Construction of the Real Numbers

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  • Опубликовано: 7 ноя 2024

Комментарии • 102

  • @beatoriche7301
    @beatoriche7301 4 года назад +45

    This is a really good and concise explanation of Dedekind cuts - well done!
    For those who are interested, here's a bit of discussion around multiplication of cuts. (I'll let S denote the cut representing the real number s in the following.) Let's use the naive definition of cut multiplication that _S*T_ simply represents all products of rational numbers in _S_ and _T._ Now, if _s_ and _t_ are both strictly positive rational numbers, then in particular, _-2s_ and _-t_ are in _S_ and _T,_ respectively, and so their product is in _S*T_ - however, that product is _2st,_ and so it is larger than _s*t,_ which means our "cut" actually goes beyond the rational number it is meant to represent. In fact, if you've understood this step, it'll be pretty easy for you to show that the product (as we defined it) of two rational cuts greater than the zero cut will actually yield the entirety of the rationals, and so it's not even a cut - which is pretty bad because it violates the field axioms.
    Since negative numbers really mess up the whole thing, defining the multiplication of two cuts is a bit more subtle. You first start by only considering cuts greater than the zero cut (more precisely, cuts of which the zero cut is a strict subset) and only the rationals greater than 0 in these cuts. Then, the product of _S_ and _T_ may be defined as simply being the union of the zero cut and the set of all products of strictly positive elements of _S_ and _T._ Now, multiplying subsets of the zero cut can simply be defined using established consequences of the field axioms, such as _(-a)*b = -a*b._
    Related to that is a technical nuance not discussed in this video - the definition of the additive inverse itself. It's pretty natural to define _-A_ as the cut such that _A + (-A) = 0_ (the zero cut) - but that is not quite it. For instance, if you choose _A_ to be the zero cut, then _-A_ is not a cut because it has a greatest element of _0_ (straightforward exercise to the reader). However, our good friend Weierstrass comes to the rescue - instead defining _-A_ such that, given an element _q_ of _-A,_ there exists a strictly negative real number _ε_ such that, for all _p_ in _A,_ _p + q < ε_ will do the trick.
    I hope this explanation satisfies those looking for more details - though there are still some things I left out. However, I hope I gave a basic idea of how to deal with these technicalities.

    • @MichaelRothwell1
      @MichaelRothwell1 4 года назад

      How about this definition of "-A": Take the set of all rationals greater than every element of A, then take the set of opposites of this set.

    • @jonasdaverio9369
      @jonasdaverio9369 4 года назад +2

      Very good comment!

    • @kruksog
      @kruksog 3 года назад

      The anime profile pic made it hard to give the thumbs up, but you earned. (Just a joke, enjoy what you enjoy.)

  • @Domzies
    @Domzies 4 года назад +6

    You have an Interesting style of writting. You always start off with such big letters and somehow manage to write everything without cluttering it.

  • @mikhailmikhailov8781
    @mikhailmikhailov8781 4 года назад +16

    Equivalence classes of Cauchy sequences are bae and you wont convince me otherwise.

  • @gustavoorocha
    @gustavoorocha 4 года назад +6

    I am Brazilian and I study mathematics here. I love watching your videos to improve my English. Kisses dear!

  • @jonasdaverio9369
    @jonasdaverio9369 4 года назад +4

    I think you're a really good teacher for foundational mathematics

  • @theproofessayist8441
    @theproofessayist8441 Год назад +2

    I just had a crazy thought? If there was a collab with Dr Peyam and Masahiro Sakurai the universe would implode and projective geometry division by zero would become the new norm.

  • @hdheuejhzbsnnaj
    @hdheuejhzbsnnaj Год назад +1

    I think a bit more care could be taken in discussing how Dedikind is using the set of everything BELOW the irrational number itself to stand in for the number, and then defining operations on SETS (which don't include the number) to construct a field Q* that we later just swap out for R. It's a subtle point that may be lost on the viewer.

  • @subhrajitdasgupta3868
    @subhrajitdasgupta3868 3 года назад +3

    Thanks a lot professor ❤ respect!

  • @levav8
    @levav8 4 года назад +7

    this is really well done! could you do the construction of surreal numbers as well? as a tribute to John Conway, who died from covid19 a while ago.

  • @Kdd160
    @Kdd160 4 года назад +5

    Very intelligent

  • @riccardocas1239
    @riccardocas1239 2 года назад

    Thank you so much, finally understood it!!!!
    Didn’t get that the cuts are the actual reel numbers!!!

  • @balthazarbeutelwolf9097
    @balthazarbeutelwolf9097 3 года назад +1

    From a computational pov, Dedekind cuts and their arithmetic have a problem. We could implement them as a triple, consisting of a member (proof of non-emptiness), a non-member (proof of non-totality) and a characteristic function for set-membership. One can then define the characteristic function of a summation with a bit of approximation trickery, and the member/nonmember come in handy at that stage. Only problem is: this will not work completely when two irrational numbers add up to a rational one. Because in that case the constructed characteristic function will be undefined on one rational number, the actual summation result itself.

  • @MichaelRothwell1
    @MichaelRothwell1 4 года назад +1

    Here is a definition of π* that doesn't depend on any trigonometric functions:
    We know π/4=1-1/3+1/5-1/7+.... (the Leibniz formula for π, which is the Maclaurin series for arctan(1))
    We can write this alternating series as a sum of positive terms by grouping the terms in pairs:
    π/4=(1-1/3) + (1/5-1/7) + (1/9-1/11) + ....
    Or π/4=2/(1×3) + 2/(5×7) + 2/(9×11) +...
    So π=8/(1×3) + 8/(5×7) + 8/(9×11) +...
    Now that we have written π as the sum of a series of positive terms, it is the limit and also the supremum of the partial sums of the series.
    Hence π* is the union of the cuts corresponding to the partial sums of the series.
    This method can be used to define the cut for any real number expressed as the sum of a series of non-negative terms, for example for e=1+1/2!+1/3!+... .

    • @drpeyam
      @drpeyam  4 года назад +4

      Except you need limits to define series and you need Sup and real numbers to define limits :)

    • @MichaelRothwell1
      @MichaelRothwell1 4 года назад +2

      @@drpeyam Sure, you need real numbers to prove that my cut works, but the definition of my cut (as the union of the cuts of rational numbers) doesn't require any knowledge of real numbers.

  • @jayjayf9699
    @jayjayf9699 4 года назад +2

    I was looking at my old lecture notes a few months ago it would of been good if these videos were around then , I know the construction were to do with dedekind cuts

  • @rohithnarra9026
    @rohithnarra9026 4 года назад

    You the man Dr Peyam!!!!!!

  • @rabindranathghosh31
    @rabindranathghosh31 4 года назад +9

    Seems like you changed your thumbnail format. This looks colourful and attractive!

    • @Kdd160
      @Kdd160 4 года назад +4

      Yeah u are correct

    • @tlk1432
      @tlk1432 4 года назад

      I think the former thumbnails were better tbh,reminds me of BPRP's thumbs,which are great

  • @dgrandlapinblanc
    @dgrandlapinblanc 2 года назад

    Ok. Thank you very much.

  • @awesomecraftstudio
    @awesomecraftstudio Год назад

    You say that any subset of elements greater than a given subset is not a cut. However in my textbook, cuts are defined exactly the other way around, where they have no upper limit and no smallest element. It would make sense that you could define it both ways right?

  • @royeyckmans260
    @royeyckmans260 2 года назад

    Very helpful, thank you!!

  • @superlative_custard
    @superlative_custard 2 года назад

    Very helpful thank you

  • @johnnykarki6347
    @johnnykarki6347 4 года назад +3

    At 3:03, real numbers are subset of rational numbers? Isn't it the other way around?

    • @johnhippisley9106
      @johnhippisley9106 3 года назад +1

      *A* real number is a subset of the rational numbers. The reals are the set of all of those real numbers

  • @toaj868
    @toaj868 4 года назад +1

    Can C be constructed by taking the cartesian product of R with itself and then forcing the addition and multiplication properties?

    • @drpeyam
      @drpeyam  4 года назад +5

      I mean that’s kind of what C is

  • @debendragurung3033
    @debendragurung3033 4 года назад +1

    BRILLIANT, but what does all that mean?

  • @aneeshsrinivas9088
    @aneeshsrinivas9088 2 года назад

    Can every algebraic number be written as a cut in a similar fashion to sqrt(2)? If so how would you prove it?

  • @TheEarlVix
    @TheEarlVix 4 месяца назад

    As soon as you start involving infinite sets your mathematics is skating on thin ice.

  • @gavintoohey6604
    @gavintoohey6604 3 года назад

    Thanks mate :3

  • @zhouj8634
    @zhouj8634 4 года назад

    Real awesome!

  • @aniruddhnls
    @aniruddhnls 2 года назад

    at 4:45 in condition 2, the s that is a rational number less than r, is that the same S as the real number? if not, why use the same variable, and if it is, then I do not understand condition 2.

  • @anfnle4394
    @anfnle4394 4 года назад

    Hi dr.Peyam can you make a video if possible on constructing the lebesgue measure?But like without using the "caratheodory approach" i.e. restricting the lebesgue outer measure to the lebesgue sigma algebra,and instead using the riesz representation theorem approach just like in paparudin.

  • @aneeshsrinivas9088
    @aneeshsrinivas9088 2 года назад

    Is it true that for all cuts and every rational number eps>0 then there is a number x in X such that x+eps is not in X? I’m pretty sure this is true and have a proof written up but i’m just trying to make sure. So is it true dr peyam. I thank your response if you do respond.

    • @aneeshsrinivas9088
      @aneeshsrinivas9088 2 года назад

      I think you can prove this by proving that for such a set, x,x+eps,x+2eps,x+3eps,… is in here assuming x is a member of X. so basically the whole rationals are covered by cut axioms. since for any n in whole numbers x+n*eps is in here. Its a matter of using the archemedian property to conclude that any y in rationals is in this set and get a contradiction with this set being a cut.

    • @aneeshsrinivas9088
      @aneeshsrinivas9088 2 года назад

      I meant that for such a set where there is eps>0 such that x in X->x+eps in X.

  • @jayjayf9699
    @jayjayf9699 4 года назад

    Why is the cut defined from less than a number? Surely the same logic could be use to create a cut from a number greater than instead of less than to negative infinity of all rational numbers?

    • @drpeyam
      @drpeyam  4 года назад

      Yeah sure, it’s symmetric

  • @我妻由乃-v5q
    @我妻由乃-v5q 4 года назад +2

    How do you define pi with the cuts?

    • @fellipeparreiras4435
      @fellipeparreiras4435 3 года назад

      Let a_n converge to pi, with a_n < pi for all n. Then the dedekind cut that represents pi is {x in Q | x > a_n for all n}.

    • @s1mppeli
      @s1mppeli 3 года назад

      @@fellipeparreiras4435 Correct me if I’m wrong, but this is circular logic. The limit of a sequence (epsilon-delta) requires a definition of a real number, since we start from the assumption that epsilon and delta are real. ayou can’t use limits to define the reals if you’ve already used the reals to define limits.

    • @fellipeparreiras4435
      @fellipeparreiras4435 3 года назад +1

      @@s1mppeli I guess you're right. I just searched this at the time and thought it made sense. Im curious to see a solution to this. I think we just take as an axiom that the set of Dedekind cuts has as many elements as R in the naive idea of real numbers

    • @billh17
      @billh17 2 года назад

      ​@@s1mppeli said "Correct me if I’m wrong, but this is circular logic."
      Yes and no. One does not need to justify how one got to such a sequence a_n. One can write down that sequence without mentioning limits. Thus, { x in Q | x > a_n for all n} is well-defined and one can show that it is an upper Dedekind cut.
      However, all of this is moot. After one has defined the real numbers as Dedekind cuts and showed that they form a complete ordered field and that the least upper bound theorem is true and limits of real numbers can be defined using epsilon-delta, the notion of real numbers as Dedekind cuts is no longer at the forefront and the theorems from being a complete ordered field go to the forefront. That is, particular real numbers like pi and e are not defined as Dedekind cuts but instead each is defined as a limit of a particular sequence of real numbers.
      Consider the following example. One wants to solve the differential equation dy/dx = y. One finds the series solution exp(x) = 1 + x + x^2/ 2 + .... Note that exp(x) is defined for all real numbers x (not just rational numbers x). Then, e is defined as exp(1) where 1 is the real number 1 (not the rational number 1). Note that e = exp(1) is not defined as a Dedekind cut (although one can "unwind" things and work out what the Dedekind cut for e = exp(1) actually is, but this is not necessary).
      Compare with the usual definition of negative integers as being equivalence classes of ordered pairs (n, m) of natural numbers. For example, -5 = { (0, 5), (1, 6), (2, 7), ...}. After one shows that the integers form a ring, one forgets about them being ordered pairs since the negative of the integer 5 works just fine for negative 5 (and there is no longer the need for the equivalence class of ordered pairs of natural numbers).

    • @s1mppeli
      @s1mppeli 2 года назад

      @@billh17 Thanks for taking the time to reply. I'm really eager to understand this, since the definition of real numbers has been a real sore spot for me lately.
      I still can't quite seem to wrap my head around it.
      You say: "One can write down that sequence without mentioning limits", could you elaborate? I don't believe that to be true. Without limits, the only way to define such a sequence would seem to be to actually list all the terms. There is an infinity of them, so you cannot, in fact, ever list them all.
      It seems you are pretty much using the same circular logic as before with an extra step of axiomatically saying "this sequence exists" without the tools to define it unambiguously. Then using the existence of this unjustifiable sequence, you formulate tools to actually define it.
      This feels baseless. What am I missing?

  • @hansjiang373
    @hansjiang373 4 года назад +6

    Hi, this "real numbers" playlist, are the videos in the correct order?

    • @drpeyam
      @drpeyam  4 года назад +2

      Yes

    • @hansjiang373
      @hansjiang373 4 года назад +5

      @@drpeyam Wow thanks for your reply!! Youre an amazing teacher, I love all your videos

    • @drpeyam
      @drpeyam  4 года назад +3

      Thank you ❤️

  • @shawon265
    @shawon265 4 года назад

    So from the first point of dedekind cut -∞ and ∞ are not real numbers. Is my conclusion correct?

    • @drpeyam
      @drpeyam  4 года назад +1

      Yes

    • @shawon265
      @shawon265 4 года назад

      @@drpeyam Wow! You replied so quickly! Thank you so much!

  • @hexeddecimals
    @hexeddecimals 4 года назад +1

    The cut for root 2 seemed like a bit of a hack :(

  • @jarrodanderson2124
    @jarrodanderson2124 10 месяцев назад +1

    Can an irrational number times an irrational number ever be an irrational number?

    • @drpeyam
      @drpeyam  10 месяцев назад +1

      Yes, sqrt(2) x sqrt(3) = sqrt(6)

  • @ekadria-bo4962
    @ekadria-bo4962 4 года назад +1

    Is this Dedekind Cut?

  • @User_Name.1
    @User_Name.1 2 года назад

    Please don’t get it wrong but this is not a construction of R. You give examples of cuts and defined the dedekind cut but you didn’t construct R from Q. First of all you get sets and not numbers, so if you think about how you construct Q from Z then the new rational numbers look like the numbers we already knew. That’s not the case with the cuts. You actually have sets and not numbers as Elements and that’s why you can picture or visualize them good but are difficult to handle for implementing the idea in other theories. That’s also the reason why the Dedekind cuts can only be used in Dimension 1 and not higher. For calculating or implementation in other theories it makes more sense to use Cantor’s definition via Cauchy sequences. Another notice is that you don’t make a difference between the cuts of Q and R, what I mean by that is for example the rational number 2 is written as the cut r_2={r element of Q | r

    • @drpeyam
      @drpeyam  2 года назад

      ?

    • @drpeyam
      @drpeyam  2 года назад

      It’s the standard construction of R actually

    • @VS-is9yb
      @VS-is9yb Год назад +1

      ​@@User_Name.1He used only rational numbers to define Dedekind cuts

    • @User_Name.1
      @User_Name.1 Год назад

      @@VS-is9yb that’s right but it’s still not a construction of IR. First of all he defines irrational numbers as Sets of Q and there is no converting from sets to numbers! Secondly he is telling that there is no way to show pi as a set of Q like he showed for root 2. why is there no cut example for pi? Lastly there are a lot of books who are claiming to construct IR through Dedekinds cut but all of them are showing the same example of root 2 but no one have a cut for pi or e or infinite other irrational numbers. By the way Most of the books don’t show that you have to convert the sets to numbers! Some do and therefore the construction of IR is incomplete in this video. Hope it’s more clear now.

  • @papa15891
    @papa15891 4 года назад

    Looks like surreal numbers

  • @foreachepsilon
    @foreachepsilon 4 года назад +1

    I’ve always wondered about dedekind cut math. Any recommendations of online discussions on how to multiply two cuts, divide, square, etc.?

    • @billh17
      @billh17 2 года назад

      K H asked "Any recommendations of online discussions on how to multiply two cuts, divide, square, etc.?"
      A well known trick is to define just the positive real numbers (as Dedekind cuts of positive rational numbers).
      If R1 is a (positive) Dedekind cut and R2 is a (positive) Dedekind cut, then:
      a) R1 + R2 = { x in Q+ | (Ea)(Eb)( a in R1 and b in R2 and x = a + b) }
      b) R1 * R2 = { x in Q+ | (Ea)(Eb)( a in R1 and b in R2 and x = a * b) }
      c) R1 / R2 = { x in Q+ | (Ea)(Eb)( a in R1 and b in R2 and x = a / b) } Note: not sure about this one.
      d) R1 * R1 = { x in Q+ | (Ea)(Eb)( a in R1 and b in R1 and x = a * b) }
      Then, the negative real numbers can be defined in a similar way that negative natural numbers are defined (as ordered pairs of natural numbers).

  • @nathanisbored
    @nathanisbored 4 года назад +2

    im a little confused, if we're allowed to define a real number as a set of other numbers, then why couldnt we have done that when defining the rational numbers themselves? instead of inventing an equivalence class, why not just say 1/2 = {(1,2), (2,4), (3,6)...} isnt that basically like defining the number as a set of a bunch of pairs of other numbers? how is that different than the idea of a cut for example?

    • @drpeyam
      @drpeyam  4 года назад +3

      Hmmmm, not sure, but I think it’s because if you just defined rational numbers as pairs of integers, then addition wouldn’t be well-defined. It’s by taking equivalence classes that you can define [(a,b)] + [(c,d)] = [(ad+bc,bd)]. In general for sets you wouldn’t be able to do this (although again I might be mistaken).

    • @drpeyam
      @drpeyam  4 года назад +2

      And the idea of cuts wouldn’t work for rational numbers because you’d be able to define rational numbers as cuts of integers, but the set of integers < 1/2 is the same as the set of integers < 1/3, so it wouldn’t distinguish 1/2 from 1/3 really.

    • @akashpremrajan9285
      @akashpremrajan9285 4 года назад

      Aren't equivalence classes special case of sets? With the notion of equality added in: so that [1/2] = [3/6]

    • @iabervon
      @iabervon 4 года назад

      You'd have to do something to avoid including as rational numbers {(1,2), (2,5), (4,7), ...} or {(1,2), (2,4)}. Once you've worked out which sets are rational numbers and which don't follow the rules, you've made an equivalence class by hand. Like, your a/b is {(c,d), for c,d in Z where bc=ad}, which is just putting the equivalence relation definition inside the definition of an equivalence class. An equivalence class for (S, ~=) equivalent to r is {s in S where s ~= r}, and defining the rationals as equivalence classes just lets you skip proving the properties of equivalence classes specifically for the rationals.

  • @stefanofalero
    @stefanofalero 4 года назад

    its pronounced, precisely, “King DeeeeeeDeeeeeDeeeeee”

  • @icew0lf98
    @icew0lf98 4 года назад +1

    try prooving 0

    • @aneeshsrinivas9088
      @aneeshsrinivas9088 2 года назад

      0.5 isnt in the 0 cut, but anything in the 0 cut is in the cut representing 1, since 1>0>anything less than 0. So anything in the 0 cut is in the 1 cut.

  • @oximas
    @oximas 3 года назад

    umm what
    ???

  • @mmmhhhrrrddd6580
    @mmmhhhrrrddd6580 2 года назад

    Are you Iranian?

  • @DiegoTuzzolo
    @DiegoTuzzolo 4 года назад

    how do I formally show that pi is a cut? because it is transcendental, so we want use a logic sentence to construct this number

    • @drpeyam
      @drpeyam  4 года назад

      It’s the cut that contains rational numbers like 3, 2.5, 3.1, 3.1415, -2, 0.01 (basically all rational numbers < pi)

  • @eliyasne9695
    @eliyasne9695 4 года назад +1

    17:09
    ...smaller then the area of a unit circle.

  • @perfectly_cromulent
    @perfectly_cromulent 4 года назад

    Love the video, just a bit too quiet

  • @txikitofandango
    @txikitofandango 4 года назад

    my prediction: this will have to do with the fact that, between any two real numbers, there exists a rational number. So, if you change a real number even a little bit, you will jump over at least one rational number. so the set of rationals associated with the real number will change, and so the rational numbers associated with the real number is what matters

    • @drpeyam
      @drpeyam  4 года назад

      Actually not?

  • @donmcrapsavier4463
    @donmcrapsavier4463 4 года назад

    NaMIstAyscr DR

  • @dimitrisnatsios8409
    @dimitrisnatsios8409 4 года назад

    Dr Peyam I love you more than my drumset.

    • @Kdd160
      @Kdd160 4 года назад

      Drumset??

    • @drpeyam
      @drpeyam  4 года назад +1

      The ultimate compliment, thank you 🥰

  • @thomasjefferson6225
    @thomasjefferson6225 Год назад

    I don't know why, but the more I watch this and the more I think about it the more I think this is nonsense and the real number line makes no sense to me.
    I'll write this for an exam sure, but I'm not believing this made up construction 😅
    For example you say we can do addition and multiplication to real numbers.
    What's e + pi + sqrt 2 then?
    Oh yeah you gotta use these cuts that make no actual Computational sense.
    This seems so flawed

  • @nadiyayasmeen3928
    @nadiyayasmeen3928 4 года назад

    O hi! I am the only recent reply?

  • @theproofessayist8441
    @theproofessayist8441 Год назад

    Ha King Dededede: Dedekind Cuts - punny. Dr. Peyam if you ever have free time try looking at Kirby lore. It's actually pretty damn crazy. Kirby is a god that fights against eldritch abominations.

    • @drpeyam
      @drpeyam  Год назад

      I love the Kirby games, they’re my fave 😁