OP is wrong. A truly random real number does have a much higher probability of being an irrational number or repeating rational number, but it is certainly not the case that a truly random number “will be” one of these two as terminating rational numbers are still possible to select.
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
OP is wrong. A truly random real number does have a much higher probability of being an irrational number or repeating rational number, but it is certainly not the case that a truly random number “will be” one of these two as terminating rational numbers are still possible to select.
There an infinite number of numbers that have infinite length and are not irrational or repeating. Infinity is larger than youre imagining.
Are you referring to arbitrarily large numbers? Still essentially the same as decimals in the other direction.
Do you have a mathematical proof for the OP’s claim that a truly random number must have infinite digits?
you’re claiming OP is wrong, you need the proof homie
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:
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.
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)
Probability of both being 0:
then for n 0s
Pn = (1/10)^n
as n -> inf, Pn -> 0
put another way, (1/10)^inf = 0