Infinitesimal Monad (extra footage)

[@sserpxeadnap]

Please please do more videos with Professor Wood!! Extremely enjoyable

[@calebritchie1069]

She’s so engaging!!

[@cmd2tuts]

Professor Carol Wood is a fascinating person, it would be great to see more of her in the future.

[@lethargogpeterson4083]

I love Brady's "No!"

[@mahdibnhussein4882]

Lethargo G Peterson Oh I understand what do do you mean 👌

[@reallyWyrd]

I gotta go study the hyper-reals so I can start messing with people. I want to draw little funnel shapes over each integer on the number line and then be like, "and here, you see this funnel shape? This is the whirlpool draining away your sanity."

[@raywilliams6717]

Omg that's great.

[@mondoke]

I just love how after years of talking about science with lots of scientists for a living, there are still topics that make blow Brady's mind.

[@ffhashimi]

I think this woman is an amazing teacher

[@unclvinny]

Dr. Wood is great, hope you can do more videos with her.

[@garethdean6382]

Glorious. I want to print this video onto a custom cake and eat it.

[@fantiscious]

Me: *finishes watching the previous video, sees this one* Me: Oh, another one! Professor Wood: **immediately starts with** "Well it's worse than that" Me: _OH NO_

[@SpamDestroyer]

I love how this immediately starts off with "It's worse than that!"

[@VladVladislav790]

-"even K!" - brain exploded

[@mahdibnhussein4882]

So watch the video "How to count past infinity" for Vsauce channel

[@spencerhixonauthor]

But that would also mean "1/K" has ITS OWN MONAD! The number line just became an infinite line of infinite infinitessimals.

[@km1dash6]

I don't know why, but I really like infinitesimals.

[@BewilderedBird]

I really, really like this approach to calculus. In fact, I think the deltas and epsilons are much less intuitive and contribute to much of the confusion to calculus students. This infinitesimal approach is much closer to what Leibniz was thinking when he discovered/invented calculus and makes more intuitive sense.

[@okoyoso]

Do a video on the smooth infinitesimal analysis too! It's similar to non-standard analysis in that it uses infinitesimals.

[@levinb1]

1:11 Her thesis advisor was ABRAHAM ROBINSON!!!!! No wonder!!!

[@thevastidity888]

I LOST IT WHEN I HEARD THAT.

[@micknamens8659]

Positive infinity is usually considered as the limit of N, which is supposed to be unique. What is the relationship between infinity and K, or K+1 ?

[@firstlast-wg2on]

I picture numbers in this context like this; "1" is just a letter describing the peak of a 'concentration' of many little unique entities that we label "numbers", "1" does not refer to a singular object (like one donut) but rather the highest 'concentration' of unique entities. For example, you can have those infinitesimals close or far away from "1" on the number line, all that really means is the entity is bigger, or as I call it more concentrated. If you applied this constantly then numbers are more like bumps than a long sequence, you know like 1 2 3 4, it's more like a series of completely unique patterns that peak at certain points. This post literally means nothing but it's a cool idea, it could be shifted into a 3D representation of numbers too.

[@firstlast-wg2on]

I suppose you could draw from that and say that the "numbers" don't have any order, one is no longer bigger than the other, and rather that it has a different identity to the previous.

[@GorduneDelaine]

Great speaker! I'd love to hear more

[@sinecurve9999]

njwildberger just flipped out....

[@Alex-mg6vn]

+sinecurve9999 If he finds real numbers appaling, hyperreals and infinitesimals must be his worst enemy!

[@zachsmith9841]

[@FishKungfu]

The infinitesimal monad set is really cool and I like it, but it agitates me at the same time.

[@thedoublehelix5661]

fun fact: its technically not even a set. Its too big to be a set! Its something called a class

[@TheBasikShow]

Funner fact: That’s not even necessarily true! There exist surreal number systems which have exactly as many elements as the real numbers, even though they also have all the infinite galaxies and infinitesimal monads that you could ever want.

[@Kram1032]

What happened there at 4:00 - 4:02 ? It's just blacked out and you begin to say something mid-sentence after that?

+Kram1032 Illuminati confirmed.

[@cavalrycome]

+Kram1032 They had to censor the dark math.

[@bgezal]

+Kram1032 Brady blacked out from sheer insight. --Oh my god it's full of stars.

[@effortless35]

+cavalrycome In 2015 after decades of searching we finally found the dark mather.

[@Kram1032]

Bas Keurprins it doesn't look intentional though: When Brady's voice comes back, it's also cut-off, starting mid-sentence. It rather looks like an accidental imprecise cut.

[@mushkamusic]

this is bloody fascinating!

[@pbj4184]

Oh that Abraham Robinson was her advisor. Wow!

[@hreader]

Love it! The best presenter I've seen for quite some time! Also, thinking the infinite number of infinite collections of infinitesimals (i.e. next to every real number) I would suggest is more rewarding than taking illegal substances! I've recently been looking into this topic, and it seems to me to be a very good intuitive way of explaining calculus at 'A' Level. All those infinitely small dx's, 2dx's, 2xdx's etc etc make one hell of a lot more sense. Although there have been challenges to the hyperreal numbering system as well. Maybe a future video could put Prof. Wood head to head with someone with another perspective on infinity. Her other video shows the first order statement 'There exists a number K greater than all N' which unlocked a mental block I have (or had) and probably other people do as well which is that for any N there's always an N+1 - and that's it. You can't get beyond the natural or real number lines. Suddenly though the idea of another order of infinite numbers (and their infinitesimal reciprocals) became reasonable. Much to ponder here. I recommend doing a search on 'compactness theorem' or 'compactness theorem first-order logic'.

[@JA-cn6vu]

Let R be the reward set obtained for thinking about these numbers while in normal consciousness, and let R* be the reward set obtained for taking illegal substances. It may be true that card(R)>card(R*), but unless either R* is a proper subset of R, or one or the other of them is of infinite cardinality, it would also then be true that card(R U R*)>card(R). :-)

[@stevenlu7324]

@@JA-cn6vu This is fantastically eloquent and is in line with my life philosophy. Nothing wrong with the substances provided you provide adequate compensation like having a safe space.

[@Scy]

So the 1/x graph where the line never crosses x=0? Can you explain that with these monad thingies?

[@brendawilliams8062]

I know you enjoyed that study. Thankyou.

[@SunnyTheGentleFox]

So basically, it's another way to think about Calculus.

[@guildpilotone]

Looking at the Real number line with monads, and the numbers in the monads having their own monads and so forth, is like visually drilling down through the Mandelbrot set or some other fractal image… it just goes on forever.

[@MozartJunior22]

+guildpilotone What numbers would be in this monad?

[@bgezal]

+guildpilotone Sierpinski was a greater set theorist than Mandelbrot :)

[@jdferreira]

+guildpilotone The numbers in a monad don't exactly have their own monads, since the monad is the set of infinitesimal numbers around a number. So, for example, consider 1/k. Its "monad" would contain number such as 1/k + 1/k = 2/k = 1/(k/2), 1/k + 1/k^2 = (k + 1)/k^2 = 1/(k^2/(k+1)) [notice that k^2/(k+1) is an infinite number]; all these numbers are already part of the monad around 0.

[@MozartJunior22]

+jotomicron Here is a suggestion - to add a monad over a monad, you have to define a new "number", L, which will work like K but is "bigger" than K, in the sense that for every natural number n, L>K+n. It's a new sort of number that will now create monads infinitely smaller then the K-monad. This way, the number 2+1/K+1/L is inside the monad of 2+1/K which is in the monad of 2. In fact, instead of calling it K and L and use up letters, we can have K(1),K(2),K(3),K(4)... and those numbers will be subscript. now for 2 natural numbers a>b and for every natural n, K(a)>K(b)+n. And while we're at it, we can consider the concept of K(K1), or K(K(K(K(K(K....)))) K1 times. My mind is about to explode so I will stop here.

[@jdferreira]

+MozartJunior22 I hadn't considered that! I'll have to stop for a second and think about the implications!

[@ChristOfSteel]

Having a background in category theory and their monads, I wonder how these monads relate to this monads?

[@ThegamerIdiotgenius]

Wow... what kind of classes greater than calculus can I take to learn about this stuff? (I'm taking AP Calc right now.)

[@ffhashimi]

my question.. is there a " connection point" between these infinitesimal and the reals? from 0 we begin with the infinitesimal who are smaller than any of the reals so we have to move from 0 infinitely and we won't reach the "smallest" real but is there a way to "make a jump" from the infinitesimal to the reals ? if not .. how we consider them ( infinitesimal and reals) as "living" in the same line ? they will look as they live in a defferent layers of the number line .. or is there a gap in our understanding of both infinity and number line? any idea ?

[@JLConawayII]

It gets worse?!!

[@thomas.edgerton]

+JLConawayII I would say better !

[@MozartJunior22]

1:25 Why would you ditch epsilon and delta? They work great in exactly defining the limit, without "intuition". Why would mathematicians go back to the messy definitions? By the way, does 1/K have a monad? Or am I stretching this too far?

[@kuroninja90]

+MozartJunior22 Because it's not messy it is equivalent to the epsilon and delta definition and just as rigorous. Google hyper reals if you want to know more

[@MozartJunior22]

+kuroninja90 I've heard of hyperreals But the fact is, you can define limits in the real numbers without having to add this new type of numbers.

[@mattcay]

+MozartJunior22 They wouldn't go back to the messy definition. If you'd like to to prove the result of some derivative for example, you'd probably use the simple (or at least simpler) epsilon-delta. But before you formalize your theory you start with thinking and in this thinking infinitesimals are very useful. For example: What is the length of an arc? Of course it is the sum of the lengths of infinitesimally small line segments -- Integral[dS]. Now, what is dS? Hmm... let's try "Pythagorean Theorem" -- dS^2 = dx^2 + dy^2 => dS = sqrt(dx^2 + dy^2). Now, let's factor dx^2 out of the sqrt and we get dS = sqrt(dx^2(1+(dy/dx)^2)) => dS = sqrt(1+(dy/dx)^2) * dx. So finally we got that the arc length is simply S = Integral[sqrt(1+f''(x)) * dx]. As you can see that way we obtained the formula in a very simple and intuitive way -- we used some basic geometry to obtain a complicated formula. Of course this is not a rigorous proof, but making it so is often the last step you make.

[@LeonhardEuler1]

+MozartJunior22 The definitions aren't really messy -- they can still be expressed in precise language (a monad on a category C is defined as a triple of data -- an endofunctor F on C, as well as natural transformations from the identity functor in C to F and from F^2 to F), which satisfy certain coherence conditions. There are many reasons for coming up with new definitions like this. Mathematicians are often just interested in generalizing results we are familiar with to some broader context. However, this can also be useful in understanding something about the original definitions and the context they arose in. For example.... well.... I don't know what sorts of math classes you have taken, but you've likely heard of the quadratic formula. This formula is great -- if I give you any quadratic equation whatsoever, ax^2+bx+c=0, then you can give me back the two solutions to it using the quadratic formula, which is determined just by the numbers a, b, and c. What you may not have seen is that there is also a cubic formula giving all 3 solutions to equations of the form ax^3+bx^2+cx+d=0 in terms of a, b, c, and d, as well as a quartic formula giving all 4 solutions to equations of the form ax^4+bx^3+cx^2+dx+e=0 in terms of a, b, c, d, e. (Look up the formulas and you'll see why they aren't taught, they are AWFUL!) It turns out, however, that it stops there. There IS NO quintic formula, nor any formula for higher degree polynomials. I don't mean that we haven't figured it out what the formulas are, I mean that it is impossible to write one down -- we can prove it is impossible. So what does this have to do with what I mentioned above? Well, unlike something like the quadratic formula that you can prove using completing the square (so you only need to know about ordinary arithmetic for real numbers), the way this is proven is by using a bunch of stuff related to ring extensions called Galois theory. This stuff is, more or less, a fancier way of writing down what addition, multiplication, the distributive property, etc are, which makes sense in a larger context, and by understanding those new, fancier definitions of what we all learned in elementary-high school, you are able to get the result I mentioned above. Short version: sometimes a new perspective on old definitions not only allows you to prove more general results about other "number systems", but sometimes provides the clarity needed to understand ordinary numbers better.

[@bgezal]

+MozartJunior22 A monad is a set and epsilon is a quantity?

[@kuro68000]

The sound is badly messed up on this video.

[@TheLemmingeater]

does this mean you could have an monad within a set of monads, for example if you had a set of numbers h that where larger than all of k and then take 1/h?

[@liviumarginean1474]

I have one question. second 34 when she says even K has infinitesimal buddies. is K+1/k infinitesimal close to K? Or in order to find the number infinitesimal close to K should we create another line called R** where there is a number K* bigger than all the numbers k on the R*? and we would have k+1/k* is infinitesimal close to k.

[@b43xoit]

I'm not an expert, but my reading from these videos is you do that in R*. You don't need an R** to understand K + 1/K.

[@coreyredmon5611]

My question: is 0.999... (0.9 repeating) equal to 1-1/K?

[@Yomama5923]

+Corey Redmon You aren't going to like the answer, but (I believe) it's a no. 0.999… = 1. ViHart and, I think, Numberphile did a video on this before. Here's why. To understand 0.999… set it equal to x. 0.999… = x Multiply both sides by 10. 10(0.999…) = 10(x) Subtract original equation. 9.999… = 10x -(0.999… = x) ------------------ 9 = 9x Divide by sides by 9. x = 1 Since x originally equaled 0.999... therefore 1 = 0.999...

[@coreyredmon5611]

That is what I thought because I saw that video and I am convinced that 0.999...=1. This means that 0.999 cannot equal 1-1/K because if it did using substitution you come to the solution that 0=1/K which she established is not true.

[@Hecatonicosachoron]

+Corey Redmon If you have an infinite number of infinitessimals that are ordered with respect with each other (e.g. so that 2/K > 1/K) then even if you chose to say that 0.999... does not equal 1 then it is also not uniquely defined.

[@TrimutiusToo]

What about infenitesimal fraction? (like rational, but not natural)

[@gryg666]

Hmm, so if we are thinking of "space" (or area) of numbers in infinitesimal to some number, does those numbers have also similar area? Looks like some kind of fractal.

[@frosch03]

wow, first time i heard about monads, without the topic being functional programming... so are these monads in any way connected to the ones found in functional programming?

[@garthgoldwater5256]

frosch03 not an expert but I think the monads in functional programming have more to do with the concept of a monad in category theory. they were both inspired by leibnitz (spelling?) but pretty far apart theoretically. also, the programming monad and the category theory monad are not strictly equivalent

[@adarshmohapatra5058]

The only connection that I can think of is that they come from the Greek word Monas/Monad meaning unit. A lot of English terms in Science & technology come from Greek.

[@aaronjones553]

Seems like an extension of this to the complex plane would be interesting and extremely brainbending.

[@ianprado1488]

WHAT IS THE CARDINALITY OF R*???

[@mantkcmantk8717]

Certainly bigger than the cardinality of R. But there are several R*'s. Each may have a different cardinality, although all larger than the cardinality of R. I do not know the smallest passible cardinality of R*, it may depend on the set theory.

[@km1dash6]

Do infinitesimals have even smaller infinitesimals surrounding them?

[@b43xoit]

No.

[@darkaquatus]

It's funny if you try to approach this with rational thinking. Most people would get and accept the idea of infinite numbers, either positive or negative. But when you start thinking about infinitesimals, things get tricky. You see, these are actual numbers you interact with and touch on a daily basis. When you draw a line between 0 and 1, you have actually touched and drawn over these infinitesimals. Yet, they have their own little reality; their own little universe, making it impossible to pinpoint them down in your own reality. There's something pretty odd about the thought of being able to interact with and touch these numbers as long as you're moving. But the moment you stop moving and look down, this specific value will have suddenly disappeared. It's like you can see it as long as you have a blurred vision of it, but the moment you focus, it will melt away like snowflakes in the sun. Naturally, I understand why this is, but there's something strangely, philosophically appealing to the idea.

[@PianoMastR64]

Maybe it's because it touches on the idea that there are all kinds of realities out there, especially the ones we haven't thought of before.

[@GenericInternetter]

Do you realise that you’re making a very accurate description of subatomic physics? For example, check out electrons, especially supersymmetry. en.m.wikipedia.org/wiki/Supersymmetry

[@angelmendez-rivera351]

Generic Internetter Not at all. There are no infinitesimals in sub-atomic physics specifically.

[@ianmoseley9910]

Angel Mendez-Rivera uncertainty principle

[@angelmendez-rivera351]

Ian moseley That is not atomic physics per se, and the uncertainty principle has nothing to do with infinitesimal numbers. In fact, the statistics of the uncertainty principle come from real-valued mathematics with no infinitesimal numbers in it.

[@scathiebaby]

How does "K" relate to transfinite ordinals such as ω ? Because, the definition is similar.

[@MuffinsAPlenty]

K relates to neither ω nor ℵ₀. ω is an ordinal number, whose purpose is to describe order type of a well-ordered set, ℵ₀ is a cardinal number whose purpose is to describe the "amount of elements" in a set, and K is a transfinite hyperreal number... It doesn't have a big purpose besides that it simply exists. Here's a nice way to show that these three numbers are very different: try subtracting 1. What happens? K − 1 is defined and K − 1 < K. Simple. ℵ₀ − 1 is not _technically_ defined, but we can still make sense of it. If you take any set of cardinality ℵ₀ and subtract any set of cardinality 1, you will always get a set of cardinality ℵ₀. So while subtraction is not defined as an operation on cardinal numbers (A set of cardinality ℵ₀ minus a set of cardinality ℵ₀ could be a set of any cardinal number less than or equal to ℵ₀, depending on the context, so ℵ₀ − ℵ₀ cannot be defined), any reasonable meaning for ℵ₀ − 1 is that ℵ₀ − 1 = ℵ₀. ω − 1... Much like with cardinal numbers, subtraction is not defined on the ordinal numbers. But, in some contexts you could still make sense of it, possibly. Many ordinal numbers describe the "position" of an element in a well-ordered set. So if you have an ordinal number α which describes the position of an element in the set, then α − 1 would describe the position of the element immediately previous to the element in position α. Now, a well-ordered set is said to have order type α if the _last_ element in the set is in position α. But there's a problem: some well-ordered sets don't have a "last element," so their order type is represented by a limit ordinal. ω is the first limit ordinal: it represents the order type of the set of natural numbers: {0 < 1 < 2 < 3 < 4 < ...} If ω − 1 were to have meaning like α − 1 has meaning, then ω − 1 would represent the position of the "second-to-last" natural number, which is a nonsensical notion. So ω − 1 has no meaning whatsoever. So we have: K − 1 is defined and K − 1 < K ℵ₀ − 1 is not technically defined, but it makes sense to say ℵ₀ − 1 = ℵ₀ ω − 1 is not defined and one cannot make sense of the expression in any way Of course, a caveat to all of this: if one works with the surreal numbers, depending on how you define K, you could very well have that K = ω. The surreal numbers are weird, though.

[@MuffinsAPlenty]

***** Well, there's probably not a nicer way to write it than k/x. But I can tell you that k/x is greater than every real number, for sure. If x > 1, then k/x < k, and if 0 < x < 1, then k/x > k. This all follows from a weird logical rule that R* is elementarily equivalent to R. This gives us the transfer principle. In other words, any first order statement that is true about R corresponds to a first order statement that is true about R*. We have to be a little bit careful, though, that we can actually write the statement in first order terms. But for things like "for every positive real number y, if x > 1, then y/x < y" can be written in first order terms, and then corresponds to the statement that "for every positive hyperreal number y, if x > 1, then y/x < y." Similarly, one gets the statement, "for every positive real number y, if 0 < x < 1, then y/x > y," so by the transfer principle, we get "for every positive hyperreal number y, if 0 < x < 1, then y/x > y." The fact that k/x is greater than every real number cannot be stated by way of the transfer principle, since there is no way to come up with a first order statement like this using just the real numbers. But, one can still prove that k/x is greater than every real number pretty easily. Suppose k/x = y. Then since R* is a field, xy = k. But x is finite, so there exists a real number M such that x < M. If y were also finite, then there would exist a real number L such that y < L. Then xy < ML. But ML < k, since ML is a real number and k is greater than all real numbers. Therefore, we would have xy < k, which contradicts that xy = k. Therefore, it must be the case that y is greater than all real numbers.

[@MuffinsAPlenty]

***** Well, I am studying to get a PhD in mathematics. Now, logic and set theory are not my area of study, but I like to look at these ideas on the side sometimes.

[@notoriouswhitemoth]

Between every two possible integers, there are an infinite number of rational numbers. Between every two possible rational numbers, there are an infinite number of irrational algebraic numbers. Between every two possible algebraic number, there are an infinite number of transcendental numbers. Between every two possible transcendental numbers, there are an infinite number of possible hyperreal numbers. The part where it might get confusing: between every two possible hyperreal numbers, there are an infinite number of possible hyperreal numbers - because when you're dealing with infinities, your most basic assumptions about the finite break down.

[@angelmendez-rivera351]

notoriouswhitemoth Between any two hyperreal numbers, there are infinitely many surreal numbers. Between any two surreal numbers, there are infinitely many surreal numbers, because the set of surreal numbers has the biggest cardinality of any ordered field and the longest measure. The cardinality of the surreal numbers is the same as that of the universe of set theory.

[@angelmendez-rivera351]

Giannhs Polychronopoulos You are asking if it is possible? Possible it is, but very difficult. I think the key to the fact is that the surreal numbers technically do not form a set. They’re too much to be able to form one. Instead, they form what is called a class. Though you cannot have a set of all sets, you can have a class of all ZFC sets, for instance. These classes form a more general type of object than a set. It is in the study of clases that you can prove a statement such as the cardinality of the class of surreal numbers is that of the set theoretic universe. Edit: I know I did use the term set in my first comment, which I realize is misleading. I did not want to delve into the technicalities initially, and using the word “collection” instead does not feel better, though.

[@angelmendez-rivera351]

Giannhs Polychronopoulos ah, I don’t the know the specifics, to be honest. I watched John Conway’s lecture about games and how his theory of games led to his discovery of the surreal numbers. However, the class of games is a super-class of the surreal numbers and is said to be “bigger” in some sense. In this lecture, he did say specifically the cardinal number describing the class of surreal numbers is that of the set theoretic universe. He did not specify an explanation of proof of this, which I imagine he intended for us to take for granted in the lecture, since the lecture is not about classes and cardinalities in general. I have read online that the surreal numbers form a class, not a set, but I don’t know of any specific books to read. Actually, I’m looking for books of my own myself to read about it too. I can link you his lecture, though, if it helps

[@b43xoit]

@@angelmendez-rivera351 Either set theory doesn't have a universe, or I missed something. Russel paradox?

[@y2536524]

Can we have K divided by K in N star , is it one if defined?

[@joeybf]

+y2536524 Division is not defined in *N, or in N for that matter.

[@ekaingarmendia]

if the units of all the reals are u, and the unit of k is Mu (maxi unit) and ( a u + k Mu = k Mu ), then what is the unit of log(k)? It supposedly is u, which means it belongs in the real numbers, but ( log(k) > x k > y^x and because y^x is a real number, then k > any real ). Does that mean that there are maxi numbers greater than any real number but smaller than every maxi number? Am I right? or is this just some sort of sorcery?

[@inthefade]

This reminds me of the first time I dropped acid.

[@zachariahhanson1792]

How many infinitesimals are there? Is this number bigger than k? and how many infinitesimals are after infinity but before k-n? Is it half the size of the number of infinitesimals around 2.337?

[@thestarjon]

In |N*, there are exactly as many elements as in |N (both are countable infinite). That may sound counter intuitive, but it definitley is the case. There also is the exact same amount of numbers in |R and |R*, although both contain more elements than |N or |N* (|R is uncountable infinite). If you are interestest, I can prove it to you. :) EDIT: Acutally, I'm not quite sure that |R* contains as many elements as |R because of the monads, but with |N and |N* it's definitly true.

[@zachariahhanson1792]

Starjon go on... Bear in mind that i have only done up to college level maths, which was essentially just mechanics, so... Yeah ;-)

[@MuffinsAPlenty]

I'll just use R to represent real numbers, C to represent complex numbers, R* to represent a hyperreal number construction, and N to represent the natural numbers. +scpDgJ |R| = |C| = |R*|. All three sets have the "same number" of elements, even though R is a proper subset of both C and R* and even though C and R* are incomparable by way of set containment. +Starjon You are correct. The cardinalities of R and R* are the same, though it isn't a trivial proof. First, note that |R| ≤ |R*| is obviously true since R is a subset of R*. Now, in order to show that |R*| ≤ |R|, we have to look at how R* is constructed. R* is constructed as equivalence classes of sequences of real numbers (by definition, a sequence is indexed by N). Since R* is the collection of _equivalence classes_ of sequences of real numbers, R* can be identified with a subset of the set of sequences of real numbers, denoted R^N. Thus, |R*| ≤ |R^N|. Hence, it suffices to prove that the set of sequences of real numbers R^N has cardinality ≤ |R|, i.e., to prove that |R^N| ≤ |R|. Sequences of real numbers is simply R^N, and so, by definition, |R^N| = |R|^|N| = (2^(ℵ₀))^(ℵ₀) = 2^(ℵ₀ * ℵ₀) = 2^(ℵ₀) = |R| Thus, we have |R| ≤ |R*| ≤ |R^N| ≤ |R|, giving |R| = |R*|.

[@MuffinsAPlenty]

***** Yep. I mean, I probably couldn't show you an explicit one-to-one correspondence, but using other, more general facts about cardinality, I can prove that a one-to-one correspondence must exist.

[@MuffinsAPlenty]

***** It certainly is a surprise! Although this isn't entirely rigorous, you can think of infinities as growing exponentially. The amount of extra stuff you need to go from one infinity to another increases dramatically as these infinities get larger.

[@Hythloday71]

Prof. Norman J Wildberger will be most cross !

[@twwc960]

+Hythloday71 He doesn't get cross. He just sort of rolls his eyes and smirks.

[@trissylegs]

+Hythloday71 I was just thinking that.

[@MauveTendingToBeige]

+Hythloday71 Maybe Wildberger can give his vision on Numberphile. Would be cool...

[@twwc960]

MauveTendingToBeige He lives in Sydney, so next time Brady goes back home to Australia, maybe he can film a segment or two.

[@weezhaochuakhoo1032]

What would he be crossed about?

[@captainharris8980]

Cool stuff. I've always wondered why the extremely small fractions weren't studied more and if they had some use. Very interesting. Thanks for the vid.

[@weakamna]

Ok.. so are there monads in the monads? taking any of those 1/k numbers, and add another 1/k on _those_, do you get a monad for those numers then? something like (a + (1/k)) + (1/k)?

[@TheEternalVortex42]

Not really, since that's just a + 2/k. There is only one level of monads

[@willmartin7108]

So is this like a bridge between discrete and continuos math?

[@b43xoit]

What you mean "bridge"?

[@BluRey100]

She is correct. This lines up with Fractals, Omega Code, Menger Cube and String Theories. She deserves an Honorary Cosmological Degree.

[@delwoodbarker]

Does R* model R-squared, that is, do they model complex numbers? If so, they form an ordering on the complex plane, which is elegant and satisfying.

[@xpeson9]

I broke the mandelbrote shape by watching a video from numberphile "2"... And I dont regret the consequences yet.

[@fould13]

"I have 3 sheep" "I have $20 of gas in my car" "I really wish I had K things" Even K is useful.

[@RolandHutchinson]

But if you had K things, where would you keep them? Even if you rented out every room in Cantor's hotel...

[@scathiebaby]

The cardinality (set size) of these numbers might be larger than the cardinality of the continuum !

[@HighKingTurgon]

1/k is real, I can intuit its existence on the number line, great. But is K a real number? Is it countably infinite if we have to hop forward past all n element of the natural numbers?

[@jesusnthedaisychain]

God damn the man who invented everything to the right of the decimal point!

[@Ensivion]

I wonder what would happen if you extended the mandelbrot set to the hyper-complex plane.

[@nishadseeraj7034]

so does that mean 1/k has a monad around it too?...what about the numbers within the monads?

[@xxX_420BlazeIt_Xxx]

1/k will not have a monad. For it to have a monad would imply numbers even smaller than the infinitesimals. If you did this then the inverse of those numbers would be even bigger than all of the hyper reals.

[@nishadseeraj7034]

makes sense...thanks!!

[@indigodarkwolf]

But just as we can choose some number K > r (for all R), couldn't we choose some L > k (for all K)? Because we're choosing a field of hyper-reals or monads for which the same rules apply as with the reals, why can't there be an infinite number of fields of increasingly infinite or infinitesimal size?

[@fabricenonez9098]

In a literal sense, not only 1/K has a monad, but it's just the same as 0's monad. Because, u is infinitely close to 1/K iff u-1/K is infinitesimal, and since the sum of two infinitesimals is still infinitesimal, that happens iff u=u-1/K+1/K is infinitesimals. The monad is well defined for any hypperreal, and it's just the monad of 0 translated to the hypperreal. But if your question was about infinitesimals but relative to the hypperreals, well there aren't any in the hypperreals for the same reason there are none in the reals (0 0

[@angelmendez-rivera351]

FBI Wrong. 1/k does have a a monad, and the monad is given by the set of all numbers of the form 1/k + r/k^2, where r is a real standard number.

[@_abdul]

why not ?

[@vs-cw1wc]

Question: Is this idea capable of talking about divergent series?

[@randomperson10082]

To me it seems like 'the monad of a number in a monad' would just be the rest of the numbers in the same monad, so it's not fractal in that sense.

[@ZardoDhieldor]

+TheZanyCat You're right! As far as I can tell...

[@joeybf]

+TheZanyCat Actually not. For example 1/K itself has a monad around it, made up of all the standard real multiples of 1/K^2. Similarly 1/K^2 has a monad, made of all the multiples of 1/K^3, and so on. Then you can start talking about things like 1/K^K and 1/K^(K^K), and it gets even more fun.

[@danielfinegold1989]

+TheZanyCat The monads are all infinitely small so there wouldn't be any overlap between them.

[@BluRey100]

+TheZanyCat fractals repeat the same constant as they get bigger and smaller infinitesimally. think of the Menger Cube...it's always a Cube but yet it's a Fractal. Divine!

[@nayutaito9421]

+Joey Beauvais-Feisthauer 1/K^2 is also included in the monad of zero, right?

[@549231]

Is 1/k bigger or smaller than 1-0.999999... ?

[@omp199]

+Nic Szer I will assume that you mean K, not k. 0.999... = 1, so 1 - 0.999... = 1 - 1 = 0. And 1 / K > 0.

[@549231]

But is 0.999...=1 in R*? At least with limits to approach 0.999..., 1/K is smaller, since whatever number you use, K will be bigger, and doesn't that mean that, in R*, 1/K is smaller than 1-0.999? I mean, if with limits it is smaller, where does it make the transition fron smaller to bigger when you replace the limit with 0.999... . Doesn't this mean that it is at most equal, but mean bigger. (?)

[@WarmongerGandhi]

+Nic Szer 0.999... isn't infinitely close to 1. 0.999... doesn't approach 1 in the limit. It is another way of writing 1, in the same way that 2/2 is another way of writing 1. 0.999... =2/2 = 0.5 + 1/2 = 5^0 = e^(2πi) = 5-4 = ln(e) = (-1)^2 = i^4 = 1. 0.999... = 1 doesn't change in R* (or any other set), nor do any of those other ways of writing 1 (except, obviously, that e doesn't exist in N, etc.). If any of them did, that would mean 1 ≠ 1, and all of math is broken. However, 1/k IS less than 1 - 0.9...9 for any finite number of 9s. I.e., 0.1 > 0.01 > 0.001 > 0.0001 > ... > 1/k > 0 = 1 - 0.999...

[@omp199]

Nic Szer To say that 0.999... = 1 is just to say that the limit of the partial sums of the series 9/10^1 + 9/10^2 + 9/10^3 + ... is equal to 1. That limit is exactly equal to 1. If you have an alternative way of interpreting 0.999... other than as a limit, then I would be interested to hear it.

[@549231]

Thanks for the explanation

[@Vladi1295]

Where is n/k then ?

[@mantkcmantk8717]

It is n times 1/k. If n is a natural number, n/k is still an infinitesimal.

[@kalleguld]

Is 0.9̅ still equal to 1, or is it 1 - 1/k ?

[@MuffinsAPlenty]

+Kasper Guldmann It depends on how you define "K" and "0.999..." Usually, when people are claiming that 0.999... = 1, they are working in R (not R*). In R, there is only one really natural definition for what infinite decimal notation means, and so we define infinite decimal notation to be the limit of the sequence of the first n decimals. So, 0.999... = lim{0.9, 0.99, 0.999, 0.9999, ...}. It is based on this definition that 0.999... = 1. In the the hyperreal numbers R*, there are multiple natural ways you could define infinite decimal notation. In some of them, 0.999... does not equal 1 but is infinitely close to 1 (here, the difference would be some "rational" infinitesimal number: depending on how K is defined, the difference could very well be 1/K, but may not be). In other possible definitions, 0.999... is still equal to 1.

[@BeCurieUs]

More like this, more like this more like this!

[@ARBB1]

Very interesting.

[@KarlFFF]

Does all numbers in the Monads have their own Monads? - I think the answer would be yes, I can't see why it shouldn't.

[@aman-qj5sx]

No. -2+1/k does not have a monad because -2+1/k+1/k is just another member of the same monad.

[@Zafoshin]

That would be N** prolly

[@CosmiaNebula]

Of course, but it would be just a "submonad" of that monad.

[@ronmedina429]

Yes but it would be the same monad. Monad(b) are all hyperreals that are infinitely close to b, if we take any number in this monad it would be of the form b + e where e is infinitesimal, and Monad(b + e) are all infinitely close to b + e, ie. of the form b + e + e`, where e` is some infinitesimal. But e + e` is infinitesimal, and thus is also in Monad(b), thus Monad(b+e) is contained in Monad(b) and similarly, Monad(b) is in Monad(b+e).

[@JaapVersteegh]

I guess all hierarchies of monads are still smaller than the surreals

[@poncio2632]

Honestly I understand this really well, it's really the same and the opposite of saying there are infinite numbers between x and y, but on the number itself

[@DiamondzFinder_]

What about (1+(1/k))/k ?

[@mantkcmantk8717]

It is 1/k + 1/k^2.

[@martinshoosterman]

but you can still go in between these infinitesimals. between 0 and 1/k is (1/k)/2 infact there is even 1/k^2

[@nirhghs9783]

This brings question to my mind, does infinitesimal of infinitesimal exists, even this can be repeated infinite times. In infinitesimal circle of 1/K, does any point like 1/K on smallest infinitesimal circle can be divided into infinitesimal circles also, and can that be done infinite number of times. I mean nested infinitesimal in infinite number of times. This makes sense as you can have any point on original number system that infinitesimal circle, any point on infinitesimal circle can have another level of infinitesimal circle and this process gets repeated infinite number of times. Otherwise, if this is not possible then you are breaking mathematics laws of distance between any 2 points of number system has fixed number of points.

[@KronFox]

I wonder how p-adic numbers are related to infinitesimals and hyperreals

[@JaapVersteegh]

So, is there a relation between N* and omega?

[@b43xoit]

Omega comes from ordinals.

[@brianpark3910]

does K/K = 1?

[@BewilderedBird]

I assume no, just like how infinity/infinity is not = 1

[@MuffinsAPlenty]

K/K = 1. This follows from the transfer principle. In the real numbers, the following sentence is true: (∀x)(x≠0⇒x/x=1) Thus, by the transfer principle, it is also true in the hyperreal numbers. When we're dealing with limits using the standard real numbers, we just have two symbols: +∞ and −∞, one representing something larger than all real numbers and one representing something smaller than all real numbers. But in the hyperreal numbers, you don't just have one of each such number. K, K+1, K−π, K^2, 2K−4, etc. are all hyperreal numbers which are greater than any real number, and arithmetic on them works like ordinary arithmetic. So something like 2K/K = 2, and (K+5)−(K) = 5. Perhaps you can think of using +∞ as collapsing all of these "bigger than all real numbers" numbers into a single symbol and trying to retain its properties. Sure, some things are unambiguous, but other things are ambiguous when doing so.

[@shruggzdastr8-facedclown]

What about Nigel Tufnel's favorite number (insert long, ponderous pause... here), 11?

[@pijiu4u]

So.. one cannot fathom what is the last digit of infinity, but would the last digit of an infinitesimal be 1? as in 0.000....01?

[@angelmendez-rivera351]

Chuck Zhang No, there is no such a thing as the last digit of an infinitesimal. Infinitesimals, by construction, do not have decimal expansions. Only the real numbers do that. 0.000... 1 does not exist.

[@T3sl4]

It bothers me that "real numbers" are called as they are. Has anyone had any proposals for renaming them? The "Unreals" perhaps? It's probably caused problems to many a student, that the Complex or Imaginary Numbers as called they are (they aren't very complex at all, just made of two parts!), but this bothers me even more than that. And in a much more insidious way, because it's a very nuanced thing to understand why this should be, and this truth has, only very gradually and relatively recently, come within my understanding. There is fundamentally no way to represent a real number, in the entirety of our universe! We can only speak of impossibly boring numbers, which happen to map from the rationals, like 0.000001, or of impossibly lucky numbers, like pi, which has a rational approximation around 3.14159..., but which is infinitely richer than anything we can express. Why, then, should reals continue to be called as such!?

[@angelmendez-rivera351]

Tim Williams Because mathematicians are the worst at nomenclature.

[@fould13]

Monoternions?

[@b43xoit]

> Dedekind cuts.

[@Jopie65]

1. So if K is in R* and not in R, is 1/K in R or not? 2. x^0 = 1. INFINITE^x = INFINITE. INFINITE^0 = undefined. But how much is K^(1/K) ? (Or the K's root of K)

[@alexanderreynolds9705]

+Johan 't No 1/K is not in R. The real numbers have the property that all non-zero numbers can have their reciprocal taken. I.e. if 1/K is in R then 1/1/K = K is in R.

[@SquallLeonhart4]

+Johan 't Hart My answer are the solutions of the relative limits. Watching the video it seems she's talking about limits in a different way than the usual. The limit for k that tend to infinity of 1/k is 0. The limit for k that tand to infinity of k^x is infinity if x>0, 1 if x=0, 0 if x

[@Jopie65]

+Ivan Marino I thought of it a bit. And I also think K^(1/K) = 1. Here's my reasoning. Consider f(x) = x^(1/x). f(1) = 1 f(2) = 1,414 (or root(2) ) f(3) = 1,44 f(4) = 1,414 f(5) = 1,380 f(6) = 1,348 ... f(100) = 1,047 So it looks like it starts at 1, bumps up, and then asymptotically nears 1 again. Its probably not a formal mathematical proof, but I think this pretty much makes it clear the result is 1 :) Just to troll, consider this: f(K - 1) > f(K) but f(K -1) = 1 f(K) = 1 soooo... 1 > 1 ??? :)

[@SquallLeonhart4]

no, the mistake is in suppoding that f(k-1)>f(k). k=k-1, than f(k-1)=f(k). However you're reasoning is right about the limit. How you said isn't a math prove but you can see thebehavior of the funcion in this way.

[@HorrorFreak1408]

+Johan 't Hart close but no cigar. As 1/(k-1) is still larger than 1/k, it stands to reason that (k-1)^(1/(k-1)) is less than k^(1/k)

[@douro20]

Now we have people talking about uncertainty...it can certainly drive people mad!

[@guycomments]

So first, at the inception of the real numbers, we had mathematicians saying there were infinitely many steps in between 0 and 1. And now we are saying that there are infinitely many steps in between the infinitely many steps. It starts to seem so redundant when I try to really wrap my head around it.

[@HorrorFreak1408]

which of course leads to an infinite number of levels of infinitesimals lol

[@superbuu122]

But there aren't infinite many steps, because there's 2 destinations, 0 to 1, a beginning and an end, so at some point you begun and at some point you ended. There is A LOT of steps yes, however they are countable, be asue you have a start and end point.

[@superbuu122]

But there aren't infinite many steps, because there's 2 destinations, 0 to 1, a beginning and an end, so at some point you begun and at some point you ended. There is A LOT of steps yes, however they are countable, be asue you have a start and end point.

[@sbgirl54]

Isn't this homomorphic to R^2? to attribute a infinitesimal part, mapped to a member of R, to another real number?

[@MuffinsAPlenty]

Homomorphic? As rings? Or as something else? One thing to keep in mind is that ring homomorphisms are very strict. Even though we often thing of C and R^2 as being more or less the same thing, there is no "nice" ring homomorphism between them. C is a field, but R^2 is not even an integral domain. The same thing applies to R*. R* is a field. Or did you mean homeomorphic as topological spaces?

[@b43xoit]

R^2 doesn't have order.

[@BinaryDash]

Awesome.

[@assimptota1]

1/(2+1/K)=?? can i simplify this?

[@assimptota1]

+Gustavo Marques and is this in the "finite" part of the reals* or in the "infinite/k" part of the reals*?

[@thebil]

+Gustavo Marques What do you mean by "simplify"? If the regular algebra works, that's the same as k/(2k+1). Does that count as simplification?

[@thehackerman1234]

+Gustavo Marques it would be in the finite part. You're dividing 1 by something infinitesimally bigger that 2, so the answer is a number infinitesimally smaller than 1/2. I'm not sure if you can get an exact answer but you can use the binomial approximation and get 1/(2+1/K) ~= 1/2*(1-1/(2K)) = 1/2 - 1/(4K)

[@ediza.8485]

+Gustavo Marques Just say that 1/k=0 to make things easy for now, then the equation becomes 1/(2+0) which is 1/2. Except since 1/k is infinitely close to 0 the real answer would be infintely close to 1/2 though I am unsure how much infinitely close it would be.

[@ediza.8485]

+thebil No that does not count as simplification. If someone asks you to simplify the equation (2+2) do you say ''Yeah that's easy, the simplified version would be 2x(2+2)'' ?

[@abelgerli]

And yet another explanation of Monads that are totally different from computer science where we use them. 😯😯😯 But a nice explanation with a Monads as an abstraction.

[@tjthreadgood818]

Irresistible Fun!

[@mueezadam8438]

_(Me before watching this):_ You can’t say *“after all the natural numbers”* that makes no sense! _(Me after Professor Wood’s explanation):_ N* and monads are mathematically rigorous. Sorry, not up for debate.

[@jnm236]

Think of K as an independent dimension rather than infinity, and the confusion disappears. The theoretical information content in `a + Kb + c/K` includes the real number a, the real number b and the real number c.

[@burtonlang]

Brady makes the comment that the number of fractions in R is nothing compared to the number of fractions in R*. But there are as many elements in [0,1]_R as there are in R itself. As far as I can intuit, R* has the same cardinality as R, which would mean that |[0,1]_R| = |[0,1]_R*| and so the number of fractions is the same. Right?

[@KinataKnight]

+Alex Bicksler It's worth mentioning that R^ isn't actually a single structure (using ^ instead of * to avoid KZbin bolding). By the compactness theorem, N and R can be extended to some N^ and R^, and those in turn can be extended further to some N^^ and R^^, each of which could have served as the original N^ and R^, and so on. And by another logic theorem, Lowenheim-Skolem, there will be N-like and R-like structures of arbitrarily large cardinality.

[@bakedbreadchickenegg]

Okay, I've been reading a book called "The Infinite Book" by John D. Barrow and in it he talks about power sets of infinity being higher orders of infinity. Although he hasn't defined infinity yet (at the point I'm at in the book), he does say (I think) that the power set of countable infinity is an infinity of a higher order and that to get the next higher order, you must then take the power set of this! So, is the set of all real numbers as big as the set of the power-set of countable infinity? On another completely (sarcasm) different note, is 1 / magnitude of the set of all integers, bigger than 1 / K, or is it smaller? ...Or the same number? THANKS

[@angelmendez-rivera351]

Thomas Bridge The power set of the countable infinite cardinal K is, to be precise, 2^K, and this is strictly bigger than K. However, it is not known and it is furthermore not provable that 2^K is equal to the cardinality of the real numbers.

[@b43xoit]

In cardinal infinities, yes. But this K seems to come from a different system than them.

[@EnergyCuddles]

This upsets me. x)

[@przemekkobel4874]

Seems like even square roots went to war and sqrt(2) has taken over land previously occupied by sqrt(3).

[@km1dash6]

Weren't the Pathagerians big on infinitesimals? And didn't Zeno prove they didn't really exist?

[@kashmirha]

Besides she is a great lecturer, she can make that sound too: 4:30 :D

[@beardydave926]

Is there a set of numbers where SQRT(x)>x? They probably can't exist on the real, integer, imaginary or any other number line and have their own plane of existence. If they exist, do they have a name? Have they yet to be discovered?

[@MuffinsAPlenty]

Yes, if 0 < x < 1, then sqrt(x) > x. As an example, sqrt(1/4) = 1/2 since 1/2 * 1/2 = 1/4. And 1/2 > 1/4.