I see what you’re saying (assuming you mean a random integer from 0 to infinity), but it couldn’t really, since there’s no such thing as an integer with infinite digits - any random integer will have finite number of digits.
The real problem is there’s no way to choose a random number from 0 to infinity. Every finite number has a probability of 0, and in fact, for any number you choose, there is 0 probability that it will be less than that number. Note that 0 probability is different from “impossible” - see en.wikipedia.org/wiki/Almost_surely
are we counting an infinite number of zeroes after the decimal point, as having infinite digits? because if you specifically exclude while numbers, your output would not be truly random, though it would be essentially impossible to distinguish it from true randomness.
Furthermore, “arbitrarily large” also does not mean “infinitely large”. For example, although prime numbers can be arbitrarily large, an infinitely large prime number does not exist—since all prime numbers (as well as all other integers) are finite.
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
No, if truly random it could be any number from 0 to infinity. The randomization doesn’t impart any qualities to the selected number.
If you randomly selected numbers from the infinite range of numbers for an infinite number of time, you would get a result of “7” just as often as getting “3.456e11”.
The probability of getting a finite number is pretty much zero.
For any range [0; n], where n is finite, there are always infinitely many numbers larger than n, so the probability of getting a number in said range is n/(n+infinity). I feel very confident in saying that something with that probability will never happen.
The probability of getting any number with a given set of characteristics is pretty much 0, but that doesn’t mean the number doesn’t exist once generated.
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