algebra precalculus - how to find out any digit of any irrational number?

We know that irrational number has not periodic digits of finite number as rational number.
All this means that we can find out which digit exist in any position of rational number.
But what about non-rational or irrational numbers?
For example:
How to find out which digit exists in Fortieth position of $\sqrt[2]{2}$ which equals 1,414213.......
Is it possible to solve such kind of problem for any irrational number?

Answer

Let $\alpha$ be an irrational number. As long as there exists an algorithm the can decide whether $\alpha>q$ or $\alpha

For $\alpha=\sqrt 2$, the decision algorithm is quit simple: If $q=\frac nm$ with $n\in\mathbb Z, m\in\mathbb N$, then $\alpha0\land n^2>2m^2$.