# Mark the correct alternative in the following question:

Question:

Mark the correct alternative in the following question:

For real numbers $x$ and $y$, define $x R y$ iff $x-y+\sqrt{2}$ is an irrational number. Then the relation $R$ is

(a) reflexive

(b) symmetric

(c) transitive

(d) none of these

Solution:

We have,

$R=\{(x, y): x-y+\sqrt{2}$ is an irrational number; $x, y \in \mathbf{R}\}$

As, $x-x+\sqrt{2}=\sqrt{2}$, which is an irrational number

$\Rightarrow(x, x) \in R$

So, $R$ is reflexive relation

Since, $(\sqrt{2}, 2) \in R$

i. e. $\sqrt{2}-2+\sqrt{2}=2 \sqrt{2}-2$, which is an irrational number

but $2-\sqrt{2}+\sqrt{2}=2$, which is a rational number

$\Rightarrow(2, \sqrt{2}) \notin R$

So, $R$ is not symmetric relation

Also, $(\sqrt{2}, 2) \in R$ and $(2,2 \sqrt{2}) \in R$

i. e. $\sqrt{2}-2+\sqrt{2}=2 \sqrt{2}-2$, which is an irrational number and $2-2 \sqrt{2}+\sqrt{2}=2-\sqrt{2}$, which is

also an irrational number

But $\sqrt{2}-2 \sqrt{2}+\sqrt{2}=0$, which is a rational number

$\Rightarrow(\sqrt{2}, 2 \sqrt{2}) \notin R$

So, $R$ is not transitive relation

Hence, the correct alternative is option (a).