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Which of S (y=x+1) and T (x-y∈Z) is Equivalence | JEE

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Question
Let RR be the real line. Consider the subsets S={(x,y):y=x+1, 0<x<2}S = \{(x, y) : y = x + 1,\ 0 < x < 2\} and T={(x,y):x−y is an integer}T = \{(x, y) : x - y \text{ is an integer}\} of the plane R×RR \times R. Which one of the following is true?
ABoth SS and TT are equivalence relations on RR
BSS is an equivalence relation on RR but TT is not
CTT is an equivalence relation on RR but SS is notcorrect
DNeither SS nor TT is an equivalence relation on RR
Solution
For T (x−y∈Z)T \ (x - y \in \mathbb{Z}):
x−x=0∈Z⇒reflexivex - x = 0 \in \mathbb{Z} \Rightarrow \text{reflexive}
x−y∈Z⇒y−x∈Z⇒symmetricx - y \in \mathbb{Z} \Rightarrow y - x \in \mathbb{Z} \Rightarrow \text{symmetric}
x−y, y−z∈Z⇒x−z∈Z⇒transitivex - y,\ y - z \in \mathbb{Z} \Rightarrow x - z \in \mathbb{Z} \Rightarrow \text{transitive}
 ∴T\therefore T is an equivalence relation. For S (y=x+1)S \ (y = x + 1): x=x+1x = x + 1 is impossible ⇒S\Rightarrow S is not reflexive ⇒\Rightarrow not an equivalence relation. Correct answer: (3)
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