Question

For real numbers $x$ and $y$, we define $xRy$ iff $x-y+\sqrt{5}$ is an irrational number. The relation $R$ is

• A. reflexive
• B. symmetric
• C. transitive
• D. None of these

Answer 1

The correct option is A reflexive

$x,y\in R$
$f(x,y)=x-y+\sqrt{5}$ is an irrational number
Case1: Reflexive
$f(x,x)=x-x+\sqrt{5}=\sqrt{5}$ is an irrational number
$f(y,y)=y-y+\sqrt{5}=\sqrt{5}$ is an irrational number
$\therefore f(x,y)$ is a reflexive relation
Case2: Symmetric if $(x,y)$ is an irrational number holds $f(y,x)$ an irrational number
Let $x=\sqrt{5},y=1$
$f(x,y)=\sqrt{5}-1+\sqrt{5}=2\sqrt{5}-1$ is an irrational number
$f(y,x)=1-\sqrt{5}+\sqrt{5}=1$ is not an irrational number
$\therefore f(x,y)$ is not a symmetric relation
Case3: Transitive relation if if $(x,y)$ is an irrational number and $f(y,z)$ is an irrational number holds $f(x,z)$ an irrational number
Let $x=1,y=2\sqrt{5}$
$f(x,y)=1-2\sqrt{5}+\sqrt{5}=1-\sqrt{5}$ is an irrational number
$f(y,z)=y-z+\sqrt{5}$ can be irrational number if $z=\sqrt{5}$
$\therefore f(x,z)=x-z+\sqrt{5}\Rightarrow f(1,\sqrt{5})=1$ is not an irrational number
$\therefore f(x,y)$ is not a transitive relation