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God created the real numbers

(www.ethanheilman.com)
136 points Bogdanp | 12 comments | 29 Aug 25 15:31 UTC | HN request time: 0.421s | source | bottom
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andrewla ◴[29 Aug 25 18:33 UTC] No.45067770[source]▶
I'm an enthusiastic Cantor skeptic, I lean very heavily constructivist to the point of almost being a finitist, but nonetheless I think the thesis of this article is basically correct.

Nature and the universe is all about continuous quantities; integral quantities and whole numbers represent an abstraction. At a micro level this is less true -- elementary particles specifically are a (mostly) discrete phenomenon, but representing the state even of a very simple system involves continuous quantities.

But the Cantor vision of the real numbers is just wrong and completely unphysical. The idea of arbitrary precision is intrinsically broken in physical reality. Instead I am off the opinion that computation is the relevant process in the physical universe, so approximations to continuous quantities are where the "Eternal Nature" line lies, and the abstraction of the continuum is just that -- an abstraction of the idea of having perfect knowledge of the state of anything in the universe.

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NoahZuniga ◴[29 Aug 25 19:33 UTC] No.45068389[source]▶
You know it wouldn't be possible for us to tell the difference between a rational universe (one where all quantities are rational numbers) and a real universe (one where you can have irrational quantities).

The standard construction for the real numbers is to start with the rationals and "fill in all the holes". So why even bother with filling in the holes and instead just declare God created the rationals?

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omnicognate ◴[29 Aug 25 20:02 UTC] No.45068743[source]▶
As in why bother using real numbers in physics? Mostly because you need them to make the maths rigorous. You can't do rigorous calculus (i.e. real analysis) on rationals alone.
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Kranar ◴[29 Aug 25 22:10 UTC] No.45069997[source]▶
You don't need the full set of real numbers to do physics, only the computable subset of the real numbers. Using the full reals is mostly done out of simplicity.
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1. gpm ◴[29 Aug 25 22:15 UTC] No.45070039[source]▶
What do you mean by "the computable subset of the reals" formally?

Is sqrt(2) computable?

Is BB(777) computable?

Is [the integer that happens to be equal to BB(777), not that I can prove it, written out in normal decimal notation] computable?

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2. Kranar ◴[29 Aug 25 22:20 UTC] No.45070072[source]▶
A computable real number is a real number for which a Turing Machine exists that can compute it to any arbitrary precision.

So yes sqrt(2) is computable.

Every BB(n) is computable since every every natutal number can be computed. It's the BB function itself that is not computable in general, not the specific output of that function for a given input.

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3. skybrian ◴[29 Aug 25 23:16 UTC] No.45070478[source]▶
That doesn’t sound right to me. What about the machines that don’t halt? You can’t compute whether or not to skip them directly.

> A busy beaver hunter who goes by Racheline has shown that the question of whether Antihydra halts is closely related to a famous unsolved problem in mathematics called the Collatz conjecture. Since then, the team has discovered many other six-rule machines with similar characteristics. Slaying the Antihydra and its brethren will require conceptual breakthroughs in pure mathematics.

https://www.quantamagazine.org/busy-beaver-hunters-reach-num...

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4. Kranar ◴[29 Aug 25 23:20 UTC] No.45070501{3}[source]▶
What specifically doesn't sound right?
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5. sfpotter ◴[29 Aug 25 23:30 UTC] No.45070561[source]▶
Interesting point about BB(n)... Is it known that BB(n) is finite for every n?
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6. skybrian ◴[29 Aug 25 23:34 UTC] No.45070579{3}[source]▶
I believe it is by definition? The machines that don’t halt are filtered out. The trouble is how to do the filtering.
7. Kranar ◴[29 Aug 25 23:38 UTC] No.45070593{3}[source]▶
Yes BB(n) is always a natural number which is by definition finite.
8. skybrian ◴[29 Aug 25 23:43 UTC] No.45070625{4}[source]▶
The claim is that every bb(n) is computable but I don’t think you can compute bb(6) without knowing which machines won’t halt. That doesn’t seem like a finite calculation?

But given the answer, I suppose you could write a program that just returns it. This seems to hinge on the definition of “computable.” It’s an integer, so that fits the definition of a computable number.

My mistake.

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9. tux3 ◴[29 Aug 25 23:56 UTC] No.45070715{5}[source]▶
Yes. A valid question for a specific n would be whether you can prove the value of BB(n). If you don't care about provability, you can indeed just produce a number that happens to be the right one.

So as you noticed, it only makes sense to talk about whether a function is computable, we can't meaningfully talk of computable numbers.

10. Kranar ◴[30 Aug 25 00:00 UTC] No.45070749{5}[source]▶
Yes exactly, imagine a function HH(n) that returns 0 if the Turing machine represented by the integer n halts, and 1 if it doesn't.

Then HH the function itself is not computable, but the numbers 0 and 1, which are the only two outputs of HH are computable.

Integers themselves are always computable, even if they are the output of functions that are themselves uncomputable.

11. ◴[30 Aug 25 02:13 UTC] No.45071349{3}[source]▶
12. edanm ◴[30 Aug 25 07:52 UTC] No.45072759{5}[source]▶
The main thing to make it clear is that BB(n) for a specific n isn't a function - it's a number. Just like Mult(10,4) isn't a function, it's a number (40).

So a specific BB(n) is just a number and is computable.