Let `S` be a relation on the set `R` of all real numbers defined by `S={(a , b)RxR: a^2+b^2=1}dot` Prove that `S` is not an equivalence relation on `Rdot`

Explanation

To prove that the relation \( S \) defined by \( S = \{(a, b) \in \mathbb{R} \times \mathbb{R} : a^2 + b^2 = 1\} \) is not an equivalence relation on \( \mathbb{R} \), we need to check the three properties of equivalence relations: reflexivity, symmetry, and transitivity. If any one of these properties fails, then \( S \) is not an equivalence relation.

### Step 1: Check Reflexivity
A relation \( S \) is reflexive if for every element \( a \in \mathbb{R} \), the pair \( (a, a) \) is in \( S \).

- For \( (a, a) \) to be in \( S \), we need:
\[
a^2 + a^2 = 1 \implies 2a^2 = 1 \implies a^2 = \frac{1}{2} \implies a = \pm \frac{1}{\sqrt{2}}.
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