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:
$$
z^2=1+\cfrac{1}{1+\cfrac{1}{2}}=\frac{5}{3},
\quad\hbox{and}\quad
z=\frac{\sqrt5}{\sqrt3}.
$$
