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If AA be the set of all real numbers and R be the relation on A defined by R={(a,b):a2+b2=1,∀a,b∈A}R = \{(a,b): a^2+b^2=1, \forall a,b \in A\}, then R is

Solution

✅ Correct Option: 2

Reflexivity fails since a2+a2=1a^2+a^2=1 only for a=±12a=\pm\frac{1}{\sqrt{2}}, not for all reals. Symmetry holds because a2+b2=b2+a2a^2+b^2=b^2+a^2. Transitivity fails: (1,0)(1,0) and (0,1)(0,1) belong to R, but (1,1)(1,1) does not since 1+1≠11+1 \neq 1.

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