Too smart 😆

@@jazlearn5147he’s got you there

I change that one to a factorial

@@Ilikewooper I think you win. Here is your trophy 🏆 😀

I add one to that factorial

This comment is gold. 😂

If you picked a number truly at random then it would almost certainly be larger than Rayo’s number.

How would you pick a random number?

​@@Juguitosdemora close your eyes and choose

@@davidyoung2990 It would be a 100% percent chance that it would be bigger than the Rayo´s number

@@IlIlIlIlIlIlIlIlIlIlIlIlIl1123yes and no, there is negative numbers too

Sameeee

Correction: 11!!!!! Is 1 quintuple factorial iterated. 11!!!!! Is 66.

Anytime! Feel free to voice some more video ideas that you would like to see if you want! I'm glad I could help!

I couldn't really understand numberphile's explanation, so this is useful.

@@marty2035 Glad I could help!

​@@jazlearn5147explan f10(10) (10) (10) or largest garden largest number

TREE is finite for all finite inputs, so that is indeed finite.

And of course the same thing can likely be done with the Rayo function by stacking its own function within itself (i.e."Rayo(Rayo(Rayo...)")@@hyperpsych6483

We know TREE(n) is always well-defined (finite) because of Kruskal's tree theorem. Kruskal is basically the reason any of us are ever even talking about TREE(3) in the first place.

wtf is TREE

dude did u even watch the video? dont comment if u dont know shit@@matheuscabral9618

Crazy MASSIVE! 🤯

rayos number to septation of rayos number and double factorial x rayos number

It's very interesting. Yes, that is how he overcame it!

I'm glad you enjoyed it! 👍 😁

MASSIVE!

Indeed it is! Busy Beaver function is a Turing machine!

Thanks! 👍 Grahams Number is the smallest of the three, then the next biggest is TREE(3), and then Rayo's Number makes both those numbers look like 0 in comparison! The Busy Beaver function fits in between TREE(3) and Rayo's number 👍 The magnitude of Rayos' number is uncomprehensible. It's BIG!

but what if i lets say take Tree(googleplex)^grahams number factorial

@extazy9944 I would say Rayos number is bigger, and I would guess the busy beaver function of a googol to be bigger. These numbers are massive!

@@jazlearn5147 what about tree(busy beaver)

⁠​⁠​⁠​⁠@@extazy9944rayos number is bigger than tree(tree(tree(tree…(tree(googolplex^grahamsnumber!) with tree(3) nested trees

Thanks! 👍 😃

Anytime! If you have any other requests, feel free to voice them 👍 The Busy Beaver function is probably my favorite! If you look into it a bit deeper, you see that the reason computers can't find a value is because there is an astronomical amount of different instruction card combinations, and for each combination, you can't stop the beaver until you know it has finished or if it looping. This is hard because there are cyclic patterns that make it look like the beavers' movements are looping, but they aren't. It's an amazing computer function and crazy to think about with googol Instruction cards!

Very interesting indeed. The busy beaver function is actually uncomputable, though! Meaning that you cannot compute the value of BB(x) for any x with one program, even with infinite time. This is because turing machines are equivalent to our computers, and so you could imagine a contradiction similar to Berry's Paradox if you could compute the value of BB(x) with one program. Say, if you have the amount of states that you need to construct that turing-machine as x, you could make an x+y+1 (for some relatively small y) state turing machine which will start with every cell of the tape at zero, use y (here you can imagine why y would be relatively small: if it wasn't, BB would be slow growing!) states to set up the number x+y+2 (encoded into whatever encoding for numbers the x-state turing machine for computing BB uses), then x states for the BB program, and finally that one extra state to move right until it finds a zero, and then set that zero to one (if it's state z, then this would be 1->11z, 0->110) to increase the number. This would give us BB(x+y+1) > BB(x+y+2), which is impossible. This argument is slightly handwavy, but there is a very closely related problem known as the halting-problem with formal proofs. Please note that this is different from one program existing which computes the value of BB(x) for some fixed x, which obviously exists: it's just the winner of BB(x). Now... how was BB(4) computed? (and how is BB(5) currently being computed?) Programmers can write "deciders" which will take a turing machine as input, and either say that it definitively doesn't halt or that it doesn't know whether the machine halts or not. Now we just have to make deciders which are strong enough to decide every four or five state turing machine, and then run the ones that it doesn't know about until the machine halts. Practically, you can't really know whether the decider is strong enough, so you just have to run the turing machines until a stronger decider comes along that says it doesn't halt. Currently, a large project focused on this is bbchallenge (at bbchallenge.org)

@eig5203 very interesting. For BB(1), there are only 64 different cards to analyze, so surely this is computable??? Couldn't you do this with a pen and paper ?

@@jazlearn5147 Yes, that is essentially running a decider yourself instead of using computers. Once you get into something like BB(800) (I believe the current lower bound is BB(748)), then there aren't any deciders which will provably (in ZFC) work for all turing machines of that size. Eventually, every (recursively-enumerable) axiom system will run into this problem.

BB(5) is probably just 4098. BB(4) = 13 isn't as easy as counting. it was actually really hard to prove that every turing machine with 4 states which took longer than 13 steps to terminate, didn't terminate at all. same with BB(5)

🤯

Thanks! There is a lot more I could have said as you probably know, but I didn't want to make a 40-minute video! Lol

Thanks! Yes, I do! There is one about Googol, Googolplex, Grahams Number, and TREE(3), which are HUGE numbers, but nothing compared to Rayos Number.

@@jazlearn5147can you do USDGCS_2(k)?

Yup. You win 🏆 😆

in the rules of the battle it was stated that you can't just use the opponents ideas and build on it, each new number has to be a unique concept

Well what about infinite? By definition infinite = infinite+1

@@wolfVFV Exactly the point I was making!

@@wolfVFVinfinity is a concept, not a number. kinda like how your virtual AI girlfriend is not a real woman either.

@@wolfVFV Infinite is not a number, adding number 1 into infinity is like adding 1 into an apple it doesn't make sense

Yea. At first I did that because I was using Wolfram Alpha as my source, which told me to do it like that. But then I realized it was an ai source and thus did not fully understand what I was doing 😆 so I looked at single factorial values and the numbers would have gone off the screen real quick (because they are so big!), so I just decided to introduce double factorials for fun while keeping the numbers on the screen lol 😆

Fair enough. I tried doing my own back of the napkin math using Sterling's approximation to find the value for single factorials and the resulting power tower is a nightmare to try and simplify. Thanks for the clarification! As a very, very handwaved simplification of the real value, I got the tetration of 10 to 6, or 10⬆️⬆️6 using up arrow notation.

@connerwinder2218 Thank you! 👍 😊

a↑ⁿb=a↑↑↑...n...↑↑↑b

Appreciate YOU! 😁 💛

I haven't heard of that one! 😀 I'll have to look into it! 👍 Thanks for the sub!

Beth is transfinite. You can’t just say “infinity” when someone asks you for a large number.

@@diht that's why I said " I know some would argue it's not a number", I wouldn't consider it one, but in some ways it can be viewed as one.

@@ZephyrysBaumI would say that the Beth numbers are numbers since they describe sizes, with Beth numbers describing the sizes of Infinities. I think what needs to be established is that a number must be finite since you could just chose a Rank-into-Rank Cardinal, but that’s meaningless since most people don’t know, nor will know how those numbers are defined

@@blightborne6850 They will know if you explain it to them. But Beth numbers are the starting point to studying Strongly Limits. If you want more info, I could talk to you more about that in Orbital Nebula server.

We are on the come up! 😁👍❤️

0:58 I'm confused by double factorials. Also, many single factorials get bigger than a few double factorials. (((3!)!)!)! = ((6!)!)! = (720!)! = (2.601*10^1746)!

@-wvy_ Yes, single factorials do get bigger, but due to the way it was written on the board, we must use double factorials. I was confused at first as well, but that is what every source that I looked at said.

@@jazlearn5147 What are your sources even? Because I found an interview with Augustin Rayo on the Math Factor Podcast, which tells a completely different story: Rayo starts by writing a 1. Then Elga writes down as many 9's as he can fit on the board. Rayo counters by writing as many 1's as he can fit on the board, which resulted in a much larger number, because he could fit way more digits of 1 on the board than Elga's 9's. Then Elga changes all except the first two 1's to factorials, clearly resulting in a bigger number. There was no mention of double factorials whatsoever.

@@melooone when you write many factorials, the sources I looked at said they become double factorials. As for the other slight inaccuracies, I did that for the video sake. Didn't want it to be too long. It was more to introduce the numbers, not so much to be historically accurate.

Thank you! 😁 I'm glad you enjoyed it!

It's smaller than Rayos' number but huge. I don't think anyone can comprehend the size of Rayos' number. By using set theory language, you can go beyond anything you could imagine!

@@jazlearn5147 Ayo What about (Rayo's number)^TREE(Rayo's Number) ????

Correct!

Damn! 😆

RAYO NUMBER×2 RAYO NUMBER^2

Very interesting!

This has a definability issue (in that salad numbers aren't conclusive) but it's kinda funny to see somebody make a function which actually uses salad numbers so nice i guess

bro had enough

lets surpass f_ΒΗΟ(PLANK^googol(10↓↓3))?&PLANK^googolplex(TREE(3)) where ? is from the number BIGG

Kid named Nuclear Engine:

Well, now imagine that, PLUS ONE

What you described is not even a quark compared to rayo's number, or even to far smaller numbers such as Graham's number and TREE(3).

Check out Knuth's up arrow notation. What you've described is a Googolplex (11 up-arrows) 2. Also using a googlgolplex becomes unnecessary, just use 3's instead and add 2 more up arrows to ensure it's larger.

@@blackeyefly I was going to say that the Graham sequence is larger than this. I'd argue that the FGH would imply tree(3) is also larger than rayos number

@@jazzabighits4473 I don't think that's correct, surely you can define the TREE function with far fewer than a googol symbols of set theory

Yes, it is indeed bigger but I am pretty sure it is not as well defined as Rayos Number.

lol i came here just to comment this :D

@@jazlearn5147 ironically, googologists consider Rayo's to be ill-defined and Garden well-defined.

​@@jazlearn5147 It is much more well-defined than Rayo's Number.

@@jazlearn5147 in googology, lngn is accepted as the largest well-defined googolism and rayo is actually ill-defined according to googologists

Thank you! 😊

sure it's nearly impossible to say anything about it but so is tree(1000) using your logic and most people would say is a number even though people can barely say anything about it so it is a number just a completely useless number

Uh, so is many other "large numbers" (even, say, TREE(3)). We have no way to even find their last digits, let alone first digits, but they are considered numbers. Also, someone has made a 7901 character Rayo script which exceeds BB(2^65536-1).

So it is computable. You found a way to compute is. True, doing so in the real world is a tad impractical due to the amount of computer time required - but that's a small matter. It's finite, it's computable, it's just big.

all three of your first three replies are basically saying the same thing

@@vylbird8014 No? the rayo-string was constructed set theoretically and proven to be able to output a larger number than S(2^65536-1).

Is that actually how it works? I looked at a lot of sources, and it said the way I did it was right. Your way makes more sense, tho 😆 you are probably right. Thanks for the knowledge!

@@jazlearn5147 The number of exclamation marks is the number subtracted for each multiplication. Five exclamation marks means that the multiplication is 11×(11-5)×(11-5-5)...etc. As it happens this stops neatly at 1. In other cases, we would stop before things go to zero or below. The keyword here is "multifactorial", and Wikipedia has some information about it if you're interested. Interpreting more than two exclamation marks as a combination of double and single factorials is what the WolframAlpha website does - which might be what you used(?) - because it hasn't been programmed for triple etc. for some reason. It's not entirely certain, but in context I think the 11!!!!! on the blackboard during the competition was intended to be ((((11!)!)!)!)!. Stacking single factorials like that gives the largest possible result. There's also that 66 would have been a losing move.

I always assumed it meant ((((11!)!)!)!)! And since 66 < 1111111, I'm sure so did they. Just shows the importance of defining your functions if you're using them in an obscure way rather than assuming everyone is on the same page.

@19t2000 Yes, I believe you are correct in that assumption. I made a mistake.

Well, too late. You can't add a bigger input because that would be breaking the rules, also get it "GOOGOLogy". fitting for the largest accepted number

It's the largest because the Rayo() function is the absolute fastest growing function defined in mathematic terms we have today

@@moahammad1mohammadfunction that is used by LNGN:

@@moahammad1mohammadlarge number garden number is probably bigger Also little bigeddon is bigger

That’s definitely true. I suppose for the purpose of this battle it didn’t matter because there was an implied rule that each turn had to use a novel idea, so his opponent couldn’t use his own Rayo function against him with just a higher input. But yeah, I definitely agree, the number is arbitrary, but the function is pretty general.

rayo+1!2 < rayo^^^rayo

Very true! Thank you! 👍😁

Then it would be one hundred trillion billion million times bigger 👍 😁

a guy called harvey friedman wrote a proof of this statement in 2000

graham's number is g(64) what's g(1) then? well its just (NUMBER)↑↑↑↑↑↑(NUMBER) and g(2) is (NUMBER)(G1 ↑'s)(NUMBER) and i think you get the pattern there

On the fast growing hierarchy (FGH), Graham's number (or sequence) is equivalent to the function of ordinal omega or omega+1 (where 0 = successor/counting function, 1 = addition, 2 = multiplication, 3 = tetration......with infinity = "omega function", the next "strongest" function being omega+1) Graham's number sits between this level of function and the previous level, that is, its growth rate is 'omega', or faster than anything that can be described through any lower function (for example, you can't express graham's number as an exponent, or even as a tetration, or anything really less than its explained growth. Eventually, after 1, 2, 3,..........infinity (omega), omega+1, omega+2, you get to omega times omega, then omega times omega times omega, etc. The whole time these are describing insanely fastly growing functions. In the end, there's some ridiculous ordinal called the Rieman Zeta ordinal or something like that, describing some ridiculously fast growing function. The rate of growth of TREE(3) is higher than the Zeta ordinal. However, I've still been able to explain the "strength" or growth rate of these numbers using just the English alphabet and a few symbols (numbers, brackets, equals signs, etc.), let's say 50 symbols at most right? Rayo's Number denotes a number that is so large that you need at least 1 googol symbols to explain it (rather than the 50 I'm using right now).

I'll look into that. I'm guessing it'll be very complicated mathematics, but I'll see 👍

@@jazlearn5147 The youtube channel numberphile explains it well, check oiut their extra footage video on TREE(3)

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1

I don't even know what to say to that... it's definitely bigger, but I can't comprehend the size of that number!

@@jazlearn5147( tree (Rayo's Number To The Power Of Rayo's Number To The Power Of Rayo's Number To The Power Of Rayo's Number To The Power Of Rayo's Number To The Power Of Rayo's Number) )!

TREE(this)

IIRC the game had rules where they couldn't just say "that plus one" or "TREE(that)", because otherwise it would go on forever.

That's called the successor function, and it's not necessarily a number. But I'll allow it 👍 😆

Smart move 👌

The rules of the game were that you couldn't use naive extension such as adding one.

Go on.

The Infinity! in question (yes infinity isn't a number i know I'm just joking)

oblivion numbers arent well defined

that was a part of the contest that he didnt explaine, they could use a technique that was used before so just adding a one to rayos nummber would just be using rayos technique. but yes adding a 1 to rayos nummber would make it bigger LOL

Not a stupid question, the video maker just didn't include the important rule of "doing something new" when presenting a bigger number. That's definitely a bigger number and you could make an even unfathomably bigger number like Rayo(Rayo(10^100)), but it just uses the same function as before and is not a new creation.

It is different because you multiply every second number up to a given number rather than every number, thus making it much smaller. I was under the impression that in the actual Big Number Battle, they used double factorials, but it appears from further research that this was not so. They used single factorials.

You're right. Double factorial is a different function and shoudn't be confused with using factorial twice. There is a mistake in the video

That is just how the notation works. You are 100 percent correct, tho. If we just took factorial each time, the number would be significantly larger!

@@jazlearn5147 So, if you do not use parentheses, the notation defaults to double factorials wherever possible over a factorial of a factorial?

@secret12392 Yes, that is correct 👍

@@jazlearn5147Bruh

TREE(n) function goes 1, 3, and then huge. BB(n) remains small-ish up to 5, and then it becomes huge but probably wouldn't surpass TREE(n) until n equals 8 or 9. They do intersect, and BB grows faster overall.

It's a good formula 😆

Nice, but I add one to that and win, lol. Nah, jk, you win 🏆

😢

​@@arandomgamer3088 pentation it

Well done! 👏

utter oblivion ill defined

ABSOLUTELY MASSIVE 🫡🤯

Thank you!!!

Not quite. The busy beaver is a specific set of rules on a turing machine

Well, we only really know up to BB(4), and to do that, they just run every possible program and hope it doesn't take 6 years to spit out an answer lol

This is getting out of hand 😆 That number is MASSIVELY MASSIVE!

Love this 😆 🤣

I have already made a video on TREE(3) so I left it out of this one 👍

TREE doesn’t grow faster than BB(n). The Busy beaver function is uncomputable, the TREE function is computable. (In fact, we have found it’s growth rate.)

@@SG2048-meta We have in fact not found its growth rate. The believed growth rate is the order type of a certain related set, but Irrational Arrow Notation proves that the growth rate is not always equal to the order type.

@@zswu31416yeah sorry for that mistake.

😆 my favorite comment 💯

fr

You have to be novel and create new unique ideas. That was the rule. 👍

not even close 💀

You can define the TREE function using first order set theory in far fewer than a googol symbols. This means that Rayo's number must necessarily be more than TREE(TREE(TREE(TREE(TREE(...TREE(3)...))))) I don't know exactly how far deep it could go but I know that I couldn't write it out. Even if I used every atom in the observable universe to denote another iteration of the TREE function.

Interesting!

These numbers are A LOT bigger! Like a lot... 😆

Facts! 💯

I haven't looked into that one yet. Sounds amazing! 😁

Bowers has some interesting ideas XD

ultimate oblivion

utter oblivion, oblivion, ultimate oblivion, etc. are all ill defined croutonillion is the largest well defined number by virtue of having all ill defined steps replaced with a known well defined one, and it uses lngn extensions in its own definition croutonillion is also the largest salad and finite number known, but its not the largest valid googologism because it is a salad number.

@@worldprops333 you meant the well defined version of it?

The rules were that each player could not use naive extensions to create their next number. Adding one is considered a naive extension. Their methods of creating each new number had to be new and unique 👍 otherwise the battle would have gone on forever! 😁