I've Been Pulled Into Watching The Gotham Chess Recaps Of The International Chess Championship And Here

Euclidean Geometry in Mathematical Olympiads

Geometric proofs… might anyone have a good resource for learning them more formally? I need them for a project I’ve recently begun, and hope to find a guide stronger than those reserved for high school geometry courses haha.

Specifically, i’m looking to understand lines, line intersections, things of that type, in a deeper way.

Thank you! :)

“Science, my lad, has been built upon many errors; but they are errors which it was good to fall into, for they led to the truth.”

A Journey to the Center of the Earth - Jules Verne

Yes.

Suppose 2n/(n-2) is an integer. Call it k. Then 2n = (n-2)k. But notice also that (n-2)×2 = 2n-4. So 2n is divisible by n-2, and so is 2n-4. So their difference, 4, is also divisible by n-2.

(

To see this, subtract the two equalities above. You'll get 2n - (2n-4) = (n-2)k - (n-2)×2, or, simplifying, 4 = (n-2)(k-2), so 4 is divisible by (n-2)

)

The only (positive integer) factors of 4 are 1, 2, and 4. So n-2 has to be 1, 2, or 4, and thus n has to be either 3, 4, or 6.

The question asks only for positive integers, but it would be a mistake to exclude the negatives. If we take negatives into account also, then -4, -2, and -1 also work, for which we get n to be -2, 0, and 1. And notice here that 1, although reached via a negative, is in fact a positive integer solution.

So the only numbers are 1, 3, 4, 6.

For more information on this, this kind of question comes under a part of math called Number Theory.

Does anyone know if 3, 4, and 6 are the only positive numbers for which 2n/(n-2) is an integer or are there more?

This exactly. It feels as if the brains of my friends flow smoothly towards the shore of certainty through the choppy seas of truth- meanwhile I'm taking a raft.

there's a very specific feeling that makes it difficult to study math in an academic environment when all students around you seem like they're exponentially more intuitive than you in this subject

Things Math Professors Say

"The proof is trivial." (Oh, cool. Guess I’m just an idiot then.)

"Left as an exercise." (Translation: You’ll never solve this in a million years.)

"It’s obvious, really." (Sure, if you’re a demigod.)

"By inspection." (Stares harder at problem… still nothing.)

"For small values of epsilon." (How small? Subatomic? Microscopic? Vibes?)

"WLOG (Without Loss of Generality)." (Oh, we’re just assuming it doesn’t matter now? Alright.)

"Details omitted." (Because apparently, you don’t need to understand it.)

"By the usual argument." (Which you somehow don’t know because you weren’t born in 1702.)

"Assume the rest holds." (That’s some impressive optimism right there.)

"The usual abuse of notation." (Why does this feel like an emotional wound?)

"Almost surely correct." (But also possibly wrong? Cool, thanks for the clarity.)

"A non-rigorous approach." (I thought math was supposed to be precise?!)

"Assume it’s obvious." (Buddy, NOTHING about this is obvious.)

"The reader may verify." (No, the reader may CRY.)

"To the interested reader." (Guess I’m not interested enough, huh?)

"Well-behaved functions only." (We’re function-shaming now?)

"Obvious to the trained eye." (Guess I’ll never make it out of amateur league.)

"A trivial case analysis." (Trivial to WHO??)

"Integrate by parts, twice." (Bold of you to assume I got it the first time.)

"As you can clearly see." (Oh, I clearly see my FAILURE, alright.)

"It works in practice too." (Unlike me, who barely works at all.)

"Assume a spherical cow." (Are we doing math or abstract sculpture?)

"A standard result." (Not in my standards, pal.)

"We skip the tedious algebra." (No, no, please—I wanted to suffer MORE.)

"Assume non-zero solutions exist." (Okay, now we’re just assuming life works out.)

"The usual topology." (Bro, I don’t even know the unusual topology.)

"Finitely many cases left." (Just kidding, there’s 72.)

"By virtue of symmetry." (Virtue? I have none left.)

"Don’t worry about the constant." (The constant is probably my GPA dropping.)

"Assume continuity." (I’m assuming my brain is breaking.)

"Smooth functions only." (Guess I’ll leave, I’m clearly not smooth enough.)

"The simplest non-trivial case." (Simplest? NON-TRIVIAL? Pick a side!)

"Epsilon goes to zero." (Epsilon isn’t the only one losing it.)

"And the rest follows." (Where? Straight to my breakdown?)

I remember being in elementary school and feeling a deep alienation at people’s just vicious rejection of mathematics, the reason being, like

It’s like. A classic approach to hypothetical communication with aliens that you start with the things you know you have in common and proceed from there. Mathematics is that.

Imagine you approach someone you don’t understand with a thing specifically designed to be something they and you can agree on as a starting point for communication and they just react with “yeah i hate this”

They discovered Busy Beaver Five!

BB(5) = 47,176,870

https://discuss.bbchallenge.org/t/july-2nd-2024-we-have-proved-bb-5-47-176-870/237

mathematical revelation so great i almost became religious

This is very interesting! It makes sense that, multiplication being repeated addition, it should be able to arise from just the group structure of the integers, but it never occurred to me before reading this. Thanks for the write-up.

Okay so to get the additive group of integers we just take the free (abelian) group on one generator. Perfectly natural. But given this group, how do we get the multiplication operation that makes it into the ring of integers, without just defining it to be what we already know the answer should be? Actually, we can leverage the fact that the underlying group is free on one generator.

So if you have two abelian groups A,B, then the set of group homorphisms A -> B can be equipped with the structure of an abelian group. If the values of homorphisms f and g at a group element a are f(a) and g(a), then the value of f + g at a is f(a) + g(a). Note that for this sum function to be a homomorphism in general, you do need B to be abelian. This abelian group structure is natural in the sense that Hom(A ⊗ B,C) is isomorphic in a natural way to Hom(A,Hom(B,C)) for all abelian groups A,B,C, where ⊗ denotes the tensor product of abelian groups. In jargon, this says that these constructions make the category of abelian groups into a monoidal closed category.

In particular, the set End(A) = Hom(A,A) of endomorphisms of A is itself an abelian group. What's more, we get an entirely new operation on End(A) for free: function composition! For f,g: A -> A, define f ∘ g to map a onto f(g(a)). Because the elements of End(A) are group homorphisms, we can derive a few identities that relate its addition to composition. If f,g,h are endomorphisms, then for all a in A we have [f ∘ (g + h)](a) = f(g(a) + h(a)) = f(g(a)) + f(h(a)) = [(f ∘ g) + (f ∘ h)](a), so f ∘ (g + h) = (f ∘ g) + (f ∘ h). In other words, composition distributes over addition on the left. We can similarly show that it distributes on the right. Because composition is associative and the identity function A -> A is always a homomorphism, we find that we have equipped End(A) with the structure of a unital ring.

Here's the punchline: because ℤ is the free group on one generator, a group homomorphism out of ℤ is completely determined by where it maps the generator 1, and every choice of image of 1 gives you a homomorphism. This means that we can identify the elements of ℤ with those of End(ℤ) bijectively; a non-negative number n corresponds to the endomorphism [n]: ℤ -> ℤ that maps k onto k added to itself n times, and a negative number n gives the endomorphism [n] that maps k onto -k added together -n times. Going from endomorphisms to integers is even simpler: evaluate the endomorphism at 1. Note that because (f + g)(1) = f(1) + g(1), this bijection is actually an isomorphism of abelian groups

This means that we can transfer the multiplication (i.e. composition) on End(ℤ) to ℤ. What's this ring structure on ℤ? Well if you have the endomorphism that maps 1 onto 2, and you then compose it with the one that maps 1 onto 3, then the resulting endomorphism maps 1 onto 2 added together 3 times, which among other names is known as 6. The multiplication is exactly the standard multiplication on ℤ!

A lot of things had to line up for this to work. For instance, the pointwise sum of endomorphisms needs to be itself an endomorphism. This is why we can't play the same game again; the free commutative ring on one generator is the integer polynomial ring ℤ[X], and indeed the set of ring endomorphisms ℤ[X] -> ℤ[X] correspond naturally to elements of ℤ[X], but because the pointwise product of ring endomorphisms does not generally respect addition, the pointwise operations do not equip End(ℤ[X]) with a ring structure (and in fact, no ring structure on Hom(R,S) can make the category of commutative rings monoidal closed for the tensor product of rings (this is because the monoidal unit is initial)). We can relax the rules slightly, though.

Who says we need the multiplication (or addition, for that matter) on End(ℤ[X])? We still have the bijection ℤ[X] ↔ End(ℤ[X]), so we can just give ℤ[X] the composition operation by transfering along the correspondence anyway. If p and q are polynomials in ℤ[X], then p ∘ q is the polynomial you get by substituting q for every instance of X in p. By construction, this satisfies (p + q) ∘ r = (p ∘ r) + (q ∘ r) and (p × q) ∘ r = (p ∘ r) × (q ∘ r), but we no longer have left-distributivity. Furthermore, composition is associative and the monomial X serves as its unit element. The resulting structure is an example of a composition ring!

The composition rings, like the commutative unital rings, and the abelian groups, form an equational class of algebraic structures, so they too have free objects. For sanity's sake, let's restrict ourselves to composition rings whose multiplication is commutative and unital, and whose composition is unital as well. Let C be the free composition ring with these restrictions on one generator. The elements of this ring will look like polynomials with integers coefficients, but with expressions in terms of X and a new indeterminate g (thought of as an 'unexpandable' polynomial), with various possible arrangements of multiplication, summation, and composition. It's a weird complicated object!

But again, the set of composition ring endomorphisms C -> C (that is, ring endomorphisms which respect composition) will have a bijective correspondence with elements of C, and we can transfer the composition operation to C. This gets us a fourth operation on C, which is associative with unit element g, and which distributes on the right over addition, multiplication, and composition.

This continues: every time you have a new equational class of algebraic structures with two extra operations (one binary operation for the new composition and one constant, i.e. a nullary operation, for the new unit element), and a new distributivity identity for every previous operation, as well as a unit identity and an associativity identity. We thus have an increasing countably infinite tower of algebraic structures.

Actually, taking the union of all of these equational classes still gives you an equational class, with countably infinitely many operations. This too has a free object on one generator, which has an endomorphism algebra, which is an object of a larger equational class of algebras, and so on. In this way, starting from any equational class, we construct a transfinite tower of algebraic structures indexed by the ordinal numbers with a truly senseless amount of associative unital operations, each of which distributes on the right over every previous operation.

born to go to beautiful libraries and study, forced to hustle in my room <3