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Last entry, I drew out the whole proof by cases to prove that if a D_x and T_y shared a common prime factor, c, then either R_x or R_y would be greater than c. Of course, we demonstrated in the contradiction, that a special case, z = 1 existed. This entry, I will extrapolate on the "things we can conclude" section on last post. Recall the following equation: And, setting aside the special case of z = 1, which we showed last time will produce a prime number, what can be discovered for when z is not 1? So, only when z = 1 is R_y +/- 2 R_x is prime. Also, if this is not the case, then cz will not be prime. And since z is not 1, and can be expressed by a fraction, then 2D_x - T_y will not be prime. It will likewise be divisible by c. Also, R_x or R_y will be greater than c. And, of course, it is also implied that R_y +/- 2 R_x will not be prime either.
Some interesting consequences, maybe applicable to the development of the argument. Until later, - AF
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Pardon for not making a timely update to the progress through JF's segment on his work through trying to make headway into Goldbach. But I do not yet totally understand the rationale for JF's looping argument. So I have been kind of stuck for the past few days, making little headway. That being said, I figured I might as well give the progress I have made so far, and if anything more comes of it, I will amend it. This way, my work does not stagnate and die. I am going to be working on the content from about 1:39:50 in the stream. JF says he would like to prove the following: Having the general equations for the D and T-sets: And supposing that D_x and T_y share a common prime factor, c, then either R_x or R_y is greater than c. The problem, JF says, is it is not true. And he is right. Supposing they did share a common factor: We will set up a proof by contradiction, and will need to work by cases. For the cases such that R_x and R_y are not greater than c, we have:
And need to check these values for R_x and R_y for +/- 2 R_x
We also can conclude that z is odd. Therefore, you could have a z such that z = 1.
This violates JFT, of course, so we can conclude the opposite, that is there will never be a z such that z = 3.
Both cases yield the same type of contradiction. Makes sense, since this is a violation of JFT.
So we now know that z != 3, and z > -1. This establishes an upper and lower bound.
If R_x and R_y = c, then z = -1. This is likewise a violation of JFT. Maybe this was not the best way to present, because we got the same idea from the previous case. But I am being thorough. So we have examples where z can definitely exist and R_x or R_y are not greater than c. But we have also shown it is for one specific case. Now we can list some important conclusion here:
That is it for now, sorry for the delay. - AF I have been getting lost in the weeds lately in JF's Goldbach conjecture argument. To such an extent that I believe there could be some fundamental issues JF may have overlooked that doom it. Or, what is probably more likely, I am missing something. After all, JF's stream was pretty rushed, demonstrating a lengthy, but elegant set of arguments. Therefore, I want to perhaps explore the T-set in more detail and maybe connect the two constraints with what we now know from previous posts. From post 2 , I proved that, for the D-set, the following logical rules were true: This can be proved identically for the T-set. Again, the assumption we have been working with must be true for the rules to be practical. The assumption being, T is not prime. For the sake of clarity, this post will seek to demonstrate to the reader the following:
By JFT, we know necessarily that T and E are coprime, so there is a contradiction.
Let us see where this leaves the rest of the proof. I have some ideas to elaborate on from here and what we already know.
- AF Last entry, I gave the expanded proof by cases of JF's second theorem, which implied that if two D's, D_x and D_y, shared a common factor, c, then at least one of the other prime factors, R_x and R_y, would be greater. For this entry, I will be skipping a section which, incidentally, JF says is very important. I plan on getting back to it later after I understand all of what JF's work does. This entry will correlate with the contents of the stream starting at about 1:29:37, which should be when the above video begins. This is where JF introduces what he calls the "T-set". Recall the equations we have been examining previously. The equations of the form: And, again, we are supposing the set of D's produced consist of prime factors which can be put back into the equation above, the reiterative search for a pair of primes. Now, we can slightly modify this equation wherein we can likewise introduce these corresponding R's into. This is what produces the T-set. Here is that equation: Now, we know by JFT for the D equation that D and E will always be coprime. As you can probably guess, it is the exact same demonstration which shows that T will likewise always be coprime to E: Like the proof before, R must be prime, which implies no R/C exists which is an integer, and since B must be an integer, we can conclude the contrary is true. Therefore, T is always coprime to E. Now it has been established that both T and D are coprime with E. The next natural question you would have to ask is whether or not T and D are coprime with each other. In the stream, JF says that they are coprime except for if they have a common factor of 3, or in the "+" case. JF has a different proof, but I will demonstrate to the reader now some of these truly interesting results which were on the stream in more detail: This will lead you to two expressions for R: Expression 1: Expression 2: Now, here is where things get interesting. We can do the same type of demonstration on expression 2 as we have been doing many times before. That is, assume that T and D have some common factor C. By letting T= AC and D = BC: Another spicy contradiction. But what does this imply? In expression 2, since R is positive, we can conclude that T > 2D. But this may not always be the true. A case in point: Suppose E = 224 and R = 17. We will use the addition side of the D equation:
So, were you to input these numbers, all of which satisfy the constraints, into expression 2, you would end up with a negative number. This means that the two results should be treated as cases. Expression 1 is the case for when 2D > T and expression 2, conversely, when T > 2D. This leads us to make some very curious implications: So, let us suppose you have case 1, where 2D < T: you know T and D necessarily share no prime factors. For case 2, we can actually use my above case in point: 129 (mod 3) = 0, so T, which is 207, will likewise share the common factor of 3. (For the record: 207 = 3 * 69). For case 3, I can offer you another case in point:
And neither 149 nor 187 are divisible by 3. But 2D-T will always be divisible by 3. Next, I will show you a series of three graphs: We can draw some important facts looking at these three graphs. All are of the same forms as the respective either D equations or the single T equation. Facts we have found:
That is, the only time you will ever find a common factor between a D_x and a T_x is when D_x is found from: E/2 + R_x = D_x. And R will only be in a domain of (0, E/4). Not only will those two facts be true, but the only common factor they will ever have will be 3. Wow. That does not imply all primes in this domain yield D's and T's divisible by 3, but if they exist, they can only be found there. A similarly important fact which is not implied here is that a D_x cannot share any common factors with a T_y. I say again, that is not necessarily true, they could potentially share common factors in that case. Maybe D_2 shares a common factor with T_4. That is a post for another time, though.
I was proud of this post. Seems like some good information could be within it. Until later, - AF Last entry, we showed that JF's R-circles method could be represented in the form of a bipartite graph consisting of two subsets: D and R. The next theorem JF gives us can be derived, actually, using JFT. Because Weebly seems to be incapable of giving you the exact time I stamp in the YouTube URL, JF starts work on theorem 2 at 1:20:27. For a prime factor shared between D_x and D_y, C, either R_x or R_y is greater than C. I will put this also in the LaTeX: Supposing D_x and D_y have a shared prime factor of C, and taking the difference between the two equations (JF does D_y minus D_x, so I will follow this convention), we get: Since we know the result will be an integer, let everything in the parentheses on the left of the equation be z: Now, for the sake of thoroughness, I will do the proof by cases more in-depth than was presented on JF's stream. We will again set up a proof by contradiction by supposing the three following cases: Case 1: Same result as in post 1: JFT. We know already that z must be an integer, and R_x is a prime. Case 2: I added the equation just for the sake of redundancy. This results in the same quagmire as case 1. Case 3: This likewise leads to a contradiction. Therefore, at least one of the R's must be greater than C. Amor Fati's Interjection:You might have realized a possibility is R_x = R_y = C as a case. But if they were equal, then there would be no difference, and therefore no point to the theorem. In addition, we already know from JFT that there exist no cases ever such that D_x has a factor or R_x. These cases were added for the sake of redundancy. Also, this theorem supposes that there exists a common factor between two D's. For what it is worth, I am not totally convinced such a thing is possible. Granted, I have no proof. That is part of the reason I went into so much detail about the bipartite graphs, since there is a corollary to bipartite graphs which goes something like: No bipartite graphs consist of an odd cycle. That is, all bipartite graphs are even cycled. But this does not quite pan out as well as I had expected. Because you could assume that something like this bipartite graph occurs: Now, I still left the rigorous explanation for the aforementioned reader who is serious about utilizing JF's work. My initial line of reasoning was that an even number of D's leads to an odd number of R's, and if there is an odd number of R's and they cycle, then there would be an odd number of vertices and that would prove that no D can reference to a previous R. But that was really naive and a foolish instinct, because you end up with just the type of graph above: and even number of R's to D's, and, what's more, an even number of even R's to D's. So you cannot really make any progress there, at least I cannot. But, hey, I leave that stuff in for if there is some progress to be made in something I had already thought about. This is supposed to be a resource you can get JF's work from, and some potential avenues for progress on a yet unsolved problem.
That is all for now, -AF Last entry, I laid out some ground rules I said would become useful for JF's next claims. Specifically, for the things stated at 1:18:22 into his stream. The above video should start precisely then, so the reader can follow along well enough. JF's "R-circles":Again, for the sake of all of the following, we are assuming: JF says that R_2 is a prime factor generated from D_1, and it is put back into the search. This was something you might recall from the first post on JF's Goldbach work. The set of logical rules now comes into play. By rule 3, no R_x = R_x+1 and, at 1:18:49, JF says we cannot create a "self-conducting line". That is to say, an R which refers to itself. In graph theory, such a feature is referred to as a loop. Next, by rules 1 and 2, it can be deduced that all D's produced refer to a respective R. This corroborates JF's "R-circles". Amor Fati's Interjection:Now the R-circles is a really good development because it allows you to graphically understand that a D points to some R. We also know, as JF says, that "we cannot create a D that references it's own R". But, I believe I can give you a better representation with graph theory. For this, I will create what is called a bipartite graph. For a bipartite graph, all vertices of a graph, G, must be grouped into one of two sets, R and D, with no overlap. Further, no edges, E, in G can occur such that they exist between vertices of the same subset. Summed here: An important note is that all bipartite graphs are simple graphs. No simple graphs contain loops or multiple edges between vertices. This is where the rules in part II become useful. By rule 3, multiple edges cannot exist. We can deduce by the same rule that no loops may exist. Therefore, such a graph would be simple. By rule 2, we know that we can group vertices within two subsets, R and D, without any overlap. And again, by rules 3 and 4, we know that no edges will exist between vertices of the same subset. Hence, G can be represented as a bipartite graph: This will allow us to progress onto JF's Theorem II. So there is still more to come
- AF Last entry, I made an organized display of JF's Theorem (henceforth on here "JFT"). I said he continued his proof, and I wanted to make a series of entries here so that the reader wasn't inundated with information. Next, I want to expand on what we have already learned before we get to theorem two in JF's livestream. The next thing JF says is you cannot infinitely iterate D_x's. This is true, since E is a finite number. He makes some claims next that I will not verify in this post. But I think I can use some simple logical rules to lay a groundwork for his next theorem he presents in the stream. This is the case for when: And here are the logical rules I can give you:
To prove no R_x = R_x+1, you would have to suppose D_x was coprime with E, but not prime. (JFT) Suppose also that D_x has factors, A and C. If an R_x+1 could be generated in such a case, then R_x+1 would have the same value as R_x. For a proof by contradiction, one of the prime factors would have to be R_x. To put it more generally: So let A = R_x, and we can solve for C: And how does such a result relate with what we already know? Well, by the constraints, we know that E and R_x are coprime. We also know that C is an integer. The only way the left side of the equation will yield an integer is when R_x = 1. You are probably saying the proof looks familiar, because it does. This is just JFT, applied with a prime factor of R_x.
Proof of 4: Consequence of 3. Since, if, besides 1, no D_x and D_x+1 share common prime factors, then D_x and D_x+1 cannot be equal. I think this can lay the groundwork for some of the claims JF makes next. -AF I wanted to expand on some of the mathematics covered on one of JF's livestreams. That way, there is at least a more, uh, clean presentation of what JF wrote down. I am just clarifying it in my own way, maybe making some interjections of my own, and so on. The video is here: So, the Goldbach Conjecture: "Every even integer greater than 2 can be expressed as the sum of two primes." -From Wikipedia And I will list the constraints to JF's theorem: Now, at around 1:14:07 JF's writing gets messy, but he says the numbers produced by the following equation: are prime pairs which add to E. But, as he later elaborates on, this is not always true. A case in point is let E = 28, and let R_1 = 5. This produces 9 and 19. If you used bigger numbers, then maybe you could get a pair from R_1 and one of the D's produced, here you cannot. You can see here, 9 is not a prime number. But he does this reiterative approach. So we know that 9 is coprime, and obviously consists of a pair of prime factors. In fact, if we take that pair of prime factors, 3 and 3, we can let R_2 = 3 here: And JF's algorithm nails it this second time: since The D_2's generated are 17 and 11! So we reach a conclusion for the first iteration, that is the pair of D_1's: Either the D pairs produced are prime and sum to E, or they consist of prime factors, which could be put back into the search, using JF's algorithm. I cannot go so far as to say that the prime factors produced will lead to a pair of primes which sum to E, since that would be a proof of the Goldbach conjecture, but I can say, at least, that the prime factors can go in, since, by definition, they would be prime, and they would be less than R_1, and consequently do-able as R_2. All of that is well and good, but just what is JF's Theorem, or maybe, if you prefer, Gariepy's Theorem? It is for the general equation: Given the aforementioned constraints, (E divisible by 4, R is a prime where E is coprime to R), JF's theorem says that D is also coprime with E. There exists a proof, which can be demonstrated by contradiction. JF has his own basically identical approach, but I can give you a similar idea here:
Now we will multiply both sides by the reciprocal of c: But by definition, R is prime, a and b are also integers, so there is a contradiction. Therefore, we can say that JF's Theorem, that is, for the general equation, D is always coprime with E, is true. Congratulations, JF! You have your own theorem. That is actually pretty cool.
We also know, incidentally, that D will be an odd number, since E is divisible by 4, and there can be no sum of two prime numbers in the range we are looking for (i.e. E < 4) where the sum can be expressed by 2 + x. That is the proof of JF's Theorem. I know that is not all of his work he presented, but this entry was long, so I want to add a second entry later. Just to make it more discrete of a resource, and not a mountain of information. Sorry about the LaTeX being shown in gifs, also. Weebly is pretty messy at formatting, and images were to only clean way I found for putting the equations on here. - AF Or Stefan's ParadoxThis problem has really been gnawing at me, and of course Stefan did not respond to the email I sent him, so I cannot work out the issue with the actual originator of the argument. The issue is that it makes logical sense, of the same form as one of my favorite arguments demonstrating the existence of an objective truth: "If the statement, 'no truths exist' is true, then truth exists." You elegantly demonstrate that there is a set called "objective truths" and it always has to be occupied, therefore proving it exists. But Stef's still feels right, but it does not add up on paper. So let us recap: "If the statement 'UPB is wrong' is true, then UPB exists because the statement demonstrates a preference for UPB." Maybe it is a fallacy of four terms:
U <=> UPB A <=> Arguments against UPB D <=> Things which demonstrate UPB V <=> Valid things If this is the case, I can contrast it to the argument I am so partial to: If S, then T S, therefore T. Where: S = Statement of no truths existing T = Set of true things In any case, you can demonstrate that there is a set of truths called objective with an argument of sets:
That is what confuses me. Because, even if I shifted the goal posts to say Stefan's argument contains the fallacy of four terms, we can break it down into agreed upon syllogisms and get the same result from yesterday:
The issue here is that you arrive at a contradiction when you suppose U is V. But you do not get a contradiction assuming the opposite, and the proposition that for some A, V:
I do not have an answer. Because, by implication, it seems that Stefan is relying on the second premise being true to prove the first premise. But the second premise being true would imply that the first premise is untrue. What do you think?
"Oi mate, I figured out how bloomin' morality works, yeah? So you know this problem what wifs been stumping the greatest moinds for millenia, yeah? I solved it, an' all it is is jus a fouckin' 2x2 square an' all. Good fing I'm brilliant enough to figur dis fing out, you're welcome."
If you think you have solved one of mankind's greatest problems, you probably have not. Think of all the crazy technical problems in mathematics. Men spend their entire lives on one field of math, probably never solving that one problem they struggle over. Think about how specific mathematics are, and then compare it to the monumental task of something as ethereal and soft as morality. Let us not, then, assume that our two years of arguing on the internet allows us to make any serious judgement on big problems like this. |
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