Let R be the real line. Consider the following subsets of the plane $R\times R$.

$\begin{array}{l}S=\left\{\right(x,y):y=x+1\text{ and }0<x<2\}\\ T=\left\{\right(x,y):x-y\text{ is an integer }\}\end{array}$

Statement-1 : T is an equivalence relation on R but S is not an equivalence relation on R. 

Statement-2 : S is neither reflexive nor symmetric but T is reflexive, symmetric and transitive

detailed solution

Correct option is A

Since x ≠ x + 1 (x,x)∉S , so S is not reflexive Next x,y∈S,⇒y=x+1⇒x=y−1⇒(y,x)∉S so is not symmetricSince x – x = 0 0 is an integer (x,x)∈T∀x∈TAgain (x,y)∈T⇒x−y is an integer⇒ y – x is also an integer ⇒ (y, x) ∈ T So T is symmetric. Also , (x,y)∈T,(y,z)∈T⇒ x – y and y – z are integers ⇒ x – z = (x – y) – (y – z) is also an integer⇒ (x,z)∈Tso T is s Transitive .Which shows that statement-2 is true and hence statement-1 is also true.