Consider the relations R _ 1 and R _ 2 defined as a R _ 1 b ⇔ a ^ 2...

Given,

$a{R}_{1}b ⇔ {a}^{2} + {b}^{2} = 1; a, b \in R$

${R}_{1}$ is not reflexive as \left(a, a\right) \notin a{R}_{1}b$</p><p>So, it is not equivalence.</p><p>Now, solving</p><p>\left(a, b\right){R}_{2}\left(c, d\right) ⇔ a + d = b + c; \left(a, b\right), \left(c, d\right) \in N$

Reflexive: $a + b = b + a \to True$

Symmetric: $\left(a, b\right){R}_{2}$\left(c, d\right)$</p><p>$⇒ a + d = b + c$</p><p>$⇒ d + a = c + b$</p><p>$⇒ c + b = d + a$</p><p>$⇒ \left(c, d\right)${R}_{2}\left(a, b\right)$

Transitive: \left(a, b\right)${R}_{2}\left(c, d\right)$ ⇒ a + d = b + c . . . \left(i\right)$</p><p>\left(c, d\right){R}_{2}\left(e, f\right) ⇒ c + f = d + e . . . \left(ii\right)</p><p>Now, adding above equation we get,</p><p>$⇒ a + f = b + e$</p><p>$⇒ \left(a, b\right)${R}_{2}\left(e, f\right)$

So, ${R}_{2}$ is reflexive, symmetric and transitive

Hence only ${R}_{2}$ is equivalence relation.