Question

Let $R$ be the set of real numbers
Statement-1:
$A=\{(x,y)\in R\times \{R:y-x$ is an integer $\}$ is an equivalence relafion on $R$
Statement-2:
$B=\{(x,y)\in R\times R:x=\alpha y$ for some rational number $\alpha \}$ is an equivalence relation on $R$.

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

Answer 1

Answer: (D)

Statement- 1 is true
We observe that
$\underline{\text{ Reflexivity }}$
$xRx$ as $x-x=0$ is an integer, $\forall x\in A$
Symmetric
Let $(x,y)\in A$
$\Rightarrow y-x$ is an integer
$\Rightarrow x-y$ is also an integer
$\underline{\text{Transitivity }}$
Let $(x,y)\in A$ and $(y,z)\in A$
$\Rightarrow y-x$ is an integer and $z-y$ is an integer
$\Rightarrow y-x+z-y$ is also an integer
$\Rightarrow z-x$ is an integer
$\Rightarrow (x,z)\in A$
Because of the above properties $A$ is an equivalence relation over $R$
Statement 2 is false as ${}^{\prime }{B}^{\prime }$ is not symmetric on $R$ We observe that
$OBx$ as $0=0x\forall x\in R$ but $(x,0)\notin B$