# Let S be a relation on the set R of all real numbers defined by

Question:

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.

Solution:

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.