That’s not how that works.
this post was submitted on 12 Nov 2023
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Showerthoughts
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Doesn't it depends on whether we are talking about real or integer numbers?
EDIT: I think it also works with p-adic numbers.
No. In the set of real numbers it is still very possible to randomly select a number that can be written with finite digits.
op is right, infinity is larger than you're imagining
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:
- 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
No it wouldn't. You don't understand how random works
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.
If there's any probability of an event, on an infinite timeline, it occurs infinite times.
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 https://en.wikipedia.org/wiki/Almost_surely
I think it's right
Edit:
TIL: when saying random numbers, some people think to integers, others to real numbers.
I also think that's correct... if we are talking about real numbers.
People are probably thinking about integers. I'm not sure about OP.
EDIT: I think it also works with p-adic numbers.
Yes real numbers, but as far as I'm aware it'll happen for integers too almost surely
I think you're confusing "arbitrarily large" with "infinitely large". See Wikipedia Arbitrarily large vs. (...) infinitely large
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.
For integers I disagree (but I'm not a mathematician). The set of integers with infinite digits is the empty set, so AFAIK, it has probability 0.
Wait, are you high in the shower? Because that's a little different.
why?
Oh my God it's spez
Is he here to fuck lemmy up too?
👋 👋
Fuck you!
fuck you
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.
Why would it have infinite digits? It is just as like to be 0.
yes, don't listen to the biologists who tell you otherwise.