Question

Let $R$ be a relation from $R$ to $R$ the set of real numbers defined by $R=\{(x, y): x, y \in R$ and $x-y+\sqrt{3}$ is an irrational number $\}$. Then, $R$ is

• A. an equivalence relation
• B. reflexive
• C. transitive
• D. symmetric

Answer 1

Answer: (B)

The given relation is $R=\{(x, y): x, y \in R$ and $x-y+\sqrt{3}$ is an irrational number $\}$.
Then, $x R x \Rightarrow x-x+\sqrt{3}=\sqrt{3}$
which is an irrational number.
$\Rightarrow R$ is reflexive.
Now, $ 2 R \sqrt{3} \Rightarrow 2-\sqrt{3}+\sqrt{3}=2$
which is not an irrational number.
and $\sqrt{3} R 2 \Rightarrow \sqrt{3}-2+\sqrt{3}=2 \sqrt{3}-2$
which is an irrational number.
$\therefore 2 \not R \sqrt{3}$ but $\sqrt{3} R 2$
$\Rightarrow R$ is not symmetric.
Also, $-\sqrt{3} R \sqrt{5}$ and $\sqrt{5} R 2$
As $-\sqrt{3}-\sqrt{5}+\sqrt{3}=-\sqrt{5}$ is an irrational number and $\sqrt{5}-2+\sqrt{3}$ is an irrational number.
But, $-\sqrt{3}-2+\sqrt{3}=-2$ is a rational number.
$\therefore-\sqrt{3}$ is not related with 2 .
$\Rightarrow R$ is not transitive.