Correct option (d) T is an equivalence relation on R but S is not
Explanation :
We have
T = {(x,y): x - y ∈/}
As 0 ∈/ T is a reflexive relation. If
x - y ∈ / ⇒ y - x ∈/
Then T is symmetrical as well.
If x - y = I1 and y - z = I2, then x - z − = (x - y) + (x - z) = I1 + I2 ∈I;
therefore, T is transitive as well.
Hence, T is an equivalence relation. Clearly,
x ≠ x + 1 ⇒ (x,x) ∉ S.
Therefore, S is not reflexive.