Way Bigger Than Graham's Number (Goodstein Sequence) - Numberphile

[@numberphile]

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[@MichaelKah4712]

Wait, this got EVEN MORE interesting right when it stopped! We need a video about the speed limit!

[@robcoh]

I can't derive 55!

[@hans_bier]

Absolutely!

[@MrPlasterbrick]

Seconded

[@CodingDragon04]

It was briefly mentioned in the Tree(g(64)) vs g(Tree(64)) video with Tony I think.

[@robertschreur5138]

Agree

[@AlexanderEVtrainer]

Definitely need to hear more about that arithmetic speed limit. It's hard to imagine why a finite operation (even a stupidly big one) would increase too fast for basic math laws to keep up with.

[@alexritchie4586]

I guess there's a couple of reasons. Firstly that the universe puts a hard limit on how fast information can be transferred, so if the number was so enormous that relaying the amount of information contained within it would exceed the predicted lifetime of the universe, actually calculating it before the end of all existence would have to exceed the information speed limit. Secondly, Landauer's Principle states that computation has a maximum efficiency in proportion to the temperature at which those computations are performed, so it may be the case that the number could never be calculated by any computer we can currently imagine that would operate at a temperature higher than absolute zero.

[@jerrr-c-squared]

@@alexritchie4586 I don't think that's what this is referring to, pure math couldn't care less about the physical limits of reality.

[@convindix9638]

@jerr-c-squared True, and another thing is there's nothing special about Peano arithmetic. There's are a ton of formal systems used in foundations of mathematics (PA, KP, ZFC, RCA_0, ATR_0, etc.), each with their own speed limits, some slower than Peano's, some faster. There wouldn't be any reason for Peano arithmetic to be the special one with connections to the physical universe

[@Rubrickety]

@@alexritchie4586These things are true, but aren’t part of the unprovability results, which hold even if the universe lives arbitrarily long. They’re related to topics like primitive-recursive functions. I agree a video on this would be interesting. It’s a challenging topic though.

[@adb012]

How it takes "infinity" machinery to proof facts about finite numbers kind of reminds me of how it takes calculating negative square roots (and hence complex numbers) to find REAL roots in polynomials with REAL coefficients. The imaginary parts always magically cancel leaving you with a real number, but a real number that you could not access unless you take a detour through the complex world.

[@soranuareane]

I wouldn't mind a proper deep dive into Goodstein's own proof, including all of the fancy (and ugly!) ordinal analysis. Please do a video on infinite regress; it's such a satisfying idea!

[@Xnoob545]

The proof for goodstein sequences always dying off involves replacing the "base" numbers in the power towers with omegas (like the ordinal infinity), and using rules of ordinals to prove it

[@lynxfl]

The TREE(g64)th element of the Meta Goodstein Sequence is still finite

[@jamestappin4741]

But is it bigger than Rayo's number?

[@taxicabnumber1729]

The TREE function grows much faster than the Goodstein Sequence, which grows much faster than the G function. So TREE(TREE(3)) >>> Goodman(TREE(G64)). TREE also has a big brother called SCG, which is based on graphs rather than trees and grows much faster than TREE (or any salad number composed of nested TREE functions, etc). And Rayo's number is much, much larger than any of those.

[@shophaune2298]

@@jamestappin4741 No. Because of order of operations here, MGS(TREE(G(64))) ~= TREE(G(64)). It only takes ~7300 rayo symbols to define BB(n), so BB(BB(2^65536)) >> TREE(G(64)) takes around 15000 symbols. Rayo's number is the largest number you can make with 10^100 symbols.

[@Einyen]

@@taxicabnumber1729 So GMS(12) > G(64) "GMS = Goodstein Meta-Sequence", what is the smallest x so that GMS(x) > TREE(3) ?

[@david-melekh-ysroel]

​@@Einyen it would be GMS(o(4)) Where o(4) is my friend's omicron function of 4, it's valued at 4^^4, in other words : 4 tetrated to 4

[@ffc1a28c7]

We covered in my logic class showing that the Ackermann function (essentially the same idea of extending exponentiation and tetration) is what we call not "primitive-recursive". This is essentially saying that in Peano arithmetic, we can't prove anything actionable about it.

[@lkruijsw]

If your logic is restricted for Sigma 1 expressions (so no 'For All') for the induction condition you can't proof that Ackermann is terminating. This is a simpler example of incompleteness than Goodstein. The input of an Ackermann function can also be written as a very simple ordinal, a x epsilon + b.

[@Xnoob545]

If anyone's watched the extra videos about TREE vs. Graham and is familiar with fast-growing hierarchy, Goodstein sequences are approximately f_ε_0 (n) level (that's epsilon nought) Way way faster than Graham's omega + 1, but WAY WAY WAY slower than TREE (that one goes way beyond even gamma_0)

[@alansmithee419]

That's a capital gamma btw, for the feffermann-schute ordinal. There's a lower case gamma_0 much much earlier in the hierarchy (depending on notation), which comes shortly after epsilon_0.

[@miannekahkol9556]

I love when sequences start off looking normal and then zoom off to absurd numbers at even more absurd paces!

[@TheTaxiDriver1729]

there is a great video about this topic by PBS Infinite Series called How Infinity Explains the Finite

[@murmol444]

yeah, I remembered a similar problem about cutting Hydra's heads and thought it was on Numberphile. But it was PBS for sure

[@FloydMaxwell]

3^^3^^3 gets "so big" because of the eval order. If it was 27^^3, that is 19,683. But it is 3^^27 (=7,625,597,484,987).

[@shophaune2298]

I think you mean to have single ^ arrows there. 3^^27 >>> 7.6trillion

[@FloydMaxwell]

@@shophaune2298 Apparently you are correct. Was just going from 40+ year old memories.

[@NekuraCa]

Love the gyrobifastigium on the shelf in the background

[@QuantumHistorian]

Always fun to see the incompleteness theorem coming up. I knew of other examples, but none in strictly finite mathematics. Just really hammers home that what is "true" in mathematics depends entirely on the axioms you decide to work from, rather than some intrinsic thought-independent reality.

[@AFastidiousCuber]

It's funny you say that because Godel interpreted the incompleteness theorems as evidence in favor of platonism.

[@j0nasbs]

That escalated quickly

[@zaffyr]

This reminds me of the ant on a rubber band: If you stretch a rubber band at any speed you want, if an ant walks across it at a costant speed, it will eventually reach the other end, no matter how fast you stretch. This comes from the fact that the amount of length to go gets increased just as much as the ant's progress. In the goodstein sequence, the ant is represented by the -1 math is so interconnected; it's beautiful

[@christiandior8726]

what a beautiful mind you are!

[@mathematicalmusings]

Only works if you stretch at a constant rate. If the ant is going at 1m/s and has 1 metre to go, you can clearly always stretch the band in the next second so it has more than 1 metre to go in the next second so it will never reach the end.

[@maksymisaiev1828]

@@mathematicalmusings no, it will eventually. As long as distance behind ant will grow, it will reach the end. The ant analogy doesn't work on non stretchable band, where you just add distance, but on stretch band, distance behind the ant is growing in the relation to stretching speed.

[@mikekeenanphd]

Why is the video stopping before it even covered the topic properly?

[@platinumpengwinmusic5564]

Fun fact! Context on the example number ³4 which comes to 1.3 x 10¹⁵⁰, the number of possible states of a 7x7 Rubik's cube is ~ 1.95 x 10¹⁶⁰.

[@franklehman8677]

Surprised the proof via strictly descending infinite ordinals wasn't at least briefly sketched. It's fairly intuitive I think, once you get the swing of those omegas, how they can stand for simple integers. Notating using infinite ordinals also make the role of that "-1" much more obvious than a kind of handwavy description of gradually eating into blue blocks.

[@lkruijsw]

Then you first need to proper introduce the concept of ordinals.

[@koudyhoShorty]

1:25 - "Wow. Why is that so big?" - "Because of the + 3."

[@codacoder]

Also very cool to hear axiom independence/provability mentioned! Meta-mathematics is underdocumented!

[@smylesg]

That -1 changes everything😮

[@xinpingdonohoe3978]

Certainly seems to. Interesting.

[@robcoh]

Negative.

[@francoi8]

This is my favorite topic in mathematics. More of this!

[@fussyboy2000]

Is there a reasonable proof why meta Goodstein(12) >> g64? How many more steps are there to get beyond TREE(3)?

[@AgentM124]

Can we know the last 5 or 10 steps for numbers like 12 or 13 or more? Or can you only compute from the start?

[@jimmyh2137]

The last steps are always a sequence of 1+1+1+1... You'll have a single "big block" that gets converted.

[@AgentM124]

@jimmyh2137 I guess you're right and the length of that sequence is just... Long, before you get to the first power tower.

[@jimmyh2137]

​@@AgentM124 ​ you can break the tower relatively quickly, for example at 2:00 (the 19) you're breaking the 4^4^4 tower only three steps later. It would then be 8^8^8 -1 and you'll have to recalculate just like at 9:20

[@poppyseedsnuranium]

This's the kinda math that really gets my juices flowing. (I know that's weird.)

[@Wecoc1]

Might have to see a doctor about that one.

[@robcoh]

This puts you in a Ricci flow.

[@WRSomsky]

To paraphrase D.Adams: It's big. Really big. You won't believe how vastly, hugely, mind-bogglingly big it is. I mean, you may thing Graham's Number is big, but that's just peanuts to this! Listen! 😁

[@lkruijsw]

To proof that the Goodstein sequence finishes you have to use induction, but with the induction condition containing a generalization over a function. Peano Arithmetic can't generalize over functions.

[@Exkingcss]

Could the 3n+1 also be an unprovable theorem? How do you determine whether or no its unprovable? To prove the unprovableness? At what point do you stop trying to prove and say "nah, this is unprovable".

[@Xnoob545]

It might be unprovable and we might never know

[@alansmithee419]

It is possible that it's unprovable. However it's also possible that its unprovability is unproveable.

[@maksymisaiev1828]

I think for 3n+1 it would be better to go from pattern way. In other words, show that any number can go into one or ottther pattern and as quantity of prime numbers reduces over time, chances to not bump into some specific case is going to 0. Problem is to find those patterns for bigger numbers. But in most cases, it should be decently big number that after that we just don't have resources to prove. Not sure, but there are pretty few theorems which create exception on really big numbers.

[@gdclemo]

It's possible. There are variants of the 3n+1 problem, some of which are unprovable. We can't know which ones, we can only know that out of the set, there must be some that are true but impossible to prove. I'm not sure how we know this but I imagine that it involves some translation of the halting problem.

[@quinn7894]

I like the Johnson solids in the background

[@rtpoe]

I like the 1x1x1 Rubik's Cube!

[@BrianMPrime]

What really pushes it over the line is that minus one. Graham's number could never

[@bernhardkrickl3567]

To paraphrase XKCD: What if we used more axioms?

[@convindix9638]

Goodstein's theorem is true but not provable in Peano arithmetic Kruskal's tree theorem is true but not provable in the stronger system called ATR_0 The Robertson-Seymour theorem is true but not provable in an even stronger system called Π^1_1-CA_0 A theorem called "Borel determinacy" is true but not provable in the even stronger system Zermelo set theory For more math like this (even including weaker systems than Peano arithmetic too), what these theorems are, and what these systems mean, Harvey Friedman's book has a nice long introduction chapter (250 pages!) with all this and more in it

[@wfcyellow]

No, the biggest number is 57. I think.

[@javen9693]

What about 56

[@abdulwasiq2056]

No, 42 it is.

[@not1not2but3]

Hmmm I come up with (4^3)-9=55….wait. Dang it. Thought I had it.

[@ThisIsAYoutubeAccountAsd]

I tested all integers from -(10^10) to 56 and your claim holds, so it's very likely that you're correct.

[@xinpingdonohoe3978]

Seriously? How can you possibly think that? That's ridiculous. Numbers like 24 exist. Did you just happen to forget about them?

[@Paul71H]

It would be nice if he showed an example of how a large number could get smaller from one step to the next using this method, since it seems like each step makes the number not just bigger, but ridiculously bigger. But evidently it can happen, if all of these sequences eventually go to zero. Update: I watched further, and I see that he sort of gave an example of this, but I think it would have been helpful to go a little further with the example.

[@hypnogri5457]

there comes a point where writing the previous number in the new B=n+1 base results in it being written as a*B^n_1 + b*B^n_2 + ... with all the n_1, n_2,... in the respective power towers being smaller than the base B (which means that those numbers will not be incremented in the next step). At that point, the number will just get smaller 1 by 1 at each step and never increase ever again. It basically reaches the top and then it gets smaller exactly 1 every step

[@Bobbynou]

Love the video! One question I had is how does the curve look when it hits the maximum value? Is it an even bell curve, or do the ones eradicate the number very quickly towards to end, or does it go big very quickly and the ones start eating away the number over a very long period of time? And does it depend on the initial number?

[@shophaune2298]

The value of any given sequence peaks when there's no more of the base to bump up anymore - it's just a long list of 1+1+1+... at that point. So once it hits the peak it just chews through the number one 1 at a time.

[@CantEscape1.4M]

Just add 11!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! arrows and put two Graham's numbers in between

[@Unlimit-729]

Why are you yelling

[@hypnogri5457]

11!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! is equal to 11 if you use the definition of the generalized double factorial

[@patrickobrien8520]

Wonder what the bell graph of goodstiens for each number looks like?

[@FranklinLee-t3k]

For the sequence starting with the number 4, the largest term in the sequence is 201326592^201326591 - 1, which ends in the digits 4457003007. And that is when you finally start subtracting 1 until you get to zero.

[@cheeseparis1]

Best occasion ever to create an AI-generated video that reads all the numbers of the sequence, for us to watch while doing the dishes.

[@JonahNeff]

It's hard to imagine a more incomplete and thus confusing way to demonstrate the nature of the goodstein sequence.

[@charlievane]

14:09 but isn't the question about a potentially infinite sequence of operations, to be proved that it's actually finite?

[@100beep5]

Wouldn't it make more sense to count the meta-Goodstein sequence from when it's all 1s rather than when it hits 0?

[@Calmerism]

My mind was blown by the fact that minus one eventually beats some special kind of x to the x function 🤯

[@maksymisaiev1828]

that is for this case. And that is not that hard to see. As we see in case of starting 4, when we reach number 26, the outcome number is smaller than 3^3^3, and any next calculation by will for sure be less than sequence rule (for example, the next arithmetic representation will never be bigger than 4^4^4, 5^5^5 and so on). In short, the representation of sequence will always have some limit, which creates lower limit and than lower limit until you have just 1s. Same with 3n+1 series. It looks like multiplication by 3 should give big result over and over despite dividing by 2.

[@ParkerShimoda]

I'm surprised you haven't done a video on this yet

[@ParkerShimoda]

also it' s not "unprovable", the proof with ordinals shows it eventually decreases, and if you know the max, you can just replace the increasing values with a bit more that and create a strictly decreasing sequence that is equal to or larger than the original sequence.

[@jlivewell]

Nothing better than an unproven theorem and a large number on a Tuesday! 😂

[@DavidWang-v7u]

You’re not keeping me silent with freebies and gifts and four drinks that are free as my enemy for sure..

[@cheeseburger118]

How long until busy beaver outpaces?

[@hypnogri5457]

you can make an upper bound by making a program that simulates this, compile it into a n-band turing machine and then convert that into a 1-band turing machine. I think there are programming languages available that compile to turing machines

[@convindix9638]

It's known there is a 51-state Turing machine that halts after >f_ε_0+1(8) steps on blank input tape, so probably at most around 51

[@FuncleChuck]

Do we need numbers that big? Even theoretically?

[@Hagalaz_Drums]

hear me out. so if starting this sequence with 12, takes more than graham's number of iterations to resolve. what if the base number to do this sequence to is graham's number? how would you even begin to express that? why stop there? why not meta-meta-sequences of using the resulting meta sequence as the next base number?

[@shophaune2298]

The trouble is that doesn't actually noticeably change the growth rate of the function. The Goodstein sequence grows at around f_e0(n) in the Wainer fast-growing hierarchy. Feeding the Goodstein sequence into itself is equivalent to f_e0(f_e0(n)), or f^2_e0(n). This is a long way below f_e0+1(n), which you get if you take the meta-meta-meta-meta-meta-....-meta-sequence, where the number of meta-s is equal to your starting number. so to move up one rung on the "ladder" of fast growing functions you have to do this over and over and over, whereas getting substantially higher (say, f_e1(n)) is going to need an entirely new function.

[@robcoh]

@Hagalaz_Drums never meta fast growing hierarchy he didn't like!

[@benzimmermanmusic]

That thumbnail looks threatening!

[@bookert2373]

Obviously 42 (reference: Hitchhiker’s Guide to the Galaxy). QED

[@pinkraven4402]

TREE(N) finiteness is also unprovable with Peano axioms right?

[@AbbeyRoad69147]

No idea, but TREE(N) is provably more awesome, of course.

[@aikumaDK]

8 minutes in and the main question on my mind is "how do we know how long the meta sequences are?" or at least where did the estimates come from?

[@alansmithee419]

This is something that's always confused me. No matter how much I look into googology I can never figure out how people actually work with these objects.

[@chicojcf]

great video.

[@AbbeyRoad69147]

TREE(N) is larger in awesomeness by far.

[@hypnogri5457]

it also is larger when it comes to growing faster than goodstein

[@thecakeredux]

I accept the knowledge, but I can't say I understand it. You'd think that you could conceptually (not in reality) proof this by exhaustion, if you could perform infinitely many arithmetic operations, so I don't quite understand how it is outside of arithmetic axioms, then. Or did I just sneak in infinity in a way that violates some principles here?

[@valentinziegler1649]

The topic is interesting and would have been a great chance to introduce ordinal numbers, and show the actual proof that the sequences terminate. Unfortunately, the speaker chose to make a mystery out of this. I don’t know why. Search KZbin, there are much better videos about Goodstein Sequences.

[@rev6330]

But now I wonder: Back in the days you said about TREE(3) that it COULD in theory be proven to be finite by finite arithmetics only. The proof would need more symbols than what would fit into the universe (2 quad-arrow 10 if I remember correctly), but still, that qualifies as provable to me. Same for TREE(n) for all n, just with even more ridiculously long proofs. And now you're saying the Goodstein Meta sequence is completely unprovable in finite arithmetics? So does this mean it has to eventually outgrow TREE(n)?

[@valentinziegler1649]

For every specific value of n, Peano Arithmetic can prove that g(n) terminates. You can’t prove the general statement that for all n, g(n) terminates

[@shophaune2298]

TREE(n) might be provable in finitary arithmatics, but not necessarily using Peano's axioms - you'd need a stronger set of finitary axioms. For reference we know that Goodstein's Sequence grows around f_e0(n) in the Wainer fast growing hierarchy, while the weak tree(n) function grows faster than f_SVO(n), and the normal TREE(n) function grows notably faster still.

[@Xnoob545]

Sadly Goodstein does not outgrow TREE its pathetically small in comparison to TREE

[@luczeiler2317]

Now I'm left wondering how big the speed limit is?

[@shophaune2298]

Peano's axioms will struggle with any function that grows at or above f_e0(n) in the Wainer fast growing hierarchy - the speed limit mentioned in this video and the growth rate of the Goodstein meta-sequence. What does that mean in concrete terms is hard to say, but to give a sense of how these ordinal-indexed functions grow: Graham's function (of which his number is G(64)) ranks around f_w+1(n) in the hierarchy, where w is the least transfinite ordinal omega. Bower's linear array notation caps out around f_(w^w) (n). f_e0(n) diagonalises across tetration of w, so f_e0(1) = f_w(1), f_e0(2) = f_(w^w)(2), f_e0(3) = f_(w^w^w)(3), etc.

[@Xnoob545]

f_ε_0(n) That's epsilon null plugged into the fast growing hierarchy

[@Xnoob545]

​​@@shophaune2298 i dont think the fundamental sequence of e0 is usually w, w^w, w^w^w... Usually it goes 1, w, w^w, ... Edit: ok this is very debatable but I'm saying I usually see it starting with 1

[@shophaune2298]

@@Xnoob545 e0[0] would be 1 yes

[@shemmo]

What is the practical use for this?

[@hypnogri5457]

He mentioned it in the video. Constructing sequences like this can proof things about the mathematical systems we work in. Its a very abstract use. You might not directly need this to build a car, but things like this might be indirectly linked to things like provability, computability, NP=P, time complexity of algorithms, etc.. through abstract means, which might or might not be useful in fields like cryptography, information theory, etc... (very far reach because even though it might be indirectly linked, it is only a small piece in a puzzle of thousands of other more relevant proofs to the field. But random discoveries like these can be unexpectedly used in different fields to prove things that might prove useful later on.)

[@rykehuss3435]

anyone know how fast the meta-sequence grows in the fast growing hierarchy?

[@shophaune2298]

f_e0(n)

[@RandoBox]

What everyone is wondering is where this sequence is in the fast growing hierarchy?

[@alansmithee419]

f_{epsilon_0}(n), or thereabouts. I believe it's a bit (in googological terms - i.e. a lot) slower, but faster than any previous FGH function.

[@HunterJE]

Tortoise and the hare if instead of the hare taking breaks the tortoise is very slowly and subtly cutting away at the hare's achilles tendon

[@robcoh]

Oh snap!

[@Anthony_Stuart]

Look Around You tells me the biggest number is 45,000,000,000, although they do say some mathematicians think there may be even larger numbers

[@zaffyr]

some might say that adding 1 will make the number even larger, but that remains unproven

[@gdclemo]

@@zaffyr be careful with those large numbers around calcium. You could risk causing Helvetica Scenario.

[@AidanRatnage]

Why is n^n^n n^(n^n), rather than (n^n)^n?

[@nyphakosi]

power towers are evaluated from the top down evaluated from the bottom up would be, for example, (3³)³

[@hypnogri5457]

because we have a number n^a and then we break up 'a' into its own base representation, i.e. n^(b^c) where b^c=a. (n^b)^c would make no sense in our original n^a context, as our goal was to break down a into a = b^c

[@mydwchannel]

George Russell's older brother is decent at maths isn't he?!

[@scaredyfish]

If techniques outside of Peano’s axioms had to be used to prove it, is it possible it’s not true under Peano’s axioms?

[@convindix9638]

There is a model of Peano's axioms where it's true and a different model in which it's false. Like how you can't prove Euclid's fifth axiom from the first four, so as a result there is a model where the fifth axiom is true (Euclidean geometry) and a different one where it's false (hyperbolic geometry). On the other hand, whether the parallel postulate and Goodstein's theorem are really true "in the real world" requires taking a philosophical argument about what is true in reality

[@scaredyfish]

@ Ok, but in that case, is it correct to call it an example of something which is true but not provable?

[@convindix9638]

@@scaredyfish Yeah (specifically there is still some debate over whether things like the continuum hypothesis are really true or not, or if a statement about sets can be "really true" in any sense, but pretty much all mathematicians agree Goodstein's theorem is true)

[@lkruijsw]

@@scaredyfish It is not provable in Peano Axioms. If you allow generalization over functions in the induction condition, then it is provable.

[@wilkinsune]

Imagine the Goodstein sequence of Graham's number.

[@adb012]

Goodstein's metasequence doesn't grow THAT fast. It took Goodstein(12) to reach Graham's number. It takes the Tree function only Tree(3) to reach to... well, Tree(3), which is ridiculously bigger than Graham's number.

[@shophaune2298]

In terms of growth rates it's true that there are many functions that beat Goodstein, but this is still notable as being the first function not specifically constructed for such that was proven to be beyond Peano's axioms.

[@jimmyh2137]

You're falling on a misconception. A sequence can be faster than another one even if all terms before some point are lower ("slower start"). All terms after that point are bigger. If a sequence is like, 1 - 2 - 3 - 4 - 5 - 6 - Tree(3) - Tree(50) - Tree(800)... then it's obviously faster growing than a normal Tree sequence even if the first 7 entries don't show it. That said, Goodstein is indeed slower than Tree.

[@adb012]

@@jimmyh2137 ... I am not falling for that misconception. I was just making an assumption. I am very familiar with how, for example, X^2 beats X^4 between 0 and 1, and the slope of X^2 beats the slope of X^4 between 0 and 1/sqrt(2), but then X^4 outruns X^2 forever after that. And in this case, Goodstein also grow faster for the first couple of cases: G(1)=2 > TREE(1)=1 G(2)=4 > TREE(2)=3 But then... G(3)=6

[@hypnogri5457]

even if tree grows faster (I don't know), your argument doesnt proof that. It could be the case that after a certain value of N, Goodstein(N+a) >> Tree(N+a) for all a>=1. That would still mean that goodstein grows faster even if Tree grew faster for N steps before that (N can be billions, googols, etc..)

[@david-melekh-ysroel]

My friend Brahim's omicron function is slightly faster than this Goodstein sequence

[@dfmayes]

These number play episodes are my favorite.

[@NottoriousGG]

What the fudge?!

[@thephelddagrif2907]

Obviously Graham's number + 1

[@kbsanders]

Whatever the biggest number is, just add 1 to it and name it after me.

[@Tzedito]

Thor to the Thor to the Thor + 1 + 1 and I can only see Thors and ones. 🤭

[@RadioactiveLobster]

The -1 is entropy. It will always win.

[@richbuilds_com]

It will always wi.

[@fullfungo]

It will always w.

[@xinpingdonohoe3978]

It will always.

[@dalevross]

It will alway.

[@maxisawesome7357]

It will alwa.

[@KratosElGreatos]

Pause the video and go to the first frame 😉

[@hasidman3618]

Whatever the number is, just add 1.

[@lumi2030]

people talk so much about goodstein sequences which are obnoxious to calculate, but never about the primitive sequence system, which is much easier to implement in almost every programming language and equally strong, with a more direct correspondence to ordinals up to e_0

[@galoomba5559]

What's that?

[@lumi2030]

@@galoomba5559 search it up on the googology wiki

[@convindix9638]

​@@galoomba5559 Here is the process called expanding an array A to n: Given an array A of natural numbers, pop off its rightmost entry and call it c. If c = 0 then you are done, otherwise copy+paste everything from the rightmost c-1 to the end n times. (For example [0,1,2,3,2] expanded to 2 is [0,1,2,3,1,2,3,1,2,3].) Then, if you expand [0,1,2,...,n] to n, then expand the result to n+1, then to n+2, ..., the number of steps you can do this for is about the same as the length of a Goodstein sequence starting with 2^^n

[@convindix9638]

Numberphile in 10 years: Σ_2-STABILITY

[@9nr]

Of course it's 42

[@codacoder]

"Eventually" is a really tricky concept! Problems like knowing whether an asteroid will eventually collide with a planet in an n-body gravitational system, or a sequence or program (like the Antihydra) eventually halts very quickly defies humanities' collective understanding! Loved the notion of a speed limit :)

[@shill2920]

90 is a pretty big number

[@javen9693]

What about 89

[@shill2920]

@javen9693 mathematicians say its a little bit smaller, maybe.

[@johelsen5776]

Who cares about a has-been like Graham's Number?? It's Tree(3) or nothing, and I expect this video to go into that, too! (Taps Play in anticipation)

[@NaviafromtheSDR]

The biggest number is 1.

[@temperr.haring3508]

Nice Megaminx on the shelf

[@Monkala2]

I FOUD IT FIRSTI KNEW ABT IT FIRST RHAHAHA

[@deleted-something]

Welp 😅

[@1.4142]

⬆⬆⬆⬆⬆

[@gaxamillion_]

5

[@jort281]

Yeah, I watched that documentary too

[@illesizs]

I get why replacing numbers with bigger ones is a fun thing to do when you are bored in maths class, but why do yo subtract 1 at the end each time?

[@shophaune2298]

Without the -1, no matter what you start with it spirals off into infinity. With the -1 it EVENTUALLY ends, but still grows at a speed greater than almost any other function you could name, and that's a lot more mathematically interesting.

[@dylanbrown9241]

Tree(3)

[@javen9693]

What about Tree(3) - 1

[@dylanbrown9241]

Tree(3) is bigger then Tree(3) - 1

[@YonDivi]

But what about tree(3) -1 +1?

[@dylanbrown9241]

Tree(3) - 1 + 1 is just Tree(3)

[@not1not2but3]

Ok….but why?

[@laxmansingh4440]

Good

[@CompanionCube]

your mom is bigger than graham‘s number

[@valentinziegler1649]

Dude makes it sound more mysterious than it actually is… Transfinite Ordinal numbers are a concept that you can explain to kids who enjoy mathematics.

[@agogobell28]

My real question is, why?

[@steve_s9412]

Why what?

[@thequeenofstaplesmiles]

my life is a lie

[@cs8712]

first

[@Nah_Bohdi]

Meaningless number puzzle. 🙄

[@Unlimit-729]

No it's not

[@mobgamerz7397]

Comment no. 1 First 🎉