Infinitesimal Monad (extra footage)

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Numberphile2

Numberphile2

8 жыл бұрын

Extra footage with Professor Carol Wood.
Main video: • The Infinitesimal Mona...
Rubik's Cube videos: bit.ly/rubiksvids
Exactly a Yard: • Exactly a Yard - Objec...
NUMBERPHILE
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Пікірлер: 339
@sserpxeadnap
@sserpxeadnap 8 жыл бұрын
Please please do more videos with Professor Wood!! Extremely enjoyable
@calebritchie1069
@calebritchie1069 9 ай бұрын
She’s so engaging!!
@lethargogpeterson4083
@lethargogpeterson4083 8 жыл бұрын
I love Brady's "No!"
@mahdibnhussein4882
@mahdibnhussein4882 6 жыл бұрын
Lethargo G Peterson Oh I understand what do do you mean 👌
@cmd2tuts
@cmd2tuts 8 жыл бұрын
Professor Carol Wood is a fascinating person, it would be great to see more of her in the future.
@reallyWyrd
@reallyWyrd 8 жыл бұрын
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
@raywilliams6717 4 жыл бұрын
Omg that's great.
@ffhashimi
@ffhashimi 8 жыл бұрын
I think this woman is an amazing teacher
@mondoke
@mondoke 8 жыл бұрын
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.
@levinb1
@levinb1 4 жыл бұрын
1:11 Her thesis advisor was ABRAHAM ROBINSON!!!!! No wonder!!!
@SpamDestroyer
@SpamDestroyer 7 жыл бұрын
I love how this immediately starts off with "It's worse than that!"
@garethdean6382
@garethdean6382 8 жыл бұрын
Glorious. I want to print this video onto a custom cake and eat it.
@VladVladislav790
@VladVladislav790 8 жыл бұрын
-"even K!" - brain exploded
@mahdibnhussein4882
@mahdibnhussein4882 6 жыл бұрын
So watch the video "How to count past infinity" for Vsauce channel
@Bakkerkid
@Bakkerkid 3 жыл бұрын
But that would also mean "1/K" has ITS OWN MONAD! The number line just became an infinite line of infinite infinitessimals.
@unclvinny
@unclvinny 7 жыл бұрын
Dr. Wood is great, hope you can do more videos with her.
@BewilderedBird
@BewilderedBird 7 жыл бұрын
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.
@fantiscious
@fantiscious 2 жыл бұрын
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_
@ooos2989
@ooos2989 5 жыл бұрын
Do a video on the smooth infinitesimal analysis too! It's similar to non-standard analysis in that it uses infinitesimals.
@FishKungfu
@FishKungfu 8 жыл бұрын
The infinitesimal monad set is really cool and I like it, but it agitates me at the same time.
@thedoublehelix5661
@thedoublehelix5661 3 жыл бұрын
fun fact: its technically not even a set. Its too big to be a set! Its something called a class
@TheBasikShow
@TheBasikShow Жыл бұрын
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.
@sinecurve9999
@sinecurve9999 8 жыл бұрын
njwildberger just flipped out....
@Alex-mg6vn
@Alex-mg6vn 8 жыл бұрын
+sinecurve9999 If he finds real numbers appaling, hyperreals and infinitesimals must be his worst enemy!
@mushkamusic
@mushkamusic 8 жыл бұрын
this is bloody fascinating!
@firstlast-wg2on
@firstlast-wg2on 7 жыл бұрын
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
@firstlast-wg2on 7 жыл бұрын
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.
@JLConawayII
@JLConawayII 8 жыл бұрын
It gets worse?!!
@thomas.edgerton
@thomas.edgerton 8 жыл бұрын
+JLConawayII I would say better !
@micknamens8659
@micknamens8659 4 жыл бұрын
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 ?
@zachsmith9841
@zachsmith9841 6 жыл бұрын
@km1dash6
@km1dash6 4 жыл бұрын
I don't know why, but I really like infinitesimals.
@BluRey100
@BluRey100 8 жыл бұрын
She is correct. This lines up with Fractals, Omega Code, Menger Cube and String Theories. She deserves an Honorary Cosmological Degree.
@brendawilliams8062
@brendawilliams8062 3 жыл бұрын
I know you enjoyed that study. Thankyou.
@hreader
@hreader 5 жыл бұрын
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
@JA-cn6vu 4 жыл бұрын
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
@stevenlu7324 2 жыл бұрын
@@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.
@Hythloday71
@Hythloday71 8 жыл бұрын
Prof. Norman J Wildberger will be most cross !
@twwc960
@twwc960 8 жыл бұрын
+Hythloday71 He doesn't get cross. He just sort of rolls his eyes and smirks.
@trissylegs
@trissylegs 8 жыл бұрын
+Hythloday71 I was just thinking that.
@MauveTendingToBeige
@MauveTendingToBeige 8 жыл бұрын
+Hythloday71 Maybe Wildberger can give his vision on Numberphile. Would be cool...
@twwc960
@twwc960 8 жыл бұрын
MauveTendingToBeige He lives in Sydney, so next time Brady goes back home to Australia, maybe he can film a segment or two.
@weezhaochuakhoo1032
@weezhaochuakhoo1032 8 жыл бұрын
What would he be crossed about?
@Kram1032
@Kram1032 8 жыл бұрын
What happened there at 4:00 - 4:02 ? It's just blacked out and you begin to say something mid-sentence after that?
8 жыл бұрын
+Kram1032 Illuminati confirmed.
@cavalrycome
@cavalrycome 8 жыл бұрын
+Kram1032 They had to censor the dark math.
@bgezal
@bgezal 8 жыл бұрын
+Kram1032 Brady blacked out from sheer insight. --Oh my god it's full of stars.
@effortless35
@effortless35 8 жыл бұрын
+cavalrycome In 2015 after decades of searching we finally found the dark mather.
@Kram1032
@Kram1032 8 жыл бұрын
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.
@captainharris8980
@captainharris8980 7 жыл бұрын
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.
@pbj4184
@pbj4184 4 жыл бұрын
Oh that Abraham Robinson was her advisor. Wow!
@notoriouswhitemoth
@notoriouswhitemoth 8 жыл бұрын
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
@angelmendez-rivera351 5 жыл бұрын
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
@angelmendez-rivera351 5 жыл бұрын
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
@angelmendez-rivera351 5 жыл бұрын
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
@b43xoit 3 жыл бұрын
@@angelmendez-rivera351 Either set theory doesn't have a universe, or I missed something. Russel paradox?
@ChristOfSteel
@ChristOfSteel 8 жыл бұрын
Having a background in category theory and their monads, I wonder how these monads relate to this monads?
@darkaquatus
@darkaquatus 8 жыл бұрын
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
@PianoMastR64 8 жыл бұрын
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
@GenericInternetter 6 жыл бұрын
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
@angelmendez-rivera351 5 жыл бұрын
Generic Internetter Not at all. There are no infinitesimals in sub-atomic physics specifically.
@ianmoseley9910
@ianmoseley9910 5 жыл бұрын
Angel Mendez-Rivera uncertainty principle
@angelmendez-rivera351
@angelmendez-rivera351 5 жыл бұрын
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.
@aaronjones553
@aaronjones553 8 жыл бұрын
Seems like an extension of this to the complex plane would be interesting and extremely brainbending.
@inthefade
@inthefade 8 жыл бұрын
This reminds me of the first time I dropped acid.
@tjthreadgood818
@tjthreadgood818 3 жыл бұрын
Irresistible Fun!
@randomperson10082
@randomperson10082 8 жыл бұрын
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
@ZardoDhieldor 8 жыл бұрын
+TheZanyCat You're right! As far as I can tell...
@joeybeauvais-feisthauer3137
@joeybeauvais-feisthauer3137 8 жыл бұрын
+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
@danielfinegold1989 8 жыл бұрын
+TheZanyCat The monads are all infinitely small so there wouldn't be any overlap between them.
@BluRey100
@BluRey100 8 жыл бұрын
+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
@nayutaito9421 8 жыл бұрын
+Joey Beauvais-Feisthauer 1/K^2 is also included in the monad of zero, right?
@BinaryDash
@BinaryDash 8 жыл бұрын
Awesome.
@guildpilotone
@guildpilotone 8 жыл бұрын
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
@MozartJunior22 8 жыл бұрын
+guildpilotone What numbers would be in this monad?
@bgezal
@bgezal 8 жыл бұрын
+guildpilotone Sierpinski was a greater set theorist than Mandelbrot :)
@jdferreira
@jdferreira 8 жыл бұрын
+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
@MozartJunior22 8 жыл бұрын
+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
@jdferreira 8 жыл бұрын
+MozartJunior22 I hadn't considered that! I'll have to stop for a second and think about the implications!
@BeCurieUs
@BeCurieUs 8 жыл бұрын
More like this, more like this more like this!
@ARBB1
@ARBB1 2 жыл бұрын
Very interesting.
@SunnyTheGentleFox
@SunnyTheGentleFox 8 жыл бұрын
So basically, it's another way to think about Calculus.
@EnergyCuddles
@EnergyCuddles 8 жыл бұрын
This upsets me. x)
@KarlFFF
@KarlFFF 8 жыл бұрын
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
@aman-qj5sx 6 жыл бұрын
No. -2+1/k does not have a monad because -2+1/k+1/k is just another member of the same monad.
@Zafoshin
@Zafoshin 6 жыл бұрын
That would be N** prolly
@CosmiaNebula
@CosmiaNebula 6 жыл бұрын
Of course, but it would be just a "submonad" of that monad.
@ronmedina429
@ronmedina429 6 жыл бұрын
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
@JaapVersteegh 6 жыл бұрын
I guess all hierarchies of monads are still smaller than the surreals
@xpeson9
@xpeson9 8 жыл бұрын
I broke the mandelbrote shape by watching a video from numberphile "2"... And I dont regret the consequences yet.
@fould13
@fould13 5 жыл бұрын
"I have 3 sheep" "I have $20 of gas in my car" "I really wish I had K things" Even K is useful.
@RolandHutchinson
@RolandHutchinson 4 жыл бұрын
But if you had K things, where would you keep them? Even if you rented out every room in Cantor's hotel...
@scathiebaby
@scathiebaby 8 жыл бұрын
The cardinality (set size) of these numbers might be larger than the cardinality of the continuum !
@jesusnthedaisychain
@jesusnthedaisychain 8 жыл бұрын
God damn the man who invented everything to the right of the decimal point!
@poncio2632
@poncio2632 6 жыл бұрын
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
@jnm236
@jnm236 8 жыл бұрын
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.
@ThegamerIdiotgenius
@ThegamerIdiotgenius 8 жыл бұрын
Wow... what kind of classes greater than calculus can I take to learn about this stuff? (I'm taking AP Calc right now.)
@Scy
@Scy 8 жыл бұрын
So the 1/x graph where the line never crosses x=0? Can you explain that with these monad thingies?
@frosch03
@frosch03 8 жыл бұрын
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
@garthgoldwater5256 5 жыл бұрын
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
@adarshmohapatra5058 3 жыл бұрын
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.
@gryg666
@gryg666 8 жыл бұрын
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.
@ffhashimi
@ffhashimi 8 жыл бұрын
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 ?
@alaaanab
@alaaanab 5 жыл бұрын
nice!
@abelgerli
@abelgerli 3 жыл бұрын
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.
@TheLemmingeater
@TheLemmingeater 8 жыл бұрын
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?
@weakamna
@weakamna 8 жыл бұрын
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
@TheEternalVortex42 Жыл бұрын
Not really, since that's just a + 2/k. There is only one level of monads
@douro20
@douro20 6 жыл бұрын
Now we have people talking about uncertainty...it can certainly drive people mad!
@ekaingarmendia
@ekaingarmendia 7 жыл бұрын
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?
@vs-cw1wc
@vs-cw1wc 6 жыл бұрын
Question: Is this idea capable of talking about divergent series?
@Ensivion
@Ensivion 5 жыл бұрын
I wonder what would happen if you extended the mandelbrot set to the hyper-complex plane.
@mueezadam8438
@mueezadam8438 4 жыл бұрын
_(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.
@guycomments
@guycomments 8 жыл бұрын
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
@HorrorFreak1408 8 жыл бұрын
which of course leads to an infinite number of levels of infinitesimals lol
@superbuu122
@superbuu122 9 ай бұрын
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
@superbuu122 9 ай бұрын
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.
@burtonlang
@burtonlang 8 жыл бұрын
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
@KinataKnight 8 жыл бұрын
+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.
@przemekkobel4874
@przemekkobel4874 8 ай бұрын
Seems like even square roots went to war and sqrt(2) has taken over land previously occupied by sqrt(3).
@kasuha
@kasuha 8 жыл бұрын
I guess it puts a new spin on the old 0.99999... = 1 dispute since for any length of 0.999...999 < 1-1/K < 1.
@harktischris
@harktischris 4 жыл бұрын
4 years too late, no it doesn't seem like it would, because the fundamental thing you're missing is that .999... *is* 1. they're not distinct reals, just two different ways to write the same real, and so would not have their own separate neighborhoods of infinitesimals. ".999....999" is not a meaningful thing to write, because there is *no* terminating "9" for .99999... it just keeps going.
@kasuha
@kasuha 4 жыл бұрын
@@harktischris 0.999... is essentially a sum for n=0 to infinity of 9/(10^n). Such sum only exists if sequence of its partial sums has a limit. And in this kind of mathematics, there are no limits unless you redefine them, otherwise 1/K always lies in the way. Saying "it is 1" has no meaning, you have to prove it and you can't.
@harktischris
@harktischris 4 жыл бұрын
@@kasuha I suppose then it depends on what ".999..." notations mean, since looking around I found two different answers depending on who's talking and what definition is being used. If a decimal sequence .999... is defined as .a1a2a3...a{H}{H+1}{H+2}... where a1=a2=9, etc and a{H}=a{H+1}=9 where H is a hyperreal index then it seems obvious that even in R* .999...=1. If however .999... corresponds to a1=a2=etc=9 but a{H}=a{H+1}=0 then .999... is certainly not equal to 1 and a vast infinitesimal sea lies between them
@delwoodbarker
@delwoodbarker 8 жыл бұрын
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.
@willmartin7108
@willmartin7108 8 жыл бұрын
So is this like a bridge between discrete and continuos math?
@b43xoit
@b43xoit 3 жыл бұрын
What you mean "bridge"?
@kuro68000
@kuro68000 8 жыл бұрын
The sound is badly messed up on this video.
@km1dash6
@km1dash6 5 жыл бұрын
Do infinitesimals have even smaller infinitesimals surrounding them?
@b43xoit
@b43xoit 3 жыл бұрын
No.
@liviumarginean1474
@liviumarginean1474 8 жыл бұрын
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
@b43xoit 3 жыл бұрын
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.
@HighKingTurgon
@HighKingTurgon 8 жыл бұрын
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?
@KomnataNorth
@KomnataNorth 7 жыл бұрын
Very cool, and very logical, everything fits [(K-1)/K]-1= - (1/K)
@nirhghs9783
@nirhghs9783 3 жыл бұрын
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.
@TrimutiusToo
@TrimutiusToo 8 жыл бұрын
What about infenitesimal fraction? (like rational, but not natural)
@MozartJunior22
@MozartJunior22 8 жыл бұрын
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
@kuroninja90 8 жыл бұрын
+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
@MozartJunior22 8 жыл бұрын
+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
@mattcay 8 жыл бұрын
+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
@LeonhardEuler1 8 жыл бұрын
+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
@bgezal 8 жыл бұрын
+MozartJunior22 A monad is a set and epsilon is a quantity?
@zachariahhanson1792
@zachariahhanson1792 8 жыл бұрын
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
@thestarjon 8 жыл бұрын
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
@zachariahhanson1792 8 жыл бұрын
Starjon go on... Bear in mind that i have only done up to college level maths, which was essentially just mechanics, so... Yeah ;-)
@MuffinsAPlenty
@MuffinsAPlenty 7 жыл бұрын
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
@MuffinsAPlenty 7 жыл бұрын
***** 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
@MuffinsAPlenty 7 жыл бұрын
***** 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.
@ianprado1488
@ianprado1488 6 жыл бұрын
WHAT IS THE CARDINALITY OF R*???
@mantkcmantk8717
@mantkcmantk8717 5 жыл бұрын
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.
@kalleguld
@kalleguld 8 жыл бұрын
Is 0.9̅ still equal to 1, or is it 1 - 1/k ?
@MuffinsAPlenty
@MuffinsAPlenty 8 жыл бұрын
+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.
@shruggzdastr8-facedclown
@shruggzdastr8-facedclown 4 жыл бұрын
What about Nigel Tufnel's favorite number (insert long, ponderous pause... here), 11?
@y2536524
@y2536524 8 жыл бұрын
Can we have K divided by K in N star , is it one if defined?
@joeybeauvais-feisthauer3137
@joeybeauvais-feisthauer3137 8 жыл бұрын
+y2536524 Division is not defined in *N, or in N for that matter.
@Crazy_Diamond_75
@Crazy_Diamond_75 8 жыл бұрын
Oh no... this would be absolute insanity on a complex plane. It's fuzzy enough with just a single-component number line D:
@IsaacLevy
@IsaacLevy 8 жыл бұрын
Wesleyan mathematics dept is amazing
@DekarNL
@DekarNL 8 жыл бұрын
Numberphile once tought me that 0.99999 .... repeating = 1, so 1/K must be 0.
@km1dash6
@km1dash6 5 жыл бұрын
Weren't the Pathagerians big on infinitesimals? And didn't Zeno prove they didn't really exist?
@martinshoosterman
@martinshoosterman 8 жыл бұрын
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
@_abdul
@_abdul 3 жыл бұрын
why not ?
@scathiebaby
@scathiebaby 8 жыл бұрын
How does "K" relate to transfinite ordinals such as ω ? Because, the definition is similar.
@MuffinsAPlenty
@MuffinsAPlenty 7 жыл бұрын
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
@MuffinsAPlenty 7 жыл бұрын
***** 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
@MuffinsAPlenty 7 жыл бұрын
***** 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.
@KronFox
@KronFox 8 жыл бұрын
I wonder how p-adic numbers are related to infinitesimals and hyperreals
@bakedbreadchickenegg
@bakedbreadchickenegg 8 жыл бұрын
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
@angelmendez-rivera351 5 жыл бұрын
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
@b43xoit 3 жыл бұрын
In cardinal infinities, yes. But this K seems to come from a different system than them.
@godbelow
@godbelow 8 жыл бұрын
No, it's not what you think. It's much, much worse! - Dark Helmet
@JaapVersteegh
@JaapVersteegh 6 жыл бұрын
So, is there a relation between N* and omega?
@b43xoit
@b43xoit 3 жыл бұрын
Omega comes from ordinals.
@yongmrchen
@yongmrchen Жыл бұрын
The idea seems quite straight and intuitive, but I don’t think I get the idea about what it is even in an intuitive sense and why we do this in a purely mathematical sense.
@TheSnobar
@TheSnobar 8 жыл бұрын
This is what they teach us in analisi in italy for studying physics
@nishadseeraj7034
@nishadseeraj7034 7 жыл бұрын
so does that mean 1/k has a monad around it too?...what about the numbers within the monads?
@xxX_420BlazeIt_Xxx
@xxX_420BlazeIt_Xxx 7 жыл бұрын
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
@nishadseeraj7034 7 жыл бұрын
makes sense...thanks!!
@indigodarkwolf
@indigodarkwolf 7 жыл бұрын
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
@fabricenonez9098 7 жыл бұрын
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
@angelmendez-rivera351 5 жыл бұрын
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.
@beardydave926
@beardydave926 9 ай бұрын
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
@MuffinsAPlenty 3 ай бұрын
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.
@coreyredmon5611
@coreyredmon5611 8 жыл бұрын
My question: is 0.999... (0.9 repeating) equal to 1-1/K?
@Yomama5923
@Yomama5923 8 жыл бұрын
+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
@coreyredmon5611 8 жыл бұрын
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
@Hecatonicosachoron 8 жыл бұрын
+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.
@gideonbuckwalter4128
@gideonbuckwalter4128 8 жыл бұрын
Professor Wood's thesis advisor was Abraham Robinson!? That's so cool!
@abelincoln7473
@abelincoln7473 8 жыл бұрын
Good grief that really is a mathematical rabbit hole, particularly if numbers truly are discrete bits of information.
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