Видео

You don't like America but you want to use our website and talk in English to try to make cash here.. We need to fix 🤡's like this so it is NOT possible.

Get off our USA website .. you been told already..... You don't like the USA don't use our stuff! Do you need me to type it in Italian?

get off our website USA hater

This is a USA website you are welcome to F-O at any time buddy!

Huh ... is it really that simple? I was reading the definition and got pretty confused: ∀A(A≠∅⟹∃x(x∈A∧x∩A=∅)). And the Wikipedia article left me even more baffled.

As soon as I saw this guys face I figured he was about to tell me that I never had the makings of a varsity athlete

I thought that equal meant: all shapes had an equal amount of parts

No it does not result in a contradiction. Russell was wrong and Frege was right, the set of all sets which do not contain themselves does not contain itself. Consider….there are two sets of men in the town, those who do shave themselves and those who do not who instead, visit the barber. The barber has posted a sign which instructs that he only shaves those who do not shave themselves. The paradox arises when we consider who would shave the barber, i.e, to which set of men would he belong. But the answer is to neither. There are four criteria by which the two sets of men, necessary to the paradox, are defined as such. Three of these four are identical, that they are men, live in town and must be beardless, so the deciding factor is the fourth one, their relationship to shaving, that one shaves themselves and the other not. If this were not so, there would be only one set of men and the paradox would fail. The barber then shares these same three criteria and like with the first two sets of men, it is his relationship to shaving which is the deciding factor as to what set of men to which he would belong. But it is not the same as the other two but rather that he shaves others. If you deny this logic, you deny the means of definition of the first two sets of men and the paradox fails. If you accept it the paradox fails, because it is not a paradox. That Russell could be so far off is astonishing to me, but he was, unless anyone can argue away that I have defined above. I do not think anyone can but would be excited to see someone try. It would be a fun discussion. Any thoughts?

This whole business of axioms makes no sense to me as it relates to logic in mathematics. Consider, the statement 1+1=2 is almost tautological and true by definition. There is no reason for an axiom because the statement cannot be untrue. This is unequivocal. That the concept of 1 can be conceptualized in consequence of the perception of “a something by itself” and that label “1” arbitrarily formulated and applied to that concept means that that which was perceived is “1” something by itself. As long as all in witness agree that “1” means a something by itself, there is no need for proof. It cannot be otherwise. That another “1” or something by itself can be perceived and understood as such is by definition possible or the original something by itself could not have been perceived, which it was (and we named it). By extension then, we cannot but understand that 1 and (or +) 1 is (1 and 1) or “2”, a second concept we formulate and name. The label 2 meaning a something by itself and another something by itself is thus by definition true. Again, no need for axioms or proof of its truth for were we unable to comprehend its unequivocal truth we could not have expressed the proposition to begin with which we did. Again, by extension and the application of the same logic, we know that 2+2=4 by the same logic which we would be obliged to apply. The further extension of these simple propositions would be by the same logic, by definition true, unequivocally. What then is the point of these extensive, sometimes redundant axioms by which sophistry like that of Goedel’s claim of the “hole” in the logic of math can be imposed. Any proposition which is true by definition requires no proof and so the insistence on the imposition of some analogously wordy and thus, unnecessary scheme is actually defiant of the logic it is purported to support or validate. Any thoughts?

Se ve muy legible para 4 segmentos

Ah, you use Separation schema and Replacement schema. Some formulation use use Empty set axiom with Replacement schema instead. Because by defining a function unto itself, we can separate any element we want. Also, I think you pick an implication Zorn's lemma instead the usual Axiom of Choice. Which is quite unusual.

You make axioms like so easy, when I first entered the axioms of ZFCs, I had a lot of trouble memorizing them and understanding them. Please continue make videos on this. Especially on the axiom of choice, Zorn's lemma, that I still don't quite get it.

A number is a linguistic designation for a position of quantity. Quantity is dividing something into equivalent portions - fungibility.

Is there a Triforce made out of loads of more triforces

the math aint mathin and ik ian dumb buh idk if sum goin on up there cuz yu not even breakin it down for sum of the ppl who actually dnt kno math buh the ones that do can see yu jus speed solvin and breakin it down

Cool video, very informative and clear

If the two jars are separated in space, they are differently located.

awesome topic

Greaaat

I like this channel

Great! I love your revisualizations at the end. To see such structure emerging in the simple act of counting.... Quibble: at 6:26: you define the "<" relation as the proper subset relation, but this is not correct without qualification, since there are many subsets of an ordinal that are not ordinals at all. The more natural relation (which is part of what makes von Neumann ordinals so cool) is simple set-membership: x<y iff x is a member of y. Then you wouldn't need to qualify "x is an ordinal" -- and in an introduction like this you surely wouldn't want to: the whole point is to show how ordinals ordinals to "emerge" (almost by magic!) from the recursive A-union={A} operation. One other thing: you write "si" for 'if" there, which makes me happy but might confuse some people! 🙂)

how do you find the 6 first numbers ? That is the interesting problem.

axiom is what we can't do any better

Tremaine Roads

Your videos helped so much! But can you please share axioms 7,8 and 9 too?😢

You are making simple things complicated.

Please try to put the shape alone in the picture when u ask us to pause.

1st is Axiom of Extension. Halmos, Naive Set Theory.

Fantastic video! I've been self studying set theory and was having trouble understanding the notation for this axiom, but this video cleared it up fantastically. Thank you!

Nice solution

why was i tagged?

The barber shaves all men who do not shave himself. Who shaves the Barber?

Thank you, Professor. Will you make the video for axioms 7,8, and 9?

Chau

Nice intro dude 😂

Women do shave

Meh, 7 segments is nice and easy, no need to do 4

I find in many cases 6 segments is the lower bound for making the digits normal. It would also be the minimum for display all 26 letter and numbers distinctly in one case. 6 is just enough segments to display all in capital and lowercase as well as numbers, but a lot of 7-segment displays don’t have support for an M or W. Hence why 14 and 16-segment displays have become so common.

I don't understand what you mean by quarter circle? But the ant will turn 5 times (R,L,RL,R) before it gets back to A.

It's the Fibonacci numbers

It's ( 2n^3 + 5n^2 + n )/8 1^2+2^2+3^2+...+(n-1)^2+n^2 - [ (1^2)/4+(2^2)/4+(3^2)/4+...+((n-1)^2)/4 ] n^2 + SUM i=1 to n-1:(i^2) - 1/4*SUM i=1 to n-1:(i^2) n^2 + 3/4*SUM i=1 to n-1:(i^2) n^2 + 3/4*[ SUM i=1 to n:(i^2) - n^2 ] n^2 - 3/4*n^2 + 3/4*SUM i=1 to n:(i^2) 1/4*n^2 + 3/4*[ n*(n+1)*(2n+1)/6 ] 1/4*n^2 + 3/24*[ n*(n+1)*(2n+1) ] 2/8*n^2 + 1/8*[ n*(n+1)*(2n+1) ] 2/8*n^2 + 1/8*[ 2n^3 + 3n^2 + n ] ( 2n^3 + 5n^2 + n )/8

b^0=1, because the right limit and the left limit are equal to 1 for all positive non-zero base value in R.

But is congruence identical to equality?

U Look Like Clever 😜😺

Maybe worth noting that there are non-classical, paraconsistent logics within which derivations of the sort you sketched (say, of 1=2 from an arbitrary contradiction) don't work. For example, in typical relevant logics, disjunctive syllogism is not valid.

I think it's kinda misleading to equate the special case with the axiom itself.

I understand the point of your videos. Still, I don't agree with them in terms of foundations upon which we can start building up maths. Indeed, to formulate those axioms we need something even more fundamental that we need to agree on : first order logic. You couldn't even formulate your first axioms without the "if and only if". It is so simple that it is taken for granted but it shouldn't because there I so much logic that intuition alone can provide. Hence the difference between logic and common sense. Once implication is involved, things start to become messy and without that knowledge, building up maths will quickly become a dead end. But yeah, I understand the simplification here since it's hard to start with logic without knowing a thing about sets and axioms and it's hard to introduce sets and axioms without knowing logic. I firmly believe those 3 should be taught in parallel but to grab people's interest, it's a cool place to start.

6:50 “An axiom is something that does not need a proof“ 😂😂😂😂😂 no. And axiom is a ground truth by definition. Meaning we say that it’s true. Or we assume that it’s true. And then we build the rest of our system on this and other assumptions. Its truth is assumed by definition. We might be able to prove it we might not. We just assume for now that it is true.

I love this and I find it hilarious 💀

Don't forget Terrence Howard's axiom that states : 1x1 = 2