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Every Unsolved Math problem that sounds Easy 

ThoughtThrill
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These are some of the famous and toughest math problems, which are unsolved.
Timestamps
0:00 The Kissing Number
1:16 The Goldbach Conjecture
2:25 Collatz Conjecture
3:39 The Twin Prime Conjecture
5:41 The Unknotting Problem
7:05 Pi + e
8:30 Birch and Swinnerton-Dyer Conjecture
9:14 Riemann Hypothesis
11:01 The Lonely Runner Conjecture
11:46 is γ rational?
Sources:-
/ how-to-win-a-million-d...
en.wikipedia.org/wiki/Lonely_...
www.cantorsparadise.com/the-l...
en.wikipedia.org/wiki/Millenn...
Domain of Science
- DISCLAIMER -
This video is intended for entertainment and educational purposes only. It should not be your sole source of information. Some details may be oversimplified or inaccurate. My goal is to spark your curiosity and encourage you to conduct your own research on these topics.

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25 июн 2024

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Комментарии : 334   
@NotBroihon
@NotBroihon 7 дней назад
I have proven the Riemann Hypothesis but I'm currently busy feeding my cat.
@ThoughtThrill365
@ThoughtThrill365 7 дней назад
😂 🐈
@enderyu
@enderyu 6 дней назад
I have a truly marvelous demonstration of the Riemann Hypothesis which this comment is too small to contain.
@rtfgx
@rtfgx 6 дней назад
I have a proof of Goldbach but am too lazy to write it down.
@benjamintrent5309
@benjamintrent5309 5 дней назад
Did you adopt your cat from Schrodinger? We may have another problem.
@mexicano1891
@mexicano1891 5 дней назад
I found the Schrödinger cat but my 1nt3rnet 1$ sufeπ1ng fr0m sup3π∆÷×∆÷
@exor6100
@exor6100 7 дней назад
I remember in my intro to proofs class seeing Goldbach’s conjecture on a problem set and being really frustrated I couldn’t figure out a super simple looking problem. I looked it up just to find I’d spent over an hour on an open problem lol
@ThoughtThrill365
@ThoughtThrill365 7 дней назад
Oh wow!
@eprjct
@eprjct 7 дней назад
I did that with Collatz. I thought my prof said it was an open problem bc it's too ez and nobody wants to solve it.
@ratzou2
@ratzou2 7 дней назад
Teacher's just trying to see if one of you gets it so he can solve it for himself haha
@leif1075
@leif1075 6 дней назад
​@@ThoughtThrill365Thanks for sharing..are you a mathematician?
@shauas4224
@shauas4224 4 дня назад
I spent around a month banging my head on it. Got some interesting ideas but lost interest after that Maybe will try again some time in the future
@DestroManiak
@DestroManiak 7 дней назад
riemann hypothesis definitely does not sound easy.
@Gl-ml4ii
@Gl-ml4ii 7 дней назад
I highly recommend you to watch the 3blue 1brown video it will sound super easy after a day or two of just thinking about it
@senco445
@senco445 7 дней назад
Is that not subjective? Since the video title itself is subjective, he can do whatever he wants in this case.
@artophile7777
@artophile7777 7 дней назад
​@@senco445 Sure, easy for a once in a millennium genius.
@Aedi
@Aedi 7 дней назад
it looks solvable, not trivial. With the right amount of knowledge it can start to look like you should be able to solve it by working through it enough. But it inevitably hits a blockade if you try any of the common ways to test it
@docsigma
@docsigma 6 дней назад
I think “sounds easy” is supposed to mean “easy to describe”, not “easy to solve”. Maybe?
@EastBurningRed
@EastBurningRed 6 дней назад
yeah, apparently the birch conjecture was so easy and trivial that the statement of it is left as an exercise to the viewer
@wicowan
@wicowan 5 дней назад
yea proof is easy too, but the margin here is too small sorry
@maniamapper3202
@maniamapper3202 4 дня назад
It was the only one I could prove in my head, the others I had to use a sticky note to help write it out. But don’t worry I won’t spoil them all for you guys so you can have a chance to solve them yourselves! ☺️
@victorfunnyman
@victorfunnyman День назад
@@maniamapper3202 how did you solve it tho
@space_twitch1926
@space_twitch1926 6 дней назад
When Euler says he cannot prove something, the math community shivers.
@evynt9512
@evynt9512 2 дня назад
Is Omega computable?
@iankrasnow5383
@iankrasnow5383 7 дней назад
My favorite unsolved problem in mathematics is the Moving Sofa Problem. Say you need to move a couch (or any 2d shape) around a corner in an L-shaped "hallway" of unit width. What is the maximum area of the couch and what is its shape? This problem was proposed in the 60s and we have a very good approximation, but no exact solution yet, at least not one that has been proven.
@ayushjha3716
@ayushjha3716 5 дней назад
There is a similar problem in Project Euler that is solvable and fun
@ffc1a28c7
@ffc1a28c7 7 дней назад
<a href="#" class="seekto" data-time="332">5:32</a> btw, it's been proven that Yitang's method will not work to bring it down to 2. 6 is the proven minimum of the method.
@cubicinfinity2
@cubicinfinity2 5 дней назад
Interesting.
@fredfeinberg3995
@fredfeinberg3995 5 дней назад
And I believe that requires the Elliott-Halberstam conjecture.
@zanti4132
@zanti4132 2 дня назад
Yes, and I don't believe the assumption of the Elliott-Halberstam Conjecture (alluded to in the video as "subtle technical assumptions") makes the minimum claim of 6 valid. Without the conjecture, the minimum gap is currently 246.
@bartolhrg7609
@bartolhrg7609 7 дней назад
<a href="#" class="seekto" data-time="510">8:30</a> You literally forgot to say what is Birch (...) conjecture You just said it has something to do with elliptic curves
@_unkown8652
@_unkown8652 7 дней назад
It’s an extremely complicated statement about the number of rational solutions (i.e. , pairs of numbers (x;y) of the form (a/b;c/d) such that y^2=x^3+tx+s) to the equation of the curve, I don’t really know why he included it since I haven’t seen anyone explain it in less than 20 minutes
@adizcool2-655
@adizcool2-655 6 дней назад
It is a bit trickier to explain than other conjectures so maybe that's why he didn't? It basically says that the elliptic curves he described have infinitely many rational solutions only if a particular function called Hasse-Weil L-function gives the value of a variable S as 0. If this function gives the value of S as 1, then there will be finite number of rational solutions to the elliptic curve equation.
@d4slaimless
@d4slaimless 6 дней назад
@@adizcool2-655 so it was actually quite easy to explain: you just did it in small simple comment. Title of the video "Every Unsolved Math problem that sounds Easy". So the problem should be explained (easy), or shouldn't be included at all.
@ianstopher9111
@ianstopher9111 5 дней назад
The thing with BSD is it takes quite a bit of unpacking. It is not just the elliptic curves but knowledge of the L-function. The equation for the first Taylor coefficient around s=1 looks like voodoo, but it has been shown to be correct in so many examples. It is my favourite of the Millenial Problems.
@ianstopher9111
@ianstopher9111 5 дней назад
For the kissing numbers, I assume the reason 8 and 24 is because Maryna Viazovska (and others) solved the sphere-packing problem in those dimensions using the E8 lattice and the Leech lattice back in 2016 - it made the news. A well-deserved Fields medallist.
@felipegiglio2047
@felipegiglio2047 5 дней назад
2 small mistakes that I noticed: 1 - recent research has shown that there are infinetely many primes with distance at most 2 houndred and something (cant remember exactly). But they havent proved for 6 yet. We can only lower the constant to 6 if the Riemman Hypothesis is true. And according to Terence Tao, we're not close to improving this result. Any further result would need a huge math breakthrough 2 - It's not really a mistake, but something important that you missed. You don't need to go as far as algebraic numbers when the matter is pi+e. We don't even know if they are rational, and neither pi.e or any pretty expression that you could make with them
@iantino
@iantino День назад
I mean, being rational is a stricter case than being algebraic.
@daniestrijdom8248
@daniestrijdom8248 7 дней назад
Leon-hard? Leon... HARD? Is that how he said his name? No really, this is more important than the unsolved mysteries of math
@daniestrijdom8248
@daniestrijdom8248 7 дней назад
great video though
@ThoughtThrill365
@ThoughtThrill365 7 дней назад
😂😂
@tuppelkneinhoftsnaak
@tuppelkneinhoftsnaak 6 дней назад
I believe it's actually quite close. I'm not German (I'm Dutch), but the 'h' is supposed to be heard I think.
@codycast
@codycast 6 дней назад
Settle down, nerd
@grmpf
@grmpf 6 дней назад
Believe it or not, the closest way of writing it out with English sounds would be "lay on hard".
@magicmulder
@magicmulder 4 дня назад
Collatz conjecture is basically a race. It’s all about proving you always hit a power of 2, and while there’s infinitely many of them, they also get spaced out up the number line while 3n+1 tries to keep up.
@boium.
@boium. 6 дней назад
Every time again and again I see a picture of Hermann Grasmann when people talk about Goldbach. Goldbach wrote to Euler, yet we have a photograph of Goldbach and not one of Euler. That's because the first photocamera wasn't invented for like 60 years after Goldbach and Euler died. We do not know of any depiction of Goldbach, and people just take the first image that pops up when you google Goldbach and are like "Yup, that's him. No more research needed." It's so sad to see it happen every time :(
@EnderSword
@EnderSword 5 дней назад
"Math problems that sound easy" Problem #1: So... There's these 24 dimensional Spheres...
@weirdo911aw
@weirdo911aw 7 дней назад
whats crazy is that theres a chance that some of these theorems are completely unprovable but are, regardless, true. so yes, euler was a (leon)hard mf
@asagiai4965
@asagiai4965 7 дней назад
A lot of these are actually provable in a sense it is just hard. Like you if you discover a function that when you input any number that can easily know if it is prime or not. You can probably answer goldbach.
@jerry3790
@jerry3790 6 дней назад
Ironically, a proof that they are unprovable also serves as a proof that they are true, for some of these theorems. For example, if you can prove that the goldbach conjecture is unprovable true or false, that means a counter example cannot exist and thus it is true.
@ha-kx9we
@ha-kx9we 6 дней назад
A function is not nearly precise enough. A function defined as giving 1 for primes and 0 for non primes does not help at all
@andrewluo3792
@andrewluo3792 6 дней назад
@@jerry3790 I had to ponder about this since my mind jumped to incompleteness, but yes you are right, a dis-proof of the goldback conjecture is just to find a counterexample. If one doesn't exist then it is true...
@sstadnicki
@sstadnicki 6 дней назад
@@jerry3790 Nitpickery: a 'standard' counterexample. The unprovability of the conjecture would mean that there are models of the natural numbers where Goldbach is false and therefore there's an n with no primes p and q s.t. p+q=2n; the only problem is that this n doesn't exist in the standard (minimal) model of the natural numbers, so in some sense it's an infinite (and/or 'unknowable') natural number. (That said, mind, this is still a great point to bring up and applies to any number of unprovable statements.)
@klausolekristiansen2960
@klausolekristiansen2960 7 дней назад
Famous conjectures which are easy to understand: There are infinitely many perfect numbers All perfect numbers are even
@oro5421
@oro5421 6 дней назад
*perfect numbers are the ones, equalling to the sum of their smaller divisors. Such as 6, 28, 496…
@barakeel
@barakeel 5 дней назад
There are infinitely many primes of the form x^2+1.
@kruksog
@kruksog 7 дней назад
You kind of brushed past one of the things that is so valuable in the Riemann zeta hypothesis. Solving it gives us some more structure in the primes. It has been shown already that zeta(s) is also equivalent to the infinite product, over all primes p, of ( 1/(1 - p^(-s)) ). Edit: noting that by setting this equal to the infinite sum fornula for zeta given in the video, we get a correspondence between the natural numbers and the primes.) Solving Riemann could give real insight into the distribution of the primes. As I tried to point out above, there is a very direct link between the Riemann zeta function and the distribution of primes. Cool video. Enjoyed it a lot.
@ha-kx9we
@ha-kx9we 6 дней назад
Which can be quite easily proven with probabilities ! I think its great how you can use a seemingly unrelated field to get such a quick proof
@wicowan
@wicowan 5 дней назад
yea but this is only one of its applications. Proving the extended version of the Riemann hypothesis proves A LOT of unsolved theorem
@diogoandre756
@diogoandre756 7 дней назад
The riemann hypothesis and *especially* the birch swinnerton dyer conjecture are completely impossible to understand for 99% of people. They definitely are not simple looking :|
@tommyrjensen
@tommyrjensen 6 дней назад
There are versions of RH that can be explored using a calculator. Such as the conjecture stated as σ(n) < exp(γ)·n·log(log(n)) (n > 5040) on the Wikipedia page, where σ is the divisor sum function (e.g. σ(12)=1+2+3+4+6+12=28) and γ is the Euler-Mascheroni constant ≈ 0.5772.
@ianstopher9111
@ianstopher9111 5 дней назад
You can check some of the results for BSD using a few lines of python (and some libraries) but that only gives you insight that it mostly checks out for small examples. As far as 99% I think that is an exaggeration. If you want to understand the machinery of BSD many people can achieve that, but it is not achievable in a 10-minute video. If you are interested enough and devote some time you can get to have a workable knowledge of BSD. As a start, Álvaro Lozano-Robledo's Elliptic Curves, Modular Forms, and Their L-functions I found great from taking up to the statement of BSD conjecture in Chapter 5, and all in less than 200 pages. You are going to need some familiarity with undergrad maths in places, but he tries to make the climb as easy as possible.
@andrewparker8636
@andrewparker8636 4 дня назад
Definition 1.0: We define the phrase "sounds easy" to mean "doesn't sound easy".
@ere4t4t4rrrrr4
@ere4t4t4rrrrr4 2 дня назад
<a href="#" class="seekto" data-time="518">8:38</a> "it's the only other one we can remotely describe in plain English" P vs NP can be easily described: P is the set of decision problems that can be solved in polynomial time and NP is the set of decision problems whose solutions can be verified to be correct in polynomial time; the problem asks whether those two sets are identical. The only further issue is how to define "decision problem" (easy, it's a problem with yes/no answer) and how to define "polynomial time" (a bit harder: a solution that runs in polynomial time has a number of steps proportional to a polynomial P(s) where s is the size of the input in bytes; here "input" is, for example, for the unknotting problem described previously, a description of the knot you want to decide whether it's identical to the unknot. We can easily verify that a given way of moving the folds proves that a knot is the unknot, so this problem is in NP)
@samuelabreu4349
@samuelabreu4349 День назад
I mean, for Computer Science people its easy to feel what Polynomial time is. For other people I dont think its That simple. But yeah, much easier than the Riemann Hypotesis
@MasterHigure
@MasterHigure 6 дней назад
<a href="#" class="seekto" data-time="415">6:55</a> Matrix multiplication is not n^3 complexity. The naive implementation is, but the best currently known is about n^2.37.
@UnknownString88
@UnknownString88 7 дней назад
No unsolved math problem sound easy to me considering the geniuses that already tackled them 😅
@clintonfarrier244
@clintonfarrier244 6 дней назад
Well, you can easily solve two thirds of the collatz conjecture. If you repeatedly divide any even number by two, you will inevitably get an odd number as a result, and by definition 3n+1 cannot be divisible by 3 so threeven numbers will only ever appear at the very start of the function. Therefore, as long as the collatz conjecture is true for all numbers which do not have 2 or 3 as factors, then it must also be true for those that do. The rest is a bit trickier, but that's a pretty solid start, we've already completed 66% of the proof just like that.
@barakeel
@barakeel 5 дней назад
@@clintonfarrier244 If collatz conjecture is true for multiple of 3 then it's true for all numbers.
@magicmulder
@magicmulder 4 дня назад
One big problem of knot theory was solved by a mathematician from another field who worked on it for two weeks in her spare time. A problem from graph theory that was open for 30 years was solved by a Chinese mathematician on half a page, and a middle schooler can understand the proof. Sometimes the solution is easier than we think.
@mathgeniuszach
@mathgeniuszach 7 дней назад
A lot of these problems seem to revolve around finding a way to relate sums to primes. I wouldn't be surprised if one day someone found a way to relate them and quickly solved many of these problems using it.
@asagiai4965
@asagiai4965 7 дней назад
More like an easy way to find prime is the key
@Aedi
@Aedi 7 дней назад
a greater understanding of primes would make a lot of these much easier
@mrbutish
@mrbutish 6 дней назад
Prime number is a metaphor. It represents a computationally expensive object. Similar to how calculus is around slopes and areas when you zoom in. We would not know how to deal with circular like objects in calculus for example which is a pain. I am talking about how the surface area and volume and directions can be negative and absolute value is forbidden for example.
@chatzaght
@chatzaght 6 дней назад
The Lonely Runner Conjecture sounds as if it can incorporate the aspects of music theory in order to solve it. That is, each runner can be given a set BPM. Another idea could be equally separated notes in terms of rhythm.
@kristinborn8882
@kristinborn8882 7 дней назад
obviously gamma is rational, It's just gamma/1
@bred4ev3r
@bred4ev3r 7 дней назад
well yes but actually no Just like how pi is rational. It’s circumference/diameter or whatever it is idk
@VictorHugo-xn9jz
@VictorHugo-xn9jz 4 дня назад
lmao
@douglasstrother6584
@douglasstrother6584 3 дня назад
It's kind of crazy that the difference between two divergent functions is "a skooch over a half".
@victorvega4945
@victorvega4945 7 дней назад
great stuff, thank u for posting! <a href="#" class="seekto" data-time="105">1:45</a> leon-hard lol
@ianmoore5502
@ianmoore5502 7 дней назад
10:29 bern-hard
@venerable_nelson
@venerable_nelson 7 дней назад
Two things that make these unproved math conjectures so captivating. 1) Godel's incompleteness theorem suggests that there exist true conjectures that cannot be proven. ('suggests', there may be alternative systems that can prove a conjecture that is unprovable in a different system). 2) The Poincaré conjecture was one of the millennium prize problems that did get proven, which does demonstrate that not all of these conjectures are unprovable. We'll really never know if these problems are unprovable. We only know that they are not unprovable once they are proven.
@udasai
@udasai 7 дней назад
It is possible to prove something unprovable. You demonstrate that the problem is independent of standard axioms (that its truth or falsity does not lead to contradiction either way), and thus is also independent from their consequences (other proven theorems). You then need novel axioms, which is what the completeness theorem suggests is always eventually necessary. The issue is that you can always add an axiom that states "proposition X is true", which is vacuous, although it can be amusing to consider the consequences of such spurious addenda to the proof system, such as with the Banach-Tarski Paradox.
@robertroach9157
@robertroach9157 6 дней назад
​@@udasaiThere's a mind-bending but valid way of ACTUALLY proving a conjecture true if it's been shown to be indeterminable from the axioms: If a conjecture is shown to be indeterminable from the axioms, for instance the Riemann hypothesis, then that would mean that there's no counterexamples to said conjecture, hence the conjecture is correct (No counterexamples could exist, since the existence of one would mean that the conjecture isn't indeterminate, but rather is provably false.)
@lecdi6062
@lecdi6062 6 дней назад
@@robertroach9157 So I presume that means that in the case of the Riemann Hypothesis specifically, there is no way of proving that it is unprovable even if it is unprovable... Because that would prove it, and hence it would not be unprovable, leading to a contradiction.
@ha-kx9we
@ha-kx9we 6 дней назад
Im not an expert, but that depends on what logic you use. In intuitivitionist (not sure about the translation) that wouldnt work
@sya8002
@sya8002 6 дней назад
@@robertroach9157 this is not how indeterminacy works. If one would show that RH is undecidable (from an axiom set, say ZFC), that would mean that there is a model of ZFC (=set theoretic universe) where you have a counterexample to RH, and one where you can't have one. As ZFC does not admit some form of "canonical" model (as opposed to, say, PA), the problem does not have a definitive answer within the chosen system.
@kephalopod3054
@kephalopod3054 2 дня назад
Goldbach conjecture: 2n = p + q could be restated as n = (p + q) / 2, n >= 2, p and q primes: every integer greater than 1 is the average of two primes.
@johnredberg
@johnredberg 4 дня назад
Regarding lowering the upper bound on prime gaps: Assuming a major unsolved conjecture to be true is not a "subtle technical assumption". A technical assumption in mathematics is a restriction of the general problem to a subset of special cases (which can range from covering only a few finitely many cases up to "almost all cases") for which the problem can be explicitly proven. Their unconditional bounds are an astounding achievement. Going from 246 down to 6 on the other hand is "nice" and may point to new ideas, but may just as well turn out to be completely hopeless. BIIIIG IF.
@theflaggeddragon9472
@theflaggeddragon9472 6 дней назад
Bro you cant even state the Birch and Swinnerton-Dyer conjecture without technical definitions so i don't know if it sounds easy to anyone. For the curious, it says that the rank of an elliptic curve of a number field is the order of vanishing of its L function. For this to even make sense, you need to knoe the Mordell-Weil theorem (not too hard) and the Modularity Theorem which Wiles proved to conclude Fermat's last theorem. BSD is very deep
@jodfrut771
@jodfrut771 7 дней назад
leonHARD Euler
@jodfrut771
@jodfrut771 7 дней назад
well turns out that is the correct way to say it lmao
@swqb
@swqb 6 дней назад
and the eu in euler is pronounced oi
@rosiefay7283
@rosiefay7283 7 дней назад
If you think we wouldn't understand the Birch_Swinnerton-Dyer conjecture, fine, but just leave it out. No point in starting an explanation, and saying that elliptic curves have certain properties, but stopping there as if you've taught us anything. We're none the wiser.
@ferlywahyu342
@ferlywahyu342 4 дня назад
It seems like trying to prove the Colatz conjecture will only get us stuck with circular arguments
@stefanbergung5514
@stefanbergung5514 7 дней назад
Thank for saying, what the Birch and Swinnerton Dyer Conjecture is. And so easily to understand too.
@Pan_Tarhei
@Pan_Tarhei 7 дней назад
I have been a subscriber for three days and I really like your content ;)
@ThoughtThrill365
@ThoughtThrill365 7 дней назад
❤️❤️
@nintendoswitchfan4953
@nintendoswitchfan4953 5 дней назад
A lot of the problems seem to revolve around the wierdness of the integers. For the real numbers we have things like calculus which look at functions over the real numbers and things like limits and derivatives. But for the integers we have a field studying them called number theory however many unsolved problems come from there (including goldbach and collatz) as we don't have much knowledge on that field and don't understand things like the prime factorisation of n vs n+1 or the distribution of primes.
@tahsinabrar-sl7yq
@tahsinabrar-sl7yq 7 дней назад
This dropped exactly when I needed it. Subscribed immediately. Excellent work.
@ireallydontknow3299
@ireallydontknow3299 2 дня назад
One interesting thing about Goldbach's conjecture is that a slightly weaker form of it has been proven. That is, every odd number bigger than 5 can be written as the sum of 3 (not necessarily distinct) primes. This follows easily from the normal ("strong") version of the conjecture, because every odd number is just an even number + 2. So if you can write every even as p + q, you can write every odd as p + q + 2. Going from the weak form to the strong form is difficult, though, and an unsolved problem.
@zanti4132
@zanti4132 День назад
In the table at <a href="#" class="seekto" data-time="52">0:52</a> for the kissing number problem, the correct numbers are known for dimensions 1, 2, 3, 4, 8, and 24; also, the numbers for dimensions 5, 6, and 7 are believed to be correct, though not proven. Noting that the numbers grow exponentially and never more than double, isn't it safe to say the lower bound for dimension 23 can be increased to 98,280? I get this number by taking half of the number for dimension 24, which is known to be correct.
@ottolulau6403
@ottolulau6403 6 дней назад
very great video Idea! I've been searching for such a list!
@tommyrjensen
@tommyrjensen 6 дней назад
The table at <a href="#" class="seekto" data-time="420">7:00</a> is confusing. The decision versions of the problems on the right are NP-complete, in particular this means that it is an open problem (called NP=P?) whether they can be solved in polynomial time. Also technically in Computational Complexity Theory, problems are "Exponential Time" if they can be solved quickly enough, in time at most exp(p(n)) for some polynomial p, where n is the size of the input. A problem that cannot be solved in polynomial time is superpolynomial, and a problem that cannot be solved in exponential time is superexponential.
@Hlebuw3k
@Hlebuw3k 6 дней назад
The Goldbach Conjecture solution - with some assumptions, 3 probably being the deal breaker: 1. Any number (and by extention, any even number) can be described as an addition of 2 numbers, with either: both of them being exactly half of the sum; or one of them being smaller than half the sum, and the other being larger than half the sum (this includes one being equal to the sum and the other being zero, and one of them being larger than the sum and the other being negative) 2. There are infinitely many prime numbers. 3. There does not exist a "gap" in prime numbers, where there is no primes between half the sum, and the sum. 4. Two odd numbers always add up to an even number. 5. All primes larger than 2 are odd.
@Hlebuw3k
@Hlebuw3k 4 дня назад
Additionally, we can simply subtract the largest prime that's less than the sum, from the sum, and check if it results in a prime. If it does not, subtract the next largest prime. And I see it now, if this series does not result in a prime until half of the sum is reached, then the conjecture is broken, but apparently we just can't figure out if that's the case?
@anthonybernstein1626
@anthonybernstein1626 5 дней назад
The P vs NP problem doesn't sound too difficult either: give an example of a task that can't be computed quickly but if someone gives you a solution, it can be verified quickly (quickly == the number of steps is a polynomial of the input size).
@BedrockBlocker
@BedrockBlocker 6 дней назад
Great vid! Love how you broke it down.
@chsovi7164
@chsovi7164 4 дня назад
you can't take a math problem like the birch and swinnerton-dyer conjecture, decide it's too hard to state, then put it in a video like this anyway lmaoo?????? god that blindsided me, gave me a good laugh thanks dude there's actually quite a number of decent unsolved problems that would be better to mention, such as whether it's possible to make a 3x3 magic square out of 9 distinct square numbers. not nearly as famous as anything on this list
@clarencejohncabahug5466
@clarencejohncabahug5466 6 дней назад
Honlestly, you only need the names Erdos and Euler and you get a truckload of easy looking open problems.
@TheKlikluk
@TheKlikluk 7 дней назад
Honestly great video and explanation of the various problems. Many I had not heard of before. You did it a little bit with the Collatz conjecture and the Kissing number but it would have been awesome to get a little bit more insight into why these are actually also interesting problems or what benefit solving them could have for the real world.
@danbromberg
@danbromberg 3 дня назад
Bravo! for putting this together...
@user-cg5pu7bl1c
@user-cg5pu7bl1c 7 дней назад
Great video! Topics were explained really well!
@susanafaciolince7755
@susanafaciolince7755 7 дней назад
this is the most american way of pronouncing euler-mascheroni lol
@ThoughtThrill365
@ThoughtThrill365 7 дней назад
😄
@ianweckhorst3200
@ianweckhorst3200 7 дней назад
Man I love the Euler masheroni constant, but I had no idea it wasn’t known if it was rational or not, but has some fun relationships with sums and proven irrational numbers
@colin351
@colin351 7 дней назад
They haven't taught ai voice synthesizers how to pronounce mathy words
@dbliss314
@dbliss314 6 дней назад
Oily macaroni
@M1Miketro
@M1Miketro 6 дней назад
WHEELER MASKERONI
@aaaabbbbccccddsf
@aaaabbbbccccddsf 7 дней назад
damn that is gamma a rational number hit hard, recently learned about it in my discrete maths class and would never suspect it to be such a mystery, great video
@ToguMrewuku
@ToguMrewuku 7 дней назад
umm did you juat skip over swinsgrns dyer conjecture? not explaining it?
@ThoughtThrill365
@ThoughtThrill365 7 дней назад
its very technical and not easy to put it in simple words. maybe i'll try in future videos.
@InterPixelYoutube
@InterPixelYoutube 3 дня назад
Great video - I may have been you 5,000th subscriber as when I hit subscribe it went from 4.99K to 5K!
@stibiumowl
@stibiumowl 3 дня назад
As much interesting this quest of search for answers is for our curiosity, we should stop asking for "is this and that linear combination of transcendental numbers an algebraic number or not?" because already there are countably infinte many ways to linearly combine Pi and e, so as soon we add even more Irrationals, there are even more linear combiations of them, all are infinitly many. So this question can go forever if we not stop asking if these and that linear combinations of Transcendentals yield rational or irrational numbers. Side note: In one of your in-video-pictures, you show <a href="#" class="seekto" data-time="469">7:49</a> and thanks for the fact that 3,3030 0300 0300 0030 0000 3000 0003... is non-repeating and therefore irrational, I had not realized until now that EVEN if we know all of the infinite digits of a number, it can still be without repeating digit-sequences. Now I know that we can just as easily (as we create or think of resp. Rationals) create Irrationals with also knowing all digits by just defining a non-repeating and infinite digit sequence like "each 3 you pass, put one more 0 between it and the next 3 with increasing string-length of just FINITELY repeating 0-s" and its non-repeating, analogously with any other strictly monotonous, whole-number-for-y-yield function of x="number of 3-s before the given digit in total" and y="number of 0-s until the next 3" and then expanding by exchanging 0 and 3 with one of the 44 other pairs of nonequal decimal digits or puting finite amount of other digits inside the above-created irrationals. (The monotony part is not needed if we still can be certain that no y value does ever repeat i.e. for a injective function, but monotony can more easily be understood, most easy for f(x)=x) Or just doing sums again, the all-infinite-digits-known irrationals should be more easy to decide if their linear combinations are rationals or not. For example, adding 4,4404 0040 0040 0004 0000 0400 0000 4... to the above irrational with only 3s and 0s, we get 7,7434 0340 0340 0034 0000 3400 0003 4000 0003 4000 0000 3400 0000 0034 0000 0000 0340 0000 0000 034... that is again irrational, as it does thesame with 34 as was with 3. Even though we could use any injective-whole-value-function for the amount of 0-s, using monotonous functions makes it more obvious to aknowledge this function by looking at the number, most easily with f(x)=x as I used above. Even more easily, we take only the injective function and instead of transforming it into amount of 0-s, just write all consecutive values as their finte digit-sequences and than back to back as decimal digits of a single number. This leads to, for f(x)=x+1, to my new favorite number. It is 1,23456789101112131415161718192021222324252627282930313233343536373839... as it lists all digit-sequences of natural numbers as its own digits, but is itself all-digits-known-irrational. But I again not know if its also transcendtal or algebraic. PS: You said there are more Irrationals than Rationals, is there a mathematical proof or could the Cardinality of both sets be the same?
@MagicGonads
@MagicGonads День назад
Regarding the PS: Yes, using Cantor's diagonal argument A) The rationals (Q) are an equivalence class over order pairs of integers (Z^2) (separated by the `/` symbol) subject to the equivalence relation `~` that a/b ~ p/q iff aq = bp That means they are a subset of Z^2 which has the same cardinality as N (countably infinite), so |Q| is at most |N|. But we can also show that Q is a superset of Z which has the same cardinality as N (simply choose z in Z to get z/1 in Q, for each z this is a distinct element in Q), so |Q| is at least |N|. So ultimately, |Q| = |N|. B) Cantor showed generally that the power set of any set has a strictly greater cardinality than the original set. In detail for any set S, P(S) is its power set, and |S| < |P(S)|. Even when applied to infinite sets. The power set is notated `2^S`, as it is the set of subsets of S, because a subset is isomorphic to a characteristic function that chooses a binary (0 or 1) choice for each element of S, so a mapping from S to a set of 2 elements. So |S| < |2^S| for all sets S. C) 2^N can be interpreted as the set of infinitely long (but countable) strings of binary digits, otherwise known as the 'binary representations of the fractional parts of real numbers'. And obviously, the fractional parts of real numbers is a subset of the real numbers (R). So |R| is at least |2^N|. D) Taking |Q| = |N| from point A, |N| < |2^N| from point B, and |2^N|
@guti9709
@guti9709 3 дня назад
<a href="#" class="seekto" data-time="564">9:24</a> man i must get that one million reward, I’ve always loved that fragrance!!
@StoicTheGeek
@StoicTheGeek 7 дней назад
For all those asking about the Birch & Swinnerton-Dyer conjecture, there is an excellent book that explains what it is, assuming you only know high school maths. It is Elliptic Tales, by Avner Ash and Robert Gross.
@StoicTheGeek
@StoicTheGeek 7 дней назад
tl;dr. The rational points on an elliptic curve form an abelian group. BSD relates the rank of that group to the value of the Hasse-Weil L-function for the curve at a certain point.
@StoicTheGeek
@StoicTheGeek 7 дней назад
tl;dr. The rational points on an elliptic curve form an abelian group. BSD relates the rank of that group to the value of the Hasse-Weil L-function for the curve at a certain point.
@dAni-ik1hv
@dAni-ik1hv 5 дней назад
I have found a proof for the Riemann hypothesis, however discovering my proof is left as an exercise to the reader.
@__christopher__
@__christopher__ 3 дня назад
I've heard the proof is in the pudding. Unfortunately that was after I've eaten the pudding, so I guess the proof no longer exists.
@paulgoogol2652
@paulgoogol2652 День назад
I imagine somewhere a brain surgeon rolling up his sleeves.
@SirLightfire
@SirLightfire 5 дней назад
The thing about irrational and trancendental numbers is, their definition is "complementary" or rather, x is irrational if it's *not* rational We don't have methods for identifying irrational or trancendental numbers, so each number has to be proven not rational or not algebraic
@mrbutish
@mrbutish 6 дней назад
Watching this video makes me happy
@chaosredefined3834
@chaosredefined3834 7 дней назад
The reimann-zeta function is only defined for that when re(s) > 1. Outside of that, it's just the magical "analytical continuation" that might be a bit much for this video...
@tommyrjensen
@tommyrjensen 6 дней назад
Not too hard though. It is a smooth function defined on all of ℂ\{1} that agrees with the Riemann sum for s > 1.
@maxdragonis1304
@maxdragonis1304 5 дней назад
For 3x+1 you can simply understand it tends too sink lower and higher spurts of going up are more common as you got higher in numbers but go back down too 4 2 1 If you th8nk ab9ut how our counting system is infinite you realize thats the going upwards in numbers will be more common And going back down will be less common eventually probably wnding at omega which is infinity+ 1 And in order too get a pair too do so would have too reach at a minimum of (W+1)/2 So in order too solve 3x+1 you would just have too solve omega Using other counting systems like binary the equivalent tok omega would be an infinite number of 1 in digits adding 1 would eventually make it equal 0 through basic computation and can be prooven that W = 0 Which through this would state That negative numbers are past(W+1)/2 in our counting system making number lines really just being a portion of a circle and probably making our graphs a 4d sphere How ever the point still stands If there us ever another looping chain in the 3x+1 problem it would be literally impossible too find due too it would have too cross the (W+1)/2 border both ways and have too loop Doesnt mean it doesnt exist but i believe there is a strong arguement too the fact that there is not a nother chain in the if odd you do 3x+1 and if even you divide x by 2 Than 4 2 1 aswell as the 3 other prooven chains due too it gravitating towards 0 and W which is 0 from the context of negative numbers in all retrospect.
@firstname4337
@firstname4337 4 дня назад
I solved most of these back in high school
@HAYTAMIBN-mz9ic
@HAYTAMIBN-mz9ic 4 дня назад
don't you dare to stop. you are doing great
@jamesknapp64
@jamesknapp64 4 дня назад
I thought Tao basically showed there is No unbounded sequence of Colletz numbers, but that doesnt eliminate the chance that there is a loop
@chixenlegjo
@chixenlegjo 7 дней назад
What’s the most efficient way to pack unit spheres in 4d space? That’s also unsolved, surprisingly.
@kephalopod3054
@kephalopod3054 2 дня назад
Collatz conjecture could be restated as: for every integer, every Collatz sequence reaches a power of 4.
@kephalopod3054
@kephalopod3054 2 дня назад
If you don't start with a power of 2, you can only reach a power of 4 because 2^k - 1 is divisible by 3 only when k is even.
@kshitijthakkar8074
@kshitijthakkar8074 7 дней назад
Hello Everyone, As a part of my Masters Programme me and my friend had written down a proof of Goldbach's conjecture using Graph Theory , the panel that we presented it too did not challenge it, but I'd really welcome and appreciate someone looking through.
@lex224ification
@lex224ification 6 дней назад
that's the wonkiest pronunciation of "leonhard" i've heard yet
@holdthatlforluigi
@holdthatlforluigi 3 дня назад
Well-made. No intro makes it feel like long-form tiktok haha
@ianstopher9111
@ianstopher9111 5 дней назад
Related to kissing numbers (very vaguely): what is the smallest square that can contain N non-overlapping squares of unit size? We don't know what is the minimal size length for N=11,12,13 and most higher N.
@genghiskhan6688
@genghiskhan6688 3 дня назад
I feel like the use of comic sans in this video was a deliberate and calculated act of evil.
@PieterMante
@PieterMante 7 дней назад
I like that some of these math problems are covered by Varitasium. If you want a deeper explenation you should definitely visit his channel.
@DrR0BERT
@DrR0BERT 5 дней назад
Random personal fact: Swinnerton-Dyer is my mathematical grandfather on the Mathematical Genealogy Project.
@asmithgames5926
@asmithgames5926 6 дней назад
Another famous one: We don't know whether Pi^(Pi^(Pi^Pi)) is a whole number or not.
@wicowan
@wicowan 5 дней назад
yea or for example we know e^pi is transcendental but we don't know if pi^e is ahahah, or even worse we know e^e and e^(e^2) can't be both not transcendals, i.e. either one of them or both are transcendental
@ianstopher9111
@ianstopher9111 5 дней назад
We don't even know if pi is normal.
@alantew4355
@alantew4355 3 дня назад
If a problem is fed to a supercomputer or AI, who then claims that it’s solved (proven or disproven) but yet also claims that humans would not be able to understand the proof, for some reason (such as reading the proof would take more time than a person’s lifespan). In that case, is the problem considered solved?
@JackOHaraEngineering
@JackOHaraEngineering День назад
I didn’t know that progress had actually been made on the twin prime conjecture, also if Euler said he can’t prove it, it may be over
@ArbitraryCodeExecution
@ArbitraryCodeExecution 6 дней назад
was literally playing bach's a minor violin concerto right before watching this
@monishgokulsing5903
@monishgokulsing5903 5 дней назад
The goldbach conjecture sound weirdly very easy tho like. Knowing that all prime number after 2 are odd meaning that they could be written as (2n+1) so it would be logical to assume that the sum of any multiple of 2 which is essentially 2p. So when adding two prime number (2n+1)+(2n+1), it would just equal 4n+2 or 2(n+1). Actually nvm i understood why its really difficult, i felt like an idiot writting this
@maqrux
@maqrux 6 дней назад
Fantastic subject matter!
@ceulgai2817
@ceulgai2817 6 дней назад
<a href="#" class="seekto" data-time="707">11:47</a> Damn, the way you asked "Is Gamma Rational?" made it sound like a creepypasta from the late noughties.
@user-dk8mj4uo9c
@user-dk8mj4uo9c 3 дня назад
I can create such problems, for example, how many different ways to place sea battle ships in n by n square exist?
@razaam1699
@razaam1699 6 дней назад
Can u make a video on the abc conjecture?
@joebazooks
@joebazooks 4 дня назад
how do we know that some of these so-called problems or any popular problem in general are not just necessary or natural features of a base 10 system?
@zxidenbel7000
@zxidenbel7000 6 дней назад
<a href="#" class="seekto" data-time="392">6:32</a> Imagine someone solving what is essentially The Computer Science Problem (P = NP) just to quickly figure out whether two strings were knotted.
@trillex1861
@trillex1861 7 дней назад
my math prof in the second semester put some of theese as task on our Worksheets.
@bratpeki
@bratpeki 6 дней назад
Is this channel in any way associated with The Paint Explainer? Great video!
@wicowan
@wicowan 5 дней назад
I feel like the way we do maths is not suitable for number theory, like damn how can easy problems like these be so damn hard to prove but we can prove so much things for example in higher dimensional geometry. Mindblowing...
@jossdeiboss
@jossdeiboss 3 дня назад
I don't understand what "lonely" means in mathematical terms. Does it mean further than a certain distance?
@canaDavid1
@canaDavid1 6 дней назад
<a href="#" class="seekto" data-time="415">6:55</a> prøving this would imply that P≠NP, solving a Millenium problem and giving you a lot more fame than the unknotting problem alone (in addition to a million dollars)
@ianstopher9111
@ianstopher9111 5 дней назад
Unsolved Math problem that sounds Easy: are there any Lychrel numbers in Base 10 aka the 196-problem? It is quite simple to state but we simply don't know if 196, 295 etc. are Lychrel numbers.
@nathanbeer3338
@nathanbeer3338 7 дней назад
Can I prove the knotting problem? Well, I can knot...
@samrees4648
@samrees4648 7 дней назад
Great video
@Nightcake
@Nightcake День назад
whats the song in the background? i know its fur elise but its different too
@ThoughtThrill365
@ThoughtThrill365 День назад
did you like it?
@festusmuldoon
@festusmuldoon 4 дня назад
The definition of the Riemann Zeta function you gave is only correct for real part greater than 1. It would have been nice if you had defined it for 0 < Re(s) < 1 using the alternating zeta function.
@seba7475
@seba7475 7 дней назад
I tought you had more subcribers, this is video is really well made. Keep up the good work
@MOSMASTERING
@MOSMASTERING 5 дней назад
I always wondered what would happen if you gave these super simple questions (as in - super simple to STATE, not to solve) and slipped them into a high-school or lower exam.. Either they'll never get it, because it's just that difficult, if not impossible.. or, they'll see it from a completely unseen angle and make progress on it.
@ironfoot1938
@ironfoot1938 7 дней назад
Ah yes as I am certain that there is no poly time algorithm for unknotting I'm quicky gonna prove P≠NP. Thanks for the tipp.
@diribigal
@diribigal 6 дней назад
This video made it sound like Yitong Zhang proved that there's a particular number around 70million where we know there are infinitely many primes with that separation, which is not true.
@gigabytemon
@gigabytemon 6 дней назад
Ah, that's easy! The answer is yes times maple syrup.
@keyboard_toucher
@keyboard_toucher 4 дня назад
<a href="#" class="seekto" data-time="730">12:10</a> "Meet the Eeler-Mascheroni."
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