Hacker News Reader

Comment by option | Hacker News Reader

Note that the probability of randomly picking a rational number from [0,1] is exactly 0. That is, with P=1 you will end up with an irrational number, not representable in any modern computer.

View on HN

It seems paradoxical, but that's why we have the notion of "almost surely". https://en.wikipedia.org/wiki/Almost_surely

The event set is non-empty (i.e. there are rational numbers between 0 and 1) but the probability of picking one at random is zero.

That's a much less profound statement, though, because you're talking about an implementation detail of "modern computers".

First of all, the set of values in the range [0,1] that can be represented by a computer is not at all the same as the set of rationals. If you're talking about fixed-point or floating-point data types, then there are only a finite number of bit patterns (corresponding to dyadic rationals), so obviously the probability of hitting one of them precisely is zero.

You may think I'm just being pedantic here, but I'm really not. The theorem you described is only interesting because there are an infinite number of rationals between zero and one, and yet the set still has measure zero. But if you're willing to say that any rational number is theoretically "representable" in a computer, then you're already assuming that we can go beyond the natively-available data types, and come up with our own representations -- e.g., a tuple of numerator and denominator. And in that case, we can also come up with representations of irrational numbers; for example, you could have a datatype that can finitely specify any algebraic number, such as sqrt(2) or the golden ratio. So the fact that a typical CPU happens to only be able to natively deal with rationals is irrelevant.

The beauty of the diagonalization argument described in this paper is that it applies to all possible discrete representations, no matter how weird or complex, without having to make any assumptions.

Is concept of "infinity" viable at all considering that number of states of our Universe is huge but finite. And so is any vocabulary for number forms - it is finite.

"probability of randomly picking a rational number from [0,1] is exactly 0."

If to consider that as a lemma only ...

You cannot and will never be able to prove that, right?

This is a mind-blowing. So rational numbers are an invention of humans. That is to say, natural numbers simply don't exist unless and until they are explicitly defined. Am I understanding correctly?

The answer to your question kinda gets into question-begging about what you mean about "invention". But, nominally, all numbers you know are inventions of humans. Exactly how well they correspond to the real universe is still an open problem. Obviously in many cases the answer is "well enough that you need not worry about the deviations very often", but things like "Is the universe continuous or discrete at the core, or some hybrid?" are open questions, which has direct implications on the question of how "real" any of the numbers are. Even the naturals may be quite artificial.

Someone else responded to me in a recent Hacker News discussion that really clarified this in my head: A real number essentially has an infinite number of digits after the decimal place - the difference between a rational and an irrational number is that the digits end up in a repeating pattern in a rational number (you can think of rationals that terminate as really having an infinite number of zeros, e.g. 1.5 as 1.5000000...). Thus, to randomly pick a real number, you randomly select digits an infinite number of times after the decimal. Clearly there is no way a random process will produce an infinite number of repeating digits, thus the probability of picking a rational would be 0.

Wouldn't it be simpler to explain it this way: there are infinitely many reals between any two different reals, and thus the theoretical probability of picking any number between them at random is 1/infinity, which we think if as 0.

But in that case, why is is more probable to pick an irrational number?

I don't think it's simpler that way at all, mainly because the ways in which some "infinities" are larger than others, which explains why it's more probable to pick an irrational number. The rational numbers are countable, while the real numbers are not.

The most mind-blowing part is that P(E)=0 does not imply E is impossible. My statement above is just a consequence of that. See my Quora's answer to related question https://www.quora.com/Is-there-a-point-where-statistical-imp...

"God created the integers, all else is the work of man." - Leopold Kronecker

Really? Interesting. I would have expected it to be something like epsilon.

I'm not sure if this is entirely mathematically valid, but it's at least intuitively valid. Imagine we're in decimal (for familiarity). Imagine we're generating our random number by drawing digits out of a bag containing the ten digits, then replacing and drawing again. This is of course a magical perfectly uniformly distributed bag.

In order to draw a rational number, you must randomly select a number that has a repeating pattern in decimal. Imagine that for the sake of argument that we're in this "repeating pattern" and, amazingly, we've already drawn it 1000 times! What a roll we're on! In order for the number to be rational, how many more times must we draw this pattern? Infinitely many. What is the probability that in one of those infinite fair random draws, the repeating pattern gets broken? 1. It's just not possible for infinitely many fair draws to produce a repeating pattern; that would be proof that the process generating the number was in fact not random in the first place. (I can't be that sloppy if we're doing finitely many draws but I believe that's justifiable at infinity.)

Imagining through this process of drawing a random number I think makes this more intuitively obvious than imagining being presented with a completed number and trying to figure out if it's rational.

it is entirely mathematically valid. See here for some math https://www.quora.com/Is-there-a-point-where-statistical-imp...

The catch, here is "selecting a number randomly" - that isn't constructively defined. And to me this looks kind of similar to the notion of "measurement" in quantum theory - not precisely defined either.

The measure of epsilon is 0.

Proof: what else could it be? If it's not 0, there's a smaller number, contradicting your (intuitive) definition of epsilon.

Wouldn't it be more accurate to describe epsilon as an infinitesimal?

There are no infinitesimals in the reals. You can define a mathematical structure containing infinitesimals [0], but it's not the reals.

[0] Specifically, the dual numbers: https://en.wikipedia.org/wiki/Dual_number

Interesting. How do you describe the difference between two successive real numbers?

> How do you describe the difference between two successive real numbers?

There is no such things as «two successive real numbers».

Yeah, there aren't even two successive dual numbers. What's after 1? If it's 1 + epsilon, then what about 1 + epsilon / 2?

Epsilon is a variable, not a number, so that wouldn't make sense. The probability is less than epsilon for all epsilon > 0. The only non-negative real (and we know the probability must be some non-negative real) satisfying that is 0.

Reply on news.ycombinator.com