If \( A \) be the set of all real numbers and R be the relation on A defined by \( \mathrm{R}=\{(a,b):a^2+b^2=1,\forall\ a,b\in\mathrm{A}\} \) , then R is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → symmetric but neither reflexive nor transitive
Reflexive: (0, 0) is not in R, since \( 0^2+0^2\ne 1 \). So R is not reflexive.
Symmetric: if \( a^2+b^2=1 \) then \( b^2+a^2=1 \). So R is symmetric.
Transitive: (0, 1) and (1, 0) are in R, but (0, 0) is not. So R is not transitive.
Hence R is symmetric but neither reflexive nor transitive.
NCERT (2026-27 edition):
Types of Relations
