Pentagonal Numbers
6:12
5 жыл бұрын
Triangular Numbers
6:10
5 жыл бұрын
Surreal Numbers
20:55
6 жыл бұрын
Field Extensions and the Hyperreals
6:35
Field Theory - an Introduction
8:45
6 жыл бұрын
Group Theory - an Introduction
9:15
6 жыл бұрын
Пікірлер
@perpetualrabbit
@perpetualrabbit Ай бұрын
I feel stupid but I still don't see the difference between Aleph numbers and Beth numbers.
@thrax4939
@thrax4939 6 ай бұрын
I find it difficult to come to terms with that zero is required in the group to satisfy the axioms yet can’t be used to prove the additive and multiplicative qualities. Is there any way to rationalize that?
@ПендальфСерый-б3ф
@ПендальфСерый-б3ф 6 ай бұрын
What are the additive and multiplicative qualities?
@LarghettoCantabile
@LarghettoCantabile 10 ай бұрын
I'm not sure how the continuum hypothesis might not be true. Represent [0, 1] by the set of subsets of N, each represented by an infinite bitmap, which can also be interpreted as a list of binary decimals. There will be collisions in the interpretation, with for instance .10+ being the same real number as .01+ (i.e. a zero followed by an unbounded list of 1's); but we will still have an onto function; and we can presumably build a bijection from it. Our number of bitmaps is 2 to the aleph null; so this is presumably the cardinality of [0, 1]. Now, define a bijective function between [0,1] and R, for example by considering a semicircle of size 1, with a tangent at midpoint, and mapping each semicircle point to a tangent point by drawing a radius. This proves [0, 1] and R have the same cardinality, which is presumably the cardinality of our list of decimal numbers, i.e. beth_1.
@HPTopoG
@HPTopoG 10 ай бұрын
That doesn’t prove that the continuum hypothesis is true because it doesn’t say anything about the actual value of the continuum. What’s important to understand when dealing with CH is that we don’t know which ℵ number 2^ℵ0 actually evaluates to. It’s a bit like if somebody asked you to calculate 2×3 but you didn’t know what 3 meant. The reason we know CH is unprovable is that we can describe mathematical universes in which the continuum is exactly ℵ1 and other universes in which the continuum is provably larger than that. The former is true in Gödel’s constructible universe L and the latter is true in Cohen’s forcing extension. Both are somewhat difficult to understand.
@christophersedlak1147
@christophersedlak1147 Жыл бұрын
thanks
@christophersedlak1147
@christophersedlak1147 Жыл бұрын
thanks
@rrr00bb1
@rrr00bb1 Жыл бұрын
I had read through On Numbers And Games (Conway), and Surreal Numbers (Knuth). Being fundamentally a binary definition is interesting; and it's not taken far enough. The mechanical definition of the <= operator is kind of amazing. But this whole business of just having sequences of numbers with no definition of the patterns is infuriating. I much prefer the explanatory power of 2s complement; and binary numbers extended fully in both directions. I would much rather see recursion used to completely replace these definitions that use "..." with sequences defined without having to write down the programs that generate the sequences. For example, repeating decimals can be replaced with recursion: A = 0.9 + 0.1 A. = 0.99 + 0.01 A. ... " = 0.99999....". 10 A = 9 + A. 9 A = 9. A = 1.
@electroflame6188
@electroflame6188 Жыл бұрын
What happens when you divide by zero in these?
@moshenazarathy6211
@moshenazarathy6211 Жыл бұрын
the same like in every field - you can't -since divide by zero is multiply by the reciprocal of 0, but the reciprocal of zero is undefined (since 0 * any = 1 means 0 = 1 means FALSE)
@nickosc88
@nickosc88 Жыл бұрын
You can’t measure the size of an infinite set, by definition it has no size. You can’t add or subtract objects that have no size. Again, that is a contradiction in terms. Transfinite arithmetic is at best a fanciful speculation (or rather an example of the limits of mathematics) and at worst meaningless jiberish
@olbluelips
@olbluelips Жыл бұрын
Me when I can’t find 2-3 because it’s impossible to subtract 3 from 2 and trying to do so would be meaningless speculation
@fullfungo
@fullfungo Жыл бұрын
“[…] by definition it has no size”. Yeah, no, this depends on the definition of “size”, or rather “cardinality”. If we define it in such a way that it makes sense for infinite sets, then by definition they will have a size.
@markosskace514
@markosskace514 Жыл бұрын
Most mathematics is "fanciful speculation" and also "meaningless jiberish". For instance, there isn't any infinities in the physical universe. Mathematics IS NOT about physical universe, it is about relationships between abstract entities - it doesn't need to have connection to anything real.
@tomkerruish2982
@tomkerruish2982 7 ай бұрын
There are more things in Heaven and Earth (and Mathematics), @nickosc88, than are dreamt of in your philosophy.
@Anders01
@Anders01 Жыл бұрын
Amazing concept. That the surreal numbers include things like infinitesimals and still with the same arithmetic as the real numbers.
@musimedmusi8736
@musimedmusi8736 Жыл бұрын
Nice concise basic review
@kolzh7985
@kolzh7985 2 жыл бұрын
i don't get how the formula for the addition is x+y = (XL+y, x+YL | XR+Y, x+YD}, there's an addition there
@tomkerruish2982
@tomkerruish2982 11 ай бұрын
It's a definition by recursion; every addition in the definition involves an earlier created number. Eventually, you'll get to 0, which has no options, and the process stops. An explicit example: 1+ 1 = {0|} + {0|} = {1+0,1+0|}; 1 + 0 = {0|} + {|} = {0+0|}; 0 + 0 = {|} + {|} = {|} = 0. Thus 0 + 0 = 0; 1 + 0 = {0+0|} = {0|} = 1; 1 + 1 = {1+0|} = {1|} = 2. A bit lengthy, but way shorter than Principia Mathematica. The surreals are said to be well-founded, in that it is impossible to find an infinite sequence a0, a1, a2,..., where a1 is an option of a0, a2 is an option of a1, etc. (There are plenty of surreal numbers where you can construct such a sequence of any given finite length, but never one of infinite length.) This allows us to make definitions such as this one for addition, provided that we make sure to always involve earlier created numbers. For more information, look up transfinite recursion and transfinite induction, both of which are used extensively with the surreals. Edit: It's not transfinite recursion and induction, but e-recursion and e-induction, where 'e' should be the 'is an element of' (epsilon) symbol, which my phone keyboard lacks.
@wdfusroy8463
@wdfusroy8463 2 жыл бұрын
Like every other video I have ever seen on the topic of transfinite ordinals we are told that [1 + ω] is not = to [ω +1] because the latter expression adds the 1 to the end of the already specified ordinal ω, while the former merely throws another non-transfinitely ordered number, [i.e. 1] into the already ordered number ω. O.K., that makes sense. But no one ever goes on to explain the analogous difference between [1 - ω] and [ω 1 exactly that is supposed to mean! Worse still, if we include extensions of the integers into the hyperreals of Non-standard Analysis, or the Surreal Numbers, we can easily generate new numbers like ω/2, or sqrt[ω], or even lnω and other such odd beasties. But what is never made clear, at least that I have ever seen, is why these new and smaller infinite ordinals do not contradict the original definition of ω which specifies that ω represents the "first transfinite integer. I hope someone can explain why such a contradiction does not occur, if indeed it does not. Thanks!
@tomkerruish2982
@tomkerruish2982 11 ай бұрын
I don't know enough about the hyoerreals to answer your question concerning them. As for the surreals, omega is still the smallest infinite ordinal, where ordinals are defined to be surreal numbers with no right options.
@linuxp00
@linuxp00 10 ай бұрын
Well, in hyperreal setting. It's really a relabeling of lemniscata symbol ∞ for ω, given that a concrete definition, that is its the ordinal representing the cardinality of ℕ. Said that, they are just different relative to others transfinite numbers of the same order, that is, ω + 1 < ω + 2 < ω², or their reciprocal ε, such that ε = 1/ω. But compared to the finite numbers, is just as ∞ in the standard analysis. Example: (ω²+2ω)/ω = ω+1 -ω/√ω = -√ω (3ε²+2ε)/ε = 3ε+2 5ε/ε = 5 When taken the "standard part" that will be simplified to a finite number or a regular old infinity. The examples above have std part equal to: ∞, -∞, 2, and 5, respectively.
@Fircasice
@Fircasice 2 жыл бұрын
I do not understand why m^lambda = n^lambda = 2^lambda.
@HPTopoG
@HPTopoG 10 ай бұрын
As an easier example, let m=3 and n=2. Then an element of m^λ is a ternary sequence f of length λ and an element of n^λ is a binary sequence g of length λ. Take any ternary f and reinterpret it in binary to get a binary g. Go from binary to ternary to get the other map direction. If you can show these are both injective or both surjective, then the Cantor-Bernstein theorem implies they have the same cardinality. Alternatively if n<m, the easiest map is just the inclusion map from n^λ into m^λ that does nothing to f. Then all you have to do is show the map in the other direction is an injection. Another way, pick any fast growing function s from λ to itself. Then for a given f∈m^λ, for every ordinal α<λ, 1. if f(α)<2, make h(s(α))=f(α) and h(β)=0 for everything else. 2. if f(α)≥2, then set h(s(α))=0 and h(s(α)+1)=f(α)+1 3. h(β)=0 for everything else. Show this is an injection. It is basically just carrying like in basic arithmetic but with enough gaps to avoid any infinite carrying.
@youtubeuserdan4017
@youtubeuserdan4017 2 жыл бұрын
Great video. Very concise and to the point.
@bobross7005
@bobross7005 2 жыл бұрын
Thank you for your channel!
@PunmasterSTP
@PunmasterSTP 2 жыл бұрын
Hyperreals? More like "Incredible videos that bring the feels"...of intrigue! Thanks again for making this series.
@PunmasterSTP
@PunmasterSTP 2 жыл бұрын
Field theory? Also, "Fantastic videos for me!" Thanks for making them.
@PunmasterSTP
@PunmasterSTP 2 жыл бұрын
Group theory - an introduction? More like "Great video, and now I want to watch another one!"
@jafarbarmaki6668
@jafarbarmaki6668 2 жыл бұрын
God bless you wherever you are. Thank you
@marijnstollenga1601
@marijnstollenga1601 2 жыл бұрын
Nicely explained, with pauses where you get at least a second to think about it.
@gaylynnblanchard9682
@gaylynnblanchard9682 2 жыл бұрын
What comes after Beth numbers,ordinals ?
@sharpnova2
@sharpnova2 2 жыл бұрын
the way you pronounce omega is just.. hilarious and weird.
@sharpnova2
@sharpnova2 2 жыл бұрын
you sound completely out of breath in this video... gasping for air at the start of every sound bite. additionally, there were several points in the video where you'd list a few ideas/elements/etc. with nothing related to them on the screen. you really place a high cognitive load on the viewer when you present things this way. regardless, thanks for talking about the surreal numbers. as someone who has taught about them in the past, it was nice to see a video that covered most of the details.
@alephomega955
@alephomega955 2 жыл бұрын
(if I'm wrong, feel free to correct me.) Using the proof on how you cannot list all real numbers, the set of real numbers should be equal to the set of natural numbers. You can list the irrational numbers then make one that doesn't appear on the list, but you could still pair up each number you created with each natural number, you just start with 1 on the new one you created and label them from there on. It's the same thing as adding to an infinite amount, which doesn't change the value. You could describe another infinite amount of numbers using the same method and still not change the value, because you'll never run out of naturals to pair it up with.
@MikeRosoftJH
@MikeRosoftJH 2 жыл бұрын
Obviously, there are uncountably many irrational numbers. There are countably many rational numbers, but uncountably many real numbers; and union of two countably infinite sets can't be uncountable. It's also fairly easy to construct an explicit one-to-one mapping between real numbers and irrational numbers.
@linuxp00
@linuxp00 10 ай бұрын
It's possible to construct an uncontable infinite set from a cartesian product between two countable infinite sets. Though, it's not in any way mappable to a countable set. My argument in favor of the cardinality: |ℝ| > |ℤ| is that you can imagine the real numbers as the Cartesian product between two natural sets, but with a key twist. While the "integer part" of the reals grows in absolute value: {X ∈ ℕ | 0 ≤ |X| ≤ ...99999} (where we can pair numbers from the interval ]-∞, +∞[ to the interval [0,+∞[ ), the "decimal portion" decreases: {Y ∈ ℕ | 1 > 0.9999.... ≥ Y ≥ 0} (also mappable to the naturals, but read backwards). If we plotted this Cartesian product X × Y, we would have an infinite set of decimals for each of the infinite real integer values, i.e., ∞ ^ ∞ (which is mappable to 2^∞, the power sets of the naturals). However, it is still impossible to list all real numbers, as we can always add a new digit to the left in the integer part or to the right in the decimal part. Moreover, even if you try to "move the decimal point" all the way to the right, attempting to transform reals into integers, it is always possible to add a decimal interval after this integer, turning it back into a real number. This is trying to illustrate, in an intuitive way, that you already "exhaust" the set of naturals/integers just to count between 0 and 1, thus the cardinality of the reals is greater than that of the integers and naturals.
@MikeRosoftJH
@MikeRosoftJH 10 ай бұрын
@@linuxp00 "It's possible to construct an uncountable infinite set from a cartesian product between two countable infinite sets." No, that's not possible. Cartesian product of two countably infinite sets is countably infinite; and this result doesn't require axiom of choice. (As opposed to the more general statement that union of a countably infinite collection of countably infinite sets is countably infinite, which *does* require axiom of choice, or a weak variant thereof - axiom of countable choice. For example, it's consistent in absence of axiom choice that the set of real numbers is a union of countably many countable sets; and that doesn't contradict Cantor's result that there are uncountably many real numbers.) There's no such natural number as ...99999. Any natural number is finite in magnitude, and therefore its decimal (or base-n) expansion has finitely many digits. And that's by definition; a set is finite, if its number of elements is equal to some natural number. (It is an empty set, or it has exactly one element, or it has exactly two elements, or it has exactly three elements, or... and so on. Of course, the "and so on" part is a bit involved; you can't have a formula of an infinite length.) The mathematical structure towards which you are heading are the n-adic numbers; in this structure, any base-n expansion (infinite sequence of base-n digits) is an element of this structure, and then you define mathematical operations on these base-n expansions. For example, the number ...999 is equal to -1 - you can verify that indeed ...001+...999=0. (Usually, n is taken to be a prime number, because otherwise you have divisors of zero - a pair of non-zero numbers a and b such that a*b=0.) But the problem is that there are uncountably many n-adic numbers - any infinite sequence of digits corresponds to a unique n-adic number. (From Cantor's diagonal proof it follows that n-adic numbers can't be mapped one-to-one with natural numbers; however, they can be mapped one-to-one with real numbers.)
@linuxp00
@linuxp00 10 ай бұрын
@@MikeRosoftJH I'm inspired by adics, but I'm not using an actual adic arithmetics. I'm just using naturals as infinite strings of digits in reverse. Naturals are infinitely countable, so are their digits. So, as I see it the whole set of Naturals are contained in the interval [0,1[. Though, it's the same for all the gaps between two consecutive integers, so it is impossible to count the reals. No need for Diagonalization, just Zenon's Paradox in it's essence.
@MikeRosoftJH
@MikeRosoftJH 10 ай бұрын
@@linuxp00Again, every natural number is finite in magnitude, and therefore has finitely many non-zero digits. If you reverse the decimal expansion of a natural number, you get the set of all real numbers in an interval whose decimal expansion has finitely many digits after the decimal point. And that there are countably many such numbers is not a new result; all such numbers are rational, and that there are countably many rational numbers was proven by Cantor himself. And that one specific function from natural numbers to real numbers doesn't cover all reals doesn't by itself prove that real numbers are uncountable. It's a well-known result that an infinite set can be mapped one-to-one with its strict superset or subset. (Precisely, this is true for every set which has a countably infinite subset.) For example, consider the function n->2*n on natural numbers; or, conversely, the same function on an interval from 0 to 1. So in order to prove that set A can't be mapped onto set B, it doesn't suffice that one specific function between the two sets doesn't cover all elements of B; you have to prove that no function does. (Such as by using the diagonal proof, or the earlier Cantor's proof by nested intervals.) In set theory, an infinite sequence is not a process. It's a function whose domain is the set of natural numbers. And a function is itself a set: set of ordered pairs. For example, the function f: x->2*x on the set {1,2,3} is the set {[1,2], [2,4], [3,6]}; if the pair [a,b] is an element of the set f, we by convention say that f(a)=b. Zeno's paradox isn't a theorem of mathematics. In fact, Zeno was heading towards perfectly well-defined concepts of calculus, such as limit of a sequence, infinite sum, or derivative of a function. (Except for that he wasn't able to understand the infinite case, and so he instead concluded that the infinite case is impossible.)
@MarcusAndersonsBlog
@MarcusAndersonsBlog 2 жыл бұрын
Sorry to butt in to your university assignment, but @2:30, and by definition of LIMITS to infinity of an infinite sum, the real number 5.000... of that LIMIT is NOT =4.999... Furthermore because I can write the Surreal number { 4.999... | 5.000... } then, by definition, the real 4.999... CANNOT equal the real 5.000... because I have just defined the number halfway between the two of them. What is it with you brainwashed people who want to persist with this insanity that 0.999... = 1.000... ? Even in the Surreal number system the proof they are different is PATENTLY CLEAR !!!! The consequence of this cardinal error of rigor is that you make the same error employed by Zeno's paradoxes. At infinitesimal precision the unit interval becomes zero and the decimal positional number system fails. You cannot count if the unit interval is zero (at infinite precision). Zeno paradox relies on ultimately counting with a unit interval of zero, and that is exactly what you are doing by erroneously claiming 4.999... = 5.000... You can defend decimal 9 recurrence to the death if you want, I dont care, but its not going to alter the fact that 4.9999... does not equal 5.000..., and now I have just proven it with Conway's Surreal notation, as per my 1 line proof above.
@xxLittleAmishxx
@xxLittleAmishxx 2 жыл бұрын
I didn’t know this was disputed, but I was mind blown when my teacher showed me a proof for 0.999 repeating equaling 1.0 but honestly I just accepted it because I figured he knew more about math than I did. Now you’ve got me questioning the proof
@adamkolany1668
@adamkolany1668 2 жыл бұрын
Why the condition in the definition of inequality is so weird? that is why everywhere the write: "there is no x_L in X_L that Y ≤ x_L" but "Y ≤ x_L " means that "for all y in Y (y ≤ x_L)", so that means "there is no x_L in X_L that (for all y in Y (y ≤ x_L) )" which is equivalent to "for all x_L in X_L there exists y in Y that it is not (y≤x_L) ) " which by linearity of the order reads "for all x_L in X_L there exists y in Y that (x_L < y) ) " which is much simpler to comprehend . where do I make a mistake ??
@tomkerruish2982
@tomkerruish2982 11 ай бұрын
You're assuming that surreal numbers are necessarily ordered; that, for any surreal a and b, exactly one of a>b, a=b, a<b holds. This is true, but needs to first be proven.
@bananamanjunior7575
@bananamanjunior7575 2 жыл бұрын
I still don't understand X_L•Y_L, X_R•Y_R, X_R•Y_L and X_L•Y_R Are we multiply these set together?
@BarnabyFWNightingale
@BarnabyFWNightingale 2 жыл бұрын
What is set A at 2:52?
@arkeusalexander9054
@arkeusalexander9054 2 жыл бұрын
First time i encountered this set (learned maths in french). So i know : N, Z, D, Q, R, C
@Xnoob545
@Xnoob545 2 жыл бұрын
Algebraic numbers
@BarnabyFWNightingale
@BarnabyFWNightingale 2 жыл бұрын
@@Xnoob545 Thanks!
@aslpuppy1026
@aslpuppy1026 3 жыл бұрын
If the continuum hypothesis can’t be proven false that means that you can’t construct a set with cardinality between aleph_0 and beth_1. This is because constructing a set with cardinality between aleph_0 and beth_1 would disprove the continuum hypothesis. So my question is, if it is impossible to construct a set with a cardinality between aleph_0 and beth_1, wouldn’t that mean that the continuum hypothesis is true.
@MikeRosoftJH
@MikeRosoftJH 2 жыл бұрын
You can't prove (without additional axioms) that there is such a set whose cardinality is strictly between Aleph-0 and P(Aleph-0); and you can't prove that there isn't such a set. In other words: there's a model of set theory where the set in question doesn't exist (such as the constructible universe); and there is a model in which the set does exist. (The proof uses the technique of forcing: start with some model of set theory, such as the constructible universe, and then add elements to it so that the new model still satisfies the axioms of set theory, and also satisfies the proposition whose consistency with the rest of set theory axioms we are trying to prove; for example, continuum hypothesis is not true, or real numbers can't be well-ordered, or the like.)
@linuxp00
@linuxp00 10 ай бұрын
Well, according to my current understanding, any real number can be represented as a triplet: [a sign (-1 or 1), a natural number for the integer part (digits to the left of the decimal point), and a natural number for the decimal part (digits to the right of the decimal point)]. So, between any two consecutive natural or integer (naturals + sign) numbers, there are infinite natural numbers as decimal digits, which hint at a discretization, but that are uncontable because just as integers grow unbound, decimals shrink in the same manner, making their count unfeasible. Because of that I think that if continuum hypothesis is set to false, that also allow us to define infinitesimals conceptually, with their fractal nature (i.e. it's possibly to do magnifications on the real number line and it will self similiar, either by zoom in or zoom outs). P.S: Writing π as 3 + 0.14596... = (+1, 3, ...69541). It is possible to have the infinite digits of π, or any irrational.
@MikeRosoftJH
@MikeRosoftJH 10 ай бұрын
@@linuxp00 It's indeed possible to define infinitesimals, by creating an alternate model of first-order theory of real numbers (i.e. the hyperreals). But continuum hypothesis has nothing to do with it; on real numbers there are no non-zero infinitesimals, regardless of whether or not continuum hypothesis is true. Likewise, fractals have nothing to do with this.
@linuxp00
@linuxp00 10 ай бұрын
@@MikeRosoftJH yeah, I mistook my words on it. But, surreals have this fractal design with their higher-orders of infinities and infinitesimals. May it help consider them in the CH problem.
@masonmireles9295
@masonmireles9295 3 жыл бұрын
Can anybody please dummify this down? I really want to understand, but this is difficult. What are the implications to the reals world? Where can I use this?
@boubakersoltani4566
@boubakersoltani4566 2 жыл бұрын
This is just pure math, no real world implications at all. Basically, mathematicians said: what if we pack all numbers in one black box and name this box "omega"? One can object: but numbers are infinite, so how can one fit them all in a box? But we should not forget that this is just math, it is all only about being consistent with your first assumptions no matter whether they are from reality (like biology and physics) or just from mere mind. So, by using this box "omega" as a new unit, we can now count the boxes: first omega box, second omega box, ... millionth omega box and go on. Here again you can see that the series of boxes is endless, so why do not pack all these boxes in a new bigger box? this new box has "omega" number of smaller boxes, which by turn has "omega" numbers each. The new box is called then "Omega times Omega". And we continue counting this way, and getting more bigger embedding boxes. This recursive process is what we call "ordinal generation".
@masonmireles9295
@masonmireles9295 2 жыл бұрын
@@boubakersoltani4566 Thank you so much for helping me to better understand this! It truly means a lot! I am currently pursuing a major in mathematics. Is there any chance I could email you to ask about this further please?
@boubakersoltani4566
@boubakersoltani4566 2 жыл бұрын
@@masonmireles9295Hi Mason! I wanted to really thank you for your trust, but indeed I am not that expert in set theory. Maybe you won't gain much from me.
@masonmireles9295
@masonmireles9295 2 жыл бұрын
@@boubakersoltani4566 Okay. Thank you anyways for your help! I truly appreciate it! Not a lot of people go out of their way to help others out anymore. You helping me truly made my day.
@boubakersoltani4566
@boubakersoltani4566 2 жыл бұрын
@@masonmireles9295 glad to here that ^^ I wish you all success.
@jimtwisted1984
@jimtwisted1984 3 жыл бұрын
Do surreal numbers include the Gödel number of its non derivable sentence?
@manicmath3557
@manicmath3557 3 жыл бұрын
Miss your videos man
@crazye7132
@crazye7132 4 ай бұрын
same
@smellybathroom
@smellybathroom 3 жыл бұрын
very well done. thank you
@kennyearthling7965
@kennyearthling7965 3 жыл бұрын
Ok I was not ready for this, need to rewatch. Heavy stuff, very interesting
@brookeostrowski7348
@brookeostrowski7348 3 жыл бұрын
thank you for such a simple explanation! i’m writing my undergrad thesis on the hyper reals w/little abstract algebra background and appreciate the clarity.
@PunmasterSTP
@PunmasterSTP 2 жыл бұрын
I just came across your comment and was curious. How'd the thesis go?
@NoNTr1v1aL
@NoNTr1v1aL 3 жыл бұрын
Absolutely amazing! May I ask for a book which contains solved examples for fields and maybe even modules?
@biggerthaninfinity7604
@biggerthaninfinity7604 3 жыл бұрын
5:13 w1 is the first uncountable ordinal. What is e0?
@hypercosmologyexplanations5178
@hypercosmologyexplanations5178 3 жыл бұрын
e0 is omega to the power omega to the power omega...[repeat omega times]. e0 is much much smaller than w1
@arkeusalexander9054
@arkeusalexander9054 2 жыл бұрын
@@hypercosmologyexplanations5178 You mean greater as it is an infinite tetration/Hyper"w" ?
@vishalarora6599
@vishalarora6599 4 жыл бұрын
√26 us not whole
@Sportliveonline
@Sportliveonline 4 жыл бұрын
we see infinity in surreal numbers
@Sportliveonline
@Sportliveonline 4 жыл бұрын
UNREAL if u ask me ~~~~is there a purpose in all of this ~~~
@frankfruit
@frankfruit 4 жыл бұрын
i have a question about the diagonal argument, why can we use the same argument for the natural number and say that there is always a number not on the list. but is it cause the all naturals number s are list able that why no matter what number we make we can find it on the list thus the argument fails for the natural numbers. if so then wouldn't that apply to real numbers too or there is something i am not able to understand about this argument.
@ReinLie
@ReinLie 4 жыл бұрын
@@bakert11 wonderful explanation.
@Prasen1729
@Prasen1729 3 жыл бұрын
Hi, the idea is you assume that the set is listable/countable. Set of natural numbers, N is listable as you can write them down : 1, 2, 3, ..... Given any natural number, I can find it in the above list. I can do the same with Z and Q. But not possible with R. Simple reason for me is, N, Z, Q are all discrete sets whereas R is continuous, so if you suppose it is listable, put them in an order no matter how, I can always construct a new real number which was not in the above list. The same way interval (0, 1) is not listable. Here is a try : Suppose the open interval (0, 1) is countable. So lets list them. Here are the decimal numbers. a_11, a_12 etc ... are the digits. A1 : 0.a_11 a_12 a_13 .... a_1n ... A2 : 0.a_21 a_22 a_23 .... a_2n ... A3 : 0.a_31 a_32 a_33 .... a_3n .... ..... An : 0.a_n1 a_n2 a_n3 .... a_nn .... Now if we can find a decimal number which is outside the above list, then our assumption is wrong. Can we find a new number ? Yes we can. Consider this element A = 0.a_1 a_2 a_3 ... a_n ... with the condition that a_1 ≠ a_11 , a_2 ≠ a_22, a_3 ≠ a_33 .... a_n ≠ a_nn. The 1st inequality above proves A ≠ A_1, 2nd inequality proves A ≠ A_2 etc ... nth inequality proves A ≠ A_n etc ... Hence A is not equal to any of the numbers above. Therefore A was not listed and hence the list was not complete. If (0, 1) is not countable then R can't be either as (0, 1) ⊂ R. Here is video on Numberphile useful. kzbin.info/www/bejne/m53ZgI2jZclnfpI
@MikeRosoftJH
@MikeRosoftJH 3 жыл бұрын
The reason is: every natural number is finite, and so it has finitely many digits. (And that's by definition; a set is finite, if its number of elements is equal to some natural number - it's an empty set, or it has exactly one element, or it has exactly two elements, or ... and so on.) If you try to apply the diagonal procedure on the sequence of all natural numbers, you get a sequence which has infinitely many non-zero digits, which doesn't represent any natural number. Likewise, when you apply the diagonal procedure to a sequence of all rational numbers - and such a sequence exists - then the resulting number must be irrational.
@linuxp00
@linuxp00 10 ай бұрын
My argument in favor of the cardinality: |ℝ| > |ℤ| is that you can imagine the real numbers as the Cartesian product between two natural sets, but with a key twist. While the "integer part" of the reals grows in absolute value: {X ∈ ℕ | 0 ≤ |X| ≤ ...99999} (where we can pair numbers from the interval ]-∞, +∞[ to the interval [0,+∞[ ), the "decimal portion" decreases: {Y ∈ ℕ | 1 > 0.9999.... ≥ Y ≥ 0} (also mappable to the naturals, but read backwards). If we plotted this Cartesian product X × Y, we would have an infinite set of decimals for each of the infinite real integer values, i.e., ∞ ^ ∞ (which is mappable to 2^∞, the power sets of the naturals). However, it is still impossible to list all real numbers, as we can always add a new digit to the left in the integer part or to the right in the decimal part. Moreover, even if you try to "move the decimal point" all the way to the right, attempting to transform reals into integers, it is always possible to add a decimal interval after this integer, turning it back into a real number. This is trying to illustrate, in an intuitive way, that you already "exhaust" the set of naturals/integers just to count between 0 and 1, thus the cardinality of the reals is greater than that of the integers and naturals.
@denonepson
@denonepson 4 жыл бұрын
love listening to you. You're sharp like a blade. Helps a lot. love from Bangalore
@joeyhardin5903
@joeyhardin5903 4 жыл бұрын
and this is all just on the real axis. bruh imagine this with complex numbers, no, quarternions, no, infinite dimensions, no, UNCOUNTABLY INFINITE DIMENSIONS
@ranjitsarkar3126
@ranjitsarkar3126 4 жыл бұрын
There ARE actually surcomplex numbers.
@joeyhardin5903
@joeyhardin5903 4 жыл бұрын
@@ranjitsarkar3126 damn
@ManojKumar-yw9kj
@ManojKumar-yw9kj 4 жыл бұрын
Good video
@ManojKumar-yw9kj
@ManojKumar-yw9kj 4 жыл бұрын
Yes dude👏👏👏👏
@wdfusroy8463
@wdfusroy8463 4 жыл бұрын
So where is the promised "continuation"?? All of the stuff in this video is very basic.
@philosopherhobbs
@philosopherhobbs 4 жыл бұрын
Great video. What about division?
@MkhanyisiMadlavana
@MkhanyisiMadlavana 4 жыл бұрын
4:45 weren't you supposed to say 'not necessarily bigger' instead of just 'not bigger' because it depends on the order type?
@PhilBagels
@PhilBagels 4 жыл бұрын
Typo at 15:48 : the capital Y by itself in the second line, should be a lower-case y.
@joelproko
@joelproko 4 жыл бұрын
Does anyone know whether the surreal numbers can represent i in their {...|...} notation? It can't be {0|0}, {-1|1} or {-x|x}, because those can't be negated, but i can. The definition of √ω as {0,1,2,3,...|ω,ω/2,ω/4,ω/8,...} suggests something like {0,1,2,3,...|-1,-1/2,-1/4,-1/8,...} or {-1,-1/2,-1/4,-1/8,...|0,1,2,3,...}, which on one hand makes sense (of course it wouldn't converge onto a single number), but on the other hand it seems impossible to derive recognizable complex numbers from these suggestions.
@eldritchcookie7210
@eldritchcookie7210 4 жыл бұрын
if you mean i as in the imaginary unit then no, however according to wikipedia there is an extension of the surreal numbers called the surcomplex numbers, although the surcomplex numbers aren't a proper class
@WWLinkMasterX
@WWLinkMasterX 4 жыл бұрын
naively, I would think you could just invent a new system where you "rotate" about the left and right "handed-ness" of this system.
@joelproko
@joelproko 4 жыл бұрын
Does anyone know whether the surreal numbers can represent *i* in their {...|...} notation? It can't be {0|0}, {-1|1} or {-x|x}, because those can't be negated, but *i* can. The definition of √ω as {0,1,2,3,...|ω,ω/2,ω/4,ω/8,...} suggests something like {0,1,2,3,...|-1,-1/2,-1/4,-1/8,...} or {-1,-1/2,-1/4,-1/8,...|0,1,2,3,...}, which on one hand makes sense (of course it wouldn't converge onto a single number), but on the other hand it seems impossible to derive recognizable complex numbers from these suggestions.
@donaldb1
@donaldb1 3 жыл бұрын
I don't know for sure, but I suspect not. I think the surreals can be ordered, as positive or negative, obeying the usual arithmetical rules (negative times negative equals positive, etc), but the complex numbers cannot be. For instance, the usual laws of positive and negative arithmetic would suggest that if i.i=-1 then -i.i=+1, but it doesn't. It just equals -1 again. However, you can obviously extend the surreal numbers to surreal complex numbers by adding i in.
@joelproko
@joelproko 3 жыл бұрын
@@donaldb1 Uhm... -i.i equals +1...
@donaldb1
@donaldb1 3 жыл бұрын
@@joelproko Oh yes, so it does. Haha. Um.
@donaldb1
@donaldb1 3 жыл бұрын
What I _meant to say_ is, if you can have positive or negative then, 1) For any number x either x or -x (but not both) should be greater than zero 2) the product of any two numbers greater than 0 should also be greater than zero So, by 1) either i or -i should be greater than zero, in which case their square should be greater than zero, by 2). But they both square to -1. So they can't be positive or negative. Is that better?