how to find out any digit of any irrational number?

Question

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 1

Let $\alpha$ be an irrational number. As long as there exists an algorithm the can decide whether $\alpha>q$ or $\alpha<q$ for any given rational $q$, you can obtain arbitrarily good rational approximations for $\alpha$. Especially, you can find upper and lower bounds good enough to uniquely determine any desired number of decimals.

For $\alpha=\sqrt 2$, the decision algorithm is quit simple: If $q=\frac nm$ with $n\in\mathbb Z, m\in\mathbb N$, then $\alpha<q\iff n>0\land n^2>2m^2$.

Answer 2

In some cases, yes. See spigot algorithms and digit extraction algorithms. In most cases however, no. Most irrational numbers have no such algorithm.

Answer 3

You can use continued fraction approximations to find rational numbers arbitrarily close to any irrational number.

For $\sqrt 2$ this is equivalent to the chain of approximations $\frac 11, \frac 32, \frac 75, \frac {12}{17} \dots$ where the fraction $\cfrac {a_{n+1}}{b_{n+1}}=\cfrac {a_n+2b_n}{a_n+b_n}.$

The accuracy of the estimate at the $n^{th}$ fraction is approximately $\left|\cfrac 1{b_n b_{n-1}} \right|$ - so you go far enough to get the accuracy you need to identify the decimal digit you want from the rational approximation.