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God created the real numbers

(www.ethanheilman.com)
136 points Bogdanp | 3 comments | | HN request time: 0s | source
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andrewla ◴[] 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 ◴[] 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 ◴[] 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 ◴[] 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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gpm ◴[] 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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Kranar ◴[] 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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1. sfpotter ◴[] No.45070561[source]
Interesting point about BB(n)... Is it known that BB(n) is finite for every n?
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2. skybrian ◴[] No.45070579[source]
I believe it is by definition? The machines that don’t halt are filtered out. The trouble is how to do the filtering.
3. Kranar ◴[] No.45070593[source]
Yes BB(n) is always a natural number which is by definition finite.