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

Incompleteness*

Gödels silly theorems lol

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

@@micayahritchie7158 yeah

🫡🫡🫡

Yep, exactly!

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.

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

What are you on about?

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

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

My favourite part of a comment "section"

@@SheafificationOfG a mment cosection

Thanks!

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

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.

Merci !

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.

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

@@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.

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

Taking a dip in the Xeeleeverse Logic Pool, eh?

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.

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

@@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

@@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).

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!

@@SheafificationOfG yeah it does thanks bro

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-

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

@@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.

Didn't expect to see you here

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!

@@SheafificationOfG ty so much

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?

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!

DAMMIT

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.

@@ahoj7720 was talking about the continuum hypothesis.

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

​@@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?

​@@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?

successor of first

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

@@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.

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

@@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.

WTF?

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

Bait ☝️😂

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.