- #1

asdf60

- 81

- 0

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?