Let S be a relation on the set R of all real numbers defined by
$S=\left\{(a, b) \in R \times R: a^{2}+b^{2}=1\right\}$
Prove that S is not an equivalence relation on R.
We observe the following properties of S.
Reflexivity :
Let $a$ be an arbitrary element of $R$. Then,
$a \in R$
$\Rightarrow a^{2}+a^{2} \neq 1 \forall a \in R$
$\Rightarrow(a, a) \notin S$
So, $S$ is not reflexive on $R$.
Symmetry : Let $(a, b) \in R$
$\Rightarrow a^{2}+b^{2}=1$
$\Rightarrow b^{2}+a^{2}=1$
$\Rightarrow(b, a) \in S$ for all $a, b \in R$
So, $S$ is symmetric on $R$.
Transitivity :
Let $(a, b)$ and $(b, c) \in S$
$\Rightarrow a^{2}+b^{2}=1$ and $b^{2}+c^{2}=1$
Adding the above two, we get
$a^{2}+c^{2}=2-2 b^{2} \neq 1$ for all $a, b, c \in R$
So, $S$ is not transitive on $R$.
Hence, S is not an equivalence relation on R.