Let S be the set of all real numbers. Show that the relation

Solution:

Given that, $\forall a, b \in S, R=\left\{(a, b): a^{2}+b^{2}=1\right\}$

Now,

$\underline{R}$ is Reflexive if $(a, a) \in \underline{R} \underline{\forall} \underline{a} \in \underline{S}$

For any $a \in S$, we have

$a^{2}+a^{2}=2 a^{2} \neq 1$

$\Rightarrow(a, a) \notin R$

Thus, $\mathrm{R}$ is not reflexive.

$\underline{R}$ is Symmetric if $(a, b) \in \underline{R} \Rightarrow \underline{(b, a)} \in \underline{R} \underline{\forall} \underline{a, b} \in \underline{S}$

$(a, b) \in R$

$\Rightarrow a^{2}+b^{2}=1$

$\Rightarrow b^{2}+a^{2}=1$

$\Rightarrow(b, a) \in R$

Thus, $R$ is symmetric.

$\underline{R}$ is Transitive if $(a, b) \in \underline{R}$ and $(b, c) \in \underline{R} \Rightarrow(a, c) \in \underline{R} \forall \underline{a}, b, c \in \underline{S}$

Let $(a, b) \in R$ and $(b, c) \in R \forall a, b, c \in S$

$\Rightarrow a^{2}+b^{2}=1$ and $b^{2}+c^{2}=1$

Adding both, we get

$a^{2}+c^{2}+2 b^{2}=2$

$\Rightarrow \mathrm{a}^{2}+\mathrm{c}^{2}=2-2 \mathrm{~b}^{2} \neq 1$

$\Rightarrow(\mathrm{a}, \mathrm{c}) \notin \mathrm{R}$

Thus, $R$ is not transitive.

Thus, $R$ is symmetric but neither reflexive nor transitive.