Let N denotes the set of natural numbers and r is a relation in N×N. Which of the following is not an equivalence relation in N

Let (a, b),(c, d)∈N×N.The relation R defined by (a, b) R (c, d)⇔a + b = b + c is an equivalence reaction.The relation R defined by (a, b) R (c, d)⇔ad = bc is an equivalence relation.Let R = {[(a, b), (c, d)]:ad (b+ c)                         = bc (a + d), (a, b), (c, d) ∈N×N}                      ad (b + c) = bc (a +d)⇔b + cbc=a + dad⇔1c+1b+1d+1a⇔1a-1b=1c-1dLet (a, b) ∈N×NWe have,                         1a-1b=1a-1b    ∴                            (a, b)R (c, d)R is reflexiveLet                      (a, b)R (c, d)∴                        1a-1b=1c-1d⇒      1c-1d=1a-1b⇒         (c, d) R (a, b)    R is Symmetric.Let                (a, b) R (c, d) and (c, d) R (e, f)Let                 (a, b) R (c, d) and (c, d) R (e, f)∴                            1a-1b=1c-1dand                      ⇒                        1a-1b=1e-1f                              (a, b) R (e, f)⇒ R is transitive.∴R is equivalence relation.Let                       R={[(a, b), (c, d)] : bc (b + c)                                = ad ( a + d) , (a , b), (c ,d )bc N×N}Let                             (a, b)∈ N×NWe have, b a(b +a) =  ab ( a +b) R is reflexive.Let (a, b) R (c, d)∴      bc (b +c) = ad (a +d)⇒        ad (a + d) = (b + c)⇒        da(d + a) = cb (c + b)⇒         (c, d) R (a, b)⇒ R symmetricLet (a, b) R ( c, d) and r (e, f).∴ bc (b +c) = ad (a +d) and de ( d+ e) =  c f (c, f).These equations need no imply be (b + e) = a f (a + f).∴ (a, b) may not be R - related to  ( c, f)∴R is not transitve.∴R is not an equivalence relation.