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Syzygies · a month ago
Mathematicians get enamored with particular ways of looking at things, and fall into believing this is gospel. I should know: I am one, and I fight this tendency at every turn.

On one hand, "rational" and "algebraic" are far more pervasive concepts than mathematicians are ever taught to believe. The key here is formal power series in non-commuting variables, as pioneered by Marcel-Paul Schützenberger. "Rational" corresponds to finite state machines, and "Algebraic" corresponds to pushdown automata, the context-free grammars that describe most programming languages.

On the other hand, "Concrete Mathematics" by Donald Knuth, Oren Patashnik, and Ronald Graham (I never met Oren) popularizes another way to organize numbers: The "endpoints" of positive reals are 0/1 and 1/0. Subdivide this interval (any such interval) by taking the center of a/b and c/d as (a+c)/(b+d). Here, the first center is 1/1 = 1. Iterate. Given any number, its coordinates in this system is the sequence of L, R symbols to locate it in successive subdivisions.

Any computer scientist should be chomping at the bit here: What is the complexity of the L, R sequence that locates a given number?

From this perspective, the natural number "e" is one of the simpler numbers known, not lost in the unwashed multitude of "transcendental" numbers.

Most mathematicians don't know this. The idea generalizes to barycentric subdivision in any dimension, but the real line is already interesting.

afiori · a month ago
Using this representation can one "efficiently" sum or multiply numbers? I was under the impression that this was its main drawback
brianberns · a month ago
I read this with pleasure, right up until the bit about the ants. Then I saw the note from myself at the end, which I had totally forgot writing seven years ago. I probably first encountered the article via HN back then as well. Thanks for publishing my thoughts!
xnorswap · a month ago
The ants argument feels rather like a retelling of Zeno's Paradoxes.

https://en.wikipedia.org/wiki/Zeno%27s_paradoxes

mg · a month ago
Three surprising facts about transcendental numbers:

1: Almost all numbers are transcendental.

2: If you could pick a real number at random, the probability of it being transcendental is 1.

3: Finding new transcendental numbers is trivial. Just add 1 to any other transcendental number and you have a new transcendental number.

Most of our lives we deal with non-transcendental numbers, even though those are infinitely rare.

canjobear · a month ago
> 1: Almost all numbers are transcendental.

Even crazier than that: almost all numbers cannot be defined with any finite expression.

dwohnitmok · a month ago
This is not necessarily true. It is possible for all real numbers (and indeed all mathematical objects) to be definable under ZFC. It is also possible for that not to be the case. ZFC is mum on the issue.

I've commented on this several times. Here's the most recent one: https://news.ycombinator.com/item?id=44366342

Basically you can't do a standard countability argument because you can't enumerate definable objects because you can't uniformly define "definability." The naive definition falls prey to Liar's Paradox type problems.

zeroonetwothree · a month ago
Maybe it would be better to say almost all numbers are not computable.
dinosaurdynasty · a month ago
Leads to really fun statements like "there exists a proof that all reals are equal to themselves" and "there does not exist a proof for every real number that it is equal to itself" (because `x=x`, for most real numbers, can't even be written down, there are more numbers than proofs).

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bjourne · a month ago
Really? Which number can't be defined with a finite expression?
sorokod · a month ago
By common definition of "almost all", 1 == 2

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testaccount28 · a month ago
how can i pick a real number at random though?

i tried Math.random(), but that gave a rational number. i'm very lucky i guess?

andrewflnr · a month ago
You can't actually pick real numbers at random. You especially can't do it on a computer, since all numbers representable in a finite number of digits or bits are rational.
tantalor · a month ago
Pick a digit, repeat, don't stop.
mg · a month ago
How did you test the output of Math.random() for transcendence?

When you apply the same test to the output of Math.PI, does it pass?

kridsdale1 · a month ago
Use an analog computer. Sample a voltage. Congrats.
tzs · a month ago
If we are including numbers that aren't actually proven to be transcendental but that most mathematicians think are, I'd put Lévy's constant on the list.

It is e^(pi^2/(12 log 2))

Here's where it comes from. For almost all real numbers if you take their continued fraction expansion and compute the sequence of convergents, P1/Q1, P2/Q2, ..., Pn/Qn, ..., it turns out that the sequence Q1^(1/1), Q2^(1/2), ..., Qn^(1/n) converges to a limit and that limit is Lévy's constant.

barishnamazov · a month ago
Don't want to be "that guy," but Euler's constant and Catalan's constant aren't proven to be transcendental yet.

For context, a number is transcendental if it's not the root of any non-zero polynomial with rational coefficients. Essentially, it means the number cannot be constructed using a finite combination of integers and standard algebraic operations (addition, subtraction, multiplication, division, and integer roots). sqrt(2) is irrational but algebraic (it solves x^2 - 2 = 0); pi is transcendental.

The reason we haven't been able to prove this for constants like Euler-Mascheroni (gamma) is that we currently lack the tools to even prove they are irrational. With numbers like e or pi, we found infinite series or continued fraction representations that allowed us to prove they cannot be expressed as a ratio of two integers.

With gamma, we have no such "hook." It appears in many places (harmonics, gamma function derivatives), but we haven't found a relationship that forces a contradiction if we assume it is algebraic. For all we know right now, gamma could technically be a rational fraction with a denominator larger than the number of atoms in the universe, though most mathematicians would bet the house against it.

servercobra · a month ago
Both Euler's and Catalan's list "(Not proven to be transcendental, but generally believed to be by mathematicians.)". Maybe updated after your comment?
gizmo686 · a month ago
> Essentially, it means the number cannot be constructed using a finite combination of integers and standard algebraic operations (addition, subtraction, multiplication, division, and integer roots)

Slight clarification, but standard operations are not sufficient to construct all algebraic numbers. Once you get to 5th degree polynomials, there is no guarantee that their roots can be found through standard operations.

hidroto · a month ago
I am no mathematician, but i think you may be overstating Galois result. it says that you cant write a single closed form expression for the roots of any quintic using only (+,-,*,/,nth roots). This does not necessarily stop you from expressing each root individually with the standard algebraic operations.
mjd · a month ago
I'm glad that someone decided to be "that guy". Putting the Euler-Mascheroni constant third the list was a very questionable choice.
zkmon · a month ago
If a number system has a transcendental number as its base, would these numbers still be called transcendental in that number system?
moefh · a month ago
Yes. A number is transcendental if it's not the root of a polynomial with integer coefficients; that's completely independent of how you represent it.
gizmo686 · a month ago
The notion of transcendental is not related to how we right numbers. However, in abstract algebra, we generalize the notion of algebraic/transental to arbitrary fields. In such a framework, a number is only transental relative to a particular field.

For instance, the standard statement that pi us transcendental would become the pi is transcendental in Q (the rational numbers). However, pi is trivially not transcendental over Q(pi), which is the smallest field possible after adding pi to the rational numbers. A more interesting question is if e is transcendental over Q(pi); as far as I am aware that is still an open problem.

frutiger · a month ago
I think the elements of the base need to be enumerable (proof needed but it feels natural), and transcendental numbers are not enumerable (proof also needed).
JadeNB · a month ago
I think your parent comment was speaking of a "base-$\alpha$ representation", where $\alpha$ is a single transcendental number—no concerns about countability, though one must be quite careful about the "digits" in this base.

(I'm not sure what "the elements of the base need to be enumerable" means—usually, as above, one speaks of a single base; while mixed-radix systems exist, the usual definition still has only one base per position, and only countably many positions. But the proof of countability of transcendental numbers is easy, since each is a root of a polynomial over $\mathbb Q$, there are only countably many such polynomials, and every polynomial has only finitely many roots.)

jibal · a month ago
> I think the elements of the base need to be enumerable (proof needed but it feels natural)

Proof of what? Needed for what?

The elements of the number system are the base raised to non-negative integer powers, which of course is an enumerable set.

> transcendental numbers are not enumerable

Category mistake ... sets can be enumerable or not; numbers are not the sort of thing that can be enumerable or not. (The set of transcendental numbers is of course not enumerable [per Georg Cantor], but that doesn't seem to be what you're talking about.)

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senfiaj · a month ago
> Euler's constant, gamma = 0.577215 ... = lim n -> infinity > (1 + 1/2 + 1/3 + 1/4 + ... + 1/n - ln(n)) (Not proven to be transcendental, but generally believed to be by mathematicians.)

So why bring some numbers here as transcendental if not proven?

rkowalick · a month ago
As far I know, Euler's constant hasn't even been proven to be irrational.
auggierose · a month ago
Because it still might be transcendental. Just because you don't know if the list is correct, doesn't mean it isn't.
senfiaj · a month ago
Yes it's "likely" to be transcendental, maybe there are some evidences that support this, but this is not a proof (keep in mind that it isn't even proven to be irrational yet). Similarly, most mathematicians/computer scientist bet that P ≠ NP, but it doesn't make it proven and no one should claim that P ≠ NP in some article just because "it's most likely to be true" (even though some empirical real life evidence supports this hypothesis). In mathematics, some things may turn out to be contrary to our intuition and experience.
loloquwowndueo · a month ago
So it’s like “15 oldest actors to win an Oscar” and including someone who’s nominated this year but hasn’t actually won. But he might, right?

No, my dudes. Just no. If it’s not proven transcendental, it’s not to be considered such.

nuancebydefault · a month ago
I would have expected more numbers originating from physics, like Reynolds number (bad example since it is not really constant though).

The human-invented ones seem to be just a grasp of dozens man can come up with.

i to the power of i is one I never heard of but is fascinating though!

SOTGO · a month ago
To prove something is transcendental we would need to know how to compute it exactly, and I’m struggling to see how that would come up frequently in a physics context. In physics most constants are not arbitrary real numbers derived from a formula, they’re a measured relationship, which sort of inherently can’t be proved to be transcendental
longemen3000 · a month ago
some constants that may or may not be transcendental: - Percolation Thresholds: https://en.wikipedia.org/wiki/Percolation_threshold - Critical scalings in 3d: https://en.wikipedia.org/wiki/Universality_class#Ising_model
cozzyd · a month ago
Yeah I'd expect Bessel function zeroes and such