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

Home » Let be the set of all real numbers. Show that the relation is symmetric but neither reflexive nor transitive.

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

Solution:

$\mathrm{R}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}_{2}+\mathrm{b}_{2}=1\right\}$ and $\mathrm{a}, \mathrm{b} \in \mathrm{S}$ (As given)

Non-Reflexivity:
Assume $a$ be an arbitrary element of $S$
$\mathrm{a} 2+\mathrm{a} 2-2 \mathrm{a} 2$ , which is not equal to 1
$\Rightarrow(\mathrm{a}, \mathrm{a}) \notin \mathrm{R}$
Therefore, $R$ is not reflexive.

Symmetric:
Assume $a$ and $\mathrm{b} \in \mathrm{S}$ , such that $(\mathrm{a}, \mathrm{b}) \in \mathrm{R}$
$\Rightarrow \mathrm{a} 2+\mathrm{b}_{2}=1$ $\Rightarrow \mathrm{b}_{2}+\mathrm{a}_{2}=1$ (As addition is commutative) $\Rightarrow(\mathrm{b}, \mathrm{a}) \in \mathrm{R}$
Therefore, $\mathrm{R}$ is symmetric.

Non-Transitivity:
Assume $\mathrm{a}, \mathrm{b}$ and $\mathrm{c} \in \mathrm{S}$ , such that $(\mathrm{a}, \mathrm{b}) \in \mathrm{R}$ and $(\mathrm{b}, \mathrm{c}) \in \mathrm{R}$ $\Rightarrow \mathrm{a}_{2}+\mathrm{b}_{2}=1$ and $\mathrm{b}_{2}+\mathrm{c}_{2}=1$
On adding both the equation, we obtain
$\begin{array}{l} \Rightarrow \mathrm{a}_{2}+\mathrm{b}_{2}+\mathrm{b}_{2}+\mathrm{c}_{2}=1+1 \\ \Rightarrow \mathrm{a}_{2}+2 \mathrm{~b}_{2}+\mathrm{c}_{2}=2 \end{array}$
$\Rightarrow \mathrm{a}_{2}+\mathrm{c}_{2}=2-2 \mathrm{~b}_{2}$ , which is not equal to 1
$\Rightarrow(\mathrm{a}, \mathrm{c}) \notin \mathrm{R}$
Therefore, $R$ is not transitive.
As a result, $\mathrm{R}$ is symmetric but neither transitive nor reflexive.