Let R be a relation on [R], given by [R=(a, b): 3 a-3 b+√7] is an irrational number [].

Question

Let $R$ be a relation on $$\mathbb{R}$$, given by $$R=\{(a, b): 3 a-3 b+\sqrt{7}$$ is an irrational number $$\}$$. Then $R$ is

Accepted Answer

For reflexive :

$3 a-3 a+\sqrt{7}$ is an irrational number $\forall a \in R R$ is reflexive

For symmetric :

Let $3 a-3 b+\sqrt{7}$ is an irrational number

$\Rightarrow 3 b-3 a+\sqrt{7}$ is an irrational number

For example, Let $3 a-3 b=\sqrt{7}$

$\sqrt{7}+\sqrt{7}$ is irrational but $-\sqrt{7}+\sqrt{7}$ is not.

$	herefore R$ is not symmetric

For transitive :

Let $3 a-3 b+\sqrt{7}$ is irrational and $3 b-3 c+\sqrt{7}$ is irrational.

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

For example, take $a=0, b=-\sqrt{7}, c=\frac{\sqrt{7}}{3}$

$R$ is not transitive.

Solutions · 1