 As I make this video, I am on the traditional unceded territories of the Comox and Qualicum First Nations. I acknowledge that and I thank them for the use of their lands. A couple of months ago, I made a video about the Colatz Conjecture. And in that video, I essentially proved that if you had an integer that was within the tree, it was within the tree and it was connected to all the other integers that were within the tree. Which is great but it doesn't really address the Colatz Conjecture because the problem with the Colatz Conjecture is you're trying to prove all integers are in the tree. And if you just prove that the integer you're looking at is in the tree, you're not actually proving it for all the other integers. This was pointed out to me through comments by a number of individuals. One in particular goes by the name of Scruffy Egg who was particularly good at explaining to me where my deficiencies were and how I could improve them. And what we came to at the end was that I should probably be looking for loops. I should be looking for ways to prove that all integers were in the tree. Okay, the whole plan here is we're going to construct the Colatz Tree from integer equivalence groups of modulo 6 and show that the only loop in the tree is 1, 4, 2, 1. And that the tree contains the set of all positive integers. So starting off with the Colatz even rule. If you were wondering before about what equivalence classes are, well 2n is the even equivalence class. 2n plus 1 is the odd equivalence class. You can divide things into even or odd depending on whether they have a remainder of 1 or no remainder when you divide by 2. That's what an equivalence class does. We'll look at that for a modulo 6 in a second. Right now we're just looking at the Colatz even rule. So Colatz even rule says if I've got something that's even, 2n, I'm going to divide by 2 and get n. That's pretty straightforward. But I want to construct a tree so I want an inverted Colatz even rule and what that says is if I start with n which is any integer I do the inverse, I'm going to multiply by 2n, not divide by 2n, I'm going to multiply by 2n, and I'm going to end up with 2n. So n through the inverse inverted Colatz even rule becomes 2n. We'll do the same thing for the Colatz odd rule. Here you see the 2n plus 1. That's the equivalence class for odd numbers, remainder of 1 when you divide by 2. The Colatz odd rule says I'm going to multiply the starting number by 3. So I've got 3 times 2n plus 1 which is 3 times 2n which is 6n plus 3 times 1 which is 3. So I get 6n plus 3 and then I'm going to add 1 to that and I get 6n plus 4. So if I start with any odd number I'm going to end up with an equivalence class of 6n plus 4. 6n is part of the Modulo 6 equivalence class. But again because I'm trying to construct a tree I don't want to do the direction this way. I want to reverse it so I'm building up from the bottom. And that means my steps would be I would subtract 1 from 6n plus 4 and then I divide by 3 and I'm going to get 2n plus 1. Well that tells me that 6n plus 4 equivalence class integers are the only ones that are going to create odd integers because 2n plus 1 is every odd integer. So 6n plus 4 is going to have a precedence of 2n plus 1. So from these two rules we can see that any integer has even predecessors. So any integer, all of the integers will have an even predecessor but only integers of 6n plus 4 can have odd predecessors. So there we are right now. Remember that. This means that 6n plus 4 integers are the only ones that can have two distinct predecessors which is a requirement for initiating a loop in a tree. Well okay so think about what a loop in a tree would look like. It's going to have two paths that join. And at the lowest point that joining point will have to be something like this. It will have to have two predecessors. So the only integers that we have found that can have two predecessors they have an even and they have an odd, that is the 6n plus 4 integers. And so 6n plus 4 can have an even predecessor and an odd predecessor and that's what we're looking for to initiate a loop. These others can be any modulo class but they're not going to initiate a loop. They might be part of a loop but what I'm looking for is where loops can be initiated because if you can't start a loop in a given path there is no way you're going to have a loop in a given path. So that's kind of my structure here. So now we're going to look at equivalence classes of modulo 6. So if you were confused about modulos I'm going to show you now how they work and how they work is I'm going to put up these numbers. Now I want to sort these numbers into the different equivalence classes and the way I do that is I'll take each of these numbers and I'll divide them by 6 and I'll look at the remainder. So 18 divided by 6 is 3 with no remainder. That means it's going to show up into this equivalence class. 135 divided by 6 is 22 plus 3 which means it's going to come down here into this equivalence class because the 6n refers to I'm dividing it by 6. If you don't care about the 22 it's the 3 that's going to make it into the equivalence class. So now I'll just do a little bit of fancy animation. So we already know that there is a loop that occurs at 1 which means that we will construct the tree based on the value of n greater than 0. We will show later that there's only one loop where n equals 0 and we'll do that by inspection. So what's this about the n greater than 0? If n was 0 then one of the integers I'm going to be looking at is 1. Well I know there's a loop at 1 so I'm not going to get very far proving that there are no loops in the tree when I know there's a loop in the tree. So as a result to get around 1, that 1 that I know has a loop I'm going to take n greater than 0 so if n equals 0 you can see down the side here. If n equals 0 this will be 0, 1, 2, 3, 4, 5. So I'm going to do those ones by inspection later on. But I am going to look at n greater than 0 so that would start at if n was 1 it would be 6, 7, 8, 9, 10, 11. And then so on if n equals 2 then 6n becomes 12 and it keeps going on and on for all the integers that there are. So now I'm going to look at the even predecessors of module 6 and even predecessors are really simple it was what we were talking about. You just take the number and you double it and then you take that number and double it and double it and double it and double it. And what this table kind of shows it shows a kind of some interesting things that you might not really notice right away. One is sort of by definition each of these bottom equivalence classes is going to be distinct. They have to be distinct. That's how they're defined. If you divide a number by 6 you're not going to get a remainder of 1 and a remainder of 5. It doesn't work that way. You get one or the other. That's why it breaks up that way. So these are all distinct across the bottom. But when I double them now I've got double here. Notice here these are all distinct across this row and when I go up the columns they're distinct as well because I'm doubling each time. They're never going to be equal to what the 6n is. So what I'm looking at right now are these values. I could substitute any number in for n and then suddenly this would be populated with actual numbers and they would all be distinct and I'll be looking at the even paths going for module 6. So let's take a look at this because I can find out a little bit more about this right now. Just looking at this a path at a time. If I've got 12n there's another way I can write 12. I can write 12 as 6 times 2. So 6 times 2n, 6 times 4 and 6 times 8n, 6 times 16n. These are not equal values. I'm not claiming that. But they are all going to be in the same equivalence class. So all of these will be 6n equivalence class. Everything in this even path, even Colatz path is going to be a 6n. Do the same thing for 6n plus 1. So look at this 12n plus 2. It becomes 6 times 2n plus 2 and I double it and I get 6 times 4n plus 4 and I double this and because 4 doubled is 8 and 8 is 6 plus 2 I end up having to put a 1 in here for that 6. So it's 6 times 8n plus 1 plus 2 and then I double again and 2 doubled 4 doesn't get me out into the 6s again so I just have to double all these up. So again I end up with the equivalence classes from module 06. 6n plus 2, 6n plus 4, 6n plus 2, 6n plus 4 it continues on that way. So I started with 6n plus 1 then I got to 6n plus 2 and then the pattern starts. We can probably guess if I start with 6n plus 2 I'm going to go to 6n plus 4. I did there, I'm going to do it here. I'm not claiming these values are the same, they're not but they are the same equivalence classes. Something interesting happens with 6n plus 3. I double 6n plus 3 and I get 12n plus 6 but in a 6 module equivalence class I'm going to get 6 times 2n plus 1 and so 6 times that, doubled all the way up all those are just equivalence class 6n. 6 times a number is going to, there's no addition there, there's no remainder that I'm going to be dealing with. It's going to be a 6n equivalence class. So it's like this 6n shows up over here and that's all I get. Once I get to a 6n it repeats to infinity. Once I get to a 64 or get to a 62 it alternates with 64, repeats to infinity. I see that happen here. Now you can see I sort of gave away the game didn't I? So if I do the same thing with the modular classes here 6n plus 4 is going to go to 6n plus 2, 6n plus 4, 6n plus 2 and again on up to infinity. And here, what's interesting here is when I double it I get 12n plus 10 but if I want to change that to 6n 10 is 6 times 1 plus 4. It's going to be a remainder of 4. So I can't do 6n, I can't do 6 times 2n plus 10 I'm going to have to do 6 times 2n plus 1 plus 4 and then of course 4 goes to 2 goes to 4 goes to 2 so I still end up with that same pattern in here. So you've got this really interesting thing happen. These all go to infinity. These are all distinct values. When I put any n in here all these will become distinct values and because I can put any n in this represents all the integers. So remember I'm looking for loops. Well, right away if I'm only dealing with the even predecessors I have no loops. They're just straight lines. They just go vertically up doubling each time. That's how they work. So in the case of these ones, these even ones all they can ever be is there can't be loops in them. They're just going to be every time I hit an even I'm just going to divide by two. It's just on to the next one, on to the next one. I'm not going to be generating an odd if I'm in an even. Now if I'm in an odd that changes things a bit. So here are my even predecessors of module 6 but only the odd ones. So together these represent all the possible odd numbers. 6n plus 1, 6n plus 3, 6n plus 5 you do those equivalence classes you have all the odd numbers. All the odd remainders that you can have. You can't have more. There's 6n plus 2, that's even. 6n plus 4, that's even. 6n plus 7, that's not module 6 anymore. 6n plus 7 is actually 6n plus 1. So I have all the odd numbers represented here at this level. Now let's take a look at the and they go on infinitely. The little dots up here. They go on infinitely. So now let's look at the equivalence classes that they represent. We already figured out that 6n plus 3 is nothing but 6n's. 6n plus 5 alternates between 6n plus 4 and 6n plus 2. 6n plus 1 alternates between 6n plus 2, 6n plus 4. So I have some patterns here. What's really interesting is I have no 6n plus 4s in 6n plus 3. An odd number that's divisible by 3 will have no 6n plus 4s in its preceding even path. With no 6n plus 4s it can have no odd predecessors. These guys over here and these guys over here can. This we'll see in a second. This will generate an odd predecessor. There can be no odd predecessors for odd numbers that are divisible by 3. Just straight lines straight up. No odd numbers. Because there's no 6n plus 4s there can be no initiation of loops. There's no loops with 6n plus 3. Get rid of it. So now I've got 6n plus 1, 6n plus 5. Let's look at what happens at those 6n plus 4 points here. So I'm going to convert back to the values. Again I can sub any n in that I wanted and I'm going to get real values up here. These are all just multiplying by 2. It's all straightforward. When I get to here I'm going to subtract 1 and divide by 3. That's what I'm doing here. Now notice I can go as far up here as I want. This is to infinity. Which means this is to infinity as well. And this is to infinity. And this is to infinity. So what I've got here is I'm looking at 6n plus 1 and 6n plus 5 which are the only initiating classes that can have loops in them. And now I'm looking for where could the loops be. So the construction from any odd integer to the preceding odd integer is really central to this investigation. And I'm going to refer to it in the future as a precedent step. So I'm not going to worry about going past that next odd integer. When I get to a preceding odd integer I'm not going past it. I'm stopping there and I'm only starting from one odd integer and going to the preceding odd integer but as we can see there is a whole host of preceding odd integers. There's infinite preceding odd integers here. And so I'm just looking for the step from here the originating to here and this includes all of these. It has infinite preceding odd integers. So let's take a look at that. Let's take a look at the equivalence classes. The equivalence classes I start from 6n5 6n plus 4 2n plus 1 I'm getting an odd number. 6n plus 4 I'm always getting an odd number. There's always going to be an odd number here. I'm not going to get some other number. They're always going to be odd here. This is significant in the next couple of steps. So what I'm doing here is I'm just swapping out graphically because it's easier to show just how things work than it is to put all the numbers in places. So just to show you this can go on and on and on and on and on and on and on and each step from this one I'm going to get many preceding odd integers. So this is a one to many relationship. The one of course is the originating integer at the bottom. The many are the preceding odd integers up the side. Key point here is that in this precedent step there are no loops. I have a straight line up through the even precedence. I have a single edge across to the preceding odds. There are no loops here. There can be no loops here. If I stay within that precedent step get outside the precedent step I can't say but inside the precedent step there are no loops. Let's flip this on its side because this is where the fact that these are all odd comes in and I'm just going to look at one of them. I could look at either because they're both really based on having an odd a 6n plus 4, 6n plus 2, 6n plus 4 as I go through that's what's generating these odd numbers but this is what's really important. I've generated this odd numbers and there are no loops in here. It's a one to many. There are no loops in here. I stop here but say I wanted to say well what if this one was 6n plus 3? I'll put it there. Now I've added 6n plus 3 I've introduced no loops. There's no possible loops here because there's no loops here and then this is distinct these numbers will all be distinct and there's no loops. Okay, so what happens with 6n plus 5? Well, same thing no loops in here on this base I'm just putting an odd number in here it could be any odd number but right now I've shown if it was 6n plus 3 there's no loops if it's 6n plus 5 again I just showed there's no loops because it's one to many all the way up here and to wrap things up pop 6n plus 1 and same thing one to many with no loops based on no loops each of these and 6n plus 1, 6n plus 5 and 6n plus 3 that's all the odd numbers so any odd number put into this tree that has no loops, this one to many tree is another one to many tree or it's just a list of even numbers if it's 6n plus 3 so essentially I've shown that there are no loops using Agilo 6 equivalence classes to construct the tree but there is a loop at one so I will need to show that there are no loops for 6n, 6n plus 1, 6n plus 2 6n plus 3, 6n plus 4 and 6n plus 5 when n equals 0 I'm going to do this by inspection well before I even inspect for 6n, if n is 0 I get starting point of 0 and 0 actually isn't defined for the Colatz conjecture that's how I get rid of that 6n for n being 0 it's just not defined but let's take a look at 6n plus 1 that gives me 1 6n is 6 times 0 is 0 plus 1 I'm starting at 1 so from 1 I know I can generate all these even powers of 2 up above me I did that before this is just the Colatz conjecture going up the string here now there's my loop so this is just by inspection I just look at it and I go okay so if I started with 4 now this I'm working my way down the tree not back up the tree but if I start from 4 I go to 2, 1 and I go back up to 4 well when I'm building the tree I go 1, 2, 4 takes me back around to 1 so there is the one loop I've got but I'm okay with that because I knew it was there and I just have to prove by inspection that the other numbers, the other odd numbers are not involved in loops I'm not really worried about 2 or 4 and here are some of the other odd numbers I have to get to the odd numbers I can only in this case get to the odd numbers by going up the chain and using these higher 6n plus 4 to get to these other odd numbers notice that this is a 6n plus 5 this is a 6n plus 3 this is a 6n plus 1 this is a 6n plus 5 and these actually repeat all the way up but it depends on where you start from so we're just concerned the fact that these are odds so let's go looking next for the next odd in the initial sequence and that's 3 so I get to 3 by way of 5 and down to 3 so now I've got 3 and 3 of course is 6n plus 3 and that means that it only has even numbers above it so no loops no loops with 3 now I just have to check out 5 so I look at 5 and all the even numbers up above it all the 6n plus 4s generate odd numbers but again all of these numbers and in this case the number is 3 and then I go up this chain here the same way here my first number was 1 and then I went up this chain so this kind of throws you off a little bit as does this but actually if you're actually thinking about the precedent step you're just going up next odd up next odd up next odd and each time I do that precedent step I'm not introducing any loops all these numbers are distinct now by proof by induction for the 1 to many branching precedent steps in the Colatz tree the base case of n equals 0 was just shown and now the inductive steps so I've shown this tree just as a 1 to many and in the case I think depending on it's 1 or 3 or 2 depending on each step I'm I if I really want to do a Colatz tree each of these levels branching points would be infinite and then this branching point would be infinite again and this branching point would be infinite again well in terms of just showing how it works I don't really need to go into the infinities and they're hard to draw so what I'm going to do instead is just show this schematic to give you an idea so I can select any odd p that's not within a loop now in the Colatz conjecture that I'm dealing with 3n plus 1 there are no odd numbers within the loop the loop is 1, 4, 2, 1 1 is an odd number but I'm starting with 1 so I don't have to worry about that but at the end we'll see with other 3n minus 1 for instance which is the negative Colatz tree there are multiple loops so if I select the p and select it within a loop and that p has to be connected back to 1 but basically what I'm saying is if I'm applied m applications of the precedent step from 1 I'm going to end up at this p and this p could be any p in the tree that's not it's odd and it's not in a loop do that and gold so the next step I apply the precedent step now I've applied the precedent step to all these and they go on to infinity all these guys up here they go on to infinity the point is from this point on the precedent the odd numbers that are proceeding this odd number are going to be disjoint from all these other numbers because this number is disjoint from all these other numbers and when you apply the Colatz rules they're very determinative they will only apply the numbers you can't have multiple results from the same number you give it one number you're only going to get one result so now I apply the precedent step at m plus 1 time bang there I am there's my m plus 1 the resulting all predecessors are distinct from the rest of the tree and I've introduced no loops so the precedent step can be done an infinite number of times and that could create a tree with infinite distinct integers a tree with infinite distinct integers contains all the integers so all integers converge to the 1, 4, 2, 1 loop and that's pretty much what the Colatz conjecture says so there you have it that's what I found by constructing the tree and essentially the Colatz tree ends up as a whole series of n to infinity trees that have disjoint edges stacked on each other that means there's no loops and because you're doing it inductively it means all the integers are there at least that's my understanding of what I've done now I may be wrong, I'm not a mathematician so certainly make any comments that you want about whether you think I've gone wrong no, maybe I got it right I think I'll be honest I think it's probably too simple to be right but I can't find holes in it so if you find holes in it please let me know one of the things that came up on the previous video was negative numbers if I put negative numbers in the Colatz formula the 3n plus 1 or the even 1 dividing by 2 put negative numbers in I do get distinct disjoint trees I get a number of loops and that is true but when I do that I'm actually creating a new tree a new form that's equivalent to the 3n minus 1 the tree shapes are the same for 3n plus 1 with negative numbers and 3n minus 1 with positive numbers so what that means is this is actually a different tree than the Colatz tree and that different tree this the logic I've used still applies to it except you have to use modulo 18 because modulo 18 gets you up to cover all the initiation of all the loops the biggest loop is at minus 17 or 17 if it was 3n minus 1 you have to get above that so modulo 18 gets that to you and then the same arguments apply with one other exception that other exception is now you've got a loop when you do your induction step you can't be inside of a loop because in the induction step you're saying these are separate there's no overlap but if you're in a loop some of your successors are going to be equal to your predecessors because you're in a loop you come around and they're going to be there again so that induction step doesn't happen until you get above where the loops are and as soon as you get above the loops are the logic stays the same and it all works the same way but that's the difference so for 3n plus 1 which I just showed 3n minus 1 which I just talked about it seems to me that this works find holes in it let me know I'm absolutely open for comments this has been a very fascinating journey this Colatz conjecture is well in some ways it's so difficult but in other ways it's so interesting and I've learned so much from it by playing around with it so thank you for your attention and I look forward to your comments