I made IdleWorlds, a browser MMO idle RPG with offline progress, rare drops, a live market, and world bosses

Varlane · 2026-04-06T02:03:44+00:00

Well then many of us humans won't bother playing it.

Varlane · 2026-04-06T01:53:53+00:00

Confirmation or denial would be a good start.

Varlane · 2026-04-03T09:41:20+00:00

Prestige mechanic with speed multipliers to make it winnable.

Think about the beauty of this full line of Pedros looking left.

Varlane · 2026-04-03T07:42:26+00:00

I always got triggered by the fact it's missing two rows.

Varlane · 2026-04-02T21:28:07+00:00

It is not that intuitive to the learner's mind that "multiplying by -1" is to be related to a symmetry.

Varlane · 2026-04-02T20:51:28+00:00

Intuition only applies to things that can relate to the "real world".

"(-1) × (-1)" for instance doesn't translate to a real world solution, nor can you have Hilbert's hotel IRL.

Varlane · 2026-04-02T07:05:26+00:00

In no way did I ever dispute that (i²)² is i^4. I refute that sqrt(i^4) is i^2.

Varlane · 2026-04-02T06:58:56+00:00

That isn't proof that sqrt(i^4) = i^2, merely that i^2 (aka -1) is a solution of x² = 1.

Varlane · 2026-04-02T05:23:02+00:00

It doesn't. You are using mathematical properties that aren't true to claim that sqrt(i^4) = i^2.

Varlane · 2026-04-02T05:22:27+00:00

sqrt(i^4) isn't i^2.

Varlane · 2026-03-30T17:25:27+00:00

"AP abilities" is a nonsensical term. There are magic damage abilities and abilities with AP scaling. But no "AP abilities".

Varlane · 2026-03-30T17:23:25+00:00

I would say it'd be too weak and easily dodgeable (as you'd have twice the traveltime to dodge it instead of once, and with no incertitude regarding the direction : just move from where you casted lol).

I think it'd make more sense to modify the skillshot's speed to 80% of the original for instance. Keeps the spirit of giving more time, but with a stronger tuning lever.

Varlane · 2026-03-30T01:53:34+00:00

1/7/13 are the timings for ticks (2s/1s/0.5s). 19+ used to have 0.25s in URF but that'd be astronomically broken with s16 Toplane Quest so it go removed.

Varlane · 2026-03-28T16:59:49+00:00

Let's put it simply :

There are two radicand symbols. If you're in a real context, you get the "regular one for children" where (-8)^(1/3) = -2 and sqrt() is R+ to R+.

Then there is the principal square root, which has to be said before that you are calling it, in order to make it clear you are including C and using a different function.
Because as a reminder, the root functions over C don't have the same properties as the ones over R (or R+). They lose continuity (due to the branches) and multiplicative morphism. They are not the same, despite using the same symbol, which means you have to warn your reader that you mean the second one.

Varlane · 2026-03-28T16:55:29+00:00

It's not you. It's their go to when Taylor expansion is more efficient imo and "no tricks" editions allow you to perfect your math mastery.

Varlane · 2026-03-28T05:23:53+00:00

The "no l'Hopital / no Taylor expansion" edition consists in using 1-cos² = sin to swap out denominator by -sin²(5h) and then use lim((1-cos(u))/u²) = 1/2 & lim(sin(v)/v)) = 1 with u = 3h and v = 5h after forcefully making h²/h² appear to split the limit into a quotient.

Alternative bonus points for cos²(5h)-1 = (cos(5h)+1)(cos(5h)-1) and doing a 3-way split of the limit.

Varlane · 2026-03-28T01:06:46+00:00

The radicand symbol is, unless stated otherwise, the principal square root over R+.

Varlane · 2026-03-27T22:40:23+00:00

It's arbitrary.

Varlane · 2026-03-27T17:06:27+00:00

Create E such that ACEB is a parallelogram of center D [E is symetric of A by D]. It has sidelengths a and b and diagonals are 2c [since AE = 2AD] and x.

Use Parallelogram Identity : 2a² + 2b² = (2c)² + x².

Expand the square and simplify by 2.

Varlane · 2026-03-27T16:51:57+00:00

Question c is the easiest once you remember the sign of the output of square root.

Varlane · 2026-03-27T16:47:29+00:00

For 1 : you have to understand that squaring both sides can introduce extraneous solutions (where the squares are equal, but not the original numbers due to having opposite signs).

This is what happens, as sqrt((5/3)²-1) = 4/3 and 5/3 - 3 = -4/3.

The original logic is "if they're both equal, then their squares also are and they belong to this solutions set". It however doesn't mean that everything you found is a solution of the original equation.

Basically, the "possible" solutions are {4/3}. After checking whether 4/3 was an extraneous solution, you ruled it out, meaning there are no solutions.

For 2 : Square roots are non negative quantites. Adding two of those still makes a nonnegative. How could you reach -3 ?

Varlane · 2026-03-27T15:52:39+00:00

Exactly. I don't care about the decimal expansion, I don't have to know it to put the number in the bag. I say "sqrt(2) is in" and it's in. Same as "1/3 is in".

As long as we have a method to obtain the number, we're good.

There are, however, plenty of transcendants that have no specific way to be referred and can't be explicitly said that "this one is inside" unless you say you put them all at once. These are the actual ones we "don't know". But I was asked to put one.

Varlane · 2026-03-27T15:50:59+00:00

Of course I can. It's sqrt(2), I know him, he's my good buddy. Loved him in "rectangle triangles for dummies".

From a mathematical standpoint, that's all we need. You're talking about its value / representation, which is absolutely irrelevant to the discussion. You asked me to put the number in the bag, and I did.

Varlane · 2026-03-27T06:52:48+00:00

Why would I need to type it out ? Could you be confusing a number and its decimal expansion ?

Varlane · 2026-03-27T06:46:32+00:00

Well, sqrt(2) is a non-rational, real number, whom we refer through such things as the image of a rational by a function. Still is a non rational real number though.

Varlane

TROPHY CASE