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
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! ☺️
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.
@@miloyallbecause of the Elliot-Halberstam conjecture, weren't you listening 😂 Jk, but also I would be unsurprised if the experts' answer was basically that, or an elaboration of that conjecture
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.
I’m just someone on the internet, but I absolutely get why that’s really, really hard to find a proof for. A brute force method would be to literally name every shape, and the very few hard limits on what is possible don’t make intuition a viable option.
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
@@radielkill You're not far off.....in 3 dimensions, where it's already been solved. In higher dimensions, specifically multiples of 8 you need modular forms.
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
My bet is on transcendental for every arithmetic combination of the two. But proving transcendence is absurdly hard. Even for pi and e themselves it was pretty much dumb luck that we even found the proofs. And they're hard. Regarding irrationality I feel like at least pi/e should be easy (or e/pi, doesn't matter), as it ultimately amounts to the impossibility of p*pi = q*e for natural numbers p and q.
@@lonestarr1490 the only pretty expression (and any derivation of it) that we know of is e^pi. Theres a theorem that says that the only cases where algebraic^algebraic is algebraic are the trivial ones (when the base is 0 or 1, or when the exponent is rational). So, as (pi^e)^i = -1, and -1 and i are algebraic, therefore pi^e cant be
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.
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
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.
@@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.
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.
Video title: Unsolved maths problems that sound easy Narrator of video: "This conjecture is the only one we can explain in plain English" *doesn't explain the conjecture* *moves on* Side note, I'm pretty sure P = NP is a fairly easy conjecture to explain. The existence and smoothness of the Navier-Stokes equation isn't super easy, but you don't need a ton of technical details to explain what it is and why it hasn't been solved yet.
You can make a game out of the Collatz conjecture. 1. include a chain counter 2. add a leaderboard 3. get a ton of people on it so that they disprove it while trying to get a high score
Most of them are very easy to state compared to 99% of open problems in mathematics that aren't number-theoretic. Except Birrch and Swinnerton-Dyer. That one is totally out of place lol
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.
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.
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.
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.
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.
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.
@@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...
@@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.)
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.
8:38 "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)
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
It's actually important that the ""input size"" is measured in bytes (or rather: in bits) not e.g. in size of a natural number. These grow differently, as n bits can represent 2^n natural numbers.
(I say this because you are the first person I ever heard who explicitly mentioned that the input size has to be measured in bits/bytes. This is usually always glossed over. Note also that the difference between bits and bytes is not important here, because they are related by a constant factor (8) which is ignored for usual measures of time complexity.)
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.
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.
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.
@@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.)
@@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.
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.
In the table at 0:52 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. * * * * * This addendum is coming a few months later. For dimensions 1 through 8 and 24, where the kissing number is known or believed to be correct, it is noted that the dimension is always even and a factor of the kissing number; also, the prime factors for the kissing number are always less than the dimension. Adding to that the assumption that the jump to the next dimension never more than doubles the next kissing number, and the lower bound for dimension 23 becomes 98,532. (98532 = 2² × 3² × 7 × 17 × 23.) Of course, this is pure speculation and proves nothing.
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.
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.
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
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.
Another fun open problem sinilar to π+e is whether or not Catalan's constant is rational. This constant comes up in almost every combinatorial problem and has some neat connections with the zig zag numbers
Hats off to all the absolute geniuses that manage to solve some of the most difficult problems in mathematics. But I feel like that's basically the definition of looking for a solution to a self-imposed problem. What value do we garner from finding a solution to the question whether there are infinite amounts twin primes or not? The answer is either yes or no and neither answer would change anything about how the world or universe behaves or how we understand it. I consider myself somewhat of a nerd but these problems are so nerdy and seemingly irrelevant I wonder if the time spend on those problems couldn't be put to better use in other areas of mathematics.
Math major here. 1. Knowledge for knowledge's sake is good. 2. Whenever a 100 year old open problem is solved, it usually isn't because someone just used current math tools in a clever way, but rather invented new tools that then go on to be super useful for more math. For simple algebra and number theory problems, ring theory was invented. At the time it seemed hopelessly abstract, but it lead to algebraic geometry, which is not only one of the most promising areas of mathematics when it comes to unification of math, which is like our version of quantum gravity, an ultimate goal, but also is used in computer graphics. Every single shape on display, unless it's bitmap, including every letter on screen is an algebraic variety carved out by polynomials. Group theory and compex numbers were ones considered the mathematical equivalent of science fiction and are now absolutely integral in all things physics, electricity and fluid motion. 3. They are already useful, regardless. The distribution of primes is immensely important in cryptography. Both Goldbach and twin prime conjecture are about that and the Riemann Hypothesis basically solves all questions about prime numbers by explicitly giving us a formula. Birtch and Swinnerton-Dyer is even more ambitious for cryptography and algebraic geometry, whose importance I already explained. Knotting is important in biochemistry and physics. Lonely runner is part of Ergodic theory, which is super useful everywhere from chemistry, to probability, to fluid and aerodynamics. Sphere packing is an optimization problem. Optimization problems are always useful one way or another for speeding up computation, optimizing profit or minimizing energy use or material waste on assembly lines and the like.
The table at 7:00 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.
Another unsolved math problem that I like that sounds simple is the Perfect Cuboid problem where proving all side lengths, face diagonals, and interior diagonal of a rectangular prism is an integer for the equation a^2+b^2+c^2=d^2. It is comprised of a set of pythagorean theorems and even though it's not a $1mil question it's still a fun little equation to look at.
If you consider the quadratic x^2 - (pi+e)x+ex=(x-e)*(x-pi). then you immediately get that pi+e and pi*e cannot both be algebraic, since they would be coefficients to a quadratic with transcendental roots. Similarly, if a and b are both trancendental, then at least one of a+b and ab must also be.
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.
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?
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 :|
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.
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.
kissing numbers are usually called coordination numbers in material science, which is also where the body-centered and face-centered structures come from.
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).
Goldbach conjecture is tantalizingly true, because if you simulate a LARGE grid the pattern becomes pretty clear, so it’s just a matter of proving why that pattern continues. For the Lonely Runner, clearly we’re asking if a set of arbitrary frequency will always at some point yield zero, ie destructively interfere, this might be possible if we can define the the lonely runner in terms of all the other runners.
the Goldbach conjecture is a tricky one to prove, but with the aid of simulation, it becomes much clearer. I agree with your assessment that the pattern will continue as the grid becomes larger, but we need to find a way to prove it. As for the Lonely Runner problem, your assertion about destructive interference is quite intriguing. If we can define the "lonely runner" in terms of all the other runners, we might be able to prove that the set of arbitrary frequencies will eventually yield zero.
The smallest positive integer multiple of 6 that isn't adjacent to a prime number is 120 and nearly half the ones up to that are adjacent on both sides
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.
these ones you can’t officially *prove* because you would have to test every single number is literally a manifestation of my greatest, most debilitating paranoia and anxieties… the fact there’s a non zero chance of this really specific terrible thing happening keeps me up at night and even though it’s not scary when it’s maths, i feel scared.
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
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
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
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.
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.
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.
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.
Really cool video! I'm so glad this got recommended to me! I'll definitely check out more of your channel. Btw 2:26 If you want words like "if" in latex, I recommend using \text{if} so it doesn't look like imaginary i * function f
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...
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 7:49 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?
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|
If you proved pi + e was algebraic, that would also determine whether all other linear combinations are, since algebraic plus transcendental is transcendental I think
2:25 yes, it will always hit a number that is just 2 elevated by another number and then it will land in 1, this is because there are infinite numbers Pd:sorry for my English
That’s not a proof. If I keep multiplying 3 by itself, it will never hit a power of 2. Even though there are infinite numbers. Your argument doesn’t work.
And that is great but by any means or intents the main goal on the youtube is to grow and that's the quickest way for you it seems just putting it out there. Have a beautiful day. :)
Correction goldbachs conjecture .. evry even no greater than 2 is sum of 2 primes OR The sum of 1 and another prime. The vedio missed the second part Edit . Gold Bach conjecture is also proved for nos greater than 10^1346
Goldbach conjecture could be rewritten does X exist for all integers greater than 1 such that N +/- X are both prime? I prefer this framing because it becomes a problem about structural patterns within the primes and makes it feel more connected to Terrance Tao and other work around patterns in the primes. Edit: thinking about it this way we could then say are all numbers the center point of a pair of primes and we might also ask about special properties of those center points that could then possibly lead to a solution maybe in higher dimensions those center points are neatly arranged. Edit 2: We don't have the best bounds but we have proved that for any number (greater than 25) there will always be a next prime of size 6N/5 or smaller. There are better results but my maths isn't good enough to understand them but if we could prove a tight enough bound on the frequency of pairs of a certain size and maximum size of prime gaps a proof would be in sight.
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 :(
lets imagine any multi of 6. 6x 2 less would be 6x-1 2 less would be 6x-2 which is equal to 2(3x-1) which is divisible by 2, so non-prime 3 less would be 6x-3 which is equal to 3(2x-1) which is divisible by 3, so non-prime 4 less would be 6x-4 which is equal to 2(3x-2) which is divisible by 2, so non-prime 5 less would be 6x-5 (or 6y+1 ie: y=x-1) 6 less would be 6y it covers all numbers, above 6, hence, all prime numbers can only be in form of 6n+1 or 6n-1
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.
I have never once heard the Riemann Hypothesis explained in a way that makes it "sound easy," as claimed in the video title. After watching this video, that still holds 🤨
It's not complicated. The sum is just single variable calculus - the simplest form of math of all. The Zeta function is just an analytic continuation of that sum, which is a fancy way of saying a smooth function that agrees with the sum on all the places where the sum is well-defined. Then the question is if all nontrivial zeros of the function have real part 1/2. Yes or no? That's it. The implications are huge, though. If we have all the nontrivial zeros, which are known to be countably many, we have an explicit formula for the nth prime. Nearly all of number theory is solved then, which is the are of mathematics that is known for producing the most amount of unsolved problems and the biggest amount of those "simple to state, elusive to prove" statements.
"Math problems that sound easy"
Problem #1: So... There's these 24 dimensional Spheres...
😂😂
You have never heard a hard problem ….
Wait until you hear about spheres with infinite number of dimensions...
just wait till its a non integer amount of dimensions
@terriplays1726 what do you mean he's never heard a hard question, he just watched a video full of questions harder than you've ever solved
When Euler says he cannot prove something, the math community shivers.
Is Omega computable?
He just needed more grease.
Put him and Gauss in a room, they'll finally figure anything out.
@@scottrackley4457 the Euler-Gauss solution of the collatz conjecture
Dig him up, clone him, give him access to a modern computer.
I have proven the Riemann Hypothesis but I'm currently busy feeding my cat.
😂 🐈
I have a truly marvelous demonstration of the Riemann Hypothesis which this comment is too small to contain.
I have a proof of Goldbach but am too lazy to write it down.
Did you adopt your cat from Schrodinger? We may have another problem.
I found the Schrödinger cat but my 1nt3rnet 1$ sufeπ1ng fr0m sup3π∆÷×∆÷
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
Oh wow!
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.
Teacher's just trying to see if one of you gets it so he can solve it for himself haha
@@ThoughtThrill365Thanks for sharing..are you a mathematician?
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
My kissing number is 0
😘
@@javen9693 Wow one
:(
I'm 0
Me too lol
yeah, apparently the birch conjecture was so easy and trivial that the statement of it is left as an exercise to the viewer
yea proof is easy too, but the margin here is too small sorry
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! ☺️
@@maniamapper3202 how did you solve it tho
the creator is such a pvssy for liking this comment
5:32 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.
Interesting.
And I believe that requires the Elliott-Halberstam conjecture.
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.
Very interesting. Dare I ask why 6?
@@miloyallbecause of the Elliot-Halberstam conjecture, weren't you listening 😂
Jk, but also I would be unsurprised if the experts' answer was basically that, or an elaboration of that conjecture
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.
There is a similar problem in Project Euler that is solvable and fun
PIVOT!!!
@@laszlovincze5095 okay, good to know you got the reference covered 😂
I’m just someone on the internet, but I absolutely get why that’s really, really hard to find a proof for. A brute force method would be to literally name every shape, and the very few hard limits on what is possible don’t make intuition a viable option.
I think the problem is referenced in The Long Dark Tea-time of the Soul.
"We can't figure out Pi+E."
"Pie."
No that's pi x e :)
@Anev-ms5bn You weirdly solved an unrelated problem for me, by proxy of pointing that out.
@@bobSeigar it's computer scientist vs mathematician 😂 - "pi" + "e" vs pi x e
riemann hypothesis definitely does not sound easy.
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
Is that not subjective? Since the video title itself is subjective, he can do whatever he wants in this case.
@@senco445 Sure, easy for a once in a millennium genius.
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
I think “sounds easy” is supposed to mean “easy to describe”, not “easy to solve”. Maybe?
so math nerds can't figure out kissing, big news
I feel that the problem is related to percentages areas like platonic solids in some ways.
@@radielkill You're not far off.....in 3 dimensions, where it's already been solved. In higher dimensions, specifically multiples of 8 you need modular forms.
@@radielkillplatonic… that’s the problem 😉
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
I mean, being rational is a stricter case than being algebraic.
My bet is on transcendental for every arithmetic combination of the two. But proving transcendence is absurdly hard. Even for pi and e themselves it was pretty much dumb luck that we even found the proofs. And they're hard.
Regarding irrationality I feel like at least pi/e should be easy (or e/pi, doesn't matter), as it ultimately amounts to the impossibility of p*pi = q*e for natural numbers p and q.
@@iantino that is true lol, dont know what went through my head at the time
@@lonestarr1490 the only pretty expression (and any derivation of it) that we know of is e^pi. Theres a theorem that says that the only cases where algebraic^algebraic is algebraic are the trivial ones (when the base is 0 or 1, or when the exponent is rational). So, as (pi^e)^i = -1, and -1 and i are algebraic, therefore pi^e cant be
11:54 And 9.94687 is rational.
"You probably haven't heard of knot theory."
You clearly don't know that I'm subscribed to Stand-up Maths.
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.
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
yea but this is only one of its applications. Proving the extended version of the Riemann hypothesis proves A LOT of unsolved theorem
8:30 You literally forgot to say what is Birch (...) conjecture
You just said it has something to do with elliptic curves
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
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.
@@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.
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.
Video title: Unsolved maths problems that sound easy
Narrator of video: "This conjecture is the only one we can explain in plain English"
*doesn't explain the conjecture*
*moves on*
Side note, I'm pretty sure P = NP is a fairly easy conjecture to explain. The existence and smoothness of the Navier-Stokes equation isn't super easy, but you don't need a ton of technical details to explain what it is and why it hasn't been solved yet.
Leon-hard? Leon... HARD? Is that how he said his name? No really, this is more important than the unsolved mysteries of math
great video though
😂😂
I believe it's actually quite close. I'm not German (I'm Dutch), but the 'h' is supposed to be heard I think.
Settle down, nerd
Believe it or not, the closest way of writing it out with English sounds would be "lay on hard".
You can make a game out of the Collatz conjecture.
1. include a chain counter
2. add a leaderboard
3. get a ton of people on it so that they disprove it while trying to get a high score
Not a bad idea, unless it is true and the game never ends 😂
Wow, didn’t expect to see you here!
Famous conjectures which are easy to understand:
There are infinitely many perfect numbers
All perfect numbers are even
*perfect numbers are the ones, equalling to the sum of their smaller divisors. Such as 6, 28, 496…
There are infinitely many primes of the form x^2+1.
@@oro5421just to clarify
For 6 you would sum
1+2+3
And for 28 you would sum
1+2+4+7+14
Is that it for the definition?
R A C I S M
@@RadeticDaniel Yes
Definition 1.0: We define the phrase "sounds easy" to mean "doesn't sound easy".
Most of them are very easy to state compared to 99% of open problems in mathematics that aren't number-theoretic. Except Birrch and Swinnerton-Dyer. That one is totally out of place lol
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.
More like an easy way to find prime is the key
a greater understanding of primes would make a lot of these much easier
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.
@@mrbutishThis is meaningless
so what I'm hearing is "just solve Riemann"
No unsolved math problem sound easy to me considering the geniuses that already tackled them 😅
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.
@@clintonfarrier244 If collatz conjecture is true for multiple of 3 then it's true for all numbers.
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.
6:55 Matrix multiplication is not n^3 complexity. The naive implementation is, but the best currently known is about n^2.37.
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
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.
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.
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
@@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...
@@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.)
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.
My gf helped me to solve the balls kissing problems. I can now confidentily say that its 2
8:38 "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)
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
It's actually important that the ""input size"" is measured in bytes (or rather: in bits) not e.g. in size of a natural number. These grow differently, as n bits can represent 2^n natural numbers.
(I say this because you are the first person I ever heard who explicitly mentioned that the input size has to be measured in bits/bytes. This is usually always glossed over.
Note also that the difference between bits and bytes is not important here, because they are related by a constant factor (8) which is ignored for usual measures of time complexity.)
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.
Well obviously, that’s not exactly a useful insight. Plenty of people have tried that approach hahah.
@@avy1 nah give bro a couple more days, he'll solve it. trust.
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.
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.
@@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.)
Im not an expert, but that depends on what logic you use. In intuitivitionist (not sure about the translation) that wouldnt work
@@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.
@lecdi6062 No, proving that it is unprovable would not prove it, because a statement may be both unprovable and false.
great stuff, thank u for posting!
1:45 leon-hard lol
10:29 bern-hard
Yes, that's how their names are pronounced. You pronounce the H in German.
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.
In the table at 0:52 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.
* * * * *
This addendum is coming a few months later. For dimensions 1 through 8 and 24, where the kissing number is known or believed to be correct, it is noted that the dimension is always even and a factor of the kissing number; also, the prime factors for the kissing number are always less than the dimension. Adding to that the assumption that the jump to the next dimension never more than doubles the next kissing number, and the lower bound for dimension 23 becomes 98,532. (98532 = 2² × 3² × 7 × 17 × 23.) Of course, this is pure speculation and proves nothing.
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.
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.
Thank for saying, what the Birch and Swinnerton Dyer Conjecture is. And so easily to understand too.
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
1:45 Nobody: Leon-hard Euler
I'm surprised he didn't say Yooler lol.
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.
Another fun open problem sinilar to π+e is whether or not Catalan's constant is rational. This constant comes up in almost every combinatorial problem and has some neat connections with the zig zag numbers
Hats off to all the absolute geniuses that manage to solve some of the most difficult problems in mathematics. But I feel like that's basically the definition of looking for a solution to a self-imposed problem. What value do we garner from finding a solution to the question whether there are infinite amounts twin primes or not? The answer is either yes or no and neither answer would change anything about how the world or universe behaves or how we understand it. I consider myself somewhat of a nerd but these problems are so nerdy and seemingly irrelevant I wonder if the time spend on those problems couldn't be put to better use in other areas of mathematics.
Math major here.
1. Knowledge for knowledge's sake is good.
2. Whenever a 100 year old open problem is solved, it usually isn't because someone just used current math tools in a clever way, but rather invented new tools that then go on to be super useful for more math. For simple algebra and number theory problems, ring theory was invented. At the time it seemed hopelessly abstract, but it lead to algebraic geometry, which is not only one of the most promising areas of mathematics when it comes to unification of math, which is like our version of quantum gravity, an ultimate goal, but also is used in computer graphics. Every single shape on display, unless it's bitmap, including every letter on screen is an algebraic variety carved out by polynomials. Group theory and compex numbers were ones considered the mathematical equivalent of science fiction and are now absolutely integral in all things physics, electricity and fluid motion.
3. They are already useful, regardless. The distribution of primes is immensely important in cryptography. Both Goldbach and twin prime conjecture are about that and the Riemann Hypothesis basically solves all questions about prime numbers by explicitly giving us a formula. Birtch and Swinnerton-Dyer is even more ambitious for cryptography and algebraic geometry, whose importance I already explained. Knotting is important in biochemistry and physics. Lonely runner is part of Ergodic theory, which is super useful everywhere from chemistry, to probability, to fluid and aerodynamics. Sphere packing is an optimization problem. Optimization problems are always useful one way or another for speeding up computation, optimizing profit or minimizing energy use or material waste on assembly lines and the like.
I have been a subscriber for three days and I really like your content ;)
❤️❤️
You can proof the Goldbach conjecture by just opening you eyes. The proof is obvious and left as an exercise for the reader.
The table at 7:00 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.
This dropped exactly when I needed it. Subscribed immediately. Excellent work.
Another unsolved math problem that I like that sounds simple is the Perfect Cuboid problem where proving all side lengths, face diagonals, and interior diagonal of a rectangular prism is an integer for the equation a^2+b^2+c^2=d^2. It is comprised of a set of pythagorean theorems and even though it's not a $1mil question it's still a fun little equation to look at.
If you consider the quadratic x^2 - (pi+e)x+ex=(x-e)*(x-pi). then you immediately get that pi+e and pi*e cannot both be algebraic, since they would be coefficients to a quadratic with transcendental roots. Similarly, if a and b are both trancendental, then at least one of a+b and ab must also be.
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.
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?
Indeed, 3 is the deal breaker. The other four are not assumptions, but proven facts.
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 :|
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.
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.
I have found a proof for the Riemann hypothesis, however discovering my proof is left as an exercise to the reader.
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.
I imagine somewhere a brain surgeon rolling up his sleeves.
guess we doin circles now
1:46 That man is not Goldbach. He is Hermann Grassmann, the inventor of exterior algebra.
kissing numbers are usually called coordination numbers in material science, which is also where the body-centered and face-centered structures come from.
The sum of all twin primes (5,7) and larger is not just divisible by 6 but even 12.
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).
leonHARD Euler
well turns out that is the correct way to say it lmao
and the eu in euler is pronounced oi
@@jodfrut771 According to Wikipedia, it's more like LAY-own-hart.
9:24 man i must get that one million reward, I’ve always loved that fragrance!!
Goldbach conjecture is tantalizingly true, because if you simulate a LARGE grid the pattern becomes pretty clear, so it’s just a matter of proving why that pattern continues.
For the Lonely Runner, clearly we’re asking if a set of arbitrary frequency will always at some point yield zero, ie destructively interfere, this might be possible if we can define the the lonely runner in terms of all the other runners.
the Goldbach conjecture is a tricky one to prove, but with the aid of simulation, it becomes much clearer. I agree with your assessment that the pattern will continue as the grid becomes larger, but we need to find a way to prove it. As for the Lonely Runner problem, your assertion about destructive interference is quite intriguing. If we can define the "lonely runner" in terms of all the other runners, we might be able to prove that the set of arbitrary frequencies will eventually yield zero.
Every number theory problem in existence.
I swear even complex analysis problems become way more approachable, once you have the mathematical knowledge.
The smallest positive integer multiple of 6 that isn't adjacent to a prime number is 120 and nearly half the ones up to that are adjacent on both sides
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.
Ah so there’s an infinite number of sexy primes
these ones you can’t officially *prove* because you would have to test every single number is literally a manifestation of my greatest, most debilitating paranoia and anxieties… the fact there’s a non zero chance of this really specific terrible thing happening keeps me up at night and even though it’s not scary when it’s maths, i feel scared.
Well there are ways to mathematically prove things without trying every case
There are other proof methods other than proof by exhaustion, such as direct proof, contradiction, contraposition.
I tought you had more subcribers, this is video is really well made. Keep up the good work
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
very great video Idea! I've been searching for such a list!
Exactly what I was looking for, thanks!
Another famous one: We don't know whether Pi^(Pi^(Pi^Pi)) is a whole number or not.
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
We don't even know if pi is normal.
this is the most american way of pronouncing euler-mascheroni lol
😄
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
They haven't taught ai voice synthesizers how to pronounce mathy words
Oily macaroni
WHEELER MASKERONI
Great vid! Love how you broke it down.
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.
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.
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.
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.
Really cool video! I'm so glad this got recommended to me! I'll definitely check out more of your channel. Btw 2:26 If you want words like "if" in latex, I recommend using \text{if} so it doesn't look like imaginary i * function f
I knew about knots in math and how to determine if they’re solvable from the video “how to turn a sphere inside out”
You know the one
Great video - I may have been you 5,000th subscriber as when I hit subscribe it went from 4.99K to 5K!
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...
Not too hard though. It is a smooth function defined on all of ℂ\{1} that agrees with the Riemann sum for s > 1.
8:43 I think we can't call the elliptic curve a function since some y is corresponding to more than one x
Anyway good video
Yeah, more like an algebraic variety, set, graph of polynomial, or topological space.
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 7:49 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?
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|
If you proved pi + e was algebraic, that would also determine whether all other linear combinations are, since algebraic plus transcendental is transcendental I think
I have discovered a single truly marvellous proof of all of these, which the comment section is not large enough to contain.
It seems like trying to prove the Colatz conjecture will only get us stuck with circular arguments
Nice pun!
2:25 yes, it will always hit a number that is just 2 elevated by another number and then it will land in 1, this is because there are infinite numbers
Pd:sorry for my English
That’s not a proof. If I keep multiplying 3 by itself, it will never hit a power of 2. Even though there are infinite numbers. Your argument doesn’t work.
Just found you through algorithm and the video is awesome but you gotta do something with that intro and outro for real.
most of my subscribers like it without intro and outro.
And that is great but by any means or intents the main goal on the youtube is to grow and that's the quickest way for you it seems just putting it out there. Have a beautiful day. :)
addition is actually so complex under the hood
we just think of adding little natural numbers and get the wrong idea
It's difficult when we try to associate it with the distribution of prime numbers, which seems completely uncorrolated.
Correction goldbachs conjecture .. evry even no greater than 2 is sum of 2 primes OR The sum of 1 and another prime. The vedio missed the second part
Edit . Gold Bach conjecture is also proved for nos greater than 10^1346
I thought Tao basically showed there is No unbounded sequence of Colletz numbers, but that doesnt eliminate the chance that there is a loop
That's pretty formidable progress, though.
Nice AI voice
Random personal fact: Swinnerton-Dyer is my mathematical grandfather on the Mathematical Genealogy Project.
11:47 Damn, the way you asked "Is Gamma Rational?" made it sound like a creepypasta from the late noughties.
Bravo! for putting this together...
Yitang Zhang the GOAT…one time i gave a report to him in a research group and he fell asleep…absolute legend
Goldbach conjecture could be rewritten does X exist for all integers greater than 1 such that N +/- X are both prime?
I prefer this framing because it becomes a problem about structural patterns within the primes and makes it feel more connected to Terrance Tao and other work around patterns in the primes.
Edit: thinking about it this way we could then say are all numbers the center point of a pair of primes and we might also ask about special properties of those center points that could then possibly lead to a solution maybe in higher dimensions those center points are neatly arranged.
Edit 2: We don't have the best bounds but we have proved that for any number (greater than 25) there will always be a next prime of size 6N/5 or smaller. There are better results but my maths isn't good enough to understand them but if we could prove a tight enough bound on the frequency of pairs of a certain size and maximum size of prime gaps a proof would be in sight.
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 :(
that also happens with kant and jacobi lol
lets imagine any multi of 6.
6x
2 less would be 6x-1
2 less would be 6x-2 which is equal to 2(3x-1) which is divisible by 2, so non-prime
3 less would be 6x-3 which is equal to 3(2x-1) which is divisible by 3, so non-prime
4 less would be 6x-4 which is equal to 2(3x-2) which is divisible by 2, so non-prime
5 less would be 6x-5 (or 6y+1 ie: y=x-1)
6 less would be 6y
it covers all numbers, above 6,
hence, all prime numbers can only be in form of 6n+1 or 6n-1
Honlestly, you only need the names Erdos and Euler and you get a truckload of easy looking open problems.
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.
3:40 THY END IS NOW
WEAK
USELESS
I have never once heard the Riemann Hypothesis explained in a way that makes it "sound easy," as claimed in the video title.
After watching this video, that still holds 🤨
It's not complicated. The sum is just single variable calculus - the simplest form of math of all. The Zeta function is just an analytic continuation of that sum, which is a fancy way of saying a smooth function that agrees with the sum on all the places where the sum is well-defined. Then the question is if all nontrivial zeros of the function have real part 1/2. Yes or no? That's it. The implications are huge, though. If we have all the nontrivial zeros, which are known to be countably many, we have an explicit formula for the nth prime. Nearly all of number theory is solved then, which is the are of mathematics that is known for producing the most amount of unsolved problems and the biggest amount of those "simple to state, elusive to prove" statements.
Ah, that's easy! The answer is yes times maple syrup.
8:45 elliptic curves aren't functions, they map some inputs to multiple outputs
Solving the Collatz is currently my retirement plan - wish me luck haha
same goes for finding perfect traingle numbers.. we only found 4 till this day and number 5 seems to be impossible to find
ahh yes, I remember learning the formal definition of "almost all", miss those days