Question

If $R$ is the set of real numbers. Then,

Statement I: $A=\{(x,y)\in R\times R:y-x\text{is an integer}\}$ is an equivalence relation on $R$.

Statement II : $B=\{(x,y)\in R\times R:x=ay\text{for some rational number}a\}$ is an equivalence relation on $R$.

• A. Statement I is correct, Statement II is correct; Statement II is not a correct explanation for Statement I
• B. Statement I is correct, Statement II is incorrect
• C. Statement I is incorrect, Statement II is correct
• D. Statement I is correct, Statement II is correct; Statement II is a correct explanation for Statement I

Answer 1

The correct option is B

Statement I is correct, Statement II is incorrect

Explanation for the correct option:

Step 1: Verify the equivalence relation on $R$ for statement I,

A relation $R$ on a set $A$ can be considered as an equivalence relation only if the relation $R$ will be reflexive, symmetric, and transitive.

Reflexive: a relation is said to be reflexive, if $(a,a)\in R$, for every $a\in A$.

Symmetric: a relation is symmetric, if $(a,b)\in R$, then $(b,a)\in R$.

Transitive: a relation is transitive if $(a,b)\in R$ and $(b,c)\in R$, then $(a,c)\in R$.

From statement I: Given $A=\{(x,y)\in R\times R:y-x\text{is an integer}\}$

As $y-x$ is an integer,

Since $x-x=0$ is an integer, which implies $A$ is reflexive, as $(x,x)\in R$

Assume $y-x$ is an integer, then $x-y$ is also an integer, which implies $A$ is symmetric, as $(x,y)\in R$ then $(y,x)\in R$

Again assume $y-x$,$z-y$ is an integer, then adding both we get

$y-x+z-y=z-x$, which implies $A$ is transitive as $(x,y)\in R$ and $(y,z)\in R$ then $(x,z)\in R$ .

Therefore, $A$ is an equivalence relation.

Hence, statement I is correct.

Step 2: Verify the equivalence relation on $R$ for statement II,

From statement I: Given $B=\{(x,y)\in R\times R:x=ay\text{for some rational number}a\}$

Given $x=ay\Rightarrow \frac{x}{y}=a$ is a rational number

Since $\frac{x}{x}=1$ is a rational number, which implies $B$ is reflexive, as $(x,x)\in R$

Assume $\frac{x}{y}=a$ is a rational number, then $\frac{y}{x}=a$ may not be a rational number,

because for $(0,y)\in R$ then $\frac{0}{y}=a\Rightarrow a=0$ is rational but for $(y,0)\in R$ then $\frac{y}{0}$ is undefined that is irrational,

Therefore, $B$ is not symmetric.

Again assume $\frac{x}{y}=a$,$\frac{y}{z}=b$ are a rational number, then multiplying both we get

$\frac{x}{y}\times \frac{y}{z}=ab\Rightarrow \frac{x}{z}=ab$, which implies $B$ is transitive as $(x,y)\in R$ and $(y,z)\in R$ then $(x,z)\in R$ .

Hence, $B$ is an not an equivalence relation as it is not symmetric.

Hence, the correct option is (B).