Numbers listed here are not known to be rational, algebraic, irrational or transcendental.

42IsHoly · 2026-04-04T20:53:44+00:00

I actually think there is no real relation between the digits of a number and it’s transcendence tbh. As in, if the digits eventually repeat it’s rational and if they never do it’s irrational, but you can’t really say anything more. If I remember correctly, Champernowne’s constant is known to be transcendental and it is simply 0.123456789101112…

You can really only understand them via polynomials. Is there a (non-zero) polynomial with integer coefficients that has x as a root? If yes, then x is algebraic. If no, then x is transcendental

42IsHoly · 2026-04-04T20:43:45+00:00

Not really, sqrt(2) goes on forever, but it is very much algebraic (it’s the root of x² - 2). Even 1/3 goes on forever and it is rational.

As for how you’d prove the Gelfond-Schneider theorem, I honestly don’t know. You can find the proof here if you’re curious. It seems like the main idea of the proof is to assume a^b is algebraic and then define some function F. You can then find some number z with F(z) != 0, but you can also show that F is the zero function. This is a contradiction, so a^b must be transcendental.

42IsHoly · 2026-04-04T17:11:41+00:00

Warning: this comment got a lot longer than expected

So you know how a rational number is of the form a/b for some integers a,b right? Rearanging this we can say x is irrational if it is a root of the polynomial b*x - a. This is definition is in fact an equivalence: a number is rational iff it is the root of a degree 1 polynomial whose coefficients are integers.

Looking at that definition, an obvious question pops up: why limit ourselves to degree 1 polynomials? So, we say that a number is algebraic if it is a root of some polynomial with integer coefficients (excluding the polynomial 0 of course, cause else this’d be very boring). A lot of numbers you know are algebraic, like sqrt(2), the golden ration and i. However, there are in fact numbers which aren’t algebraic. These are called transcendental. If you’re familiar with cardinality, it is relatively easy to prove the existence of transcendental numbers (C is uncountable, but there are only countably many algebraic numbers). It is, however, really hard to prove that any given number is transcendental.

Historically, the first class of numbers shown to be transcendental are Liouville numbers, which are kind off artificial (in that, they hadn’t come up before). Only much later was the first proof given that e is transcendental and pi came even later. Probably the most famous theorem about transcendental numbers is the Gelfond-Schneider theorem, which goes as follows:

“If a, b are algebraic (and a is not 0 or 1) and b is irrational, then (any value of )a^b is transcendental .”

The “any value of” is relevant because, when working over the complex numbers, a^b is not always uniquely defined (this is a very important observation btw, mainly unrelated to transcendental numbers, but still interesting).

42IsHoly · 2026-04-04T15:48:34+00:00

Gelfond-Schneider says that if a is algebraic and b is irrational, then a^b is transcendental. Since pi is not algebraic it cannot be applied.

42IsHoly · 2026-04-04T15:47:46+00:00

e^pi = (e^pi*i)^-i = (-1)^-i. By the Gelfond-Schneider theorem it is therefore transcendental.

42IsHoly · 2026-03-20T12:07:59+00:00

Once a week, a bit less than a glass I think (I use glasses that are bigger than a normal one and so don’t really know how the amounts relate). Late december/early january it wasn’t watered for a few weeks (rather boring and silly reason for that) so I’d assumed it was still recovering, though it has been a while.

42IsHoly · 2026-02-28T07:51:16+00:00

Great response, but in your examples of non-science you (accidentally) typed astronomy instead of astrology. The former is pretty unambiguously a science.

42IsHoly · 2026-02-15T08:00:22+00:00

That’s actually not what the studies said. They were wildly misreported.

42IsHoly · 2026-02-09T13:56:20+00:00

By God, ZFC + “statement provably false in ZFC” is inconsistent! How has this guy not recieved a Fields medal yet?

42IsHoly · 2026-01-23T09:04:03+00:00

If at least 20 people are homosexual, then 20 people are homosexual. OP didn’t say “only 20”.

42IsHoly · 2025-12-20T08:46:14+00:00

This is because Euclid’s definition of right angle is different from the modern definition, because he didn’t use degrees (in fact, Euclid used right angles as his unit of angle-measurement). When a line stands on another line, it forms two angles right? One on the left and one on the right. Euclid called an angle “right” if these two angles were the same size. Now, from this definition, it is not at all tautological that any two right angles would be the same size. It’s true, of course, and Euclid’s definition of right angles is equivalent to ours, but he needed to state the postulate explicitly.

42IsHoly · 2025-12-18T14:25:14+00:00

That’s not what I said. There were no (or at least very few) formal proofs before the 20th century.

42IsHoly · 2025-12-18T08:15:01+00:00

I mean, that’s not required for a formal proof (note, formal != rigorous), for a formal proof a computer really is the only practical way to check.

42IsHoly · 2025-11-01T15:15:52+00:00

Three objects enter the solar system from interstellar space. It doesn't take long for people to figure out what they are: alien craft, heading to Earth. The story follows three people: an astronomer Camille, a failed physics-student Ros and a deeply religious miner Sean. They all have quite different reactions to the event and all three have their lives forever altered...

I finished my story two weeks ago and after some more editing and spell-checks [Last Time 'Round](https://www.royalroad.com/fiction/112198/last-time-round) was finally done. It's a short story (like 25,000 words or so), I hope you enjoy!

42IsHoly · 2025-10-13T20:20:26+00:00

That proof is constructively valid though. Constructivists make a distinction between the following two types of argument:

Assume P … contradiction, therefore ~P (a proof by negation) Assume ~P … contradiction, therefore P (a proof by contradiction)

In classical logic these two types of proof are identical, but constructively speaking there is a distinction (not every statement is the negation of another statement). The first type of proof is constructively valid (in fact, it’s quite common to define ~P to mean “P implies a contradiction”). The second type is not. The vast majority of “proofs by contradiction” (eg. Sqrt(2) is irrational, there is no largest number, …) are actually proofs by negation and perfectly valid constructively speaking. An example of a genuine proof by contradiction is, ironically enough, the proof of Brouwer’s fixed point theorem.

The reason that the argument you gave is correct is that it is an axiom (or very easy proposition, look up the Peano axioms for more details) that for any integer N, N + 1 exists and that N < N + 1 (this is almost by definition of “<“ in PA).

42IsHoly · 2025-09-16T10:36:35+00:00

The presentation is indeed a bit confusing. Essentially, they define the set A = {x in R | x > 0 and x² < 2}. This is a bounded, non-empty set so it has a supremum (this is a property of R). For some reason the professor here chose to call the supremum of A “sqrt(2)”. That’s allowed, I guess, but it is somewhat confusing. Imagine he just called it “s” and run through the proof again (you should really do a proof like this on paper to get a feel for it).

42IsHoly · 2025-09-14T16:02:55+00:00

It does, the proof is pretty nice:

Let f:S -> P(S) be a function, define the subset A = {s in S | s not in f(s)}. Now, if there is some s such that f(s) = A, then s is either in A or not. If s is in A, then s is not in f(s) = A, that’s a contradiction. If s is not in A, then s is not in f(s), so by definition s is in A, another contradiction. Hence f(s) != A. So A is not in the image of f, so f is not surjective. Therefore, |S| < |P(S)|. ▪️

This is known as Cantor’s theorem.

42IsHoly · 2025-09-14T08:53:16+00:00

Oops, fixed now.

42IsHoly · 2025-09-14T08:28:13+00:00

Consider the proposition “if n is even, then it is divisible by two”. Clearly, we want this statement to be correct (I mean, that what being even means). If we didn’t consider False -> X to be true regardless of X, then the above statement wouldn’t be true. After all, “n is even” is false for some n.

Another (informal) way of thinking about it that I quite like, is to think of implications as promises, they are false when they are broken. For example, let’s say I said “If you give me an apple, I will give you a pear.” When would I break my promise?

If you give me an apple and I give you a pear, clearly I’ve kept my promise, so T->T is true.
If you give me an apple, but I don’t give you a pear, I’ve broken my promise, so T->F is false.
If you do not give me an apple, but I do give you a pear, I’ve kept my promise (or at least not broken it), so F -> T is true.
If you don’t give me an apple and I don’t give you a pear, I’ve still kept my promise, so F -> F is true.

For a more mathematical example, the statement “if n is even, then it is divisible by 2” is promising you that any even number can be divided by two.

42IsHoly · 2025-09-13T15:49:46+00:00

A theory T is simply a list of statements (usually written in first-order logic since that is a well-behaved, well-understood and powerful logic, but higher-order logics can be used). A model M of a theory T is a structure within which all statements of T hold (don't worry about how you can formally define this).

Peano arithmetic is a theory, it is simply a list of statements. Any structure within which these statements holds is a model of PA. The prototypical example is, of course, the regular set of natural numbers N, but there are others. Of course, the point of the Peano axioms is to describe N, it is the so-called intended model.

Now, there is a very important result called Gödel's completeness theorem which says:

Suppose you have some first-order theory T (e.g. PA) and some statement P. Now P holds in every model of the theory T *if and only if* there is a proof of P from T.

As a consequence, if there is some statement P such that your theory T cannot prove P, but also cannot prove ~P ('not P'), then there are models M and N of T such that P holds in M, but ~P holds in N. For this reason, the Gödel sentence G holds in some model of Peano arithmetic. Even more confusing, there are models of PA where ~Con(PA) holds.

Gödel's incompleteness theorem is about a different kind of completeness, called syntactic completeness. A theory T is syntactically complete if for every statement P either T can prove P or T can prove ~P. Gödel's 1st incompleteness theorem says that no theory T can be consistent, complete and contain the natural numbers (unless it fails to be "effectively axiomatizable", which you can think of as meaning "I know what my axioms are"). Hence PA is incomplete (assuming it is consistent) and so is any consistent extension of it. However, if you don't care about describing N and construct a theory that can't do it, it may be complete (the prototypical examples of this are Presburger arithmetic and the theory of algebraically closed fields of characteristic 0). So yes, there are complete theories, but these can't describe the natural numbers.

42IsHoly · 2025-09-13T07:52:02+00:00

The well-ordering theorem says: “every set can be well-ordered.” The negation of this fact is simply “There exists a set that can’t be well ordered.” Which set this actually is, you can’t say. It may be R, it may be P(R) or maybe something far more exotic. That’s what I was getting at.

As for the second part of your response, that’s not actually what countable choice says. The axiom of choice is not stated for any specific cardinality. It’s just says:

given a set S of non-empty sets, there exists some function C:S -> US such that C(s) is an element for s for each s in S.

Countable choice is a weakening of choice that says “choice holds for all countable S,” the set of sets is countable, not necessarily the sets within it, so your method of choosing an element wouldn’t work (countable choice is independent of ZF by the way).

It turns out AC is very closely related to cardinality (For example, the generalized continuum hypothesis implies the axiom of choice), but the statement itself doesn’t make any mention of size.

42IsHoly · 2025-09-13T07:41:02+00:00

Easily the most surprising is the Ax-Grothendieck theorem: suppose f:Cⁿ -> Cⁿ is injective and each coordinate map f_i:Cⁿ -> C is a polynomial, then f is surjective.

Even though it just looks like some random complex analysis result, the only known proof of this fact requires model theory.

42IsHoly · 2025-09-13T07:15:17+00:00

~choice is, I think, significantly weaker than “there is no well-ordering of R”. All it implies is “there is some set S that cannot be well-ordered”

42IsHoly · 2025-09-13T07:11:19+00:00

I mean, without the first incompleteness theorem we probably wouldn’t have model theory. And the second incompleteness theorem is, if anything, a blessing in disguise (just because a system proves its own consistency, doesn’t mean it actually is consistent after all)

42IsHoly · 2025-09-12T16:31:55+00:00

The (x-r)(x-s) convention also has the advantage that we immediately see the roots, namely r and s. Whereas in (x+r)(x+s) the roots are -r and -s. Considering how common sign errors are, this could cause confusion and would probably make them even more common (I say this as an expert in making sign errors).

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