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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?

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  • $\begingroup$ you can always try it out : you need a number that times itself is between 1 , followed by 79 nines and 2 followed by 79 zero's $\endgroup$ – Willemien Aug 23 '13 at 12:10
  • $\begingroup$ You may be interested in the idea of a computable number. Almost all numbers are not computable. $\endgroup$ – user1729 Jul 10 '14 at 14:29
  • $\begingroup$ The Bailey–Borwein–Plouffe formula can be used to find arbitrarily deep digits of $\pi$, but such a formula does not exist for most irrational numbers. You usually have to just calculate it out to the necessary accuracy and look at the digits. $\endgroup$ – NovaDenizen Jul 10 '14 at 15:29
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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 $\alpha<q\iff n>0\land n^2>2m^2$.

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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.

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In general, no.

Suppose that for every irrational number $r$ there were an algorithm that takes a natural $n$ as input and returns the $n$-th digit of $r$. The possible algorithms are countable, all the irrationals are not, hence it is not possible to have such algorithms for every irrational.

However, such algorithms do exist for the so-called computable number.

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

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