TITLE: How to find Fraction from Decimal?
QUESTION [6 upvotes]: Problem: 
A decimal number $z$ of 8 digits after decimal Point is given.How can  $x,y$ be found  such that the first 8 digits of  $\frac{x}{y}$  are same as  $z$ (one may ignore digits after first 8 digits ) ?
Decimal Point is a point or dot used to separate the whole number part from the fractional part of a number.
Restrictions are-

$x,y$ have no rational or irrational number common as a factor in $x,y$ , i.e. their Greatest Common Divisor is $1$. 
$x,y$ could be irrational*.
$y$ is not a power of $10$.
find the smallest possible $x,y$ , if possible.

Example:  Consider the number, $0.29411764705882352941176470588235$ as $z$.
Here,  The first 36 digits of $\frac{5}{17}$  are $29411764705882352941176470588235$. 
So, $x=5, y=17$. So, if $0.29411764705882352941176470588235$ is given, is there an algorithm which can find $5,17$ ?.
*Note:  The problem considers the first 8 digits after decimal point. So, if $\frac{x}{y}$ is  infinite after decimal point,  but has first 8 digits same as $z$, after decimal point , the solution is acceptable, that is why $x,y$ could be irrational.
Related Post: This post is related but does not help much.

REPLY [4 votes]: As Raymond Manzoni suggested in his comment, the best approximation of a decimal number can usually be achieved by transforming it into a continuous fraction. I'll try to explain how that works with an example.
Let's take $z=0.29411764$, as you suggested, and compute its reciprocal $1/z=3.4000000816$. The integer part of this number ($3$) is the first term in the continuous fraction expansion, and $z=1/3$ is the first approximation. 
Consider now the fractional part of $1/z$, that is $0.4000000816$, and compute its reciprocal $2.4999994900001$. The integer part ($2$) is the second term in the continuous fraction expansion and $z=1/(3+1/2)$ is thus the second approximation. The remaining fractional part is $0.4999994900001$, whose reciprocal is $2.00000204000168$: this gives $2$ as the third term in the continuous fraction expansion, leading to $z=1/(3+1/(2+1/2))$. The fractional remainder $0.00000204000168$ is very small: that means that the approximation we have reached is very good. Indeed you may check that
$$
{1\over\displaystyle3+{1\over2+{1\over2}}}={5\over17}.
$$
This example suggests a possible strategy: compute the continuous fraction expansion of $z$ until you get a small remainder. This works well if $z$ is the truncated decimal representation of a fraction having small integers as numerator and denominator. 
If, on the other hand, $z$ is the square root of such a fraction, then no small remainder will be obtained, and the terms of the continuous fraction will show a periodic pattern. In that case one may compute the continuous fraction expansion of $z^2$ to get an approximated fraction $a/b$, so that $z=\sqrt{a}/\sqrt{b}$.
Example: $z=1.29099445=1+0.29099445$. The reciprocal of the fractional part is 
$3.43649165817424=3+0.43649165817424$ and again the reciprocal of the fractional part is $2.29099452709545=2+0.29099452709545$. Notice that the last remainder is very close to the first one: if you continue with the expansion you'll get a (quasi-) periodic continued fraction:
$$
z=1+\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{2+\cfrac{1}{3+\dots}}}}
$$
This is not exactly periodic because of the limited precision, but you are not going to obtain a small remainder in a few cycles.
Let's try then $z^2=1.6666666699308=1+0.6666666699308$. We get in turn $1.4999999926557=1+0.4999999926557$ and $2.0000000293772=2+0.0000000293772$. The fractional remainder is small, so that we can stop with:
$$
z^2=1+\cfrac{1}{1+\cfrac{1}{2}}=\frac{5}{3},
\quad\hbox{and}\quad
z=\frac{\sqrt5}{\sqrt3}.
$$