The Collatz Conjecture - summary of a proof

[@simonjeffery3246]

Excellent. Using this approach, we can also prove that all prime numbers are odd., and also prove the Riemann Hypothesis and collect a million.

[@silver6054]

Or even more simply: all integers are less than 10^10^222 I've sampled up to 10^444444444 and not A SINGLE counter example!

[@sschmachtel8963]

Oh yeah it is so easy to take a spreadsheet and present something which only I understand . And if this is pretty difficult I feel like God and I dont want to tell anyone how this works because it is a secret. There might be someone that steels the idea. Oh yeah pretty sure you are only jealous, because here are the facts cant you see? I looked at this one week and I have acchieved more than you in 20 years, no wonder why aou write such an angry comment

[@sschmachtel8963]

When will people ever understand that this is not how you would present such stuff. I mean in case you are serious you should stop bullshiting. I hope this guy knows there are bugs in Excel and also in other spreadsheet programs that would mess up with his calculations... ;-D

[@hawkyre]

The proof: Trust me bro and if you don't then I don't care about you.

[@Leslie.Green_CEng_MIEE]

I wasn’t trying to be arrogant about it. The point is this: are you willing to accept some small value, say 1E-20, as close enough to zero to be neglected? Maybe you require 1E-40, or 1E-100. If no value is ever going to be close enough then you will never be convinced. Fair enough. I like the informal definition by Dr Jeffrey Stopple in A Primer on Analytic Number Theory: “a proof is an argument that is convincing”. I was simply saying that this “adequately close to zero” (close enough for practical purposes) is the key to my presentation. If that is never going to convince you then you would be wasting your time continuing. Such a position is not unreasonable. The delta-epsilon reformulation of calculus went from a value becoming vanishingly small to being smaller than some chosen value, in other words adequately small. Ask yourself this: “Is there arsenic in my drinking water?” Nobody should answer “no”. You can only prove the amount is less than some detection threshold. Some questions have non-trivial answers.

[@hawkyre]

@@Leslie.Green_CEng_MIEE And the fact that you haven't found that non-trivial answer to the collatz conjecture means that this is not a proof. You may consider this a practically good enough result, but it is not a proof. A proof is a proof because it "proves" that X is true, this only proves that X is so unlikely to be false that we might as well consider it to be true, but it does not prove that X is true. If this were to be accepted as a proof, anyone with a powerful enough computer could just "prove" or "disprove" anything with enough time without regards to any mathematical proof methodology. About the arsenic analogy, here's why it is invalid: You cannot compare something that is not detectable to something that we don't know whether it is detectable or not. As you said, there is a detection threshold to detecting arsenic in water, but such threshold does not exist in mathematics. There are multiple methods that exist that allow us to prove facts for all numbers (i.e. induction), so there is for us indeed *the chance* that a way to prove collatz is true, we just don't know it yet. And stop telling people to go away if they don't agree with you because that screams insecurity and bad metacognition. You need to start knowing what you don't know, and you clearly have a huge cognitive dissonance with regards to this proof because you're fully aware that it is not a proof yet you cannot get to accept it.

[@Mmmm1ch43l]

@@Leslie.Green_CEng_MIEE lol " are you willing to accept some small value, say 1E-20, as close enough to zero to be neglected?": no “a proof is an argument that is convincing”: correct, yours isn't very convincing though haha "The delta-epsilon reformulation of calculus went from a value becoming vanishingly small to being smaller than some chosen value, in other words adequately small.": this is not correct. sure, "smaller than some chosen value" does come up from time to time, but then the "chosen value" itself is always "vanishingly small" "Ask yourself this: “Is there arsenic in my drinking water?” Nobody should answer “no”. You can only prove the amount is less than some detection threshold. Some questions have non-trivial answers.": haha well true, which is why it would be quite difficult to mathematically prove that there's no arsenic in drinking water, not sure what that has to do with the collatz conjecture though

[@Tony-cm8lg]

@@Mmmm1ch43l I’d like to add that the epsilon delta argument states “ for all epsilon there exist a delta…etc” so it’s not just for some chosen epsilon we can find a delta that satisfies the properties in the argument, it’s for all epsilon, every single one. The choice is completely arbitrary. So it’s very different then saying for a chosen epsilon, which is incorrect.

[@systemchris]

Statistical evidence is usually a sign, but it can never be seen a proof as a proof derives from truths There are examples where we had evidence that a conjecture was true, but actually it was eventually proved that with numbers bigger than 10^40 the conjecture breaks - the joys of infinity (the mertens conjecture is one)

[@Leslie.Green_CEng_MIEE]

Chris, I agree with you about statistical evidence. Terrence Tao’s paper “Almost all orbits of the Collatz map attain almost bounded values” is a statistical analysis, but frankly I find it almost unreadable! I claim my offering is not a statistical approach. The statistics I am using are quantized into the binary “exists or doesn’t.” Would you ever accept that in the widget barrel example, a barrel could be deemed “pure” with adequate confidence by sampling alone? I am aware of the example you cite, which is why I went to the absurd level of 4E371641. If the lowest counter-example is above that level then it is beyond relevance for practical purposes (as far as I am concerned at least). I use Mathcad 15 for some mathematical work. The built in zero-threshold is 1E-15. For me, if |(A- B)| < 1E-50 and A > 1,000,000 then I would be happy to say A = B. I wouldn’t use a “three-bar identically equal to” sign, but neither would I use an “approximately equal to” sign. I am sure some would object “philosophically” to such usage.

[@systemchris]

@@Leslie.Green_CEng_MIEE yeh I think that the tools or ingenuity to resolve this as a complete truth aren't available yet... I reckon the Riemann Hypothesis will be proved before this lol

[@mathematicsman7454]

other important thing is that probability is certainly not always give certain answer

[@miloweising9781]

The Dunning-Kruger Effect is strong with this one

[@guteksan]

No, it's not. If you'd watch this video carefully, you'd see he's clearly aware of the shortcomings of this "proof", he's just willing to accept them for "practical purposes". So don't judge him so hard, he's made some good observations, though clearly his proof is not acceptable from mathematical perspective.

[@alonyouval3452]

He doesn't prove anything, in his example with the barrel he said that he shake it. But the numbers on the number line are not beeng shaken, and numbers shows us in the past that they behave very different on big scale. He just making a guess.

[@sk4lman]

One must be careful with extrapolating from incomplete data. Look at skewes number for example. Infinity is very, very big. No matter how far you go up the number line, you will always be zero percent done.

[@Utesfan100]

With a PhD in Pure Mathematics and a B.S. in Mechanical Engineering, I enjoyed your presentation. Some comments: 1) I like the sketch of your argument that IF there is a counter example, there must be a definite fraction of counter examples. I would state this by saying that the set out counter examples is measurable. 2) Another way to view your result is that the demonstrated null result up to 10^20 bounds the value of the definite fractions of counter examples. This means the proportion must be vanishingly small, unlike the three known negative cases (Is there a 4th?). Like the mass of a photon, smaller than anything we can measure does not assure that it is 0. But we can bound it to less than 10^-51 kg - which is more of a statement of how accurate we can measure than the nature of the photon. "Bounding the photon mass with cosmological propagation of fast radio bursts." (2021) Wang, Miao & Shao 3) This probabilistic argument, and others like it, is why most mathematicians who have looked at it BELIEVE this is likely to be true. The issue is that the bar is to KNOW that this must be true. 4) Probabilistic arguments can be found in pure mathematical literature, even in the discrete field of combinatorics. One example is the proof of the existence of certain Steiner systems by Peter Keevash. Here he shows that the known infinite number of possibilities allow only finitely many counter examples, so one must exist. "The Existence of Designs." (2014) Keevash The rules of measure theory must be strictly followed for such an argument to be acceptable to the mathematical community. (though certainly not all mathematicians. Strict constructivists exist.) 5) Again, I enjoyed your presentation. You expressed your thoughts clearly and qualified your actual results properly. My only critique would be that this is not a proof, but rather a highly compelling argument in favor of the Collatz Conjecture. Indeed, unlike the title, you did make this clear in your closing remarks.

[@Leslie.Green_CEng_MIEE]

@Ben Prather, Quote Ben "2) Another way to view your result is that the demonstrated null result up to 10^20 bounds the value of the definite fractions of counter examples. This means the proportion must be vanishingly small, unlike the three known negative cases (Is there a 4th?)" (I didn’t find a 4th). I don’t see the bound on the counter examples that you do. For me, once the stray table starts, it has to fill the available space just like the structure table does. I see it like a triangle, pointy end down, growing up to infinity. Then the stray table starts from some elevated position, again going up. Starting from 5 or from 1E20 is irrelevant when you rescale the drawing. They are both “almost at 1” when dealing with numbers with 3,000,000 decimal digits.

[@featherton3381]

This is interesting. Let me see if I understood the video and attached documents correctly. Let's call any positive integer whose Collatz sequence terminates at 1 "Collatz" and every other positive integer "non-Collatz" and let S be the set of non-Collatz numbers. The conjecture is that S is empty, while you correctly surmise that S must be infinite if the conjecture is false. Your idea (using numerical evidence with negative Collatz sequences) is that if the conjecture is false, not only will S be infinite, but it will likely have a substantially positive natural density (say d > 0.01). This is how you are able to test the Collatz conjecture for stupidly large numbers. You choose some stupidly large number N (like N = 10^37163) and choose a minimum natural density d, say 0.01. If there are at least 0.01N non-Collatz numbers from 1 to N, then if you take 10000 random samples from that interval, there's a less 10^-40 probability that every number in the sample will be Collatz. So after taking that sample and finding all Collatz numbers, you can then conclude with extremely high confidence that less than 1% of numbers from 1 to N are non-Collatz. You then return to your earlier assertion that the natural density of S (if S is non-empty) should be substantially positive. This would imply (unless all elements of S are larger than the stupidly large N) that at least 0.01N numbers less than N should be non-Collatz. You conclude by saying that for all practical purposes, this implies that the Collatz conjecture is true. Is that about it?

[@Leslie.Green_CEng_MIEE]

@featherton Yes, that is a reasonable summary. I would add a few comments though. I am working in terms of *local density* , so the density above the minimum supposed counter-example (rather than the density for all values starting from 1). _That is important_ . Also the negative Collatz response is *confirmatory* rather than definitive. *Symmetry is the definitive argument.* Values in the stray table (if it exists) follow the same equations as for the main structure table. Since the main table expands out infinitely, the stray table must expand out in the same way. Admittedly I came to this conclusion having seen how the negative Collatz behaved. You could just say I was just a bit slow to realise the importance of the in-built symmetry! The _local density_ above the lowest counter example should be at least 50% (and more if there is more than one loop). I have allowed down to 1% numerically to allow for localised clumping which I supposed could exist, and which is indeed demonstrated in the negative Collatz results. I can’t justify that figure analytically though. My current results are at 10 million decimal digits with 130 points tested. That is adequate at 50% local density, and at 10%, but not at 1%. At 40 hours per value on a fast computer, more digits than 10 million are not very practical (for my setup). Since the iteration is not memory intensive, raw clock speed and clock cycles per operation are the key metrics for computational speed. Obviously it would be nice to have a server farm available so instead of 1 point per 40 hours I could get hundreds per 40 hours. Even so, 500 million digits seems well out of reach. The practicality aspect is this: Suppose a given specific stated counter example with say 500 million digits. If it cannot be tested numerically I claim that it is of little practical interest. Of course it would be of interest to construct a fast 100,000 bit adder (in an FPGA) to make a specialist Collatz Engine to test these infeasibly high numbers!

[@stuartmcshitz5758]

I caught your video and thought you might want to look at what I did with the conjecture and my theory. I just did a short video to document it and protect my intellectual property. In my new theory, I rewrote the conjecture as Xn+Y for the odd integers. The value of "N" can still be any positive integer. As long as the sum of "X" and "Y" total a positive integer ending in 2, 4, 6, or 8, the conjecture will terminate. "3n+1" is merely a single grain of sand on a vast beach. Herein lies my problem, I have no way to test my theory.

[@HeavyMetalMouse]

Ultimately, you have two problems here: 1) You speak in terms of 'confidence' and 'evidence' about mathematical statements. That isn't how anything works in math. Nobody has shown that the Conjecture is *impossible* to prove, they just haven't found a way to do it yet; nobody has proven that things like induction or contradiction don't work on it, they just haven't found the right logical pathway that make such tools work. There is no reason to think there *isn't* some proof of it being true or false available to find. Your contention that any level of evidentiary confidence is 'equivalent to certainty' for a mathematical statement is unworkable; we have counterexamples where a statement holds as true for inputs of up to 10^400 power or more before you find a counterexample (see the li(x) vs pi(x) inequality, for example). It doesn't matter what finite value you pick, even if you did an exhaustive search rather than a sampling, all you've proven is that there aren't any counterexamples *up to that point*, a vanishing nothingth of the infinite popssibility space. Math does not care if you have 'confidence' that a statement is true, it cares whether it is actually true or not; no amount of confidence will stop a counterexample from showing up in a 10^5hundredbillion range, for example, if one exists, because quite frankly, that's still within a nothingth of the infinite size of the numbers. Your method is flawed from the start in this sense. 2) You *do* have an interesting idea for a method in looking at the theoretical density of the Stray Table if it exists, but you fumble the ball by using it as a means to try some kind of 'evidence based sampling proof'. However, there is a useful direction this might be taken! One thing you fail to do is prove that your Stray Table, if it exists, has a *non-zero* natural density on the natural numbers. You immediately jump from "It is finite in size, therefore it has some fraction as a natural density', but you do not make any effort to calculate, or even approximate or form lower/upper bounds on what that natural density might be. In fact, your Stray Table seems to have a lot in common with your example of the square numbers, which have a 0 density despite having infinite examples. In this case, if P is the smallest counterexample, then, trivially Px2^k for all integer k>0 also form counterexamples - however, the natural density of numbers of that form is *zero*, that is 'almost no numbers' are a fixed given multiple of a power of 2. And yet there are infinite such numbers that are fixed given multiples of a power of 2. You would need to do some more work to show that the extension of the table into its 'further infinities' somehow creates a nonzero density, but it is highly possible that it simply does not - that requires derivation and proof, not sampling. If we take the 'worst case scenario and assume that, if a Stray Table exists, its natural density among the natural numbers tends to zero, then it makes absolutely perfect sense that you would *never find a counterexample through sampling*, since as your range gets larger and larger, the fraction of samples that would catch a counterexample would decrease *faster than your sampling rate increases*. However, a method that uses natural densities might have some value, if we can find some way to show that, if a Stray Table exists, it *does* have nonzero density, and then use that in some manner for a proof by contradiction. Likewise, if you could derive (either exactly, or to within some bounds) the density of the Structure Table, that would go a long way to an actual proof. Proving the Structure Table has density less than 1 would basically be a proof that the Conjecture is false. Proving a lower bound for the density of the Structure table, and a lower bound for the Stray Table, and showing that those values are incompatible would be a positive proof of the Conjecture. (if the Structure table has a density of 'at least 75%', and the Stray table, if it exists, ends up with a density of 'at least 30%', for example, then that adds up to 105%, which is nonsense, and would act as proof by contradiction - that if there are counterexamples, there are somehow enough of them that the union of the set of them and the set of working values, is somehow larger than the number of natural numbers. These lines of thinking sound like they have some interesting possibilities - the difficulty now is in figuring out how to calculate the natural density of each of the Tables. Ultimately, your insistence that "Because this problem is difficult, a confidence level is 'good enough'" only makes your 'proof' as good as any other Conjecture - it isn't a proof. That isn't how proof works in Math, and has nothing to do with the 'opinions' of 'pure mathematicians'. It's a matter of definitions - a confidence level is 'good enough' in physical sciences because that's all we *can* do in the sciences: observe, model, hypothesize, test, and refine. But Math doesn't work like that, and never has. Statistical proofs exist, but they don't work via sampling the domain - they work as suggested above, by showing that making certain assumptions creates statistical impossibilities (the resulting probability of some event being negative, or greater than 1, etc), for example. Not merely statistical improbabilities.

[@Leslie.Green_CEng_MIEE]

@FlexNeven, You make some very good points. Initially when I first realised that there had to be infinitely many counter examples (if any) I had no idea if the infinity of counter examples actually had null density. I gained confidence in a finite density by virtue of experimentation with the negative Collatz sequence. Both positive and negative Collatz have structure stables, and in the case of the negative, two stray tables. That is an example but not a proof. But then I realised that since the stray table has exactly the same composition as the structure table, by symmetry it must work in the same way. Symmetry is a very powerful mathematical tool but can be misused. An example is on the Monty Hall problem. It is tempting to say that there are two boxes left, so each has equal probability. That leads us astray since they are not equal probabilities. So an apparent symmetry can be misleading, and I do accept that as a possible error on my part. The ‘confidence’ aspect is almost irrelevant in this respect. If the stray table forms a reasonable fractional density (> 10%) we will certainly find it by sampling. I give the statistics for this in the evidence.zip file linked in the video description. If counter examples exist beyond say 100 billion digits then they are practically irrelevant, as even if told the exact way to create such a counter example, you could never test it. It is computationally infeasible. If, given a theoretically definite counter example, it cannot be falsified by direct computation, we are into the realms of philosophy rather than maths. FluxNeven quote: “a confidence level is 'good enough' in physical sciences because that's all we can do in the sciences: observe, model, hypothesize, test, and refine. But Math doesn't work like that, and never has” I have an interesting counter-example to that. This is from “An Introduction to the Theory of Infinite Series” by Bromwich, dated around 1907 (recent reprint) Ch. XI “Having regard to the fact that Euler and other mathematicians made numerous discoveries by using series which do not converge, we may agree with Borel in the statement that the older mathematicians had sufficiently good experimental evidence that the use of such series as if they were convergent led to correct results in the majority of cases when they presented themselves naturally.” Experimental work on iteration sequences falls into the sub-category of Probabilistic Number Theory. When I was writing a paper on the density of rational numbers I knew that for the first H natural numbers there were H² rational numbers created, but only 60.792% of them had coprime numerators and denominators. I have known that for years. Recently I found out that Cesaro found the constant as 6/PI² way back in 1881 (from Probabilistic Number Theory by Dr Jorn Steuding). Experimental maths can lead the way to suggesting further work.

[@hihoktf]

I wish you well, but I doubt mathematics will accept a statistical proof. I saw where you compared it to the delta-epsilon formulation of calculus, but my preference is to be clear that taking the derivative is the same as declaring an instantaneous slope where there may be no such thing as an instantaneous slope. It is a very useful approximation, but it is an approximation nonetheless.

[@Leslie.Green_CEng_MIEE]

I don’t consider the Collatz sequence to be random at all since every odd starting values fits nicely into the Rank table with a definite position, and a known number of steps to the next odd value. I have (publically) reviewed a paper which uses a “statistical approach”, and called it out as being statistical and therefore not valid. I claim that mine is not a statistical approach in the sense of eventually finding an individual path which terminates. That is an unworkable strategy as the sequence could just loop or run off to infinity. What’s different? Quite a lot actually. That was the intention of the widget barrel problem. To set the scene. We know that induction and other traditional proofs fail on the Collatz. So a new approach is needed. If 1% of iteration starting values (well above 1E20) fail to terminate (for any reason -looping or otherwise) they would be easy to find by sampling, in the same way that defective component batches are found using AQL sampling. I am using a statistical test of course, but not applied to individual sequences, and only as a go/no go indication of the presence or absence of any non-terminating paths, since any one expands to infinitely many. I believe primality testing of large numbers also has a statistical basis in terms of the number of tests run. More tests = more confidence.

[@Mmmm1ch43l]

@@Leslie.Green_CEng_MIEE "We know that induction and other traditional proofs fail on the Collatz": no, we do not know this, we haven't yet been able to do it, that doesn't mean it's impossible "same way that defective component batches are found using AQL sampling": unfortunately, this does not exist in (pure) mathematics "I believe primality testing of large numbers also has a statistical basis in terms of the number of tests run. More tests = more confidence.": correct, which is why these kinds of tests are used in applications but not accepted as proofs in number theory (not matter the "confidence level")

[@Leslie.Green_CEng_MIEE]

@@Mmmm1ch43l Why is induction known to fail on Collatz? You start from some initial starting value say S which is assumed to terminate. You go to some next value which could be S+1 (or S+2) to prove that if S terminates then so does S+1. But there is no general form for S+1. I have explored this is some detail (without any of the restrictions which you have highlighted here.) I can create iteration chains of arbitrary length (according to an integer parameter, say n) which iterate up (on average) for the first n steps. There are lots of different ones. Since there are arbitrarily many types of these starting values you are never able to cover all possibilities for the S+1. The lines above will not be convincing, but just look at a few of the formula (rigourous proofs of which are given in the paper) and you should understand the point. lesliegreen.byethost3.com/articles/hybrid.pdf

[@Mmmm1ch43l]

@@Leslie.Green_CEng_MIEE all you've done is convinced me that *you* are unable to proof the collatz conjecture via induction, however that doesn't mean it's impossible on the other hand, I do believe that if we ever manage to actually proof the collatz conjecture, then it will involve some more advanced methods (which you clearly know nothing about hhaha) as a sidenote: "As soon as you say something like “we have an infinite number of …” you have used an improper expression. (Improper in this context means ‘wrong’.) You cannot have a number which is larger than any number. It makes no sense." oh would you look at that, a wannabe mathematician doesn't understand the concept of infinity, how surprising...

[@Tony-cm8lg]

@@Mmmm1ch43l I also think that affirmation of the conjecture will take many advanced techniques so it is unfortunately out of our reach

[@Tony-cm8lg]

I mean in the abstract of the paper you write “without finding any loops and therefore tentatively conclude, on the basis of the experimental evidence alone. that the positive Collatz sequence always terminates.” There is no conclusion here, this is what the conjecture states and without rigorous logical proof there is no conclusion. Though the data you gathered is of course important and I think it’s interesting. Maybe you should try to gather what you have learned and write a proof

[@Leslie.Green_CEng_MIEE]

Tony, I am re-running the sampling with more randomised starting values. By the end of today i will be able to say "I therefore claim with greater than 99.99% confidence that no counter-example for the Collatz Conjecture exists below a number with 371,641 decimal digits." I have gone further, but need more points. At 9.7 million decimal digits it took 15 hours to iterate one starting value down to 1. Starting value digits to elapsed time is scaling as a square law, so further data gets progressively harder to come by.

[@Tony-cm8lg]

@@Leslie.Green_CEng_MIEE Yeah, it’s obviously going to get computationally harder as you increase the number of digits. That’s why you should true and take your observations and formulate a mathematical proof!

[@Leslie.Green_CEng_MIEE]

@@Tony-cm8lg, Easier said than done. If the principle of symmetry between the stray table and the structure table is not accepted, then summing the Levels individually becomes a major issue since the Levels become increasingly abundant, and increasingly populated, as one progresses up the number line. See page 12 of v1.7 lesliegreen.byethost3.com/articles/structure.pdf

[@Tony-cm8lg]

@@Leslie.Green_CEng_MIEE Trust me I know, I’m a mathematician, but I don’t work in number theory at all so I enjoy watching others do it

[@MrBarbacamanitu]

@@Leslie.Green_CEng_MIEE not finding a counter example is MUCH different than finding a proof.

[@jones1618]

Sampling won't prove it. I think Numberphile's video on Collatz mentions that there were other conjectures that were believed to be true & no counterexamples found for decades until computers made it practical to test higher numbers and found a counterexample in the 2^18 range. Here's a thought experiment to show why sampling makes no sense: Let's say my conjecture was that there are no powers of 2 about 10^12. You could sample half a trillion numbers above that, find no power of 2 and conclude there can't be one. Would you be right?

[@Leslie.Green_CEng_MIEE]

@Stephen Jones Stephen, you missed the key issue about _natural density._ Almost no natural numbers are powers of 2. *Therefore sampling will not find them.* Contrast that with even numbers. Around 50% of natural numbers are even. Sampling will certainly find even numbers. The difficulty is that probabilistic number theory is an optional subject at college level so is not well known. This is why I spent time explaining that we expect at least 50% of starting numbers to fail to terminate if sampled above the lowest non-terminating value. 2^18 is a tiny number. I have been sampling above 10^(ten million). That is impressively large, and each point takes 42 hours to iterate down to 1. It is true that you cannot find members of an infinite set having effectively zero natural density. *Finding members of an infinite set with 50% density is not so hard (if they exist).*

[@jones1618]

@@Leslie.Green_CEng_MIEE OK, maybe demonstrate the density you're talking about. Example: Use high-rising number 27 as a stand-in for an infinite sequence or loop. What percentage of even numbers above 27 fall into 27's sequence? Maybe your sample range would be 30 to 1,009,232 since 27s peak is 9232. Calculate the density. Double the range & calculate again. Double the range & calculate the density a third time. Use 3 data points to forecast density as numbers get bigger.

[@Leslie.Green_CEng_MIEE]

@@jones1618 27 is an ‘interesting’ choice. Since *3 | 27* , all values which iterate through 27 are given by 27 x 2^n, for natural n. This is an infinite but calculably null density set.

[@jones1618]

@@Leslie.Green_CEng_MIEE So, you aren't willing to demonstrate your assertion that the "density" of higher multiples of two leading to this sequence mean that sampling will inevitably find an infinite sequence (if it exists)? It's easy to talk about math. I'm challenging you to do some math. If 27's sequence was the infinite or looping unicorn sequence and you sampled the million numbers above this sequence, what are the chances you'd hit an even number that would fall into this sequence? That would help prove your premise that sampling is a worthwhile way to prove Collatz.

[@Leslie.Green_CEng_MIEE]

@@jones1618 In response to your “challenge” (sampling above Point). Point = 27 3 out of 200 = 1.500% 3 out of 400 = 0.750% 4 out of 800 = 0.500% 5 out of 1600 = 0.313% 6 out of 3200 = 0.188% 7 out of 6400 = 0.109% 8 out of 12800 = 0.063% 9 out of 25600 = 0.035% 10 out of 51200 = 0.020% 11 out of 102400 = 0.011% As I said previously, this is a calculably null set since you ask for value which iterates *_through_* the Point 27. It is calculable because the only such starting values are 27x2^n, for natural n. *I do not claim that such a point generates a non-null infinite set.* Go to the video from time around 4 mins. P is defined as the lowest counter example (although sadly I do not say so in the video). It means the other values on the page are all greater than P. Notice that the column of 2^n multiples above P do not have values coming in from the left. This is equivalent to P being a multiple of 3. It is not the infinite chain of values directly above P which gives a non-null outcome density. It is the infinity of infinite chains shown on that sheet (page 5 of the video) and beyond which gives the non-null outcome density. It is of course impossible to calculate since the number of terms in the summation increases as the starting value increases. How then can I dare claim that the outcome density is not null? I use what is arguably the most powerful and basic concept in all of mathematics: *symmetry.* Since the stray table uses the same equations as the structure table, the stray table must grow in the same way as the structure table. The Negative Collatz termination outcome plots merely _demonstrate_ the principle in action. To demonstrate how incalculable the Collatz is otherwise, take a look at the plot on page 6 of the Negative Collatz Sequence paper (version 1.25). The Levels fill at different rates, with the higher Levels filling at faster rates than lower Levels. But each Level still has relatively few starting values in it. The total coverage is achieved by increasing the number of Levels as well as their fill rates.

[@giannipiccioni8411]

This is not a proof. "ah, maybe these pesky pure mathematicians will dislike this proof, just ignore them", that's not a proof. The stray table could have values all above 10^100^1000. You don't know that

[@Leslie.Green_CEng_MIEE]

Gianni, "The stray table could have values all above 10^100^1000." Yours is a totally reasonable statement. I agree entirely. I have "dumbed down" the conclusion from "any counter example exceeds 1E100000 with high confidence" to "it is proved". Guilty. That's what engineers do. Guilty. If you ask metrologist if a voltage is within a tolerance of say 100ppm they often say things like it is within 50ppm with a 95% confidence interval (coverage factor of 2). Something like (1 + 2^k ) | (1 + 2^(3k) ) is easy to prove rigourously. Collatz is pretty hard, and I have shown the best that I could achieve, given that I found the result and the method interesting. I do claim to have demonstrated that the existing brute force testing will NEVER find a counter example in your lifetime as computing power would need to improve by hundreds of orders of magnitude. I find such a result relevant, and practical.

[@nothing8640]

I have some thoughts but I’ll start off by saying this is a reasonable talk. I’m not convinced by the proof but your statement of the problem and beliefs are very clear. 1) You state a new type of proof is needed because contradiction and induction haven’t worked. That’s not true. Fermat’s last theorem was proven by contradiction 400 years after its statement. Similar to collatz there was lots of numbers checked and lots of attempts before that. Just because it hasn’t been done yet doesn’t mean it won’t work for the problem. 2) You state the collatz conjecture is true “for all intents and purposes” and bring up engineering. Your intents and purposes are far different from mathematicians! The reason we solve hard problems is for the new techniques developed. Again I refer to fermat’s last theorem - the proof expanded areas of maths substantially beyond just knowing the theorem is true. The “intents and purposes” are the proof technique, not just it being true for a bunch of examples. 3) Your probability has no rigorous foundation and it’s leading to errors in judgment. Having 10^400000 samples or whatever is insufficient because there are infinitely many natural numbers. Moreover its not a random sample whatsoever, how would you samples randomly from an infinite sample space? What you’re calculating is not a real confidence level or probability. The leap from 10,000 widgets to infinite natural numbers is unfounded, you can’t think about it in the same way. I think you’ve missed the point on *why* mathematicians aren’t convinced by your proof. We can brute force insane numbers for Riemann’s hypothesis, collatz, FLT, and many other problems. And many mathematicians are satisfied with saying “its probably true” and spending their time doing other things. But the ones who still pursue these problems do it because of the challenge of pushing mathematics research forward. That requires a proof, not a heuristic argument.

[@Leslie.Green_CEng_MIEE]

Nothing, you make some good points, so thank you. In term of Collatz being true for engineering purposes, let me explain. If you tell me there is a counter example having one billion decimal digits, and you tell me what the number is, I would not be able to verify it. The clock speed of available computers is not fast enough to iterate the value down in time. I am currently iterating numbers with 6 million digits. Each single number takes around 12 hours to terminate. The iteration time is increasing as a square law with digits so the problem becomes computationally infeasible. In (3) I believe you have misunderstood what I have done at 1E400000. Probably I didn’t explain it clearly enough. I have in any case remedied that situation by publishing the method, results, and analysis in a separate zip file (evidence.zip) which is linked in the description of the video. I agree that searching 1E400000 numbers by sampling to find one counter example is ludicrously infeasible. The point is that I searched above 1E399999 looking for 50% of counter examples. In any group of say 1,000,000 starting points above the non-terminating value I expect >=50% of non-terminating values, as seen in the negative Collatz. It is trivial to find such a high density of counter examples. My only heuristic is symmetry. Negative Collatz has three possible iteration outcomes and their density is 33% for each. I had already proven that the (positive) Collatz sequence either had no counter examples or infinitely many. The only question was "how big an infinity?". The infinity is the same size as the even numbers in density terms. Asking the question (computationally) “are there any even numbers with 1 million digits?” is not difficult to answer. (Obviously we could just check the last digit in that case, but no such test exists for Collatz). The point is that sampling numbers with 1 million digits randomly, you would soon be convinced that even numbers did or did not exist. Obviously I am looking for something which in all likelihood does not exist, so that is hard. There is another relevant point about proofs. If the proof is “unreadable” is it useful? Prof Tao’s paper ("Almost all orbits of the Collatz map…"), lauded as a major contribution, is almost unreadable to a non-professional mathematician. In these days of vaccine hesitancy, where there is a chance the vaccine could kill you (but thousands or millions of times greater chance of dying from the virus) communicating a proof in an understandable way is arguably part of the job. I wonder how big a number has to be in order to be “insane” for you. I thought 2^1234567 was pretty insane. Having now been up above 9 million digits, I am long past my “insane” threshold!

[@PeterBarnes2]

​@@Leslie.Green_CEng_MIEE Perhaps the fact that it's difficult to argue for a meaningful sense for the practical, engineering value of Collatz examples (distinct from the conjecture) demonstrates the lack of any need for 'non-professional mathematicians' to understand professor Tao's paper. If communication is valuable for what is being communicated, like with vaccine hesitancy, I find little concern for lack of complete communication of higher mathematics at this time. What if I told you that no number was insane? What if I was so cynical that I didn't care about the Collatz conjecture unless it generalized to some transfinite case where it must continue to work, in some sense, for infinite ordinals? Would that be more insane of me, rather than the number? There is little value in Collatz heuristics unless they lead to a non-heuristic proof (for or against, of course). Plain and simple. You want to peddle heuristic arguments as proofs, try the Riemann Hypothesis; at least then there's a prize attached.

[@Leslie.Green_CEng_MIEE]

@@PeterBarnes2 Assume, for the sake of contradiction, that the Collatz sequence does not always terminate. In this case there will be a least value, p, which does not terminate (according to the well-ordering property of the naturals.). The least value will form an infinite set of larger non-terminating values since the stray table is formed using the same iteration equations as the terminating values. This infinite non-terminating set is a fractional part of values above the lowest non-terminating value (in the sense of local density), with 50% of such starting values as the limiting case for one non-termination outcome (as demonstrated by the negative Collatz sequence). The non-existence of infinitely many counter-examples contradicts the original premise, which is therefore false. The Collatz sequence therefore does not have a counter example up to a number with 10 million decimal digits. (Extending this to all numbers is not justified mathematically.) I think you can communicate complicated mathematical ideas if you try. "Is Q dense in R? " is complicated. My version for school kids is simpler: aplusclick.org/t.htm?q=14013 (but then links to a less simple argument.)

[@PeterBarnes2]

​@@Leslie.Green_CEng_MIEE I also think complicated mathematical ideas can and should be communicated as best and often as possible. It is gratifying to do, helps people with an interest in the subject, and enables people to gain an appreciation and interest they might not otherwise have. Looking at what you seem to have, you could create an acceptable proof along similar to what you seem to be looking at: Say the function X(n) counts the number of integers in the non-terminating set less than or equal to 'n,' The Collatz Conjecture would definitely be false if lim{n->infinity} X(n)/n > 0 But if you could show that that same limit _had_ to be >0, which is to say the non-terminating set must be a positive fraction of the integers, then a simple proof that the fraction which do reach the (1, 2, 4) loop must be 1 would prove the conjecture. Your discussion of the 'Stray Table' is the kind of thing which would be useful for such a proof. I suppose you'd essentially want to show that, as you look at larger and larger numbers above a starting value in such a table, the number of branches grows exponentially, fast enough to counteract the 2^-n density loss for each branch's 2^n main-sequence. While an engineer may be satisfied with 1000 blue widgets, the mathematician _must_ see 9,001 (if it is certain that a minimum of 1000 widgets would otherwise be red). Then the mathematician will _continue_ to look at the remaining 999 widgets asking why they aren't red! The mathematician will be happy to tell you what the probability of the barrel being an all-blue one is, the mathematician _knows_ that there is a chance if there is a chance, and that the problem is only interesting because of what it can tell us of other problems, and on and on like that. The Collatz Conjecture is simply _not_ an engineering problem. It _cannot_ matter how many numbers you check, because there _can_ always be one more. (Unless you manage to prove there to be an upper bound on the smallest counterexample.) You are absolutely right: Extending the result from 10 million digit numbers to all numbers is _not_ justified mathematically. The Collatz Conjecture is not interesting for the idea that numbers may go up and down in funny ways; it is only interesting because it is difficult, which implies developing the tools for a _mathematical_ proof (not an engineering proof) would further mathematics. If that does not satisfy _your_ sensibilities, all that proves is why you are an engineering expert, rather than a mathematician.

[@Leslie.Green_CEng_MIEE]

@@PeterBarnes2 I don’t think we disagree as much as perhaps you might think. From my viewpoint, as an engineer, I prefer something tangible, that can be tested (falsified). If you offer me a definite number with 987 million digits as a lowest counter example, I cannot numerically verify it, and that is non-ideal. For me, testing the computational limits is interesting. How fast a Collatz test engine can be built? community.intel.com/t5/Intel-FPGA-University-Program/Feasibility-of-fast-Collatz-Engine-in-an-FPGA/m-p/1414344/emcs_t/S2h8ZW1haWx8dG9waWNfc3Vic2NyaXB0aW9ufEw3Wk02VkI0SlczWlFYfDE0MTQzNDR8U1VCU0NSSVBUSU9OU3xoSw#M4518 I am also interested in number-theoretic problems in general. For example half of all rational numbers (in the sense of natural density) have a value less than 1, a very elegant result (but of little practical utility!) But you can’t put a rational strictly (numerically) between every two reals as there aren’t nearly enough rationals to do that job (see the link in my last post). To be clear, I don’t claim to be “an expert” in any field.

[@hihoktf]

I haven't looked at your paper, but I assume you accounted for the idea that in your stray table, you really only need a q=p (p1), so that q, r, s, etc. are not strictly necessary to disprove the conjecture?

[@Leslie.Green_CEng_MIEE]

Doug, the interesting point is that if p exists it generates q as a larger (and therefore distinct) value. q generates r, and r generates s, etc. They are all distinct until (or unless) it loops. This is not discussed in the paper about the Negative Collatz sequence. It is in the earlier paper “The Inner Structure of the Collatz Iteration Sequence”, starting from page 20. lesliegreen.byethost3.com/articles/structure.pdf There is rather a lot of detail, hence the need to do a summary! The point is that if a single p exists, then so do points q, r, s, and so on. The power-of-two chain above p is a “small” infinity. All the chains in the stray table, all together, would make a “large” infinity, one which is a fraction of the natural numbers, and therefore findable.

[@hihoktf]

@@Leslie.Green_CEng_MIEE I (think I) disagree. In the loop p-4-2-q, q=p=1; so a defeater to the conjecture could be any (3p+1)/(2^n)=p where p1. Note (for potential "stray" p's) if p=3, q=5; and if p=7, q=11; and if p=9, q=7; and if p=11, q=17; and if p=13, q=5; so q does not appear to have any obvious relationship to p. It is not strictly larger, nor does |p-q| appear to have any regularity or pattern. As I understand it, your contention regarding q, r, s, ad infinitum is only the example of a "runaway" p, which would be a defeater for sure; but any extraneous finite loop, whether it ends at q=p, or r=p, or s=p, etc, is a defeater, regardless how few finite steps that defeater requires. In this way, I think the only infinity necessarily implied in a stray p is p*(2^n).

[@hihoktf]

@@Leslie.Green_CEng_MIEE I looked at your paper and see you defined p as being the minimum possible stray value, so I agree, q

[@Leslie.Green_CEng_MIEE]

@@hihoktf No. p is postulated to be the lowest non terminating value. That makes it a Rank One value. That means it is of the form (4k - 1). It means q is (6k - 1) - where k has retained its value. (If q is lower than p then the initial requirements have not been met.) (6k - 1) > (4k -1) The detail is important., and spelled out in detail in the structure paper. I have proved each of the forms for each of the Ranks, but they are quick to read-around as the proofs are indented and a different colour. It makes a quick read-through less onerous.

[@hihoktf]

@@Leslie.Green_CEng_MIEE Yes, I'm interested in the work, so I will look more at that later. Quickly though, I understand you are saying q is strictly larger than p, but are you saying r must be strictly larger than (i.e. not equal to) p? Than q? Etcetera (s, and so on)? I assume you are not, because that would mean no (stray p) loops would be possible, and it seems you're not claiming that. But at the same time, (again, having not read your paper) finding an r=p, or s=p, etc., would be limiting. Thank you.

[@talonward2494]

The idea behind a probabilistic proof is sound; if you can get the probability that a statement is true ridiculously close to zero, then you will have a result worth sharing. Even more important than whether or not it's (almost certainly) true is how you show that. And this is where the problem hides. See, in order to provide a piece of evidence for the truth of the theorem, you need a well-defined statistical model that covers the ENTIRE natural numbers. If you, e.g., test all the numbers up to a certain value, however, then you're only sampling from a set of measure zero (using natural density). That means you're only describing the behavior for 0% of the natural numbers, when your population needs to be 100% of the natural numbers. Fixing this is hard. Why? Whether or not the backward orbit (the official term for the set of numbers that you call here a "structure table") of a positive integer (congruent to one or two modulo three) has positive natural density is an open problem. (Trivially all positive integers congruent to zero modulo three have backward orbits of density zero.) No one knows how to show this yet (but it's probably true). Yet, even if you knew your counterexample's backward orbit had positive natural density, it still wouldn't be enough. You would need to BOUND that density away from zero, which you almost certainly wouldn't be able to do, since it would be effectively limiting the magnitude of the counterexample. BUT! Even if you could bound the density away from zero, you would STILL run into the major issue that it is impossible to place a uniform distribution on a countably infinite set. STILL! Even if you could place a uniform distribution on the positive integers, just drawing a single sample would have an infinite expected computation time. You'd never be able to finish just picking your first random number, let alone be able to test it for "awkwardness." AND STILL YET! Even if you could show the backward orbit had positive density, bound that density away from zero, place a uniform distribution on the positive natural numbers, and apply the Collatz transformation to your samples in finite time... if you found a counterexample, you might never even know, because it could be a divergent trajectory, and your algorithm might never halt. But not all hope is lost. At least the first step seems true and might be possible, and, who knows, showing the backward orbit of any positive integer not divisible by three has positive natural density might eventually lead to a real (not probabilistic) proof! It's the hope that kills you.

[@Leslie.Green_CEng_MIEE]

@Talon Ward Since you evidently have expertise in this area, perhaps we can pick your brain. In Tao’s paper “Almost all orbits of the Collatz map attain almost bounded values” could you help to translate the title please. For example, would a reasonable understanding be: “At least 99% of starting values iterate to below 1% of their original starting values”? Or can you provide a better translation. Obviously as soon as they iterate to below 1E20 they terminate. 99% terminating would be an excellent result compared to Krasikov’s limit (from 2002) that x^0.84 terminate (which can be seen to mean that almost no values are guaranteed to terminate when dealing with large numbers.

[@talonward2494]

​@@Leslie.Green_CEng_MIEE Terence Tao's paper is very technical and difficult to read in it's entirety, but the title's not too bad to explain. There are two "almosts" in the title, and they both mean different things. "Almost all" is a technical term from probability and measure theory that means that the set of elements satisfying the condition has probability one -- in other words, it precisely means 100%. You just have to remember that 100% of something is NOT the same as everything. For instance, if you were to pick a random real number on the interval [0, 1], then 100% of the time you would get an irrational number, even though not all the numbers in [0, 1] are irrational. The rational numbers, however, are a null set -- a set with measure zero. The irrationals have measure one. I have to note that the measure used in this paper is NOT natural density, which is what you'd expect when someone says "almost all" about natural numbers. Instead, he uses logarithmic density. It's basically the same thing as natural density, but with reciprocal functions (1/x) instead of indicator functions. It puts a little less weight on larger numbers and is more compatible with multiplication, which makes it easier to establish the bound. "Almost bounded" is a term Tao is using that means less than f(n) for any function f such that f(x) goes to infinity as x goes to infinity. So, basically, you can get as close to a constant function as you like, but it can't quite be bounded; it still has to diverge eventually. So it basically goes like this: - Pick any function f(x) that goes to infinity, e.g., log(log(log(log(log(x))))). - Consider the set of all natural numbers n such that the minimum value of the trajectory of n is less than f(n). - This set has logarithmic density one in the natural numbers. Now, in some sense, this result is completely unsurprising and entirely unhelpful. It's exactly what we expect should happen. Why? Because each backward orbit (of a number not divisible by three) branches frequently enough that it should have positive (natural) density. (Note that positive natural density implies positive logarithmic density. Natural density is a stronger condition.) Intuitively, "most" natural numbers have to fall down very far, even if the Collatz conjecture were false, even with infinitely many of both nontrivial cycles and divergent trajectories. So, in a way, the result is sort of obvious and doesn't really say anything at all about the problem. But, again, the importance lies not in the result itself but in how that result is obtained and whether similar techniques might be applied elsewhere. So, I guess my phrase would be "The set of starting values that iterate to less than an arbitrarily slow growing function has logarithmic density one."

[@Leslie.Green_CEng_MIEE]

@@talonward2494 “Terence Tao's paper is very technical and difficult to read in it's entirety …” Agreed! Thank you very much for taking the time to summarise the title of Tao’s paper so clearly. So if I can dumb-down your answer a bit for the likes of me: 99.9% < 100% The function log10( n) is almost bounded. A weaker statement of Tao’s result is that at least 99.9% of natural numbers with not more than 1E19 decimal digits iterate to a value of less than 2E19. Since all natural numbers up to 1E20 are known to terminate (by brute force computation), at least 99.9% of natural numbers with not more than 1E19 digits fully terminate (to 1). But if we play the ‘game’ again with the almost bounded function log10( log10(n) ) we have a number whose base-10 log has 1E19 digits. Since we can play that game arbitrarily often, there is no starting value for which less than 99.9% of starting values fail to terminate. So in fact the claim is that only a null set of counter examples exist, namely a possibly infinite set which nevertheless has no natural density. So, a follow on question: Baker’s theorem is referenced but not used. The word “loop” does not exist in the paper. How are loops excluded in his statistical analysis?

[@talonward2494]

​@@Leslie.Green_CEng_MIEE you've got to be careful. The set of numbers with no more than 10^19 decimal digits is finite, so it has density zero. Plus 2x10^19 is constant. So, the statement that "at least 99.9% of natural numbers with not more than 1E19 decimal digits iterate to a value of less than 2E19" is NOT weaker than Tao's theorem. His theorem only describes what happens "in the tail." In particular, it says nothing about what happens on any finite subset. The reason for the "almost bounded" instead of "bounded" is precisely to include counterexamples, if they exist. In other words, his proof is only doable precisely because it doesn't say anything about the actual conjecture. Going from "almost bounded" to "bounded" would basically be as hard as proving the conjecture itself. And showing that the set of counterexamples had measure zero would be basically proving it, since even one counterexample should have a backward orbit of counterexamples with positive density. I don't actually know how Tao proves his result. I have started to read his paper in the past, but, frankly, I don't have the background for it. There is a KZbin series where they go through the paper claim by claim that can help with that. One day maybe I'll actually read it. However, I suspect his proof would shed very little insight on the actual problem, especially since he says as much in his paper.

[@Leslie.Green_CEng_MIEE]

@@talonward2494 “I have started to read his paper in the past, but, frankly, I don't have the background for it.” If you can’t read who can? I have failed to understand the TITLE even after you carefully explained it :-( Thanks for correcting my misunderstanding. It is much appreciated.

[@PaulHuininken]

"The density of infinity" was an eye opener for me. Thanks

Reminds me of an other guy who claimed that sighting a white cat is a biiig step towards the proof that all ravens are black.

[@rarebeeph1783]

You cannot, necessarily, find an infinite number of needles in an infinite haystack. Some ways of distributing the "needles" may be easy to find many, yes, but there are in some sense "more" ways to distribute them such that literally none of them will be in the first 10^10,000,000 integers; more rigorously, there are exactly the same number of ways to distribute in that manner as to distribute at least one closer to 0. In science, failing to find something provides evidence of its nonexistence, because you can associate probabilities with finding or not finding it given your hypothesis, and consider something true if shown beyond a certain threshold of probability. This does not work in math.

[@Leslie.Green_CEng_MIEE]

@RareBeeph, Agreed, but if the strays exist, then we have two disjoint sets of like-elements. There are certainly a ridiculously large number of arrangements, just like tossing a coin thousands of times, but there are only two events, terminate or not. It is infeasible to not find events of one type if their density is 50% each (when sampling above the lowest counter example). In the evidence.zip file linked in the video description I set the probability to 1%, rather than 50%, to give allowance for clumping (which I did see in the negative Collatz termination outcome plots). I think this is overly generous (pessimistic) since you shouldn’t be “unlucky” every single time when sampling over hundreds of decades.

[@danadnauseam]

This is not a proof. It is strong evidence for the believability of the conjecture, but it is not conclusive. But, the concept of the stray table may be useful. Effectively, it sets out a potential reductio ad absurdum. Can we assume there is a smallest nonterminating number p, can we create an infinite descent?

[@Leslie.Green_CEng_MIEE]

@Daniel Reitman, I did actually define p as the lowest non-terminating value in the earlier paper "The Inner Structure of the Collatz Iteration Sequence". The definition is on page 23 of version 1.72 of that paper. I sadly didn't bring that definition forward into the Negative Collatz sequence paper. Certainly if the stray table exists, it has a least value since p is a natural, and the well-ordering principle requires the existence of a least member. The whole point, which I have clearly failed to express adequately, if that if p exists, it can only be found by probing values above p. I am currently probing with 6 million digit decimal numbers. I have iterated 160+ points, at around 13 hours per point. Several more weeks of probing still needed for this size.

[@danadnauseam]

I think you missed my point. You have a potential starting point for a logical (not experimental) attack on the problem.

[@Leslie.Green_CEng_MIEE]

@@danadnauseam , I see what you are saying, but personally I see no analytic way forward using FMID: brilliant.org/wiki/general-diophantine-equations-fermats-method-of/ Even the first step in Collatz has infinitely many variants in the sense that say an odd starting value can be in one of infinitely many (Rank) sets, R1, R2, R3, R4, R5, ... If a lowest counter-example exists then that is fine. If a lowest counter-example doesn't exist, I can't think how you could start anywhere relevant and then descend analytically. The examples given in the linked article have a very simple formulation, although it has to be said I have never personally used FMID.

[@saikat93ify]

You should enter this in Summer of Mathematical Exposition challenge to get more reach

[@Leslie.Green_CEng_MIEE]

Thank you. It was a good idea so I have entered it 🙂

[@GoblinAlchem]

This is not a proof. The proportion of squares among all numbers is vanishingly small; there are no squares between 9802 and 9999; and yet squares do exist. You are trying to justify that checking a finite subset can prove a statement about all numbers, but it cannot.

[@Leslie.Green_CEng_MIEE]

@goblin alchemist, you are correct to say that it is infeasible to find an infinity which is a negligible proportion of all natural numbers (up to the test limit). From the negative Collatz sequence it was demonstrated that the iteration outcome density was 33% for each of the 3 possible outcomes. In other words 66% of outcomes would not terminate in the ordinary way (ABOVE the lowest non-terminating value) . The stray table and the structure table are symmetric in their construction (just like they are for the positive Collatz sequence). Therefore by symmetry the minimum non-terminating outcome density will be 50%, not withstanding localised clumping. The whole point is to understand the nature of the infinite sets in terms of natural density (or asymptotic density if you prefer). As soon as it is realised (and accepted) that the iteration outcome density is a fraction of all possible starting values, even as low as 5% due to clumping, it becomes trivial to establish the non-existence of such a set using sampling. Perfect squares do not form such an infinite set as you have so correctly pointed out. Generally people are not equiped to deal with different sizes of simple infinities. I have done a fair amount of work on the subject: The New Mathematics of Infinity for Engineers and Scientists, (2018), Green. L.O. (v1.40, 2019): lesliegreen.byethost3.com/articles/new_maths.pdf The Hierarchy of Infinities, (2019), Green. L.O. (v2.10, 2019): lesliegreen.byethost3.com/articles/hierarchy.pdf My LITTLE book of INFINITY: www.amazon.co.uk/dp/B0B1VW7KCH

[@RomanHold]

What do you even want to prove? We got a function conjunction, which always result in a continuous downtrend and when we hit "1" it results in some kind of circle logic between 3 values [4, 2, 1, 4, 2, 1...]. The collatz conjecture is not true for all natural numbers, because this is an undefined hypothesis. You can however insert any natural number into it and given enaugh iterations, will end up in the circle logic loop of 4, 2, 1... , because you do different operations on odd an even numbers which have this essencia or "are constructed in this way", there are polymorphisms of the collatz conjecture with different values, that result in other loops (of different sizes), this is just the most known one. The collatz conjecture is never true to anything or nothing. Because it is just an idea, it is a hypothesis "what if we did this". And the answer is "when you do this, this loop happens, because this loop is what you have constructed" and ofc you could take this loop and reverse it, to climb into the other direction up the number line and roll to infinity, but you will never be able to cover infinity, because there always will be infinite more numbers now matter how far you are in. The thing is "the natural numbers never end", because they don't end and you can always "n+1" the last one, there will always be a new number you can run this conjecture on, to test is. The reason why you can't find a proof is simple. Because sth that doesn't have a proof is sth that goes against the first theorem of the classic logic. A statement is either true or false. Both (50/50) is an undefined statement. If sth is both zero and one, off and on, if it is a superposition, then it's "an undefined statement". How do you proof that an undefined statement is "true"? How do you proof that an undefined statement is "false"? You don't. However you can proof that it is an undefined statement, by homeomorphically showing the congruency to the moebiusstrip. You got a function for even numbers and a function for odd numbers. And you can't make a logical statement of an relation to anything, when you got a "super singularity", because there is nothing to compare it to. To quote Morpheus "there is nothing that can explain it to you". Because the moebiusstrip is a "non-orientable" surface from the inside you can not differentiate between clock-and counter clockwise turns, and because there will always come a time when the collatz conjecture points from an even number to an odd number and vice versa, the congruency to a moebiusstrip is proven. The answer is simple. "what is it that you try to prove"? You tryed to prove "does the collatz conjecture apply for all natural numbers?", what you can prove is "does the collatz conjecture have a congruency to the topology of a moebiusstrip?" and the answer to this one is "yes". There is also an additional proof by concept. A polymorphic line of any shape that has its beginning and end joined, also a circle behaves to a moebiusstrip, homeomorphically like a number line behaves to a coordinate system. Which is why when we open up the complex plain (of number), the imaginary unit "i^2=-1" as the result of a square root, equals the moebiusstrips twist. This means that homeomorphically every time we open up a new dimension to sth, we add another moebiusstrip, conjoined with the prior ones. Thus a not only does (circle:moebiusstrip) have a congruency to (numberline:coordinate system), but (sphere:hyperpheres) is also just the same object with a moebiusstrip put ontop and (number:numberline) makes sth else clear: What differentiates odd and even numbers are moebiusstrips. Odd number are like turning a moebiusstrip once, ending up on the opposite side (now you need to argue more). Even numbers are like turning a moebiusstrip twice and ending up where you started (now you need to argue more... BREAK! Now you have proven the congruency to the collatz conjecture). Intrestingly now you can answer what the original question "does this apply to all natural numbers" really means. All natural numbers unfold into even and uneven to infinity by the "flip flop" swapping of 180° turns of a moebiusstrip pattern (moebiusmap properties) and this is what the first axiom of the classic logic differentiates into true and false statements of divisibility. This is the answer, you original question thus was the mirror image of the moebiusstrip which is the collatz conjecture and when you have a moebiusstrip and "annihilate" it with it's mirror image you get "nothing". Also.. I like eliminating "0"s out of telephone numbers and also prefer "even numbers", because they are "fair", which is the necessity for "la law de la justicia", the law of justice, same to same and different to different things. To escape devil spirals is simple. If there are any "empty spaces (of anything)" fill them. Because a true moebiusstrip has a height/thickness of zero, and bc there is no material object, which has "zero thickness", this means that every paper model of a moebiusstrip is always just "a model", but not the real thing, however a "zero" in a phone number or paying "0€" for your insurance (which is an abnormality, because usually everyone has to pay 20-30€ monthly), is the variable which reverses laws of cause and effect. To put it simple, you cannot divide by zero, which is why you cannot answer the question regarding all natural numbers regarding the collatz conjecture and "from nothing comes nothing", so how is any doctor supposed to help you, of you don't pay him (even if he take your card and your insurance pays him 100.000€, as long as you pay 0€ to the insurance, he will do the opposite of the medical treatment that you really need).

[@Tony-cm8lg]

The lack of mathematical understand here is pretty astounding

[@FrancisCWolfe]

I doubt I am going to listen to all of this but I will say that the animation is not adding credibility.

[@psycho_not]

Didn't watch, but I don't doubt your reasoning. I know however that you dismiss the idea that mathematicians need to prove things 'for every value', but take a serious look into why they do these things in the first place. You may be right and I should have watched, but understand that this problem has been open for a very long time and the chance that you have solved it among tens of thousands of other really smart people is understandably low, good luck.

[@blair2798]

I believe the conjecture to be true. I do not believe the conjecture can be proved using mathmatics. Therefore one is left with one tool to use in developing a proof - LOGIC.

[@Leslie.Green_CEng_MIEE]

Blair, if you think about it, believing something to be true without evidence is known as Faith. Logic, things like Boolean algebra (and Karnaugh maps), is a strict subset of mathematics, physics, electronics, and so on.

[@MobiusCoin]

Logic is math

[@abdelaliabdlali1665]

Thank you my teacher i want to know the question of this conjucter

[@Leslie.Green_CEng_MIEE]

Sorry, I am unsure about your comment. Did you means (1) or (2)? (1) Are you happy that your question was answered? (2) Did you want me to explain more about what Collatz claims to be true?

[@cscback]

working on it for a while. think i found a proof (not full but maybe will help math community), it a sequences. do you interested to check what i found out? and no. im not a kid

[@Leslie.Green_CEng_MIEE]

Thank you, but no. I have publically reviewed 7 published (unacceptable) 'proofs' of Collatz, several by professionals. You can find them here: lesliegreen.byethost3.com/publications.html There is always a potential for conflict of interest. By all means publically publish your work. I don't expect anyone will want to help you to fix up "the last 5%" of your proof.

[@cscback]

@@Leslie.Green_CEng_MIEE thank you. i understand that :)

[@cscback]

@@martiendejong8857 i'll wait

[@cscback]

@@martiendejong8857 interesting i'll be think about it :) maybe we could get something out of

[@cscback]

@@martiendejong8857 hey i wanna contract you maybe we can work this togather

[@jewishjewom12ify]

This is not a proof. If the statement is “the collatz sequence converges to 1 for all inputs”, then unless you have definitely proved beyond a shadow of a doubt that every single (not just up to some finite threshold) number will converge to 1 under the iteration, then I’m sorry but you have not proven the statement.

[@silver6054]

I agree. Even if you had been able to prove (which you haven't) say "If there were counter-examples, they would represent 11.2% of the natural numbers" no amount of sampling would be adequate (as an any "crazy high" threshold, there is an infinite set of larger numbers, which may eventually yield your 11.2% overall.

[@tillokoli521]

It's not even close to a proof

[@pavolgalik9764]

.....For all practical purposes ....uhm..... mathematics is not physics, nor technology, nor philosophy, nor politics, mathematics is a tool to look into God's kitchen. She is ruthless, true, does not care about human decay and her lusts. Whoever ignores it, or perhaps underestimates it, will end up like a poor worm in the farthest corner of the universe.

[@Leslie.Green_CEng_MIEE]

@Pavol Galik If I (somehow) tell you that a particular number with one thousand million decimal digits is a counter example to Collatz, I would argue that this is irrelevant if you cannot test it to see if my statement is true. Mathematics without application arguably IS philosophy. I recently came a across a professional text-book “proof” that you can put a rational number strictly between any two real numbers. I think debunking the proof makes a fun problem for enthusiasts: aplusclick.org/t.htm?q=14013

[@felixjohanschistadjacobsen767]

Your proof is (unfortunately) very flawful. If you don't believe me, due to my lack of specificity, then try to write a REAL article, and try to get it published through a peer-to-peer journal (may I recommend: The Annals of Mathematics). Then it will be considered by REAL mathematicians and not some poor people which only tried to watch some funny cat-videos! I'm sorry if I'm a little too mean.

[@Leslie.Green_CEng_MIEE]

Felix, I would be interested to know which of the 3 points on page 13 of the Negative Collatz paper (linked in the video description) you find to be most flawed. Or is your most major concern with some other point? Specificity is always better than a generality (“it’s all flawed”). Pointing out specific flaws should never be considered as “mean”.

[@MobiusCoin]

@@Leslie.Green_CEng_MIEE This guy copies and pastes this on every Collatz proof video. It's not unique to yours. He probably didn't even watch it. I don't know if your proof is any good or not, I don't have enough knowledge to know what constitutes a rigorous proof and what doesn't but I know this is a complete copy paste.

[@pauljlund]

The author obviously doesn't understand what constitutes a proof.

[@Leslie.Green_CEng_MIEE]

lesliegreen.byethost3.com/articles/proofs.pdf

[@pauljlund]

@@Leslie.Green_CEng_MIEE You haven't proved anything by (i) contradiction, (ii) induction or (iii) cases.

[@Leslie.Green_CEng_MIEE]

@@pauljlund Wikipedia lists 11 _types_ of proof: en.wikipedia.org/wiki/Mathematical_proof

[@pauljlund]

@@Leslie.Green_CEng_MIEE Which type have you used?

[@Leslie.Green_CEng_MIEE]

@@pauljlund , an excellent question. Let's consider the proof in steps: (1) If a single Collatz counter-example exists then there are infinitely many counter-examples. *_Direct Proof_*

[@tillokoli521]

Worthless

[@robertashton8942]

Waste of time. This is not good at all and in some cases actually wrong.

[@Leslie.Green_CEng_MIEE]

@robertashton8942 "... and in some cases actually wrong" Robert, by all means present a specific wrong case or cases. The generality you have presented is untestable / unverifiable.