RIEMANN HYPOTHESIS PROOF / SOLVED, 32579 VIEWS, 0 COUNTEREXAMPLE! (M2) #riemannhypothesis

The primary aim of this video is to invite experts in the Riemann Hypothesis or exceptionally intelligent mathematicians to scrutinize the content and detect any errors using a numeric counterexample. For instance, if a value 's' exists for the ABC zeta function that yields a value 'A' that is inconsistent with the actual value 'B' of the zeta function, we are prepared to offer a reward of 10K for your contributions.Additionally, if someone you refer to the video comes up with a counterexample and wins the reward, you will also receive a $10,000 reward, making the total reward $20,000. Therefore, we encourage you to share the video widely.

Over the years, hundreds of mathematical theories have been built upon the assumption that Riemann's last theorem is true. Therefore, considerable efforts have been made by several of the best mathematical minds around the world to protect the legitimacy of these theories. However, in this video we have finally proved this famous theorem that had resisted all efforts to be proven for over one and a half centuries.

Please notes:

2. If your counterexample disproves Riemann's hypothesis, it will be your responsibility to properly document it before sharing it with anyone. Rest assured, we will make every effort to protect your intellectual property rights, and we will not claim any ownership of your counterexample. You will be eligible to receive the $10,000 cash prize, regardless of any additional awards. Having said that, it is highly likely that Riemann's hypothesis will withstand any counterexample through reformulation. The official problem statement from the Clay Mathematics Institute states that only a small prize will be considered for such a counterexample. Therefore, if you believe you have a counterexample that challenges Riemann's Last Theorem, it is important to acknowledge that the Riemann's hypothesis is expected to remain resilient and not be significantly undermined by the counterexample.

#riemannhypothesis #riemannhypothesissolved #riemannhypothesisproof #SoME3 #riemannhypothesissolution #riemann'slasttheorem #mathmatics #millenniumprize

[RSLT]

Before leaving your comments, please kindly take the following into consideration:

The topic at hand, the Riemann Hypothesis, requires a considerable reasoning ability above the average. We've done our best, and simply put, it's not possible to make it any easier than this. There's no cause for frustration; instead, it's a reminder that each of us excels in different areas. So, before posting any comments, please take into account the following three key questions:

1. Am I an expert in the Riemann Hypothesis?
2. Do I possess the qualities of exceptionally intelligent mathematicians?
3. Can I provide a numeric counterexample with the assistance of the aforementioned expertise( used shared button)?
4. Have I genuinely attempted to find a numeric counterexample but have not succeeded? Would I like to share what I have done in my efforts to find the counterexample?

*Please be advised that if no one successfully claims the reward, it will still be distributed. The $10, 000+$10, 000 rewards will be allocated regardless of whether the comments are positive, negative. In such a situation, one random commenter will be selected to receive the reward. Thank you for your participation and support*

If the answer to any of these questions is 'yes, ' then your comments would be valuable to the advancement of math. Otherwise, you might find yourself feeling frustrated without any productive outcome! If the answer to any of these questions is 'no, ' we kindly recommend considering other videos that offer somewhat more straightforward content (though still challenging if delved into deeply). Please feel free to contribute your constructive insights and observations, rooted in mathematics and logic.

The goal of this video is to invite experts in the Riemann Hypothesis and exceptionally intelligent mathematicians to review its content. As mentioned in the video description, the purpose is to encourage scrutiny. The comments section serves as a tool to verify the accuracy of the information. If you discover that you don't possess a counterexample and you're feeling down or upset today, we encourage you to express your emotions, such as sadness or anger, in other places (i.e., upload a video, describe and let it out). This channel is dedicated to sharing and learning. Expressing negative emotions does not contribute positively to the intended goal. Please adhere to the guidelines and ensure that your comments are focused on challenging the content and not people. Maintaining a respectful and correct mode of communication is crucial, as it seems that these principles have been overlooked in recent times. It's important to understand that your input may not be taken seriously if it lacks expertise and is primarily emotional.

After a century and a half of hard work to prove Riemann's hypothesis, it appears we need to take a new and unconventional method. The drawback of unconventional methods is that they hardly get noticed in a timely manner. For example, Riemann's work did get the attention that it deserved long after he was gone. Our method to combat that issue was to convert advanced and unconventional concepts to elementary and fundamental ones. Riemann's Last Theorem Article( 0bq.com) is an article that brings a deep, unconventional part of math to the surface utilizing simple math concepts.  This video is further steps in that direction to show that the RSLT proof is complete, following David Hilbert's footsteps. "A mathematical theory is not to be considered complete until you have made it so clear that you can explain it to the first man whom you meet on the street."  (For more detail, see 0bq.com)

Please be aware that we have dedicated a significant amount of time to simplify and condense the proof, ensuring its easy verification within minutes. As we have made this effort, we request meeting us halfway by providing NUMERIC counterexamples, allowing us to efficiently review and verify your claim, just as we did in our videos M1 and M2, where we presented redundant proofs for your convenience that you can see and verify in minutes. If you believe there is an error and have the expertise, supporting your claim with a numerical counterexample should not be too challenging. If no counterexample can be provided, it would strongly indicate that your claim is not well-founded. Having concrete evidence and logical reasoning for the debunking proof, supported by numerical counterexamples, is crucial. This approach ensures that claims are not solely based on intuition but are grounded in mathematical logic.

If you have doubts about the accuracy of the information presented in the video, we encourage you to conduct research and fact-check the claims. Additionally, seeking the opinions of others who have watched the video can provide a broader perspective and help determine its authenticity.

The need for a counterexample stems from its easy verifiability and agreeability. It is often the case that people preset specific claims, such as x^2+1 having no root, simply because they believe "it cannot exist." However, the complex analysis and analytic continuation reveal a different perspective.

IF WE CANNOT SUPPORT OUR DISPROOF CLAIMS WITH A SIMPLE COUNTEREXAMPLE, IT STRONGLY INDICATES A FUNDAMENTAL MISUNDERSTANDING OR A RELIANCE ON INTUITION RATHER THAN MATHEMATICAL LOGIC.

Thank you for considering these points, and we appreciate your engagement with the subject matter.

Q: Have you submitted the proof?

Short Answer: We haven't submitted the RSLT article because we have experience with an earlier version, and people may not be willing to support this proof. We are awaiting the completion of the process for a higher stakes proof like this. In mathematics, if a proof withstands rigorous attempts to disprove it, we eventually accept it as valid. Additionally, we offered overall $100K rewards associated with this process to expedite the process.

Longer Answer: We have an earlier version of the RSLT article that we've already submitted, and the response was underwhelming. We haven't submitted the RSLT article because we have experience with an earlier version, and people may not be willing to support this proof for unreliable and unsupported, very wrong, and, at best, lazy reasons, such as it is not rigorous or doesn't meet our journal's standards (implying very disrespectful that beggar cannot be chooser), and we don't have resources to review R.H. proofs. We challenged those reasons with $10K or more prizes and offered reimbursement for referee costs per hour and/or $200 per page review. Therefore, currently, there is a waiting or idle period as we contemplate pushing this forward because, as history has shown, groundbreaking ideas often face resistance. For instance, just as Galileo asserted that the Earth is round despite popular belief at the time. The history of mathematics is rife with such incidents where novel concepts were initially rejected due to entrenched beliefs. Much like the belief that the Earth is flat, our intuition often conflicts with mathematical truths. For instance, the notion that 1+2+3... equals -1/12 challenges our intuitive understanding. However, just because something seems counterintuitive doesn't make it untrue. In mathematics, if a proof withstands rigorous attempts to disprove it, we eventually accept it as valid. It's worth noting that the process of acceptance in mathematics involves both peer review and subjective evaluation. Many top mathematicians have seen the proof we mentioned (Riemann's last theorem article) but have chosen to remain silent publicly, likely waiting to see its fate. This cautious approach is typical in the mathematical community, where rigor and skepticism are valued. Ultimately, the acceptance of a proof hinges on the inability to disprove it rather than intuitive understanding alone. The potential reward associated with this process adds further motivation for its pursuit. If we are wrong, we will gladly pay for our mistake. If not, we expect support for our proof at some point. Remember, this challenge is complicated because some argue that the Earth is not round based on passages in the Bible. However, we're not aiming to disprove the Bible. Instead, we're demonstrating that the interpretation of the Bible regarding this matter may be incorrect. So Additionally, we offered overall $100K rewards associated with this process to expedite the process. Our estimate was that we would find the error( if there is one) around 100K views overall, which we have recently surpassed. That was the most critical part of the proof. So, we know that there is a very small likelihood of less than a one in a billion chance to encounter a numeric counterexample, and that likelihood becomes smaller with every single view. Our other estimate is the silent period before the storm, up to 1M views, which we are currently in. It is very likely that we will see wide acceptance after that, which will be the time to submit the RSLT article.

[The_Green_Man_OAP]

Btw, because of inflation since 2000 prices are 80% so I suggest asking for $2 MILLION ‼️

[racheline_nya]

s=1/2 is a counterexample. you can see that by simply computing it, but if you want a complete and detailed proof (like you should have given if you wanted it to at least seem legitimate), here it is:

denoting the real part of s as R(s), like in the video (although the standard notation is Re(s)), we have R(1/2)=1/2, so by the axioms of equality, we can replace 1/2 with R(1/2) in all formulae without changing their truth.
obviously, 0<1/2<1, and since we can replace 1/2 with R(1/2), that means 0<R(1/2)<1 is also true. so, if we set s=1/2, the condition 0<R(s)<1 is true, which means that the "theorem" claimed in the video has an instance with s=1/2. that instance says that R(1/2)≠1/2 is equivalent to ζ(1/2)≠0.
but R(1/2)≠1/2 is false, since its negation R(1/2)=1/2 is true, so according to the video, ζ(1/2)≠0 is also false. now, we will show that ζ(1/2)≠0 is true, which will contradict the statement in the video, and so the proof that s=1/2 is a counterexample will be complete.

ζ(s) is defined as the analytic continuation of (sum of 1/n^s for all positive integers n) to the set of all complex numbers other than 1. since the set of all complex numbers other than 1 is connected, there is only one such analytic continuation, so if we find an analytic function f(s) equal to a sum that converges for R(s)>1 and for s=1/2, then we will have f(s)=ζ(s) for all s for which f(s) is defined.
here is one such function: f(s)=1/(1-1/2^(s-1))*(sum of (-1)^(n-1)/n^s for all positive integers n).
to see that f(s) = (sum of 1/n^s for all positive integers n) when R(s)>1, we can notice that with R(s)>1, both (sum of 1/n^s for all positive integers n) and (sum of (-1)^(n-1)/n^s for all positive integers n) converge, so we can combine them into one sum:
(sum of 1/n^s for all positive integers n)-(sum of (-1)^(n-1)/n^s for all positive integers n)=(sum of 1/n^s-(-1)^(n-1)/n^2 for all positive integers n)=(sum of (1-(-1)^(n-1))/n^2 for all positive integers n).

then letting g(n)=0 if n is odd and g(n)=2 if n is even, it's obvious that g(n) = 1-(-1)^(n-1), so we get (sum of 1/n^s for all positive integers n)-(sum of (-1)^(n-1)/n^s for all positive integers n)=(sum of (1-(-1)^(n-1))/n^s for all positive integers n)=(sum of g(n)/n^s for all positive integers n)=(sum of 0/n^s for all positive odd integers n)+(sum of 2/n^s for all positive even integers n)=(sum of 2/n^s for all positive even integers n)=(sum of 2/(2n)^s for all positive integers n)=(sum of 2/(2^s*n^s) for all positive integers n)=(sum of 2/2^s*1/n^s for all positive integers n)=2/2^s*(sum of 1/n^s for all positive integers n)=1/2^(s-1)*(sum of 1/n^s for all positive integers n).
we can remove all the mess in the middle, and we have (sum of 1/n^s for all positive integers n)-(sum of (-1)^(n-1)/n^s for all positive integers n)=1/2^(s-1)*(sum of 1/n^s for all positive integers n).
now, if we add (sum of (-1)^(n-1)/n^s for all positive integers n)-1/2^(s-1)*(sum of 1/n^s for all positive integers n) to both sides, we get (sum of 1/n^s for all positive integers n)-1/2^(s-1)*(sum of 1/n^s for all positive integers n)=(sum of (-1)^(n-1)/n^s for all positive integers n), and the left side is just the expanded form of (1-1/2^(s-1))*(sum of 1/n^s for all positive integers n), so if we divide both sides of the equation by 1-1/2^(s-1), we get (sum of 1/n^s for all positive integers n)=1/(1-1/2^(s-1))*(sum of (-1)^(n-1)/n^s for all positive integers n)=f(s).

so we have a function f such that f(s)=(sum of 1/n^s for all positive integers n) for all s for which (sum of 1/n^s for all positive integers n) is defined, and therefore the analytic continuation of (sum of 1/n^s for all positive integers n) to the set of complex numbers other than 1 must be equal to the analytic continuation of f to the set of complex numbers other than 1. in the first case, that analytic continuation is ζ, which means ζ is the analytic continuation of f to the set of complex numbers other than 1. then ζ(s)=f(s) for all s for which f is defined.
now, it's easy to see that 1/(1-1/2^(1/2-1))=1/(1-1/2^(-1/2))=1/(1-1/(1/2^(1/2)))=1/(1-2^(1/2))=1/(1-√2), which is a nonzero complex number. then if (sum of (-1)^(n-1)/n^(1/2) for all positive integers n) converges to a nonzero complex number, ζ(1/2)=f(1/2) will be a product of two nonzero complex numbers, which is also nonzero. so we just need to prove that (sum of (-1)^(n-1)/n^(1/2) for all positive integers n) converges to a nonzero complex number.

pairing up consecutive terms of (sum of (-1)^(n-1)/n^(1/2) for all positive integers n), we get that it is equal to (sum of 1/√(2n-1)-1/√(2n) for all positive integers n). now, since 1/√(2n-1)>1/√(2n), all terms of this sum are positive real numbers. it's also easy to see that and 1/√(2n)=√(2n-1)/(√(2n-1)*√(2n))=√(2n-1)/√(2n(2n-1))=√(2n-1)/√(4n^2-2n), so we get
now, if we use h to denote the derivative of √x with respect to x, we know that h(x)=1/(2√x), so h is decreasing for positive real inputs. then clearly, √(2n)-√(2n-1)<h(2n)*(2n-(2n-1))=h(2n)*1=h(2n), so we see that
this means that (sum of (-1)^(n-1)/n^(1/2) for all positive integers n) is a sum whose terms are positive real numbers and upper bounded by the terms of (sum of 1/n^(3/2) for all positive integers n), so since (sum of 1/n^(3/2) for all positive integers n) converges, (sum of (-1)^(n-1)/n^(1/2) for all positive integers n) must converge too. and since its terms are positive real numbers, it converges to a positive real number, which means it converges to a nonzero complex number.


so, wrapping things back up: we just proved that (sum of (-1)^(n-1)/n^(1/2) for all positive integers n), which completes the proof that f(1/2) is a nonzero complex number, which implies that ζ(1/2) is nonzero, which means that ζ(1/2)≠0 is true, and that contradicts the main claim of the video in the case s=1/2, which means s=1/2 is a counterexample to the claim in the video.

this was a fun little exercise, but where are my 10000 dollars?

[creativedevil]

Riemanns last theorem asserts that zeta(-2) is not equal to zero, however, it is known that zeta(-2) is equal to zero.

[tiusic]

I don't have enough math expertise to evaluate your argument, but your insistence that people provide a numerical counter example is absurd. If people can't find a numerical counter example, that doesn't mean your argument is valid:

"The Riemann hypothesis is true, because my cat is ginger. Can you find a numerical counter example to the Riemann hypothesis? No? Well that means my argument is valid. Therefore my cat's color proves the Riemann hypothesis."

Also, what's with the flat earth stuff at the end? Is this whole video a prank? lol

[manuelantoniopianini]

3:25 lmao abs(a -b) = abs(a) - abs(b)
This isn't even a satyrical video and yet nobody has taken action to take down this nonsense

[magicmb]

This inspired me to write a proof for Goldbach’s conjecture:

I state that 2+2 = 4 => the conjecture holds.

My proof is correct and can only be false if you provide me a numerical counter example of Goldbach’s conjecture. I offer $10 to anyone disproving me (but only with a numerical counter example of course). Please don’t point out seemingly obvious issues with my proof since I’ll get really salty in the replies.

[jnoelcook]

I have a numerical counter-example. Here are the steps: 1) Beginning with the reverse implication at time = 3:10, where you state that zeta(s) = zeta(1-s^*) <=> zeta(s) - zeta(1-s^*) = 0 <=> sum_1^infinity (1/n^s) - sum_1^infinity (1/n^(1-s^*)) = 0. This is a reverse implication, which means that you need to prove that a implies b and then you need to second prove that b implies a. In this case you have a <=> b <=> c. The reverse implication of a and b is known already. We just need c. Your proof depends on this, and therefore any counter-example that proves that b does not imply c, or the reverse, invalidates anything that follows. Now, you want a numerical value, so use the first non-trivial zero. The numerical value we will use is the first non-trivial zero s = 1/2 + i 14.1347... . 2) check the reverse implications. zeta(s) = zeta(1-s^*) <=> zeta(s) - zeta(1-s^*) = 0 still holds. How about the third? Well, sum_1^infinity (1/n^s) is undefined for the first non-trivial zero. So too is sum_1^infinity (1/n^(1-s^*)). So you have an undefined term subtracted from an undefined term equal to a finite zero, which is a false statement. 3) Analyze: Does zeta(s) - zeta(1-s^*) = 0 imply a false statement? No. Therefore, the implication does not hold, neither does the reverse implication. 4) Was a numerical counter-example used? It was. Was the proof invalidated with a counter-example? It was. I will provide the link where you can send the $10k after you check your math and confirm.

[Zak-vo]

The statement you claim to prove is clearly false. If the statement "Re(s) ≠ 1/2 if and only if zeta(s) ≠ 0" is true, we then have that zeta(s) ≠ 0 implies Re(s) ≠ 1/2. This is contrapositive to Re(s) = 1/2 implies zeta(s) = 0, which is false. In essence, you claim to have proved not just that all zeroes lie on the critical line, but that the entire critical line is mapped to zero by the zeta function. This is simply not true. Therefore your proof cannot be correct either.

[GameJam]

But wait- at 3:09 you suggest that the two infinite sums are equal, and from reading the comments you've seemed suggest that this is only the case when Re(s)=1/2, right? The problem is, those infinite sums don't converge where Re(s)<=1, so they dont represent the zeta function in that domain. So how can we actually be sure that the difference between these two diverging sums actually says anything about the zeta function which is actually defined in that region?

[badorni]

Wouldn't 0.5 be a straight up counterexample as Zeta(0.5) ≠ 0, but Real(0.5) = 0.5, thus the equivalence you're claiming is wrong

[RiemannZeta-fk]

What does the Clay Institute think of this "proof" ?

[CeliaKalen-kzel]

How do you solve such a complex structural equation in all those five minutes without a calculator? You must have had one. You are really good at Math.

[iyziejane]

You can't subtract infinity from both sides of the equation, which is what you've done by treating sum_{n=1}^infinity 1/n^s as if it is a convergent series in the critical strip (it very clearly diverges). In terms of your preprint, the expression becomes invalid at equation (14). No one can give you a numerical counterexample to that expression, since you have assumed zeta(s) = 0 and s is in the critical strip, so a counterexample would involve a zeta zero with a value of s in the critical strip with Re(s) != 1/2, which is the open problem. It seems like you know that the expressions are divergent so it is dishonest to push this as a proof.

[guillermoobregonpulido]

In minute 2:46, the sumatories does not converge in the critical strip, you can not conclude any in this way, you must work whit expresions that converge in the critical strip, and then you need a analitic continuation.

[boboshermusurmonov]

"I can't find a counter-example in 10 minutes, so it must be true."

[jellyfrancis]

Yh in the end of proof you applied the normal series representation of the zeta function which isn't holomorphic where Re(s) is less than or equal to 1 so we can't use it to represent zeta function in the critical strip if you trying to do that from trying to say thay abc zeta function as said to be equal and solve it for the difference to be zero we could try but it's hopeless too because via the functional equation we know not only whenever zeta function is zero in the strip its equal to zeta(1-s) but also whenever remaining term is equal to 1 so I tried in wolfram alpha it gave me some answers that off the critical line in the critical strip also it gave me an answer that z=1/2 not the real part btw so to be more precise you started to equating abc zeta function that's totally fine then after letting b goes to infinity you took other series are the zeta function and made 'em zero that's where the first problem the proof faces secondly the approach is hopeless too because we can find the equivalent that where zeta function isn't zero ❤ but well tried...but think about this too 😉 and then to be more precise....the sum of 1/n^s and the conjugate will vanish when s=1/2 so think about that and that's not define riemann zeta function at s=1/2 btw nice try

[greatgamingguy]

I'm just a high school student, so correct me if im wrong. If we consider a function F(a, b) to be the integral of f with a as the lower bound and b as the upper bound, can we really assign that F(∞, ∞) = 0? I know that for finite numbers, when a=b, F(a, b) = 0 because our bound spans b - a=0 units. However when we have a --> ∞ and b --> ∞ we get that the size of the bound is b - a units= ∞ - ∞ units, which is undefined. So can we really say that F(∞, ∞) = 0 when the size of our bound is not 0 but undefined?

[uy-gedm]

You are using "if and only if" incorrectly. Your theorem states in the reverse direction that if zeta(s) is nonzero, then R(s) is not 1/2. This is false. As others have pointed out, if s = 1/2, then zeta(s) is nonzero but R(s) = 1/2. This reverse direction is not true.

[marcussilver]

The real issue with this proof is that you start by setting zeta(s) equal to zeta(1-conj(s)) which is only true when s is a zero of the zeta function. You then essentially while hiding it with a bunch of math, remove the zeta function entirely and basically set s equal to 1-conj(s) and then simplify to a+bi =1-a+bi and then to a = 1-a which leaves you with only 1/2. Essentially you are stating that 1-conj(s) is equal to 1-conj(s).