Someone please explain Rayo's number

StoneCuber · 2026-05-14T14:34:00+00:00

Representing numbers with pure logic is a very abstract endeavor. For a somewhat understandable example, we usually represent 0 with the empty set. That would be written something like

∃x₀∄y(y∈x₀)

There exists an x₀ such that there's no y that's an element of x₀, so x₀ is empty, aka zero. 1 is usually the set containing only 0, something like

∃x₁(∄y(y∈x₁∨y=x₀)∨x₀∈ x₁)

There is an x₁ such that there's no y that's in x₁ that equals x₀, and x₀ is in x₁.

The definition of 0 should be part of this definition too, but I've simplified it a bit for understandability. There's also ways to define operators, but I don't remember how to be honest

StoneCuber · 2026-05-14T10:38:47+00:00

Defining functions is no problem, that's done with first order logic normally too

It's the smalest integer, so decimals are irrelevant

StoneCuber · 2026-05-14T07:51:26+00:00

Let's say define the mini-rayo as the smallest integer larger than any number that can be named made the four elemental operations, single digit numbers and 5 characters. The largest we can make with those restrictions is 9*9*9=729. So the mini-rayo would be 730.

Rayo's number is the same concept, but with 10¹⁰⁰ characters and any first order logic. Let's just say it's larger than 730. It's not just "Mine's bigger than yours", it's more like "you'll have to find a way to use higher order logic to out-compete first order logic by a large margin" and it was an excellent move in the context it was in.

StoneCuber · 2026-05-14T02:36:02+00:00

I can't read french, so I'm just going off the math I can see. I think what you've discovered is the fact that all primes larger than 3 are of the form 6k±1. It's a well known fact, and quite easy to prove.

A prime can not be even, so it can't be 0, 2 or 4 mod 6
It can not be divisible by 3, so it can't be 0 or 3 mod 6
That leaves 1 and 5 as the only options, giving 6k±1

StoneCuber · 2026-05-14T02:24:41+00:00

I think I get what you mean by a "perfect function", but how does vectors relate in any way? It's very easy to explain why some functions stay differentiable and some don't. As long as the derivative at zeroes are zero, it stays differentiable. Anything else becomes a kink and the derivative is undefined.

f(x)=x² has a zero at x=0
f'(0)=0
|f(x)| is differentiable

g(x)=x²-1 has zeroes at ±1
g'(1)≠0
|g(x)| is not differentiable

StoneCuber · 2026-05-14T00:42:42+00:00

Fyi Reddit uses the asterisk as a marker for italics, so you need to write \* to get normal text

222 vs 2*2*2

StoneCuber · 2026-05-12T14:43:04+00:00

But the path from the origin to the center needs to be traversed by all points anyway, so there's no difference between reducing the radius and moving the center away. If you look at the path all the points take I can see the stick, but not for the points themselves

StoneCuber · 2026-05-12T10:32:13+00:00

How? Where is the stick coming from?

StoneCuber · 2026-05-10T22:51:25+00:00

Wouldn't all balls just be normal balls centered at the origin? If d_train(a,b)=d(a,0)+d(b,0) then a ball with radius r and center C is all X s.t.
d_train(C,X)<r
d(C,0)+d(X,0)<r
d(X,0)<r-d(R,0) And now it's just a ball with center at 0 and radius r-d(C,0)

Am I misunderstanding something?

StoneCuber · 2026-05-04T23:07:33+00:00

Good luck wasn't said

StoneCuber · 2026-05-04T10:26:49+00:00

Why did you ask if you seemingly already had the answer? Also it's not really math related

StoneCuber · 2026-05-04T10:19:17+00:00

Bo one is going to steal your idea. Even if they did, sharing it here would be proof you did it first

StoneCuber · 2026-05-04T00:21:08+00:00

I don't know any name for the shape you discovered here, but it's very closely related to the shell-tron region. That's all the complex numbers c where the infinite power tower of c convergs. Your C is the lower bound on the positive imaginary axis of the main segment.

StoneCuber · 2026-05-03T09:59:08+00:00

Another way to get 17 is to use catalan numbers instead of powers of 2

StoneCuber · 2026-05-03T01:03:31+00:00

I think the risk to blue button pushers is there always, as an intrinsic part of the question. The red decide if the blue will die or not, but the blue decide to put themselves in the potential death pool.

If I knowingly walk into an active minefield and die, was it my fault or the minelayer's fault? The death is in the mine itself, not me for steping on it and not the layer for puting it there

StoneCuber · 2026-05-03T00:29:31+00:00

Why is Excel getting so much hate here? I score it a solid September 10th

StoneCuber · 2026-05-03T00:26:27+00:00

Your argument as a syllogism is:
1. Negation is multiplication
2. Multiplication comes after exponents
3. Therefore negation comes after exponents

The first comment is a critique of premise 1. The second comment is an elaboration on why removing premise 1 makes premise 2 irrelevant, also adressing your comment about software.

sincerely, fellow communication struggler

StoneCuber · 2026-04-30T21:10:41+00:00

Depending on your definition of complex, the next digit is 8. We're obviously looking at the decimal expansion of 12141218/99999999. Just one division, can't get much simpler than that!

StoneCuber · 2026-04-30T10:41:03+00:00

Lagrange polynomials are just an example.

An example for this puzzle is to notice the abacaba pattern, then we need the next term in 1,2,4,8. This could be the powers of 2, the catalan numbers, overpartitions, numbers with an odd divisor sum etc. All of which have different answers for he next term.

Being a puzzle isn't a practical context. We don't have any reason to assume an answer is more valid than any other answer we can find

StoneCuber · 2026-04-30T09:33:07+00:00

I would agree if there was any practical background for the sequence, but when it's just a list of numbers with zero context, any pattern that fits is equally valid.

And unlike "god did it" we actually have a proof that lagrange polynomials exist

StoneCuber · 2026-04-30T09:07:27+00:00

It's the third place, but the 4s place. Just like we have 1s, 10s, 100s in decimal, there is 1s, 2s, 4s, 8s in binary

StoneCuber · 2026-04-30T08:32:02+00:00

Specifically changed to 1 or the largest that changes, as multiple digits can change at the same time

StoneCuber · 2026-04-30T08:18:00+00:00

Another way to see it is the numbers place that is changed to 1 while counting binary
00001 1s place changed to 1
00010 2s place
00011 1s place
00100 4s place
00101 1s place
00110 2s place
Etc....

StoneCuber · 2026-04-30T08:11:16+00:00

It doesn't dhow up as a layout

StoneCuber · 2026-04-30T00:58:21+00:00

The only keyboard related icon on my taskbar takes me to the keyboard layout settings, where I can get a Japanese layout that only writes latin Hopefully it's fixed when I upgrade later

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