[nyrikki]

Almost all real numbers are normal numbers, which don't even have a finite representation.

Sure you can assign them to an arbitrary set, but you don't have access to the value.

It is a hay in the haystack problem, where you really only have access to the needles, not the hay.

[gowld]

It's even more extreme than that!

Take the (uncountable) set of Real numbers. Remove the normal numbers, which is almost all of them in the sense that the probability that "a uniformly randomly chosen real number is normal (and therefore also undescribable)" is 1. The remaining set of numbers, which has measure 0 in the Real numbers, is still uncountable, meaning that the proability of randomly choosing a describable number in that set is again 0.

I'm not sure how deep this chain can go. Google AI says "only 1 steps" but it's not admiting the case described in this comment.

[LegionMammal978]

> Almost all real numbers are normal numbers, which don't even have a finite representation.

Plenty of normal numbers have a finite representation from which digits can be efficiently extracted. E.g., Champernowne's constant (in any base) is normal, and you can find its digits with a relatively simple algorithm.

All computable reals can similarly have their digits extracted by some algorithm or another, even though it may take a long time. I wouldn't call that "not having access to the value". Of course, uncomputable numbers are a different story, but they have nothing to do with normality in any base.

And of course, radix representations are not the only way to evaluate real numbers. E.g., you could represent them with simple continued fractions (which would still allow addition, multiplication, comparison, etc.), and then you could write out any quadratic irrational with a periodic expansion.

[nyrikki]

"almost all" or "almost everywhere" is in italics because I mean it in the measure theory sense.

Meaning it holds for all elements of a set except for a subset that has measure zero.

Yes some normal numbers are in the constructable reals, but it is a measure zero subset.

You are putting your hand in the haystack and only finding needles, finding the hay in the haystack is the problem here.

[LegionMammal978]

Almost all reals are uncomputable. Also, almost all reals are absolutely normal. But the two concepts have nothing to do with each other.

> Yes some normal numbers are in the constructable reals, but it is a measure zero subset.

Almost all irrational computable reals (in the sense of natural density) will be normal, for any sane enumeration. Just because a real number is computable doesn't mean it's less likely to be normal.

[nyrikki]

Note I intentionally didn't invoke uncomputable, because computable is a highly constrained definition, and in this case is a circular refrence.

> All computable reals can similarly have their digits extracted by some algorithm or another, even though it may take a long time.

It was an intentional abstraction to avoid a self referencing claim, not to say that they are equivalent to anything.

The nice thing about constructible reals is after the construction you can typically forget how you constructed them, be that through Axioms, Cauchy sequences, Dedekind cuts etc...

The computable reals by definition can be computed to within any desired precision by a finite, terminating algorithm. That is why I said it is begging the question.

> Almost all irrational computable reals (in the sense of natural density) will be normal, for any sane enumeration. Just because a real number is computable doesn't mean it's less likely to be normal.

While some have Conjectured claims close to this, looking into why there have been no proofs for even a single number that was not explicitly created to be normal in any base may be a good lens in to the hay-in-the-haystack problems I was refrencing.

From: "Distribution Modulo One and Diophantine Approximation (Cambridge Tracts in Mathematics, Series Number 193)" Page 81, section 4.1 "Equivalent definitions of normality"

> Lemma 4.3: Let b and r be integers greater than or equal to 2. If a real number is simply normal to base b^r, then it is simply normal to base b.

That there may exclude many of what appear to be simply normal numbers from actual ones. As a graduate level text that book may be a bit expensive for what it is, but a good reference in my experience.

The 'Normal' property that is the full measure set of the reals is far more constrained than natural density. Which is why it is so surprising that is is the property of almost all of them.

That is why it was offered as a lens, specifically one that was worked on before computability as a subject, as an intentional way to gain distance from the almost intractable polysemy problems there.

[LegionMammal978]

If you originally meant that "almost all real numbers are normal numbers without a finite representation," then I have no disagreement with you. I'd read your comment as saying that "[normal numbers] don't have a finite representation", which clearly has several counterexamples (Champernowne's constant & co.). I apologize if that was a misinterpretation.