To analyze the relation p given by xpy iff x−y+√2 is an irrational number, let's evaluate whether the relation is reflexive, symmetric, transitive, or an equivalence relation. Reflexive : A relation is reflexive if every element is related to itself. For xpx to hold, we need x−x+√2 to be irrational. Simplifying, x−x=0, and 0+√2=√2, which is indeed irrational. Therefore, the relation p is reflexive. Reflexive: Yes Symmetric : Consider (√2,1)∈p⇒√2−1+√2=2√2−1 is irrational number. but (1,√2)p as 1−√2+√2=1 is not an irrational number. Thus p is not symmetric. Symmetric : No Transitive : (√2,1)∈p as √2−1+√2 is irrational. (1,2√2)∈ρ as 1−2√2+√2=1−√2 is irrational. Now √2−2√2+√2=0 is not irrational. ⇒(√2,2√2)∉p ⇒p is not transitive relation. Transitive: No Equivalence Relation : A relation is an equivalence relation if it is reflexive, symmetric, and transitive. We've established that the relation p is reflexive but not symmetric not transitive. Therefore, the relation p is not an equivalence relation. Equivalence Relation : No Thus, the correct answer is : Option A : Reflexive