/  
(∃x) 2 ˃ x² ˃ 2 ‒
1
10n

Nur durch Induktion kann sich das ergeben



{
    



1² ˂ 2 ˂ (1 +
1
10⁰

A² ˂ 2 ˂ (A +
1
10⁰

(A +
a
10n + 1
)² ˂ 2 ˂ (A +
a + 1
10n + 1
     A²      (A +
1
10n + 1
)²     (A +
1
10n




               ︸
                                       ︸
                                  0 ≤ a ≤ 9
(Ƒ)

A² ˂ (A +
1
10n + 1
)² ˂ (A +
2
10n + 1
)² ˂ (A +
3
10n + 1
)² ‒ ‒ ‒ ˂ (A +
1
10n