Addition, multiplication, ... what comes next? (It's not exponents)

[@paulbloemen7256]

I’m 73 now. When 13 I asked my math teacher if there was an operation next in line in the series multiplication, addition, so in the other direction. I finally got my answer here.

[@nzqarc]

Succession is usually considering the 0th hyperoperation, its basically just counting

[@kales901]

@@nzqarc what is the -1st

[@nzqarc]

@@kales901 doesn't really exist

[@ramsey2155]

I fail to imagine an old man at 73 watching math videos and commenting with a youtube account, time is passing too fast

[@esphix]

​@@nzqarcThe problem with that consideration, though, is that it's not even a binary operation

[@kjetil1845]

these are called the "commutative hyperoperations". i think you should add that name somewhere in the description or the title, so that this video shows up when people search for it, because this is definitely the best explanation i've seen of them. i remember seeing another cool overview of them a few years ago, that went in a different direction than this video. unfortunately i'm no longer able to find it, because again, they didn't include the name anywhere.

[@dangnabbit1379]

I think I saw this with 3 blue 1 brown

[@Archetype784]

This has happened to me a few times before: when you independently discover an interesting field of mathematics that is obscure enough for you to think that you’ve done something new, but it turns out that it was all formulated by someone else already.

[@dangnabbit1379]

@@Archetype784 Is this not true of all features we now experience as our culture? The difference is in how they are implemented and perpetuated. To learn is to explore and discover creation, whether it is yours or someone else's.

[@michaelasmitty]

Are commutative hyperoperations defined in the negative n direction as done in the second half of the video? If not, you should absolutely write up and try to publish this! Very cool!

[@paulkanja]

did you eventually find it?

[@d3tcovax]

This video feels like discovering math

[@udveetpatil8002]

This sentence reminds me of that one 3blue1brown video 'What does discovering Maths feel like'

[@sebastianlenzlinger9291]

@@udveetpatil8002that’s because the sentence contains „discovering math“ 🫢

[@Fire_Axus]

your feelings are irrational

[@cmdlp4178]

​This might just be a new field of mathematics

[@CasualLifeExperiencer]

​@@cmdlp4178Could well be, yeah

[@musicarroll]

I wrote a paper on this back in 2001: " The Natural Chain of Binary Arithmetic Operations and Generalized Derivatives." Bennett also discussed this in a 1915 paper, "Note on the operation of the third grade."

[@makhnoboi1996]

I've actually been doing a similar thing with the derivative over the past few days. Do you know if the field has changed since 2001 and would it be possible to make a meaningful contribution as an amateur?

[@musicarroll]

@@makhnoboi1996 There are a few citations of my paper since then, but very few. I have another one in the works.

[@notpea]

sounds like a weird sequel to kubrick's movie

[@timseguine2]

In case you didn't realize K[-1] is also known as the log semiring, and it comes up when talking about log probabilities which are all of the nonpositive reals. It is a convenient way of dealing with numerical issues while dealing with large categorical distributions, since it more efficiently uses the available floating point bits.

[@unaryplus]

Thanks, I've never heard of the log semiring. It looks like the log semiring has the same operations as K{-1}, but a different underlying set. (The wikipedia definition includes +∞ as well as -∞, but doesn't include logs of negative numbers.)

[@timseguine2]

@@unaryplus There are some slight differences based on your construction, but this is mostly explained by your specific choice of equivalence classes, and some details with respect to the exact definition of the operators.

[@BosonCollider]

@@unaryplus You may also be interested in the tropical numbers, which are the limit of the tropical numbers when the base goes to infinity, so that plus is replaced by max or min, depending on sign conventions. The log semiring is a quantization of the tropical number, with the tropical numbers only capturing the order of magnitude arithmetic aspect This is something I'd love to see extended to K{-1} . Do the K{-n} in general have a dequantization?

[@JerkoFlapdoodle]

You're the first to go BELOW -∞ in a one-dimensional manner@@unaryplus. Most treatments of ∞ ultimately invoke projective geometry. (Alexandroff Extensions, Points at Infinity, Clifford Algebra). But here you're creating "data-states" that map to other data states, with an ordering relation that evolves naturally from your criteria about distribution/linearity. You in fact ITERATED linearity, hence you always permute the same underlying set (Reals: you always get a bijection from log(log(log...(x)...))) to the Real Line, which we call "x", which itself emerges because it is symmetric with e^e^e^e^...(x) which your curves beautifully display. I'm sure you've heard about these, but I think of Ordinals, Surreals, and CGT values like "Tiny" as possibly in bijection with one another. If that's the case, then log/exp are the true foundational operations of mathematics, as they map you between Sets and Powersets, and your curves show that all numbers on the real line can be seen as an infinite collection of Sets (logs) and Powersets (exponentials) in a way that is commutative or "grouplike" in 1-dim.

[@demonicdrn3037]

This was such a cool video. I cannot express how exciting it is to learn all of this; it feels like discovering the complex numbers again. You totally deserve to win #SOME3

[@unaryplus]

Wow, thank you! Sadly, I didn't finish the video in time.

[@GrifGrey]

@@unaryplusname it SoME4 so you'll be the first submission when that happens

[@a1-h8]

Agreed this is by far the best SoME3 video I've seen! I loved every second.

[@firefly618]

Agreed. Awesome idea, elaboration, and presentation. This is the coolest new concept I've seen in years.

[@yumyumyucky6313]

@@unaryplusthat’s tragic

[@pontusloviken94]

Note that when you take the operations smaller than addition to negative infinity K[-infty] you get: a*b = max(a,b)

[@Nonkel_Jef]

The vast majority of numbers have over 1.65 million digits, so it’s still practical.

[@janthran]

i'm thinking if there's a good way to search for primes with this it might help with computation at least a little bit

[@alephnull4044]

This comment makes no sense. By that logic, a bound of any size is still practical. And then "practical" has no meaning.

[@SomebodyHere-cm8dj]

that is the joke​@@alephnull4044

[@حارثسہی]

@@alephnull4044it’s a joke

[@WAHa06x36]

The real joke is calling the reals the "reals", when they are the most absurdly unreal thing in maths.

[@pocarski]

This is very interesting if we look at it in complex numbers. In C, log(z) is defined for all values except 0, so +_n can be used on anything except 0. If we define log(0) as inf, and exp(inf) as 0, then we get a consistent projective plane-like system without any weird points. The "number line jumping and/or sharing" described in the video fits very neatly into the multivalued nature of the complex log. A natural idea is to try and find a continuous extension of +_n. This can probably be done by first deriving log^1/N and exp^1/N (functions that when applied N times become equal to log and exp) and then extrapolating onto the rest of the rationals and then the reals. I will look into this later and will edit this comment after I do

[@popularmisconception1]

on complex numbers, log(z) is multivalued. e.g. for negative -z without imaginary element, log(-z) = real_log(z) + i*pi*(2k+1), while for complex log for z is log(z) = real_log(z) + i*pi*(2k). So a single number in K_-1 really correspondes to infinitely many numbers in C. Further applying this complex version of logarithm to all of the results of previous logarithms will give us "infinity-squared" number of results. Applying the logarithm infinitely many times, as we do for exponential numbers, might actually give us the whole complex plane. Or it might not; some log(|z|) + i*pi*(imaginary number) might be missing here. IDK, I'd like to see a proof. Also I'd like to see a proof showing that further applying the logarithm will always end up in new numbers, because... maybe not, maybe some exponential numbers are not unique and overlap in complex plane.

[@viliml2763]

@@popularmisconception1 The periodicity of the exponential function cancels out the multivaluedness of the logarithm. exp(log_a(z) + log_b(w)) = exp(log_0(z)+log_0(w)+(a+b)*2pi*i)=exp(log_0(z)+log_0(w)), where I named the k-th branch of the logarithm log_k

[@kappasphere]

​@@viliml2763It's true that complex log being multi-valued cancels out with the periodicity when chained together directly, but it doesn't work like that in all cases, such as in the operation discussed in this video: If you calculate -1 `star` e^½, that's defined as exp(ln -1 * ln e^½) = exp(iπ(2k+1) * ½) = exp(iπk + iπ/2) ={i if k even, -i if k odd} So in this case the multi-valued result really doesn't cancel out.

[@MattMcIrvin]

That was my first thought--if you take the complex plane plus a point at infinity, all of these number lines embed in it in strange curved ways. I wonder how they would look plotted on a Riemann sphere.

[@popularmisconception1]

@@viliml2763 yeah exactly. That's why I don't like when they teach that negatives numbers don't have logarithms. You can just invent them analogically to how negatives can be invented. As long as you don't multiply them among each other before you exponentiate them back, that causes problems I could not get over in my efforts, that's the point when the k-th branch matters.

[@krakow10]

I had the same thought like 10 years ago and came up with the log exp solution forwards and backwards, but I couldn't imagine how it could be useful so I stopped playing with it. You really carried the idea as far as you could without expectation for a reward, you must have true mathematicans' blood!

[@VaradMahashabde]

Man, casually giving a polylog and polyexp classifying function as an exercise is too much. Amazing video!

[@AJMansfield1]

Your audio is mixed rather quiet. YT allows replacing audio tracks after the fact IIRC; if you can, see if you can replace the audio with a re-render with the master gain set 20 dB higher. In general, when mixing audio for KZbin, your target should be mix to somewhere above -16 LUFS. Anything louder than that will be automatically normalized to the same volume as other videos when you upload, but KZbin doesn't automatically increase the volume of videos that are quieter than that the same way.

[@unaryplus]

You're right, it is very quiet. I'll try to fix that if I can. I've never recorded anything like this before.

[@veeseir]

theres also some weird low frequency bumps and pops

[@sanes4402]

You made me feel as if everything i've learned in math classes has lead me to understand this video. Thank you

[@twixerclawford]

Trying to imagine how this all extends into the complex numbers makes my brain break

[@MagicGonads]

They don't, since exp is not bijective in C but it is in E (by definition)

[@DrunkenUFOPilot]

It's been done. It depends on generalizing exp^n and log^n to any real n and the dealing with Abel's equation, and... well there's a bunch of stuff. But it's been done.

[@juchemz]

I really loved this video. I can tell you understand this number system very deeply. It all clicked for me seeing each Kn with its own number line, really highlighting the isomorphisms with the real numbers. I appreciated the pacing too, covering all the fundamental ideas without over explaining. You deserve many more views. Thank you for sharing this.

[@atimholt]

I had this thought a couple months ago. Didn't take it anywhere-nice to see people thinking about the implications.

[@lock_ray]

That's funny... I had this exact train of thought at some point, to an uncanny degree. Even went as far as thinking about "subaddition" log(exp x + exp y), though this is where our tracks diverge - the exponential numbers system is new to me :) An interesting tidbit I came across on my own track is: if you switch to base 2 rather than base e, you get a pretty neat relationship between operator n and operator n+2. Namely, in any base the "square" function x -> x •_n x has the form x-> x •_(n+1) c_n for some constant c_n depending on n. If the base is 2 then we get the nice property that c_n is the identity element for •_(n+2). Example: if ° represents subaddition then x°x = x+1 so that c_(-1) = 1, or the multiplicative identity. Similarly x+x = x•2 so c_0 = 2 or the identity of •_2 and so on... A way to express this fact formulaically is x [n] (x [n+1] y) = x [n+1] (y [n-1] y) where [k] is the kth operator. In case n=0 this becomes x + xy = x(y°y)

[@davejones7610]

Yes, base 2 is very interesting -- and potentially more practical.

[@popularmisconception1]

You can come to this from the side of algebra: Think of a field (R,+,•) which has two special elements: ZERO (a + identity and • annihilator, 0) and UNITY (a • identity, 1). But on a two-operation algebra, you could also define another kind of special non-zero element x such that x + x = x • x. Let's call that DEUCE, since in real numbers there is only one and it is 2. Precisely since x + x = 2 • x. You could also define it as a sum of two multiplicative identities, DEUCE = UNITY + UNITY, since these equalities will always hold: x + x = 1•x + 1•x = (1+1) • x = x • x, therefore x = 1+1. Your thing works nicely, since in base-two exponentiation exp^2(ZERO) = DEUCE (exp(ZERO) = 2^0 = 1 = UNITY, and exp(exp(ZERO)) = exp(UNITY) = 2^1 = DEUCE). Now think about a structure, where the basis of exponentiaton of K_n would not be fixed, but would always be a DEUCE of (K_0, +_0, •_0). The next base (BASE_n+1) would always be DEUCE_n ^ DEUCE_n. Wonder how this would go backward to negative n, but it would probably stop at base 1 logarithm :D so negative extensions in this way would be probably limited, but maybe for an extra nice property.

[@kjetil1845]

> if ° represents subaddition then x°x = x+1 so that c_(-1) = 1, or the multiplicative identity. something neat about this is that s(x) is also x+1, and s(x) is what we usually use for "subaddition", so using base 2 connects this approach with the standard approach

[@Xonatron]

You created the successor operator. Like multiplication is repeated addition, addition is repeated succession. I view everything as shortcuts to manual succession.

[@BS-bd4xo]

Or x•(x*y) = x*(y+y) Aka x•(x^log(y)) = x^log(y+y) ??? No that can't be. Huh?

[@mintsjams8862]

I haven't watched the full video yet, but I took your advice and wow, this is only of my favorite problems I've ever solved. I'm embarrassed how long it took to figure out (5 hrs over the course of a weekend), but when it finally clicked it was so satisfying.

[@cparks1000000]

20:37 That escalated quickly. A wild category theory has appeared.

[@rauljvila]

Love these "generalization of simple operations"! Great video, thanks!

[@kannix386]

have you considered using √2 as the base instead of e to conserve the property 2*2=4 for all operations *?

[@unaryplus]

Nice observation. I will talk about alternative bases in my next video.

[@ゾカリクゾ]

satisfying.

[@duskyrc1373]

Essentially you can define an infinite number of ☆ operations (and by extension ◇, etc.), determined by the base used. All to properties you discuss hold regardless of what number you choose. Edit: in that case, is this really a binary operation? Is it not a trinary operation where you have preselected one of the inputs as e?

[@eladto]

1984 reference? :)

[@kannix386]

wait, what?@@eladto

[@soupisfornoobs4081]

When the channel name mentions unary, you know it's gonna be a banger

[@luker.6967]

I love functional analysis focusing on identities like the ones you use. Great stuff.

[@a1-h8]

Absolutely incredible for your first ever maths video! This is way better quality than most well-established maths channels. Well done and thank you!

[@G.Aaron.Fisher]

This reminds me a lot of the research I did after learning what a slide rule was. Turns out everything is addition all the way down.

[@RealNoobKing]

Guys, I’m 163, when I was 1 day old, my great grandmas hamster said I could never find the true theorem of neurons secret multiplication. But because of this beautiful video, the 162 year and 364 day wait is finally over. Thank you unary plus

[@TeaTimeee]

7:36 shouldn't it say "*(n - 1)" instead of just "*"? For example the star operation uses multiplication but the diamond operation uses star, so each consecutive operation uses the previous one to define this property x *(n) y = exp^n(log^n x *(n-1) log^n y)

[@alejrandom6592]

4:33 I find it very elegant that here "e" (2.71828...) is the identity, which is also the symbol that is generally used in group theory for the identity element.

[@IroAppe]

This video and all the ideas in the comments are very inspiring. It reminds me back to the point made in the beginning of university mathematics: You can think of anything you want, Math is all imaginary (at least initially). As long as it is logically consistent, it's usable. The courage to not stop at exactly 10:35: "But this not true for plus zero, because there's no such thing as dot negative one, and if there were, its domain K negative 1 would have to be larger than the real numbers." Most people would immediately stop there. However, it takes exactly that courage in order to advance, to explore the possibilities there are, and find new ways to work with mathematical concepts. At first, it seems wrong. But if you find a way to keep that new system logically consistent with itself, and also even tie it in logically consistently with the existing Mathematics, you've done nothing wrong. It's a logically consistent extension set of Mathematics. Now with refining, for example trying different bases as the comments did suggest, it might become something. I wonder if that is how the most advanced Mathematicians work. They want to solve a problem, but the existing Mathematics provide no solution. So they work with different ideas and try new things that miiight not be existing Mathematics, but are still logically consistent, and invent new Mathematics on the way to solving that problem. Adding to the kit of tools we have to solve problems. It's still mindblowing to me how this just works. We invent completely imaginary structures by seeking out new possibilities, and as long as we can make it logically consistent, there might come along an application that proves that this is how reality works after all.

[@alejrandom6592]

I liked this video when I first watched it, but now that I know some basic group theory, I love it!

[@geekjokes8458]

might seem off-topic, but i think this video highlights one of the paradoxes in teaching mathematics: on one hand, it's nice that you construct the operations and the notation (and the visualization ideas) piece by piece, just like ideas usually form, instead of making a _definition_ upfront, seemingly out of nowhere, about how to talk, write, and even think, about this new concept the viewer wouldnt know about - which was something i always hated and thought "how am i supposed to derive this if i dont already know the answer?" on the other hand, the constant change of notation you made made it kind of difficult to follow the first time i watched (just after it came out), which is softened by the fact that i can rewatch, rewind, etc, which is something i obviously cant do in a real classroom

[@theendofemeralds]

I love this, I have to find a way to share it with someone who will find ot as interesting as I do. Good job with the title and thumbnail, had they been less interesting, because I found this right at the end of my lunch break, I wouldn't have saved it to my watch later, I would never have given a view to this master piece.

[@alejrandom6592]

Damn that escalated quickly. Not a second wasted. Great video

[@StainlessHelena]

This video makes me feel like the man in the Flammarion engraving sticking his head through the firmament to see the mechanics of the spheres. Great work!

[@agfd5659]

This is a really cool concept and a very natural explanation of how you discovered these sets!

[@WithinEpsilon]

Ill be damned if this doesn't make it to the top 10 entries! #SoMe3

[@lollertoaster]

This video brings me back to my childhood, when I would also invent completely new maths just for fun.

[@AJMansfield1]

This is the best #SoME3 entry I have seen so far! You've done a remarkable job keeping me hooked and inviting me to try to work out the next step myself. I was already considering extending the (K0,+) group to the infinite dihedral group to allow (K1,·) to cover negative integers -- which is EXACTLY the group structure you ended up with for the K-1 integers. And now I'm left wondering how I could define analogous countable groups for more negative K -- and if the limit of these groups analogous to E would still be countable.

[@luiz00estilo]

My man just nonchalantly invented a new number system damm

[@AJMansfield1]

1:30 My attempt before watching: a🟊b = exp( log a · log b ) Not only is it left and right distributive over multiplication, it's also commutative! Commutativity: a🟊b = exp( log a · log b ) = exp( log b · log a ) = b🟊a Left Distributivity: a🟊(x·y) = exp( log a · log(x·y) ) = exp( log a · (log x + log y) ) = exp( log a · log x + log a · log y ) = exp( log a · log x ) · exp( log a · log y ) = (a🟊x)·(a🟊y) Right Distributivity: (x·y)🟊a = a🟊(x·y) = (a🟊x)·(a🟊y) = (y🟊a)·(x🟊a) [from the two previous properties]

[@theobscure]

Literally amazing video!! How does it not have millions of views?? I’m sure that this is gonna change.

[@Fire_Axus]

it forgot the complex numbers

[@antoniusnies-komponistpian2172]

Almost only math students can understand this

[@octavianova1300]

This is honestly an incredible original mathematical exploration and it's astounding this is your first video!

[@flavertex658]

THANK YOU! I got weirdly obsessed with this question and put a lot of effort into it, but had to give up before reaching a satisfying answer to completely generalizing evening. This is exactly the train of thought i was pursuing and you've pursued it to the horizon - thank you so much!

[@jucom756]

The integer repeated exponentiation function can be generalised to the reals differently depending on the conditions you set, if the only condition is continuity it's as easy as linearly connecting 0 to 1 to get a continuous solution.

[@terdragontra8900]

The function you get that way is not infinitely differentiable, and finding a closed form for one that is seems impossible due to a theorem i came across, but there may be an infinite series or something I'm not sure.

[@DrunkenUFOPilot]

Yes, indeed! And functions can be defined that aren't piecewise, but smooth and natural (i sound like a beer ad...) Look up the works of Australian mathematician George Szekeres (1961), and German mathematician Helmut Kneser (1950).

[@yigitrefikguzelses291]

This is a wonderful video! I am very happy that i found this channel!!!

[@angeldude101]

At first I was wondering how this idea would react to the ℂomplex numbers, since it allows for logarithms of negative numbers. That said, (or perhaps fittingly given what came later), the logarithm of a ℂomplex number isn't unique. Instead, it repeats every τ units along the "imaginary" axis, which I suppose isn't really that different to how every K_n number is duplicated as a K_m number where m < n. That said, it doesn't give a defined logarithm for 0 beyong "-∞," nor does it really give an idea of how to treat ln(-∞). So like how your ⋆ operator isn't completely unrelated to powers, the Exponential numbers have some connections to the ℂomplex numbers but aren't quite the same thing.

[@unaryplus]

The logarithm of an exponential number is unique. It can be a bit confusing, since the way I displayed the exponential numbers in the last part of the video involved drawing each number more than once.

[@itsmeagain1415]

@@unaryplus I get the part you mean that every log is unique and I think (in some weird 500page paper illustration) that maybe you can unravel the periodicity of complex functions to be formulated in terms of some more regular (non-periodic) functions on the exponential numbers ... wait, did you try to explore if these numbers form rings? fields? groups? or some abstract algebraic entity with unique properties?

[@TheOneMaddin]

I agree with the other comments. This should win Some3. It demonstrates that originality and clarity are more important than flashiness and pretentious complexity. Better than recent 3B1B videos.

[@DarkPortall]

Sadly it wasn't submitted on time

[@kolskytraveller1369]

It isn't very useful on its own, and there were more enjoyable SoMe3 submissions

[@coopergates9680]

Most underrated content I found today. It's easy for conventions to clash, eh? What I see at 7:07 is often used (in my experience) for notating the whole logarithm raised to a power. Such as log^4(n) being (log n)^4. I'm sure you've seen that done for trig functions as well. cos^3(n) = (cos n)^3

[@cmilkau]

Came to the same conclusion when stumbling over a^(log b) = b^(log a), which is exactly this commutative exponentiation exp((log a)(log b)), and always wondered why there isn't more attention to this sequence of operations. (You can continue this for commutative tetration and so on, and even in the other direction beyond addition.)

[@angelmendez-rivera351]

The reason it hasn't gotten much attention is that, while it makes for neat trivia, it's largely inapplicable to almost everything we do.

[@viliml2763]

17:30 Another way you could change the visualization is by using a double infinite stack of half-lines. The real interval (0, 1] would be mapped to the K1 interval (-inf, 0], (1, e] to K2 (-inf, 0] etc

[@blackestbill7454]

There once was a language so neat, Where numbers and patterns did meet. Math spoke with such glee. And all came to see, As existence danced to its beat.

[@madddd1]

Ever watched a math related video as a detective thriller? Then this is a video. Smelled log of negative numbers from a mile away but still didnt expect the final result

[@didles123]

*The gist of this video is:* Exponentiation treats multiplication as repeated addition and generalizes it through the Ackermann function a +ₙ b = φ(a, b, n). The video instead treats multiplication as a ⋅ b = exp(ln(a) + ln(b)) and generalize it through function composition a +ₙ b = exp⁽ⁿ⁾(ln⁽ⁿ⁾(a) + ln⁽ⁿ⁾(b)), since this preserves commutativity and distributivity. The operator after multiplication in this generalization is a ⋆ b = a +₂ b = exp⁽²⁾(ln⁽²⁾(a) + ln⁽²⁾(b)). Note: + = +₀, ⋅ = +₁, ⋆ = +₂ Furthermore it considers +ₙ for negative n, and defines the non-standard exponential numbers. Complex numbers extend real numbers by defining i² = -1 (this leads to i = √(-1) and Euler's identity exp(τi/2) = -1). Exponential numbers extend real numbers first by defining the set K₋₁ by defining exp(~0) = -1. (this leads to ~0 = ln(-exp(0)) = ln(-1)). Taking the ln of numbers in K₋₁ leads to K₋₂, and exponential numbers are E = K₋ₙ as n → ∞. *Some of my thoughts on the video:* In the interest of time the video seems to leave out a lot of stuff that I had to look into to make sense of it. For example, the ~ operator confused me at first. I think the general unary inverse operator made it easier for me to understand: -ₙ x = exp⁽ⁿ⁾(-ln⁽ⁿ⁾(x)). This makes it clear that products ⋅₀ uses -₀ to extend to negatives, that -₁ is the multiplicative inverse, and that ⋅₋₁ would need to use -₋₁. It also makes it easy to prove for all integers n that -ₙ(-ₙ x) = exp⁽ⁿ⁾(-ln⁽ⁿ⁾(exp⁽ⁿ⁾(-ln⁽ⁿ⁾(x)))) = exp⁽ⁿ⁾(--ln⁽ⁿ⁾(x)) = exp⁽ⁿ⁾(ln⁽ⁿ⁾(x)) = x. Thus ~ can't be composed to express K₋₂ exclusive numbers, even though they could be expressed with -₋₂. I was also confused because not many concrete examples of exponential numbers were given, and a standard notation wasn't provided. Complex numbers can be denoted as a + bi, but it was unclear how this worked for exponential numbers. I realize they can be decomposed into the tuple (a, b) ∈ [0,1)×ℤ. Perhaps this was avoided since it would make the exercise at the end obvious. Maybe there is a better decomposition. Either way having some concrete examples of them in a standard notation makes it much easier to understand the video past the K₋₁ part. Other commenters have noticed that since a ⋅ b = 2^(log₂(a) + log₂(b)) also maintains the distributive property, and so it works for other bases. I guess we will have to wait for a video about other bases, but it would be nicer if it was pointed out that e was not the only choice. One of the advantages I see for using base e is how calculus deals with it. For positive n, [exp⁽ⁿ⁾(x)]' = Π exp⁽ⁱ⁾(x) for i = 1 → n and [ln⁽ⁿ⁾(x)]' = 1/(Π ln⁽ⁱ⁾(x)) for i = 0 → n. It also helps to define an operator Eₙ such that Eₙ(f) = exp⁽ⁿ⁾ ∘ f ∘ ln⁽ⁿ⁾. An interesting property is Eₙ(f ∘ f) = Eₙ(f) ∘ Eₙ(f), and in general Eₙ(f⁽ⁱ⁾) = [Eₙ(f)]⁽ⁱ⁾. That is to say this generalization distributes over function composition. For constant functions f, Eₙ(f) = f. This generalizes our operations. Eₙ(+) = +ₙ, Eₙ(⋅) = ⋅ₙ, and my suggested operator Eₙ(-) = -ₙ. This can also generalize averages. Suppose for a finite list of numbers X, we have a function avg(X) which returns the arithmetic average. We can define a general averaging function avgₙ = Eₙ(avg). The function avg₁ is the geometric mean. Note that the harmonic mean is g ∘ avg ∘ g where g(x) = 1/x. The final thing I noticed is that it can generalize derivatives using Dₙ(f) = Eₙ(D(f)) where the normal derivative D(f) = f'. These general derivatives maintain the property Dₙ(f +ₙ g) = Dₙ(f) +ₙ Dₙ(g). They also compose nicely such that (Dₙ)ⁱ = (Dⁱ)ₙ. They also seem to maintain the property of throwing away constant functions: for any constant function g(x) = C, then Dₙ(f +ₙ g) = Dₙ(f). More specifically D₁ throws away multiplied constants while D throws away added ones. If f(x) = nx then (D₁ f)(x) = exp(1/x) which is independent of n.

[@DrunkenUFOPilot]

About averaging... the additive mean, the multiplicative-based "geometric" mean... start with two numbers, compute both of these means, then take those two numbers, repeat... it rapidly converges to a value related to elliptic integrals. What happens if we use means based on the additive and the star operations?

[@annaclarafenyo8185]

Your construction on the real numbers is a pale version of "Riemann sheets", which instead of doubling number lines, creates a 2-dimensional complex manifold to represent the multivalued function. You should do the same thing in the complex plane, I also had this thought about iterated operations at some point, it's really interesting that it isn't discussed anywhere in the literature.

[@jakobr_]

For the first problem, there’s the super-logarithm, an inverse of tetration. Take your exponential number, put it into slog base e, and add 1. It’s sloppy but I’m not sure if there’s any other way to do it, and I’m also not sure if slog is even defined yet for non-integer outputs.

[@youknowwhenitsreal7]

You're pretty much right. For an element x which is in K_n but not K_(n+1) apply log n times, then add n. This map takes K_n \ K_(n+1) to the interval [n,n+1). Of course, if n is negative then really you're applying the exponential function n times

[@toasteduranium]

I want this to win #some3 and perhaps see some more professional investigation here. Also, I recall reading a Medium article on a similar concept, as well. It said something about how the operation they discovered (I think referred to as softplus or something) worked in one particular probability case better than the traditional format. Edit: it wasn’t soft plus. I can’t recall what it was called. It was someone’s personal endeavor, and the graph of it was smoothly curved to look like the corner of a squircle. It used a logarithm.

[@iancampbell4394]

So glad I found this. I had been working on a paper on this topic a couple of months ago, but couldn't finish what I wanted to do. You can actually reconstruct most of calculus in terms of these general "dot" operations, but I couldn't figure out how to prove Taylor's Theorem in a general setting. You can derive it and all the underlying rules hold in the proof, but showing the remainder term vanishes is a head-scratcher.

[@Sup3rdud4]

I had been wondering for a while if there was such a thing as exponential numbers. I had seen hyperbolic and hexagonal numbers. Seeing how they define number systems in abstract algebra. I didn't expect it to be so robust

[@unaryplus]

I've never heard of hexagonal or hyperbolic numbers before. Sounds interesting. No idea if there's any relationship with the exponential numbers as I've defined them in this video.

[@amruthchangappa]

@@unaryplus hyperbolic numbers are better known as split-complex numbers, which is probably a more well known name? Hyperbolic numbers is a way better name for them though

[@applimu7992]

I have no idea about the first exercise (something about tetration?), but the infinite chain of the set of real numbers w/ logs between them *has* to be the diagram in which E is a colimit over. Overall I loved this video btw!!

[@unaryplus]

Yes, the diagram is ...

[@eladto]

most exciting video I've seen in a long while 🎉 respect!

[@ErikLeppen]

Whenever I noted the "indexed dot" operators, I was expecting negative values of n to appear some time which would swap the log and exp operators. I didn't expect the extended domains. Intriguing journey into new mathematics, and all explained very clearly and concisely.

[@nrrgrdn]

Thank you! I have been obsessed about what comes after addition and multiplication

[@robkb4559]

Brilliant concept, beautifully paced and animated. Bravo!

[@sonicwaveinfinitymiddwelle8555]

This makes me think more and more of moving away from number lines and adapting values in their true forms because it's hard to see patterns in number lines without complex equations which take time to express

[@K_Huss]

Definitely top 5 math videos ive seen. I'm curious however, is this something you came up with in your free time/for a research project or there literature about and its applications? I would love to learn more

[@unaryplus]

I came up with it, essentially by the process described in the video. I have no idea what the applications could be, if there are any. I will probably make a follow-up video sometime in the next few months.

[@Chewxy]

​@@unaryplus IINM you described a structure that is isomorphic to the "native" structure for computer functions. At any rate, subscribed and bell'd for the next video on this!

[@Arsonade]

​@@unaryplusI'm very unfamiliar with this kind of math but I kept thinking throughout the video about how multiplication is used in regression for coefficients and often interpreted in terms of associative effects (addition being used more for overall bias). I keep trying to think of what * would mean in this context

[@francescosorce5189]

This video is an absolute gem, especially the construction of E. I would have loved a little more discussion of what you suggested as exercises, because I feel this is one of the most fun examples of a colimit I've seen so far.

[@AlanKey86]

This was awesome... can't say I understood it all, but it was awesome to see it all unfold :D

[@FunMaker39]

Wow, this was great. My googological brain just begs to consider a ⋅ω operation where x ⋅ω y = expʸ(logʸ x ⋅ logʸ y) that should outpace all the previous ones. We could probably go further with larger ordinal numbers. I wonder what properties such operations would have and if this way we could construct a sort of continuous fast growing hierarchy this way.

[@mysteriousgrimreaper]

I was literally just thinking about what a binary operation version of exponentiation would look like and then I found this video, thank you for satisfying my curiosity 😅

[@meganton9417]

What a great video, my favourite entry to Some23 so far!

[@angelmendez-rivera351]

0:47 - 1:24 Another issue is, exponentiation is not exactly well-defined as a binary operation on most number systems we use. Also, even if you were to go and use monoid morphisms to construct an exponentiation-like operation, it turns out exponentiation is constructed from monoid homomorphisms from (Z, 0, +) to (Z, 1, •), rather than from monoid endomorphisms on (Z, 1, •), which is why the property x ^ (y + z) = (x ^ y) • (x ^ z) is instead treated as the defining property, rather than using the distributive property.

[@aryaghahremani9304]

you can increase the the domains and ranges of the higher order n operations to include all positive numbers by using something like (1+)^x where 1+ is like 1+epsilon and log_1+(x) instead of exp and ln

[@Izzythemaker127]

This is the first video I've seen on your channel, but this was amazing, I subscribed. Also it feels weird to think this has the same cardinality as the real numbers, but if a bijection between them exists it must.

[@notwithouttext]

yo it's izzy the maker

[@Izzythemaker127]

@@notwithouttext yo it's not without text

[@notwithouttext]

yo it's @@Izzythemaker127 replying to not without text

[@huhneat1076]

I just wanted to multiply repeatedly but I accidentally found the minus levels of the number line

[@landy4497]

this was AMAZING. fantastic subject, everything was incredible. absolutely loved it

[@odiloncintra1377]

I have felt dumb before, but this video , somehow, make my stupidity hurt more... But I'm glad that people capable of creating or understanding this level of math exist.

[@smiley_1000]

The construction of the exponential numbers reminds me of a direct limit: We have the inclusions as bonding maps and natural injections from each object. Edit: I think this is just what you mean by constructing E as a colimit at the end

[@Catman_321]

These numbers are very interesting! Makes me wonder a lot of things: Can these numbers be mapped to the complex numbers? The log is a multivalued function in the complex numbers so maybe there is a way to map them for each K_n. Alternatively, is there some way to extend these numbers to a continuous grid? For example, in the example you showed us, each K number line is separate from each other, and the points in between arent defined. Is it possible to define non-integer K systems such as K_0.5 or K_2.71828? Is it even possible to take half a log? Also, since you created this exponential number system, does that mean all *n and +n functions can be defined for all numbers in E?

[@Djake3tooth]

Try googling "half-iterates" and read the wikipedia, it seems like you could use this technique

[@BS-bd4xo]

I think yes cuz log of any complex number (including negatives) results in a complex number. Since log(r•exp(θi)) = log(r) + θi. So applying log to a negative number multiple times gives you some complex number. I'd love to know tho how such a graph looks like. I believe that K_0.5 is not allowed. This is because expⁿ and logⁿ is about repeatedly applying a function, which you can't do a half times. Unfortunate.

[@quantranhong1092]

@@BS-bd4xoI mean we have half derivatives and half integral, surely we can have a "half applied" function right ?😂

[@crimsondragon2677]

Yeah, sure. I’ll add this to my worldview.

[@lexyeevee]

it was never stated explicitly but i'm enjoying the fact that a^ln b = b^ln a, something i'd never thought about before

[@carsonlopez8284]

This is amazing- the fact that there exists hyperoperations that are abelian blows my mind- I thought there would never be such operations. But the fact that they all involve the functions exp and logarithms means that particular forms of exponentiation, tetration, and beyond are abelian. While I've been thinking about hyperoperations, everyone has been questioning whether or not π tetrated to 4 is an integer. I'm interested to see if I could do that.

[@luisvasquez5015]

Instant subscribe. Well done. Pedagogical presentation and just the right amount of good rigour 👏👏👏👏

[@tricky778]

You've done the right thing using superscript for repeated application, I've always wondered why people use superscript on an operation to indicate a power of its result when it's obvious it should be this instead.

[@jpphoton]

Excellent. No superflousness. just what it is - no more - no less. Jedi gemz. Thank you

[@DrunkenUFOPilot]

Nicely made video! I enjoyed the motivation at the start and the development of the topic. 1:25 - pause and try to find a solution myself.... LOL! I already did that over forty years ago! I wondered the same things way way back in high school, wrote a paper in college that got me into a few minor math conferences, and it's been a fun topic to watch ever since. But I also wondered - any time I see integers, I wonder what about fractional values, real values. Is there an operation +[1/2] halfway between adding and multiplying? Sure there is! All you gotta do is define exp^{1/2}, no problem... that's where things get interesting.

[@gregc517]

I've asked the initial question, but instead of all of the real numbers, I wanted to do it over the integers/natural numbers, and came up with an operation, which when given 2 positive natural numbers as lists of prime powers that make up the number (going infinitely to the left with 0, for example 12=3*2^2 would be ...0,0,1,2), multiplies them as if it was an infinitary number (so no carries), and puts them back together into a new positive integer (for example 3 star 15 = 35, because 3 is ...0,0,1,0; 15 is ...0,1,1,0; and 35 is ...0,1,1,0,0). This is commutative and associative because it is just normal multiplication over a large enough base, and it distributes over normal multiplication because normal multiplication is just addition of those prime power lists.

[@DavidColbyCode]

Correction--0,0,1,0 star 0,1,1,0 would be 0,0,1,0 ; which is 3 yet again, not 35.

[@gregc517]

@@DavidColbyCode no, I mean it to work like normal multiplication, just without carrying, so since 10*110=1100 and doesnt use carrying, 3 star 15 is 35

[@gregc517]

The commas are there to separate the infinitary digits, and so star, even though it doesn't use carry, still isn't simply a dot product between the lists (for the same reason that 12*21 is not 22). This way it distributes over normal multiplication (which just adds the corresponding numbers in the list), is associative and commutative, and all of this is easily shown using the fact that regular multiplication has those properties over regular addition, and that for the algorithms of multiplication by hand and addition by hand, the only part in which the base is relevant is when carrying, and for big enough bases and the same input digit lists, the resulting digit list is the same. All I'm doing is choosing a big enough base to interpret the powers of primes as digits in so that there will be no carries in the process, which can be interpreted as just doing those algorithms in infinitary. More examples to show what I mean: 16 star 16 = 2^16 = 65536, because ...0,4 multiplied by ...0,4 is ...0,16; 48 star 81 = 5^4*3^16 = about 2.7*10^10, because ..0,1,4 multiplied by ..0,4,0 is ..0,4,16,0; 6 star 6 is 90 because ..0,1,1 multiplied by ..0,1,1 is ..0,1,2,1

[@Lokdora]

You can also expand the subscript notation to -1, x + y = log(exp x • exp y), so + is •_{-1}

[@bjornfeuerbacher5514]

That's already in the video, around 14:50.

[@notwithouttext]

while watching this already cool video i thought "well addition doesn't HAVE an identity, what do you do there?" but negative infinity makes a lot of sense! it's very satisfying how it extends and even adds new numbers that can correspond to the real numbers!

[@FZs1]

If I understand correctly, the exponential numbers can actually be bijected to the reals! For each exponential number _x_ , there is exactly one line _n_ , on which _x_ falls in the range [0;1[. Let the value of _x_ on the line _n_ be _v_ . Then, the real value _r_ corresponding to _x_ is _r=n+v_ . In reverse, the real number _r_ is mapped to the exponential number having value _v_ on line _n_ , where _v_ is the fractional part of _r_ and _n_ is the integer part. So, the exponential numbers have a number line!

[@unaryplus]

This is the correct answer to problem 1 at the end of the video.

[@FZs1]

@@unaryplus Oh indeed it is! I somehow completely missed those two problems. After watching the video, I just noticed this relation and wanted to throw it in as a cool fact...

[@Tim0feyK]

Me while watching the video 0:00 Ow, cute video about multiplication, nice! 2:45 An isomorphism's just dropped? Intriguing... 4:24 Look, new identity elements! 4:32 I smell a ring... 5:44 I see what's coming. 5:50 Yep. 7:00 New notation, fi-i-ine... 7:20 Already lost it. 8:50 Grows fast indeed... 9:44 Ok, now I really lost it. 10:45 Stop us from What? 10:56 What? 11:12 Negative what? 12:10 So, it's a wheel now. 12:16 Oh, here we go again... 13:40 At this point I have no idea what's going on 14:54 Btw, haven't we just went beyond real numbers? 15:43 Infinitely many times? 16:33 E 19:05 Summary, I guess... 20:05 Exercises for viewers. Man, this makes experience of the video complete.

[@aleph0540]

Bahahahha similar experience here. It goes from look at this cool thing I found to you know nothing mere mortal.

[@macronencer]

OK, I'm pretty tired this morning and I'll be honest, I lost focus during some of this so I'll have to go back and watch again at some point. However, I see pretty much what you've done here, and it's wonderful!

[@IronFairy]

Ok, I didn't expect you to answer what comes BEFORE addition as well. Neat!

[@cynical5062]

The e^e^e... operations, can be nicely represented using Knuth's Up-Arrow Notation. With that, we have dom(⋅ₙ) = e ↑↑ (n - 2) (This only works for n ≥ 2 though, but for n=1 and n=0 you can just display them as constants. I vaguely remember there existing some operation similar to Knuth's uparrows with 0 and 1 as the first two values, but I don't remember the details. If it comes back to me I'll leave a comment here.)

[@strangeWaters]

Another fun exercise is to try and go in the opposite direction. Is there anything *below* addition? It turns out that addition distributes over the "max" operation, and you end up in an interesting area called "tropical geometry".

[@DvdAvins]

I'm not sure one should define ~x as less than negative infinity. Discontinuities often break monotonicity. But it does seem to work.

[@angelmendez-rivera351]

3:26 - 3:55 It should be noted this can be done with any monoid isomorphism, not just exp. If you have a field (F, 0, 1, -, +, •), then you can consider the monoids (F, 0, +) and (U, 1, •), where U is some multiplicative subset of F. If f is a monoid isomorphism from (F, 0, +) to (U, 1, •), then the operation x ✦ y = f(f^(-1)(x) • f^(-1)(y)) distributes over •. This is because x ✦ (y • z) = f(f^(-1)(x) • f^(-1)(y • z)) = f(f^(-1)(x) • (f^(-1)(y) + f^(-1)(z))) = f(f^(-1)(x) • f^(-1)(y) + f^(-1)(x) • f^(-1)(z)) = f(f^(-1)(x) • f^(-1)(y)) • f(f^(-1)(x) • f^(-1)(z)) = (x ✦ y) • (x ✦ z). exp just happens to be continuous (it respects the field topology), so it is a convenient choice.

[@notwithouttext]

i looked it up and, at least for the real numbers, +-1 is the "smooth maximum", which makes sense looking at the graph. interestingly, the normal maximum function also satisfies the distributive properties, and also has negative infinity as identity, but structurally different

[@november666]

I’ve been exploring these concepts in the context of wheel theory and how they apply to parallel addition. It’s all very interesting

[@yassine-sa]

His voice feels like he'll never stop generalizing whatever he finds

[@circuitcraft2399]

I recall seeing your * operator in "Complete Sets of Postulates for the Theory of Real Quantities" (1903) by Edward V. Huntington. He mentions in passing that the positive reals under multiplication are isomorphic to the reals under addition, and writes down the formula for *, only using base 2. (This appears in §2, Theorem I')