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Do I see it right that a randomly chosen number $a$ of the form $ a = 0.abc...xyz$ with random digits $a,b,c,x,y,z$ an approximation of a transcendental real number is?

It is an approximation of a transcendental number to any finite precision, I assume.

Now, is it approximating a transcendental number for sure?

I think it is, because there are infinitely more transcendental numbers than algebraic numbers. So the chosen number is either algebraic, or between two algebraic numbers. As there are infinite numbers between, the number is one of them.

With probability $ p = 1 $, and not in any way $ p < 1 $.

That would mean $a$ is transcendental for any practical purpose $u$.

And $a$ is transcendental for any purpose $u$, for any finite definition of $u$, which may not be practical.

It is even transcendental for anything that is not a purpose. Any definition of $u$ is finite. Because the visible universe is finite.

Now, is that right?

(Obviously, $a$ is not strictly transcendental, because it is finite. But in which sense is it not transcendental - it is chosen, and the chosen number can be any number whatsoever. A chosen number must be encoded, and it would be nice if all mass-energy of the earth, excluding us, is enough. Or say our galaxy, and Andromeda, to be generous. But feel free to encode with the visible universe. Without us.)

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    $\begingroup$ This is not clear. Every finite decimal is arbitrarily close to a transcendental. If your string has length $N$, just take $\frac {\pi}{10^{N+2}}$ or such. $\endgroup$ – lulu Sep 9 '18 at 14:26
  • $\begingroup$ Oh, that is a shorter approach! But I think I have to calculate all digits up to the digits I need, right? $\endgroup$ – Volker Siegel Sep 9 '18 at 14:34
  • $\begingroup$ @lulu I think you mean $\frac{abc\cdots xyz00+\pi}{10^{N+2}}$. Just $\frac\pi{10^{N+2}}$ on its own it's usually way too close to $0$. $\endgroup$ – Arthur Sep 9 '18 at 14:35
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    $\begingroup$ @Arthur. Right. I should have added....since there are transcendentals arbitrarily close to $0$ there are transcendentals arbitrarily close to every rational (since a rational plus a transcendental is again transcendental). $\endgroup$ – lulu Sep 9 '18 at 14:40
  • $\begingroup$ @Arthur Aceppted , that creates infinitely many transcendentals. I'm curious whether the method in the question works too. $\endgroup$ – Volker Siegel Sep 9 '18 at 15:36
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Yes.

Actually, more is true: practially all of such generated numbers can't be expressed by any finite formula.

Here "practically all" means, if you would have a machine capable to generate infinite many digits, and then analyze that the number is transcendent, you would have 100% probability that it would be.

The proof is very simple: all the formula we use are a finite combination of finitely many symbols. All the programs we develop to generate a number, are a finite combination of finite symbols. Thus, the count of anything what we can express, is $\aleph_0$. But the count of the real numbers are $\aleph_1$. They are "inifitely times more".

Transcendence only reduces the possible expressions to the equations with integer coefficient.

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