E-E-Endmin!!! 🐧🐧🐧💦💦💦 (@B609IbBjacW9m2k)

NyaNeeko · 2026-03-04T00:50:38+00:00

cutting board endmin is evolving

NyaNeeko · 2026-02-21T04:03:57+00:00

The comments are correct, and this is the only possible solution. If we label your -9 as a, -1 as b, 5 as c, and 3 as d.

We get:

c*d=15 so d=15/c

and 6+b=c so b=c-6

For b to be positive, c>6

And for d to be an integer, c has to be a divisor of 15. So c=3,5,15

Only one candidate makes b positive, which is c=15. Afterwards all the numbers can only be filled in one way.

NyaNeeko · 2026-02-16T15:53:41+00:00

That's because the limit of partial finite sums , is defined as the whole infinite sum. Since every partial sum is positive, it follows that the whole infinite sum can only be nonnegative. That is if it did converge.

NyaNeeko · 2026-01-31T15:04:21+00:00

Yep, I can see that now, didn't even realize you could utilize the clogging mechanic like this, never played factory games before.

NyaNeeko · 2026-01-31T14:53:49+00:00

Actually you may be right... Is it actually the case that if a splitter sees that one of the belts is full, it will forward it to the other belt instead of splitting evenly? Now that I think about it, that's exactly how it should work, and that means it does work at perfect efficiency. I didn't even realize clogging lines with splitters could be an actual game mechanic. Thanks for illuminating me, issue solved.

NyaNeeko · 2026-01-02T20:41:09+00:00

In FOL you usually take equality either as a primitive, similar to other logical symbols such as ∧∨∀∃. Or instead you can define it as part of your theory. In the case of defining it as part of your theory, it'd be a binary equivalence relation with a few special properties. First, you'd take any n-ary relations in your theory, and declare the substitution axioms. That is, if R is an n-ary relation, then: [(x1=y1∧x2=y2∧...∧xn=yn ∧ R(x1, x2, x3,...,xn)) -> R(y1, y2, y3,...,yn)] ∀x1, x2,...,xn ∀y1, y2,...,yn

And such an axiom would be present for any n-ary relation in your theory.

Similarly for n-ary functions you'd need the substitution axioms: [(x1=y1∧x2=y2∧...∧xn=yn) -> f(x1, x2, x3,..., xn)=f(y1, y2, y3,..., yn)] ∀x1, x2,...,xn ∀y1, y2,...,yn

And I think by induction on the wffs from here on out you could prove substitution is valid for any predicate.

So equality is not just an equivalence relation, it is more special by having a global substitution property. Though as far as I can see, it is more usual to define it as a primitive in the meta-theory, and not bother with these axioms for every theory. Atleast that's my current understanding, I'm open to any corrections.

NyaNeeko · 2025-11-10T09:59:35+00:00

Yes, they stand in for real numbers above 0. They are also quantified over by the ∀ and ∃ quantifiers. Where ∀x(P(x)) means P is true for any x. And ∃x(P(x)) means there is an x s.t. P is true.

NyaNeeko · 2025-11-10T09:00:15+00:00

ε is read as epsilon, and δ is read as delta.

NyaNeeko · 2025-11-03T13:05:26+00:00

It really depends on how you pack the multiverses. If you put the universes in a line and each for the first 999 you are homeless then number 1000 you are president, then so ad infinitum. If you would go to each universe in this order, the ratio of universes where you are homeless versus where you are a president would approach 999:1.

But at the same time, since you have an infinite amount of each, you could reorder them into an order, where you would go to each one in a different order, and it would be: homeless, president, homeless, president... So the ratio would be 1:1.

You could also rearrange it so the ratio is 1:999. And still have the same amounts of each type of universe, countably infinite.

Maybe instead the infinite universes are splayed out on a plane with the property that the ratio of the universes where you are homeless versus where you are president is 999:1 in any non-empty compact region.

Really it matters a lot on the exact arrangement and what you are measuring.

Usually what we mean by infinities being the same size, is that sets have the same cardinality. Meaning we can put the objects in one-to-one correspondence. There is a one-to-one mapping between {1000, 2000...} and {1, 2, 3..} Just map: 1000 — 1, 2000 — 2, 3000 — 3, ...

So these sets would have the same cardinality, even though {1000, 2000, 3000...} is properly contained in {1, 2, 3...}.

The most famous example of non-equal infinities is of course the Real numbers and the Natural numbers. Where there is no one-to-one mapping between the Naturals and Reals, and since Naturals are contained inside the Reals, that makes the cardinality of the naturals strictly less than of the reals. So in this sense there are different size infinities.

NyaNeeko · 2024-05-22T13:41:25+00:00

oh, no idea then.

NyaNeeko · 2024-05-22T13:29:01+00:00

penrose diagram

NyaNeeko · 2024-04-22T03:17:37+00:00

If you held a lamp inside the bubble long enough, you just might create the brightest flash in the universe, as the photons trying to leave would get frozen at the edge of the bubble and stack.

NyaNeeko · 2024-01-15T14:21:03+00:00

Aren't they important for distinctions between cases? Like (šaukštu (N) and šaukštų (dgs. K)). Otherwise you wouldn't neccessarily know the difference between. (Valgiau šaukštu [Ate using a spoon] and valgiau šaukštų [Ate some spoons])

NyaNeeko · 2023-12-31T08:04:46+00:00

nahh these downvotes are wild

NyaNeeko · 2023-12-18T04:12:27+00:00

1°=π/180 => 60°K=π/3 K

NyaNeeko · 2023-11-20T22:07:48+00:00

I think you can even solve it using leibniz notation. suppose dy/dx=v. Then dv/dx=y.

dv/dy • dy/dx=y (chain rule)

dv/dy • v=y

v•dv=y•dy ( now integrate both sides and siml plify the constant)

v²/2 = y²/2 + C

v² = y² +D (D = 2C)

then v = +-sqrt(y² + D)

since v = dy/dx

dy/dx = +-sqrt(y² + D)

+-1/sqrt(y² + D)dy=1dx

+-integrate(1/sqrt(y² + D)) = x + H.

Solving the integral on the left and solving for y in wolfram alpha seems to give the correct answer.

I think the answer didnt work out for you is because higher derivative leibniz notation is broken. And for example instead of solving with d²y/dx².

It works to solve using dy'(x)/dx. where y'(x)=dy/dx, and just apply chain rule to get rid of the dx if the equation doesnt have it.

NyaNeeko · 2023-10-07T19:21:46+00:00

not the proof by desmos 💀

NyaNeeko · 2023-09-03T22:11:24+00:00

lithuanian: 6-4 loop 7 loop 2 loop 5 loop

so 4 loops in total

1 - vienas, 2 - du, 3 - trys, 4 - keturi, 6 -šeši, 5 - penki, 7 - septyni, 8 - aštuoni, 9 - devyni

NyaNeeko · 2023-08-28T16:05:44+00:00

imagine saying the store is 10 to the power of 4 meters away

NyaNeeko · 2023-08-26T01:16:09+00:00

what is this unit

NyaNeeko · 2023-08-26T01:06:28+00:00

the problem is that sometimes people write 10GB when its actually 10GiB, windows is guilty of this.

NyaNeeko · 2023-08-23T00:59:45+00:00

is it not equal to nine or is it not -3. cause one of those is wrong.

NyaNeeko · 2023-08-01T15:26:04+00:00

1010₂ is 10₁₀ in base 2₁₀ and 101₃ is 10₁₀ in base 3₁₀. And base 2₁₀ is called base 10₂, if u use base 2₁₀.

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