I'm taking a course in rigorous calculus, using the famous calculus textbook by Tom Apostol. I'm required to prove that if x is real and satisfies 0 <= x < h, for all positive real h, then x = 0. Here is my 'proof': if x >= 0, then either x > 0 or x= 0. i'm going to prove that x > 0 leads to contradiction. if x>0, then let h = x/2 > 0. then x-h = x-x/2 = x/2 > 0, and therefore, h < x...which contradicts the h > x for all h. So x=0. Is this proof all right and sufficiently rigorous?