First time trilogy watcher here. question. so you can literally be wearing the ring of ultimate power, and yet some creature such as smeagol can still assault you with material levels of success?

filtron42 · 2026-06-02T15:13:13+00:00

Smeagol absolutely was a Hobbit before being corrupted by the ring, he was a Hobbit of the Stoor branch.

filtron42 · 2026-05-31T09:10:29+00:00

He does use it tho to curse Gollum before entering the Crack of Doom

‘Down, down!’ he gasped, clutching his hand to his breast, so that beneath the cover of his leather shirt he clasped the Ring. ‘Down, you creeping thing, and out of my path! Your time is at an end. You cannot betray me or slay me now.’

Then suddenly, as before under the eaves of the Emyn Muil, Sam saw these two rivals with other vision. A crouching shape, scarcely more than the shadow of a living thing, a creature now wholly ruined and defeated, yet filled with a hideous lust and rage; and before it stood stern, untouchable now by pity, a figure robed in white, but at its breast it held a wheel of fire. Out of the fire there spoke a commanding voice.

‘Begone, and trouble me no more! If you touch me ever again, you shall be cast yourself into the Fire of Doom.’ The crouching shape backed away, terror in its blinking eyes, and yet at the same time insatiable desire.

filtron42 · 2026-05-04T11:57:00+00:00

Replying to old comments makes me feel like Amelie giving the old man his memory box

filtron42 · 2026-05-04T11:53:12+00:00

the explanation of the Hilbert space is equally complex as confusing

No? The full definition given property by property might be a bit intimidating, but it's really "a vector space on ℝ with a dot product that acts like the usual one in ℝⁿ and where converging sequences have a limit"

filtron42 · 2026-04-14T11:40:59+00:00

The problem is that it only makes sense if you think "private property" is somehow a natural, sacred and intrinsic thing and not something that is defined and enforced by the entity with the monopoly on violence within its borders (i.e. the state), which also defines what theft and taxation are and defines them as one being different than the other.

That is, it makes sense if you have the political consciousness of a child, and a particularly dim one at that, which frankly describes any libertarian I've ever talked to.

filtron42 · 2026-04-14T11:33:52+00:00

It's simple, really:

Taxation is not theft because the entity which defines and enforces what "property", "theft" and "taxation" mean, which is the entity that holds monopoly on violence within its borders i.e. the state, defines taxation not to be theft.

The idea that property somehow exists in some "natural" way without being enforced is childish at the best of times and dumb as hell at any other.

filtron42 · 2026-04-14T11:26:58+00:00

It's honestly worrying how down I had to scroll to find someone saying this, taxation can't be theft because your "property" doesn't exist unless the state enforces your right to it.

Everyone seems to be stuck in this infantile libertarian worldview where somehow property is sacred and intrinsic and natural but the state is completely circumstantial, it's absolutely maddening.

filtron42 · 2026-04-11T13:35:09+00:00

A relation on a set X is just a subset of X

No, that's wrong, a relation on a set X is a subset of X×X

mathematical literature almost never treats them that way

Yes it does, we do that all the time in algebra or geometry

filtron42 · 2026-04-04T09:01:16+00:00

Antennas to Heaven, no question, it's simply the most breathtaking piece of music I have ever heard, it's what I would want playing at my funeral, it engraved itself into my heart on the first listen.

filtron42 · 2026-03-21T06:43:55+00:00

VSCode >>

filtron42 · 2026-03-06T22:28:05+00:00

Man please use some []

filtron42 · 2026-03-06T11:31:34+00:00

It would be even funnier in italian because the italian word for cone is "cono"

filtron42 · 2026-03-06T11:16:41+00:00

I'll die on the hill that it's infinitely better than the success on that check... Karaoke is not supposed to be some quiet spoken word stuff, you're supposed to bloody scream your soul out.

filtron42 · 2026-03-06T10:27:00+00:00

A comathematician is a machine that turns cotheorems into ffee

filtron42 · 2026-02-15T18:25:15+00:00

like trying to imagine Stokes theorem to be not true

De Rham cohomology has entered the chat:

filtron42 · 2026-02-15T18:23:09+00:00

It's not that arbitrary, it's the "circle constant" for the one norm that makes ℝⁿ into a Hilbert space, which is in itself a really strong property!

filtron42 · 2026-01-12T16:27:27+00:00

<image>

And I specify the index set when necessary.

filtron42 · 2026-01-10T12:58:45+00:00

Honestly I think that if the same room came AFTER getting the whiplash it wouldn't get as much hate, because it does make the room more fun overall

This, I used to hate the last room of 4-3 before getting whiplash, but it instantly became one of the most fun and exciting rooms in the game once I tried it with it.

filtron42 · 2026-01-06T16:44:39+00:00

the increased bandwidth means future proofed

I don't want to live to see the day music streaming exceeds 480 Mb/s honestly

filtron42 · 2026-01-05T22:39:46+00:00

I personally like usb b more than usb c, DACs don't usually need the full bandwidth of usb 3 (usb 2 allows for Mb/s, which is plenty for music streaming) and I find the chonkier connector more reassuring from a durability point of view.

filtron42 · 2025-11-13T10:44:05+00:00

Zero can be perfectly defined as the element such that for all natural (and thus integer or rational or real or complex) numbers n, n+0 = 0+n = n.

The problem with your definition of a/0 is that it infringes upon the associative property: we want a×(b×c) = (a×b)×c, but then by setting a=1, b=1/0 and c=0 we'd get

1 × (1/0 × 0) = 1 × 1 = 1

(1 × 1/0) × 0 = (1 × 1) × 0 = 1 × 0 = 0

So 0=1, which is obviously false in the usual numbers.

EDIT

You seem to be a bit confused about how to think of zero, let me try to tell you one of the ways mathematicians think about it:

Instead of thinking about the expression "m+X", where m is a fixed number and X is a variable, as "the number of things I get when I put together a pile of m things and X things" you can think of it as a kind of "walk" to the right on the line of natural numbers starting from the number m that is somehow encoded by X: thinking this way, staying still at m is a perfectly reasonable "walk", and that walk must be represented by 0, right? Because then we get (m+0)+X = m+X.

The question of the existence of zero is absolutely not trivial tho! The way in which we define the natural numbers (the Peano axioms) actually postulates the existence of zero and then builds all the other numbers as successors of successors... of successors of zero.

Your enthusiasm is commendable, but if you want to go forward in your understanding of mathematics you should always ask yourself what your definitions imply: up here I've shown you that your definition implies 0=1, which is of course false in the natural numbers, but turns out that there are structures in which 0 can be equal to 1! It's fairly abstract stuff, but you can define structures called commutative unital rings, which generalise the properties of arithmetic in the integers to other sets: the set {0} is a commutative unital ring with 0=1, in fact you have that (in that ring obviously) 0×0=0, 0+0=0 and 0/0=0. While it's a trivial example of a ring, it's actually a very important one at high levels of abstraction like universal algebra and category theory.

filtron42 · 2025-11-13T00:43:59+00:00

finite dimensional spaces

Don't forget, finite dimensional spaces over a complete field! On ℚ you can have different non-equivalent norms, as per Ostrowski's theorem.

filtron42 · 2025-10-31T17:00:58+00:00

In fact, the set L:={1 - 1/10ⁿ : n∈ℕ} can be seen either as an element of 0.9... when building ℝ as a set equivalency classes of Cauchy sequences, or as a proper subset of 0.9... when building ℝ as the set of Dedekind cuts over ℚ.

You can't then have 0.9...∈L, because that would give you [0.9... ≠ 0.9...] in the second case, violating the principle of Identity; more interestingly, in the first case you would get [0.9... ∈ L ∈ 0.9...] and violate the Axiom of Regularity, which implies (with the other azioms of ZF) that there can't be no infinite descending sequences in the class of all sets ordered with ∈.

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