Consider the relations R 1 and R 2 defined as a R 1 b ⇔ a 2 + b 2 = 1 for all a , b , ∈ R and a , b R 2 c ,…

Given,

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

$R_1$ is not reflexive as $\left(a,a\right)\notin a{R}_{1}b$

So, it is not equivalence.

Now, solving

$\left(a,b\right)R_2\left(c,d\right)\iff a+d=b+c;\left(a,b\right),\left(c,d\right)\in N$

Reflexive: $a+b=b+a\to \mathrm{True}$

Symmetric: $\left(a,b\right)R_2\left(c,d\right)$

$\Rightarrow a+d=b+c$

$\Rightarrow d+a=c+b$

$\Rightarrow c+b=d+a$

$\Rightarrow \left(c,d\right)R_2\left(a,b\right)$

Transitive: $\left(a,b\right)R_2\left(c,d\right)\Rightarrow a+d=b+c ...\left(i\right)$

$\left(c,d\right)R_2\left(e,f\right)\Rightarrow c+f=d+e ...\left(ii\right)$

Now, adding above equation we get,

$\Rightarrow a+f=b+e$

$\Rightarrow \left(a,b\right)R_2\left(e,f\right)$

So, $R_2$ is reflexive, symmetric and transitive

Hence only $R_2$ is equivalence relation.