AIEEE 2008 Solved Paper

Given S={(x,y):y=x+1
and 0<x<2}
As ‌‌x≠x+1
for any ‌‌x∈(0,2)⇒(x,x)∉S
∴S is not reflexive.
Hence S in not an equivalence relation.
Also T={x,y):x−y is an integer }
as x−x=0 is an integer ∀x∈R
∴T is reflexive.
If x−y is an integer then y−x is also an integer
∴T is symmetric
If x−y is an integer and y−z is an integer then (x−y)+(y−z)=x−z is also an integer.
∴T is transitive
Hence T is an equivalence relation