Question

Let $R$ be the set of real numbers.
Statement-l:$A=(x,y)\in R\times R$: $y-x$ is an integer is an equivalence relation on R.
Statement-II:$B=\left\{\right(x,y)\in R\times R:x=\alpha y$ for some rational number $\alpha$} is an equivalence relation on R.

• A. Statement-1 is true, Statement-2 is true;Statement-2 is a correct explanation for Statement-1.
• B. Statement-1 is true, Statement-2 is true;Statement-2 is not a correct explanation for Statement-1.
• C. Statement-1 is true, Statement-2 is false
• D. Statement-1 is false, Statement-2 is true.

Answer 1

Answer: (C)

Statement-1 is true, Statement-2 is false

$x-y$ is an integer
$x-x=0$ is also an integer
$\therefore A$ is reflexive
$y-x$ is also an integer. Hence, it is symmetric.
$x-y$ and $y-x$ are both integers, then sum of them is also an integer. Hence, transitive.
If a set is symmetric, transitive and reflexive then it also has equivalence relation.
Clearly, A is an equivalence relation
Now B,
$\frac{x}{y}=\alpha$ is rational but $\frac{y}{x}$ need not be rational i.e. $\frac{0}{1}$ is rational but $\frac{1}{0}$ is not rational. Hence, B doesn't show equivalence relationship