Well to begin with this is irrelevant nonsense. The
finite decimal representations, i.e. those representations
that stop somewhere, have the form:
d/10^n
So they are a subset of the rational numbers. And the
irrational numbers, are all those real numbers, which
are not rational numbers.
So their union gives the reals:
I union Q = R
And they are disjoint:
I intersect Q = {}
So they form a partition of the real numbers into
two kind of numbers. Which can be also written
as follows:
R \ Q = I
Now the algeraic irrational numbers (AI) and the
transcendental numbers (T) form a further partition,
of irrational numbers I. We can write:
R \ Q \ AI = T
Or if we use algebraic numbers (A) as an umbrella
for algeraic irrational numbers (AI) and rational
numbers (Q), we get:
R \ A = T
Now R is uncountable, A is countable, so T will
be uncountable as well. So the probability, that
a prefix belongs to a transcendental number is:
The probability is 0
Hint A is countable because the polynomials
Z[X] are countable. Now P(B | C) = P(B n C) / P(C) =
P(B n C) / (P(B n C) + P(~B n C)). We can simulate
the event that we fix a prefix,
by using C=[0,1), this is the same as a prefix "0.",
Further use Rn=[-n,n) and B_n = Rn intersect T, we
then have:
P(B_n | C) = 1/(2n) / (1/(2n) + (2n-1)/n)
= 1/2n
So that we have for lim n->oo Rn = R, finally:
P(T | C) = lim n->oo 1/2n = 0
The problem is that we allow an arbitrary transcendental
number. We have also readily from algebraic numbers the
same result, on other ground that P(B n C) = 0 already:
P(A | C) = 0
So it doesn't say something about transcendental numbers.
If we restrict ourselfs from the beginning to lets say
transcendental numbers in the interval [0,1) different
probabilities.
https://math.stackexchange.com/q/37121/4414
https://math.stackexchange.com/q/2070246/4414