Behauptung
Beweis
IA ()
IS ()
\begin{alignat*}{3}
&\quad&a_{n+1} &> \sqrt{x_0} &\quad& \\
\stackrel{\text{Def.}}{\iff} &&
\frac{1}{2} \Big(
a_n+\frac{x_0}{a_n}
\Big) &>
\sqrt{x_0} \\
\iff &&
\frac{1}{2} \Big(
a_n+\frac{x_0}{a_n}
\Big) &>
\underbrace{
\sqrt{
a_n \cdot \frac{x_0}{a_n}
}
}_{=\sqrt{x_0}} \\
\stackrel{\text{1.9}}{\impliedby} &\quad \rlap{
\frac{1}{2}(a+b) \ge \sqrt{ab} \quad \wedge \quad
\text{LS} \neq \text{RS, da } a_n \neq \frac{x_0}{a_n} \text{, da }
} \\
&\quad \rlap{
a_n = \frac{x_0}{a_n}
\iff a_n^2 = x_0
\iff a_n = \sqrt{x_0} \ \unicode{x21af} \ \text{(IV)}
\ \blacksquare
}
\end{alignat*}
Latex align testing
split
multiline
\begin{multline}
a = b + c + d + e + f + g + h + i + b + c + d + e + f + g + h + i \\ + b + c + d + e + f + g + h + i + b + c + d + e + f + g + h + i
\end{multline}
\begin{multline}
a = a + b \\
\shoveleft{b = c} \\
\shoveright{c = d} \\
d = e
\end{multline}
gather & split
align
alignat
flalign
\begin{flalign}
x&=y & X&=Y\\
x’&=y’ & X’&=Y’\\
x+x’&=y+y’ & X+X’&=Y+Y’
\end{flalign}