b>0 does not imply 2+b/a>0 (e.g., a=1, b=−1/3).
T: (a,b)$\in$T $\Rightarrow$ a²−b²=I$\in$Z, then b²−a²=−I$\in$Z $\Rightarrow$ (b,a)$\in$T. So T is symmetric. S: 2+a/b>0$\Rightarrow$a/b>−2. If (b,a)$\in$S then 2+b/a>0, but b/a<−1/2 is possible, so S is not symmetric. Answer: (2)
Correct Answer: 2