In a sense, concatenation is "the" associative operation. Every equation satisfied by concatenation is satisfied by every associative operation. This is easy to understand keeping in mind that associative operations allow you to drop parentheses, so you can write terms just by writing sequences of arguments to the operation.
Not exactly, concatenation is reversible up to associativity, unlike most of the other operations, they make a free monoid, but the other monoids differ from the free one
I find it to be more a matter of semantics rather than anything else. The "and then" is pretty much precisely how one would understand what the order of operations is, thus it seems wrong to claim, that the description of associativity as the ability to change the order of the operations is faulty. I guess, sure, if this helps someone to understand the concept, which seems to be the case scrolling through the comments, it certainly is useful, but I would not claim it is anything more than either a more in-depth explaination or a slightly alternative approach to look at associativity.
I agree with pretty much everything you say. The "and then" take has order sort of built into it. Here is my response to a similar comment: "You are right to protest that you will never find a definition that is completely unrelated to ordering or grouping. This is because any alternative definition of the associative property can always be used to derive the syntactic rule of associativity, which can then be interpreted as being about order all along. The challenge was issued more out of a desire to nudge people's conception of associativity out of its comfort zone and see what creative ideas people come up with. And I gotta say, you all did not disappoint!" I also agree that a notion "and then" is necessary to understanding order of operations. I further agree that there is nothing faulty about the usual definition of associativity (after all, it is the defining property of associativity). That said, I would also add that the usual definition leaves much to be desired. Questions like "Why is associativity everywhere?", "Why is it important to study?", "Why is it baked into physics?", and "What do we study when we study semigroups?" are better answered by a perspective shift to the "an operator is associative iff it is _and then_" definition of associativity.
Yeah same. But after watching I remembered learning about matrix multiplication and actually working with matrices for marcov chains, and I remembered having a similar epiphany back then. I think it's useful to at least work with one or two groups that are not commutative in order to really "get" associativity. And that usually doesn't happen. Where I live at least, once you learn about the first non commutative group (matrix multiplication), it's too late and many students stopped really thinking about the stuff they are forced to learn in math lessons. And that's really a waste of potential because it is not a hard concept to understand.
Be careful what you dismiss. What are abstract non-formal ideas today are new and rigorous theories tomorrow. Finding a set of semantics which make sense to you personally and are not just dryly adopted because "they work" is part of being good, confident, and fluid with math. Moreover, I believe that good math, what math is, is neither a pure manipulation of symbols nor specific example of something in the real world. It is about abstractions of things in the real world, such as perfect circles, perfect lines, and perfect planes, perfect cubes, functions which behave perfectly like polynomials, etc, which do not exist in the real world but we can reason about them anyways, precisely because of our experiences in the real world. But also our experiences with language and semantics influence this too, I believe. A simple notation is often suggestive of new truths, but some notations can conceal them. Fortunately, if we choose notations and words to understand things or seek to find/see meanings in the phrases and notations established by others, and truly understand those people who made them as we would understand ourselves, connecting them to our abstract and perfect world of mathematics in a way that often seems "silly" or "non-rigorous", we are doing the work of mathematics still. So-called "Silly" and "non-rigorous" mathematics, not formalisms and symbols, is the heart of math. The philosophy of mathematical formalism does not distinguish - or at least does not need to - between the statements "a divides b" and "not a=0 and there exists an integer n such that a times n equals b", and also between the statements "e^(i*pi)=-1" and "(well, it's too much to write out, but like sum from 0 to infinity of blah blah Taylor Series w/ definitions of sums and limits and things, etc.)" but the former is essential to understanding number theory and the latter to complex analysis. Yes, to a formalist everything is "just semantics", but to a human these are not obvious statements connecting our mental images of a perfect, abstract mathematical world to previous definitions. And these things took humans hundreds of years to discover and understand, yet a formalistic textbook calls them "semantics" and says "let x be defined as y" without giving any further thought to it all, ignoring the brilliance and meaning of the statements, and with it everything that makes math interesting. my point is: math can progress without formalism and definitions (it has done so at the least for thousands of years). It can't progress without "silliness" and "semantics" and "alternative explanations", however.
@@bcthoburn Thank you for your reply. Your 1st point would indeed be reasonable if we didn't have a way of making precise statements, which we could use for rigorous definitions. But since we do, any set of semantics serves only as the means for us to interpret and understand these concepts and, unless incorrect, each holds no more and no less value than any other alternative. As for the 2nd, my original comment does mention, that the idea presented in the video can be useful, as every person may find a certain explanations more helpful than others. What I tried to point out is the sensationalized nature of the title, which, I tried to argue, gives a false expectation of the content of the video: that is, there is little difference between the explanation given and the viewpoint it is said to replace.
Thank you for this great video. I think the idea of object-action duality exposes something fundamental that is missing in elementary math education. As some of your early examples show, we commonly switch between object-object interaction and object-action interaction without ever acknowledging the distinction. Sometimes this leads to confusion for students, especially those without a good sense of math intuition. My daughter struggles with math (she almost certainly has dyscalculia, a math version of dyslexia), and, after watching this video, it seems clear that part of her struggles stem from the way we casually interchange object-object and object-action models. A good example is negative numbers: we treat -1 as both an object and an action without ever acknowledging that is what we are doing. This actually became a stumbling block for my daughter in an Algebra 1 problem. She had to simplify something like 5 - (x + 3). Distributing the subtraction, turns the action of subtraction, into negative objects. She had trouble with the this (partly because of the implicit 1 as a coefficient), and I was at a loss to explain it to her, and I now I see that it is exactly because of the action-object switch. I'm still not sure how to explain object-action in a way that would make sense to young students, but it seems important. Another great example is fractions. Is a fraction a number or an operation? We use it as both interchangeably and almost never acknowledge the shift. I wonder if this is part of the reason so many kids have trouble with fractions. I would be very interested to read a math educator's/researchers view on this.
It is not taught because it is detrimental. Thinking of numbers as abstract objects is a very important skill, it leads to confusion at the start because abstract thinking is a difficult skill but one of the most important ones that math teaches. And regardless, the distinction between object and action has no meaning in mathematics, and is more a philosophical matter than anything, so there is no reason to teach it in the first place, and teaching it would probably do more harm than good, as it circumvents the process of abstraction.
Easy. Just define "a - b" as a shorthand for "a + (-b)". The unary minus being the additive inverse. Same can be done for division/fractions = multiplication by the multiplicative inverse.
Excellent presentation and unique interpretation. Just one suggestion : Counter examples like Sedenions and the interpretation of this difference would have been appreciated.
Great suggestion! The original script had a non-associative example (division) that was regrettably cut to shorten the video. Here's a condensed summary of that take: Division is not associative ((8÷4)÷2 is 1 but 8÷(4÷2) is 4) and as such cannot not be an "and then" operation. If we divide by 4 and then divide by 2, this is not the same as dividing by (4÷2). While insightful, this observation alone isn't enough to prove that division is not an "and then" operation because I interpreted the action 𝑥 narrowly to mean divide by 𝑥. Suppose there was some other action interpretation of numbers where division meant "and then". Well, in that case division would satisfy the "and then" property: s(𝑏÷𝑐)=(s𝑏)𝑐. Which was shown to imply division is associative. But wait! Division is not associative. To avoid the contradiction we must reject our supposition that division can ever be an "and then" operation.
@@linguamathematica2582 this argument seems kinda circular. like you dont really give intuition for why divisioon isnt 'and then'. you just defined 'and then' to mean associative.
@@_okedata My mistake, let me attempt a clearer approach: 1. Assume division is "and then" 2. As explored in 14:46 - 15:16, the definition of an "and then" operation '&' is one which obeys s(𝒂&𝒃)=(s𝒂)𝒃 3. Since division is presumed to be "and then", it by definition obeys s(𝒂÷𝒃)=(s𝒂)𝒃 4. As explored in 16:15 - 17:45, s(𝒂÷𝒃)=(s𝒂)𝒃 implies that (𝒂÷𝒃)÷𝒄=𝒂÷(𝒃÷𝒄) 5. But (8÷4)÷2≠8÷(4÷2), contradicting (𝒂÷𝒃)÷𝒄=𝒂÷(𝒃÷𝒄) The contradiction forces us to reject assumption 1, meaning division is not "and then". (Sorry if I'm still bungling this argument up 👉👈)
@@linguamathematica2582 so in order to show that division in not associative you assume that it is an "and then" operation, then say that "and then" operations must be associative and finally provide a counterexample for division being an associative operation? Either I didn't understand something, or checking andthenity of operation is no more intuitive than checking its associativity
@@murmol444 Yes, my argument is very convoluted if the goal was to demonstrate that division is not associative. A more direct way to do that would be to just use step 5. The conclusion I had in mind was that division is not an "and then" operation. In my response to Farhan K, a proof by contradiction was used to show that. As for whether checking "and then"ity is more intuitive than checking associativity, I believe that depends on the application. In the group example in 21:27 - 23:23, it was far easier to verify that the group law is an "and then" than that it is associative. In contrast, in multiplication it is easier to verify that it is associative by constructing a 3D box with dimensions a x b x c and demonstrating that the volume is the same whether doing (a x b) x c or a x (b x c). The occasional ease of checking is really just a bonus, I find that the more profound finding is that the associative property actually says something intelligible about the operations that obey it: that they are all "and then" operations. The property (ab)c=a(bc) fails to give me an intuition about why it is so ubiquitous, important, or interesting and fails to give me an intuition about when I should evoke, expect, or observe the property. These are questions that are more easily answered by the "and then" interpretation. For an example, you can check out my answer to "why is associativity important?" here: math.stackexchange.com/a/4270764/977481
This was an amazing video, however, I think there was one change you should have considered making. As you mentioned in the description, this video was essentially an intuitive proof of Cayley’s theorem for semigroups, a theorem I was not familiar with before watching this video. After googling it, I found the formal statement of the theorem made the video make a lot more sense to me. Maybe it would have been a good idea to throw in this formal statement of the theorem at the end? I hope you don’t see this as me saying “this video needs more rigor! Get rid of the intuition.” Because personally, I don’t think I would have understood how meaningful cayley’s theorem for semigroups was without the intuitive arguments given in the video. I just think maybe 1 minute at the end of the video that explained how all of this translated to rigorous mathematics would have been nice. All in all, this is definitely one of my favorite SoME videos. Will def be subscribing (and checking out that poll you mentioned at the end of the video.)
I hate to be this guy, but even "and then" doesn't work properly. You can use "and then" with division. Divide by 7 and then divide by 3 and then divide by 2, but 7/2/3 is not the same as 7/(2/3).
If by "that guy" you mean inquisitive, constructive, and mentally venturesome, then you totally are. Please don't change :) I fully agree with you. You can use "and then" with division just as you've pointed out. More generally, you can "and then" any operation. However, this has no bearing on the arguments in the video since the claim there was that associative operations are precisely those operations that mean "and then" and not those operations that merely can be "and then"ed. So what exactly does it mean to be "and then"? An "and then" operation takes two actions and combines them into a two step action "do a and then do b". Since "and then" operations are always associative, division cannot be such an action. But how do we know that being "and then" implies associativity? If an operation * meant "and then", this would imply that (a*b)*c would mean "do a and then do b and then do c" and that a*(b*c) would mean "do a and then do b and then do c". Taken together, we could write (a*b)*c="do a and then do b and then do c"=a*(b*c) . Whatever the starting operation, being "and then" implies being associative. That said, your observations are very profound. You can always "and then" any operation. Let's probe deeper and ask the question: what operation "and then"s division actions? That is we want an operation "?" that satisfies a/(b?c)=(a/b)/c. This is multiplication! Multiplication is how we "and then" division! And, true to form, multiplication is associative.
@@linguamathematica2582 I would say "divide by 7 and then divide by 3 and then divide by 2" is giving clear instructions which just happen to be very different from "7/3/2". The action "divide by 7" is perhaps better expressed as a function, f(x) = x / 7. The action "divide by 3" is g(x) = x / 3, and of course "divide by two" is h(x) = x /2. This way of thinking virtually replaces the whole "and then" concept with functions, and obviously works the same after that. I think replacing "and then" with functions is much clearer, because for me, watching the video and reading your explanations in the comments, any time you try to define "and then" it just feels like you're defining associativity, and the distinction feels slippery so that the "aha" moment of the theorem is hard to grasp. Whereas, by using functions, there's clearly (for me) a nice thing happening where three things (for example the three things "one", "plus", and "one") are becoming two things, a function and an argument ( f of 1). And the function is clearly an action. All of this amounts to saying, I think of function composition as the definitive associative operator. But because of this, I have trouble accepting an "object/action" duality for function composition itself. IE, the function is the thing the machine does, not the act of hooking the machine to another machine.
How about “a closed operation on a set S is associative if the functions f_a for each a in S defined as f_a(x) := (x ¥ a) (where ¥ is the operation in question) satisfy f_{a ¥ b)(x) = f_b(f_a(x)) ”
Yours is the core of the video: Cayley's theorem for semigroups. It's interesting that function composition, when read from left to right is more naturally translated as "but first" rather than "and then", but otherwise it's the same idea. It's even more interesting that if you generalize the operation to a partial operation, and the functions to partial functions you get something very close to the Yoneda lemma
First time I feel such confusion after watching a math video. I have been able to follow the most "confusing" 3b1b videos, but although this explains something so fundamental, I feel like I am introduced to something totally new. I probably have to rewatch this a couple if times to fully grasp the concepts, but all in all this is truly well done. If I had to change one thing on this video is the pace. Some times I feel something is being rushed, while others are being explained too much for no reason. It might be me, I don't know.
It's probably not you. I am very new to making videos and do not have this all worked out yet. I would appreciate a breakdown of which chapters felt rushed/slow. No pressure.
@@linguamathematica2582 so, the intro and the first chapter, up until the 13 minutes mark is very well made, the introduction is on point and you clarify what you want to explain setting the basics. Now when you are explaining what associativity is, I feel like the video gets a bit slow, but without reason. Also at 15 minutes almost no-one is concentrated enough to get through new, potentially difficult concepts. You can give us a mental break, maybe change the visuals, or something that shows the video is progressing. At about 18 minutes I took a break and then kept watching and the video felt much more enjoyable (you can even advise the viewer to stop/ rewatch, or ask questions to help him see if they are ready to continue). After that the group theory part was a nice example and the conclusion really does it's job well, it revisits what we saw and helps close the video smoothly. And lastly, when I was referring to "rushed" parts at 18-20 minutes you seem to wrap up a very important idea in just 2 minutes. If you want my tips as a viewer I can list them: 1) long videos=hard videos to follow, keep the viewer engaged with questions and make it feel like we are doing actual progress (maybe split the video into more parts so that I can see where we are and where we are going) 2) the visuals are great, but the glow thingy for me was a bit hard to follow. Do an underscore, a different capitalization, or different colour next time. 3) plan exactly how much time something needs to be explained based on how important and how difficult it is, before writing your script, this will help a lot with ups and downs in speed. Generally speaking too hard or too unimportant parts can be completely skipped or shown as honourable mentions. That's all from me, I am looking forward for your next projects and videos.
One thing I feel is absolutely worth noting, possibly even necessary to note, is not only that the associative property applies to operations that operate both on objects and actions, or that it can be viewed as an 'and then' operation, but that when the operation is associative, _the 'and then' works the same way for both objects and actions._ After all, you can take non-associative operations like (say) subtraction or floating-point addition, and those operate both on objects and actions, just like regular addition. Moreover, unless you _define_ 'and then' to only apply to associative operations (which I can only assume is not what you're trying to do as it kinda defeats the idea that this is a new way to think about associativity), it's perfectly legitimate and intuitive to say 'a minus b, and then minus c' or 'a float-add b, and then float-add c'. Without any additional qualifiers, you could therefore apply subtraction to the above construction: if s is your object, sx indicates s-x, and x&y indicates x+y, the equation s(x&y) = (sx)y holds even though subtraction is clearly not associative. (It's still subtraction on the object; 'x&y == x+y' is just defining how subtraction works on the action, and we're free to define it differently here because it's in a different context.) But if you include the qualifier that 'and then' should work the same way for both objects and actions, then it starts looking like associativity. Subtraction is non-associative not because it can't be expressed as an 'and then' operation, but because if you do, it works differently between the action and the object. Acting on an object looks like s-x, but acting on an action looks like y+x. A similar argument can be made for float addition and other non-associative operations. The way I see it, that's truly where the distinction between associative and non-associative operations lies.
Nice breakdown! Yes, this is exactly what was meant by associative operations are "and then" operations. It does not mean that you _can_ "and then" them (any operation can do this), it means that they _are_ the "and then" operation. In your subtraction example, you observed that to "and then" subtraction actions, you use addition. Using the conclusions in the video, this makes addition associative without the need to check anything further.
Well, you've touched upon three ways of looking at things: two objects combine to form another object, two actions combine to form another action, and an action transforms an object into another object. But there is a fourth one! An OBJECT transforms an ACTION into another ACTION. For example, a pair of apples transforms the addition of three apples into the addition of five apples.
I definitely liked the production values and some of the points of this video. That said, I wanted to share my struggles with this video. On my first viewing, I made it about a third of the way in and stopped watching because I couldn't tell where it was headed (and when I went back to finish, it seems like the main "and then" idea didn't need the flexibility of all three object-object/object-action/action-action interpretations, and so the video felt a little out of order). After finishing, the main thing I struggled with was the lack of clear signposts/separation of the math from the philosophy. It made it hard to decide when I should be thinking of what was said as a simplified description of a theorem, and when it as just a helpful interpretation to keep in mind. Finally, the line "Its meaning can remain hidden, even after centuries of instruction" was extremely offputting. Even if these weren't your intention, it suggests 1. You might be unaware of the formal notion of group actions, or how group theory developed historically (this is only a problem since you're making a claim about centuries of history). 2. You think that your philosophical ideas here are brand new (and with no provided justification in the video or description that you've "done your homework") . 3. You think that your interpretations are not merely good and deserving of spreading, but the sole definitive "meaning" of associativity. But to be clear, I really agree with you about, say, how groups should ideally be introduced to students, and the benefits of thinking about an operation from those three perspectives, and appreciate you making a video emphasizing those things.
Thank you for this. Giving clear signposts is something vital to do in the classroom and in presentations, but I'm still trying to figure out its role in videos (where I believe, perhaps in error, that the right amount of uncertainty can create intrigue that pulls the viewer in). That said, your response tells me that my clumsy effort to draw you in only pushed you away. I'll try to add more clarity in the future and see how it plays out. As for the line "Its meaning can remain hidden, even after centuries of instruction", I am sorry that it was offputting. I did not mean that it is hidden from _mathematicians_ but from those who've learned it. It followed my discussion of pedagogy and I meant _instriction_ to be an operating word there. Something like "after centuries of instruction, its meaning still isn't taught." That said, I can really be in my own head when script writing and your interpretation now seems more natural to me than my own. After all, remaining hidden seems to imply that no-one at all has discovered it. I'm glad that despite my amateur mistakes you still appreciate the video :)
"Assoociativity is not merely a property, but also an idea". Thanks for putting the grandiloquous word salad at the beginning of the video. Saved me about half an hour.
I felt like watching a 3B1B video, others have given some advices and they definitely seem good, but I just wanna say I am really impressed by the feel of video and how you go about it just like grant! I hope you are here to stay :-)
I think that even though overall the video is great it doesn't hammer down a very important point - associativity means that objects of operations and operations themselves are identical. Or in a mathematical language arguments are isomorphic to functions. Take for example expressions 5+3+2 and 5-3-2, they can be written as 5(+3)(+2), meaning acting of an operation +3 to 5 and then acting of +2 to the result, and 5(-3)(-2) (same meaning) or as (5+3)+2 and (5-3)-2 respectively, meaning combining a pair of objects via + or - , the former is associative, the latter isn't, because (-3)(-2)≠(3-2)=(+1). There is no correspondence between objects themselves (think about apples in the example) and the operation (removing the apples) in subtraction, which exists with the sum. Whenever you can treat objects and operations as one and the same, you will have associativity. It all boils down to that moment of s(a&b)=(sa)b and just omitting & for notation sake is harmful, IMO. It should've been made more clear, that only when you can use "a" both as an action, like "b", and an object, just like "s", you will have associativity, which is expressed in the omitting of & in the expression. Otherwise you may have to combine a&b in a strange manner and hence you don't get associativity.
I prefer to think of addition and multiplication as the fundamental, associative "binary or more" operations, and subtraction and division as unary operations, that way I don't have to worry about the processes not being associative 5-3-2 = 5 + (-3) + (-2), is this not easier to work with, and less ambiguous than "5-3-2"? I get confused with "5-3-2" because of the non-associativity, 5-(3-2) != (5-3)-2 and all that, and I prefer forcing subtraction to be "associative" via reducing it to a unary operation: 5 + (-3) + (-2) 5/3/2 = 5 * (1/3) * (1/2), this one's a bit tougher to demonstrate because while -3 ("negative three") stands on its own fine, /3 ("reciprocal of three") does not and you gotta use 1/3 to get the point across
Thank you for your thoughtful response! The idea that associativity means that objects of operations and operators themselves are identical is a great start, but I’d like to see if we could improve it. It turns out that just being an operation (even a non associative one) means that objects of an operation can be interpreted as operators. For instance, in subtraction we can write 3−2 or we can write 3(−2), in division 3÷2=3(÷2) and in general, 𝑎∗𝑏=𝑎(∗𝑏). The mapping 𝑏↦(∗𝑏) maps objects to actions. But wait, you observed that there seems to be no objects that correspond to (−3) in the apple scenario. That doesn’t seem to fit with what I just said. I would argue that the lack of an object representation of (−3) is a consequence of anti-apples being hard to find irl and not of some mathematical limitation. So, let’s shift to a different representation of addition, adding 1D vectors. In this domain, the object corresponding to (−3) is an arrow with the same length as 3 but pointing in the opposite direction. Well, if object-action mapping is a property of any operations, what is associativity then? Well, your proposal that associativity means there is an isomorphism between arguments and functions was on the nose! To fully unpack your insight, we can specify the binary relations that the isomorphism should preserve. The full isomorphism is written (∗(𝑎∗𝑏))=(∗𝑎)&(∗𝑏), and the conclusion is that an associative is isomorphic to the “and then” operation. Finally, by noting that the object-action mapping 𝑏↦(∗𝑏) need not be injective, we relax our statement to “associativity means that there is a homeomorphism from ⟨𝑋,∗⟩ to ⟨(∗𝑋), &⟩”. Very nice!
@@linguamathematica2582 Thanks for the response. I think I should apologize for omitting a very important point that you indeed can make any operation associative via homomorphism b→*b, as you have pointed out, but not only that you need to additionally create an operation with a property a°(*b)=a*b as well, by which I mean ° is a binary operation acting on a (which here would become *c for some c) and *b. And by construction ° would be associative. So yes, even for non-associative operations it is possible to make an interpretation that would be associative. However, it would require this ° operation, which in a sense would do for objects the same thing that a composition would do for functions or "and-thenning" them. Just as with -3 we replace everything with addition and for ÷2 we replace everything with multiplication both of which take the role of of this ° operation. ...I think that, even though it probably was unintentional, without a small clarification of this the video is harder to appreciate for some, because many people who are not familiar with this idea will come up with many operations that feel like "and then", but aren't associative, e.g. -,÷,↑,√ and so on. And my comment was for the most part directed at them that it's not that all of this aren't "and then" operations, but that they don't have an object in the specified domain which would correspond to the operation. And you are 100% right that it's not a problem of math, but our inability to have real tangible interpretations in this situations, so we choose not to do them. Anyway, as I said, great video, it still warms me to see that something like this is on the YT and people can see it, I hope my comments, previous and current, haven't created an expression that I am not simply amazed by your work. I very much am.
@@linguamathematica2582 I was confused about why subtraction is not associative. Can't I subtract, and then substract? But here you say (-(a-b)) is not equal to (-a)&(-b). I am still confused, and hopeful you can use graphics to explain this in a future video. At any rate, your video is great.
11:11 first one: connect factory line g and f together second one: have a factory line programmed as g and edit it to incorporate f third one: program g into a factory line then program f incorporated into it
Great work on this video! I find the slight pauses in audio between topics to really help segment the lesson. A little pushback on the classification of groups as “reversible procedures”, if you don’t mind… The “and then” operation of taking a step on an infinite sidewalk with the elements of -1, 0, or 1 steps is always reversible, but does not define a group. Cannot forget the importance of closed-ness :)
Thank you! After some thought, I completely agree with you. I hope it didn't take away from the rest of the video. I clumsily tried to sidestep closed-ness by using the word "procedure" to include actions generated by "and then"-ing. At the very least, I should have made that explicit. In the sidewalk example, I would classify 3, for example, as the procedure 1 & 1 & 1. And since it is a procedure it needs to be included in the group of "reversible procedures".
@@linguamathematica2582 Unfortunately, including all “procedures” doesn’t seem to work either. For example, in the video you presented the actions “open the fridge”, “find the milk”, and “pick up the milk”. As “find the milk” is difficult to reverse, we can make a new collection of actions as follows: a: open the fridge b: take out the milk c: close the fridge d: put the milk back e: do nothing Also include all procedures, like “get the milk” = a&b. This collection of actions also satisfies your definition of a group; it uses an “and-then” operation, has an identity, and every action is reversible. However, as a collection of procedures, it’s nonsense. Consider, for example, c&d&d&c&a. “Close the fridge and then put the milk back and then put the milk back and then close the fridge and then open the fridge.” That’s the ravings of a madman. So, what you really want is not only to include all procedures, but also that any action can be taken at any time. One of the reasons that these fridge-milk operations do not define a group is that you can’t do any of the milk operations when the fridge is closed. This, by the way, is why groups are almost always introduced in terms of “symmetry”. In order to have any action be available at any time, something about the world must remain as before after it is changed, ready to be acted on again. I was fortunate enough to learn group theory in a way that emphasized this, so it was made clear that all symmetries are built on groups and all groups can be thought of as symmetries. If you explain in this way, all four group axioms (closure, identity, inverses, and associativity) seem not only justified but also necessary.
@TheBasikShow Fantastic! Thank you for this thorough and illuminating breakdown. I completely agree with you and perhaps "reversible procedures over a state space" would have been more accurate, although far less memorable and punchy. This was probably the biggest mistake in the video, and I apologize if it took away from the main thrust of the video. At the very least I should clarify how I sleep at night. In my own (often heretical) thinking, I extend the domain of actions to include the "bad" cases. The core question we need to answer is: "what do you do when you perform an action that is not afforded by the current state?" I've found 3 answers that are universally applicable: 1. You don't: if this ever happens, you've done something illegal (in programming for example you might throw an exception, in mathematics you might say that it's the ravings of a madman) 2. Return an error state: pick some value to return that is disjoint from the outcomes of afforded actions. Often, actions on error states are always presumed to yield another error state (the set-theoretic view of functions returns the empty set when outside the domain. See also how infinity acts as an error state in arithmetic on the extended reals: en.wikipedia.org/wiki/Extended_real_number_line#Arithmetic_operations) 3. Do nothing: whatever the input state, return that (In the fridge example, opening an opened fridge would leave the fridge opened, so the output state is unchanged). If you chose (1), as you probably should to stick to mathematical convention, then your arguments follow. In my thought processes, I often choose (3) by reasoning that if you can't perform the action then you don't change the state. Now I just extend the actions to fit as many states as needed. I've found there is an elegance to this because it mirrors reality (after all, reality doesn't dissolve when you attempt an impossible action), although this is definitely not anywhere near an accepted method. In the fridge example, extending using (3) does not yield a group because closing a closed fridge yields the same state as closing an open one and the action is not invertible.
I am so sad I found this video two years too late. That 24:37 was a holy-s*** moment and I will never un-see what it means now. You just got one more sub. I hope I can see more of your work.
Great video! One think that could have made it better, is if you had explored how non-associative operations fail to be "and then" operations. You understand something a lot better if you know when it does not apply.
The "object-action duality" in this video seems to be equivalent to "every semigroup (set with associative operation) is isomorphic (structurally equivalent) to the set of endomorphisms (functions) on it given by f(x) = s + x, for each s in the semigroup, with the operator given by (function) composition." We can also give the monoid and group laws like so: "in the action-dual of a monoid, the action of the identity element is the identity function, f(x) = x," and "in the action-dual of a group, the action of an inverse element is the inverse function of the element's action." That last property also means that all endomorphisms of that form are injective, and since we can combine an inverse element with any other element, surjective. Therefore, the action of every element of a group is a bijection.
This video is brilliant! Object-action duality explains something I have been wrestling with for quite some time. The difference between an operation and the result of an operation. The Schrödinger equation of Quantum Mechanics consists of the Hamiltonian operator applied to the Psi function having the same result as multiplying the psi function with the energy eigenvalue. The Hamiltonian operator is then the action, and the eigenvalue is the object that has been measured. The equation is then a question about what the psi functions are given the boundary conditions, and what the possible energy eigenvalues are. Or, in general, any quantum measurement is represented by an operator operating on some psi function, whereby the eigenvalues are the objects, the possible results of the measurements. In other words, in quantum mechanics the object-action duality is equivalent to the eigenvalue and operator 'duality'. At present, I am using this approach of object-action duality,which I called object - operator connection, to demystify the appearance of the complex number *_i_* in quantum mechanics. Your explanation of group theory as the study of reversible procedures is quite interesting! Given the fact, that all fundamental laws of physics are reversible, it suggests that the study of laws of physics is not only helped by group theory, but it might even be the case that the study of physics _is_ the study of group theory, applied on energy!
Um, yeah, modern physics *is* group theory. That's why the standard model is U(1)×SU(2)×SU(3) and why things like momentum, energy, and electric charge are conserved (Nöther's Theorem). A bunch of String theory stuff is also just group theory (like E8×E8).
How does this and-then interpretation and the object-action duality compare to monads? It feels like you're talking exactly about monads, but I'm not sure, since monads necessarily form monoids (i.e. associative with a neutral element), but not vice versa... I think. See the wikipedia page on Monad (functional programming) - section "Definition". It's simpler than the category theory page. Note that the bind operation is often called "then" or "and then". In functional programming, the bind operation accepts a monadic context with objects of a specific type (e.g. a possibly empty list of partial solutions to some problem, like placements of queens on a chessboard that don't attack each other), and an operation that transforms such objects and places them in monadic context (e.g. gives a possibly empty list of solutions that satisfy an additional constraint: a queen is added to the board without being attacked by another queen).
Excellent example of "and then" in context. As you pointed out, monads form a monoid over the bind operation and thus bind must mean "and then" in some way. In the standard definition of monads, the bind operation is indeed an "and then" of the first and second inputs. That said, I believe that monads could equally be defined so that they are "but first" composition operations (yielding a stack instead of a queue of functions) if you define the bind as ≪=:[𝑇→𝑀 𝑈]×𝑀 𝑇 →𝑀 𝑈 instead of the usual ≫=:𝑀 𝑇×[𝑇→𝑀 𝑈]→𝑀 𝑈. Also, I love your chess example with the queens for explaining monads. As for object-action duality, its inherent in lambda calculus in general. Note that the 𝑀 𝑇 argument for the bind can be an object (values of a type) as well as an action (a lambda expression that returns a value of type 𝑀 𝑇). The bind can also be seen as a gluing together of actions into a pipeline. It's even in the name "bind", which is a synonym for gluing.
This is exactly the association I had too, especially since that way of phrasing it as "and then [do]" was what actually made monads click for me back when I learned about them. Especially once the talk about more physical actions started I couldn't help but think of the IO monad in Haskell.
This is an excellent video. From a pedagogical viewpoint this changes the teaching of associativity from procedural to contextual. This change is the pinnacle of maths teaching as it allows for the building of larger schema and interconnectivity of subjects. Again, excellent video.
I think the explanation is made less clear by introducing "and then" as something destinct from function composition. By defining "sa" to mean "apply function a to s" you get new notation for "a(s)". By defining "s(b&c) = (sb)c" it's equivalent in standard notation to "(b&c)(s) = c(b(s))" but c(b(s)) = (c°b)(s): (b&c) is (c°b). The "&" and "sa" notation just lets you write the functions in the same order that they would appear as opperations since c(b(s)) = (s*b)*c. Introducing it as notation would save you the trouble of reproving that function composition is associative.
Really amazing explanation. I never thought of it this way, especially how you applied it to Group theory. I'm not a math student, but this makes me want to be.
Thank you, this is a reasonable concern. Here are my responses to similar comments: "This is a distinction that probably should have been made more explicit in the video. Associativity does not mean that you _can_ "and then" the actions (as you've pointed out, this is always the case). Instead, associative operations _are_ "and then" operations. This is precisely what I hoped to convey with "associative = and then" Division is not associative since (8÷4)÷2 is 1 but 8÷(4÷2) is 4 and as such should not be an "and then" operation. Indeed, observe that if we divide by 4 and then divide by 2, this is not the same as dividing by (4÷2). While insightful, this observation alone isn't enough to prove that division is not an "and then" operation because I interpreted the action [𝑥] narrowly to mean divide by 𝑥. Here is a more general proof: 1. Assume division is "and then" 2. As explored in 14:46 - 15:16, the definition of an "and then" operation '&' is one which obeys s(𝒂&𝒃)=(s𝒂)𝒃 3. Since division is presumed to be "and then", it by definition obeys s(𝒂÷𝒃)=(s𝒂)𝒃 4. As explored in 16:15 - 17:45, s(𝒂÷𝒃)=(s𝒂)𝒃 implies that (𝒂÷𝒃)÷𝒄=𝒂÷(𝒃÷𝒄) 5. But (8÷4)÷2≠8÷(4÷2), contradicting (𝒂÷𝒃)÷𝒄=𝒂÷(𝒃÷𝒄) The contradiction forces us to reject assumption 1, meaning division is not "and then". "
Great video! An example of a non-associative operation would've been nice for contrast. EDIT: Like what exactly goes wrong in terms of associativity when you subtract 1 from 2 and_then subtract that from 3 compared to subtracting 2 from 3 and_then subtracting 1 from that. 3-(2-1) != (3-2)-1
3 + (-2 + -1) = (3 + -2) + -1 I've thought about this in the past, by rephrasing the subtraction as addition of negative numbers, the operations now are associative. What changed? In my opinion but ultimately I think it comes from the "object" idea of the numbers. You can also think of numbers themselves as an operation or function. In the object conception the numbers are being operated on, and so the "who does what to who" matters quite a bit. In the function conception of numbers, there are no objects being operated on, and so the "difference" isn't derived as a relation between two things. No idea if that made any sense to anyone, and to make things worse I'll go deeper. In one perspective there are no nouns, only verbs. In another perspective nouns and verbs exist, and nouns verb other nouns. The object mindset highlights a the continuity of individual patterns through transformations, but obscures the transformations. The function mindset highlights patterns of transformation, but obscures the continuity of patterns through transformation. Or to paraphrase (butcher) the words of the philosopher Dogen. When firewood is burned, it becomes ash, there is no wood, only ash. In the object conception, the firewood is still there, and this could be a problem for us if we viewed not being burned as a definition of firewood. Because the behavior and information are coupled it's hard to say what we can or can't do to a piece of firewood. In the same way an improperly applied function mindset would cause us to lose our children when they grow one inch taller. We've lost the continuity of a pattern through a transformation.
Subtraction is not considered an operation. And in case it is referred as one in a general sense, it is not a binary operation in a set of natural numbers. 2 is a natural number, 3 is too, but 2-3 isn't. So it isn't associative, as applying "and then" doesn't make sense when the operation is not binary itself.
@@roseCatcher_ But subtraction absolutely is a binary operation in the set of integers, or rational numbers, or real numbers, or complex numbers, or .... I find your comment too dismissive, as there are there a plethora of contexts in which subtraction is a bona fide non-associative binary operation, and in which op's question is _completely reasonable._
2:17 I would try to define like this: Associativity means that an operation which takes in two inputs and returns one output of the same kind can be uniquely extended to an operation which takes in three inputs and returns one output, and further, can be uniquely extended to an operation taking in an arbitrary finite number of inputs.
I don’t think it’s possible to explain it without the use of the word "order", its synonyms or recreating its definition (e.g. using the "and then" from the latter part of the video). Because associativity means "order of constituent [actions|operations|etc] doesn’t matter". It’s *about* the order.
All associative operations are instances of function composition. If * is associative on X then the corresponding function is f:X -> X^X defined by f(x)(y)=x*y where X^X is the set of functions from X to itself.
It took a little long to get to the point IMO, but I still learned something new! I didn't really think of associativity as a big ball of actions combined together, rather as the state combined together, but that actually makes a lot of sense once you pointed out that function composition is associative too (and the fridge example)
ehen you multiply a*b*c, you create a new number, that already fully described by this expession. when you get the number in another expression you can multiply given numbers, and in the end get the expression of number that please you as much as you need. for example 4*2*2 can be a square, a multiple of 2 or 8, or even power of 2 in alternative expressions, but still will have all of this properties in all of expressions
I'm a mathematician who studies algebra, and non-associative (among other more exotic forms) of algebra have been recently important in my work. I found your video very insightful. It even has helped me better understand something that's come up recently in my research. You really hit the nail on the head with your understanding of associativity and "object-action duality". Your main point can be stated concisely as follows: Let A be something where it makes sense to talk about algebra (i.e. a set, a vector space, etc). Donote the set of functions A->A by Fun(A). This naturally comes equipped with an associative operation given by function composition. Now consider an operation on A as a function m : A×A -> A. For clarity, I'll write m(x,y) as just xy. This naturally induces a map (the so called "action map") a : A -> Fun(A) where a(x) is the function f(y) = xy. That is, it takes the obect x to the action given by multiplying x (on the left). Here's the key point: The operation m is associative if and only if the action map takes the operation m to the composition operation. That is, if a(xy) = a(x)°a(y). An operation satisfying this property is what you call an "and then" operation. Indeed, one can prove that this is equivalent to associativity, and in this way all associativity derives from the fact that function composition is associative. Proof: Note that a(xy) is the function where, for any z in A, a(xy)(z) = (xy)z. Likewise (a(x)°a(y))(z) = a(x)(a(y)(z)) = a(x)(yz) x(yz). Thus a(xy) = a(x)°a(y) if and only if, for every z, (xy)z = x(yz).
I answered: "If I apply a procedure to thing 1 and thing 2, computer their result and apply a procedure to that result and thing 3. That is the same as applying a procedure to thing 2 and thing 3, computing their result and then applying a procedure to thing 1 and that result." where a procedure takes 2 things and returns 1 thing.
@@ApesAmongUs well, order is very important to associativity. How else do you say "the order of inputs matters, but the order of execution does not matter" which isn't just obscuring "order" under descriptions of what "order" means in this context? Is associativity not stating that A☆B☆C might not = B☆A☆C, but that A☆(B☆C) does = (A☆B)☆C?
@@ferociousfeind8538 I agree, but that was the challenge he made in the video - define it without using those words or equivalents. I think his point might have been that it isn't possible.
I think one problem with this outlook (and all outlooks have problems, so this isn't a knock) might lie in describing what it means for a group to not have the associative property. In the case of the cross product in R^3, I'm not sure I can even define what A x (BxC) even means in this context of actions. Maybe I could say that they don't have the object-action duality, but I'm not sure what that actually tells me about the objects I'm working with.
Thanks! I'm the same way with the 1.5x or higher. Once you get used to it, you can understand everything being said. I downloaded a chrome extension that lets me speed it past 2x for the slow talkers.
My best shot at defining associativity without the words order, brackets etc. An operation is associative if and only if, given mathematical objects within the domain of the operation, the result of applying the operation between all of the given mathematical objects is defined for only one value or no values.
What a wonderful idea! Associate means to "join together". In this sense it is a synonym of gluing. So I guess that would mean that a valid answer to the challenge might be "associative operations are those that associate objects together"
As far as I can see there's a better word for what we call associativity: I'd call it Sequenciality That's because if A,B,C,D and E are associative transformations, then ABCDE=(AB)CDE=(ABC)DE=... (and so on) So the sequence of transformations uniquely determines the overall resulting transformation. However if the transformations aren't associative, then ABCDE not equals (AB)CDE... (and so on) then, if associative, the sequence of transformations uniquely determines the overall one if not associative, you can partition subsequences of transformations and get to other results Then associativity is the property of a transformation to be decomposed into a sequence of other transformations while being sensitive only to the sequence of these transformations. That's why I'd call it Sequenciality.
I keep imagining the transformation of the left side of the equation into the product and then back into the right side of the equation and here's my intuition and thought process on the 2:17 challenge: associative operations are undoable the input of the associative operation can be the output of an other application of that operation - a combination, if you will when multiple instances of the operation are chained together, they can be recombined thanks to that "undoability" and "combinability." *It feels like an associative operation IS a combination of actions or IS combination itself. * Anyway, idk if that definition properly excludes non associative operations and I think I have come too close to using the word "group" and I think I've just spent a lot of time rewording that old grade school explanation, oh well.
Your train of thought was so fun. I like how in tune you are with your intuition. Like, the way you extract structure from your intuitions is pretty unique and inspiring
My answer to the challenge at the beginning: "When an operation is associative, it means that when applying that operation to 3 or more numbers, you will always get the same result by applying it to any two of those numbers, and then applying the operation to that result and any other of the remaining numbers, until you have applied to operation to all of the numbers you wish." I think I got it but I may have missed something subtle Edit: I was missing something, although one may argue whether it was subtle or not. I think I just described commutativity...
This is an excellent description of the "jumbling property" evoked in 1:20. It is a property you get when both associativity and commutativity are satisfied.
"Rasie to the exponent" is an "and then" function, it even obeys your defining rule s(b&c) = (sb)&c but it is not associative (a^b)^c =/= a^(b^c) Therefore "and then" does not imply associativey
I feel this video is a _bit_ tricky in its presentation. I don't think the whole "and then" thing is clearly explained in full detail. Start with a set S and a binary operation ★. You can then get a natural function φ : S → Functions(S,S) given by φ(s) = s★−. (This is the whole shifting from "object" view to "action" view.) Then ★ is associative if and only if φ is a "homomorphism". That is, ★ is associative if and only if φ(s★t) = φ(s)∘φ(t). Written in this way, it's a bit obvious because (s★t)★− = φ(s★t) and φ(s)∘φ(t) = s★(t★−). The issue here is that you can think of a^b = d as object-action-object, but not as action-action-action, since raising a number to the a power and then to the b power is not the same thing as raising it to the d power. Nonetheless, the action-action-action viewpoint is very important for modern algebra if you ever need to use groups or modules to study something.
The states and actions are a very interesting visualisation of associativity. The way I personally think about it is that associative operations can be merged into variants of them with an arbitrary amount of inputs and also broken apart again while remaining equivalent. For example it's possible to concatenate 5 symbols, compose 5 functions, calculate the greatest common divider between 5 whole numbers or add any 5 real numbers, but not to divide 5 real numbers or take the arithmetic mean of 5 numbers while still being able to split it into multiple parts at any point.
Nice, under this interpretation of associativity being "and then" operations one could think of + being nothing more than the successor operation, and something like +3 could now be interpreted as meaning +^3 or +++. While it doesn't naturally extend to concatenation (you'd have each ASCII char represent a special catenation function), it extends to function application quite nicely and it also makes a case for postfix notation: xf and by extension: xf^n. What is more, it also extends quite very naturally to computational processes, best exemplified by concatenative languages like Forth and pipelining on a unix shell. The idea is so fertile an enlightening all this comes from just looking at the & interpretation.
If yours was the only comment I got on this video, I would be satisfied. I picked this topic because of how fruitful it was in thinking about addition and the successor function, concatenation and writing, and computational processes. To see you share this sentiment was, frankly, touching. Its fun to think of addition as an abelian group and consider how to incorporate commutativity into "and then" operations. By looking at a few ways of doing so, I ended up generating some nice models for different types of addition.
Perhaps I misunderstood the aim of the video, but it's 𝘯𝘰𝘯-associativity that I interpret as being "and then" operations. For example (12/6)/2 doesn't equal 12/(6/2). The first is divide 12 by 6, and then by 2. The second is divide 6 by 2 and then divide 12 by that. On the other hand 12x6x2 doesn't require any "and then's". Declaring that multiplication is associative, or that (12x6)x2 = 12x(6x2) merely amounts to ruling out the need for any "and then", for any particular sequence to the operations.
I like this is essentially saying group and group action is two identical/important aspect of group theory I’m however really surprised that you said “the more I study groups, the more empty, the more joyless, the more opaque I feel” because that doesn’t sounds like someone making this kind of video will feel. I do agree that math textbooks are usually extremely polished that, most of the time it doesn’t make any sense why are we studying math. Unfortunately that’s just what we mathematician always do and we find beauty in doing so. If you have gone this far realizing the reason behind group theory, which is usually referred as theory of symmetry not reversible procedure, I would expect you find learning more of group theory gives more and more endless insight behind it. It’s definitely true for everything I learned with math.
I think there's something really important missing from the explanation. When you have an operation like "a✳️(b✳️c)" then a and b are _both_ starting-states Around 16:00, the video talks about the difference between (1) having a state s, an action b, and using "apply action" to connect them, vs (2) having an action a, another action b, and using "and then" to connect them The top row at 16:00 covers two scenarios: State1 ( Action1 Action2 ) -vs- (State1 ) Action1 Action2 But that's missing the scenario that we actually need: State1 ( "operator" (State2 Action 1 ) ) That last one might not be very clear, but that's because it doesn't fit very well into this narrative. But it's the scenario we actually deal with when we're looking at associative operations Consider "3×(4×5)". The starting state is the number three, followed by an action of "multiply by 20". That action is made up by using a second starting number (4), and the action "multiply by 5". We have a number that you make using "(start(4) -> action(×5)". And somehow we need to show that the action of multiplying by that number, is the same action as "action(×4) action(×5)". I don't think the video does that For comparison, consider division: " 3 ÷ (4 ÷ 5) ". If we try and do the same thing for division, then we end up claiming that "action(÷ 0.8)" is the same thing as "action(÷4) action(÷5)". What's different between multiplication and division? I don't see why the argument in the video should be valid for one and not valid for the other.
This is a great video. I really love this type of deep insights into the inner workings of mathematics. However, there are some things that might need more explanations or can be improved in the video: 1. At 19:17 it is not entirely clear why “b” and “c” become actions, from what we are presented up to this point in the video it makes sense only to make “(b*c)” an action, not “b” and “c” individually. 2. At 17:27 if “s” would be 1 and the actions “(ab)c” and “a(bc)” turned out to be “add 1” and “multiply by 2” respectively, then we could not say that they are the same actions. It can be remedied thought by saying that “s((ab)c)=s(a(bc))” holds for EVERY “s”, which would make the two actions act the same on EVERY object, which is the defining property of being the same action (at least for functions). 3. You gave “gluing” an interpretation of, loosely speaking, “combining action-snapshots in time”, but at 27:18 you use it to “glue together the apples”, which is a completely different interpretation that is only related, because “gluing” has both physical and temporal meaning (in English). Overall, though, I really enjoyed this video because it is not just a philosophical discussion of associativity, but a useful idea to have; like when you use it to prove associativity of group actions without checking each combination of its members. This is one of the best maths videos I have seen in quite some time.
Thank you so much for the kind words and the specific and constructive criticism. Especially as a new creator I need as much as I can get. I'll try to fill in the gaps in the video by addressing your concerns here: 1. The idea was to use the construction at the top line of 19:17 while specifying different values for s and x. If I notate actions as primed variables, then a*b=ab' is true by specifying s -> a and x -> b. Then I just substitute. 2. I absolutely agree with the importance of making sure it's true for every state and, although I was not nearly as thorough or clear as you were in explaining it, I did attempt to mention it at 17:27 "and finally, since the two actions change every possible starting state into the same end state, the two actions are equivalent" 3. Yes, I do rely on the ambivalence of "gluing" in my video. The aspect of "gluing" that I'm referencing is about linking objects together in a sequence, whether in time or space. Thanks also for pointing out what you enjoyed about the video (philosophy and useful mathematical ideas), it's a huge help for planning the future direction of the channel.
Great video! I do think it's missing a simple example of a non-associative algebra, e.g., 3D real vectors equipped with the cross product operation. This would highlight the importance of reversibility and strengthen your insight about groups and symmetry (in a very evocative and geometrically intuitive way).
2:18 challenge: Addition is associative because when I am adding, let's say, what I have sold every day, the total physically exists, it's can't be two different things, thus it does not matter how I add it as long as every day gets the same weight. Maybe it sounds odd but it's as far as I have thought about it
there’s this pipe operator in the R programming language that lets you write function application linearly, e.g. x |> mean() |> round(2). In the community, people read the pipe aloud as “then”. “Take vector x then compute the mean then round to two digits.” I think another language (JS?) has an explicit .then() method for chaining operations together too. “then” is such a good word.
I think the difference between object and action can be made clear with the concept of operators. I had much trouble in school to remember to take the minus sign into account in my arithmetic calculations. I so often forgot the simplest rules about it and made error over error. Well somehow I made it into university still. In quantum mechanics they introduced the concept of an operator, and suddenly everything made sense, in the way that the minus sign always BELONGS to the number, the operator and the object operated upon go together. So in your example and with superfluous but illuminating bracketing: (a) + (b) = (c) is the object-object approach, (a) (+b) = (c) is the object-action approach, and (+a)(+b) = (+c) is the action-action approach. In the object-object approach addition is seen as something connecting two objects but not being a property of the objects itself. In the action-action approach addition is a property of each object already. Maybe one could even write (+a) + (+b) = (+c). Finally it is interesting to note different ways of notation, such as reverse polish notation. One can write addition also as a b + then or also +(a,b) like a real function of two arguments, and I wonder if showing all these alternatives (infix, prefix, postfix notation) early on in school would help students to get a better understanding of the concept? Because all of these notations occur in mathematics over and over again and it is often just for historical reasons why one happens to be the standard notation and the other ones aren't.
A lot of intriguing points here. Especially what I took to be your main point: introducing many notational conventions. I wonder how that would play out. Operators gave me the same clarity as they've done for you. I am even currently working on a paper that essentially boils down to: if you introduce operators your equations become way cleaner. You're the second commenter to bring up rpn. Makes me want to play with the notation now.
@@linguamathematica2582 Now that I thought again about reverse polish notation I remembered that it might match the topic of associativity in a deeper way still, since it is a way to write calculations without any bracketing and in some way it even introduces a natural order in which to perform operations thereby in some way making the associativity rule not necessary anymore. Of course your point was that it is not really only a rule but some deeper principle or property, so one still wants to keep that, naturally. It would not be a deeper principle if it vanishes by a change of notation of course. Btw. quantum mechanics also made me understand commutativity. And dual vectors too. Maybe they should teach it in schools already. Another thing that came to my mind again is that John Baez was looking for a good layman example for associativity. To quote him: “[...] while it’s very easy to imagine noncommutative situations-putting on shoes then socks is different from socks then shoes-it’s very difficult to think of a nonassociative situation.” If, instead of putting on socks then shoes, you first put your socks into your shoes, technically you should still then be able to put your feet into both and get the same result. “The parentheses feel artificial.” Associativity is involved in the number systems of the reals, the complex numbers, the quaternions, but not the octonions. It is like creating numbers according to the same scheme ( en.wikipedia.org/wiki/Cayley%E2%80%93Dickson_construction ) but always losing some algebraic property on each step, first order, second commutativity, third associativity. This appears kind of as a hierarchy of properties.
My favorite example of your point is exponentiation, something that is not associative. 2^(3^4) does not equal (2^3)^4, but you kindof can create an exponential group if you make each element, n, of a group the act of raising something to the power n. Now 2&3 means raising something to the second power then raising it to the third power. 2&(3&4) means squaring something then cubing it then raising it to the power of four. If that is the case the exponent group is associative. Squaring and then (cubing and then raising to the 4th) is the same thing as squaring then raising to the 12th or raising to the 24th. Likewise (squaring and then cubing) and then raising to the fourth is the same thing as raising to the 6th and then to the 4th or raising to the 24th.
When I heard the challenge at 2:22, I was expecting something about binary vs ternary operations, and on to higher numbers of operands, with something like the idea that lumping arbitrarily many things together is an N-ary operation where all the sub-lumpings are somehow interchangeable. But I couldn't figure a way of saying that without the forbidden words (or their implicitly forbidden equivalents). A process with steps and sub-steps is one way of looking at it, but it doesn't feel natural in the familiar cases of addition and multiplication.
Sounds like confirmation bias, especially since none of the non-associative algebras are considered. Octonions are a good case in point. But for just a very primitive example, consider the case of (a*b)+c vs a*(b+c). These are two different operations, but the logic discussed in this video really doesn't rely on anything that would depend on whether the operations are identical or not. Otherwise, that would clash with discussion of function compositions, seeing how addition and multiplication are both simply different functions from RxR to R. Understanding where this all breaks down leads to understanding why associativity is an axiom of the group (well, smigroup) and why that matters.
Is this notation consistent with all this? I think it is. * a b c d where * is an associative action and a,b,c,d... are objects that can be acted on by * means - pick any two consecutive objects (call then N and M) and allow * to act on them - replace N M with the result of N*M - repeat until the set (a,b,c,d ...) is now one value reflective of the combination of all the initial objects
Excellent! I well hooked onto your content! I am eagerly awaiting your next upload. Ps: I like that you cover such seemingly mundane topics in si much depth.
I think one of the huge things you missed here was why the associative property exists in the first place and why it needs to be pointed out. Mathematical binary operations (+,-,×,÷, ect.) take exactly two operands, no more or less. And there need to be a reconciliation between that fact and the desire to use more than one operation (binary or not) and more than 2 operands in an expression.
Associativity is a property (a unary relation for functions). Associativity is a property belonging to Categories (more generally a Semi-Groupoid or Semi-Precategory). The morphisms of the structure, map between its elements, and connect in a directed fashion (morphisms are referred to as arrows in a category)- (it is also common to say categories are related to Quivers / Multi-edge directed Graphs)… Anyway, we can draw a category with a directed multigraph- I would argue this is a form of abstraction, but, it doesn’t matter. The resulting graph shows the connectivity of the arrows, which allows us to visualize some properties. Regardless. Morphisms inherit the property of transitivity, which is what allows us to chain arrows together. Morphisms are not limited to mapping between elements of the structure; morphisms can map between the morphisms themselves (not technically). Those morphisms are referred to as Functors (or “cell-1” arrows I think…). Lastly, a Homomorphism is a morphism is a morphism of a specific form. F( x # y ) = F( x ) • F( y ), # and • are arrows that relate the elements, while F is a functor that transforms them. (or I got that backwards). Anyway, homomorphisms allow for us to have a transformation applied before ~OR after a relation (like an operation), and end up with the same result: Which is more or less Associativity. Functions are homomorphisms, and Operations are Functions- meaning, that we have access to that form in one way or another. Which is why “+” , “*” , … whatever operation, might have some associativity. Not all operations are associative I believe: remember a Homomorphism allows for “#” and “•”; whereas, if it were the associative property - those would probably be the same base operation. Example: If F(a) -> a, then: ( x * y ) = ( x ) * ( y ), “*” happens to be associative, but there might be additional requirements (perhaps domain/ range specifications). Example 2: If F(a) -> log(a) Log( x * y ) = Log( x ) + Log( y ), While this is a homomorphism, there is no clear associative property. This is better as a test or example, than a formal definition of the associative property. I don’t know the formal definition, but I bet it’s pretty cool- probably not as round-about. But yeah, associativity is totally a property. It’s used to name Magmoids / Magmas, and its importance is significant. Cheers
First axiom: A is absolutely identical to A. If p is to q and q is to r then p is to r. Logic and basis of union/intersect as property of sets i.e. the abstraction of the member/arrangement of a set. The way math is taught benefits from a primer on the logic of arithmetic.
Doing the challenge before finishing the video An associative operation is any operation wherein any object can be described to be composed by some sub objects via the associative operation, which can either be described compositionally and or are considered base units of the operation, an operation is associative iff some object is described by a composition of only base elements, all elements of the composition are exchangeable with any other base element in the composition so long as each base element never occurs the same number of times in every composition This is probably wrong but should serve as a distinct definition of how I currently think of the property given the provided restrictions before watching the video I might have broken some of the rules or at least their spirit by using the idea of compositions and objects with sub objects which is somewhat synonymous to groups but it’s the best I got lol
There appear to be a few inaccuracies. The linkages that lead to "getting the milk" are associative if you think of them as connecting film clips together. But if you understand them as real actions, the execution is not possible. But associativity actually means that you should be able to do the subsequent linkage before the first one in order to achieve the same result. In addition, in my opinion it is more interesting to also consider non-associative linkages, especially commutative, non-associative linkages. For example, if you define: "(x linked to y) is equal to ((x to the power of y) times (y to the power of x))" then this is a commutative linkage, but it is not associative. On the other hand, if you define: "(x linked to y) is equal to ((x divided by 2) times (y divided by 2))" it is both commutativ and associative, although both differs from usual multiplication.
I disagree with your formulation of the object-action duality. When you transformed an object to an action, you didn't use the operator, them you ignored the operator, but it was still there. I think that it would be better to partially apply the operator to the object to create an action. So the expression “2 + 3” could be interpreted as “2 (+3)”, the object 2 and the action of adding 3. Instead of object-action duality, more accurate would be partial application, that a binary operator can be applied to a single object to create an action on another object.
Wait i have an idea on how to describe the associative property to someone , if they know how to find the area of square and rectangle , Imagine a sheet of paper that has side lendth of B cm and C cm wide the area of this rectangular sheet of paper is BC now imagine if now the same paper was stacked on top of each other to make its height A then the volume occupied by the stack will be ABC, now image a sheet of paper placed vertically and has side lendth of A and C now area of this will be AC now stack them to besides each other such that the lendth of the stack is B then the volume of this will be BAC and both stack look same there for ABC=BAC.
I think the & operator makes a lot more sense when you're familiar with procedural programming. The way I've understood equations for a long time is that they represent two ideas. One is plain equality, and the other is an algorithm. One tells you two things are equal, the other tells you that you can build one side to be equal to the other using symbol manipulation. The & operator just makes it clear that it is the abstract idea that makes the algorithm move on from one step to the next step. In programming, you assume the & symbol to be the next line (Python), the semicolon (languages inspired by C), or any other terminating character. This is made a bit more clear when you get funny symbols like ++, for loops, or my favorite, the ternary operator ?: that indicate multiple instructions wrapped into one expression. That nice wrapping is associativity.
Oh, that is a wonderful take! I hadn't considered that multi-instruction syntax (like ++) is really a sort of shift of parenthesis. There is another way to think of associativity in the context of coding. The order in which you write the instructions doesn't matter. This is only the case because the concatenation of lines of code is associative.
An associative operation is one whose steps can be rearranged without affecting the outcome. I guess "step" isn't a synonym for "order" but the fact that they can be rearranged implies there is an order which doesn't matter.
Imagine you have three boxes labeled A, B, C and a machine that takes two boxes and combines them into one box. The two boxes the machine takes are inserted into two holes, one on the left side of the back of the machine, and one on the right side of the back of the machine. The machine will spit out the combined box from the front. Associativity describes a property the machine has. If you put B in the left hole and C in the right hole, you get out a box I'll call D. Then if you put A in the left hole and E in the right hole, you get out a box I'll call F. Now, let's say you put A in the left hole and B in the right hole, to get a box I'll call G. Then put G in the left hole and C in the right hole to get a box called H. To say the machine is associative is to say that F = H.
Very great for your first video! However, I find naming the operation you defined as "and then" quite misleading. First of all, "take x, and then multiply y, and then multiply z" is not sufficient information to follow instructions in non-commutative cases. Thus I wouldn't want to call multiplication an "and then" operation; to fix this, we should consider "multiply from left" and "multiply from right" as two distinct operations. But even then there are natural examples of operations which are not associative, such as multiplication of octonions. I can multiply octonions from left in succession, which I'd intuitively think as "and then":ing left multiplications together, but yet the operation is not associative. What you really defined is the idea of action, more specifically right action. This is usually defined as group acting on some set (as in your example dihedral group acting on a triangle), but the notion of action can be generalized to any associative algebra (i.e. algebra with associative operation), or even non-associative algebra as far as you require action of (xy)z to be equivalent with the action of x(yz), although I'm not aware of any natural examples of such actions. The name "action" still has the issue that some things I'd intuitively think as actions are not actually actions (e.g. left multiplication of octonion), but that's just the limitation of our words and highlights the necessity of precise definitions.
Lots of good points here! Let me try to address them individually: 1. "and then" on non-commutative operations: I agree with you that "take x, and then multiply y, and then multiply z" is ambiguous for non-commutative products. That said, this observation is not important to the "associative=and then" conclusion. The conclusion states that each associative operation is an "and then" operation on _some set of actions_ (the right actions were a constructible example), and not that the associative operation is an "and then" operation on _all actions_ . I did not make that clear in the video, apologies about that. That said, we could perform all the steps on the left actions, and this would lead to the Cayley's theorem in it's original form: that associative operations are function composition. 2. Octonions are non-associative: This was my response to a similar comment demonstrating that being "and then"able is not the same as being "and then": " This is a distinction that probably should have been made more explicit in the video. Associativity does not mean that you _can_ "and then" the actions (as you've pointed out, this is always the case). Instead, associative operations _are_ "and then" operations. This is precisely what I hoped to convey with "associative = and then" " 3. Generalizing right actions to non-associative algebras: Absolutely! That is an excellent insight. It's the reason that object-action duality is generalizable to any binary operation. Note that my response to 2 demonstrates this is not an indication that non-associative operations are "and then" operations. 4. Left multiplication of octonion isn't an action: Fortunately, this is an action! Let [a*] be the action of left multiplication by 'a'. It is an action that maps each starting state 's' to the end state s[a*]=a*s. 5. "the limitation of our words and highlights the necessity of precise definitions": it seems I need some work in this department. I think I sacrificed clarity for flow. I'm still not sure what the right balance is since I don't think I want my videos to be lectures but rather to be expeditions where we unveil some core idea. Maybe I should publish the precise definitions in a document along with the video, but then I would take even longer to get a video out 😅
@@linguamathematica2582 I'm not saying there was anything wrong with contents of your video - you even gave the formal definition of "and then" operators, which is great! - just that some clarifications and exploration to when things break down would've completed the discussion. Now I feel like the subject was slightly oversimplified, leaving the impression that "and then":ing is always associative. 4. What I meant is that left/right multiplication of octonions isn't action in the sense action is usually defined. To make this more precise, we should make the distinction between object and action more explicit. Let O be the set of objects and A the set of preactons on O (which we later turn into actions), which formally is some collection of maps a:O->O. A priori objects have no further structure, so they can't be combined/glued. If for any two preactions a,b there exists a third preaction c such that for any object x we have a(b(x)) = c(x), we can define multiplication on the set of preactions as ab=c. Preactions together with such multiplication are called actions, or as you can them in the video, and-then operations. The object-action duality you demonstrated in the video is the special case where the set of objects and the set of actions are in some sense "the same". More precisely, there should be a bijection f:O->A such that (f(x)f(y))(z)=f(x)(f(y)(z)) for every object x,y,z, in which case action a can be identified with the object f^(-1)(a) and object x can be identified with the action f(x). For example, in the case of left matrix multiplication, each matrix A is identified with the map A(B)=AB; on the left hand side we have map evaluation, on the right hand side we have matrix multiplication. Now, let's explore this duality in the case of left multiplication of octonions. From now on, I'm denoting octonions as x,y,z. The set of objects is octonions, and the set of preactons are identified with the set of octonions by left multiplication similarly to the previous example. For this to be an action, it should satisfy x(y(z))=(xy)(z) for any x,y,z. However, this would imply associativity of octonions, hence cannot be true in general. We thus conclude that left multiplication of octonions is only a preaction, not an action. As a final remark, note that the construction of the left multiplication preactons for matrices and octonions were constructed following the object-action duality, yet only for matrices the preaction turned out be an actual action.
I'll show it for division which is also non associative and easier for my keyboard. You can't just replace '/' with 'andThen', 1 & 2 & 3 doesn't make sense because 2 and 3 are not operations. What you can do is 1 & /2 & /3, and that is function composition so it IS asociative.
Thank you for this video! This discussion is a starting point for addressing problems that font designers have with *exponential work*! Do you have any ideas for making work faster or more manageable, when you have to work along 8 or more dimensions? What type of math could be used to think about this?
associativity = "left multiplication by a commutes with right multiplication by b" or simply "multiplication on the left commutes with multiplication on the right"
"An operator is associative if the result is independent of the path taken." Examples: - After concatenating, we can't tell which seam was most recently glued. - After applying rotations or flips to a triangle, group theory only cares about the orientation we end up with, and doesn't remember the path taken. - In a function application such as f(g(x)), the function f can't "reach into" g to help it make decisions, so it can't depend on what happened within g. Similarly, g can't "look outward" into f. - When applying a non-associative operator like / or √, we have to look around more, essentially locating the parentheses or other context clues like line length. The last two are perhaps better thought of as "context sensitivity / insensitivity" that "path dependence / independence", but I think they're worth considering as one concept.
My attempt: “An associative operation is a binary operation that can be applied to arbitrarily many elements by nested evaluation, with an identical result no matter how elements are paired during computation.” It still comes out sounding like commutation. lol harder than I thought.
Here's my attempt at the challenge at the start of the video before I watch more of the video: An associative operation is a binary (2-ary) operation on a set to which we can naturally associate an n-ary operation which is related, in a canonical way, to repeated application of the 2-ary operation. For example, we can define +_3 such that +(+(a, b), c) = +(a, +(b, c)) = +_3(a, b, c).
I'm not sure this video shows the essence of associativity. The essence is that associative operation is independent of the objects it acts on. It doesn't use any "information" from the objects to change what it's doing.
It's an interesting and potentially useful way of thinking about things. It was a bit of a weird video, because you invent a new framework (objects and actjons) without quite stating that it's a new framework rather than statements about existing ones. Then you explore associativity mostly in this invented framework rather than in the ones we actually use it in. I still think it's a good video, I just think you could stand to make your intent clearer and explicitly stating when we're leaving math and entering wonderland. Obviously there IS no object-action distinction in mathematics. You could separate what numbers mean into those two categories, or a different two categories, or 1 category, or n categories. Choosing objects and actions and then showing they're equivalent is kind of the logical equivalent of an identity operation :)
To answer your challenge: A binary operation • is associative if for any x when you treat it as a unary operation x• function composition corresponds to applying the operation to each x
Object-Action duality is an excellent idea indeed. However, in the presented form, it's incomplete E.g. if fails to explain why subtraction is not associative. Indeed, what stops us from interpreting 2-3-4 as takeObject(2).andThen(subtract 3).andThen(subtract 4)? The root cause of the problem is the starting object. For the operation to be associative, takeObject has to be consistent with the nature of the action (here the "action" is subtraction). Namely, takeObject(x) has to be equivalent to takeObject(identity).andThen(action(x)). In the case of 2-3-4, the action is subtraction, and the "identity" is zero, so takeObject(identity).andThen(subtract 2) does not give 2 as we hoped, which destroys all further logic. In fact, "identity" is the only object we are allowed to "take from thin air" - the rest of the objects should be obtainable from it through the defined "action". In general, this requirement can be expressed as an axiom x=identity.action(x). You can easily verify that this is the case for every associative operation considered in the video, it's just the "identity" is specific to the operation in question (0 for addition, 1 for multiplication, an empty string for concatenation etc.)
Let's take (10-2)-3=5: I have 10 apples, I subtract 2 _and then_ subtract 3 I get 5. So while it is possible to say this in words and it will make sense, this doesn't make subtraction associative: 10-(2-3)=11. What am I missing?
![The most beautiful equation in math, explained visually [Euler’s Formula]](/img/n.gif)
In a sense, concatenation is "the" associative operation. Every equation satisfied by concatenation is satisfied by every associative operation. This is easy to understand keeping in mind that associative operations allow you to drop parentheses, so you can write terms just by writing sequences of arguments to the operation.
Precisely! This interpretation is strongly related to the gluing interpretation
Not exactly, concatenation is reversible up to associativity, unlike most of the other operations, they make a free monoid, but the other monoids differ from the free one
@@user-tk2jy8xr8b That's exactly what I meant, I just didn't want to use more terms.
@@user-tk2jy8xr8b excellent point
I'd argue that substitution is the more general aspect, and that concatenation is a specific form of substitution.
I find it to be more a matter of semantics rather than anything else. The "and then" is pretty much precisely how one would understand what the order of operations is, thus it seems wrong to claim, that the description of associativity as the ability to change the order of the operations is faulty. I guess, sure, if this helps someone to understand the concept, which seems to be the case scrolling through the comments, it certainly is useful, but I would not claim it is anything more than either a more in-depth explaination or a slightly alternative approach to look at associativity.
I agree with pretty much everything you say. The "and then" take has order sort of built into it. Here is my response to a similar comment:
"You are right to protest that you will never find a definition that is completely unrelated to ordering or grouping. This is because any alternative definition of the associative property can always be used to derive the syntactic rule of associativity, which can then be interpreted as being about order all along. The challenge was issued more out of a desire to nudge people's conception of associativity out of its comfort zone and see what creative ideas people come up with. And I gotta say, you all did not disappoint!"
I also agree that a notion "and then" is necessary to understanding order of operations. I further agree that there is nothing faulty about the usual definition of associativity (after all, it is the defining property of associativity). That said, I would also add that the usual definition leaves much to be desired. Questions like "Why is associativity everywhere?", "Why is it important to study?", "Why is it baked into physics?", and "What do we study when we study semigroups?" are better answered by a perspective shift to the "an operator is associative iff it is _and then_" definition of associativity.
Yeah same. But after watching I remembered learning about matrix multiplication and actually working with matrices for marcov chains, and I remembered having a similar epiphany back then. I think it's useful to at least work with one or two groups that are not commutative in order to really "get" associativity. And that usually doesn't happen. Where I live at least, once you learn about the first non commutative group (matrix multiplication), it's too late and many students stopped really thinking about the stuff they are forced to learn in math lessons. And that's really a waste of potential because it is not a hard concept to understand.
My point is, it is a useful video for people who didn't really get to feel groups as reversable actions.
Be careful what you dismiss. What are abstract non-formal ideas today are new and rigorous theories tomorrow. Finding a set of semantics which make sense to you personally and are not just dryly adopted because "they work" is part of being good, confident, and fluid with math.
Moreover, I believe that good math, what math is, is neither a pure manipulation of symbols nor specific example of something in the real world. It is about abstractions of things in the real world, such as perfect circles, perfect lines, and perfect planes, perfect cubes, functions which behave perfectly like polynomials, etc, which do not exist in the real world but we can reason about them anyways, precisely because of our experiences in the real world. But also our experiences with language and semantics influence this too, I believe. A simple notation is often suggestive of new truths, but some notations can conceal them. Fortunately, if we choose notations and words to understand things or seek to find/see meanings in the phrases and notations established by others, and truly understand those people who made them as we would understand ourselves, connecting them to our abstract and perfect world of mathematics in a way that often seems "silly" or "non-rigorous", we are doing the work of mathematics still. So-called "Silly" and "non-rigorous" mathematics, not formalisms and symbols, is the heart of math. The philosophy of mathematical formalism does not distinguish - or at least does not need to - between the statements "a divides b" and "not a=0 and there exists an integer n such that a times n equals b", and also between the statements "e^(i*pi)=-1" and "(well, it's too much to write out, but like sum from 0 to infinity of blah blah Taylor Series w/ definitions of sums and limits and things, etc.)" but the former is essential to understanding number theory and the latter to complex analysis. Yes, to a formalist everything is "just semantics", but to a human these are not obvious statements connecting our mental images of a perfect, abstract mathematical world to previous definitions. And these things took humans hundreds of years to discover and understand, yet a formalistic textbook calls them "semantics" and says "let x be defined as y" without giving any further thought to it all, ignoring the brilliance and meaning of the statements, and with it everything that makes math interesting. my point is: math can progress without formalism and definitions (it has done so at the least for thousands of years). It can't progress without "silliness" and "semantics" and "alternative explanations", however.
@@bcthoburn Thank you for your reply. Your 1st point would indeed be reasonable if we didn't have a way of making precise statements, which we could use for rigorous definitions. But since we do, any set of semantics serves only as the means for us to interpret and understand these concepts and, unless incorrect, each holds no more and no less value than any other alternative.
As for the 2nd, my original comment does mention, that the idea presented in the video can be useful, as every person may find a certain explanations more helpful than others. What I tried to point out is the sensationalized nature of the title, which, I tried to argue, gives a false expectation of the content of the video: that is, there is little difference between the explanation given and the viewpoint it is said to replace.
William Shakespear: "To be, or not to be"
Lingua Mathematics: "To be, and then to see."
14:55
The Shakespeare of mathematics.
I literally read this as he said it, weird.
Thank you for this great video. I think the idea of object-action duality exposes something fundamental that is missing in elementary math education. As some of your early examples show, we commonly switch between object-object interaction and object-action interaction without ever acknowledging the distinction. Sometimes this leads to confusion for students, especially those without a good sense of math intuition.
My daughter struggles with math (she almost certainly has dyscalculia, a math version of dyslexia), and, after watching this video, it seems clear that part of her struggles stem from the way we casually interchange object-object and object-action models. A good example is negative numbers: we treat -1 as both an object and an action without ever acknowledging that is what we are doing. This actually became a stumbling block for my daughter in an Algebra 1 problem. She had to simplify something like 5 - (x + 3). Distributing the subtraction, turns the action of subtraction, into negative objects. She had trouble with the this (partly because of the implicit 1 as a coefficient), and I was at a loss to explain it to her, and I now I see that it is exactly because of the action-object switch. I'm still not sure how to explain object-action in a way that would make sense to young students, but it seems important.
Another great example is fractions. Is a fraction a number or an operation? We use it as both interchangeably and almost never acknowledge the shift. I wonder if this is part of the reason so many kids have trouble with fractions. I would be very interested to read a math educator's/researchers view on this.
This was so touching to read (probably because I'm a father too). I love your insights about -1 and fractions. Thank you!
It is not taught because it is detrimental. Thinking of numbers as abstract objects is a very important skill, it leads to confusion at the start because abstract thinking is a difficult skill but one of the most important ones that math teaches. And regardless, the distinction between object and action has no meaning in mathematics, and is more a philosophical matter than anything, so there is no reason to teach it in the first place, and teaching it would probably do more harm than good, as it circumvents the process of abstraction.
Easy. Just define "a - b" as a shorthand for "a + (-b)". The unary minus being the additive inverse.
Same can be done for division/fractions = multiplication by the multiplicative inverse.
Excellent presentation and unique interpretation. Just one suggestion : Counter examples like Sedenions and the interpretation of this difference would have been appreciated.
Great suggestion! The original script had a non-associative example (division) that was regrettably cut to shorten the video.
Here's a condensed summary of that take: Division is not associative ((8÷4)÷2 is 1 but 8÷(4÷2) is 4) and as such cannot not be an "and then" operation. If we divide by 4 and then divide by 2, this is not the same as dividing by (4÷2). While insightful, this observation alone isn't enough to prove that division is not an "and then" operation because I interpreted the action 𝑥 narrowly to mean divide by 𝑥.
Suppose there was some other action interpretation of numbers where division meant "and then". Well, in that case division would satisfy the "and then" property: s(𝑏÷𝑐)=(s𝑏)𝑐. Which was shown to imply division is associative. But wait! Division is not associative. To avoid the contradiction we must reject our supposition that division can ever be an "and then" operation.
@@linguamathematica2582 this argument seems kinda circular. like you dont really give intuition for why divisioon isnt 'and then'. you just defined 'and then' to mean associative.
@@_okedata My mistake, let me attempt a clearer approach:
1. Assume division is "and then"
2. As explored in 14:46 - 15:16, the definition of an "and then" operation '&' is one which obeys s(𝒂&𝒃)=(s𝒂)𝒃
3. Since division is presumed to be "and then", it by definition obeys s(𝒂÷𝒃)=(s𝒂)𝒃
4. As explored in 16:15 - 17:45, s(𝒂÷𝒃)=(s𝒂)𝒃 implies that (𝒂÷𝒃)÷𝒄=𝒂÷(𝒃÷𝒄)
5. But (8÷4)÷2≠8÷(4÷2), contradicting (𝒂÷𝒃)÷𝒄=𝒂÷(𝒃÷𝒄)
The contradiction forces us to reject assumption 1, meaning division is not "and then".
(Sorry if I'm still bungling this argument up 👉👈)
@@linguamathematica2582 so in order to show that division in not associative you assume that it is an "and then" operation, then say that "and then" operations must be associative and finally provide a counterexample for division being an associative operation?
Either I didn't understand something, or checking andthenity of operation is no more intuitive than checking its associativity
@@murmol444 Yes, my argument is very convoluted if the goal was to demonstrate that division is not associative. A more direct way to do that would be to just use step 5. The conclusion I had in mind was that division is not an "and then" operation. In my response to Farhan K, a proof by contradiction was used to show that.
As for whether checking "and then"ity is more intuitive than checking associativity, I believe that depends on the application. In the group example in 21:27 - 23:23, it was far easier to verify that the group law is an "and then" than that it is associative. In contrast, in multiplication it is easier to verify that it is associative by constructing a 3D box with dimensions a x b x c and demonstrating that the volume is the same whether doing (a x b) x c or a x (b x c).
The occasional ease of checking is really just a bonus, I find that the more profound finding is that the associative property actually says something intelligible about the operations that obey it: that they are all "and then" operations. The property (ab)c=a(bc) fails to give me an intuition about why it is so ubiquitous, important, or interesting and fails to give me an intuition about when I should evoke, expect, or observe the property. These are questions that are more easily answered by the "and then" interpretation. For an example, you can check out my answer to "why is associativity important?" here: math.stackexchange.com/a/4270764/977481
This was an amazing video, however, I think there was one change you should have considered making. As you mentioned in the description, this video was essentially an intuitive proof of Cayley’s theorem for semigroups, a theorem I was not familiar with before watching this video. After googling it, I found the formal statement of the theorem made the video make a lot more sense to me. Maybe it would have been a good idea to throw in this formal statement of the theorem at the end?
I hope you don’t see this as me saying “this video needs more rigor! Get rid of the intuition.” Because personally, I don’t think I would have understood how meaningful cayley’s theorem for semigroups was without the intuitive arguments given in the video. I just think maybe 1 minute at the end of the video that explained how all of this translated to rigorous mathematics would have been nice.
All in all, this is definitely one of my favorite SoME videos. Will def be subscribing (and checking out that poll you mentioned at the end of the video.)
Oh, that is some solid advice. Would have made a nice end screen animation. Thank you!
Check out Yoneda lemma, Cayley's theorem is a special case of it
I hate to be this guy, but even "and then" doesn't work properly. You can use "and then" with division. Divide by 7 and then divide by 3 and then divide by 2, but 7/2/3 is not the same as 7/(2/3).
I feel like there really aren't any proper words to describe associativity. That concept is *that* abstract. But maybe that's just me.
If by "that guy" you mean inquisitive, constructive, and mentally venturesome, then you totally are. Please don't change :)
I fully agree with you. You can use "and then" with division just as you've pointed out. More generally, you can "and then" any operation. However, this has no bearing on the arguments in the video since the claim there was that associative operations are precisely those operations that mean "and then" and not those operations that merely can be "and then"ed.
So what exactly does it mean to be "and then"? An "and then" operation takes two actions and combines them into a two step action "do a and then do b". Since "and then" operations are always associative, division cannot be such an action.
But how do we know that being "and then" implies associativity? If an operation * meant "and then", this would imply that (a*b)*c would mean "do a and then do b and then do c" and that a*(b*c) would mean "do a and then do b and then do c". Taken together, we could write (a*b)*c="do a and then do b and then do c"=a*(b*c) . Whatever the starting operation, being "and then" implies being associative.
That said, your observations are very profound. You can always "and then" any operation. Let's probe deeper and ask the question: what operation "and then"s division actions? That is we want an operation "?" that satisfies a/(b?c)=(a/b)/c. This is multiplication! Multiplication is how we "and then" division! And, true to form, multiplication is associative.
@@linguamathematica2582 I would say "divide by 7 and then divide by 3 and then divide by 2" is giving clear instructions which just happen to be very different from "7/3/2". The action "divide by 7" is perhaps better expressed as a function, f(x) = x / 7. The action "divide by 3" is g(x) = x / 3, and of course "divide by two" is h(x) = x /2. This way of thinking virtually replaces the whole "and then" concept with functions, and obviously works the same after that.
I think replacing "and then" with functions is much clearer, because for me, watching the video and reading your explanations in the comments, any time you try to define "and then" it just feels like you're defining associativity, and the distinction feels slippery so that the "aha" moment of the theorem is hard to grasp. Whereas, by using functions, there's clearly (for me) a nice thing happening where three things (for example the three things "one", "plus", and "one") are becoming two things, a function and an argument ( f of 1). And the function is clearly an action.
All of this amounts to saying, I think of function composition as the definitive associative operator. But because of this, I have trouble accepting an "object/action" duality for function composition itself. IE, the function is the thing the machine does, not the act of hooking the machine to another machine.
This is one of the moments when I remember what I love math for. Thank you for this amazing video!
How about “a closed operation on a set S is associative if the functions f_a for each a in S defined as f_a(x) := (x ¥ a) (where ¥ is the operation in question) satisfy f_{a ¥ b)(x) = f_b(f_a(x)) ”
Yours is the core of the video: Cayley's theorem for semigroups. It's interesting that function composition, when read from left to right is more naturally translated as "but first" rather than "and then", but otherwise it's the same idea. It's even more interesting that if you generalize the operation to a partial operation, and the functions to partial functions you get something very close to the Yoneda lemma
First time I feel such confusion after watching a math video. I have been able to follow the most "confusing" 3b1b videos, but although this explains something so fundamental, I feel like I am introduced to something totally new. I probably have to rewatch this a couple if times to fully grasp the concepts, but all in all this is truly well done. If I had to change one thing on this video is the pace. Some times I feel something is being rushed, while others are being explained too much for no reason. It might be me, I don't know.
It's probably not you. I am very new to making videos and do not have this all worked out yet. I would appreciate a breakdown of which chapters felt rushed/slow. No pressure.
@@linguamathematica2582 so, the intro and the first chapter, up until the 13 minutes mark is very well made, the introduction is on point and you clarify what you want to explain setting the basics. Now when you are explaining what associativity is, I feel like the video gets a bit slow, but without reason. Also at 15 minutes almost no-one is concentrated enough to get through new, potentially difficult concepts. You can give us a mental break, maybe change the visuals, or something that shows the video is progressing. At about 18 minutes I took a break and then kept watching and the video felt much more enjoyable (you can even advise the viewer to stop/ rewatch, or ask questions to help him see if they are ready to continue). After that the group theory part was a nice example and the conclusion really does it's job well, it revisits what we saw and helps close the video smoothly. And lastly, when I was referring to "rushed" parts at 18-20 minutes you seem to wrap up a very important idea in just 2 minutes. If you want my tips as a viewer I can list them:
1) long videos=hard videos to follow, keep the viewer engaged with questions and make it feel like we are doing actual progress (maybe split the video into more parts so that I can see where we are and where we are going)
2) the visuals are great, but the glow thingy for me was a bit hard to follow. Do an underscore, a different capitalization, or different colour next time.
3) plan exactly how much time something needs to be explained based on how important and how difficult it is, before writing your script, this will help a lot with ups and downs in speed. Generally speaking too hard or too unimportant parts can be completely skipped or shown as honourable mentions.
That's all from me, I am looking forward for your next projects and videos.
@@giannisr.7733 Wow, amazing advice! Thank you so much for this thorough breakdown. I'm going to meditate on this for a while
One thing I feel is absolutely worth noting, possibly even necessary to note, is not only that the associative property applies to operations that operate both on objects and actions, or that it can be viewed as an 'and then' operation, but that when the operation is associative, _the 'and then' works the same way for both objects and actions._
After all, you can take non-associative operations like (say) subtraction or floating-point addition, and those operate both on objects and actions, just like regular addition. Moreover, unless you _define_ 'and then' to only apply to associative operations (which I can only assume is not what you're trying to do as it kinda defeats the idea that this is a new way to think about associativity), it's perfectly legitimate and intuitive to say 'a minus b, and then minus c' or 'a float-add b, and then float-add c'.
Without any additional qualifiers, you could therefore apply subtraction to the above construction: if s is your object, sx indicates s-x, and x&y indicates x+y, the equation s(x&y) = (sx)y holds even though subtraction is clearly not associative. (It's still subtraction on the object; 'x&y == x+y' is just defining how subtraction works on the action, and we're free to define it differently here because it's in a different context.)
But if you include the qualifier that 'and then' should work the same way for both objects and actions, then it starts looking like associativity. Subtraction is non-associative not because it can't be expressed as an 'and then' operation, but because if you do, it works differently between the action and the object. Acting on an object looks like s-x, but acting on an action looks like y+x. A similar argument can be made for float addition and other non-associative operations. The way I see it, that's truly where the distinction between associative and non-associative operations lies.
Nice breakdown! Yes, this is exactly what was meant by associative operations are "and then" operations. It does not mean that you _can_ "and then" them (any operation can do this), it means that they _are_ the "and then" operation. In your subtraction example, you observed that to "and then" subtraction actions, you use addition. Using the conclusions in the video, this makes addition associative without the need to check anything further.
Well, you've touched upon three ways of looking at things: two objects combine to form another object, two actions combine to form another action, and an action transforms an object into another object.
But there is a fourth one! An OBJECT transforms an ACTION into another ACTION. For example, a pair of apples transforms the addition of three apples into the addition of five apples.
I definitely liked the production values and some of the points of this video. That said, I wanted to share my struggles with this video.
On my first viewing, I made it about a third of the way in and stopped watching because I couldn't tell where it was headed (and when I went back to finish, it seems like the main "and then" idea didn't need the flexibility of all three object-object/object-action/action-action interpretations, and so the video felt a little out of order).
After finishing, the main thing I struggled with was the lack of clear signposts/separation of the math from the philosophy. It made it hard to decide when I should be thinking of what was said as a simplified description of a theorem, and when it as just a helpful interpretation to keep in mind.
Finally, the line "Its meaning can remain hidden, even after centuries of instruction" was extremely offputting. Even if these weren't your intention, it suggests 1. You might be unaware of the formal notion of group actions, or how group theory developed historically (this is only a problem since you're making a claim about centuries of history). 2. You think that your philosophical ideas here are brand new (and with no provided justification in the video or description that you've "done your homework") . 3. You think that your interpretations are not merely good and deserving of spreading, but the sole definitive "meaning" of associativity.
But to be clear, I really agree with you about, say, how groups should ideally be introduced to students, and the benefits of thinking about an operation from those three perspectives, and appreciate you making a video emphasizing those things.
Thank you for this. Giving clear signposts is something vital to do in the classroom and in presentations, but I'm still trying to figure out its role in videos (where I believe, perhaps in error, that the right amount of uncertainty can create intrigue that pulls the viewer in). That said, your response tells me that my clumsy effort to draw you in only pushed you away. I'll try to add more clarity in the future and see how it plays out.
As for the line "Its meaning can remain hidden, even after centuries of instruction", I am sorry that it was offputting. I did not mean that it is hidden from _mathematicians_ but from those who've learned it. It followed my discussion of pedagogy and I meant _instriction_ to be an operating word there. Something like "after centuries of instruction, its meaning still isn't taught." That said, I can really be in my own head when script writing and your interpretation now seems more natural to me than my own. After all, remaining hidden seems to imply that no-one at all has discovered it.
I'm glad that despite my amateur mistakes you still appreciate the video :)
"Assoociativity is not merely a property, but also an idea". Thanks for putting the grandiloquous word salad at the beginning of the video. Saved me about half an hour.
I felt like watching a 3B1B video, others have given some advices and they definitely seem good, but I just wanna say I am really impressed by the feel of video and how you go about it just like grant!
I hope you are here to stay :-)
Thank you for the kind words :)
Grant is such an inspiration! This video definitely wouldn't be here without him
your feelings are irrational
I think that even though overall the video is great it doesn't hammer down a very important point - associativity means that objects of operations and operations themselves are identical. Or in a mathematical language arguments are isomorphic to functions. Take for example expressions 5+3+2 and 5-3-2, they can be written as 5(+3)(+2), meaning acting of an operation +3 to 5 and then acting of +2 to the result, and 5(-3)(-2) (same meaning) or as (5+3)+2 and (5-3)-2 respectively, meaning combining a pair of objects via + or - , the former is associative, the latter isn't, because (-3)(-2)≠(3-2)=(+1). There is no correspondence between objects themselves (think about apples in the example) and the operation (removing the apples) in subtraction, which exists with the sum. Whenever you can treat objects and operations as one and the same, you will have associativity.
It all boils down to that moment of s(a&b)=(sa)b and just omitting & for notation sake is harmful, IMO. It should've been made more clear, that only when you can use "a" both as an action, like "b", and an object, just like "s", you will have associativity, which is expressed in the omitting of & in the expression. Otherwise you may have to combine a&b in a strange manner and hence you don't get associativity.
I prefer to think of addition and multiplication as the fundamental, associative "binary or more" operations, and subtraction and division as unary operations, that way I don't have to worry about the processes not being associative
5-3-2 = 5 + (-3) + (-2), is this not easier to work with, and less ambiguous than "5-3-2"? I get confused with "5-3-2" because of the non-associativity, 5-(3-2) != (5-3)-2 and all that, and I prefer forcing subtraction to be "associative" via reducing it to a unary operation: 5 + (-3) + (-2)
5/3/2 = 5 * (1/3) * (1/2), this one's a bit tougher to demonstrate because while -3 ("negative three") stands on its own fine, /3 ("reciprocal of three") does not and you gotta use 1/3 to get the point across
Thank you for your thoughtful response! The idea that associativity means that objects of operations and operators themselves are identical is a great start, but I’d like to see if we could improve it. It turns out that just being an operation (even a non associative one) means that objects of an operation can be interpreted as operators. For instance, in subtraction we can write 3−2 or we can write 3(−2), in division 3÷2=3(÷2) and in general, 𝑎∗𝑏=𝑎(∗𝑏). The mapping 𝑏↦(∗𝑏) maps objects to actions.
But wait, you observed that there seems to be no objects that correspond to (−3) in the apple scenario. That doesn’t seem to fit with what I just said. I would argue that the lack of an object representation of (−3) is a consequence of anti-apples being hard to find irl and not of some mathematical limitation. So, let’s shift to a different representation of addition, adding 1D vectors. In this domain, the object corresponding to (−3) is an arrow with the same length as 3 but pointing in the opposite direction.
Well, if object-action mapping is a property of any operations, what is associativity then? Well, your proposal that associativity means there is an isomorphism between arguments and functions was on the nose! To fully unpack your insight, we can specify the binary relations that the isomorphism should preserve. The full isomorphism is written (∗(𝑎∗𝑏))=(∗𝑎)&(∗𝑏), and the conclusion is that an associative is isomorphic to the “and then” operation. Finally, by noting that the object-action mapping 𝑏↦(∗𝑏) need not be injective, we relax our statement to “associativity means that there is a homeomorphism from ⟨𝑋,∗⟩ to ⟨(∗𝑋), &⟩”.
Very nice!
@@linguamathematica2582 Thanks for the response. I think I should apologize for omitting a very important point that you indeed can make any operation associative via homomorphism b→*b, as you have pointed out, but not only that you need to additionally create an operation with a property a°(*b)=a*b as well, by which I mean ° is a binary operation acting on a (which here would become *c for some c) and *b. And by construction ° would be associative. So yes, even for non-associative operations it is possible to make an interpretation that would be associative. However, it would require this ° operation, which in a sense would do for objects the same thing that a composition would do for functions or "and-thenning" them. Just as with -3 we replace everything with addition and for ÷2 we replace everything with multiplication both of which take the role of of this ° operation.
...I think that, even though it probably was unintentional, without a small clarification of this the video is harder to appreciate for some, because many people who are not familiar with this idea will come up with many operations that feel like "and then", but aren't associative, e.g. -,÷,↑,√ and so on. And my comment was for the most part directed at them that it's not that all of this aren't "and then" operations, but that they don't have an object in the specified domain which would correspond to the operation. And you are 100% right that it's not a problem of math, but our inability to have real tangible interpretations in this situations, so we choose not to do them. Anyway, as I said, great video, it still warms me to see that something like this is on the YT and people can see it, I hope my comments, previous and current, haven't created an expression that I am not simply amazed by your work. I very much am.
@@linguamathematica2582 I was confused about why subtraction is not associative. Can't I subtract, and then substract? But here you say (-(a-b)) is not equal to (-a)&(-b). I am still confused, and hopeful you can use graphics to explain this in a future video. At any rate, your video is great.
11:11 first one: connect factory line g and f together
second one: have a factory line programmed as g and edit it to incorporate f
third one: program g into a factory line then program f incorporated into it
Niiiice! Way better than mine 🤩
@@linguamathematica2582 thanks 👍
Great work on this video! I find the slight pauses in audio between topics to really help segment the lesson.
A little pushback on the classification of groups as “reversible procedures”, if you don’t mind…
The “and then” operation of taking a step on an infinite sidewalk with the elements of -1, 0, or 1 steps is always reversible, but does not define a group. Cannot forget the importance of
closed-ness :)
Thank you! After some thought, I completely agree with you. I hope it didn't take away from the rest of the video. I clumsily tried to sidestep closed-ness by using the word "procedure" to include actions generated by "and then"-ing. At the very least, I should have made that explicit. In the sidewalk example, I would classify 3, for example, as the procedure 1 & 1 & 1. And since it is a procedure it needs to be included in the group of "reversible procedures".
@@linguamathematica2582 Unfortunately, including all “procedures” doesn’t seem to work either. For example, in the video you presented the actions “open the fridge”, “find the milk”, and “pick up the milk”. As “find the milk” is difficult to reverse, we can make a new collection of actions as follows:
a: open the fridge
b: take out the milk
c: close the fridge
d: put the milk back
e: do nothing
Also include all procedures, like “get the milk” = a&b.
This collection of actions also satisfies your definition of a group; it uses an “and-then” operation, has an identity, and every action is reversible. However, as a collection of procedures, it’s nonsense. Consider, for example, c&d&d&c&a. “Close the fridge and then put the milk back and then put the milk back and then close the fridge and then open the fridge.” That’s the ravings of a madman.
So, what you really want is not only to include all procedures, but also that any action can be taken at any time. One of the reasons that these fridge-milk operations do not define a group is that you can’t do any of the milk operations when the fridge is closed.
This, by the way, is why groups are almost always introduced in terms of “symmetry”. In order to have any action be available at any time, something about the world must remain as before after it is changed, ready to be acted on again. I was fortunate enough to learn group theory in a way that emphasized this, so it was made clear that all symmetries are built on groups and all groups can be thought of as symmetries. If you explain in this way, all four group axioms (closure, identity, inverses, and associativity) seem not only justified but also necessary.
@TheBasikShow Fantastic! Thank you for this thorough and illuminating breakdown. I completely agree with you and perhaps "reversible procedures over a state space" would have been more accurate, although far less memorable and punchy. This was probably the biggest mistake in the video, and I apologize if it took away from the main thrust of the video. At the very least I should clarify how I sleep at night.
In my own (often heretical) thinking, I extend the domain of actions to include the "bad" cases. The core question we need to answer is: "what do you do when you perform an action that is not afforded by the current state?" I've found 3 answers that are universally applicable:
1. You don't: if this ever happens, you've done something illegal (in programming for example you might throw an exception, in mathematics you might say that it's the ravings of a madman)
2. Return an error state: pick some value to return that is disjoint from the outcomes of afforded actions. Often, actions on error states are always presumed to yield another error state (the set-theoretic view of functions returns the empty set when outside the domain. See also how infinity acts as an error state in arithmetic on the extended reals: en.wikipedia.org/wiki/Extended_real_number_line#Arithmetic_operations)
3. Do nothing: whatever the input state, return that (In the fridge example, opening an opened fridge would leave the fridge opened, so the output state is unchanged).
If you chose (1), as you probably should to stick to mathematical convention, then your arguments follow. In my thought processes, I often choose (3) by reasoning that if you can't perform the action then you don't change the state. Now I just extend the actions to fit as many states as needed. I've found there is an elegance to this because it mirrors reality (after all, reality doesn't dissolve when you attempt an impossible action), although this is definitely not anywhere near an accepted method. In the fridge example, extending using (3) does not yield a group because closing a closed fridge yields the same state as closing an open one and the action is not invertible.
I am so sad I found this video two years too late. That 24:37 was a holy-s*** moment and I will never un-see what it means now. You just got one more sub. I hope I can see more of your work.
Great video! One think that could have made it better, is if you had explored how non-associative operations fail to be "and then" operations. You understand something a lot better if you know when it does not apply.
Completely agree and is a point of regret for me now. check out the comment thread with austin8179 if your interested in a non-associative example
The "object-action duality" in this video seems to be equivalent to "every semigroup (set with associative operation) is isomorphic (structurally equivalent) to the set of endomorphisms (functions) on it given by f(x) = s + x, for each s in the semigroup, with the operator given by (function) composition."
We can also give the monoid and group laws like so: "in the action-dual of a monoid, the action of the identity element is the identity function, f(x) = x," and "in the action-dual of a group, the action of an inverse element is the inverse function of the element's action."
That last property also means that all endomorphisms of that form are injective, and since we can combine an inverse element with any other element, surjective. Therefore, the action of every element of a group is a bijection.
This video is brilliant!
Object-action duality explains something I have been wrestling with for quite some time. The difference between an operation and the result of an operation.
The Schrödinger equation of Quantum Mechanics consists of the Hamiltonian operator applied to the Psi function having the same result as multiplying the psi function with the energy eigenvalue. The Hamiltonian operator is then the action, and the eigenvalue is the object that has been measured.
The equation is then a question about what the psi functions are given the boundary conditions, and what the possible energy eigenvalues are. Or, in general, any quantum measurement is represented by an operator operating on some psi function, whereby the eigenvalues are the objects, the possible results of the measurements.
In other words, in quantum mechanics the object-action duality is equivalent to the eigenvalue and operator 'duality'.
At present, I am using this approach of object-action duality,which I called object - operator connection, to demystify the appearance of the complex number *_i_* in quantum mechanics.
Your explanation of group theory as the study of reversible procedures is quite interesting! Given the fact, that all fundamental laws of physics are reversible, it suggests that the study of laws of physics is not only helped by group theory, but it might even be the case that the study of physics _is_ the study of group theory, applied on energy!
Um, yeah, modern physics *is* group theory. That's why the standard model is U(1)×SU(2)×SU(3) and why things like momentum, energy, and electric charge are conserved (Nöther's Theorem). A bunch of String theory stuff is also just group theory (like E8×E8).
I'm thinking of a deterministic state machine.
Associativity means that the set of states is equal to the set of inputs.
How does this and-then interpretation and the object-action duality compare to monads? It feels like you're talking exactly about monads, but I'm not sure, since monads necessarily form monoids (i.e. associative with a neutral element), but not vice versa... I think.
See the wikipedia page on Monad (functional programming) - section "Definition". It's simpler than the category theory page. Note that the bind operation is often called "then" or "and then".
In functional programming, the bind operation accepts a monadic context with objects of a specific type (e.g. a possibly empty list of partial solutions to some problem, like placements of queens on a chessboard that don't attack each other), and an operation that transforms such objects and places them in monadic context (e.g. gives a possibly empty list of solutions that satisfy an additional constraint: a queen is added to the board without being attacked by another queen).
Excellent example of "and then" in context. As you pointed out, monads form a monoid over the bind operation and thus bind must mean "and then" in some way. In the standard definition of monads, the bind operation is indeed an "and then" of the first and second inputs. That said, I believe that monads could equally be defined so that they are "but first" composition operations (yielding a stack instead of a queue of functions) if you define the bind as ≪=:[𝑇→𝑀 𝑈]×𝑀 𝑇 →𝑀 𝑈 instead of the usual ≫=:𝑀 𝑇×[𝑇→𝑀 𝑈]→𝑀 𝑈. Also, I love your chess example with the queens for explaining monads.
As for object-action duality, its inherent in lambda calculus in general. Note that the 𝑀 𝑇 argument for the bind can be an object (values of a type) as well as an action (a lambda expression that returns a value of type 𝑀 𝑇). The bind can also be seen as a gluing together of actions into a pipeline. It's even in the name "bind", which is a synonym for gluing.
This is exactly the association I had too, especially since that way of phrasing it as "and then [do]" was what actually made monads click for me back when I learned about them. Especially once the talk about more physical actions started I couldn't help but think of the IO monad in Haskell.
This is an excellent video.
From a pedagogical viewpoint this changes the teaching of associativity from procedural to contextual.
This change is the pinnacle of maths teaching as it allows for the building of larger schema and interconnectivity of subjects.
Again, excellent video.
I think the explanation is made less clear by introducing "and then" as something destinct from function composition. By defining "sa" to mean "apply function a to s" you get new notation for "a(s)". By defining "s(b&c) = (sb)c" it's equivalent in standard notation to "(b&c)(s) = c(b(s))" but c(b(s)) = (c°b)(s): (b&c) is (c°b). The "&" and "sa" notation just lets you write the functions in the same order that they would appear as opperations since c(b(s)) = (s*b)*c. Introducing it as notation would save you the trouble of reproving that function composition is associative.
This is a good point! I could have just shown that '&' inherits associativity from function composition
Really amazing explanation. I never thought of it this way, especially how you applied it to Group theory. I'm not a math student, but this makes me want to be.
How about division? Seems like it would be an "and then" operation, but is not, in general, associative.
Thank you, this is a reasonable concern. Here are my responses to similar comments:
"This is a distinction that probably should have been made more explicit in the video. Associativity does not mean that you _can_ "and then" the actions (as you've pointed out, this is always the case). Instead, associative operations _are_ "and then" operations. This is precisely what I hoped to convey with "associative = and then"
Division is not associative since (8÷4)÷2 is 1 but 8÷(4÷2) is 4 and as such should not be an "and then" operation. Indeed, observe that if we divide by 4 and then divide by 2, this is not the same as dividing by (4÷2). While insightful, this observation alone isn't enough to prove that division is not an "and then" operation because I interpreted the action [𝑥] narrowly to mean divide by 𝑥.
Here is a more general proof:
1. Assume division is "and then"
2. As explored in 14:46 - 15:16, the definition of an "and then" operation '&' is one which obeys s(𝒂&𝒃)=(s𝒂)𝒃
3. Since division is presumed to be "and then", it by definition obeys s(𝒂÷𝒃)=(s𝒂)𝒃
4. As explored in 16:15 - 17:45, s(𝒂÷𝒃)=(s𝒂)𝒃 implies that (𝒂÷𝒃)÷𝒄=𝒂÷(𝒃÷𝒄)
5. But (8÷4)÷2≠8÷(4÷2), contradicting (𝒂÷𝒃)÷𝒄=𝒂÷(𝒃÷𝒄)
The contradiction forces us to reject assumption 1, meaning division is not "and then". "
The concept of the cornerstone is so cool, hope here will be more video like this in the future…
Great video! An example of a non-associative operation would've been nice for contrast. EDIT: Like what exactly goes wrong in terms of associativity when you subtract 1 from 2 and_then subtract that from 3 compared to subtracting 2 from 3 and_then subtracting 1 from that. 3-(2-1) != (3-2)-1
3 + (-2 + -1) = (3 + -2) + -1
I've thought about this in the past, by rephrasing the subtraction as addition of negative numbers, the operations now are associative. What changed? In my opinion but ultimately I think it comes from the "object" idea of the numbers. You can also think of numbers themselves as an operation or function. In the object conception the numbers are being operated on, and so the "who does what to who" matters quite a bit. In the function conception of numbers, there are no objects being operated on, and so the "difference" isn't derived as a relation between two things.
No idea if that made any sense to anyone, and to make things worse I'll go deeper. In one perspective there are no nouns, only verbs. In another perspective nouns and verbs exist, and nouns verb other nouns. The object mindset highlights a the continuity of individual patterns through transformations, but obscures the transformations. The function mindset highlights patterns of transformation, but obscures the continuity of patterns through transformation. Or to paraphrase (butcher) the words of the philosopher Dogen. When firewood is burned, it becomes ash, there is no wood, only ash. In the object conception, the firewood is still there, and this could be a problem for us if we viewed not being burned as a definition of firewood. Because the behavior and information are coupled it's hard to say what we can or can't do to a piece of firewood. In the same way an improperly applied function mindset would cause us to lose our children when they grow one inch taller. We've lost the continuity of a pattern through a transformation.
Subtraction is not considered an operation. And in case it is referred as one in a general sense, it is not a binary operation in a set of natural numbers. 2 is a natural number, 3 is too, but 2-3 isn't. So it isn't associative, as applying "and then" doesn't make sense when the operation is not binary itself.
@@roseCatcher_ But subtraction absolutely is a binary operation in the set of integers, or rational numbers, or real numbers, or complex numbers, or .... I find your comment too dismissive, as there are there a plethora of contexts in which subtraction is a bona fide non-associative binary operation, and in which op's question is _completely reasonable._
2:17 I would try to define like this: Associativity means that an operation which takes in two inputs and returns one output of the same kind can be uniquely extended to an operation which takes in three inputs and returns one output, and further, can be uniquely extended to an operation taking in an arbitrary finite number of inputs.
I don’t think it’s possible to explain it without the use of the word "order", its synonyms or recreating its definition (e.g. using the "and then" from the latter part of the video). Because associativity means "order of constituent [actions|operations|etc] doesn’t matter". It’s *about* the order.
All associative operations are instances of function composition. If * is associative on X then the corresponding function is f:X -> X^X defined by f(x)(y)=x*y where X^X is the set of functions from X to itself.
exactly.. 30 minutes which could be summarized in one sentence
Fantastic video! I've been wanting something like this for ages!
Same!
It took a little long to get to the point IMO, but I still learned something new! I didn't really think of associativity as a big ball of actions combined together, rather as the state combined together, but that actually makes a lot of sense once you pointed out that function composition is associative too (and the fridge example)
ehen you multiply a*b*c, you create a new number, that already fully described by this expession. when you get the number in another expression you can multiply given numbers, and in the end get the expression of number that please you as much as you need. for example 4*2*2 can be a square, a multiple of 2 or 8, or even power of 2 in alternative expressions, but still will have all of this properties in all of expressions
I'm a mathematician who studies algebra, and non-associative (among other more exotic forms) of algebra have been recently important in my work.
I found your video very insightful. It even has helped me better understand something that's come up recently in my research.
You really hit the nail on the head with your understanding of associativity and "object-action duality". Your main point can be stated concisely as follows:
Let A be something where it makes sense to talk about algebra (i.e. a set, a vector space, etc). Donote the set of functions A->A by Fun(A). This naturally comes equipped with an associative operation given by function composition.
Now consider an operation on A as a function
m : A×A -> A.
For clarity, I'll write m(x,y) as just xy.
This naturally induces a map (the so called "action map")
a : A -> Fun(A)
where a(x) is the function f(y) = xy. That is, it takes the obect x to the action given by multiplying x (on the left).
Here's the key point: The operation m is associative if and only if the action map takes the operation m to the composition operation. That is, if
a(xy) = a(x)°a(y).
An operation satisfying this property is what you call an "and then" operation. Indeed, one can prove that this is equivalent to associativity, and in this way all associativity derives from the fact that function composition is associative.
Proof: Note that a(xy) is the function where, for any z in A,
a(xy)(z) = (xy)z.
Likewise
(a(x)°a(y))(z) = a(x)(a(y)(z))
= a(x)(yz)
x(yz).
Thus a(xy) = a(x)°a(y) if and only if, for every z,
(xy)z = x(yz).
I'm not sure if I just take this information for granted, or if I don't understand the value. It didn't feel like this really said anything, to me.
I answered: "If I apply a procedure to thing 1 and thing 2, computer their result and apply a procedure to that result and thing 3. That is the same as applying a procedure to thing 2 and thing 3, computing their result and then applying a procedure to thing 1 and that result."
where a procedure takes 2 things and returns 1 thing.
That seems to be just sidestepping using the word "order" by describing in detail what "order" means in context.
@@ApesAmongUs well, order is very important to associativity. How else do you say "the order of inputs matters, but the order of execution does not matter" which isn't just obscuring "order" under descriptions of what "order" means in this context?
Is associativity not stating that A☆B☆C might not = B☆A☆C, but that A☆(B☆C) does = (A☆B)☆C?
@@ferociousfeind8538 I agree, but that was the challenge he made in the video - define it without using those words or equivalents. I think his point might have been that it isn't possible.
@@ApesAmongUs hmm, maybe? I got the impression the point was "those are surface features, and what associativity actually is id deeper than that"
@@ferociousfeind8538 Then can you meet the challenge and define it without referencing order or any analog of "order" (before, after, etc.)?
I think one problem with this outlook (and all outlooks have problems, so this isn't a knock) might lie in describing what it means for a group to not have the associative property. In the case of the cross product in R^3, I'm not sure I can even define what A x (BxC) even means in this context of actions. Maybe I could say that they don't have the object-action duality, but I'm not sure what that actually tells me about the objects I'm working with.
Cool stuff! Appreciate how well broken down this is
Glad you enjoyed it!
Great video! Thank you! I usually watch some sections of math videos at 1.5x or higher. Here I had to rewatch some sections.
Thanks! I'm the same way with the 1.5x or higher. Once you get used to it, you can understand everything being said. I downloaded a chrome extension that lets me speed it past 2x for the slow talkers.
My best shot at defining associativity without the words order, brackets etc.
An operation is associative if and only if, given mathematical objects within the domain of the operation, the result of applying the operation between all of the given mathematical objects is defined for only one value or no values.
Where I stopped trying to compose a simple explanation: “If you have an ordered list of values of some type, and a 2-ary endofunction on that type...”
Could you explain what is ASSOCIATIVE literally / how term applies in this situation.. It would be worth and enlightening ..!!
What a wonderful idea! Associate means to "join together". In this sense it is a synonym of gluing. So I guess that would mean that a valid answer to the challenge might be "associative operations are those that associate objects together"
Excellent. Absolutely excellent. I love that you don't shy away from groups once they are well justified!
As far as I can see there's a better word for what we call associativity: I'd call it Sequenciality
That's because if A,B,C,D and E are associative transformations, then
ABCDE=(AB)CDE=(ABC)DE=... (and so on)
So the sequence of transformations uniquely determines the overall resulting transformation.
However if the transformations aren't associative, then
ABCDE not equals (AB)CDE... (and so on)
then, if associative, the sequence of transformations uniquely determines the overall one
if not associative, you can partition subsequences of transformations and get to other results
Then associativity is the property of a transformation to be decomposed into a sequence of other transformations while being sensitive only to the sequence of these transformations. That's why I'd call it Sequenciality.
I keep imagining the transformation of the left side of the equation into the product and then back into the right side of the equation and here's my intuition and thought process on the 2:17 challenge:
associative operations are undoable
the input of the associative operation can be the output of an other application of that operation - a combination, if you will
when multiple instances of the operation are chained together, they can be recombined thanks to that "undoability" and "combinability."
*It feels like an associative operation IS a combination of actions or IS combination itself. *
Anyway, idk if that definition properly excludes non associative operations and I think I have come too close to using the word "group" and I think I've just spent a lot of time rewording that old grade school explanation, oh well.
Your train of thought was so fun. I like how in tune you are with your intuition. Like, the way you extract structure from your intuitions is pretty unique and inspiring
the wedge product is associative but not undoable
My prefered "solution" to 2:15 is "for an operation '+' on a structure A, with a, b, c elements of A, a+b+c is well defined"
My answer to the challenge at the beginning: "When an operation is associative, it means that when applying that operation to 3 or more numbers, you will always get the same result by applying it to any two of those numbers, and then applying the operation to that result and any other of the remaining numbers, until you have applied to operation to all of the numbers you wish."
I think I got it but I may have missed something subtle
Edit: I was missing something, although one may argue whether it was subtle or not. I think I just described commutativity...
This is an excellent description of the "jumbling property" evoked in 1:20. It is a property you get when both associativity and commutativity are satisfied.
"Rasie to the exponent" is an "and then" function, it even obeys your defining rule
s(b&c) = (sb)&c
but it is not associative
(a^b)^c =/= a^(b^c)
Therefore "and then" does not imply associativey
I feel this video is a _bit_ tricky in its presentation. I don't think the whole "and then" thing is clearly explained in full detail.
Start with a set S and a binary operation ★. You can then get a natural function φ : S → Functions(S,S) given by φ(s) = s★−. (This is the whole shifting from "object" view to "action" view.) Then ★ is associative if and only if φ is a "homomorphism". That is, ★ is associative if and only if φ(s★t) = φ(s)∘φ(t).
Written in this way, it's a bit obvious because (s★t)★− = φ(s★t) and φ(s)∘φ(t) = s★(t★−).
The issue here is that you can think of a^b = d as object-action-object, but not as action-action-action, since raising a number to the a power and then to the b power is not the same thing as raising it to the d power.
Nonetheless, the action-action-action viewpoint is very important for modern algebra if you ever need to use groups or modules to study something.
The states and actions are a very interesting visualisation of associativity. The way I personally think about it is that associative operations can be merged into variants of them with an arbitrary amount of inputs and also broken apart again while remaining equivalent. For example it's possible to concatenate 5 symbols, compose 5 functions, calculate the greatest common divider between 5 whole numbers or add any 5 real numbers, but not to divide 5 real numbers or take the arithmetic mean of 5 numbers while still being able to split it into multiple parts at any point.
Nice, under this interpretation of associativity being "and then" operations one could think of + being nothing more than the successor operation, and something like +3 could now be interpreted as meaning +^3 or +++.
While it doesn't naturally extend to concatenation (you'd have each ASCII char represent a special catenation function), it extends to function application quite nicely and it also makes a case for postfix notation: xf and by extension: xf^n.
What is more, it also extends quite very naturally to computational processes, best exemplified by concatenative languages like Forth and pipelining on a unix shell.
The idea is so fertile an enlightening all this comes from just looking at the & interpretation.
If yours was the only comment I got on this video, I would be satisfied. I picked this topic because of how fruitful it was in thinking about addition and the successor function, concatenation and writing, and computational processes. To see you share this sentiment was, frankly, touching.
Its fun to think of addition as an abelian group and consider how to incorporate commutativity into "and then" operations. By looking at a few ways of doing so, I ended up generating some nice models for different types of addition.
Perhaps I misunderstood the aim of the video, but it's 𝘯𝘰𝘯-associativity that I interpret as being "and then" operations. For example (12/6)/2 doesn't equal 12/(6/2). The first is divide 12 by 6, and then by 2. The second is divide 6 by 2 and then divide 12 by that. On the other hand 12x6x2 doesn't require any "and then's". Declaring that multiplication is associative, or that (12x6)x2 = 12x(6x2) merely amounts to ruling out the need for any "and then", for any particular sequence to the operations.
I like this is essentially saying group and group action is two identical/important aspect of group theory
I’m however really surprised that you said “the more I study groups, the more empty, the more joyless, the more opaque I feel” because that doesn’t sounds like someone making this kind of video will feel. I do agree that math textbooks are usually extremely polished that, most of the time it doesn’t make any sense why are we studying math. Unfortunately that’s just what we mathematician always do and we find beauty in doing so. If you have gone this far realizing the reason behind group theory, which is usually referred as theory of symmetry not reversible procedure, I would expect you find learning more of group theory gives more and more endless insight behind it. It’s definitely true for everything I learned with math.
I think there's something really important missing from the explanation. When you have an operation like "a✳️(b✳️c)" then a and b are _both_ starting-states
Around 16:00, the video talks about the difference between (1) having a state s, an action b, and using "apply action" to connect them, vs (2) having an action a, another action b, and using "and then" to connect them
The top row at 16:00 covers two scenarios:
State1 ( Action1 Action2 )
-vs-
(State1 ) Action1 Action2
But that's missing the scenario that we actually need:
State1 ( "operator" (State2 Action 1 ) )
That last one might not be very clear, but that's because it doesn't fit very well into this narrative. But it's the scenario we actually deal with when we're looking at associative operations
Consider "3×(4×5)". The starting state is the number three, followed by an action of "multiply by 20". That action is made up by using a second starting number (4), and the action "multiply by 5".
We have a number that you make using "(start(4) -> action(×5)". And somehow we need to show that the action of multiplying by that number, is the same action as "action(×4) action(×5)". I don't think the video does that
For comparison, consider division: " 3 ÷ (4 ÷ 5) ". If we try and do the same thing for division, then we end up claiming that "action(÷ 0.8)" is the same thing as "action(÷4) action(÷5)".
What's different between multiplication and division? I don't see why the argument in the video should be valid for one and not valid for the other.
This is a great video. I really love this type of deep insights into the inner workings of mathematics. However, there are some things that might need more explanations or can be improved in the video:
1. At 19:17 it is not entirely clear why “b” and “c” become actions, from what we are presented up to this point in the video it makes sense only to make “(b*c)” an action, not “b” and “c” individually.
2. At 17:27 if “s” would be 1 and the actions “(ab)c” and “a(bc)” turned out to be “add 1” and “multiply by 2” respectively, then we could not say that they are the same actions. It can be remedied thought by saying that “s((ab)c)=s(a(bc))” holds for EVERY “s”, which would make the two actions act the same on EVERY object, which is the defining property of being the same action (at least for functions).
3. You gave “gluing” an interpretation of, loosely speaking, “combining action-snapshots in time”, but at 27:18 you use it to “glue together the apples”, which is a completely different interpretation that is only related, because “gluing” has both physical and temporal meaning (in English).
Overall, though, I really enjoyed this video because it is not just a philosophical discussion of associativity, but a useful idea to have; like when you use it to prove associativity of group actions without checking each combination of its members. This is one of the best maths videos I have seen in quite some time.
Thank you so much for the kind words and the specific and constructive criticism. Especially as a new creator I need as much as I can get.
I'll try to fill in the gaps in the video by addressing your concerns here:
1. The idea was to use the construction at the top line of 19:17 while specifying different values for s and x. If I notate actions as primed variables, then a*b=ab' is true by specifying s -> a and x -> b. Then I just substitute.
2. I absolutely agree with the importance of making sure it's true for every state and, although I was not nearly as thorough or clear as you were in explaining it, I did attempt to mention it at 17:27 "and finally, since the two actions change every possible starting state into the same end state, the two actions are equivalent"
3. Yes, I do rely on the ambivalence of "gluing" in my video. The aspect of "gluing" that I'm referencing is about linking objects together in a sequence, whether in time or space.
Thanks also for pointing out what you enjoyed about the video (philosophy and useful mathematical ideas), it's a huge help for planning the future direction of the channel.
Great video! I do think it's missing a simple example of a non-associative algebra, e.g., 3D real vectors equipped with the cross product operation. This would highlight the importance of reversibility and strengthen your insight about groups and symmetry (in a very evocative and geometrically intuitive way).
2:18 challenge:
Addition is associative because when I am adding, let's say, what I have sold every day, the total physically exists, it's can't be two different things, thus it does not matter how I add it as long as every day gets the same weight.
Maybe it sounds odd but it's as far as I have thought about it
there’s this pipe operator in the R programming language that lets you write function application linearly, e.g. x |> mean() |> round(2). In the community, people read the pipe aloud as “then”. “Take vector x then compute the mean then round to two digits.” I think another language (JS?) has an explicit .then() method for chaining operations together too. “then” is such a good word.
I think the difference between object and action can be made clear with the concept of operators. I had much trouble in school to remember to take the minus sign into account in my arithmetic calculations. I so often forgot the simplest rules about it and made error over error. Well somehow I made it into university still. In quantum mechanics they introduced the concept of an operator, and suddenly everything made sense, in the way that the minus sign always BELONGS to the number, the operator and the object operated upon go together. So in your example and with superfluous but illuminating bracketing: (a) + (b) = (c) is the object-object approach, (a) (+b) = (c) is the object-action approach, and (+a)(+b) = (+c) is the action-action approach. In the object-object approach addition is seen as something connecting two objects but not being a property of the objects itself. In the action-action approach addition is a property of each object already. Maybe one could even write (+a) + (+b) = (+c). Finally it is interesting to note different ways of notation, such as reverse polish notation. One can write addition also as a b + then or also +(a,b) like a real function of two arguments, and I wonder if showing all these alternatives (infix, prefix, postfix notation) early on in school would help students to get a better understanding of the concept? Because all of these notations occur in mathematics over and over again and it is often just for historical reasons why one happens to be the standard notation and the other ones aren't.
A lot of intriguing points here. Especially what I took to be your main point: introducing many notational conventions. I wonder how that would play out.
Operators gave me the same clarity as they've done for you. I am even currently working on a paper that essentially boils down to: if you introduce operators your equations become way cleaner.
You're the second commenter to bring up rpn. Makes me want to play with the notation now.
@@linguamathematica2582 Now that I thought again about reverse polish notation I remembered that it might match the topic of associativity in a deeper way still, since it is a way to write calculations without any bracketing and in some way it even introduces a natural order in which to perform operations thereby in some way making the associativity rule not necessary anymore. Of course your point was that it is not really only a rule but some deeper principle or property, so one still wants to keep that, naturally. It would not be a deeper principle if it vanishes by a change of notation of course. Btw. quantum mechanics also made me understand commutativity. And dual vectors too. Maybe they should teach it in schools already.
Another thing that came to my mind again is that John Baez was looking for a good layman example for associativity. To quote him: “[...] while it’s very easy to imagine noncommutative situations-putting on shoes then socks is different from socks then shoes-it’s very difficult to think of a nonassociative situation.” If, instead of putting on socks then shoes, you first put your socks into your shoes, technically you should still then be able to put your feet into both and get the same result. “The parentheses feel artificial.”
Associativity is involved in the number systems of the reals, the complex numbers, the quaternions, but not the octonions. It is like creating numbers according to the same scheme ( en.wikipedia.org/wiki/Cayley%E2%80%93Dickson_construction ) but always losing some algebraic property on each step, first order, second commutativity, third associativity. This appears kind of as a hierarchy of properties.
@@scepticalchymist Hmm, I like this idea of the hierarchy of properties. Thank you for this wonderful read
My favorite example of your point is exponentiation, something that is not associative. 2^(3^4) does not equal (2^3)^4, but you kindof can create an exponential group if you make each element, n, of a group the act of raising something to the power n. Now 2&3 means raising something to the second power then raising it to the third power. 2&(3&4) means squaring something then cubing it then raising it to the power of four. If that is the case the exponent group is associative. Squaring and then (cubing and then raising to the 4th) is the same thing as squaring then raising to the 12th or raising to the 24th. Likewise (squaring and then cubing) and then raising to the fourth is the same thing as raising to the 6th and then to the 4th or raising to the 24th.
When I heard the challenge at 2:22, I was expecting something about binary vs ternary operations, and on to higher numbers of operands, with something like the idea that lumping arbitrarily many things together is an N-ary operation where all the sub-lumpings are somehow interchangeable. But I couldn't figure a way of saying that without the forbidden words (or their implicitly forbidden equivalents). A process with steps and sub-steps is one way of looking at it, but it doesn't feel natural in the familiar cases of addition and multiplication.
Sounds like confirmation bias, especially since none of the non-associative algebras are considered. Octonions are a good case in point. But for just a very primitive example, consider the case of (a*b)+c vs a*(b+c). These are two different operations, but the logic discussed in this video really doesn't rely on anything that would depend on whether the operations are identical or not. Otherwise, that would clash with discussion of function compositions, seeing how addition and multiplication are both simply different functions from RxR to R. Understanding where this all breaks down leads to understanding why associativity is an axiom of the group (well, smigroup) and why that matters.
If you apply a to b applied to c, then it's the same as when you first apply a to b, and then apply that to c.
Is this notation consistent with all this? I think it is.
* a b c d
where * is an associative action and a,b,c,d... are objects that can be acted on by *
means
- pick any two consecutive objects (call then N and M) and allow * to act on them
- replace N M with the result of N*M
- repeat until the set (a,b,c,d ...) is now one value reflective of the combination of all the initial objects
Excellent! I well hooked onto your content! I am eagerly awaiting your next upload.
Ps: I like that you cover such seemingly mundane topics in si much depth.
Thank you so much! I definitely enjoy uncovering the inner depths of the seemingly mundane.
I think one of the huge things you missed here was why the associative property exists in the first place and why it needs to be pointed out. Mathematical binary operations (+,-,×,÷, ect.) take exactly two operands, no more or less. And there need to be a reconciliation between that fact and the desire to use more than one operation (binary or not) and more than 2 operands in an expression.
Associativity is a property (a unary relation for functions).
Associativity is a property belonging to Categories (more generally a Semi-Groupoid or Semi-Precategory). The morphisms of the structure, map between its elements, and connect in a directed fashion (morphisms are referred to as arrows in a category)- (it is also common to say categories are related to Quivers / Multi-edge directed Graphs)… Anyway, we can draw a category with a directed multigraph- I would argue this is a form of abstraction, but, it doesn’t matter. The resulting graph shows the connectivity of the arrows, which allows us to visualize some properties.
Regardless. Morphisms inherit the property of transitivity, which is what allows us to chain arrows together.
Morphisms are not limited to mapping between elements of the structure; morphisms can map between the morphisms themselves (not technically). Those morphisms are referred to as Functors (or “cell-1” arrows I think…).
Lastly, a Homomorphism is a morphism is a morphism of a specific form.
F( x # y ) = F( x ) • F( y ),
# and • are arrows that relate the elements,
while F is a functor that transforms them. (or I got that backwards).
Anyway, homomorphisms allow for us to have a transformation applied before ~OR after a relation (like an operation), and end up with the same result:
Which is more or less Associativity.
Functions are homomorphisms, and Operations are Functions- meaning, that we have access to that form in one way or another.
Which is why “+” , “*” , … whatever operation, might have some associativity.
Not all operations are associative I believe: remember a Homomorphism allows for “#” and “•”; whereas, if it were the associative property - those would probably be the same base operation.
Example:
If F(a) -> a, then:
( x * y ) = ( x ) * ( y ),
“*” happens to be associative, but there might be additional requirements (perhaps domain/ range specifications).
Example 2:
If F(a) -> log(a)
Log( x * y ) = Log( x ) + Log( y ),
While this is a homomorphism, there is no clear associative property.
This is better as a test or example, than a formal definition of the associative property. I don’t know the formal definition, but I bet it’s pretty cool- probably not as round-about.
But yeah, associativity is totally a property. It’s used to name Magmoids / Magmas, and its importance is significant.
Cheers
a(bc) = (ab)c
first, b times c. Then multiply that product by a; this is equivalent to saying, first a times c, then multiply that product by c.
First axiom: A is absolutely identical to A.
If p is to q and q is to r then p is to r.
Logic and basis of union/intersect as property of sets i.e. the abstraction of the member/arrangement of a set.
The way math is taught benefits from a primer on the logic of arithmetic.
It is interesting that if the operation of acting on a state would be division by a number, then the & operation would turn out to be multiplication.
Doing the challenge before finishing the video
An associative operation is any operation wherein any object can be described to be composed by some sub objects via the associative operation, which can either be described compositionally and or are considered base units of the operation, an operation is associative iff some object is described by a composition of only base elements, all elements of the composition are exchangeable with any other base element in the composition so long as each base element never occurs the same number of times in every composition
This is probably wrong but should serve as a distinct definition of how I currently think of the property given the provided restrictions before watching the video
I might have broken some of the rules or at least their spirit by using the idea of compositions and objects with sub objects which is somewhat synonymous to groups but it’s the best I got lol
There appear to be a few inaccuracies. The linkages that lead to "getting the milk" are associative if you think of them as connecting film clips together. But if you understand them as real actions, the execution is not possible. But associativity actually means that you should be able to do the subsequent linkage before the first one in order to achieve the same result.
In addition, in my opinion it is more interesting to also consider non-associative linkages, especially commutative, non-associative linkages. For example, if you define: "(x linked to y) is equal to ((x to the power of y) times (y to the power of x))" then this is a commutative linkage, but it is not associative. On the other hand, if you define: "(x linked to y) is equal to ((x divided by 2) times (y divided by 2))" it is both commutativ and associative, although both differs from usual multiplication.
I disagree with your formulation of the object-action duality. When you transformed an object to an action, you didn't use the operator, them you ignored the operator, but it was still there. I think that it would be better to partially apply the operator to the object to create an action. So the expression “2 + 3” could be interpreted as “2 (+3)”, the object 2 and the action of adding 3. Instead of object-action duality, more accurate would be partial application, that a binary operator can be applied to a single object to create an action on another object.
I this commented is underestimated.
Wait i have an idea on how to describe the associative property to someone , if they know how to find the area of square and rectangle , Imagine a sheet of paper that has side lendth of B cm and C cm wide the area of this rectangular sheet of paper is BC now imagine if now the same paper was stacked on top of each other to make its height A then the volume occupied by the stack will be ABC, now image a sheet of paper placed vertically and has side lendth of A and C now area of this will be AC now stack them to besides each other such that the lendth of the stack is B then the volume of this will be BAC and both stack look same there for ABC=BAC.
I think the & operator makes a lot more sense when you're familiar with procedural programming. The way I've understood equations for a long time is that they represent two ideas. One is plain equality, and the other is an algorithm. One tells you two things are equal, the other tells you that you can build one side to be equal to the other using symbol manipulation. The & operator just makes it clear that it is the abstract idea that makes the algorithm move on from one step to the next step. In programming, you assume the & symbol to be the next line (Python), the semicolon (languages inspired by C), or any other terminating character. This is made a bit more clear when you get funny symbols like ++, for loops, or my favorite, the ternary operator ?: that indicate multiple instructions wrapped into one expression. That nice wrapping is associativity.
Oh, that is a wonderful take! I hadn't considered that multi-instruction syntax (like ++) is really a sort of shift of parenthesis. There is another way to think of associativity in the context of coding. The order in which you write the instructions doesn't matter. This is only the case because the concatenation of lines of code is associative.
An associative operation is one whose steps can be rearranged without affecting the outcome.
I guess "step" isn't a synonym for "order" but the fact that they can be rearranged implies there is an order which doesn't matter.
Imagine you have three boxes labeled A, B, C and a machine that takes two boxes and combines them into one box. The two boxes the machine takes are inserted into two holes, one on the left side of the back of the machine, and one on the right side of the back of the machine. The machine will spit out the combined box from the front. Associativity describes a property the machine has. If you put B in the left hole and C in the right hole, you get out a box I'll call D. Then if you put A in the left hole and E in the right hole, you get out a box I'll call F.
Now, let's say you put A in the left hole and B in the right hole, to get a box I'll call G. Then put G in the left hole and C in the right hole to get a box called H. To say the machine is associative is to say that F = H.
Very great for your first video! However, I find naming the operation you defined as "and then" quite misleading. First of all, "take x, and then multiply y, and then multiply z" is not sufficient information to follow instructions in non-commutative cases. Thus I wouldn't want to call multiplication an "and then" operation; to fix this, we should consider "multiply from left" and "multiply from right" as two distinct operations.
But even then there are natural examples of operations which are not associative, such as multiplication of octonions. I can multiply octonions from left in succession, which I'd intuitively think as "and then":ing left multiplications together, but yet the operation is not associative.
What you really defined is the idea of action, more specifically right action. This is usually defined as group acting on some set (as in your example dihedral group acting on a triangle), but the notion of action can be generalized to any associative algebra (i.e. algebra with associative operation), or even non-associative algebra as far as you require action of (xy)z to be equivalent with the action of x(yz), although I'm not aware of any natural examples of such actions. The name "action" still has the issue that some things I'd intuitively think as actions are not actually actions (e.g. left multiplication of octonion), but that's just the limitation of our words and highlights the necessity of precise definitions.
Lots of good points here! Let me try to address them individually:
1. "and then" on non-commutative operations: I agree with you that "take x, and then multiply y, and then multiply z" is ambiguous for non-commutative products. That said, this observation is not important to the "associative=and then" conclusion. The conclusion states that each associative operation is an "and then" operation on _some set of actions_ (the right actions were a constructible example), and not that the associative operation is an "and then" operation on _all actions_ . I did not make that clear in the video, apologies about that. That said, we could perform all the steps on the left actions, and this would lead to the Cayley's theorem in it's original form: that associative operations are function composition.
2. Octonions are non-associative: This was my response to a similar comment demonstrating that being "and then"able is not the same as being "and then":
"
This is a distinction that probably should have been made more explicit in the video. Associativity does not mean that you _can_ "and then" the actions (as you've pointed out, this is always the case). Instead, associative operations _are_ "and then" operations. This is precisely what I hoped to convey with "associative = and then"
"
3. Generalizing right actions to non-associative algebras: Absolutely! That is an excellent insight. It's the reason that object-action duality is generalizable to any binary operation. Note that my response to 2 demonstrates this is not an indication that non-associative operations are "and then" operations.
4. Left multiplication of octonion isn't an action: Fortunately, this is an action! Let [a*] be the action of left multiplication by 'a'. It is an action that maps each starting state 's' to the end state s[a*]=a*s.
5. "the limitation of our words and highlights the necessity of precise definitions": it seems I need some work in this department. I think I sacrificed clarity for flow. I'm still not sure what the right balance is since I don't think I want my videos to be lectures but rather to be expeditions where we unveil some core idea. Maybe I should publish the precise definitions in a document along with the video, but then I would take even longer to get a video out 😅
@@linguamathematica2582 I'm not saying there was anything wrong with contents of your video - you even gave the formal definition of "and then" operators, which is great! - just that some clarifications and exploration to when things break down would've completed the discussion. Now I feel like the subject was slightly oversimplified, leaving the impression that "and then":ing is always associative.
4. What I meant is that left/right multiplication of octonions isn't action in the sense action is usually defined. To make this more precise, we should make the distinction between object and action more explicit. Let O be the set of objects and A the set of preactons on O (which we later turn into actions), which formally is some collection of maps a:O->O. A priori objects have no further structure, so they can't be combined/glued.
If for any two preactions a,b there exists a third preaction c such that for any object x we have a(b(x)) = c(x), we can define multiplication on the set of preactions as ab=c. Preactions together with such multiplication are called actions, or as you can them in the video, and-then operations.
The object-action duality you demonstrated in the video is the special case where the set of objects and the set of actions are in some sense "the same". More precisely, there should be a bijection f:O->A such that (f(x)f(y))(z)=f(x)(f(y)(z)) for every object x,y,z, in which case action a can be identified with the object f^(-1)(a) and object x can be identified with the action f(x). For example, in the case of left matrix multiplication, each matrix A is identified with the map A(B)=AB; on the left hand side we have map evaluation, on the right hand side we have matrix multiplication.
Now, let's explore this duality in the case of left multiplication of octonions. From now on, I'm denoting octonions as x,y,z. The set of objects is octonions, and the set of preactons are identified with the set of octonions by left multiplication similarly to the previous example. For this to be an action, it should satisfy x(y(z))=(xy)(z) for any x,y,z. However, this would imply associativity of octonions, hence cannot be true in general. We thus conclude that left multiplication of octonions is only a preaction, not an action.
As a final remark, note that the construction of the left multiplication preactons for matrices and octonions were constructed following the object-action duality, yet only for matrices the preaction turned out be an actual action.
A gift of seeing further then the rest of us ... I will be waiting for new insights from you ... Thank you!
I do not see why 3D vector cross product would not be an "action" or an "and then", but it is not associative...
I'll show it for division which is also non associative and easier for my keyboard. You can't just replace '/' with 'andThen', 1 & 2 & 3 doesn't make sense because 2 and 3 are not operations. What you can do is 1 & /2 & /3, and that is function composition so it IS asociative.
Thank you for this video! This discussion is a starting point for addressing problems that font designers have with *exponential work*! Do you have any ideas for making work faster or more manageable, when you have to work along 8 or more dimensions? What type of math could be used to think about this?
associativity = "left multiplication by a commutes with right multiplication by b" or simply "multiplication on the left commutes with multiplication on the right"
I really wish I saw this before my abstract algebra midterm. It could have saved me pages of scrap work.
"An operator is associative if the result is independent of the path taken."
Examples:
- After concatenating, we can't tell which seam was most recently glued.
- After applying rotations or flips to a triangle, group theory only cares about the orientation we end up with, and doesn't remember the path taken.
- In a function application such as f(g(x)), the function f can't "reach into" g to help it make decisions, so it can't depend on what happened within g. Similarly, g can't "look outward" into f.
- When applying a non-associative operator like / or √, we have to look around more, essentially locating the parentheses or other context clues like line length.
The last two are perhaps better thought of as "context sensitivity / insensitivity" that "path dependence / independence", but I think they're worth considering as one concept.
My attempt: “An associative operation is a binary operation that can be applied to arbitrarily many elements by nested evaluation, with an identical result no matter how elements are paired during computation.” It still comes out sounding like commutation. lol harder than I thought.
Can I get another video on the I-LATE rule of the integration?
Here's my attempt at the challenge at the start of the video before I watch more of the video:
An associative operation is a binary (2-ary) operation on a set to which we can naturally associate an n-ary operation which is related, in a canonical way, to repeated application of the 2-ary operation. For example, we can define +_3 such that +(+(a, b), c) = +(a, +(b, c)) = +_3(a, b, c).
*Freud* "all dreams have a meaning" *my* *dreams* : 12:30
I'm not sure this video shows the essence of associativity. The essence is that associative operation is independent of the objects it acts on. It doesn't use any "information" from the objects to change what it's doing.
It's an interesting and potentially useful way of thinking about things. It was a bit of a weird video, because you invent a new framework (objects and actjons) without quite stating that it's a new framework rather than statements about existing ones. Then you explore associativity mostly in this invented framework rather than in the ones we actually use it in.
I still think it's a good video, I just think you could stand to make your intent clearer and explicitly stating when we're leaving math and entering wonderland. Obviously there IS no object-action distinction in mathematics. You could separate what numbers mean into those two categories, or a different two categories, or 1 category, or n categories. Choosing objects and actions and then showing they're equivalent is kind of the logical equivalent of an identity operation :)
My explanation would be something like this:
The relationship between any pair of variables has the same importance as any other pair of variables.
To answer your challenge: A binary operation • is associative if for any x when you treat it as a unary operation x• function composition corresponds to applying the operation to each x
Would be cool to look at some counter examples that are not associative
Wonderful. A very great video :)
Thank you! And nice video collection you have there 🤩
@@linguamathematica2582 Thanks as well :)
this video is GOATed
this is a fairly natural way to think of binary operators with a background in functional programming, its basically considering currying!
Yaaaaaaasss!! Currying is such a profound idea
What an amazing video! This will definitely help me gain an intuition for my abstract algebra class
Object-Action duality is an excellent idea indeed. However, in the presented form, it's incomplete E.g. if fails to explain why subtraction is not associative. Indeed, what stops us from interpreting 2-3-4 as takeObject(2).andThen(subtract 3).andThen(subtract 4)? The root cause of the problem is the starting object. For the operation to be associative, takeObject has to be consistent with the nature of the action (here the "action" is subtraction). Namely, takeObject(x) has to be equivalent to takeObject(identity).andThen(action(x)). In the case of 2-3-4, the action is subtraction, and the "identity" is zero, so takeObject(identity).andThen(subtract 2) does not give 2 as we hoped, which destroys all further logic.
In fact, "identity" is the only object we are allowed to "take from thin air" - the rest of the objects should be obtainable from it through the defined "action". In general, this requirement can be expressed as an axiom x=identity.action(x). You can easily verify that this is the case for every associative operation considered in the video, it's just the "identity" is specific to the operation in question (0 for addition, 1 for multiplication, an empty string for concatenation etc.)
Let's take (10-2)-3=5: I have 10 apples, I subtract 2 _and then_ subtract 3 I get 5. So while it is possible to say this in words and it will make sense, this doesn't make subtraction associative: 10-(2-3)=11. What am I missing?