Ignoring the difference of concentrations of reactants. Equilibrium Recently at school, while studying equilibrium we learned that in problems where the initial concentrations of reactants and the equilibrium constant are given, there is a special formula to determine wether the change in reactants can be ignored where C is the concentration of the reactant with the smallest concentration and K is the equilibrium constant: L=C/K. We learned that if L is bigger than 500, we can consider the change in the reactants negligible. However this is where my problem lies: i do not see why the C is included, doesn't the size of K by itself tell us wether a little or a lot of reactants is transformed? Not to mention, some cases turn out flat out wrong. For example: 
A+B=(reversible) 2C. 

Initial concentrations: A=10^-20M
B=10^-20M.  C=0 M

K=10^-8

By the equation we learned, L=10^-12 and thus the variation in the reactants should not be neglected. 
However, after doing the math, x~10^-24 which is 4 orders of magnitude below the beginning concentration of the reactants and completely negligible, is something wrong with my logic? Or did i discover a flaw in my textbook :)

Thanks
Let's look at your example and set up the ICE (intial-change-equilibrium) problem. Then I'll give you more inclusive of thumb. 

Let's say reaction equation is

$\ce{A + B &lt;=> 2C}$$

The law of mass action is:

$$K = \dfrac{[\ce{C}]^2}{[\ce{A}][\ce{B}]}=10^{-8}$$

Let's set up the ICE matrix. The change in concentration $x$, will be subtracted from $[\ce{A}]$ and $[\ce{B}]$ and added to $[\ce{C}]$.

$$\begin{array}{|c|c|c|c|}\hline
\ &amp; [\ce{A}]\ (\mathrm{M}) &amp; [\ce{B}]\ (\mathrm{M}) &amp; [\ce{C}]\ (\mathrm{M}) \\ \hline
I &amp; 10^{-20} &amp; 10^{-20} &amp; 0 \\
C &amp; -x &amp; -x &amp; +2x \\
E &amp; 10^{-20} - x &amp; 10^{-20} -x &amp; 2x \\ \hline
\end{array}$$

Now we pug in to the law of mass action:

$$10^{-8} = \dfrac{x^2}{\left(10^{-20} -x \right)^2}$$

Since all numbers have to be positive, let's take the square root of both sides:

$$10^{-4} = \dfrac{x}{10^{-20} -x }$$

Now we have a ratio that says that $x$ is four orders of magnitude smaller than $[\ce{a}]_i -x$. I could have easily used the initial concentration of $\ce{B}$ Put another way:

$$x \approx  K\cdot [\ce{A}]_i$$

If this value is much less than $[\ce{A}]_i$, then we can simplify.