b

a_{n + 1} := \frac{1}{2} (a_{n} + \frac{x _{0}}{a _{n}}), n \in N, x_{0} \geq 0

a_{1} > x_{0}

z.z.: (a_{n})_{n \in N} ist monoton fallend

a_{n} monoton fallend ⟺ a_{n} \geq a_{n + 1}

⟺ ⟺ ⟺ ⟺ ⟺ ⟸ a_{n} a_{n} 2 \cdot a_{n} a_{n} a_{n}^{2} a_{n} a_{n} \geq a_{n + 1} \geq \frac{1}{2} (a_{n} + \frac{x _{0}}{a _{n}}) \geq a_{n} + \frac{x _{0}}{a _{n}} \geq \frac{x _{0}}{a _{n}} \geq x_{0} \geq x_{0} > x_{0} bewiesen in (3a) ■

🪴 Quartz 4.0