Hi All, Question : Prove that xn<yn , given that x<y and n is odd . Attempt at solution : Assumptions: y-x>0 y2>0 x2>0 So y2x<y3 & x3<x2y So i need to prove that x2y<y2x i.e need to prove then y2x-x2y>0 then yx(y-x)>0, from assumptions y-x>0 so i need to prove that yx>0, so i have 3 cases Case 1 : y and x are both positive so the inequality is true and everything is true. Case 2: y and x are both negative so the inequality is true again and everything is true Case 3 : y is positive and x is negative (as y>x), then the above inequality is not true, but i go back knowing that y is positive and x is negative, then y>0 and x<0 so y3>0 and x3<0 so y3>x3 At the end x3<y3 and by repetition xn <yn End. Is this proof true ? is there an easier way. Thanks.