(Provably) Unprovable and Undisprovable... How??

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  • เผยแพร่เมื่อ 13 ธ.ค. 2024

ความคิดเห็น • 167

  • @tom.prince
    @tom.prince 6 หลายเดือนก่อน +159

    It doesn't detract from the video, but for the pedants, I'll point out that there are various points in the video where the logic depends on the hidden assumption that system of axions is consistent. In particular, the claim that a statement implying a contradiction means that the statement is unprovable depends on the system being consistent.

  • @ModusTollendoTollens
    @ModusTollendoTollens 7 หลายเดือนก่อน +88

    when 0 deploys his integral domain expansion, every stranger loses it's cero-divisor qualities, even forcing them to have inverse in the finite domain

    • @micayahritchie7158
      @micayahritchie7158 7 หลายเดือนก่อน +10

      Is this a jjk reference that I don't get?

    • @gbnam8
      @gbnam8 7 หลายเดือนก่อน +4

      @@micayahritchie7158 yeah

  • @MrSamwise25
    @MrSamwise25 6 หลายเดือนก่อน +19

    A+ video. It took a while for me to understand the difference between "ZFC as foundations" and "ZFC as an object of mathematical study".

  • @thesecondderivative8967
    @thesecondderivative8967 7 หลายเดือนก่อน +46

    Oh this makes sense. What you're saying is that it is similar to the parallel postulate or ZFC? Where something is unprovable and not unprovable from a given set of axioms because it is possible to construct a system where it can work and where it doesn't from said axioms?

    • @SheafificationOfG
      @SheafificationOfG  7 หลายเดือนก่อน +25

      Yep, exactly!

    • @gabitheancient7664
      @gabitheancient7664 3 หลายเดือนก่อน +3

      I think the case of the parallel postulate is more interesting because it's more similar to actual ZFC modelling in a way
      since you can, for example, directly construct spherical geometry FROM the geometric axioms, which doesn't satisfy the parallel postulate

  • @ishtaraletheia9804
    @ishtaraletheia9804 7 หลายเดือนก่อน +19

    This made things click in my brain! Multiple examples/realizations of a definition with different properties are extremely common in mathematics. There being different models of ZFC is only slightly weirder than that there are several different vector spaces. Gödel simply proved can NEVER pin things down to a single "true" model...

    • @aaronbredon2948
      @aaronbredon2948 6 หลายเดือนก่อน +9

      Gödel proved that there are unprovable true statements only for sufficiently powerful mathematical systems.
      He used Peano Arithmetic for his proof, and his proof applies to any system that is ω-consistent.
      Later extensions proved that it is true for any consistent system (which doesn't have much effect because almost all material systems that are consistent are also ω-consistent).
      It is possible for simpler systems (like those that lack the concept of a "next number", or that have no reliable way to go from x to x+1) to be complete and consistent. Of course, such systems have extremely limited applicability.

  • @Qreator06
    @Qreator06 หลายเดือนก่อน +5

    This is the first video where I think I understood everything.

  • @Gordy-io8sb
    @Gordy-io8sb 7 หลายเดือนก่อน +49

    7:34 Rings with characteristic 2: Allow us to introduce ourselves.

  • @awesomethegreatamazing2651
    @awesomethegreatamazing2651 7 หลายเดือนก่อน +39

    I’m very happy see content like this. Please continue to make content.

  • @amari343
    @amari343 7 หลายเดือนก่อน +92

    Wow! I've been interested in Godel's silly theorems for a while, and this was a great explanation of unprovability! Thank you!

    • @Gordy-io8sb
      @Gordy-io8sb 7 หลายเดือนก่อน +7

      Incompleteness*

    • @benjamindrhee
      @benjamindrhee 6 หลายเดือนก่อน +3

      Gödels silly theorems lol

  • @fireclub493
    @fireclub493 7 หลายเดือนก่อน +7

    The last line was great lmao. Your channel is a gem!

  • @jon2422
    @jon2422 7 หลายเดือนก่อน +4

    best math channel in the game

  • @znon701
    @znon701 7 หลายเดือนก่อน +6

    This really fibers my bundles, thank you.

    • @SheafificationOfG
      @SheafificationOfG  7 หลายเดือนก่อน +2

      My favourite part of a comment "section"

    • @98danielray
      @98danielray 6 หลายเดือนก่อน +1

      @@SheafificationOfG a mment cosection

  • @RhydianJenkins-s1p
    @RhydianJenkins-s1p 6 หลายเดือนก่อน +10

    Just discovered your channel, giving you a well deserved subscriber!!
    Only improvement i can recommend for the future is the audio quality 🙂

    • @SheafificationOfG
      @SheafificationOfG  6 หลายเดือนก่อน +5

      Great to hear that audio is the only major issue! (Exploring options to varying levels of success on that front)

  • @KnufWons
    @KnufWons 4 หลายเดือนก่อน +1

    Wow this shattered my illusions of mathematics

  • @IsaacDickinson-tf8sf
    @IsaacDickinson-tf8sf 6 หลายเดือนก่อน +1

    I just proved for odd q that (2^a(q-1/2) -1)/q that it is always an integer when a(x) is the sequence A002326 where a(0) is 1. This can be used to easily check Mersenne primes by setting it equal to 1. Sometimes it breaks like for M11 because multiple values of q work. To fix this, use the smallest value of 2^a(q-1/2) -1 for a chosen q. If it is 1, it is prime. If it isn’t, it is not prime.

  • @choiin-seoq7422
    @choiin-seoq7422 7 หลายเดือนก่อน +1

    this channel is gold, love the digestable videos, especially for someone like me with only an applied mathematics background (i barely know any pure math)

  • @dwf2606
    @dwf2606 7 หลายเดือนก่อน +3

    I like the noise. It's authentic.

  • @TheSandkastenverbot
    @TheSandkastenverbot 7 หลายเดือนก่อน +2

    Can't wait for your next videos. This one was excellent!

  • @sstadnicki
    @sstadnicki หลายเดือนก่อน

    Minor nitpickery: at 5:28 there's arguably the additional step of using distributivity to show that x.y+x.y=2.x.y; it's easy but not free. I initially thought you were going to go a different direction, and look at the question 'there exists an a such that 2.a=1' which does a fine job of differentiating the integers from the rationals (or at least the dyadic rationals) but the example that you used has enough remove from being a plausible axiom itself that it feels better as a theorem than the existence of 1/2.

    • @SheafificationOfG
      @SheafificationOfG  หลายเดือนก่อน

      Yeah you're absolutely right about there being a couple of steps behind "a + a = 2a". I can't remember anymore but I thought there was a comment much earlier about that

  • @u03b5
    @u03b5 7 หลายเดือนก่อน +32

    these are actually good wtf thanks

  • @derendohoda3891
    @derendohoda3891 4 หลายเดือนก่อน +1

    Very practical demonstration.

  • @JakubWaniek
    @JakubWaniek 7 หลายเดือนก่อน +3

    This has been hinted at in another comment (I think?) but your demonstration at 8:56 only illustrates that our ring must have characteristic 2, not the much stronger claim that it's Z/2Z

    • @SheafificationOfG
      @SheafificationOfG  7 หลายเดือนก่อน +9

      Right, if you just take the axioms of an integral domain and tack on "2 =0" you get integral domains of characteristic 2. I didn't stress the point, but I mentiom prior to the construction that the idea behind "forcing" starts with picking an existing model. So, in my case, I was starting with the model Z (usual integers) and then asserting 2 = 0.
      I was trying to mirror Cohen's approach to demonstrating the unprovability of CH: he started with a model where CH was true, and then used forcing to extend it to a model where CH failed.

  • @davethesid8960
    @davethesid8960 7 หลายเดือนก่อน

    I'm currently taking an undergrad level logic course and this video is super helpful. Thank you!

  • @asdfghyter
    @asdfghyter 4 หลายเดือนก่อน

    excellent and very accessible explanation of these concepts! as a bonus it also made it a lot clearer for me what the difference is between the single and double turnstiles, despite you not explicitly explaining them. thanks!

  • @asneecrabbier3900
    @asneecrabbier3900 7 หลายเดือนก่อน

    I wanna thank you for popping up at the right time. As a physics student who's just completed my sophomore year, and started getting into math, this is both just advanced and simple enough for me to actually be able to learn from. Keep up the good work.

    • @SheafificationOfG
      @SheafificationOfG  7 หลายเดือนก่อน +1

      Thanks fam! Maybe with more nudging you'll switch programs 🙃

    • @asneecrabbier3900
      @asneecrabbier3900 7 หลายเดือนก่อน

      @@SheafificationOfG who knows lol. Definitely gonna teach myself some abstract algebra though. It's super interesting.

  • @Satokaさとか
    @Satokaさとか 4 หลายเดือนก่อน +1

    The reasoning at 8:16 feels wrong - To me this sounds like “This statement is disprovable if we add this one rule, so if we remove this rule, we are no longer able to verify it”, like, if I want to prove if Goldbach Conjecture is unprovable, then I add a rule say that “Goldbach is false”, which it obviously is in that ruleset, and then I remove that rule, thus complete the proof that Goldbach is unprovable?
    Thank you.

    • @lerarosalene
      @lerarosalene 4 หลายเดือนก่อน +2

      See pinned comment. You can add a rule "Goldbach is false", but then you also need your new set of rules to be consistent. And proving it is a hard task. With ring of modulo 2 it's easy - there are only 2 elements. You can check that this new rule never contradicts any previous ones, but I fear proving that your new rulesets with "Goldbach is true"/"Goldbach is false" rules are consistent would be as hard as actually trying to prove/disprove it.

  • @Gordy-io8sb
    @Gordy-io8sb 7 หลายเดือนก่อน +3

    2:36 There is a 3rd. Abstract algebra (specifically field theory/group theory).

    • @biblebot3947
      @biblebot3947 7 หลายเดือนก่อน +3

      What are you on about?

  • @alansmithee419
    @alansmithee419 4 หลายเดือนก่อน +1

    I've usually taken unproveable and undisprovable to mean that we can't determine whether something is true, but is decidedly either trrue or false - we just can't prove that - but this video seems to suggest instead that the problem simply has no inherent truth or falsity at all. Am I understanding correctly?

    • @kikivoorburg
      @kikivoorburg 4 หลายเดือนก่อน +2

      In mathematics, “true” and “false” are relative to the axioms we work with. The continuum hypothesis is neither true nor false from ZFC, because its axioms simply do not force an answer. This means we can extend ZFC to make it either true or false. Basically, answering the question requires a new axiom, and valid axioms exist to give either answer.
      In science, we consider something “true” if it correctly predicts the behaviour of the universe. This is what most people mean when they say “X is true”.
      Technically, these two definitions of “truth” are _entirely independent!_ Mathematics is not formally obliged to match reality.
      However, the axioms mathematics uses today are not arbitrary!! We picked them because they make mathematics match reality, and thus makes it useful. So in most cases, when ZFC says “X is true”, you can go do an experiment and find that X is true experimentally.
      For example, “0 =/= 1” (in the integers) is true in ZFC, but also true in reality because if I have 0 apples it’s not the same as having 1 apple.
      ZFC also makes some statements we can’t experimentally verify. For example, ZFC happily deals with notions of “infinities”, even though everything we’ve ever observed in reality is finite. Statements about infinities may be true in ZFC, but are unknown in reality.
      Now we get to the continuum hypothesis. As mentioned, ZFC simply can’t answer the question, and we can extend ZFC to make it either true or false. Your question about it being “really true” is asking _whether it’s true in reality._ However, the continuum hypothesis is a question about infinities, while we have only ever done experiments on finite things. Thus, we cannot answer any question about infinities in an experimental manner!
      If some day we find a way to do science on infinities - or find that infinities are useful for predicting the behaviour of finite objects which we can perform experiments on - we may eventually find that our universe either does or does not respect the continuum hypothesis. If that happens, we will extend mathematics with the relevant axiom required to make it match the universe. However, until that day comes there is no reason to prefer one answer over the other, so we just leave it at “we don’t know, and can’t know just yet”.

  • @Diogenes_ofSinope
    @Diogenes_ofSinope 7 หลายเดือนก่อน +3

    Your pronunciation of "Gödel" is very good. I've never heard an american pronounce such a beautiful "ö".

    • @SheafificationOfG
      @SheafificationOfG  7 หลายเดือนก่อน +1

      Glad to hear I didn't butcher the name! (Btw, the only "American" I am is "North American" 🍁)

    • @Diogenes_ofSinope
      @Diogenes_ofSinope 7 หลายเดือนก่อน +1

      @@SheafificationOfG no you didn't at all. Oh I'm sorry, it is sometimes hard for me to recognize the difference between Canadian accents and US accents. I guess people are right in saying assumptions can make a fool out of you.

    • @SheafificationOfG
      @SheafificationOfG  7 หลายเดือนก่อน

      @@Diogenes_ofSinope Nah, don't worry about it; I don't think my accent is very distinct from the neighbours down south!

  • @Lucas-pj9ns
    @Lucas-pj9ns 7 หลายเดือนก่อน +17

    oh cool so that's what a model is 😃 thanks

  • @monsterhunter8595
    @monsterhunter8595 7 หลายเดือนก่อน

    Very nice video! I just discovered your channel and can't wait to watch your other ones!

  • @brentweichel932
    @brentweichel932 7 หลายเดือนก่อน +5

    Middle dot (U+00B87) for multiplication? Everyone knows that the standard symbol for multiplication is Double curly loop (U+27BF)!

  • @joehead4081
    @joehead4081 7 หลายเดือนก่อน +3

    You must have fucked up on this one because my dumb ass finally understood it

  • @MusicEngineeer
    @MusicEngineeer 7 หลายเดือนก่อน +4

    Interesting! Are there any other commonly used axiomatic systems (other than ZFC, I mean) of set theory where the continuum hypothesis *can* be proved (or disproved)? And what do we get, if we just add the trueness or falseness of the continuum hypothesis as yet another axiom to the existing ZFC axioms? We should get two different mathematical universes, right? Then maybe we could just explore both of them and use the one that we like better :-D ...I'm pretty sure, this must have been done already?

    • @recursiveslacker7730
      @recursiveslacker7730 7 หลายเดือนก่อน

      Taking a dip in the Xeeleeverse Logic Pool, eh?

    • @SheafificationOfG
      @SheafificationOfG  7 หลายเดือนก่อน +4

      There are some interesting consequences of CH or its negation (for example: if the continuum hypothesis is true, then there exists a function f(x, y) where the integrals f(x,y)dxdy and f(x,y)dydx over the unit square give different values; in general, the existence of such a function is independent of ZFC, but I'm not sure if it's equivalent to CH).
      However, there hasn't been anything so earth-shattering that makes the mathematics community at large decide if CH is true or otherwise.

    • @MusicEngineeer
      @MusicEngineeer 7 หลายเดือนก่อน

      @@recursiveslacker7730 I didn't know about the "Xeeleeverse" - but looked it up. Looks interesting.

    • @MusicEngineeer
      @MusicEngineeer 7 หลายเดือนก่อน

      @@SheafificationOfG Wow - that's interesting! Can you point me to any resources where I can learn more about this fact? In that case, I think, I'd opt for the math universe where the CH is false. ...I kinda *want* these two integrals to give the same value - that somehow seems to be "intuitively" true to me. I can "see" it - we are computing the volume between the graph of f(x,y) and the xy-plane in both cases - that should not depend on how we compute it ....at least for continuous functions. For more erratic functions...like - say - some sort of 2D version of the Dirichlet function - hmm... I don't know if the integrals *should* still be equal...maybe not...dunno

    • @SheafificationOfG
      @SheafificationOfG  7 หลายเดือนก่อน +3

      @@MusicEngineeer Sorry, I lied to you: the existence of a function where the two iterated integrals over the unit square differ is not *equivalent* to CH, but it is a consequence of CH; I've edited the original comment.
      You can get more information starting from en.wikipedia.org/wiki/List_of_statements_independent_of_ZFC#Measure_theory and following references, I imagine. (The whole article has a bunch of independence results, some of which are tied to CH, so it's also an interesting read).

  • @leo848
    @leo848 4 หลายเดือนก่อน +1

    I don't see how 1 = 0 follows from 4 = 3, as you have not shown an inference rule from a = b => a + c = b + c. Or am I missing something?

    • @SheafificationOfG
      @SheafificationOfG  4 หลายเดือนก่อน +2

      I didn't fully formalise the ambient system (for brevity). That a = b implies a + c = b + c can be chalked up to how addition defines a function (and thus preserves equality), or something like that.
      The axioms listed are built on implicit axioms for the primitives used.

  • @AlisterCountel
    @AlisterCountel 3 หลายเดือนก่อน

    One thing I’m still a bit confused on is how this all plays with the law of excluded middle. If we take it as axiom, then these statements have truth value. Should be think of them being simultaneously true and false (until we create a model where we have chosen a specific value)?

  • @CognitiveOffense
    @CognitiveOffense 7 หลายเดือนก่อน

    I enjoyed this and found it actually illuminating. Thanks.

  • @user-wv2jc4xm7r
    @user-wv2jc4xm7r 7 หลายเดือนก่อน

    4:26 We need a video about 0-based indexing.

  • @stanleydodds9
    @stanleydodds9 5 หลายเดือนก่อน

    Proving that something is false doesn't prove that the original statement is unprovable, because the system of axioms might be inconsistent (there may be some statement(s) that can be proven and disproven, or equivalently, it is possible to prove the negation of some axiom).
    For example, if I just have a simple set of axioms like:
    A) There exists an element 0.
    B) There exists an element 1.
    C) 0 ≠ 1
    D) 0 = 1
    Then I can prove that 0 ≠ 1 directly from axiom C, but this does not prove that I cannot prove the negation, that 0 = 1. And to demonstrate this, I can in fact prove that 0 = 1, directly from axiom D.
    Long story short, usually you need the caveat that says something like "if ZFC (or whatever axiom system you are talking about) is consistent, then...".

    • @SheafificationOfG
      @SheafificationOfG  5 หลายเดือนก่อน +1

      Yeah I was working with the hidden assumption that the ambient theory (in this case, the axioms of an integral domain) is consistent. Tom Prince made a similar remark in another comment iirc.

  • @loicboucher-dubuc4563
    @loicboucher-dubuc4563 7 หลายเดือนก่อน

    Awesome video

  • @HazhMcMoor
    @HazhMcMoor 4 หลายเดือนก่อน

    If i summarize correctly, you make two models which are "subset" (?) of the model you want to use, then make them each prove a statement and its negation. This makes the statement unprovable and undisprovable.
    If so, good job for the video!

    • @SheafificationOfG
      @SheafificationOfG  4 หลายเดือนก่อน

      Close, but these models aren't subsets of each other.
      You have a theory, and a statement you're interested in.
      To prove the statement is independent of the theory, you need two models of the theory (meaning all axioms of the theory hold in the model), where the statement is true in one model, and false in the other model.

    • @HazhMcMoor
      @HazhMcMoor 4 หลายเดือนก่อน

      @@SheafificationOfG "model of the theory" means the model is subset of the theory right? Or whatever the definition is where the model is the original theory + some additional axioms.

    • @SheafificationOfG
      @SheafificationOfG  4 หลายเดือนก่อน +1

      @@HazhMcMoor A theory is a set of rules, whereas a model of a theory is an actual thing that satisfies those rules.
      The integers do not form a "subset" of the axioms of an integral domain.

  • @notfancy2000
    @notfancy2000 7 หลายเดือนก่อน

    Came for the memes, stayed for the footnotes.

  • @xLuki
    @xLuki 2 หลายเดือนก่อน

    7:37 but 2 = 2 • 1 and 2 and 1 are non-zero, so 2 = 2 • 1 can't be zero by the last axiom

    • @SheafificationOfG
      @SheafificationOfG  2 หลายเดือนก่อน

      Your argument is circular; you can't assume that 2 != 0 while trying to prove 2 != 0.
      2 = 0 doesn't contradict the last axiom of an integral domain: 2 is *not* nonzero, so 2*1 need not be nonzero!

    • @xLuki
      @xLuki 2 หลายเดือนก่อน

      @@SheafificationOfG oh yeah you're right thanks. It is really confusing to see that 2 might equal 0 indeed haha

  • @wave_s6782
    @wave_s6782 7 หลายเดือนก่อน +1

    so good!!

  • @davidebic
    @davidebic 7 หลายเดือนก่อน

    During my 2am wikipedia scrolling a while back I stumbled on Gödel's incompleteness theorems. I stopped reading at 4am still stuck on paragraph one. Thanks to you I understood 5% more of that paragraph. Now the question is, how do you prove that a set of axioms is consistent or worse w-conxistent. And also what does it mean to have recursively numerable theorems, and why is it important??? Brain exploding just reminiscing...

    • @SheafificationOfG
      @SheafificationOfG  7 หลายเดือนก่อน +1

      Glad I could make a 5% dent!
      The thing about consistency is that, if your theory is consistent, you won't be able to use your theory to prove its own consistency (unless it is a really weak theory). Therefore, we can only prove *relative* consistency results: "If theory A is consistent, then theory B is consistent." In this case, you prove theory B is consistent (relative to the consistency of theory A) by using theory A to build a *model* for theory B.
      Recursive enumerability of theorems is to say that you can write an algorithm down that generates all theorems of the theory (in the sense that any theorem of your theory will eventually be generated by the algorithm in finite time). Very roughly speaking, it means that there are "essentially finitely many" axioms and laws of deduction; you don't have an infinite collection of wildly different axioms that would make reasoning with the theory virtually impossible.

  • @dreadmirror985
    @dreadmirror985 6 หลายเดือนก่อน

    Why did you use the one thinker statue that got blown up? Genuine question.
    Also great vid, love your content from the few minutes I have seen of it :)

    • @SheafificationOfG
      @SheafificationOfG  6 หลายเดือนก่อน +2

      I.... didn't actually know that, just thought it was a picture of The Thinker!

  • @98danielray
    @98danielray 7 หลายเดือนก่อน +3

    4:54
    you forgot the parenthesis on ((1+1)•(x•y)). rookie mistake.

  • @sleepykitten2168
    @sleepykitten2168 7 หลายเดือนก่อน +6

    Very good video!

  • @Jean_Dupon
    @Jean_Dupon 7 หลายเดือนก่อน

    Superbe vidéo

  • @39santia
    @39santia 6 หลายเดือนก่อน

    Nice video men !

  • @SuryaBudimansyah
    @SuryaBudimansyah 6 หลายเดือนก่อน

    Thanks for the timestamps and proper subtitle, the intro is so boring for this interesting topic at the title

  • @ario999
    @ario999 7 หลายเดือนก่อน

    I find the use of a joke paralleling this funny but since I don't feel i have a full grasp. I feel I only half get the joke. But it's a great incentive to watch the video s few times to get it lol.

  • @crix_h3eadshotgg992
    @crix_h3eadshotgg992 6 หลายเดือนก่อน

    The main thing I’ve got from this is that you could make multiple models from the same set of axioms. But isn’t 2=0 another axiom, if you force it? Then it isn’t the same set of axioms… right?

    • @SheafificationOfG
      @SheafificationOfG  6 หลายเดือนก่อน

      The point is that the common way of proving some proposition P to be unprovable and undisprovable from a set of axioms is to provide two models for the axioms: one model where P is true, and another model where P is false.
      I didn't force "2 = 0" onto the axioms, but onto a *model* of the axioms (to build a new one).
      That being said, if you include "2 = 0" as an axiom to the list of axioms in our toy theory of integers, you do get a more specific theory: namely, the theory of "integral domains of characteristic 2", and it is true that there are several models for this theory as well.
      You can use these models to prove new statements to be unprovable/undisprovable. For example: the statement "x = 0 or x = 1" is unprovable and undisprovable in this theory.
      I hope this helps clarify some things!

    • @crix_h3eadshotgg992
      @crix_h3eadshotgg992 6 หลายเดือนก่อน

      @@SheafificationOfG yeah it does thanks bro

  • @cboniefbr
    @cboniefbr 7 หลายเดือนก่อน

    Awesome video!

  • @salim444
    @salim444 7 หลายเดือนก่อน

    8:54 when in doubt always bet on Z/2Z

  • @nicopb4240
    @nicopb4240 4 หลายเดือนก่อน

    Thanks

  • @newwaveinfantry8362
    @newwaveinfantry8362 6 หลายเดือนก่อน

    More set theory videos, please!

  • @racheline_nya
    @racheline_nya 7 หลายเดือนก่อน +2

    nice video! -forcing explanation when-

  • @apia46
    @apia46 5 หลายเดือนก่อน

    that was a cool video i think

  • @mrl9418
    @mrl9418 23 วันที่ผ่านมา

    There is an error with multiplicity 2 at 1:58

    • @SheafificationOfG
      @SheafificationOfG  23 วันที่ผ่านมา +1

      I must be blind. What are the errors?

    • @mrl9418
      @mrl9418 22 วันที่ผ่านมา

      ​@SheafificationOfG No, I misread 😢 so I made the error

    • @SheafificationOfG
      @SheafificationOfG  22 วันที่ผ่านมา

      @@mrl9418 haha all good then!

  • @Ethan13371
    @Ethan13371 7 หลายเดือนก่อน

    Good video!

  • @tcoren1
    @tcoren1 25 วันที่ผ่านมา

    It's not part of the axioms, but it seems you also assumed that
    If a=b, and c=d, then a+c=b+d.
    Necessary to do stuff like adding a number to both sides
    Or is it a consequence of the other axioms? I can't see it.
    I'm only asking because a core part of the video is that 2=0

    • @SheafificationOfG
      @SheafificationOfG  24 วันที่ผ่านมา

      Yeah I didn't go all the way with the axiomatisation because it gets a bit too overboard imho to mention that all operations provided are well-defined *functions* of their arguments.
      But you are right, strictly speaking.

  • @wandrespupilo8046
    @wandrespupilo8046 6 หลายเดือนก่อน

    how can you prove that a structure satisfies a given set of axioms?
    i'm a math undergrad very involved in logic, and this questions PLAGUES me, as i've searched litterally everywhere and i can't seem to find an answer, but everyone treats it as taken for granted (like lol obviously the integers satisfy these axioms, just look at them)

    • @SheafificationOfG
      @SheafificationOfG  6 หลายเดือนก่อน +2

      To prove that e.g. associativity of addition holds in the integers is to prove for all integers a, b, c that (ab)c = a(bc). This will boil down to how you decided to define integer addition.
      One way would be to define addition of natural numbers and then adjoin the negatives. In this case, addition is defined by induction, so you can prove its associativity in the natural numbers by induction. Passing to general integers can then be done with some simpler algebraic manipulations using the definition of negatives, accounting for whenever some of a, b, c are negative.
      Other axioms could be done similarly.
      Hopefully this helps if your aim is to verify the axioms of an integral domain for integers!

    • @wandrespupilo8046
      @wandrespupilo8046 6 หลายเดือนก่อน

      @@SheafificationOfG ty so much

    • @wandrespupilo8046
      @wandrespupilo8046 6 หลายเดือนก่อน

      wait. when you talk about defining integers (and operations) based on the definition of the natural numbers and that being enough for the satisfaction of the axioms, that leads me to a confusion
      isn't the definition of the naturals literally the first-order theory N (the peano axioms)? if i naively extend your reasoning, i could prove that the naturals, as defined by the peano axioms, are OBVIOUSLY a model of N, proving it's unprovable consistency.
      i've been reading the Introduction to Metamathematics, and for the first time i've seen someone (kleene) explain the satisfiability consistency thing in a different way (describing the method before the hilbert program), he explained that a model is just another theory which can embed the original theory for which consistency is trying to be proven, e. g. euclidean geometry is consistent IF the theory for reals are consistent, but doesn't actually prove the consistency of euclidean geometry, and you could only actually prove the consistency by exhaustion if you could find a finite model. i didn't end the book so i don't know if anything has changed, what i know is that the godel kinda killed the program when proving the unprovability of the consistency of N (or any theory in which you could derive the naturals and arithmetic).
      For once, how could you prove that 0 belongs to the integers?

    • @SheafificationOfG
      @SheafificationOfG  6 หลายเดือนก่อน +2

      Right, I swept the detail of "where do these models come from??" under the rug (for the sake of keeping things simple). My model "the integers" needs to be constructed somehow, and to be precise, this means I'm building the integers inside of some ambient theory; for example, perhaps my ambient theory was ZFC. As a result, my implicit claims of consistency of the axioms of an integral domain (demonstrated by providing models of an integral domain) are only *relative* to the ambient theory wherein I'm building my models.
      Hope this helps!

  • @denizgoksu9868
    @denizgoksu9868 7 หลายเดือนก่อน

    My GOAT

  • @MyAce8
    @MyAce8 2 วันที่ผ่านมา

    ok now do a video explaining sheaf theoretic forcing

  • @whynautchase
    @whynautchase 7 หลายเดือนก่อน

    Wow I always thought that independence was a much stronger statement than that. So this means that there is more than one model satisfying ZFC?

    • @fullfungo
      @fullfungo 7 หลายเดือนก่อน

      There are at least infinitely many models satisfying ZFC (assuming it does not contain contradictions).
      Assume you have a valid model M with domain D that represents sets, and a relation IN that represents set membership “∈”.
      Now take any definable rule that uniquely ”assigns” every “set” from D to every “set” from D. This rule induced a bijective relation on D, which can be seen as a permutation.
      I will denote this idea as “F(a) = b”, where “a” and “b” are “sets” from D. It should be understood as “the rule F is satisfied for the ordered pair (a,b) and «b» is the only element that satisfies it if «a» is the first element” (the uniqueness follows from the definition of F).
      Some examples are:
      1. F(a) = a, except F({})=F({ {} }) and F({ {} })={}.
      2. F(a)=a+{ {} } if “{} not IN a”, F(a)=a-{ {} } if “{} IN a”
      Now consider a “permuted model” PerM with the same domain D that represents sets and a relation PerIN that represents set membership.
      The new membership rule is
      “a PerIN b” if and only if “a IN F(b)”.
      You can prove that all axioms of ZF are satisfied in this model, except possibly Foundation.
      For example, the empty set exists. Let’s call the representation of the empty set in D as Emp. Then by bijectivity of F, we know there is an element PerEmp with “F(PerEmp) = Emp”.
      Now the statement “a PerIN PerEmp” is satisfied iff “a IN F(PerEmp)” which is the same as “a IN Emp” which is always false.
      Therefore, “a PerIN PerEmp” is always false, which means “PerEmp” is the empty set of the new model.
      The other axioms are proven similarly.
      Edit: I meant ZF-

    • @whynautchase
      @whynautchase 7 หลายเดือนก่อน

      @@fullfungo that's neat. Doesn't really feel like a different model in any substantial sense though

    • @fullfungo
      @fullfungo 7 หลายเดือนก่อน

      @@whynautchase I made a small correction to my original comment. These model ZF without C. Which means you can construct examples and counter-examples for the axiom of choice with them.
      I would call that a rather diverse set of models.

  • @Nick-go9yd
    @Nick-go9yd 7 หลายเดือนก่อน

    You’re the goat

  • @zekejanczewski7275
    @zekejanczewski7275 4 หลายเดือนก่อน

    If any set of axioms can make unprovably true statements, isn’t that saying for any set of axioms, you are missing one axiom?

    • @Solonarv
      @Solonarv 4 หลายเดือนก่อน

      Not every set of axioms has unprovable statements. For example, the system of integers from the video together with 2=0 is complete and consistent: every statement is either provable or disprovable.
      Gödel's incompleteness results state that every *sufficiently complex* axiom system is incomplete, and give a threshold for what "sufficiently complex" means. The theory of Z/2Z just isn't complex enough for the incompleteness results to apply.

  • @MilanStojanovic9
    @MilanStojanovic9 หลายเดือนก่อน

    ❤️

  • @ucantSQ
    @ucantSQ 7 หลายเดือนก่อน +2

    I love meme some mathematics.

  • @et2124
    @et2124 7 หลายเดือนก่อน

    bigboxswe of math

  • @Gordy-io8sb
    @Gordy-io8sb 7 หลายเดือนก่อน +3

    0:58 Puh-LEASE. Grothendieck was so feeble, he couldn't accept an award, and Gödel was such a specialist it hurts. Literally not even being sarcastic.

  • @noether9447
    @noether9447 7 หลายเดือนก่อน +1

    why do people have a philosophical problem with CH but not with Euclid’s fifth postulate(the parallel axiom)??
    is there any difference between the two in the sense that both are provably unprovable and undisprovable?? 🤔

    • @ahoj7720
      @ahoj7720 7 หลายเดือนก่อน +1

      Models of geometry using different versions of the parallel axiom are all useful. So all of them are actually used. Concerning the axiom of choice, adding it to ZF is more fruitful than the contrary. The model of ZFC is used extensively. Up to now no new interesting result seem to have been found using the continuum axiom nor its negation. So it stays in limbo.

    • @noether9447
      @noether9447 7 หลายเดือนก่อน +1

      @@ahoj7720 was talking about the continuum hypothesis.

  • @OldGameAcc
    @OldGameAcc 3 หลายเดือนก่อน +1

    Title sounds like this is about God 😂

  • @srghma
    @srghma 7 หลายเดือนก่อน

    Can u make a video about Your worldview? (I just write randomly)

    • @srghma
      @srghma 7 หลายเดือนก่อน

      Not randomly, but "what is in my mind, what is interesting to me"

    • @stefanalecu9532
      @stefanalecu9532 7 หลายเดือนก่อน

      ​@@srghmaStrictly speaking, he posts whatever's interesting to him (at least I hope so), so at least that's covered
      Isn't that a bit of a personal thing to request though?

    • @srghma
      @srghma 7 หลายเดือนก่อน

      ​@@stefanalecu9532 I should not ask?
      humans will join brains and become one human being using Neuralink like in Nexus trilogy. Secrets are not possible in this universe. So, why not tell?

  • @ashman9789
    @ashman9789 7 หลายเดือนก่อน +1

    first

    • @JoaoVictor-xi7nh
      @JoaoVictor-xi7nh 7 หลายเดือนก่อน +8

      successor of first

    • @mt180extras
      @mt180extras 7 หลายเดือนก่อน +3

      @@JoaoVictor-xi7nh s(s(first))

    • @Gordy-io8sb
      @Gordy-io8sb 7 หลายเดือนก่อน +1

      @@mt180extras The successor function is actually redundant. You can just add 1 to a. But if a = p-1 (mod p), where p is the characteristic, you get 0, so watch out for that.

    • @biblebot3947
      @biblebot3947 7 หลายเดือนก่อน

      @@Gordy-io8sbin PA, addition is defined in terms of the successor

    • @Gordy-io8sb
      @Gordy-io8sb 7 หลายเดือนก่อน

      @@biblebot3947 Yeah, but the successor only works for integer values. You can't iterate s x times (x being a real number, need not be natural), that's not how it works.

  • @markolson4660
    @markolson4660 7 หลายเดือนก่อน

    I'd like to like this, but all the inane pictures make it hard to pay attention to what you're saying.

    • @SheafificationOfG
      @SheafificationOfG  7 หลายเดือนก่อน +1

      Fair enough! I'm still figuring out a good style for my videos.

  • @andrewporter1868
    @andrewporter1868 7 หลายเดือนก่อน

    New Math: actually we deny the fact that objective, transcendental truths exist which underlie all logic, and that they are abstracted from the concrete, but thanks for these rules tho.
    Ontology as developed since Aristotle and thru Scholasticism: am I a joke to you?

  • @harriehausenman8623
    @harriehausenman8623 6 หลายเดือนก่อน

    Horrible audio, sloppy math and silly pictures.
    No gracias. 🙅👋

  • @kaioxys
    @kaioxys 7 หลายเดือนก่อน

    This is all so dumb. The continuum hypothesis is obviously false because a subset of a set and a set are not the same size. So clearly they cannot be correlated. mapped onto each other.
    You can map all the numbers between 1 and 2 to all the numbers between 2 and 3, but if you map between 1 and 2 to between 2 and 4, the number is twice as large + 1 (the extra 1 being 3 itself.)
    Idk how math people get themselves stuck in such ridiculously obvious questions.

    • @fullfungo
      @fullfungo 7 หลายเดือนก่อน +6

      Bait ☝️😂

    • @AnarchoKeks
      @AnarchoKeks 7 หลายเดือนก่อน +1

      you can very easily map all the numbers between 1 and 2 to the numbers between 2 and 4. The map in question is just 2 times x. And the inverse is y divided by 2.