Question

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

Answer 1

(i) R is symmetric, since

$aRb\Rightarrow a^2+b^2=1$

$\Rightarrow b^2+a^2=1\Rightarrow bRa.$

(ii) R is not reflexive, since 1 is not related to 1, as

$1^2+1^2=1$ is not true.

(iii) Clearly, $\frac{1}{2}R\frac{\sqrt{3}}{2}$ and $\frac{\sqrt{3}}{2}R\frac{1}{2}$

But, $\frac{1}{2}$ is not related to $\frac{1}{2}$ as ${\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2\ne 1.$

$\therefore R$ is not transitive.