Question

Let R be the set of real numbers A = {(x, y) $\in$ R $\times$ R : y - x is an integer} is an equivalence relation on R. B = {(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 not a correct explanation for Statement-1
• B. Statement-1 is true, Statement-2 is false
• C. Statement-1 is false, Statement-2 is true
• D. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for statement -1

Answer 1

The Correct Option is B

x - y is an integer.
$\because$ x - x = 0 is an integer $\Rightarrow$ A is reflexive.
Let x - y is an integer
$\Rightarrow$ y - x is an integer
$\Rightarrow$ A is symmetric
Let x - y, y - z are integers
$\Rightarrow$ x - y + y - z is also an integer
$\Rightarrow$ x - z is an integer
$\Rightarrow$ A is transitive
$\therefore$ A is an equivalence relation.
Hence statement 1 is true.
Also B can be considered as

xBy if $\frac{x}{y} = \alpha$, a rational number
$\because \, \frac{x}{x} = 1$ is a rational number
$\Rightarrow$ B is reflexive
But $\frac{x}{y} = \alpha$ , a rational number need not imply $\frac{y}{x} = \frac{1}{\alpha}$, a rational number because
$\frac{0}{1}$ is rational $\Rightarrow \, \frac{1}{0}$ is not rational
$\therefore$ B is not an equivalence relation.

Answer 2

Given that;
y-x is an integer
x-x=0 is also an integer
\(\therefore\) A is reflexive
x-y=integer  \(\therefore\)x-x=0
\(\therefore\) it is symmetric
x-y and y-x both are integer and sum of them is also integer
Hence, transitive.
So, it has an equivalence relation