Question

Let R be the real line. Consider the following subsets of the plane R$\times $R ,
S = { x,y} : y = x + 1and 0 < x < 2
T = { x,y} : x – y is an integer
Which one of the following is true?
(a) T is an equivalence relation on R but S is not.
(b) Neither S nor T is an equivalence relation on R
(c) Both S and T are equivalence relation on R
(d) S is an equivalence relation on R but T is not

Answer 1

Hint: In this question, first we check that S is an equivalence relation. For this, we check three properties of equivalence relation i.e. reflexive, symmetric, or transitive. Similarly, we do it for T. Then we tick that option which follows our answers.

Formula Used: In this, we use the properties of equivalence relation i.e. reflexive, symmetric, and transitive.

Complete step by step Solution:
Given S = { x,y} : y = x + 1and 0 < x < 2 ------(1)
To check whether S is an equivalence relation or not:-
 x ≠ x+1 for any $x\in (0,2)$
i.e. $\left( x,x \right)\notin S$
 i.e. S is not reflexive
Let S is symmetric
So \[(x,y)\in R and(y,z)\in R\] $(x,y)\in R$
 $\left( y,x \right)\in R$ but $\left( y,x \right)\notin (0,2)$
 S is not symmetric
I.e. S is not an equivalence relation.
To check whether T is an equivalence relation or not:-
T = { x,y} : x – y is an integer
Therefore, x-x is an integer
Hence, T is reflexive.
Let T be symmetric
If $(x,y)\in R$, then y – x is also an integer
Then $(y,x)\in R$
Hence, T is symmetric
If \[(x,y)\in R and(y,z)\in R\]
Then (x-y) and (y-z) are integers
Then x-y+y-z = x – z is also an integer.
Hence, T is transitive
Therefore, T is an equivalence relation

Hence, the correct option is (a).

Note:To solve this type of question always remember what properties we have to use to prove whether it is equivalent or not. Some students do not read the options properly. As the options are almost the same, try to read the options carefully.