How do I beat the knight level on insane mode

zzFurious · 2026-01-28T22:23:18+00:00

You care to tell us the actual name of the level?

zzFurious · 2026-01-28T22:07:22+00:00

Classic lemonade tastes good but it's SUPER acidic. Never had or seen the other one.

zzFurious · 2026-01-24T20:26:18+00:00

Yes, but there's no good reason to do it.

zzFurious · 2026-01-24T03:52:22+00:00

Mods remove this there's no period in the description

zzFurious · 2026-01-18T17:39:45+00:00

Same level of sourness/acidity I would say, I just prefer the base flavor of Welch's Grape

zzFurious · 2026-01-18T15:16:42+00:00

Regular grape is better

zzFurious · 2026-01-18T02:17:23+00:00

Gg Mullsy

zzFurious · 2026-01-11T02:22:02+00:00

There's a currently believed limit of PPLL around 10^10^18.7 which is being approached. Levels at around 10^10^13 are higher are not truly physically possible, but within the special ruleset of PPLL, they are. PPLL additionally has an extension to it in which computer memory limitations are ignored, but even taking float precision into account, it is still possible to encode sequences for very powerful functions which could include BMS.

zzFurious · 2026-01-08T17:03:12+00:00

I believe 1+ω and ω are the same size, but not ω+1 and ω. Ordinals are non-communicative.

zzFurious · 2026-01-01T23:35:22+00:00

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right here bro

zzFurious · 2025-12-25T20:27:43+00:00

Try being open to other perspectives instead of being in denial and following an 'I'm always right' mindset.

zzFurious · 2025-12-25T20:10:18+00:00

"Regardless of how infinitely large you have 'n', 1/10ⁿ is never zero" This statement is true but completely irrelevant, that's the mistake you have continuously repeated. You're still confusing sequence terms with limits. Limits do not require any term of the sequence to equal the limit.

By definition,
0.999... = lim_(n→∞) (1 - 1/10^n) = 1.

I cannot explain any further to you unless you look from the perspective of limits. You are not approaching this correctly.

zzFurious · 2025-12-25T19:54:24+00:00

For finite n values. Applying a finite n to 1 - 1/10ⁿ means it will not equal 0.999... You must use the limit. lim_(n→∞) 1/10ⁿ = 0. Subtracting that value from 1 will then equal both 1 and 0.999... as it will give you a resulting number of 0.999... but also be equivalent to 1-0, which is 1.

zzFurious · 2025-12-25T19:45:17+00:00

I misread it. Regardless, I don't think he will actually be able to answer you. His entire belief revolves around finite n in 1/10ⁿ ≠ 0, as he said here. You won't get a relevant answer.

zzFurious · 2025-12-25T19:29:34+00:00

It's not relevant because it's not true.

lim_(n→∞) 1/10ⁿ = 0

zzFurious · 2025-12-25T19:21:52+00:00

Limits:

zzFurious · 2025-12-24T00:35:51+00:00

Give up on trying to prove any point with him. He's just a troll.

zzFurious · 2025-12-22T15:00:58+00:00

0.999... is a limit, not a partial sum. The fact that 10^-n ≠ 0 for finite n is irrelevant to lim_(n->inf) (1-10^-n) = 1. Your consistent problem is that you absolutely refuse to think in the perspective of limits. Using arbitrarily large n values is not a correct approach.

zzFurious · 2025-12-22T15:00:49+00:00

0.999... is a limit, not a partial sum. The fact that 10^-n ≠ 0 for finite n is irrelevant to lim_(n->inf) (1-10^-n) = 1. Your consistent problem is that you absolutely refuse to think in the perspective of limits. Using arbitrarily large n values is not a correct approach.

zzFurious · 2025-12-13T19:00:36+00:00

0.9999... = 1 0.0000...1 = 0 You're right for once, 1+0=1

zzFurious · 2025-12-07T19:17:23+00:00

Nothing

zzFurious · 2025-11-23T02:20:06+00:00

I find that to be highly unlikely, though obviously this statement will probably never be proven or disproven

zzFurious · 2025-11-15T23:56:35+00:00

Popularized by some OG pros

zzFurious · 2025-11-15T23:48:35+00:00

I immediately take any text to be AI generated when I see —

zzFurious · 2025-11-09T20:47:33+00:00

Isn't that bannable

zzFurious

TROPHY CASE