Let R be the set of real numbers.

Question

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

Accepted Answer

An equivalence relation on a set must satisfy three properties: reflexivity (every element is related to itself), symmetry (if an element is related to a second, the second is related to the first), and transitivity (if a first element is related to a second, and the second is related to a third, then the first is related to the third).

Statement I : $A=(x,y)\in R	imes R:$ y-x is an integer .

Reflexivity : For all $x$ in $R$, $x-x=0$ which is an integer. So, every element is related to itself.

Symmetry : For all $x,y$ in $R$, if $y-x$ is an integer, then $x-y=-(y-x)$ is also an integer. So, if $x$ is related to $y$, then $y$ is related to $x$.

Transitivity : For all $x,y,z$ in $R$, if $y-x$ and $z-y$ are integers, then $(z-y)+(y-x)=z-x$ is also an integer. So, if $x$ is related to $y$ and $y$ is related to $z$, then $x$ is related to $z$.

Therefore, $A$ is an equivalence relation on $R$.

Statement II : $B=(x,y)\in R	imes R:x=\alpha y$ for some rational number $\alpha $.

Reflexivity : For all $x$ in $R$, $x=1x$. Since 1 is a rational number, every element is related to itself.

Symmetry : For all $x,y$ in $R$, if $x=\alpha y$ for some rational $\alpha $, then $y=\frac{1}{\alpha }x$. However, if $\alpha =0$, then $\frac{1}{\alpha }$ is undefined, and therefore, $B$ doesn't satisfy symmetry.

Transitivity : If $x=\alpha y$ and $y=\beta z$ for some rational numbers $\alpha $ and $\beta $, then $x=(\alpha \beta )z$. Since the product of rational numbers is rational, if $x$ is related to $y$ and $y$ is related to $z$, then $x$ is related to $z$.

Therefore, $B$ is not an equivalence relation on $R$ since it does not satisfy the symmetry property.

In conclusion, the correct answer is

Option B : Statement I is true, Statement II is false.

Solutions · 1