Theorems That Disappointed Mathematicians

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BriTheMathGuy

BriTheMathGuy

19 күн бұрын

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Пікірлер: 180
@BriTheMathGuy
@BriTheMathGuy 20 күн бұрын
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@CheckmateSurvivor
@CheckmateSurvivor 11 күн бұрын
The world is full of fake mathematical models. Just ask the politicians.
@dennismuller1141
@dennismuller1141 18 күн бұрын
The only theorem that came to my mind when I read the title was Gödel's incompleteness theorem. And now, after watching the video I also remember the problem of constructing a square and a circle with the same area
@roihemed5632
@roihemed5632 17 күн бұрын
Won't the areas be the same if the side of the square will be the square root of pi times the radius of the circle?
@gaopinghu7332
@gaopinghu7332 17 күн бұрын
​@@roihemed5632 that's true, however there is no way to construct a side that length with just a straightedge (a ruler with no numbers written on it) and a compass.
@dennismuller1141
@dennismuller1141 17 күн бұрын
@@roihemed5632 assuming you meant pi times the square of the radius: yes, the area would be the same but it is impossible to construct with the classic rules of geometry
@roihemed5632
@roihemed5632 17 күн бұрын
@@dennismuller1141 Ok I got what you meant thanks. But I think you're wrong about the length. if x is the side of the square and x² = πr² then x = r√π
@dennismuller1141
@dennismuller1141 17 күн бұрын
@@roihemed5632 Sorry for the confusion, I misread "square root of pi times the radius" as sqrt(pi * r) instead of sqrt(pi) * r, so I corrected it to sqrt(pi * r²)
@fluffysheap
@fluffysheap 14 күн бұрын
Godel's second incompleteness theorem states that in any collection of disappointing theorems, there will always be a disappointing theorem that is not included
@josepherhardt164
@josepherhardt164 13 күн бұрын
You are EVIL. Upvoted ...
@kephalopod3054
@kephalopod3054 11 күн бұрын
This video has to be incomplete, otherwise it would be inconsistent.
@KT-dj4iy
@KT-dj4iy 10 күн бұрын
Well that's very disappointing.
@chirantanbiswas9330
@chirantanbiswas9330 Күн бұрын
There is always a disappointment that you are not aware of. Disappointment never ends.
@yonaoisme
@yonaoisme 8 сағат бұрын
that's not what goedel states.
@sensorer
@sensorer 16 күн бұрын
I'd like to see someone draw the Weierstrass function without lifting their pen
@matthew-m
@matthew-m 14 күн бұрын
right lool
@Fire_Axus
@Fire_Axus 11 күн бұрын
real
@SteveThePster
@SteveThePster 10 күн бұрын
Manufacturing a sufficiently thin/fine pen would be a challenge
@sensorer
@sensorer 9 күн бұрын
@@SteveThePster if epsilon is line thickness, I think that for any epsilon greater than zero, the thickness is not sufficient
@ccbgaming6994
@ccbgaming6994 6 күн бұрын
@sensorer My pen was manufactured in Surrealia
@ndwind
@ndwind 17 күн бұрын
Was expecting the Gödel theorems on this list
@fragileomniscience7647
@fragileomniscience7647 13 күн бұрын
Would be too easy. Also, it doesn't have to be all too bad. It implies that mathematics has no end, since independent axioms are bound to just pop up. Would be pretty boring if you could do all the math using one rote algorithm.
@moskthinks9801
@moskthinks9801 17 күн бұрын
The real theorem that dissappointed me the most was the Halting Problem. Something about having undecideable problems in certain axiomatic systems, lacking computational knowledge, and not being able to compute the busy beavers (which, honestly without HP would be tremendous still) is just astounding to me
@Alexagrigorieff
@Alexagrigorieff 15 күн бұрын
The halting problem, just like its cousin - Goedel incompleteness theorem, prohibits self-referential logic.
@sebij6811
@sebij6811 11 күн бұрын
​@@AlexagrigorieffAnd Tarski's undefinability theorem.
@muskyoxes
@muskyoxes 8 күн бұрын
I still don't know how the halting problem manages to be so _practical._ All the proofs i've seen leave me with the feeling of "it's possible to construct some outlandish gigamess of a program that can't be analyzed to see if it stops." It seems like a completely separate and unexplained result that many such problems are simple to describe
@moskthinks9801
@moskthinks9801 8 күн бұрын
@muskyoxes a good intro video to watch is about "the boundary of computation" and busy beaver numbers, it's the main video that got me interested in the halting problem
@allozovsky
@allozovsky 18 күн бұрын
2:00 But Pythagoras would probably be glad to know that the square root of two has a periodic continued fraction *√2 = [1; 2, 2, 2, ...].*
@douglasstrother6584
@douglasstrother6584 17 күн бұрын
I'd bet he would get a kick out of that.
@JJean64
@JJean64 22 сағат бұрын
Wait until he learns about transcendental numbers
@douglasstrother6584
@douglasstrother6584 20 сағат бұрын
@@JJean64 "Pythagoras, check these out!"
@allozovsky
@allozovsky 16 сағат бұрын
Hm, but did Pythagoras even know anything about the number π? 🤔
@paradoxicallyexcellent5138
@paradoxicallyexcellent5138 17 күн бұрын
Abel Ruffini is kind of a blessing though. Mathematics telling mathematicians, "Guys, guys. What are you doing? Stop solving single variable polynomial equations with roots. You have better things to be doing."
@yanntal954
@yanntal954 18 күн бұрын
3:44 I think Minkowskis question mark function denoted ?(x) Is weirder than this function. It is continues and always increasing (strongly monotonic) yet its derivative is 0 almost everywhere! You'd think such a function with derivative 0 a.e. would at least be constant a.e. but nope. This function is never constant and always increases!
@abrarjahin8848
@abrarjahin8848 18 күн бұрын
Maths is beautiful
@jb76489
@jb76489 18 күн бұрын
*math ftfy
@devooko
@devooko 18 күн бұрын
meth is more beautiful
@Bodyknock
@Bodyknock 18 күн бұрын
Heh, although ironically this particular video is about instances where math refuses to be beautiful. 😄
@ibozz9187
@ibozz9187 17 күн бұрын
Regarding Arrow’s theorem, there are non-ranked systems that satisfy these criteria. Approval Voting (pick as many or as few as you like with no upper limit) and Score Voting (Score every candidate, greatest sum of scores wins) both satisfy these conditions.
@ariaden
@ariaden 15 күн бұрын
en.wikipedia.org/wiki/Duggan%E2%80%93Schwartz_theorem
@aryandegr859
@aryandegr859 17 күн бұрын
No Gödel incompleteness theorems?
@josepherhardt164
@josepherhardt164 16 күн бұрын
And wasn't there a theorem that showed that you couldn't prove that the infinity of the continuum was equal to Aleph-1?
@antoine2571
@antoine2571 13 күн бұрын
​@@josepherhardt164what you're looking for is continuum hypothesis
@luccasguth
@luccasguth 13 күн бұрын
Continuity 👏 does 👏 not 👏 mean 👏 that 👏 a 👏 function 👏can 👏 be 👏drawn 👏withouth 👏lifting 👏 your 👏 pen
@justtimo8638
@justtimo8638 12 күн бұрын
for 👏🏻 real-valued 👏🏻 functions 👏🏻 in 👏🏻 one 👏🏻 dimension 👏🏻 it 👏🏻 does
@B0bb217
@B0bb217 12 күн бұрын
​@@justtimo8638no
@lucacesarano3661
@lucacesarano3661 2 күн бұрын
No, it does not! 1/x is a continuous function. The function indeed is simply not defined in 0. Simply it can't be extended for X=0 to a continuous function, but itself is continuous.
@jorian_meeuse
@jorian_meeuse Күн бұрын
​@lucacesarano3661 1/x is not continuous in x=0. Continuity in x=a means that f(a) = lim_x->a (x). A function is considered continuous on an interval I if the function is continuous for all x in I. Since f(0) doesn't exist for f(x) = 1/x, and the limit when x goes to 0 doesn't even exist, certainly it's not continuous in x=0. Hence, f(x) is not continuous over R.
@lucacesarano3661
@lucacesarano3661 Күн бұрын
@@jorian_meeuseThere is a logical misunderstanding in the definition. How would you evaluate the predicate "1=√-1"? True or false? The point is, it does make any sense to say it's true or false, since √-1 is even not defined. You say correctly that in X=0 the function is not defined. Exactly because of it, it does not even make sense to speak about continuity in X=0. That's because of the definition of continuity: "A function f with variable x is continuous at the real number c, if the limit of 𝑓(𝑥) as x tends to c, is equal to 𝑓(𝑐)." Well, if f(c) is not defined, simply the sentence "is equal to f(c)" can't be logically evaluated as true or false. Could you logically evaluate the sentence "1 is a yellow number"? If you say it's false, then it means that the sentence "1 is not a yellow number" is true for you, and the property "not to be yellow" is properly defined. But it is not. For this reason, one can't even say that 1/x is not continuous in x=0. But it is absolutely a continuous function, since it is continuous on every point of its domain(!).
@stevenfallinge7149
@stevenfallinge7149 17 күн бұрын
The weierstrass function ties into the idea of fractals (in fact, the function itself is an example of a fractal) and the idea that most things in nature are fractals, things that are not smooth or well-behaved.
@omp199
@omp199 17 күн бұрын
"Most"?
@josepherhardt164
@josepherhardt164 11 күн бұрын
"things that are not smooth or well-behaved." So, like my wife? ;)
@douglasstrother6584
@douglasstrother6584 17 күн бұрын
This Theorem is so boring that we're calling it a Lemma.
@cmilkau
@cmilkau 17 күн бұрын
5:40 NO! I usually let this slide but this is a maths channel. The nonexistence of a universal algorithm does NOT mean that there are unsolvable instances of the problem. There could be a (different) algorithm for *every* instance of the problem, there is just not a (single, universal) for *all* of them. There may even be a way to describe strategies for all possible cases, while there is just no systematic way to select a strategy that works, so you might continue trying forever.
@jonathan3372
@jonathan3372 13 күн бұрын
At 4:18, I recall that in analysis we usually define a smooth function as one that is infinitely continuously differentiable. Did you mean to say "a continuous curve should be differentiable almost everywhere"?
@itsmxrk.9469
@itsmxrk.9469 16 күн бұрын
Not the AI thumbnail 😭
@glennjohnson4919
@glennjohnson4919 18 күн бұрын
What an interesting presentation. Loved it.
@33invasion
@33invasion 11 күн бұрын
Just a minor correction. IIA in Arrow’s impossibility theorem isn’t about whether alternatives (added or taken out of the menu of alternatives) affect binary preference relations at the aggregate level. It’s about whether changing the preference profile outside the binary comparison yields a different aggregate preference between the two.
@MathFromAlphaToOmega
@MathFromAlphaToOmega 18 күн бұрын
There are actually some really interesting ways of solving quintic equations. One way is with power series, and the coefficients involve factorials related to multiples of 5. Klein also found a way to relate symmetries of the icosahedron to solving quintics.
@FishSticker
@FishSticker 16 күн бұрын
Yeah but they don't always work
@MathFromAlphaToOmega
@MathFromAlphaToOmega 16 күн бұрын
@@FishSticker What do you mean by that? The power series has a finite radius of convergence, but there are ways to get around that. As for the other method, I believe it always works, as long as the quintic is in the right form.
@FishSticker
@FishSticker 16 күн бұрын
@@MathFromAlphaToOmega you won't get an exact answer
@FishSticker
@FishSticker 16 күн бұрын
@@MathFromAlphaToOmega also you might not know if it exists or not
@iteo7349
@iteo7349 12 күн бұрын
The Weierstrass function has a feature which is probably even more striking to non-mathematicians: it's a continuous function, which is not monotone (increasing, decreasing, or constant) on any interval, no matter how small. Seems impossible at first glance. Also, I was anything but disappointed that quintic equations aren't solvable. People need to grow out of this "solving" mindset. To solve something only means to write it in some other way that you find simpler. Guess what, some things just can't be written in simple ways according to narrow minded definitions of simple. Conclusion: expand your horizon and move on.
@rebase
@rebase 13 күн бұрын
The Abel-Ruffini theorem states that polynomials of degree greater than 4 cannot be solved *using radicals*. A radical is a solution to the equation x^n = a, or in lay terms an n-th root. E.g. sqrt(2) is a solution of x^2 = 2 If we go beyond plain radicals, and add to our toolbox the so-called Bring radicals, which are the unique real solutions to equations of the form x^5 + x = -a, then the general quintic becomes solvable! So it isn't like quintic polynomials have this infinite complexity compared to lesser degree polynomials. Rather, quintics need a single additional function (the Bring radical function) to express their solutions. Of course, Bring radicals cannot be computed exactly, only approximated, but neither can sqrt(2). Nothing special about it.
@Hadar1991
@Hadar1991 18 күн бұрын
Arrow's Impossibility Theorem states only that you cannot achieve global independence of irrelevant alternatives in raked voting. But there are raked voting methods that are locally independent of irrelevant alternatives. In rated voting you can have global independence of irrelevant alternatives. But what is more important global of irrelevant alternatives is often criticized as something you may not want to have in your election system. More interesting (and disappointing) is Gibbard-Satterthwaite theorem which states that there is no election system which isn't susceptible to tactical voting. 4:15 Smooth curve is differentiable everywhere by definition. Weierstrass function is continues, but not differentiable at any point. Weierstrass function is in C_{0} class of smoothness in all of this domain, which basically means that it is nowhere smooth. A smooth function is in C_{infinity} class of smoothness.
@drdca8263
@drdca8263 17 күн бұрын
*which isn’t susceptible to tactical voting
@Hadar1991
@Hadar1991 17 күн бұрын
@@drdca8263 Thank you, corrected ;)
@Alexagrigorieff
@Alexagrigorieff 15 күн бұрын
And then there are conjectures that are proven to be unprovable.
@zapazap
@zapazap 10 күн бұрын
All conjectures can be proven to be unprovable modulo some axiomatic system.
@arekkrolak6320
@arekkrolak6320 2 күн бұрын
Technically speaking Pitagorean Theorem said nothing about segment length, Pythagoras was only concerned with area of constructed squares
@sweettoy3824
@sweettoy3824 15 күн бұрын
We're doing AI-generated thumbnails now too?
@erroraftererror8329
@erroraftererror8329 17 күн бұрын
I think it's for the better that we don't know everything in mathematics and that not everything has a solution. A complete mathematical understanding would mean that we have no more reason to pursue the subject. How boring life would be!
@josepherhardt164
@josepherhardt164 16 күн бұрын
This is why I rejoice whenever physics gets "broken." :)
@kaminoeugene
@kaminoeugene 17 күн бұрын
"you can draw it without lifting your pen"...
@MichaelRothwell1
@MichaelRothwell1 15 күн бұрын
For those of us who can draw nowhere differentiable graphs...
@wyqtor
@wyqtor 11 күн бұрын
Gödel: hold this beer, you're gonna need it!
@MarcoMate87
@MarcoMate87 13 күн бұрын
The solution proposed at 0:16 for the general cubic equation is wrong; it's, indeed, the solution of the depressed cubic y³ + py + q = 0, where y = x + b/(3a). Thus, the general solution you proposed needs a "-b/(3a)" in the second member to be correct.
@sidimed1956
@sidimed1956 18 күн бұрын
Can you tell us how to prepare for the IMO pls
@nathancc2526
@nathancc2526 18 күн бұрын
I would also love to know hope he sees this comment
@stevenwilson5556
@stevenwilson5556 11 күн бұрын
Hippasus, not Pythatorus discovered irrational numbers and for his discovery he was tossed into the sea to drown. A true "math martyr". The Pythagoreans did not like reality to intrude into their harmonious "rational bubble"
@LeoStaley
@LeoStaley 14 күн бұрын
I want more videos like this.
@hcm9999
@hcm9999 17 күн бұрын
It is impossible to be absolutely sure about anything.
@theoneeggo4653
@theoneeggo4653 17 күн бұрын
This man needs more viewers
@roykay4709
@roykay4709 19 сағат бұрын
Technically, I would call these "conjectures".
@DrCorndog1
@DrCorndog1 7 күн бұрын
You seriously made this video without bringing up Godel.
@giorgioleoni3471
@giorgioleoni3471 6 күн бұрын
I would add the Banach-Tarski Theorem
@JonnyMath
@JonnyMath 18 күн бұрын
Solution: "Just invent new algebra!" 😅🤣 Why not!!!😉😅
@parthpandey2030
@parthpandey2030 17 күн бұрын
Euler trying to prove i = sqrt(-1): Proof: he made it up
@JonnyMath
@JonnyMath 17 күн бұрын
@@parthpandey2030 Also known as "by definition"😅🤣
@bimrebeats
@bimrebeats 17 күн бұрын
you should, just don’t “break” the old algebra 😉
@JonnyMath
@JonnyMath 17 күн бұрын
@@bimrebeats Yes this is what I'm saying!!!😅🤣
@columbus8myhw
@columbus8myhw 15 күн бұрын
Look up Bring radicals.
@columbus8myhw
@columbus8myhw 15 күн бұрын
You should cite the Math Stack Exchange post "Theorems that Disappointed Mathematicians," assuming you based this on it
@ameya0308
@ameya0308 11 күн бұрын
What is 0+0+0+0+0.... ∞ equal to? Technically it should be 0. But when you treat this series as a Geometric Progression and apply the formula for sum of infinite terms of G.P., the result comes to be 1/(1-(0/0)). How is this possible?
@JJean64
@JJean64 21 сағат бұрын
If you treat this series as 0⁰ + 0¹ + 0² + 0³ + 0⁴ + 0⁵ + ... , when apply the geometric series formula you get 1/(1 - 0) = 1/1 = 1. Which makes sense because 0⁰ is usually defined to be equal to 1, so the series becomes 1 + 0 + 0 + 0 + 0 + 0 + ... = 1.
@sabriath
@sabriath 4 күн бұрын
Arrow had it wrong anyway....that's a direct vote system and they all fail in some form, but if you were to change into an indirect vote, suddenly it solves actually quite easily. As an example, instead of voting for which person is the best candidate for position XYZ, considering that the majority of the population doesn't even pay attention to politics and are voting on mostly false assumptions......we should be voting on the policies we want to see enacted or retracted. The candidate who scores the most in comparison to their own beliefs of what should be done against the voters that cover that area, should be the winner to the vote itself. To protect the population from tyrannical measures, a simple minimum requirement of 66% vote for those policy changes are the only ones that can actually be commanded for change....everything else is void until next term to be re-evaluated. This would work for all positions, and noting "policy" as different scenarios based on the field of position XYZ (so president will most detail command of military operations, while laws will deal with legislative branch, and punishment the judicial, etc.)
@scottleung9587
@scottleung9587 16 күн бұрын
Interesting!
@josepherhardt164
@josepherhardt164 16 күн бұрын
Standard equation for the perimeter of an ellipse? That C is not = pi * f(a, b) is seriously disappointing.
@nothingok8800
@nothingok8800 17 күн бұрын
True
@JakubS
@JakubS 17 күн бұрын
that AI person in the thumbnail looks like Hugh Dennis
@omp199
@omp199 17 күн бұрын
I can't see the resemblance.
@gabitheancient7664
@gabitheancient7664 17 күн бұрын
I don't think there's any legend about pythagoras discovering the irrationality of the square root of two, it's just that this is related, but there's other greek guy everyone says discovered this tho the pythagorean theorem is related, not only we know for a fact it was not discovered by pythagoras (we don't even know if he existed) but you don't need this theorem to know about the square root of two, it's a very easy fact to discover that the diagonal of a square is the side of another square with double its area
@crix_h3eadshotgg992
@crix_h3eadshotgg992 17 күн бұрын
I recall hearing a legend about some guy (with a name starting with “H”) proving that the square root of two being irrational, and subsequently being tied up with stone, put on a boat, and thrown into the sea. Not making this up.
@gabitheancient7664
@gabitheancient7664 17 күн бұрын
@@crix_h3eadshotgg992 ye and this was surely not pythagoras, and the legend says the pythagoreans threw him in the sea I strongly believe this didn't happen tho
@BanHelsing
@BanHelsing 17 күн бұрын
AI looking ahh thumbnail
@devenyelve4905
@devenyelve4905 2 күн бұрын
Can we solve this problem (2+x)^14.33i where i is imaginary
@allozovsky
@allozovsky 15 сағат бұрын
But that's just a binomial with a complex exponent, isn't it? Simply use Newton's binomial theorem.
@allozovsky
@allozovsky 15 сағат бұрын
But the definition of complex exponentiation is even simpler: *zʷ = exp(w·Ln(z)),* where *Ln(z) = ln(|z|) + 𝒊·arg(z) + 2𝒊πk, k ∈ ℤ* is a _multivalued_ complex logarithm, so eventually you get _infinitely_ many values (and may choose one of them on a whim).
@christopherellis2663
@christopherellis2663 Күн бұрын
Nonetheless, preferential voting is truer than first past the post.
@user-gh4lv2ub2j
@user-gh4lv2ub2j 18 күн бұрын
Yes; it took all of 10s to solve that first polynomial.
@omp199
@omp199 17 күн бұрын
What's the solution, then?
@TranquilSeaOfMath
@TranquilSeaOfMath 17 күн бұрын
That's a great thumbnail!
@mkks4559
@mkks4559 13 күн бұрын
Please don't use AI-generated images.
@user-jb8yv
@user-jb8yv 15 күн бұрын
please don’t use ai generated thumbnails
@zapazap
@zapazap 10 күн бұрын
Why?
@JoelRosenfeld
@JoelRosenfeld 8 күн бұрын
It got you to click, and it got me to click. The data is what is important here more than personal opinions. I think it’s a nice dramatic picture to generate a click.
@evanmagill9114
@evanmagill9114 7 күн бұрын
@@JoelRosenfeld Any stance on this involves personal opinions to some degree. Here's some data for you: Of the comments currently present on this video 4% of them refer to the thumbnail as AI. Of those 6 comments, 1 is neutral and 5 are negative. I personally am a lot less likely to click on videos with thumbnails that are incoherent or particularly noticably AI-generated. Here's some data we don't have: How many more or less views would this video get with a human-made thumbnail that took 5 minutes to make? 3 hours to make? If an artist was paid $30 to make a thumbnail, might it result in additional revenue that would cover those $30? Will the channel's community be affected by the use of AI in thumbnails? Will the relationship with the video's sponsor be negatively impacted? I don't intend to cause you any offense, and I'm not here to try to make you feel one way or another about AI. I just wrote this comment because I know I've used naïve arguments to dismiss people, and I would rather be called out on it than not. Opinions matter, people's thoughts are worth consideration even (and especially) when they differ from your own.
@Nolys-bk4kd
@Nolys-bk4kd 4 күн бұрын
Why not? He does what he wants
@zapazap
@zapazap 4 күн бұрын
@@Nolys-bk4kd I asked 11 days ago mate. He's not gonna answer.
@amazingplayer4954
@amazingplayer4954 18 күн бұрын
Fun fact: the degree of polynomials which always have solutions goes up to 4, just like the number of dimensions, coincidence?
@Penrose707
@Penrose707 18 күн бұрын
This is true only in so far that polynomials up to degree four may have their roots solved by using a finite number of simple algebraic operations (re: summation, multiplication/division, and exponentiation/nth roots). Technically the roots of any polynomial in degree z (the "solutions") can be determined by solving a particular series of z nonlinear equations (hint: think about coefficients of algebraic polynomials as consisting as sums of combinations of those roots. Otherwise, evoke the complex domain). This is extremely difficult and so in practice we often rely on the use of approximations and perturbation methods instead
@user-gh4lv2ub2j
@user-gh4lv2ub2j 18 күн бұрын
Fun fact: the above is wrong. All polynomials have solutions; you just ignore complex solutions. People using science as religion get facts wrong: coincidence?
@James2210
@James2210 17 күн бұрын
​@@user-gh4lv2ub2jx^2+1 😳
@ethanbottomley-mason8447
@ethanbottomley-mason8447 17 күн бұрын
Yes, this is entirely a coincidence. The reason why all polynomials of degree at most 4 can be explicitly solved in terms of radicals is because every group of order at most 24 is solvable. This is not the case in general. For instance S_5 is not solvable. This is why the roots of the polynomial x^5 + x + 1 cannot be expressed in terms of radicals. The real reason is Galois theory, not some spurious coincidence that the number 4 appears in two places.
@drdca8263
@drdca8263 17 күн бұрын
@@ethanbottomley-mason8447It seems very likely to be only a coincidence, but, I’m not sure that that description of why quintic not solvable, proves that it is. I suppose for that we might need a more precise definition of what it means for it to be a coincidence
@lazyman7769
@lazyman7769 2 күн бұрын
Continous everywhere means you can draw it without lifting your pen😂 mate get straight with your concepts
@osvaldo701
@osvaldo701 14 күн бұрын
If you are disappointed by any of those theorems, you are really not understanding them
@nathancc2526
@nathancc2526 18 күн бұрын
Can I get a heart pls been following for a long time gr8 vid btw
@reaperplays6072
@reaperplays6072 18 күн бұрын
25 seconds and no views bro fell off
@yuraje4k348
@yuraje4k348 17 күн бұрын
12hours and 4k views
@creativename.
@creativename. 17 күн бұрын
15 hours and almost 5k views bro fell off
@DKAIN_404
@DKAIN_404 17 күн бұрын
19 hours and 6.4k views bro fell off
@Chocolatefudgecakeee
@Chocolatefudgecakeee 17 күн бұрын
The npc's are coming out again
@mechanicalmonkee6262
@mechanicalmonkee6262 17 күн бұрын
😐👎
@prostoopid
@prostoopid 18 күн бұрын
W
@v8torque932
@v8torque932 18 күн бұрын
IM THE 1000TH VIEWER IM FAMOUS!
@gaurang6186
@gaurang6186 18 күн бұрын
Please like this comment so that u can solve millenium prize problem
@Ostup_Burtik
@Ostup_Burtik 18 күн бұрын
Dislike, because WE NEED VIDEO ABOUT WHAT IS i^^i!
@Maths_B
@Maths_B 18 күн бұрын
Anyone does not known how I got this much likes
@AvinamGurung-nu3kl
@AvinamGurung-nu3kl 18 күн бұрын
You ain't got any likes nigga 🤡
@riothememer2029
@riothememer2029 18 күн бұрын
Very Cool
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