Pikkugnome said:
TL;DR Summary: How does an uncomputable number and its rational approximation go together?
Digits of an uncomputable number, as I understand, can't be produced.
That depends on what you mean by uncomputable. E.g. the digits of ##\sqrt{2}## can be computed, just not all of them, or not within a finite time. I assume you mean the absence of an algorithm that comes to a hold. But that leaves the question, of whether you consider ##1/3## as uncomputable.
Pikkugnome said:
However all real numbers have rational approximations. Does it mean that there exists a bound for the rational approximation.
Usually, yes. It depends on how the real number is described. We have good knowledge about ##p,q,n## in ##\left|\sqrt{2}-\frac{p}{q}\right|< \frac{1}{n}## or ##\left|\pi-\frac{p}{q}\right|< \frac{1}{n}.## I could imagine that there were real numbers that weren't defined by an
algebraic rule like ##x^2-2=0## or ##\pi=4\tan^{-1}(1) ## or any other constructive way. But I can hardly imagine that we can define such a real number in a way so that we both mean the same number without any form of construction rule. The definition itself is limited to a language, and a language obeys rules. The question of which of them are decidable, recursive countable, context-free, etc. leads us directly into the theory of
formal languages.
Pikkugnome said:
It is odd to talk about rational approximations in a non-contructive sense, but I am ok with it.
I think this would be close to impossible. What does it mean, a real number that cannot be constructed?
Pikkugnome said:
I guess the most uncomputable number would the one, which none of its digts can be calculated.
Yes, but how would you define such a number so that we can at least speak about it and both mean the same number?
There is a finite natural number ##G_{64},##
Graham's number, we don't know all digits from. We know the last 500 or so, we know the construction, but we cannot write it down since it is too big. This number is formally computable, but factually uncomputable. The difference to a particular rational number is zero.
Pikkugnome said:
That is strange, since then we can't even find its lower and upper bounds, as then we would know one of its digits. I am guessing this is true, it seems obvious. On the otherhand, if the bounds themselves are uncomputable numbers...
No, it is not. A real number is per definition the limit of a rational Cauchy sequence. This alone is a construction principle and a possibility to define lower and upper bounds of approximations to any given degree of accuracy.
It is only strange if a) you do not define
uncomputable, b) you do not define what a
real number is, c) you ignore the
Cauchy sequence, d) you do not even name the
formal language class we can use to describe a real number.
Digits are not an appropriate way to deal with such questions.
