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

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  • Опубликовано: 1 июл 2024
  • This video begins with a search for a binary operation satisfying certain properties, and ends with the creation of a new number system: the exponential numbers.
    #SoME3 (finished late)
    Although I came up with these concepts independently, a commenter has pointed out that the operations defined in the first part of the video already have a name: the "commutative hyperoperations."
    en.wikipedia.org/wiki/Hyperop...
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Комментарии • 514

  • @paulbloemen7256
    @paulbloemen7256 9 месяцев назад +26

    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.

  • @kjetil1845
    @kjetil1845 10 месяцев назад +304

    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
      @dangnabbit1379 10 месяцев назад +5

      I think I saw this with 3 blue 1 brown

    • @Archetype784
      @Archetype784 10 месяцев назад +26

      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
      @dangnabbit1379 10 месяцев назад +4

      @@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
      @michaelasmitty 10 месяцев назад +6

      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
      @paulkanja 9 месяцев назад

      did you eventually find it?

  • @d3tcovax
    @d3tcovax 10 месяцев назад +500

    This video feels like discovering math

    • @udveetpatil8002
      @udveetpatil8002 10 месяцев назад +27

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

    • @sebastianlenzlinger9291
      @sebastianlenzlinger9291 10 месяцев назад +30

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

    • @Fire_Axus
      @Fire_Axus 10 месяцев назад +8

      your feelings are irrational

    • @cmdlp4178
      @cmdlp4178 10 месяцев назад +2

      ​This might just be a new field of mathematics

    • @CasualLifeExperiencer
      @CasualLifeExperiencer 10 месяцев назад

      ​@@cmdlp4178Could well be, yeah

  • @timseguine2
    @timseguine2 10 месяцев назад +228

    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
      @unaryplus  10 месяцев назад +57

      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
      @timseguine2 10 месяцев назад +24

      @@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
      @BosonCollider 10 месяцев назад +17

      @@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
      @JerkoFlapdoodle 9 месяцев назад

      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
    @demonicdrn3037 10 месяцев назад +408

    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
      @unaryplus  10 месяцев назад +80

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

    • @GrifGrey
      @GrifGrey 10 месяцев назад

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

    • @a1-h8
      @a1-h8 10 месяцев назад +7

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

    • @firefly618
      @firefly618 10 месяцев назад +2

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

    • @yumyumyucky6313
      @yumyumyucky6313 10 месяцев назад +2

      @@unaryplusthat’s tragic

  • @VaradMahashabde
    @VaradMahashabde 10 месяцев назад +38

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

  • @pocarski
    @pocarski 10 месяцев назад +111

    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
      @popularmisconception1 10 месяцев назад +10

      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
      @viliml2763 10 месяцев назад +9

      @@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
      @kappascopezz5122 10 месяцев назад

      ​@@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
      @MattMcIrvin 10 месяцев назад +6

      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
      @popularmisconception1 9 месяцев назад +3

      @@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.

  • @AJMansfield1
    @AJMansfield1 10 месяцев назад +139

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

    • @unaryplus
      @unaryplus  10 месяцев назад +66

      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
      @veeseir 9 месяцев назад +1

      theres also some weird low frequency bumps and pops

  • @sanes4402
    @sanes4402 10 месяцев назад +18

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

  • @musicarroll
    @musicarroll 9 месяцев назад +6

    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
      @makhnoboi1996 28 дней назад +1

      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
      @musicarroll 25 дней назад

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

  • @WithinEpsilon
    @WithinEpsilon 10 месяцев назад +6

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

  • @rauljvila
    @rauljvila 10 месяцев назад +12

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

  • @juchemz
    @juchemz 10 месяцев назад +57

    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.

  • @luker.6967
    @luker.6967 10 месяцев назад +19

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

  • @twixerclawford
    @twixerclawford 10 месяцев назад +10

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

    • @MagicGonads
      @MagicGonads 10 месяцев назад +5

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

  • @lock_ray
    @lock_ray 10 месяцев назад +106

    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
      @davejones7610 10 месяцев назад +8

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

    • @popularmisconception1
      @popularmisconception1 10 месяцев назад +7

      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
      @kjetil1845 10 месяцев назад +6

      > 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
      @Xonatron 10 месяцев назад +3

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

    • @BS-bd4xo
      @BS-bd4xo 10 месяцев назад +1

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

  • @krakow10
    @krakow10 10 месяцев назад +6

    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!

  • @mintsjams8862
    @mintsjams8862 10 месяцев назад +2

    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.

  • @pontusloviken94
    @pontusloviken94 10 месяцев назад +1

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

  • @kannix386
    @kannix386 10 месяцев назад +169

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

    • @unaryplus
      @unaryplus  10 месяцев назад +108

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

    • @zokalyx
      @zokalyx 10 месяцев назад +9

      satisfying.

    • @duskyrc1373
      @duskyrc1373 10 месяцев назад +23

      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
      @eladto 10 месяцев назад +2

      1984 reference? :)

    • @kannix386
      @kannix386 10 месяцев назад +2

      wait, what?@@eladto

  • @G.Aaron.Fisher
    @G.Aaron.Fisher 10 месяцев назад +1

    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.

  • @soupisfornoobs4081
    @soupisfornoobs4081 10 месяцев назад +4

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

  • @agfd5659
    @agfd5659 10 месяцев назад +10

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

  • @aouerfelli
    @aouerfelli 10 месяцев назад +1

    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.

  • @a1-h8
    @a1-h8 10 месяцев назад +3

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

  • @octavianova1300
    @octavianova1300 10 месяцев назад +4

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

  • @TheOneMaddin
    @TheOneMaddin 10 месяцев назад +68

    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
      @DarkPortall 10 месяцев назад +8

      Sadly it wasn't submitted on time

    • @kolskytraveller1369
      @kolskytraveller1369 10 месяцев назад +1

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

  • @nrrgrdn
    @nrrgrdn 10 месяцев назад +1

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

  • @yigitrefikguzelses291
    @yigitrefikguzelses291 10 месяцев назад +1

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

  • @landy4497
    @landy4497 10 месяцев назад +1

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

  • @AJMansfield1
    @AJMansfield1 10 месяцев назад +35

    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.

  • @angeldude101
    @angeldude101 10 месяцев назад +17

    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
      @unaryplus  10 месяцев назад +9

      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
      @itsmeagain1415 10 месяцев назад +3

      @@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?

  • @robkb4559
    @robkb4559 10 месяцев назад +1

    Brilliant concept, beautifully paced and animated. Bravo!

  • @jucom756
    @jucom756 10 месяцев назад +2

    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
      @terdragontra8900 7 месяцев назад

      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.

  • @mysteriousgrimreaper7749
    @mysteriousgrimreaper7749 10 месяцев назад +1

    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 😅

  • @toasteduranium
    @toasteduranium 10 месяцев назад +4

    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.

  • @Cammymoop
    @Cammymoop 10 месяцев назад

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

  • @eladto
    @eladto 10 месяцев назад +1

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

  • @meganton9417
    @meganton9417 10 месяцев назад +1

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

  • @blackestbill7454
    @blackestbill7454 10 месяцев назад +1

    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.

  • @Sup3rdud4
    @Sup3rdud4 10 месяцев назад +13

    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
      @unaryplus  10 месяцев назад

      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
      @amruthchangappa 10 месяцев назад +10

      @@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

  • @asdfghyter
    @asdfghyter 9 месяцев назад

    Super intriguing, excellent video!

  • @secondengineer9814
    @secondengineer9814 10 месяцев назад

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

  • @macronencer
    @macronencer 10 месяцев назад

    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!

  • @AlanKey86
    @AlanKey86 10 месяцев назад +1

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

  • @IronFairy
    @IronFairy 10 месяцев назад

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

  • @viliml2763
    @viliml2763 10 месяцев назад +2

    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

  • @francescosorce5189
    @francescosorce5189 9 месяцев назад +1

    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.

  • @theobscure
    @theobscure 10 месяцев назад +5

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

  • @DanaTheLateBloomingFruitLoop
    @DanaTheLateBloomingFruitLoop 10 месяцев назад +1

    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!

  • @edenasharpie
    @edenasharpie 10 месяцев назад +1

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

  • @asanzramaxi
    @asanzramaxi 10 месяцев назад

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

  • @FlamenLion
    @FlamenLion 10 месяцев назад

    Very interesting. Looking forward to hear more

  • @Tim0feyK
    @Tim0feyK 10 месяцев назад +3

    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
      @aleph0540 10 месяцев назад +3

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

  • @IustinThe_Human
    @IustinThe_Human 10 месяцев назад

    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.

  • @General12th
    @General12th 10 месяцев назад

    This is unbelievably cool! (Also quiet.)

  • @AJMansfield1
    @AJMansfield1 10 месяцев назад +1

    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]

  • @applimu7992
    @applimu7992 10 месяцев назад +19

    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
      @unaryplus  10 месяцев назад +7

      Yes, the diagram is ...

  • @lollertoaster
    @lollertoaster 9 месяцев назад

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

  • @blackestbill7454
    @blackestbill7454 10 месяцев назад +1

    *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

  • @cmilkau
    @cmilkau 10 месяцев назад +3

    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
      @angelmendez-rivera351 10 месяцев назад +2

      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.

  • @notwithouttext
    @notwithouttext 10 месяцев назад

    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!

  • @madddd1
    @madddd1 10 месяцев назад

    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

  • @ScienceD9000
    @ScienceD9000 10 месяцев назад +1

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

  • @FunMaker39
    @FunMaker39 10 месяцев назад +2

    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.

  • @liamdoesmath2106
    @liamdoesmath2106 10 месяцев назад

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

  • @nicolasmirandacornejo2614
    @nicolasmirandacornejo2614 10 месяцев назад +1

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

  • @MisterPenguin42
    @MisterPenguin42 10 месяцев назад

    This is amazing!!

  • @coopergates9680
    @coopergates9680 4 месяца назад

    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

  • @Mr.Nichan
    @Mr.Nichan 10 месяцев назад

    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.

  • @K_Huss
    @K_Huss 10 месяцев назад +59

    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
      @unaryplus  10 месяцев назад +46

      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
      @Chewxy 10 месяцев назад +11

      ​@@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
      @Arsonade 10 месяцев назад +5

      ​@@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

  • @DavidRabahy
    @DavidRabahy 10 месяцев назад

    Best of show, in my humble opinion.

  • @matthewbird1067
    @matthewbird1067 10 месяцев назад

    I remember first being aware of this notion from the video "The Missing Operation" by "Epic Math Time" where it was called the "powerlog". Was good stuff. Since then i have taken to using very similar notation to this video, and really love the structure it creates. In my personal notes I have even taken to using overbar/underbar for exp/ln, just because of how much it turns everything into looking like De Morgan's Laws, but instead of flipping between Duals you're moving up and down the dot operation's index.

    • @matthewbird1067
      @matthewbird1067 10 месяцев назад

      Some other cute points:
      You get a natural extension to the big-sigma/pi notation. a •(n+1) expn(b) is the summation equivalent from 1 to b for the operation •n: ie a•b is a added to itself b times (aka multiplication); a••exp(b) is a multiplied by itself b times (aka exponentiation), a•••exp2(b) is a •2ing itself b times and so on. So the usual notion of hyperoperators.
      Identity for •n is always just expn(0), Inverse for •n is always just expn(-logn(a)) and behaves nicely: invn(a •n b)=invn(a) •n invn(b); inv(n-1)(a •n b)=inv(n-1)(a) •n b=a •n inv(n-1)(b) just like -(a•b)=(-a)•b=a•(-b).
      So expressing classic operations is reasonably succinct: a/b=a•inv(b); a^b=a••exp(b), b root a=a••inv2(exp(b)); logb(a)=ln(a••inv2(b)) and so on

  • @tricky778
    @tricky778 9 месяцев назад

    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.

  • @odiloncintra1377
    @odiloncintra1377 10 месяцев назад

    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.

  • @ary31415
    @ary31415 10 месяцев назад +1

    Super interesting, nice video

  • @NotesNNotes
    @NotesNNotes 10 месяцев назад +1

    Devastating you missed SOME3 but wow this is so amazing ❤

  • @luiz00estilo
    @luiz00estilo 10 месяцев назад +1

    My man just nonchalantly invented a new number system
    damm

  • @dangnabbit1379
    @dangnabbit1379 10 месяцев назад

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

  • @Catman_321
    @Catman_321 10 месяцев назад +10

    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
      @Djake3tooth 10 месяцев назад

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

    • @BS-bd4xo
      @BS-bd4xo 10 месяцев назад +2

      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
      @quantranhong1092 10 месяцев назад +1

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

  • @nif4345
    @nif4345 10 месяцев назад +2

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

  • @FZs1
    @FZs1 9 месяцев назад +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
      @unaryplus  8 месяцев назад +1

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

    • @FZs1
      @FZs1 8 месяцев назад +1

      @@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...

  • @SatanicNerfd
    @SatanicNerfd 10 месяцев назад +2

    Love the video. I think a small neat addition would have been the following. When you introduced the chain of operations, it would have been cool to just check that this property also works with addition and multiplication, to really bring this home. You did mention it sort of later when introducing +sub n but I think at the timestamp it would have added something. 6:09

  • @aryaghahremani9304
    @aryaghahremani9304 10 месяцев назад +1

    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

  • @smiley_1000
    @smiley_1000 10 месяцев назад +2

    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

  • @the4spaceconstantstetraqua886
    @the4spaceconstantstetraqua886 9 месяцев назад

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

  • @Lokdora
    @Lokdora 10 месяцев назад +2

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

  • @sorenriis1162
    @sorenriis1162 10 месяцев назад

    Wonderful! I love maths Conway style.

  • @Izzythemaker127
    @Izzythemaker127 10 месяцев назад +4

    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
      @notwithouttext 9 месяцев назад +1

      yo it's izzy the maker

    • @Izzythemaker127
      @Izzythemaker127 9 месяцев назад +1

      @@notwithouttext yo it's not without text

    • @notwithouttext
      @notwithouttext 9 месяцев назад

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

  • @didles123
    @didles123 10 месяцев назад

    *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.

  • @blackestbill7454
    @blackestbill7454 10 месяцев назад

    Beautiful

  • @kikivoorburg
    @kikivoorburg 10 месяцев назад

    Very neat!

  • @trifonmag4205
    @trifonmag4205 3 месяца назад +1

    THIS SHOULD HAVE MORE VIEWS HOLY-

  • @netcat22
    @netcat22 10 месяцев назад +3

    It would be interesting to see what would happen if you could use fractional values of n by means of fractional tetration (which, by itself is already a difficult subject to tackle) and some fractional application of repeated logarithms.

  • @HyperFocusMarshmallow
    @HyperFocusMarshmallow 10 месяцев назад +1

    Cool video!
    My immediate thought is that logarithm has a nice complex analytic form with a larger domain, though you can't avoid having branch cuts. But using that you can expand this definition to work for more numbers. You'll potentially get quite complicated branch cuts when you start nesting them, but there's nothing weird in principle if you know how to handle it. Working it out and animating the plot could be really cool. I'm not sure how easy that is to do complex color-plots using manim. I've only done similar things using just webgl. But feel free to take the idea if you like. I'd be glad to help if you're unfamiliar with such things.
    Your domain extensions are quit neat though.

  • @alessandroippoliti1523
    @alessandroippoliti1523 10 месяцев назад +2

    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?

  • @jakobr_
    @jakobr_ 10 месяцев назад +2

    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
      @youknowwhenitsreal7 10 месяцев назад

      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

  • @cparks1000000
    @cparks1000000 10 месяцев назад

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

  • @IroAppe
    @IroAppe 10 месяцев назад

    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.

  • @crimsondragon2677
    @crimsondragon2677 10 месяцев назад +1

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