Hint Resolve Qlt_plus_pos_pos.

(** This lemma also holds for q2=Zero, due to our definition of (Qinv Zero):=Zero, Perhaps it is not right to state it like this but anyways it is true *)
Lemma Qdiv_num_denom :
forall q1 q2 p : Q,
p <> Zero -> Qmult q1 (Qinv q2) = Qmult (Qmult q1 p) (Qinv (Qmult q2 p)).