Let $\mathrm{S}$ be the set of all real numbers. Show that the relation $\mathrm{R}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}^{2}+\mathrm{b}^{2}=1\right\}$ is symmetric but neither reflexive nor transitive.
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.
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