For reflexive (x,x)∈R,x∈Z ⇒x+x=2x⟶ even For symmetric of (x,y)∈R then (y,x)∈R when x,y∈Z x+y⟶ even ⇒y+x⟶ even for transitive if (x,y)∈R⇒x+y⟶ even (y,z)∈R⇒y+z⟶ even x+2y+z⟶ even ⇒x+z is even ⇒(x,z)∈R ⇒R is an equivalence relation. ⇒ Option (4) is correct.
