One of the proofs at the start of my first abstract algebra course showed that the "basic" associativity (ab)c = a(bc) implied general associativity for finite expressions via structural induction. In other words, an associative operation is one in which parentheses are no longer important, extending a binary operation into an n-ary operation as long as the sequence of object symbols is constant. So with that in mind, let's take a swing at 2:16 Suppose you have a string A1☆A2☆...☆An and a closed computation of strings of the form X☆Y. Let P be an algorithm which 1) Chooses an arbitrary substring of A1☆A2☆...☆An of the form X☆Y, 2) Computes the substring in accordance with its binary definition, 3) Substitutes the computation for X☆Y, 4a) Loops if the new string contains any ☆ symbols, or 4b) Terminates otherwise The operation ☆ is associative if and only if for every natural n>1 the string A1☆A2☆...☆An reduces to the same result for every such algorithm P meeting the above 4 properties. In fewer words, "☆ is associative if and only if well-formed formulas of n-ary instances of ☆ are well-defined under arbitrary recursive parsing algorithms which evaluate well-formed sub-formulas of 2-ary instances of ☆."
Did I get this right that this means that every example of an expression using the well-defined n-ary operator * can be parsed much like a compiler (recursive descent parsing) such that each token of the expression will be either the LHS or the RHS of potentially n other RHS of an instance of the binary operator * given a reading direction from left to right?
This comes across as more of a philosophical look at associativity than a mathematical one. I think it would be a struggle to define objects and actions in a meaningful way that provides any mathematical context, as mathematical constructs are by their very nature able to represent a great many things, far more than just the “dual” in duality. Objects and actions as described in this video are far more tangible than abstract, and lean more toward philosophical constructs than mathematical ones,
You've picked up on my penchant for philosophy 😅 My view is that philosophy and mathematics are complementary. A great philosophical argument can often be reduced to a mathematical argument and important mathematical insights often lead to great philosophical insights. Since I've leaned toward the philosophical construct in the video, let me fill in the mathematical constructs here. The mathematical context was hinted at in 13:39 - 20:00. Let's make it more rigorous by grounding objects and actions in set-theoretic constructs. Objects are sets, elements of a set, and variables; actions are functions which map input states to output states (in general, any function can be interpreted as an action). What about the object-action duality referenced in the video? Here is my response to a similar comment where I supply a mathematical description of object-action duality in addition: "We could represent real number addition as +:ℝ^2→ℝ, and express 2+3=5 as +(2,3)=5. If we peer under the hood of the object-object interpretation, this is the notion of + that is being used. We can also represent addition using currying [+]:ℝ→[ℝ→ℝ] which I'll bracket to distinguish it from the other representation of +. So the action of adding three we could write as [+](3), but that is a bit messy so I will write it as [+3]. The key is that since [+] is injective it sets up a correspondence between numbers and a subset of functions ℝ→ℝ. This correspondence is so natural that most don't bat an eye when using 2+3=5 to describe "starting with 2 apples and then adding three more to get 5 apples.” This number-function correspondence is the instantiation of object-action duality used in the video. With these definitions, we can be more rigorous about the different interpretations of 2+3=5 by using different equations for each interpretation. The more explicit equation for “Starting with 2 apples and then adding three more to get 5 apples" would be [+3](2)=5. And the more explicit equation for "adding 2 apples and then adding three more is the same as adding 5 apples" is [+3]∘[+2]=[+5]"
I’m glad I made my comment. I enjoyed the video but for me it was missing just a little something to give me the complete satisfaction I get from understanding a mathematical problem/system, and your reply tipped the scales back into balance (regarding the mathematical/philosophical approaches) and provided that little something. Awesome stuff and thanks :)
@@linguamathematica2582 that was a thorough response. Is this a new channel? I don't see you've done other videos. Which is ashamed because you're logic is sound. I wonder if what made some of the greatest mathematician great was that they understood mathematics as an extension of philosophy and could make connections between the two subjects. The first ever functor of knowledge from one subject to another. 🤷♂️
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
This is an example of a very simple idea completely losing meaning because it became a rule and people used it only for its usefulness instead of its meaning. Thank you for recovering why associativity is used in the first place.
It is not so arbitrarily, for example, the operation (a, b) = a-b, is not well defined for three objects, that is, (1,2,3) = - 4, but it can also be (1,2,3) = 2, since -4 = (1-2) -3 and 2 = 1- (2-3), That is, you operate the same objects, but they give different results
EDIT: i now realize i made a fool of myself. and this is commutativity. multiplicative: if you take a -cube- rectangular prizm with side lengths a, b, and c, no matter how you rotate it, it has the same volume additive: say you have a shelf that is packed full of books, each with different thickness [a, b, c, d, e,...] , and you take one out randomly, then need to put it back in you don't have to remember _where_ you put it, you can just make a space and jam it in there, and it is guaranteed to fit.
I think those are demonstrations of the commutative property of addition and multiplication, and not actually the associative property of addition and multiplication associativity (without commutativity) implies that a function has discrete, different first and second inputs, but in a case with three inputs A, B, and C, taking the output of A as a first input and B as a second input and feeding it in as a first input, with C as a second input, gives the same result as using B as a first input and C as a second input, and feeding that result in as a second input, with A as a first input or, simpler, A☆B is not B☆A, but A☆B☆C can be defined as either (A☆B)☆C, or as A☆(B☆C), while (B☆C)☆A does not give the same result, nor does C☆(A☆B), or (B☆A)☆C, etc etc or, associativity seems to say the order of inputs matters, but the order of operations does not it's hard to demonstrate without finding a function which is associative but not commutative, but they exist. Say, for example, concatenation. I think it's a brilliant example. (A||B)||C = A||(B||C) = A||B||C (= "ABC") but A||B != B||A ("AB" is not "BA") anyways, ramble over
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.
That was a really cool video. I'll be honest, at the beginning I thought it was all pretty obvious, but by the part of the video when you related all of it to group theory I realized just how important it was for you to really concretely define what "and then" really is. You explained everything very clearly and the visuals made it very easy to understand. I really think I understand this topic much more clearly than I did before, even though I thought that I already understood it well.
Seems odd to forbid the word "order" when challenging us to explain associativity, then you proceed to use "and then" and timelines, which are very "ordery" things. I stayed more object- than action-oriented and considered the natural input type of the variadic extension of an operation. A general operation takes a binary tree, an associative one takes a finite sequence, and an associative & commutative one takes a finite set [edit: multiset]. But I had the same criticism of this -- a sequence is an "ordery" thing.
Quoting 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 love the binary tree idea for a general operation.
I think Qhartb spotted a confusion, an interesting contradiction. I'll try and explain it myself. Faced with an expression such as 2 + 3 x 4, we see that as it stands it's 𝒏𝒐𝒕 associative, that (2 + 3) x 4 ≠ 2 + (3 x 4). But it's precisely this non-associativity that gives meaning to an "and then" device, such as parentheses, to deal with the fact that order of actions makes a difference here. The left side now says add then multiply, the right says multiply and then add, and the whole inequality says the difference in order means a difference in quantity. On the other hand the equality of (axb)xc to ax(bxc) merely shows that "and then" is redundant to an associative operation like multiplication.
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.
Formalization: every semigroup (i.e. set with an associative, binary operation) is isomorphic to some set of functions under the operation of function composition.
@@linguamathematica2582 Honestly, I'm into logic, computer science, set theory and stuff, and I still don't understand how categories can be useful or insightful. I don't really think a non-mathematicians would get something out of those.
@@An-ht8so Yeah, that makes sense. Category theory isn't for everyone much like learning set theory isn't for everyone. My personal motivation for learning CT is two-fold. The first is my immense appreciation for the power of abstraction. This was a slowly dawning appreciation that at this point is a core part of my mathematical style. It came from studying subjects like abstract algebra, topology, differentiable manifolds, Lie algebras, and wedge products. The second motivation is my love of deep learning and finding connections between seemingly unrelated ideas. Sometimes, category theory can feel like soaring over the landscape of math and seeing the landscape as a bird might. You notice patterns that you would have never noticed from walking on foot. That feeling is, honestly, exhilarating.
@@linguamathematica2582 If you were to make a video on category theory I'd watch it in a heartbeat. I've taken group theory and I can tell how the 'Application: group theory' section was essentially an intuitive version of Cayley's theorem, and I understand that the Yoneda lemma is essentially that but way more general, although I've never been able to fully wrap my head around it properly. But after watching this video, I'd imagine it'd have something to do with how categories are associative by design.
This "object-action duality" infix notation feels misdirected to me; the actions described are operator sections, such as (*2) which is shorthand for f(x)=x*2, and painting them as distinct from the operator breaks that. When painting your numbers red, "2×3=6" vs "2+3=5" has erased the operation type on the right hand side, so how were we to know those where Product 6 vs Sum 5? Even transforming the first term similarly requires an implied monoid identity element, i.e. 0 for Sum or 1 for Product, determinable only by reading ahead to the operator. I think it would make sense to transition to prefix notation first to observe that it's the same operation working on all operands.
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
Associativity is also a coherence requirement. Its meaning is that the result of calculation doesn't depend on the path taken for calculating. Or better said all paths are equivalent. As an example of a non associative process, take the transformation of velocity through frames in special relativity. With the same starting frame and velocity and the same ending frame, you would get different results depending on the intermediate frame used.
@@linguamathematica2582 my previous answers don't appear, probably because they contain URLs. I don't have a direct reference in mind for your question. Maybe some works that try to "correct" the non associativity of STR may help, such as gyrogroups for relativity or categorical relativity.
Normally I like to praise people for their stories, but as you finally noted, you shared your journey with me, where a story is merely a memory of a journey. Well done! As a computer programmer with over 50 years of experience, this was a refreshing journey that helps me better appreciate my own journey. While learning functional programming back in the 70s, it was not until 2005 and later with Scala did I start practicing functional programming. In particular, I have seen dozens of explanations of Monoid and Monad, and they rarely helped, they usually hurt. This journey has really helped me understand monoids and monads better... thank you. I hope to see some of your presentations on Functional Programming, in particular grounding like this.
Nice work. You explain it so much better than any RU-vid video I have ever seen. It’s also explained so dryly by most teachers as I think even they don’t have an intuitive grasp of what the associative property really means. Keep it up man. 👍
Very interesting, I love 'intuition first' maths. However it would have been good to talk about non-associative operations also, maybe to show how for these the 'and-then' interpretation breaks down.
@@heinrich.hitzingerMatrix multiplication is ALWAYS associative. It's a major point of frustration for me as I try to understand octonions, which are non-associative. I know that they cannot be represented with groups, functions (aka mappings), or matrices because these things are associative.
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.)
Paused at 2:20 for challenge. My take: With a general binary operation, (which combines two inputs to produce one output), you can combine many inputs together by constructing a nested expression which has the form of a tree. In general the structure (aka the choices of how you connect branches) of the tree matters quite a lot in determining the final result, but for an associative operation, it doesn’t. Any two expression trees that start with the same list* of inputs will give the same final result regardless of the structure of the tree itself.
I think it's worth it to clarify "same list of inputs" as "same leaves in a left-to-right tree traversal". It seems like this is equivalent to the top comment with the 'any algorithm' argument otherwise, except a bit clearer and more concrete.
At 20:11 when the dark background becomes white is probably the most epic thing I've ever seen! No better allegory for enlightenment than literal enlightening. But apart from that, this video is a bloody masterpiece! The first time I really started thinking about what it actually means was when I learned that the operations of a Rubik's cube have the structure of a group and I wanted to explain to someone why the associative operation actually holds, but I've never thought about it that deeply before
This video was so cool and inspiring! I'm only learning high school mathematics right now, yet this presented an entire new topic i would be interested in learning in the future. It's really a shame that this only has 1500 views. I think more people shoud see this. Thank you!
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.
"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.
3 года назад
The associative property is an attribute that certain operations possess if and only if for any family of elements whose elements are ordered in a given permutation (such a permutation indicates the order in which such elements are to be operated upon ((1) each one with the one immediately before or after it, if such an element exists)) and are part of a set where such an operation is well-defined, the result of the overall operation does not depend on the arbitrariness in the algorithmic process of achieving the final result alluded to in (1).
Ooh, I like that you brought in the idea of permutations into this! I get the feeling that you are be uncovering something pretty deep there.
3 года назад
@@linguamathematica2582 Do you have any ideas as to what these deep connections would entail with this and other properties (and the act of applying operations to the elements of an ordered set)?
@ Nope 😅, it's just an intuition. A few others have mentioned path invariance, and I felt that it's a good start but that it's missing something. When you said permutation, it went partway into filling in that hole. It is also motivated by analogy with concepts like Galois groups, symmetry groups, and fundamental groups. Finding the underlying group of symmetries seems to lead to deep insights, and the set of all permutations is a group! Seems fitting that you've used a group to describe the symmetry of associativity.
this seems to be the interpretation of a semigroup/monoid/group (G, *) identified with the actions on a set X such that for each element g in G we define an action T_g : X -> X such that composition of actions T_g ° T_h = T_(g*h). Then _any_ associative operation * on the elements of a set G can be identified with just composition of the element's action. Composition is essentially what you describe as this "and then" operation.
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
1:38 "until we meet our first non-commutative operation" So many people seem to forget that subtraction, division, exponentiation, and logarithms (where the base and the number in the base are the two ) are not associative. The first two make sense because everyone just thinks of them as addition with negative numbers and multiplication by reciprocals. I guess maybe exponentiation confuses people because it's just repeated multiplication in the case of positive integer powers (and uses reciprocals and 1 for negative integer and zero powers), and people use the notation for repetition of the binary operation in group theory and probably elsewhere in abstract algebra. I also guess logarithms are usually thought of as a set of different functions (a different function for each base) rather than as a binary operation.
2:12 Suppose you have some books that you want to stack on top of each other, all of them have different sizes and you want to stack them in a way that keeps the larger books on the bottom and smaller on the top. No matter how you choose to stack, you will end up with the same configuration in the end. This is actually both commutative and associative. To clarify further let's take an example with three books, say A, B and C. A is the smallest and C is the largest. Now whether I make the stack by comparing A and B first and piling A on top of B, and then piling both of them on C or if I pile B on C first and then I pile A on them. it wouldn't really matter in the end. Now let's take an example with of non-associative order. Suppose you choose some points on a circle. And you try to sort them by comparing them according to the clockwise direction. That is, let's say we have two points A and B on the circle, If the shortest way to reach B from A by moving along the circle is by going clockwise, you will put B before A. but if you going anti-clockwise is as short or shorter, then you put A before B. By ordering according to this, the final result would be different depending on the initial list and the sorting algorithm used.
*Great Video* . Thanks for making it. Here is the insight I had when learning about *Associativity* . *Associativity* is just a *TRINARY* operator built on top of a binary operator. Since trinary has to be made of 2 binaries, order of operation may (non-associative) or may not (associative) matter. As a kid, examples of both help - add/multiply are associative while power is not. (2+3)+4 = 2+(3+4) because both represent 2+3+4 (trinary operator ++) vs. (2x3)x4) = 2x(3x4) because both represent 2x3x4 (trinary operator xx). Power is a counter-example. (2^3)^4 = 8^4 = 4096 NOT = 2^(3^4) = 2^81 (^^ is not a trinary operator because order of binary operations matters). *Conclusion* : A binary operator is *Associative* if it works correctly as a *trinary* operator (meaning that the *order* of binary operations does *NOT* matter). This makes a binary operator take any number > 2 of operands (not just 3/trinary) and become an N-ary operator.
The associative property is what allows us to understand what doing something more than once means. Also, temperature is not a measure of heat. It is a measure of thermal energy, which is not heat. Heat is a process that is done, in the same way that work is a process that is done.
Oh, nice catch! Yes, temperature is more accurately described as a measure of thermal energy. That said, "thermal energy" lacks the same accessibility that "heat" has :( Maybe temperature is "a measure of the potential to transfer heat"? What do you think?
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 :)
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.
I feel like this video defines an operation on the object representation, an operation on the action representation, assumes implicitly that there exists a bijection between the two representations, then shows that one being associative implies the other is. But the action representation changes the objects into functions, and the "and then" operator is just functional composition (which is known to be associative). I think the important part is actually the existence of a bijection between the objects and actions. If you can show that the binary operation on your objects is equivalent to a composition of functions, then your operation is associative because functional composition is associative. It's showing that it is equivalent to a composition of functions in the first place that is important. The reason you could express the entire expression as either objects or actions was the existence of this bijection. "1+2+3" is equivalent to "s +1 +2 +3" for some initial state s (where +n can be thought of as a function that increments the value by n). If you set the initial state to the identity (0 for addition), you get the same output from both expressions. In contrast, if you try this for a subtraction operation, this won't work due to the lack of a bijection. You will find that x could map to either +x or -x depending on whether it is the first object in the expression, so there is no way to create a bijection between the objects and actions. So I guess to go back to the challnge to define associativity without reference to "order" or related words, you could say that it means "A binary operation * is associative over a set A if there exists an object-action duality for A under *". Potentially this could be expanded with a proper definition of "object-action duality" over a particular set under a binary operation.
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.
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.
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.
Another way to arrive at the conclusion at 24:38 is by Cayley's Theorem, which implies that every group is isomorphic to a subgroup of a symmetric group. In other words, any group can be thought of as a non-empty set of invertible functions that is closed under composition and inversion, with composition being the operation of the group.
Thanks for this beautifully crafted explanation. I just started a graduate course in groups and reps for physics and this shed some much needed light on the group axioms. I might even have to take a course in category theory now!
This made my day! A huge motivator for the channel is to "shed much needed light" on the topics in math that are mostly understood/explained syntactically without a sense for the deeper meaning. To see a physics grad resonate with this message is everything I hoped for
Category theory actually has an even deeper version of this called the Yoneda lemma, so if this sort of thing interests you, I'd definitely recommend learning some category theory.
Recently myself and some internet acquaintances were playing with a recreational number system with 3 signs (I call them plus, minus, and back), and quickly found that it was non-associative, both in addition and multiplication. In trying to derive some way of solving equations, I discovered that the seemingly trivial functions f(x) = x * a, for various possible "a", could be used to gain back the associative property. If a given multiplication problem is converted to repeated function application, obviously the function application itself is associative (though it gives up commutativity). That was a very vivid lesson that "associativity is action". The action of multiplying, in fact, forms a group, despite the underlying operation of multiplying not doing so. (Our multiplication has no identity element, due to the complete symmetry of the three signs with one another.)
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
Exponentiation isn't associative but it could be with a modification. First note that a*b=exp(ln(a)+ln(b)) which gives us a way to convert multiplication down to addition. Then we can look at how to convert exponentiation down to multiplication in a similar fashion. Unfortunately, it doesn't work without a modification. Specifically, a^ln(b) = exp(ln(a)*ln(b)). So now you could define a new operator # similar to ^ such that a# b := a^ln(b). You'll find that (a# b# c=#(b# c) because a^ln(b^ln(c)) = (a^ln(b))^ln(c) so this new exponent-like operator is associative. Note that it works with other bases so there may be a more useful choice of log base. I only used e because I noticed this while working with complex numbers and their polar forms. I have no idea if this is in any way useful, I just thought it was interesting. It does break the simple pattern that multiplication is repeated addition, exponentiation is repeated multiplication, tetration is repeated exponentiation and so-on. Given that, I'm not sure it makes sense to try to continue this type of modification to tetration and beyond but I'll probably attempt it with tetration just to see if there's a different pattern there that makes sense. It also makes me wonder about a multiplication like operator that would be defined such as a@b := a*(e^b). Also, I wonder what addition converts down to.
I would have just used function composition rather than this idea of "and then". As that is effectively what function composition is. But I can also see an apparent counter-example. Consider (a/b)/c vs a*b*c. This is saying take a, and then divide it by b, and then divide it by c, vs take a, and then multiply it by b, and then multiply it by c. For the latter associativity means we can either take (a and then multiply it by b) and then multiply it by c. Or, we can take a and then multiply it by (b multiplied by c). But this doesn't hold for the former, as division is not associative. I think any good video discussing associativity, especially new ways to look at it must discuss both things which are associative and things which aren't. This is to show that your alternative viewpoint easily allows you to see things which are associative as associative, as well as seeing things which aren't as not being associative.
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.
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 :-)
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.
Brilliant! This took a bunch of amorphous ideas I had (such as 'it is better to consider group members as actions rather than objects') and magnificently crystalized them. Thinking about the Rubik's cube is what led me to the idea of actions being a better (or at least alternate) view than objects. Combining two cube configurations makes no intuitive sense, but combining a set of cube operations does. Yes, please please do your object/action duality video. I suggest the Rubik's cube will be a good illustrative example.
Very cool video, but I was disappointed you didn't give any examples of non-associative operations and why they aren't "and then" operations. E.g. subtraction, division, averaging
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.
Imagine/draw two triangles with labels a, b, and c. On one draw a line from b to c (bc) in red, and a line from a to b in blue. On the other one draw a line from a to b in red, and a line from b to c in blue. In both cases, you have drawn the same shape (ignoring line colour) from a to b to c. Therefore: a(bc) = (ab)c -- QED
I love this subject. I don’t think there’s enough engaging content probing the more…discrete and logical aspects of math. Thanks for making this. I thought the climax of “Group Theory ≡ study of reversible processes” was very satisfying. While I’m sure it doesn’t totally describe the subject, it’s certainly allowed me to see its underlying structure more clearly.
Thank you! You're right that the description of group theory is imprecise (there are some technical caveats involving closure of the binary operation). A complete description of the subject is something like "study of composition of permutations" or the classic "study of symmetries". While they capture that extra closure, they fail to answer questions like: "why am I working with groups, again?", "what are groups doing here?", "should I use a group here?", etc.
This video can be summarized as "associative operations fundamentally act on sequences". The whole "and then" connection just arranges the objects into a sequence. General binary operations act on rooted trees, commutative operations act on unordered rooted trees, associative operations act on sequences and commutative associative operations act on multisets.
My hot take: Associativity of an action means that the action is context-free. It does "the same thing" no matter the state of the object. For example, consider the operation +* defined as "a+*b = a*b if a>5, otherwise a+b". So, depending on the value of a, +* is either addition or multiplication. This is not associative: ( 2+*3 ) +* 5 = 5 +* 5 = 25 != 10 = 2+*8 = 2 +* (3 +* 5)
This intuition breaks down when you look at functions that put their arguments in some sort of “container”. Let x~y := (x, y) then (a~b)~c = ((a, b), c) but a~(b~c) = (a, (b, c)). Another example would be a parenthesizing concatenation operator || where a||b = (ab). The first example is also the reason why the Cartesian Product is not associative.
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
Very interesting. Basically, "associative" just means "composition-like" or "actionizeable". You could say it like this. If * is a binary operator the set X, then there's a map actionize(*) from X to {f:X-->X} such that actionize(*)(a)(b)=a*b. Now the question is whether * acts the same way on X as the composition operator acts on the image of actionize(*). That would be equivalent to * being associative.
Yes, well put. Something I didn't mention in the video is that "and then" is function composition when read from right-to-left, so the two are very closely related. I prefer your phrase "composition-like" to "actionizeable" because it is possible to actionize using any operator (even those that are not associative) using a(b) = a*b without requiring associativity. But as you've pointed out, only associative operations will be related to the composition operator via a*b=(a)&(b)=(b)∘(a)
I watched this video first several months ago.. I really liked it then (and I still do), and I realized it was your first and only video, so I decided to wait until you upload some more videos like that, but it's still the only video, and even though it's one of the most interesting videos I ever watched in my life, I really hope you make some more. Thanks for this amazing experience, and I hope you do give us some more soon!🌺
First of all, I appreciate the video, and really liked the analogies you found between visual representations of multiplication and addition. The whole concept of visualizing the relations between actions and objects is also really interesting. Though, in my understanding, the associative property represents the opposite of an ‘and then’ operator. In a way, the associative property is saying that, when making additive operations utilizing operators that are in the same hierarchy, we don’t need to symbolize which operations must be done first. Because it doesn’t make a difference. As you know, we can optimize language so we can represent more information with less elements, which creates higher hierarchies of the language. This is indeed what multiplication is: a compound operation, that represent a series of sums. Also, we have the logical convention that defines that lower hierarchy operations must be realized first when combining multiple orders of operation. We must only use explicit symbols to determine order when we don’t want to follow the already established precedence order. The fact that you don’t look at the word ‘then’ as expressing order, felt weird to me. Cause the ‘and then’ operator implies “first a, and then b”, which necessarily implies order. Taking a more formalist approach, ‘and then’ operator has a temporal effect/meaning, which has to be on a higher hierarchy than other basic logic symbols, such as ‘and’ or ‘then’, and does not fit in first order logic or basic arithmetics, though, I guess, it would fit in some kind of temporal logic/modal logic. (It’s important to say that ‘then’ is usually used to express implication, and, in this case, does not relate to ‘and then’ which expresses order) So, such as the parenthesis, ‘and then’ is only needed to explicitly display order, because order, in this case, matters. Which, again, is the opposite of the associative property. Thanks for the video and for bringing this subject up! Cheers
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 presentation is simply genius! It puts a big chunk of important mathematical ideas into a completely new and more intuitive and general perspective
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.
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 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.
oh wow whoever would've guessed that object/action duality applies to every operation, it's almost like all math first came from the physical world, and was abstracted to make things easier.
@@milokiss8276 i am basically saying action only and object only is actually action plus object, so yes it changes your view in a way. The only difference is the object, for example "3 apple, and adding 2 apples in it" this is an operation, "3 apples with 2 apples" originally, this doesn't mean 3 + 2, it only means 3,2. Or "adding 3 apples, and adding 2 apples, this also means 3,2. The only difference in this and the object is its base form, its as different as something like "3 oranges, with 2 oranges" To do any operation, there exist object in some sort of way, and also an action in some sort of way. So yes it is directed at your point, you shouldn't believe everything after just hearing someone explain it in a cool and calm way, because the idea itself behind it is false
This is an eye opening video. It makes me think how math is related to computability. It looks like math is a reversible algorithm. Thanks for posting this!
My attempt at the challenge you posed near the start: Imagine you have some things, all in a line, and then you want to slowly move them all into one pile by use of some 'operation'. If the operation you decide to use is associative, it means that you have the freedom to pile adjacent things together as you wish, and the pile you end up with in the end is unchanged. Maybe I cheated a bit and used some synonyms, but it was my best shot :P
In my language we use the word "connectivity" instead of "associativity" which I think is more intuitive. I think the root of the problem is that when talking about a group operation, there are really two "orders" we are talking about, which get confused. First is the order of placement of elements undergoing the operation and the second is the order in which we execute the operations. So in this interpretation, if we have an operation which is associative but not commutative, meaning we have some elements placed in a fixed order, only the order in which we execute the operations doesn't matter. So e.g. having a tape with a sequence of objects written on it in a fixed order, we could glue them and cut them however we like. Another issue with understanding it is that people have problem with noticing it and using it in practice because of lack of practical examples which are really helpful, not just on paper, in theory. A good place to look for such use cases are data structures in computer science, because a lot of them can be generalized with groups making them way more powerful in solving algorithmic problems. A good starting place to look to understand it are interval trees, which get beautifully generalized with groups. When you start to notice associativity instead of classical additivity (understood as associativity+connectivity), you get a whole new dimension of tools for solving algorithmic problems.
Maybe I missed it, but did you give any examples of where associativity fails? I would be curious to hear your thoughts as to why such examples would fail to be '& then' models. For example, in terms of '& then' why does exponentiation fail to be associative? That is to say, why is a^(b^c) not necessarily (a^b)^c? (More generally one can ask a similar question for any non-associative algebra.)
I was thinking the same thing. If he'd included a single example, like subtraction, the video would have been 10x better. I'm reminded of the veritasium video "The most common cognitive bias" where he asks people to figure out the rule he's thinking of. Everyone wants to find examples that work, and noone thinks to try examples that don't work. To really explore a property, you have to look at the cases where the property fails.
You didn't miss it, I'm still kicking myself for cutting out the division example to reduce the video length 😥 Here is my response to a similar comment: " 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". " But you asked why exponentiation fails to be associative. Tbh, I'm failing to construct an example in terms of '&' that is more illuminating than the more straightforward (a^b)^c = a^(bc) ≠ a^(b^c). Sorry 😓
@@linguamathematica2582 I hate to be a nonbeliever because I loved your video, but I still can't see this perspective. Commenting on the proof outline you have above: Instead of somehow mapping "and then" to "&", one could similarly define/map “not and then” to “&” and in theory could illustrate that associativity is the “not and then”-operator by the same demonstration of divisibility failing to be associative. This is because things like division fail to be associative regardless of what notation or language we use to refer to it. We could also work with exponentiation, and since exponentiation is not associative, your proposed correspondence would suggest that we'd see ambiguous use of "and then", similar to how the product abc would be ambiguous for a non-associative action. From this, I don't think that the comparison of “and then” to associativity is compatible with how people interpret its appearance in common language. We can give operations any name, so in particular we could name exponentiation “and then” and by considering the contrapositive to your claim we would expect ambiguous language as a consequence of exponentiation’s failed associativity. To calculate a and then b and then c we perform the following steps: a and then b and then c → a^b and then c → (a^b)^c. However, (a^b)^c is certainly well defined / not ambiguous for fixed values of a, b, and c. If I misinterpreted anything let me know. I look forward to future content as well. Edit: Had to make some re-edits to keep my thoughts straight here.
10:57 Object-object: Let y=f(x), and z=g(y). Plotting y on x produces an x² curve. Plotting z on y produces a sqrt(x) curve. Plotting z on the original input x produces a |x| curve. Object-action: We once again plot y on x to get an x² curve. Now let's view g as a transformation that takes points y on that curve and moves them to sqrt(y). After applying that transformation, we end up with a |x| curve. Action-action: Now let's imagine the plot of any curve. We can apply f to transform all points on that curve into their square, then apply g to transform all of those points onto their root. We end up with the same result as if we had just applied the transformation h that transforms all points on the curve to their quantity.
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.
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.
Function Composition: Object Object: a functional key, and a functional lock Object action: a funtioinal key unlocking something action action: Spray paint key, then use it to unlock something
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.
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._
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.
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).
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
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.
So, what I got out of this is 'associativity' is a 'gluing' with direction. Text concatenation is 'gluing' but we have to remember the implied direction of 'from left to right'. Actions can be 'glued' together, but the overall 'direction' of flow must be maintained. You show we can take actions a, b and c and 'glue' actions b&c together, but when we then 'glue' on action a, we must be sure to prefix it to b&c so that a sort of 'direction' is maintained (i.e. a&b&c). And reversibility is simply reversing this 'direction'. This 'direction' might sometimes be temporally, such as your film clips being combined to show 'get the milk'. But it can be other than that such as simple addition where we can add b and c BEFORE adding a, but the 'direction' is something more subtle.
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". "
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.
Okay so as soon as I saw the equation in the thumbnail I was trying to substitute random values for a, b, c, and I deadass said to myself "let a=b" I'm fucking crying lmao
That was fascinating and enlightening! Always struggled with groups in my physics master as well. Another connected thing i still haven't solved: Why is addition and multiplication associative, but not exponentiating? (a^b)^c!=a^(b^c), even though we get multiplication from repeated addition and exponentiation from repeated multiplication?
Epic Math Time did a video on your question: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-ofy2Kw2sIZg.html Basically, what exponentiation is to multiplication is not the same as multiplication is to addition.
@@nikolasscholz7983 This is an excellent question. Here is my response to a similar comment: " 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" " It is in this second sense that exponentiation is not an "and then" action. Here is a similar proof for why division is not an "and then" action: " 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". "
[Take two apples out] and then [take three apples out] is the same as [take five apples out]. The dual operation of subtraction is addition, rather than having subtraction be its own dual. At 15:41, if you start with b and c, and put a glow around each of them, that's just a definition of the dual operation -- giving you addition as the dual, if your initial operation is subtraction. But if you put a single glow around bc, that means taking the result of your binary operation first, and then considering it as a unary operator: it means subtracting (b-c), instead of subtracting b and then subtracting c. The problem is that a glowing bc looks the same as a glowing b next to a (separately) glowing c.
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.
At 3:32 the font made you think you were about to say: "In the arithmetic number system, the operations obey two separate but equally important properties: the commutative property which changes the order of two consecutive numbers, and the associative property which allows for changing the order of the operations themselves. These are their stories."
This might not be enlightening to many people and comes close to using "semigroup", but I think it will fit your criteria: Explanation: For a set X, let ^X be the set of finite strings of elements of x and let || be the concatenation operation. A binary operation * on a set X is associative if there exists a surjective function f: ^X -> X such that f(a_1 ... a_m || b_1 ... b_n) = f(a_1 ... a_m) * f(b_1 ... b_n). Note: You can but need not (1) require that f(a) = a or (2) fully define f by f(a_1 ... a_n) = ((a_1 * a_2) * ... a_{n-1}) * a_n or some other parenthesis placement. While (2) will make it fail your criteria, (1) may actually be a help, since it makes f unique and concrete, and we see that * extends in a unique way to an "finitary" operation, i.e., n-ary for n>0. Proof: Assume * is associative, then f defined by f(a_1 ... a_n) = a_1 * ... * a_n clearly satisfies the property. Now instead assume that such an f exists. Then a * (b * c) = f(a_1 ... a_k) * (f(b_1 ... b_m) * f(c_1 ... c_n)) = f(a_1 ... a_k) * f(b_1 ... b_m || c_1 ... c_n) = f(a_1 ... a_k || b_1 ... b_m || c_1 ... c_n) = ... = (a * b) * c. You can do an equivalent thing with commutativity: For a set X, let ~X be the set of graphs that are binary trees including singleton leaves with elements of x as leaves and an additional object v at intersections (the order of two branches is not distinguished). Let ~ being the operation of connecting of two trees to a v. A binary operation * on a set X is commutative if there exists a surjective function f: ~X -> X such that f(t_1 ~ t_2) = f(t_1) * f(t_2).
Beautiful! The free object approach is so powerful. I may just have to do a video on it :) I've never seen a construction of a free commutative magma but yours was pretty logical (construct a free magma using binary tree approach and quotient out branch orders). I wonder if there is a nicer interpretation that makes commutative magma's more intuitive.
