OP actually has the burden to prove their own claim, but here you go:
Suppose we create an algorithm to generate a random number, such that:
- The first digit is the ones
- The second digit is the tenths
- The third digit is the tens
And so on. For example, if we generated the sequence 1, 2, 3, 4, 5, 6 it would represent the number 531.246.
For a number to be non-infinite, there must be at some point be a digit where all digits after it generate a 0.
For all numbers in our sequence, the probability of generating a 0 is 1/10: there is no point at which we cannot generate a 0. Furthermore, after the first 0 is generated at a, the odds of a+1 being 0 are also 1/10, as are the odds of a+2, a+3, and a+n. So we cannot identify a b, such that entry a+b must be >0, since the odds of any given a+b generating 0 are also 1/10.
Based on this, we can use induction to show that it is possible to generate a truly random number that is a terminating rational number, and indeed it is possible to show this for any specific number as well. For example, the number 2 can be generated by simply rolling “2, 0, 0, 0, 0, …” and there is no nth digit in the sequence that cannot be generated at 0, since the odds of any given n being 0 are still 1/10.
For a number to be non-infinite, there must be at some point be a digit where all digits after it generate a 0.
For all numbers in our sequence, the probability of generating a 0 is 1/10: there is no point at which we cannot generate a 0. Furthermore, after the first 0 is generated at a, the odds of a+1 being 0 are also 1/10, as are the odds of a+2, a+3, and a+n. So we cannot identify a b, such that entry a+b must be >0, since the odds of any given a+b generating 0 are also 1/10.
the odds of randomly selecting 0 exactly an infinite number of times is exactly zero which is why OP is right
Probability of a=0 is (1/10)
- Pa = (1/10)
- Pb = (1/10)
Probability of both being 0:
- Pa AND Pb =(1/10)*(1/10)
then for n 0s
Pn = (1/10)^n
as n -> inf, Pn -> 0
put another way, (1/10)^inf = 0
