R is a relation over the set of integers and it is given by (x,y)∈R⇔x-y≤1.Then R is

As (x,x)∈R ⇒ x-x≤1                                  0≤1Thus reflexive As,(x,y)∈R ⇒x-y≤1 ⇒y-x≤1⇒(y,x)∈RThus R is  symmetric Again (x,y)∈R and (y,x)∈R⇒x-y≤1 and y-z≤1⇒x-z≤1  it need not be ture  since if x=1 , y=1.9 ,z=2.8 then x-y=0.9 ≤1, y-z=0.9≤1 but  x-z=1.8≥1  Hence Not transitive