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Consider the interval $[0,1]$. What is the probability that a number chosen at random in $[0,1]$ is transcendental?

Please give me some points on how to start this problem.

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    $\begingroup$ How many algebraic numbers are there? $\endgroup$ Nov 24, 2014 at 18:07
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    $\begingroup$ Usually we'd ask "which distribution?", but if only the distribution is continuous, then a number chosen from it will be transcendental with probablity 1. $\endgroup$ Nov 24, 2014 at 18:12

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I realize this question is old but I wanted to give my explanation:

  1. The measure of the unit interval $[0,1]$ is $1$.
  2. Cantor proved that the algebraic numbers are countable, and thus the measure of the set of algebraics on $[0,1]$ is the countable union of the measures of the singletons. Since each singleton has measure zero, the set of algebraic numbers on $[0,1]$ has measure zero.
  3. Since the algebraic numbers and the transcendental numbers both partition $[0,1]$, and the set of algebraic numbers has measure zero, the set of transcendental numbers on $[0,1]$ must be of measure $1$.

So the probability of picking a transcendental number on the unit interval is $1$. Measure of a subset of $[0,1]$ more or less translates to "probability".

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Here is a point to start with:

  • There are infinitely countable many algebraic numbers in the interval $[0,1]$
  • There are infinitely uncountable many transcendental numbers in the interval $[0,1]$
  • Add your part here...
  • Hence the probability of choosing a transcendental number in the interval $[0,1]$ is $1$
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Although the other answers to your question are standard and correct, since obviously if the algebraic numbers in an interval are countable and have measure zero,and the reals are uncountable and have a measure, in this case = 1, then the transcendentals, which are simply the reals less the algebraics, must also be uncountable and have a measure = 1. As to the probability of randomly picking a transcendental over an algebraic, that would indeed seem to be 1, since the set of transcendentals has cardinality "c" while the algebraics only have cardinality "ℵ0," thus the first set should be "infinitely" more numerous, leading them to have a probability of 1, -- altho' I have actually never seen a rigorous proof of that claim.

Much more interesting than that question, however, is something that lies latent in your statement, namely; "a number chosen at random in [0,1]." That is highly problematic, at least to me, since it is not at all clear how one could possibly actually go about choosing a "random" real number at all. That is because one cannot give any finite description of any one of the uncountable transcendental numbers. All one can do is specify a small interval, say using a decimal expression, in which the number to be chosen must reside. But from any such specification an uncountable number of further and finer interval continuations exist. In fact, there is, with only a few exceptions, no way to specify such numbers uniquely at all, but unless one can do so, it is meaningless to ask whether what one cannot specify is algebraic or transcendental! Indeed, given any long string of decimal digits that terminates, someone else could reasonably argue that the string represents a rational number, since merely by taking the existing string and repeating it ad infinitum one could produce such a number. All in all then the transcendental numbers remain rather mysterious. One can infer their existence, but producing an example of a "random" one would seem to be impossible.

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  • $\begingroup$ It is possible to specify some transcendental numbers exactly, e.g. $\pi$ is (among other things) the first positive zero of the $\sin$ function. But if we require a "specification" to be expressible as a finite sequence of, say, ASCII characters, there are only countably many of those, so it's still true that almost all transcendental numbers don't have such a specification. $\endgroup$ Oct 16, 2022 at 16:13

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