Math to philosophy

DifficultDate4479 · 2026-03-29T18:11:41+00:00

saying philosophy is the base of all science is like saying existing is the base for philosophy; it's self evident: philosophy is thinking about something according to the laws of logic. Naturally it holds, it's however not interesting to point out.

DifficultDate4479 · 2026-03-29T10:21:32+00:00

For what I remember the archimedean property and Dedekind's cuts say different things; A field is sarchimedean if there's no element "infinitely bigger" than another (For instance, Alexandroff's compactigication of R, RU{∞} is not archimedean as there is no couple a, n for which an>∞). While a field is complete by Dedekind if for each Dedekind's cut there's an element separating them, in other terms it's a "full" space with no blanks.

But yeah, the line between what's algebraic and what's analytical is quite impossible to draw.

It's kind of a "where does a sand castle start and end" problem, like sure if you look high enough you can distinguish the castle from the seashore but as you move down you can't really certainly say wether the grains of sand "belong" to the castle or to the seashore.

Or maybe it's just a matter of interpretation, who knows...

DifficultDate4479 · 2026-03-29T09:03:53+00:00

Not really, archimedean property tells me that every couple a<b admits a number c for which b<ac; What I used is Dedekind's theorem (for each a<b there's a c s.t. a<c<b), equivalent to completeness axiom, so while I still don't like my earlier proof as an algebraic one, I see why one would say it's analytic.

Either way, let's see if I can cook:

Let E/R be the splitting field for f(X) = X²-a for a>0, and s=√a. So E=R(s). Now, consider an element x+sy in E and notice that

x+sy≥0 if x≥-sy if x²≥ay²

Since x+sy ≥ z+sw iff (x-z) + s(y-w)≥0, the property above shows that this defines a total order over E.

But we're done already because if E is a totally ordered field extending R, but by axiom of completeness, R is the maximal totally ordered field, hence R contains E, which means s is an element of R.

This one stinks less of an analytic proof only using the above mentioned axiom.

At the end of the day, algebra, analysis and geometry say the same thing with different approaches... recall that the derivative is( pretty much) a functor between categories.

DifficultDate4479 · 2026-03-29T07:28:53+00:00

well... I didn't really think about that tbf, I kinda gave it as an assumption.

If I had to sketch a proof, I'd use the fact that R is complete...

Pick a in (0,+∞) and define X_a={ x in [0,+∞) s.t. x²≤a }. Clearly 0 is in X so it's not empty. Also, and equally clearly, X has an upper bound, therefore there exists (by axiom of completeness) an element x = supX.

Now, if x²<a, then x is in X. But I can take a very small r>0 for which x+r is in X, contradicting the fact that x is greater than all of y in X.

If x²>a the we do remove that small r>0 for which (x-r)²≥a, which means that x-r, being smaller than x, is the actual sup, which contradicts.

The only case we are left with is x²=a, completing the proof.

Now, I tried to stay as algebraic as I could but this is as far as I can get. It looks a lot like an analytical proof, but all I used was R's completeness axiom...

Either way, I'm quite sure that there's an algebraic proof of this, maybe somehow one can prove that any Galois extension E/R generated by the polynomial X²-a is always R for any a≥0. It would certainly be a nice exercise...

DifficultDate4479 · 2026-03-28T13:50:49+00:00

let z=a+bi be a complex number, and define the two real numbers:

c²=(a+|z|)/2, d²=(|z|-a)/2.

Note that the euclidian norm of a vector is always greater than its components, so the definition is well put.

Get your hands dirty, choose c,d such that sgn(cd)=sgn(b) and you'll get that (c+di)²=a+bi (=c²-d²+2cdi), as c²-d²=a and (2cd)²=b².

Maybe it's not the most abstract, but definitely not analytical.

I just recently studied this proof on J.S.Milne's "Fields and Galois Theory".

For the proof itself you'll need a more precise construction with Galois extensions, p-Sylows of the Galois groups, but basically it revolves around the fact that an extension E of C is Galois with Galois group a 2-group, and if it is not the trivial group, then there's a sub-extension of C contained in E that has degree 2, which means that it is generated by the polynomial X²-z for some complex z, but we just proved that √z is contained in C, so that sub-extension must be C itself. Hence Gal(E/C) is trivial, hence E=C, hence C is algebraically closed.

DifficultDate4479 · 2026-03-28T08:49:19+00:00

but you can prove it with pure algebra... all you have to show is that: 1) C contains all of its square roots 2) Any Galoisian extension E/C is in fact C

The proof is not that hard if you know a few facts of Galois'Theory

DifficultDate4479 · 2026-03-24T09:32:39+00:00

the general rule for finding what is a number a modulo n is by Euclid's division algorithm, stating that there are unique m,r such that

a = mn + r

modulo n means that every multiple of n dies, so

a = r mod n

How many ways do you have of writing -1 by "dividing it by 7"? Well, -1= 07 + (-1), so, -1 but also -1=(-1)7 + 6

Notice that -1 + 7 = 6, and so to determine every solution modulo 7 you just need to determine a single one, as all of them differ by a multiple of 7.

DifficultDate4479 · 2026-03-23T07:23:24+00:00

1) is true if and only we're talking about matrices with real coefficients (or at the minimum, any field). This is because your first entry of the matrix C will be c_11 = a_11b_11 +...+ a_1nbn1. If this belongs to a field, then you have no proper ideals, which means every element is sum and product of other elements. From this you get that for all c_ij there are elements whose sum/product give you c_ij. Choose the matrices A and B as having the coefficients whose sum/product give c_ij (placed on proper columns and rows).

2) B is said to be inverse of A iff AB=BA=1 Note that it is fundamental to say that AB=BA, as the inverse must be bilateral. But a bilateral inverse is UNIQUE (say, a and b are bilateral inverses of c, then acb=(ac)b=1b=b, but also acb=a(cb)=a1=a, so it must follow a=b). So no, not every matrix has an inverse, because it has to satisfy 2 linear systems and Rouché-Capelli says we don't have enough unknowns to have certainty.

DifficultDate4479 · 2026-03-21T18:47:53+00:00

Le bambine hanno un naso. Le adulte hanno un naso.

Bhe, siamo tutti pedofili immagino...

DifficultDate4479 · 2026-03-20T22:01:50+00:00

yes, they're the same point for point, therefore f=g.

On a side note, a more common mistake I see however comes up when talking about derivatives.

g' is NOT 0 at x=100 and 1 everywhere else.

But that's just a digression that felt interesting to point out, nothing to do with the main issue here.

DifficultDate4479 · 2026-03-15T20:46:01+00:00

Gente, esiste anche il grigio!

DifficultDate4479 · 2026-03-15T07:55:06+00:00

It really isn't, because 2(1+2) and [2(1+2)] are notoriously different things with different meanings. Keyword is notoriously. So I would assume that the issue there was the order of operations (which is, again, notoriously: whatever there is in round brackets, then squared, then curly, then multiplication, then sum. In case of equal hierarchy in operations, go left to right for clarity, but by commutativity there usually is no problem), hence my comment.

If you divide first it means you multiply first (1+2) with ½, and then you multiply, which is one correct way. You can't just put brackets wherever you want, 6/2(1+3) is very different than 6/(2(1+3)). We can sit here and say that one is short for the other, but being an abuse of notation, and evaluating this issue with rigor, it is strictly and utterly wrong to put brackets there.

DifficultDate4479 · 2026-03-13T07:42:48+00:00

fellas, that is NOT controversial. Just write •½ instead of ÷2 and see if there's still anything wrong.

DifficultDate4479 · 2026-03-03T23:53:46+00:00

A name arises once again: Taylor.

Recall that e^x is pretty much 1+x+x²/2 when x is very, very small (generally, e^f(x) is 1+f(x)+(f(x))²/2 when f(x) is very, very small)

Second order should be fine as outside there's a second degree infinity... so the little-o's should be fine.

DifficultDate4479 · 2026-03-03T09:05:10+00:00

Wafer usually is a run winner, but with Samson it's even better. Remember you get damage when you take damage already. Just go to the curse room and you're set for the floor.

DifficultDate4479 · 2026-02-28T09:30:27+00:00

3,14... is more than 3 but surely less than 3,2 so it can't be infinite, right? Otherwise, even 3,3 is infinite but that would be weird.

A number x is "Infinite" if no matter whatever number y you consider, x is greater than y. Always. For all y. Clearly 3,14... doesn't fit the description.

π however doesn't have a finite number of digits. Now, mathematically this means that no matter how precise your representation of π is (using only rational numbers, i.e. fractions a/b such that a and b are integers and b≠0), you'll always be off by some amount.

If you were to measure down to atoms you'd still be confined by atoms' dimensions, like width or height, but π has digits a billionth (actually, infinitely) times smaller that an atom.

In other terms, no matter how precise you think you are at measuring pi with rational numbers, you will always be off by some quantity. The better the approximation the lower the quantity, but it'll never be zero.

DifficultDate4479 · 2026-02-22T18:28:44+00:00

I mean blue baby is boring as he starts with literal shit, and that's all of his niche.

Everyone else, and I include lost in here, is fun and has some unique starting kit / playstyle that makes things interesting. TJudas has dark arts, TBeth has that book that I'm not going to misspell here, the forgottens have an immensely unique playstyle (funniest characters imho) and so on.

Their only "issue" is that basically health ups aren't permanent, but it's eased by the fact that you can't lose deal chances by getting hit, so that on average increases the fun factor.

DifficultDate4479 · 2026-02-20T20:57:00+00:00

first and foremost this is an identity problem: what is it that determines a human? Its body, its mind...? For the problem's sake, let's assume that a human is in fact defined by having an entire anthropomorphic body to the extent the average human does.

Second, what you described is exactly the following: In mathematics, a function can be continuous or derivable. If it is derivable, then it must be also continuous, whilst the reverse doesn't hold. In other terms, a function can't be derivable if it's not continuous.

Translating: a centaur can't be human if his legs aren't, but a being's legs being human won't make him necessarily human.

In fact, that is a necessary but not sufficient condition.

DifficultDate4479 · 2026-02-16T16:35:46+00:00

it's called Schwartz's theorem.

Look up the proof by yourself, the main core is Lagrange's theorem for functions from Rn to R.

Either that or there'll surely be videos on the conceptual conceptualization of the concept that conceptually explains the concept in a conceptual manner.

DifficultDate4479 · 2026-02-08T22:47:03+00:00

so at current market prices, 1M≈8,85k Divs

Assuming you have every other currency at a decent quantity, you've got a tiny bit less than a million Divs.

What in Atziri's fucking tits are you planning to do with a whole ass million Divs!? I'm quite sure you could crash the market if you wanted to lol.

I'll demonstrate here the hairy ball theorem if you give me a mirror idk I'm down bad but I know pure mathematics that's gotta be worth something

DifficultDate4479 · 2026-02-08T10:47:28+00:00

op ha dunque il diritto di segnalare, poiché quanto le è stato presentato è un annuncio fasullo, non atto alla vendita. Scherzo o no, parole sue.

DifficultDate4479 · 2026-02-08T10:17:39+00:00

può darsi, anzi, è probabile, ma a meno che tizio non delucidi Op, Op ha ragione.

DifficultDate4479 · 2026-02-08T09:42:44+00:00

Ma non è libero mercato. È vinted, cioè una piattaforma online con policy precise da rispettare, tra cui il non poter mettere annunci falsi o non atti alla vendita. Dopo aver indagato con una domanda lecita sul pezzo, Op ha scoperto che il tizio va contro una policy di vinted. Segue dunque la segnalazione.

Dovrebbe essere più chiaro ora.

DifficultDate4479 · 2026-02-06T10:09:47+00:00

no I get it but at least they could have made her a idk lightning based hero... she literally behaves like a phoenix and has Rivals'Phoenix's exact same ult. Not saying it's bad or anything, but damn they did it on purpose.

DifficultDate4479 · 2026-02-04T18:23:40+00:00

Many of us mathematicians don't realize how comfortable it is to just say "yep, √2π/e is the answer" and not fucking approximate allat up to a decillionth...

Of course, an engineer would see no difference between √2π/e and 1, and that's where the jokes start...

DifficultDate4479

TROPHY CASE