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.

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

9 ай бұрын

I think I saw this with 3 blue 1 brown

@Archetype784

9 ай бұрын

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

9 ай бұрын

@@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

9 ай бұрын

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

9 ай бұрын

did you eventually find it?

This video feels like discovering math

@udveetpatil8002

10 ай бұрын

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

@sebastianlenzlinger9291

10 ай бұрын

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

@Fire_Axus

10 ай бұрын

your feelings are irrational

@cmdlp4178

10 ай бұрын

​This might just be a new field of mathematics

@CasualLifeExperiencer

10 ай бұрын

​@@cmdlp4178Could well be, yeah

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

10 ай бұрын

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

10 ай бұрын

@@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

10 ай бұрын

@@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

9 ай бұрын

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.

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

10 ай бұрын

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

@GrifGrey

10 ай бұрын

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

@a1-h8

10 ай бұрын

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

@firefly618

9 ай бұрын

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

@yumyumyucky6313

9 ай бұрын

@@unaryplusthat’s tragic

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

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 KZread, 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 KZread doesn't automatically increase the volume of videos that are quieter than that the same way.

@unaryplus

10 ай бұрын

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

9 ай бұрын

theres also some weird low frequency bumps and pops

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

10 ай бұрын

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

10 ай бұрын

@@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

@kappascopezz5122

10 ай бұрын

​@@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

9 ай бұрын

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

9 ай бұрын

@@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.

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

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

22 күн бұрын

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

19 күн бұрын

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

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

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.

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

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

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.

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!

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

@MagicGonads

10 ай бұрын

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

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

@unaryplus

10 ай бұрын

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

@zokalyx

10 ай бұрын

satisfying.

@duskyrc1373

9 ай бұрын

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

9 ай бұрын

1984 reference? :)

@kannix386

9 ай бұрын

wait, what?@@eladto

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

This video is the best actual mathematics video I have seen in a very long while. Most people would not be able to understand its importance unfortunately. Most people would prefer artistic visual nonsense that makes the illusion of understanding something.

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

10 ай бұрын

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

@popularmisconception1

10 ай бұрын

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

10 ай бұрын

> 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

10 ай бұрын

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

@BS-bd4xo

9 ай бұрын

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

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

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.

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

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

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

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.

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

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.

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

This is what universities are missing out when they hang up on the guy that calls to say they "have theories"

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

10 ай бұрын

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

10 ай бұрын

@@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?

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

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

7 ай бұрын

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.

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

9 ай бұрын

Sadly it wasn't submitted on time

@kolskytraveller1369

9 ай бұрын

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

Brilliant concept, beautifully paced and animated. Bravo!

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

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

@Fire_Axus

10 ай бұрын

it forgot the complex numbers

@antoniusnies-komponistpian2172

10 ай бұрын

Almost only math students can understand this

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.

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

Super intriguing, excellent video!

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

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

10 ай бұрын

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

10 ай бұрын

​@@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

10 ай бұрын

​@@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

Woah, that's so cool i didn't expect another number system!

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

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

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 😅

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

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

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!

My man just nonchalantly invented a new number system damm

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

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!

Wow! Really fun video! Everything was explained well and the visuals were super useful!

Babe, wake up, new math!

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

10 ай бұрын

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

10 ай бұрын

@@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

this vídeo is so good, keep up the great work!!

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.

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.

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.

OH my god this video tops it for me this year!

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]

Very interesting. Looking forward to hear more

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

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

10 ай бұрын

Yes, the diagram is ...

It makes sense that you're into xenharmonic music. That naturally leads to contemplating the isomorphism between addition of real numbers and multiplication of positive numbers a lot.

It is a very interested video (also quiet well made for a first video). I really hope I will see from your chanel more math content, not just for SoME.

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

10 ай бұрын

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

@BS-bd4xo

9 ай бұрын

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

9 ай бұрын

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

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

10 ай бұрын

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.

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.

This is unbelievably cool! (Also quiet.)

Leaving a comment for the algorithm because this video is genuinely amazing

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

This is amazing!!

"I'll call the next operation in this sequence diamond" funny, the word I was just thinking of was "diamond".

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

10 ай бұрын

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

Best of show, in my humble opinion.

*M*ath murmurs softly in the night's embrace *A*midst the void, stars find their place *T*ales of numbers, harmonies trace *H*eralding patterns, existences base *S*ilence surrounds, yet patterns persist *P*ulses of equations in nebula twist *E*very atom, every celestial wist *A*nswers to math, reality's gist. *K*aleidoscope of numbers, existence's list, *S*ymphony plays, as cosmos is Kissed

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

10 ай бұрын

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

If you ever wanted to extend this even further, K^1/2

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

8 ай бұрын

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

@FZs1

8 ай бұрын

@@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...

This was a great video, it is very aad that this was not finished in time for some3

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!

Super interesting, nice video

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.

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.

Devastating you missed SOME3 but wow this is so amazing ❤

Beautiful

Amazing video. I appreciate all the hard work. It really shows. Did you come up with the system or are some some papers you have read?

THIS SHOULD HAVE MORE VIEWS HOLY-

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

Very neat!

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

9 ай бұрын

yo it's izzy the maker

@Izzythemaker127

9 ай бұрын

@@notwithouttext yo it's not without text

@notwithouttext

9 ай бұрын

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

finally answering the question I've had as a kid

Despite this being something I've thought about before but never did the calculations, this was a very interesting video. I wanted to figure out the arithmetic of something similar to modular arithmetic, but instead of working with the remainder of repeated subtractions, I was working with the remainder of repeated division, which led me to logarithms. I couldn't figure out what to do with log(x+y), so I developed an operation similar to yours, ln(exp(x)+exp(y)), and started playing with it. I did discover that addition distributes over this operation, so I could write it as x+ln(1+exp(y-x)), but that's not as interesting as when you change the base. You can do log_b(exp_b(x)+exp_b(y)), but I found that it's more convenient to write it as [ln(exp(xc)+exp(yc))]/c, where c=ln(b). Now, take the limit as c goes to infinity and your operation becomes max(x,y). If you take the limit as c goes to -infinity then you get min(x,y). These are things I noticed when playing with the graph f(x,y)=[ln(exp(xc)+exp(yc))]/c, but if you center it at (0,0,0) with f(x,y)-f(0,0), now you can also take the limit as c goes to 0, and you get (x+y)/2. Fr, f(x,y)-f(0,0) is so interesting to look at as you vary c, in fact you can discover a lot from its symmetries. Another thing is f(-x,-y)=f(x,y)-(x+y), which you can find out in 3 different ways. I didn't make any progress on the arithmetic I was talking about, as I got distracted by this interesting new operation.

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

@bjornfeuerbacher5514

10 ай бұрын

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

*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.

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

Wonderful! I love maths Conway style.

I love the idea of trying to implement this as code. So far as I know only lisp can do this natively.