Mathematics - Sets and Relations Question with Solution | TestHub

Given:

$R=\left\{\left(a,b\right):3a-3b+\sqrt{7} \mathrm{is} \mathrm{irrational}\right\}$

Reflexive:

$3a-3a+\sqrt{7}=\sqrt{7}$ is irrational.

So, $\left(a,a\right)\in R$, hence $R$ is reflexive.

Given:

$R=\left\{\left(a,b\right):3a-3b+\sqrt{7} \mathrm{is} \mathrm{irrational}\right\}$

Symmetric:

Let $\left(a,b\right)\in R$, then

$3a-3b+\sqrt{7}$ is irrational.

So, $3b-3a+\sqrt{7}$ is not necessarily irrational, since if $3a=\sqrt{7}$ and $3b\in Z$, then $3a-3b+\sqrt{7}$ is irrational., but $3b-3a+\sqrt{7}$ is not irrational.

So, $\left(a,b\right)\in R\cancel{\Rightarrow }\left(b,a\right)\in R$, hence $R$ is not symmetric

Transitive:

Take $\left(a,b\right)$ as $\left(\frac{\sqrt{7}}{3},1\right)$ and $\left(b,c\right)$ as $\left(1,\frac{2\sqrt{7}}{3}\right)$

So, $\left(a,b\right)\in R$ and $\left(b,c\right)\in R$ but $\left(a,c\right)\notin R$ which means relation is not transitive.