The correct option is A reflexive
x,y∈R
f(x,y)=x−y+√5 is an irrational number
Case1: Reflexive
f(x,x)=x−x+√5=√5 is an irrational number
f(y,y)=y−y+√5=√5 is an irrational number
∴f(x,y) is a reflexive relation
Case2: Symmetric if (x,y) is an irrational number holds f(y,x) an irrational number
Let x=√5,y=1
f(x,y)=√5−1+√5=2√5−1 is an irrational number
f(y,x)=1−√5+√5=1 is not an irrational number
∴f(x,y) is not a symmetric relation
Case3: Transitive relation if if (x,y) is an irrational number and f(y,z) is an irrational number holds f(x,z) an irrational number
Let x=1,y=2√5
f(x,y)=1−2√5+√5=1−√5 is an irrational number
f(y,z)=y−z+√5 can be irrational number if z=√5
∴f(x,z)=x−z+√5⇒f(1,√5)=1 is not an irrational number
∴f(x,y) is not a transitive relation