$10^9 \times \sqrt{3}$ what are first two digits after the decimal point? Because of floating point error, my computer basically says $10^9 \sqrt{3} \approx 1.73205 \times 10^9$ so that if we ignore the numbers before the decimal point, the fractional part is:
$$\{ 10^9 \sqrt{3} \} = 0$$
That's obviously wrong.  As the algorithm progresses, the program keeps losing accuracy, effectively getting the error I just described.  if I type it directly we get the correct answer:
$$ 10^9 \sqrt{3} = 1732050807.\color{#00D000}{56}88772$$
but the computer doesn't store all these digits forever.   I could like to know how to get the first two decimal places without a calculator.
 A: You want the two digits before the decimal point of 
$10^{11} \sqrt{3} = \sqrt{3 \times 10^{22}}$.  Let's see if we can 
get a good rational approximation of that.  Let $A = 3 \times 10^{22}$.
The Newton's method iteration for $\sqrt{A}$ is
$x \to (x + A/x)/2$.  If we start with $x_0 = 2 \times 10^{11}$ and apply
this iteration, rounding each result to the nearest integer, we get
$$ \eqalign{x_1 &=  175000000000\cr
            x_2 &= 173214285714\cr
            x_3 &= 173205081001\cr
            x_4 &= 173205080757\cr
            x_5 = x_4 &= 173205080757\cr}$$
Check that $x_4^2 > A$, so what we want is $x_4 - 1 = 173205080756$, and its lowest two digits are the answer: $56$.
A: 
The actual error $997489097* \sqrt 3 = 1727701796.0$ Nothing after the decimal point. Kid you not.

You have found a "good" rational approximation for $\sqrt 3$ and in this case that's not really surprising to get a string of $0$s after the decimal point about as long as the integer coefficients.
For example, look at the numbers
$x_n = (2+\sqrt 3)^n(1+\sqrt 3)$ and $y_n = (2-\sqrt 3)^n(1-\sqrt 3)$
$x_ny_n = -2$, so while $x_n$ gets very large, $y_n$ gets very small.
Expanding $x_n$ you get $x_n = a_n + b_n\sqrt 3$ where $a_n$ and $b_n$ are some positive integers.
Then $b_n\sqrt 3 = a_n - (a_n - b_n \sqrt 3) = a_n - y_n$.
For example with $n=15$, you get
$299303201\sqrt 3 = 518408351 - y_{15} = 518408351.00000000192898\ldots$

Your example is obtained with $x = (8+15\sqrt 3)(2+\sqrt 3)^{14} = 1727701796+997489097 \sqrt 3$
so that $997489097 \sqrt 3 = 1727701796 - (8-15\sqrt 3)(2-\sqrt 3)^{14} = 1727701796.000000176824\ldots$
which is less "impressive" because $(8+15\sqrt 3)$ has norm $-611$ while $(1+\sqrt 3)$ has norm $-2$.
A: You can use the longhand procedure: http://www.homeschoolmath.net/teaching/square-root-algorithm.php
A: One approximation can start from the square of a sum formula:
$(a+b)^2\approx a^2+2ab$ when $ a >>> b$
That gives you:
$b \approx ((a+b)^2 - a^2)/2a$
So, if you have an estimate good to some number of places (say $n$), the approximation above should give you the next $(n-1)$ digits. 
