7 ms·

Can we go further and ditch the reals, relying instead on rational numbers or even IEEE floats? After all, the computers that we use for predicting empirical re

Can we go further and ditch the reals, relying instead on rational numbers or even IEEE floats? After all, the computers that we use for predicting empirical results all run on integers.

PaulDavisThe1st 5y ago
If you think you can describe the physical world without any irrational numbers, when one of the most basic (the ratio of the area of a circle to its diameter) is irrational, I think you're probably mistaken.
- Koshkin 5y ago
  Unless we live in a computer simulation, of course, in which case we have to content ourselves with the IEEE floats.
  - analog31 5y ago
    God made the IEEE floats, and all the rest is the work of man -- Apologies to Kronecker
- btilly 5y ago
  You can describe it to within measurement error without any irrational numbers. And with fewer decimal places than you'd probably imagine. See https://www.jpl.nasa.gov/edu/news/2016/3/16/how-many-decimals-of-pi-do-we-really-need/ https://www.jpl.nasa.gov/edu/news/2016/3/16/how-many-decimal... for more.
  - midjji 5y ago
    That often works, but not always, some systems generate sequences of operations which can be symbolically simplified, or equivalently could be exactly computed using reals, but which if computed using any finite precision will fail. A simple example would be solving for the position of a planet in orbit under simple Newtonian gravity, a sufficient number of revolutions latter. For any finite precision the number of orbits can always be increased until failure occurs. Its a cool how few bits of pi are required to compute the circumference of the earth to within an atoms width accuracy. But equally cool how bad it gets if you try the same to compute the position of earth 2^64 years later.
    - btilly 5y ago
      That logic is disingenuous at best. Planetary orbits are chaotic. Long before your imprecision in pi is going to significantly mislead you, shifts in mass due to, for example, earthquakes and weather patterns are going to cause orbits to be impossible to predict. There are theoretical systems where the exact value of pi matters. But no physical system is going to match that, and measurement error is going to quickly exceed calculation errors from pi.
      - TheOtherHobbes 5y ago
        Pi is only incidentally a number. Pi is an abstraction that represents a certain relationship. (Actually more of a set of relationships.) If you define pi as a specific constant with limited precision - because "that's all physical systems need" - you lose insights into the web of relationships around it. This is a bad thing and makes many kinds of math harder. It's the conceptual equivalent of lossy data compression. You don't want to do it unless you really, really need to. And if you do it, you need to be aware that you're now using approximations instead of abstractions, and those are not the same thing.
      - wruza 5y ago
        But when you remove or take these fluctuations into account, you’re still left with an error. This rational model has no chance to ever be correct computationally, unless you cheat and add more detailed ratio every time you see a loop. Also, how exactly will you define rational pi? Let’s start with 3/1, why go any further. If it doesn’t represent reality (draw a circle and measure it with a string), well, strings have vague length anyway. See where this is going? The bigger problem is analyzis will not work, probably. Can you do analyzis with rationals? E.g. f(x)=x^2 isn’t continuous at f(x)=2/1. I’d theorize that there is a way to do finite calc without giving up on “reals”, by using enough FT coefficients (and packing this complexity into sine waves), but it’s the same sort of cheating probably, unless pi is the “origin” number of all “physical” reals. (I’m not a physicist nor math guy, so one can reframe my ideas as questions instead.)
        midjji 5y ago
- MauranKilom 5y ago
  ...but Planck length!
  - EGreg 5y ago
    Perhaps that’s where quantum uncertainty reaches a threshold where you can’t make time based distinctions anymore.
- roywiggins 5y ago
  Pi is the ratio of a mathematical circle to its diameter, but there are no physical circles which have that ratio as exactly Pi, and even if one existed, you'd never be able to distinguish it from one that was merely equal to Pi to the accuracy you are capable of measuring, because you'd need to measure with infinite precision, which you can't.
  - kadoban 5y ago
    Orbits over long stretches of time seem likely to need arbitrarily high amounts of accuracy in pi. Sure you can just pick a rational number close enough for the accuracy you need, but why should the definition of pi need to change based on what you're measuring?
    - roywiggins 5y ago
      Every simulation picks some rational approximation to Pi, because they have to. Either they will run out of time or space or collapse into a black hole before needing more than a finite number of decimal places, so for all plausible purposes we can make do with the first googleplex digits (or whatever) of Pi. I guess my argument is, since you can always just pick a rational approximation to Pi, you cannot prove empirically that we live in a universe where more than a finite number of digits of Pi matter. That is, the mathematical irrationality doesn't really matter, physically speaking, since no experiment could ever prove that every digit in Pi actually contributes to the result. If the universe does have ways to do this, to mix an entire irrational number into a physical outcome, that means hypercomputation is probably possible, since Turing machines definitely can't. https://en.m.wikipedia.org/wiki/Real_computation https://en.m.wikipedia.org/wiki/Real_computation
      - zozbot234 5y ago
        > Every simulation picks some rational approximation to Pi, because they have to. Your simulation might need to ask for an increasingly tighter bound on the real value of Pi. You can totally do this with no more than the usual rational numbers, but it's not equivalent to "just picking some rational approximation" and running with it, because what accuracy/precision you pick is outcome-dependent and it's always possible to request more.
- p2t2p 5y ago
  That is replacing actual reality with math all over again. In the actual reality this ratio is concrete, limited value. I would say that our numbers and approaches are limited that they are unable to precisely define it. It just we are so used to it that no one even thinks about challenging it. And even if they do - good luck finding funding. Now we do have tools that overcome those limitations somewhat - like limits and stuff but doesn’t remove the need for better tools. So I somewhat disagree with both of you - we need new, better numbers that are further from math abstraction and closer to actual reality. Otherwise it is like this story with Poynting vector from Veritasium video - only confuses instead of explaining.
  - drdec 5y ago
    Exactly. The square root of 2 had no more basis in reality than the so-called imaginary number i. Don't confuse the "real" in real numbers with reality, at this point it is just a name.
- int_19h 5y ago
  If the physical world is entirely quantized, including spacetime itself - which, as I understand, is still considered a valid hypothesis - then wouldn't it be possible to describe it using integers, pretty much by definition?
katmannthree 5y ago
Can you? Most likely. Should you? You’ll need to reprove more than a handful of theorems, and for what? What advantage does using rationals instead of reals get you? You might enjoy taking courses in real & complex analysis, the general purpose of which is to impart upon the receiver an understanding of why we’ve constructed those particular number systems and how despite the names they both describe things which are perfectly real in the philosophical sense. Edit: recall that the rationals are simply defined as the set of numbers which can be represented in the form a/b where a is an integer and b is a nonzero integer. This isn’t some deep philosophical tie to an underlying reality, it’s just our first attempt at defining more useful numbers that lie between other useful numbers we already invented (the integers et al). The real and complex numbers are literally just the extension of that process, filling in holes between useful numbers with more numbers until the set is closed (i.e. there are no more holes, every operation between members of the set results in another member of the set). Closure, the real reason we care so much about the reals, is just a surprise tool that helps us later. For anyone who made it this far: if you find any of this interesting you should find a book or lecture on analysis. It’s not particularly difficult and presents deeper insights into the math you likely already learned.
- Vetch 5y ago
  I enjoyed your comment and agree with all your well put points except one: on the rationals tie with reality. Having implemented exact real computation to better understand reals, I think of a real number as a kind of machine that generates infinite streams. Operations on them instantiate new machines which query their real operands, computating until there's sufficient information to emit a next term of the stream. When the next term needs an infinite amount of information to decide what to spit out next, it results in an "unproductive" infinite loop. Rational numbers are interesting, more realistic, because they always terminate. In the real world, measurement tolerances and physical limits means at some point having to extract a rational. When we work with reals we are really only working with rational approximations or symbols with associated properties and relations. Reals are a powerful and elegant tool to rigorously reason about mathematical spaces and operations on algebraic objects but trying to work with them in reality in their exact form is a fun and visceral lesson on the nature of undecidability. It's hard to go two steps without tripping over a non-terminating loop (such as any operation that starts with an irrational and results in a rational or equality testing in general). This is an observation on our interface with reality and not on its true nature, which may or may not admit reals (although my non-serious guess is that black holes form whenever you try to do something that requires a proper real number).
deleted 5y ago
[deleted]
matt123456789 5y ago
Most reals are not computable anyway: https://arxiv.org/pdf/math/0411418.pdf https://arxiv.org/pdf/math/0411418.pdf
- deleted 5y ago
  [deleted]
sillysaurusx 5y ago
You can't. See "Simulating Physics With Computers," linked here: https://twitter.com/theshawwn/status/1394159744139632640 https://twitter.com/theshawwn/status/1394159744139632640 And some thoughts on the limitations with respect to creating AGI: https://twitter.com/theshawwn/status/1446261451061145602 https://twitter.com/theshawwn/status/1446261451061145602 See section 5, "Can quantum systems be probabilistically simulated by a classical computer?" > The probability that they match is eight-tenths, the probability that they mismatch is plus two-tenths; every physical probability comes out positive. But the original f's are not positive, and therein lies the great difficulty. The only difference between a probabilistic classical world and the equations of the quantum world is that somehow or other it appears as if the probabilities would have to go negative, and that we do not know, as far as I know, how to simulate. Okay, that's the fundamental problem. I don't know the answer to it, but I wanted to explain that if I try my best to make the equations look as near as possible to what would be imitable by a classical probabilistic computer, I get into trouble. When I was younger and only slightly more foolish, I wanted to spend a lot of time researching this to see if there was a way around the problem. I quickly realized that perhaps I should focus on adding value in a field that I was good at. :) Maybe one of you can try, since the only alternative is to take Feynman at his word.
- aeternum 5y ago
  The claim that discretization is not equal to quantization is a key claim and I'm not sure the paper actually proves/shows this. If it turns out that spacetime is discrete and not continuous then we should be able to simulate it. The possible states of some chunk of spacetime (given some maximum/known energy) would be non-infinite and even though it might take us years to calculate each time-step, we could still make progress. I've seen the theory kicked around before, but why couldn't the universe basically consist of plank length sized voxels and a similarly small update timestep? It would still look plenty continuous to us.
  - sillysaurusx 5y ago
    It’s because of the anisotropy. Suppose the universe was a regular grid of voxels. You could run experiments to prove that. I don’t understand the details, but that was Feynman’s counter argument. (See the “messenger lectures” series on YouTube.) If the grid isn’t regular, you run into other problems. But iirc at one point Feynman was toying with the idea that the grid might be randomly distributed.
nocturnial 5y ago
Maybe you're interested in this paper: Discretization of the Bloch sphere, fractal invariant sets and Bell’s theorem https://royalsocietypublishing.org/doi/10.1098/rspa.2019.0350 https://royalsocietypublishing.org/doi/10.1098/rspa.2019.035... There also a video where he explains the paper: https://www.youtube.com/watch?v=YglT09Korr0&t=2700s https://www.youtube.com/watch?v=YglT09Korr0&t=2700s There are some consequences by using Q instead of C that can be experimentally tested.

Can we go further and ditch the reals, relying instead on rational numbers or even IEEE floats? After all, the computers that we use for predicting empirical results all run on integers.

PaulDavisThe1st 5y ago

If you think you can describe the physical world without any irrational numbers, when one of the most basic (the ratio of the area of a circle to its diameter) is irrational, I think you're probably mistaken.

Koshkin 5y ago
Unless we live in a computer simulation, of course, in which case we have to content ourselves with the IEEE floats.
- analog31 5y ago
  God made the IEEE floats, and all the rest is the work of man -- Apologies to Kronecker
btilly 5y ago
You can describe it to within measurement error without any irrational numbers. And with fewer decimal places than you'd probably imagine. See https://www.jpl.nasa.gov/edu/news/2016/3/16/how-many-decimals-of-pi-do-we-really-need/ https://www.jpl.nasa.gov/edu/news/2016/3/16/how-many-decimal... for more.
- midjji 5y ago
  That often works, but not always, some systems generate sequences of operations which can be symbolically simplified, or equivalently could be exactly computed using reals, but which if computed using any finite precision will fail. A simple example would be solving for the position of a planet in orbit under simple Newtonian gravity, a sufficient number of revolutions latter. For any finite precision the number of orbits can always be increased until failure occurs. Its a cool how few bits of pi are required to compute the circumference of the earth to within an atoms width accuracy. But equally cool how bad it gets if you try the same to compute the position of earth 2^64 years later.
  - btilly 5y ago
    That logic is disingenuous at best. Planetary orbits are chaotic. Long before your imprecision in pi is going to significantly mislead you, shifts in mass due to, for example, earthquakes and weather patterns are going to cause orbits to be impossible to predict. There are theoretical systems where the exact value of pi matters. But no physical system is going to match that, and measurement error is going to quickly exceed calculation errors from pi.
    - TheOtherHobbes 5y ago
      Pi is only incidentally a number. Pi is an abstraction that represents a certain relationship. (Actually more of a set of relationships.) If you define pi as a specific constant with limited precision - because "that's all physical systems need" - you lose insights into the web of relationships around it. This is a bad thing and makes many kinds of math harder. It's the conceptual equivalent of lossy data compression. You don't want to do it unless you really, really need to. And if you do it, you need to be aware that you're now using approximations instead of abstractions, and those are not the same thing.
    - wruza 5y ago
      But when you remove or take these fluctuations into account, you’re still left with an error. This rational model has no chance to ever be correct computationally, unless you cheat and add more detailed ratio every time you see a loop. Also, how exactly will you define rational pi? Let’s start with 3/1, why go any further. If it doesn’t represent reality (draw a circle and measure it with a string), well, strings have vague length anyway. See where this is going? The bigger problem is analyzis will not work, probably. Can you do analyzis with rationals? E.g. f(x)=x^2 isn’t continuous at f(x)=2/1. I’d theorize that there is a way to do finite calc without giving up on “reals”, by using enough FT coefficients (and packing this complexity into sine waves), but it’s the same sort of cheating probably, unless pi is the “origin” number of all “physical” reals. (I’m not a physicist nor math guy, so one can reframe my ideas as questions instead.)
      - midjji 5y ago
MauranKilom 5y ago
...but Planck length!
- EGreg 5y ago
  Perhaps that’s where quantum uncertainty reaches a threshold where you can’t make time based distinctions anymore.
roywiggins 5y ago
Pi is the ratio of a mathematical circle to its diameter, but there are no physical circles which have that ratio as exactly Pi, and even if one existed, you'd never be able to distinguish it from one that was merely equal to Pi to the accuracy you are capable of measuring, because you'd need to measure with infinite precision, which you can't.
- kadoban 5y ago
  Orbits over long stretches of time seem likely to need arbitrarily high amounts of accuracy in pi. Sure you can just pick a rational number close enough for the accuracy you need, but why should the definition of pi need to change based on what you're measuring?
  - roywiggins 5y ago
    Every simulation picks some rational approximation to Pi, because they have to. Either they will run out of time or space or collapse into a black hole before needing more than a finite number of decimal places, so for all plausible purposes we can make do with the first googleplex digits (or whatever) of Pi. I guess my argument is, since you can always just pick a rational approximation to Pi, you cannot prove empirically that we live in a universe where more than a finite number of digits of Pi matter. That is, the mathematical irrationality doesn't really matter, physically speaking, since no experiment could ever prove that every digit in Pi actually contributes to the result. If the universe does have ways to do this, to mix an entire irrational number into a physical outcome, that means hypercomputation is probably possible, since Turing machines definitely can't. https://en.m.wikipedia.org/wiki/Real_computation https://en.m.wikipedia.org/wiki/Real_computation
    - zozbot234 5y ago
      > Every simulation picks some rational approximation to Pi, because they have to. Your simulation might need to ask for an increasingly tighter bound on the real value of Pi. You can totally do this with no more than the usual rational numbers, but it's not equivalent to "just picking some rational approximation" and running with it, because what accuracy/precision you pick is outcome-dependent and it's always possible to request more.
p2t2p 5y ago
That is replacing actual reality with math all over again. In the actual reality this ratio is concrete, limited value. I would say that our numbers and approaches are limited that they are unable to precisely define it. It just we are so used to it that no one even thinks about challenging it. And even if they do - good luck finding funding. Now we do have tools that overcome those limitations somewhat - like limits and stuff but doesn’t remove the need for better tools. So I somewhat disagree with both of you - we need new, better numbers that are further from math abstraction and closer to actual reality. Otherwise it is like this story with Poynting vector from Veritasium video - only confuses instead of explaining.
- drdec 5y ago
  Exactly. The square root of 2 had no more basis in reality than the so-called imaginary number i. Don't confuse the "real" in real numbers with reality, at this point it is just a name.
int_19h 5y ago
If the physical world is entirely quantized, including spacetime itself - which, as I understand, is still considered a valid hypothesis - then wouldn't it be possible to describe it using integers, pretty much by definition?

katmannthree 5y ago

Can you? Most likely. Should you? You’ll need to reprove more than a handful of theorems, and for what? What advantage does using rationals instead of reals get you? You might enjoy taking courses in real & complex analysis, the general purpose of which is to impart upon the receiver an understanding of why we’ve constructed those particular number systems and how despite the names they both describe things which are perfectly real in the philosophical sense. Edit: recall that the rationals are simply defined as the set of numbers which can be represented in the form a/b where a is an integer and b is a nonzero integer. This isn’t some deep philosophical tie to an underlying reality, it’s just our first attempt at defining more useful numbers that lie between other useful numbers we already invented (the integers et al). The real and complex numbers are literally just the extension of that process, filling in holes between useful numbers with more numbers until the set is closed (i.e. there are no more holes, every operation between members of the set results in another member of the set). Closure, the real reason we care so much about the reals, is just a surprise tool that helps us later. For anyone who made it this far: if you find any of this interesting you should find a book or lecture on analysis. It’s not particularly difficult and presents deeper insights into the math you likely already learned.

Vetch 5y ago
I enjoyed your comment and agree with all your well put points except one: on the rationals tie with reality. Having implemented exact real computation to better understand reals, I think of a real number as a kind of machine that generates infinite streams. Operations on them instantiate new machines which query their real operands, computating until there's sufficient information to emit a next term of the stream. When the next term needs an infinite amount of information to decide what to spit out next, it results in an "unproductive" infinite loop. Rational numbers are interesting, more realistic, because they always terminate. In the real world, measurement tolerances and physical limits means at some point having to extract a rational. When we work with reals we are really only working with rational approximations or symbols with associated properties and relations. Reals are a powerful and elegant tool to rigorously reason about mathematical spaces and operations on algebraic objects but trying to work with them in reality in their exact form is a fun and visceral lesson on the nature of undecidability. It's hard to go two steps without tripping over a non-terminating loop (such as any operation that starts with an irrational and results in a rational or equality testing in general). This is an observation on our interface with reality and not on its true nature, which may or may not admit reals (although my non-serious guess is that black holes form whenever you try to do something that requires a proper real number).

deleted 5y ago

[deleted]

matt123456789 5y ago

Most reals are not computable anyway: https://arxiv.org/pdf/math/0411418.pdf https://arxiv.org/pdf/math/0411418.pdf

deleted 5y ago
[deleted]

sillysaurusx 5y ago

You can't. See "Simulating Physics With Computers," linked here: https://twitter.com/theshawwn/status/1394159744139632640 https://twitter.com/theshawwn/status/1394159744139632640 And some thoughts on the limitations with respect to creating AGI: https://twitter.com/theshawwn/status/1446261451061145602 https://twitter.com/theshawwn/status/1446261451061145602 See section 5, "Can quantum systems be probabilistically simulated by a classical computer?" > The probability that they match is eight-tenths, the probability that they mismatch is plus two-tenths; every physical probability comes out positive. But the original f's are not positive, and therein lies the great difficulty. The only difference between a probabilistic classical world and the equations of the quantum world is that somehow or other it appears as if the probabilities would have to go negative, and that we do not know, as far as I know, how to simulate. Okay, that's the fundamental problem. I don't know the answer to it, but I wanted to explain that if I try my best to make the equations look as near as possible to what would be imitable by a classical probabilistic computer, I get into trouble. When I was younger and only slightly more foolish, I wanted to spend a lot of time researching this to see if there was a way around the problem. I quickly realized that perhaps I should focus on adding value in a field that I was good at. :) Maybe one of you can try, since the only alternative is to take Feynman at his word.

aeternum 5y ago
The claim that discretization is not equal to quantization is a key claim and I'm not sure the paper actually proves/shows this. If it turns out that spacetime is discrete and not continuous then we should be able to simulate it. The possible states of some chunk of spacetime (given some maximum/known energy) would be non-infinite and even though it might take us years to calculate each time-step, we could still make progress. I've seen the theory kicked around before, but why couldn't the universe basically consist of plank length sized voxels and a similarly small update timestep? It would still look plenty continuous to us.
- sillysaurusx 5y ago
  It’s because of the anisotropy. Suppose the universe was a regular grid of voxels. You could run experiments to prove that. I don’t understand the details, but that was Feynman’s counter argument. (See the “messenger lectures” series on YouTube.) If the grid isn’t regular, you run into other problems. But iirc at one point Feynman was toying with the idea that the grid might be randomly distributed.

nocturnial 5y ago

Maybe you're interested in this paper: Discretization of the Bloch sphere, fractal invariant sets and Bell’s theorem https://royalsocietypublishing.org/doi/10.1098/rspa.2019.0350 https://royalsocietypublishing.org/doi/10.1098/rspa.2019.035... There also a video where he explains the paper: https://www.youtube.com/watch?v=YglT09Korr0&t=2700s https://www.youtube.com/watch?v=YglT09Korr0&t=2700s There are some consequences by using Q instead of C that can be experimentally tested.