Question

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

• A. Reflexive
• B. Symmetric
• C. Transitive
• D. None of these

Answer 1

The correct option is A Reflexive

Given $x$ and $y$ are real numbers. Relation is $x-y+\sqrt{2}$ is irrational.

A relation $R$ in $A$ is said to be reflexive, if $(x,x)\in R$ for every $x\in A$.

Putting $y=x$ , $x-x+\sqrt{2}=\sqrt{2}$ which is irrational. Hence the relation is reflexive.

A relation $R$ in $A$ is said to be symmetric, if $(x,y)\in R⟹(y,x)\in R$ for $x,y\in A$.

Let $x=\sqrt{2}$ and $y=1$. Then $x-y+\sqrt{2}=\sqrt{2}-1+\sqrt{2}=2\sqrt{2}-1$. This is an irrational number.

Now put $x=1$ and $y=\sqrt{2}$, then $x-y+\sqrt{2}=1-\sqrt{2}+\sqrt{2}=1$ This is rational number.

Hence relation is not symmetric.

A relation $R$ in $A$ is said to be transitive, if $(x,y)\in R$ and $(y,z)\in R⟹(x,z)\in R$ for all $x,y,z\in A$.

Let $x=1,y=2\sqrt{2},z=\sqrt{2}$.

$x-y+\sqrt{2}=1-2\sqrt{2}+\sqrt{2}=1-\sqrt{2}$ is irrational.

$y-z+\sqrt{2}=2\sqrt{2}-\sqrt{2}+\sqrt{2}=2\sqrt{2}$ is irrational.

$x-z+\sqrt{2}=1-\sqrt{2}+\sqrt{2}=1$ rational.

Hence relation is not transitive.