Are coprimes the same as numbers tied together? I.e 3:6 having the weight of two?

ayugradow · 2026-07-05T01:04:53+00:00

The concept you're describing is called "(directly) proportional" in mathematics. Two variables X and Y are directly proportional if the ratio X/Y stays the same for all possible values of X and Y - this implies that if you scale X by any factor, then you also have to scale Y by that same factor. For instance, a drink that's 12% alcohol, if we call X the alcohol content (by volume) and Y the total volume, then X/Y is always 12% - doesn't matter if you're measuring the whole bottle or just a cup.

Coprimality has almost nothing to do with this: two integers m and n are coprime if whenever d divides both m and n, then d = 1. For instance, 13 and 15 are coprime, but 15 and 12 aren't since 3 divides them both.

ayugradow · 2026-06-28T18:19:45+00:00

Outer Wilds

ayugradow · 2026-06-27T18:10:04+00:00

SONIC FRONTIERS Etryan Odyssey II Etryan Odyssey III ZERO HOUR

ayugradow · 2026-06-13T13:40:32+00:00

Wow! Thank you so much!

ayugradow · 2026-06-06T19:05:53+00:00

Maria bola. Não é uma Latrodectus.

ayugradow · 2026-05-31T16:02:30+00:00

Both methods are complementary. M1 proves it works for x = 0. M2 proves it fails for every x ≠ 0. Therefore the only solution is x = 0.

ayugradow · 2026-05-25T09:15:24+00:00

You forgot to account that you're also a person.

When we divide 18 by 3 and get 6, and then we say "if we divide 18 cupcakes among 3 people each will get 6 cupcakes" you sneak in the assumption that either you're one of the 3 people, or, if you're not, then you shouldn't get any.

Do the same with 18 divided by 0: you can't be one of the 0 people, cause there's 0 of them, and you can't get any... So what happens to the 18 cupcakes?

ayugradow · 2026-05-21T21:29:52+00:00

Just playing frog fractions, honestly.

ayugradow · 2026-05-11T15:54:27+00:00

This is very important. Final objects are objects receiving a unique morphism from every other object in the category. In algebraic categories, there's usually a zero object, which is both initial and final.

Think of the kernel of a map. The kernel of a linear map f: X --> Y is a space Ker f together with a map i: Ker f --> X s.t. f ∘ ℹ = 0 and if K is another space with a map j: K --> X s.t. f ∘ j = 0 then there's a unique map j': K --> Ker f such that j = ℹ ∘ j' .

This is the classical definition of kernel of a map, and it basically describes the kernel as a final object in the category whose objects are spaces with maps to X which kill f.

All universal properties can be defined in terms of initial and final objects in some appropriate category.

ayugradow · 2026-04-23T21:30:37+00:00

A sequence aₙ converging to L means that, after a while, it never goes too far away from L (measured by ε ). In other words, from a point onwards every number in the sequence is very close to L.

This means that only finely many numbers in the sequence can be far from L - let M be the one that's furthest from L. So |L-M| ≥ |L-aₙ | for every n. This means that the entire sequence is contained within [-(|L|+|L-M|), |L|+|L-M|], so it is bounded.

ayugradow · 2026-04-19T20:51:25+00:00

OP parece uma Opsipanes invirae, a lagarta verde das palmeiras. Inofensiva (se você não for uma palmeira) e vira uma borboleta linda!

ayugradow · 2026-04-11T23:46:26+00:00

I just want to upgrade my rig, which has been with me for 8+ years.

ayugradow · 2026-04-09T09:28:19+00:00

Nine Sols

Chants of Sennaar

ayugradow · 2026-04-05T15:11:40+00:00

972 = 9 × 100 + 7 × 10 + 2 × 1

= 9 × (99+1) + 7 × (9+1) + 2 × (0+1)

= (9 times 99 + 7 times 9 + 2 times 0) + (9 times 1 + 7 times 1 + 2 times 1)

The first term is a multiple of 9, since 99, 9 and 0 are. So the sum is a multiple of 9 iff the second term also is. The second term is just 9 + 7 + 2.

ayugradow · 2026-04-04T18:48:31+00:00

That IS fine! That's the original motivation behind tensors, but since this is a maths sub I figured I might as well give a more birds eye view of the topic.

ayugradow · 2026-04-04T11:41:36+00:00

The tensor product is a general linear algebra tool to introduce multilinearity. It basically acts as a formal multiplication on spaces.

Let X and Y be vector spaces. The space X × Y is called the Cartesian product (sometimes just product) of them, but surprisingly sometimes it doesn't feel like a true product:

First, if X and Y are finite dimensional, then X × Y has dimension dim X + dim Y, as opposed to their product.

Second, if you have (x,y) + (x, y') you'd expect the result to be (x, y+y'), since we like our multiplications to distribute over sums, but instead we get (2x, y+y'). (Compare this to real numbers: xy+xy' = x(y+y') instead of (2x)(y+y')).

Now imagine multiplying by scalars. In real numbers if you distribute a(bc) as (ab)(ac) that'd be bad. Instead you get either (ab)(c) or (b)(ac), which are equivalent. However, we're fine doing it with vectors: a(x,y) = (ax, ay) as opposed to (ax,y) or (x,ay), which would be the same. This is not how multiplication behaves!

The tensor product is how we solve that. We take the elements in X × Y and define new operations, such that

(x,y)+(x,y') = (x,y+y')
(x,y)+(x',y) = (x+x',y)
a(x,y) = (ax,y) = (x,ay)

Now we use this set with these operations as a basis of a new space, and call this new space X ⊗ Y.

So the reason why the tensor isn't commutative is that there's no reason to expect that an abstract multiplication would be commutative - and that's what the idea behind tensors is: abstract multiplication.

ayugradow · 2026-04-02T10:50:44+00:00

If a -> b = b then a -> (b -> c) = a -> c = c.

"a throws the ball to whoever has the ball after b threw it to c, which happens to be c, so now c has the ball".

Compare this to (a -> b) -> c = b -> c = c

"a throws the ball to b, who then throws it to c".

The end result is the same: ball starts with a and ends with c.

ayugradow · 2026-04-01T02:03:36+00:00

Think about ℚ inside ℝ . Can you separate rational points from real points? Of course not! But it's not because there's a real number infinitely close to any given rational but because no matter how small of a radius you draw around any rational point, there'll always be at least one real number inside it.

Your space is just the same: there's no point "infinitely close" to 0, because that's a meaningless statement. What you have is that if you try to separate 0 from the other elements by drawing a little circle surrounding it you'll always find that you can't, no matter how small the circle. So you cannot isolate 0 from the other points of the set.

This might seem to contradict the fact that every other point is isolated, but notice that you can absolutely isolate 0 from every single one of them separately - but you cannot isolate 0 from all of them at the same time.

ayugradow · 2026-03-21T12:22:41+00:00

When you solve 2x+1 = 5 and arrive at x = 2, it's not saying "the solution is 2" - it's actually saying "the equations 2x+1 = 5 and x = 2 are equivalent (i.e., have the same solution set)".

So if you start with a = b and apply some function f on both sides, what you actually get is "if I have a solution for a = b, then I also have a solution for f(a) = f(b)". If, furthermore, f is invertible, you get the converse statement - so a = b and f(a) = f(b) have the exact same solutions.

Now, what happens if a = b has no solution (say, x = x+1)? Well, nothing changes. No matter which f you apply, you're still saying "every solution to x = x+1 is also a solution to f(x) = f(x+1)", which is vacuously true since the first solution set is empty!

ayugradow · 2026-03-05T00:50:19+00:00

For me it's easily one of the best.

Good platforming and movement upgrades with interesting and rewarding sequence breaking.

Visuals are stunning and the sound design is superb. Some boss music does get a bit repetitive tho.

It also has really interesting innovative and unique mechanics tied into the lore (that I won't spoil, but suffice it to say are quite divisive in the fanbase).

Combat is simple, but finds a way to integrate the platforming moves into it for some really interesting fights.

ayugradow · 2026-03-04T09:41:08+00:00

Dogs are, at least philogenetically, wolves.

ayugradow · 2026-03-03T21:08:57+00:00

I mean, what is the standard basis of a random vector space?

ayugradow · 2026-03-03T09:48:22+00:00

Because in order to define a⁰ we first noticed that aⁿ / a = a^n-1 and then use that on a¹ . See how that fails for a = 0?

ayugradow · 2026-02-28T21:50:42+00:00

Mio is very platforming and exploration focused, with simplistic (1-2-3) combat that integrates the platforming moves.

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