57/100<x≦q/p≦1 ∀q/p∈Q
|x−q/p|<1/p^2
→ q/p−x<1/p^2
⇔ x<q/p<x+1/p^2
⇔ x−q/p<0<(x−q/p)+1/p^2≦1/p^2
∴ x−q/p<1/p^2
∴ 0≦q/p−x<1/p^2 → 57/100<x<q/p+1/p^2