Distribution of irrational or transcendental numbers

Lets say we select a random number $$0.abc.......$$ between $$[0,1]$$ where each digit after decimal is independently drawn from a fixed discrete probability distribution ($$\sum P(X=n)=1$$ for $$n\in\{0,1,...9\}$$). Obviously, $$0=0.\bar{0}$$ and $$1=0.\bar{9}$$ are included in this range. What is the probability that a random number is irrational/transcendental? Are there infinitely more transcendental numbers than irrational numbers? Can we get a closed-form solution for uniform distribution? Is it possible to select a distribution such that it always generates an irrational or rational number? Clearly, it is possible to generate only rational numbers by fixing the distribution to a fixed point mass.

The rational numbers are merely countably infinite, while the real numbers are uncountably infinite, so the probability of a number in the interval [ 0 , 1 ] being irrational is "certainty".

The algebraic numbers are also countably infinite, so the probability of a number in that interval being transcendental is also "certainty".

• Thanks. Is there any proof of algebraic numbers being countably infinite? Commented Feb 5, 2021 at 8:21
• It is similar to the proof of countability of the rational numbers: the number of polynomials of any degree is countably infinite, since each term has an integer coefficient, and polynomials each have an integer number of terms. The number of zeroes (which are algebraic numbers) of each polynomial is an integer, so we have a "Cartesian product" of countably infinite sets.
– user882145
Commented Feb 5, 2021 at 8:25
• Makes sense! thanks a lot for simple explanation Commented Feb 5, 2021 at 8:27
• The part that always "feels weirdest" to me is that once you accept the concept of the real numbers, "almost none" of them is a rational or even an algebraic number...
– user882145
Commented Feb 5, 2021 at 8:32
• So true! Very interesting indeed. Most numbers are wierd it seems Commented Feb 5, 2021 at 8:36