Question

Let $R$ be the set of real numbers.
Statement $(1)$: $A = \{ (x,y) \in R \times R:y - x$ is an integer}
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 true, Statement-$2$ is true; Statement-$2$ is 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

Hint:In this question, we are given two statements and we have to check which one is correct and incorrect and similarly, we will use the definition of equivalence relation to check whether statement (1) and (2) are true or not.
And we know that a relation is equivalence if the relation is reflexive, symmetric and transitive, where
Reflexive if $\forall a \in S, \Rightarrow \left( {a,a} \right) \in R$
Symmetric if $\forall a,b \in S,\left( {a,b} \right) \in R, \Rightarrow \left( {b,a} \right) \in R$
Transitive if $\forall a,b,c \in S,\left( {a,b} \right) \in R,\left( {b,c} \right) \in R, \Rightarrow \left( {a,c} \right) \in R$.Using this concept we try to solve the question.

Complete step-by-step answer:
In this question, we are given that $R$ be the set of real numbers and then we are given two statements.
Now we have to check whether statement (1) and (2) are true or not.
Now firstly we will consider statement (1),
Statement$(1)$: {$A = \{ (x,y) \in R \times R:y - x$ is an integer} is an equivalence relation.
Now to check whether statement (1) is true we will consider the definition of equivalence relation.
We know that the relation $R$ is equivalence on set $S$ if the relation $R$ is reflexive, symmetric and transitive, where
Reflexive if $\forall a \in S, \Rightarrow \left( {a,a} \right) \in R$
Symmetric if $\forall a,b \in S,\left( {a,b} \right) \in R, \Rightarrow \left( {b,a} \right) \in R$
Transitive if $\forall a,b,c \in S,\left( {a,b} \right) \in R,\left( {b,c} \right) \in R, \Rightarrow \left( {a,c} \right) \in R$
Now if all the relations are valid then the relation is equivalence relation,
Now using the above definition of equivalence relation for $A$ on $R$

Now taking,
{$A = \{ (x,y) \in R \times R:y - x$ is an integer}
Now we will check that $A$ is reflexive, transitive and symmetric.
Reflexive:
$
  \forall x \in R \\
  x - x = 0 \in Z \\
   \Rightarrow (x,x) \in A \\
 $
so, $\forall x \in R$, we have $(x,x) \in A$
Hence, $A$ is a reflexive relation.
Symmetric:
Let $(x,y) \in A$
 $
  \forall x,y \in Z \\
   \Rightarrow y - x \in Z \\
  {\text{also we can say that }}x - y{\text{ is a integer}} \\
   \Rightarrow x - y \in Z \\
 $
So, we mean that $(y,x) \in A$
So, for $(x,y) \in A$, we have $(y,x) \in A$
Hence it is a symmetric relation also.
Transitive:
Let $x,y,z \in R$
$(x,y) \in A$and $(y,z) \in A$
$ \Rightarrow y - x \in Z - - - - (1){\text{ and }}z - y \in Z - - - - (2)$
Now we know that the sum of two integers is also an integer.
So, using this for (1) and (2), we get
$(y - x) + (z - y) \in Z$
Therefore $z - x \in Z$
Hence $(x,z) \in A$
Hence if $(x,y) \in A$ and $(y,z) \in A$, $ \Rightarrow (x,z) \in A$
Hence it is a transitive relation.
Hence from the definition of the equivalence relation we get that $A$ is an equivalence relation.
So, the statement (1) is correct.

Now consider statement (2) which says that$B = \{ (x,y) \in R \times R:x = \alpha y$ for some rational number$\alpha $} is an equivalence relation on$R$.
Now again proceeding with the definition of equivalence relation-
Now we will check whether $B$ is reflexive, transitive and symmetric.
Reflexive:
$\forall x \in R$
We write that $x = 1.x$ and $1$ is a rational number.
So, for $x \in R,x = \alpha x,\alpha = 1 \in \mathbb{Q}$
$ \Rightarrow (x,x) \in B$
so, $\forall x \in R$, $ \Rightarrow (x,x) \in B$
Hence, we can say that it is reflexive.
Symmetric:
Now let $B$ is a symmetric relation
So, if $(x,y) \in B$, $\forall x,y \in R$
now $ \Rightarrow \left( {y,x} \right) \in R$
now, let’s take one example
Since $(0,x) \in B,\forall x \in R$
because $0 = 0 \times x,\alpha = 0,0$ is a rational number
but here we can see that $\left( {x,0} \right) \notin R$
because $x \ne \alpha \times 0,$ for $\alpha $ be any rational number.
So, the relation is not symmetric for all $x \in R$.
hence our supposition was wrong and the relation $B$ is not symmetric relation.
Since the relation is not symmetric so no need to check further for transitivity because we have got to know that it is not an equivalence relation.
So, statement (2) is not true.

So, the correct answer is “Option C”.

Note:In these types of problems, we should keep in mind that to disprove anything we just need one counter example but to prove anything, we need to prove it for all the possible values. Like for statement (2), when we consider the symmetric relation, we let that the relation is symmetric and then by taking one counterexample for which relation is not symmetric we contradicted our assumption.