Let $A$ be the number between 0 and 1, in which 10 ones stand after the decimal point and all other digits are 0. Determine the square root of $A$ to 20 significant digits!
According to the text of the task, $A=0.1111111111=\frac{1}{9} \cdot 0.9999999999=$

$=9 \frac{1}{9}\left(1-10^{-10}\right)$, so $\sqrt{A}=\frac{1}{3} \sqrt{\left(1-10^{-10}\right)}$, which means we essentially need to determine the square root of $B=1-10^{-10}$ with high precision. The identity $(1-x)^{2}=1-2 x+x^{2}$ suggests (with the choice $2 x=10^{-10}$) that the square root of $B$ is close to $C=1-\frac{1}{2} \cdot 10^{-10}=\left(1-5 \cdot 10^{-11}\right)$, since the square of this is only $x^{2}=25 \cdot 10^{-22}$ more than $B$. Let's see how much we need to reduce $C$ to get a value less than $\sqrt{B}$. Reduce $C$ by $y$, the $C-y0, \underbrace{3333333333}_{10 \mathrm{db}} 1 \underbrace{66666666}_{9 \mathrm{db}} 5$

While the upper limit is

$$
\frac{1}{3}\left(1-5 \cdot 10^{-11}\right)=\frac{1}{3} \cdot 0, \overbrace{9999999999}^{10 \mathrm{db}} 5<0, \underbrace{3333333333}_{10 \mathrm{db}} 1 \underbrace{66666666}_{9 \mathrm{db}} 7
$$

According to these, the value of $\sqrt{A}$ to twenty decimal places