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I'm an undergrad taking some measure theory right now. This is probably not exactly the mathematical formalism they are using, but it could be similar. If I a

I'm an undergrad taking some measure theory right now. This is probably not exactly the mathematical formalism they are using, but it could be similar. If I asked you to pick a random real number from 0 to 1, what do you think the probability is that the real number is rational? A natural way to answer this question is to try to generalize the way we say the interval from 0 to 1 is length 1 to more kinds of sets. Measure theory does exactly this, but we find out that the measure of the rational numbers is 0! This means that the probability of picking a rational number is 0. But that's clearly impossible you say, because there are an infinite number of rational numbers from 0 to 1! But in a precise mathematical way, the probability is 0. Now a funny think I said was "pick a random real number". Since the computable numbers are also a measure 0 subset of the real numbers, it's literally impossible to randomly pick a real number with a computer...

darawk 8y ago
Yes, but that doesn't mean that there are only irrational numbers. There are rational numbers, they just have zero measure. Similarly, these curves come in more than two types - those other types just have zero measure in the probability space of the curves.
aportnoy 8y ago
Not only that, we should expect to always get an irrational number, but always get rationals from a finite subset of Q. I recently asked a closely related question on Math.SE: https://math.stackexchange.com/questions/2952087/sampling-from-a-continuous-distribution/ https://math.stackexchange.com/questions/2952087/sampling-fr...
cix_pkez 8y ago
Hey, non-mathematician computer science type here. If I follow correctly, the issue with randomly picking any real number in that interval is that irrational numbers would require infinite computational steps to resolve. So the probability is really 0 that you'll get an irrational. If you have a finite number of computations, you're guaranteed to resolve to a rational, while if you have an infinite number of computations, you never resolve to anything. Is that a decent lay interpretation?
- aportnoy 8y ago
  The opposite is true. Take the interval [0, 1], "sample a random real number x" from it. P(x is rational) = 0, P(x is irrational) = 1. Informally, the overwhelming majority of real numbers are irrational.
  - mavhc 8y ago
    What's weird is there's always a rational number between any 2 irrational numbers, yet there are way more irrational numbers
    - lisper 8y ago
      All this becomes pretty intuitive if you think about numbers in terms of their decimal expansions. A "random number" is one with a random digit in each of its (infinitely long) list of decimal digits. A rational number is one where, at some finite point, all of the digits start to repeat in some finite pattern. The odds of that happening by chance are zero. Likewise, if you take any two randomly generated list of decimal digits, at some point there will be a digit in the same place that is different between the two. At that point, you can construct a rational between the two by choosing the smaller digit and then adding "1000....". But yeah, it's kind of weird.
    - thaumasiotes 8y ago
      Corollary: rational numbers must serve double duty separating many different pairs of irrational numbers. You can keep going with things that sound weird: There are infinitely many rationals between any two irrational numbers. There are as many rational numbers between any two finite irrational numbers as there are between positive and negative infinity.
      - throwawaymath 8y ago
        Can you clarify that last point? I don't think it's strictly true. Aside from countable and uncountable infinities, you can have larger and smaller infinites as well. Unless every set S of all rational numbers between any irrational x and irrational y is isomorphic to the set P of all rational numbers, I don't see that this is correct. And I don't immediately see that you can put them into 1-1 correspondence.
man-and-laptop 8y ago
> Now a funny think I said was "pick a random real number". Since the computable numbers are also a measure 0 subset of the real numbers, it's literally impossible to randomly pick a real number with a computer... I dispute that. See this: http://math.andrej.com/2008/02/06/representations-of-uncomputable-and-uncountable-sets/ http://math.andrej.com/2008/02/06/representations-of-uncompu... [edit] The downvoters can't argue with facts. Let me hear them. I am not deleting this comment.
- gpm 8y ago
  That blog post was a good read. I also wanted to say that I certainly believe that you've done real math and got interesting results from the sort of model we were discussing above. It does seem like, as you say, a rich and natural theory. That said I wanted to make another counter argument here. Particularly based on the meaning of the phrase "pick a random real number". To pick a number at random is a necessarily randomized process, however it should give a classical output. After picking the number I should be able to make two copies of it, put them on ships flying away from eachother at the speed of light, and both ships should be able to perform the same computation (without entangling particle pairs beforehand). Moreover any reasonable representation of a real number should admit operations like "what is its n^th digit". Under these two constraints, you would have to flip an infinite number of coins, and it would take infinite time and space to pick a real number. Or in other words, you can't really.
  - man-and-laptop 8y ago
    The way your spaceship analogy works in TTE, is that the random real number would be some kind of radio source that gets listened to by the two ships. Also, the way TTE works, there are diminishing returns from listening to the radio source for longer and longer time - this is because all functions computable in TTE are continuous, and therefore have diminishing returns from knowing their input to greater and greater accuracy (this is exactly the difference between continuous functions and discontinuous functions). Therefore you only have to listen to the radio source for some time before saying "Yep, I'm happy, I know the output of my function to within epsilon". Even the value of Pi in physics only needs to be known to a small number of decimal places - which suggests that all functions that are "physically computable" are continuous, further suggesting how natural TTE is. Again, my comments are getting downvoted by ignoramuses. This is not the first time.
deleted 8y ago
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EGreg 8y ago
The trouble is with your definition of picking. If the selection process involves digital steps then for sure you can ONLY pick computable numbers. But if the selection process involves some analog movement then it’s an interesting open question — my guess is that because of Planck Length, you can only pick rational numbers. So in what sense do irrational numbers or transcendental numbers or imaginary numbers exist? They are useful abstractions only. Who was it that said “God created the integers, man invented everything else”?
jchanimal 8y ago
Take is why I think AI can’t have qualia. They aren’t embodiesd.
nightcracker 8y ago
The much more interesting question is, what is the density of the rational numbers in the set of computable numbers.

darawk 8y ago

Yes, but that doesn't mean that there are only irrational numbers. There are rational numbers, they just have zero measure. Similarly, these curves come in more than two types - those other types just have zero measure in the probability space of the curves.

aportnoy 8y ago

Not only that, we should expect to always get an irrational number, but always get rationals from a finite subset of Q. I recently asked a closely related question on Math.SE: https://math.stackexchange.com/questions/2952087/sampling-from-a-continuous-distribution/ https://math.stackexchange.com/questions/2952087/sampling-fr...

cix_pkez 8y ago

Hey, non-mathematician computer science type here. If I follow correctly, the issue with randomly picking any real number in that interval is that irrational numbers would require infinite computational steps to resolve. So the probability is really 0 that you'll get an irrational. If you have a finite number of computations, you're guaranteed to resolve to a rational, while if you have an infinite number of computations, you never resolve to anything. Is that a decent lay interpretation?

aportnoy 8y ago
The opposite is true. Take the interval [0, 1], "sample a random real number x" from it. P(x is rational) = 0, P(x is irrational) = 1. Informally, the overwhelming majority of real numbers are irrational.
- mavhc 8y ago
  What's weird is there's always a rational number between any 2 irrational numbers, yet there are way more irrational numbers
  - lisper 8y ago
    All this becomes pretty intuitive if you think about numbers in terms of their decimal expansions. A "random number" is one with a random digit in each of its (infinitely long) list of decimal digits. A rational number is one where, at some finite point, all of the digits start to repeat in some finite pattern. The odds of that happening by chance are zero. Likewise, if you take any two randomly generated list of decimal digits, at some point there will be a digit in the same place that is different between the two. At that point, you can construct a rational between the two by choosing the smaller digit and then adding "1000....". But yeah, it's kind of weird.
  - thaumasiotes 8y ago
    Corollary: rational numbers must serve double duty separating many different pairs of irrational numbers. You can keep going with things that sound weird: There are infinitely many rationals between any two irrational numbers. There are as many rational numbers between any two finite irrational numbers as there are between positive and negative infinity.
    - throwawaymath 8y ago
      Can you clarify that last point? I don't think it's strictly true. Aside from countable and uncountable infinities, you can have larger and smaller infinites as well. Unless every set S of all rational numbers between any irrational x and irrational y is isomorphic to the set P of all rational numbers, I don't see that this is correct. And I don't immediately see that you can put them into 1-1 correspondence.

man-and-laptop 8y ago

> Now a funny think I said was "pick a random real number". Since the computable numbers are also a measure 0 subset of the real numbers, it's literally impossible to randomly pick a real number with a computer... I dispute that. See this: http://math.andrej.com/2008/02/06/representations-of-uncomputable-and-uncountable-sets/ http://math.andrej.com/2008/02/06/representations-of-uncompu... [edit] The downvoters can't argue with facts. Let me hear them. I am not deleting this comment.

gpm 8y ago
That blog post was a good read. I also wanted to say that I certainly believe that you've done real math and got interesting results from the sort of model we were discussing above. It does seem like, as you say, a rich and natural theory. That said I wanted to make another counter argument here. Particularly based on the meaning of the phrase "pick a random real number". To pick a number at random is a necessarily randomized process, however it should give a classical output. After picking the number I should be able to make two copies of it, put them on ships flying away from eachother at the speed of light, and both ships should be able to perform the same computation (without entangling particle pairs beforehand). Moreover any reasonable representation of a real number should admit operations like "what is its n^th digit". Under these two constraints, you would have to flip an infinite number of coins, and it would take infinite time and space to pick a real number. Or in other words, you can't really.
- man-and-laptop 8y ago
  The way your spaceship analogy works in TTE, is that the random real number would be some kind of radio source that gets listened to by the two ships. Also, the way TTE works, there are diminishing returns from listening to the radio source for longer and longer time - this is because all functions computable in TTE are continuous, and therefore have diminishing returns from knowing their input to greater and greater accuracy (this is exactly the difference between continuous functions and discontinuous functions). Therefore you only have to listen to the radio source for some time before saying "Yep, I'm happy, I know the output of my function to within epsilon". Even the value of Pi in physics only needs to be known to a small number of decimal places - which suggests that all functions that are "physically computable" are continuous, further suggesting how natural TTE is. Again, my comments are getting downvoted by ignoramuses. This is not the first time.

deleted 8y ago

[deleted]

EGreg 8y ago

The trouble is with your definition of picking. If the selection process involves digital steps then for sure you can ONLY pick computable numbers. But if the selection process involves some analog movement then it’s an interesting open question — my guess is that because of Planck Length, you can only pick rational numbers. So in what sense do irrational numbers or transcendental numbers or imaginary numbers exist? They are useful abstractions only. Who was it that said “God created the integers, man invented everything else”?

jchanimal 8y ago

Take is why I think AI can’t have qualia. They aren’t embodiesd.

nightcracker 8y ago

The much more interesting question is, what is the density of the rational numbers in the set of computable numbers.