 So good morning, everyone. So until now, we have developed quite an involved machinery for studying these conjugacies and topological conjugacies in various settings. And what we've done has been non-trivial. But in some sense, the systems that we have been studying have been very simple. Not much can happen, these linear maps or these one-dimensional different morphisms. Basically, every point either ends up at infinity or ends up in a fixed point. There's not many different things that can happen. So these last couple of lectures for this part of the course, I want to change a little bit the focus. We are not going to focus so much on the issue of topological conjugacy, but I want to give a couple of more interesting examples from a dynamical point of view. And in particular, we're going to study a class of examples which falls under the category of minimal homomorphisms. So if you remember, when we have a map on a metric space, if x is in a metric space and f is a map, then we described, we said that the omega limit is a very important concept to describe the dynamics of a point. So we say that some point x in x has a dense orbit in x if omega x equals the whole space. You remember what the omega limit is? The omega limit is the set of accumulation points of the orbits. So what does it mean that the omega limit is the whole space? If this is your set, x, big x, this is our initial condition x, it means somehow that the orbit of x goes everywhere. It does not just converge to some fixed point. It must move around a lot in the space because it accumulates every point in the space. We have not yet seen some examples in which this happens. That's what I mean in some sense when we've seen very simple examples. The examples we've seen so far have been examples where basically everything converges to a fixed point or converges to infinity. There's not much happening. This already is something much more interesting in the dynamical system. We say that f is minimal if omega x equals x for all points x in x. So in general, we shall see later in the next second part of the course, for those of you who will follow the second part of the course, we will see other examples in which you have a combination of behavior. You might have some fixed points. You might have some periodic points. You might have some points which have dense orbits. Within the same system, you have this very rich behavior. At the moment, after having studied systems that have just fixed points, we are going to study systems in which every orbit has a dense orbit. And then we're going to look at systems that have a mixture of these two. So today and the next lecture, we're going to look at some examples of minimal systems. So the most standard and simple example is that of circle rotations. So what do we mean by circle rotations? Maps on the circle. So we want to define a map, F, S1 to S1. So to define this, first we just have to recall exactly what we mean by the unit circle. I think you know that the unit circle can be represented in various different ways. So what is the unit circle? One way to represent the unit circle is literally as the unit circle in the complex plane. E2 pi ix for x between 0 and 1. This is the unit circle in the complex plane. You're familiar with this way, yeah? This is the unit circle. You also know that we could write the unit circle as a quotient z. You know about this, what I mean by this? I mean that I identify. I look at the real line and identify x is equivalent to y if and only if the distance between x and y is equal to 1. So I identify all points that have distance 1, and so what I get is topologically a circle. I think you probably have done these quotient spaces in the topology course. Have you done quotient spaces in the topology course? Yes? Sorry? OK. But this is really, you think of it as the circle. Another way, so we will not use it in a fundamental way. So I will not go over it. If you're confused, you can just look up these quotient spaces in some topology book. Another way which is also exactly equivalent is to write this as the unit interval 0, 1 with the identification of 0 and 1. So with the equivalence relation, that 0 is equivalent to 1. Because the unit circle is just a circle. It's just an interval, really. But it differs from an interval topologically because 0 and 1 are identified. That gives it a different topology as a space, as a topological space. So each of these is really parameterized in some sense by the open interval 0, 1. So this is parameterized by all values of x between 0 and 1. And this also because you identify all the points whose distance is bigger than 1. So this is parameterized by 0 and 1. So the easiest way for us to work with and to think of is more or less this or any way to think of the circle as the interval 0, the open interval 0, 1. So think of S1 as the open interval 0, 1. But knowing that when you, in some sense, if you move past 1, what you're doing is you're coming back to 0. So the interval 0, 1. If this is 0, and then you think of it as having a topology so that the limit, as you converge to 1, is actually equal to 0. That's what that identification is, OK? So when we have operations or functions on S1, the most useful way to formulate them is to think of them as being on the interval, but mod 1. So we define a function, the function that we're interested in. So let me write this as a definition. Definition F S1 to S1 is a rotation or sometimes also called translation by an angle theta if it is of this form F of x equals x plus theta mod 1. So the mod 1 is just used to remind us as what it means is that you just take the non-integer part of the number. So x belongs to the circle, which means x is a number between 0 and 1. But in principle, this angle theta could be any number. But when you add, even if theta is a number between 0 and 1, the sum of these two numbers might be a number bigger than 1, OK? But you take mod 1, which you only take the non-integer part. So this corresponds to the fact that you are thinking of the interval. So if this is, let's put 0 here. So if this is x on the interval and x plus theta, if theta is big so that x plus theta is bigger than 1, so this is x plus theta, so this is theta. And then the mod 1 tells you that you have, that you're back on the circle, OK? I mean, I think it should be fairly clear, you should understand. You know what, are you familiar with this notion mod 1? Yeah, yes? OK, let me know if there's anything you don't understand about this. So these are all just little technicalities to be able to formalize in, when you look at the circle, it's very obvious what this is. This is a circle rotation, which means you take a point on the circle and you rotate it by an angle 2 pi i theta if you use this notation here, or just theta, OK? If you think of this as the unit interval, if you parameterize this by the unit interval. So what we want to study is the dynamics of these maps. So what is the dynamics? What do you think will be the dynamics of this? What happens if we iterate? What does the dynamics mean? Remember, you take a point and you iterate the point. So in this case, iterating the point means adding theta. You keep adding theta. What do you think is going to be the orbit? So what is the dynamics? So question, what is the orbit of some point x, OK? And what is the omega limit of some point x? Any ideas? It depends on the value of theta. That would make sense. So what are the possibilities? What happens if theta is 0? Always try 0 first. What happens if theta is 0? Then we have the identity, f of x equals x. And what does the identity mean? Every point is a fixed point. What if theta is not 0? Can you have fixed points? So if theta is, for example, 1 in this parameterization, then of course x1 is in our parameterization of the circle. It has length 1. So if theta is equal to 1, then you will come back to x. And you will get, again, the identity, right? Because this is mod 1, if theta is an integer, then you will still get the identity map here. That's right. So theta is an integer. So really, what we only need to consider, theta between 0 and 1, because if theta is an integer, then mod 1, this is just like adding 0. Theta mod 1 is just 0. And if theta is bigger than 1, then it doesn't matter. All it corresponds is to going around several times. And we just really need to take x plus whatever the value of theta is, mod 1. That's what we need to consider. OK, so it looks like it will be relevant whether theta is rational or irrational. How does it make a difference? We had a periodic point. That's correct. I will show that in a second. A very simple example. For example, let's say theta is 1 third. Then you can see what is this point going to do. You start with x, then you add 1 third, then you add 1 third, then you add 1 third again. And then you come back exactly to x. So x will be periodic or period 3 if x is 1 third. And you can easily, sorry, if theta is 1 third. And you can see that every point will have period 1 third. Because that's all you're doing. It's a circle rotation. You're taking the circle, rotating it by 1 third, rotating by 1 third, rotating by 1 third again. And you end up where you started. So the third iterate is the identity. So you have every point is a period 3. And the same for 1 fourth, 1 fifth, 1 sixth, and so on. So in fact, the result is this. So theta is rational if and only if. Every point is periodic. And what happens when theta is irrational? OK, there's a reason why I'm giving this example. So theta is irrational if and only if every point has a dense orbit. F is minimal. So the rational part is very easy. So suppose theta is rational. Then what is the iterates? So fn of x is equal to x plus n theta. This is what the translation is, of course. Iterates of x is x plus theta. If you apply it again, you add theta again. If you apply it again, you add theta again. So the iterates in general, this is in any case the iterates of fn of x have this form. If theta is rational, then this is just equal to x plus np over q, mod 1. Always mod 1. So what? Exactly. So if when n equals q, so when n equals q, what do we get? Then we get that np over q equals p, which is an integer. And so this mod 1, this is equal to 0 mod 1. All right, you've done this at least in the introduction to number theory course, if not in the topology course. So this is 0. So in this case, this means that fq of x equals x for all x in S1. So every point, so every x is periodic with period q. So notice we've said a little bit more just that every point is periodic, but the period is given exactly by the denominator of the rational. Assuming this rational number is in minimal terms, of course. OK, similarly, the converse is also very easy, because if there exists a periodic point, so conversely, suppose there exists one periodic point. So suppose there exists x and q, x in S1, q in Z, such that fq of x equals x plus nq equals x. So suppose x is a periodic point of period q, then this must be 2. OK, then we have that nq is equal to 0 mod 1, theta q. Sorry, yes. So theta q is equal to 0 mod 1. And so theta equals p over q for some, sorry, this should be, sorry, yes, and theta, sorry, this should be, that's what I want. So this, in particular, using this, this means that if there exists one periodic orbit of period q, then theta must be rational p over q, and therefore, every point is periodic of period q. So that's what we've shown in the statement. So theta is rational if and only if every point is periodic. So we show that if it's rational, every point is periodic. And if you have even just one periodic point, then it's rational, and therefore, in particular, every point is periodic. OK, so now the irrational case is a little bit more interesting also. So now suppose theta is irrational. Wait, let's go in the other direction first, because it's simple. So theta is rational if and only if every point has a dense orbit. So if we suppose that every point has a dense orbit, so suppose there exists a dense orbit, so if x, if there exists some x in x such that omega x is s1, then why does this imply that theta is irrational? If it's not irrational, then it's rational, and then every point is periodic, which contradicts this assumption. OK, very simple contradiction. So if x, if it's dense, then, OK, so assume by contradiction, assume by contradiction that theta is rational, OK, then every point is periodic, which contradicts that omega x equals x1 s1, OK, kind of almost trivial contradiction argument, OK? So the only real non-trivial part is that if theta is irrational, then every orbit is dense, because this, of course, does not imply that, right? It just says if it has a dense orbit, it might be that you never have a dense orbit, so this does not imply that. So now assume theta is irrational, so theta is irrational. So how do we show that something is dense? We need to take a point x here, so this is x plus theta, this is x plus 2 theta, and so on, x plus 3 theta. So it moves around the circle. We need to show that it's dense, that 0 is a limit point, OK? We can show that 0 is a limit point. How do we show that 0 is a limit point? It's an infinite set. That's a good start. And so? Yes? Yes. So have you seen this proof before? You know this proof? Or you're just thinking now? Right. Yes, yes, yes. Exactly. You can formulate it in several different settings here. OK, so let me try to formalize for everyone else this argument, right? So correctly, as Soheila mentioned, the first observation is that this must be an infinite set. So the orbit of x is infinite. Who can tell me why it's infinite? Honey, can you tell me why the orbit of x must be infinite? And so? And why is it infinite? Intuitively, it's obvious. But can you make a precise statement to justify that this is infinite? By contradiction, if it was finite, exactly. And if it's periodic, rational, exactly, OK? Simple contradiction here also. OK, so theta plus x is infinite. And so, OK, I'm going to formulate this a little bit differently from the way you described. But fundamentally, it's the same way. I want to show that it accumulates every point, right? So the way I'm going to do that is I'm going to take a cover. So we want to show let epsilon greater than 0. And cover, cover s1 with possibly overlapping length less than epsilon. So I just cover all the circle with intervals of length less than epsilon, OK? And now I'm going to show sufficient to show that the orbit of x intersects every one of these arcs, one of these arcs. Because if it intersects every one of these arcs, it means that it comes within epsilon of any given point on the circle, OK? If I can do this for arbitrarily small epsilon, then I will have shown that the orbit of x comes arbitrarily close to any point on the circle. And I will show that the orbit is dense. So all I need to show is that it intersects every element of this cover. So the fact that it's infinite, I will use in the following way. So this is a finite cover, OK, with a finite number. So I should say with a finite number of arcs of length less than epsilon. So because this is finite, since plus x is finite, there exists at least one arc. Let's call this J in S1, OK? With the length of J is less than epsilon, because all of those arcs have less than epsilon, such that the orbit of x intersects J in at least two points. You agree, Mariam? You agree with this? Yes? Sorry? Since the orbit of x is infinite. Thank you. Was that the problem? Since the forward orbit of x is infinite and you've covered the circle with a finite number of arcs, clearly there must be one arc that contains at least two points. In fact, an infinite number of points, but at least two points, it's obvious. This is sometimes called the pigeonhole principle, but it's kind of a trivial thing, right? You have a certain number of boxes. If you have more things than you have boxes, at least one box must contain two things. So suppose xl, which is equal to fl of x, fl of x belongs to J and xm equals fm of x belongs to J, with let's suppose that xl equals fl. So this is equal to x plus l theta. And this is equal to x plus m theta. And we can suppose that m is greater than l. One of them has to be greater than the other. So what does this mean? This means that both of these points, so this is one of these little arcs. We have these two points here. One is xm and one is xl. So what this means is that x plus l theta minus x plus m theta, which is just xl minus xm, is less than epsilon, because the length of J is less than epsilon. And so this means that this is mod 1. So this means here that l theta, l theta minus m theta, well, we assume, OK. So for simplicity's sake, assume that m is greater than l. I'm going to write it like that. So I'm going to write m theta minus, well, I don't know actually. No, I don't know exactly where. So I can write it like this. m minus l theta mod 1 is equal to xl minus m, which is less than epsilon. OK, so this is just to say that if, so what this is saying is that even though m and l might be completely different, so it could be that, so I look at the orbit of x, and after 1,000, I land here, l is equal to 1,000. And then I have to wait a very long time. I keep going around and around, and I keep missing it. But then after 1 million, it's whatever, I fall in here again. OK, so what can I say about the relation between m and l? Well, m and l have no relation whatsoever. But because of this property, because I know that there is a relation that is given by the fact that this point and this point are close. And this point is of the form x plus l theta. And this point is of the form x plus m theta mod 1. This is all mod 1. This is crucial, of course. Mod 1, then it means that m minus l theta mod 1 is less than epsilon. OK, so now what this means is I can take the iterates. So let theta tilde equals m minus l theta mod 1. This is less than epsilon. And now consider the iterates f. Also, maybe this is not really what I want to say. So what I want to say is now we take the iterates f n m minus l of x. OK, sorry, I don't really want to introduce. Now consider the iterates f n m minus l of x. So what is that going to give us? This is going to give us x plus n m minus l theta mod 1. So maybe I did want to introduce it. So this is like equals x plus n theta tilde n theta tilde. Maybe I did want to introduce that. Where theta is less than epsilon. Theta tilde is less than epsilon. So what am I saying here? That if I take multiples of the map, this here is like saying f m minus l is my new map. And I iterate this n times. This is really what I'm saying. So I'm taking multiples as iterates of x. I'm taking multiples of m minus l. So it's like I'm taking iterates of this map here. And iterating this map here corresponds to a rotation by an angle theta tilde, which is less than epsilon. So these iterates here of x, if I start now with this point x, and I iterate by multiples of this, then each time I shift by theta tilde, I rotate by theta tilde. I have got a new rotation by an angle theta tilde by taking these multiples. So I take x, and then the image of x will be at a distance theta tilde from here, which is less than epsilon. And then when I apply again, I will get another, I will rotate again by an angle theta tilde, which is less than epsilon. So the distance between each point in this image will be less than epsilon. And therefore, after a while, I will have to necessarily intersect every single element of the cover that has length epsilon. So iterating f m minus l corresponds to a rotation. I should have said here probably equal epsilon here, length equal epsilon, corresponds to a rotation of length theta tilde equals m minus l theta mod 1 less than epsilon. And so the iterates of f m minus l of x of x will intersect every element, the epsilon. And this completes the proof, because epsilon is arbitrary and I can do the same argument so I can show that some iterate of x will intersect every element of this cover. Any questions? So this is an example of a map which is minimal, so every orbit is dense in here in the circle. So for the rest of this lecture, I just want to do a little parenthesis before we go on to the next example of a minimal system. This was the first example that I wanted to give you was irrational circle rotations. And this lemma actually has a very interesting application in number theory. And so I think I would like to describe that before we go on to the other example. So let me finish by giving you this application, which is quite surprising because it doesn't seem to have anything to do with circle rotations. So application. OK, this is a question in number theory. As you know, number theory is fairly pure and fairly abstract. But you can ask these sort of questions. So you can say, for example, take a sequence. So OK, let me just write the proposition which explains the proposition. So let k in n, not a power of 10, then for any given finite sequence a0, a1, a2, aL, with ai being just a digit. For any given finite sequence, for any, maybe I should say for any given finite sequence, there exists n in n such that the initial digits of k to the power n coincide exactly with the given sequence. So k is here. So k is any given integer. a0, a1, a2 is any given sequence. I guess what I was a little bit stuck there on is I guess we exclude the fact that it starts with 0. So we have a i belongs to 0, 1, 9. And a0 is different from 0. Otherwise it doesn't really make sense. There exists n in n. So what this says is look at the sequence k, k squared, k cubed, k to the fourth, and so on. This is an infinite sequence. And k is any integer number. So k could be 4, 5, 6, 13, whatever. You take the powers of k and you're interested for whatever reason at the initial digits of this term, k to the n. And I claim that if you look at n sufficiently large, then you can get any finite initial sequence of digits to occur as the initial digits of k to the n. Is the statement clear? It's not clear. Tell me what. Try to formulate it yourself. Initial digits. So k is an integer. So k to the n is an integer. So what does an integer look like? k to the n will be equal to a0, a1, aS. This is the integer. The this is the number, where ai belongs to 0, 9, and a0 is different from 0. This is how you write this integer as a sequence of 0s and 9, a sequence of digits. So I claim that the first l terms are those that I were given. So I can find for any sequence, you give me the sequence. You say, OK, the sequence. So let's just write down an example here. So example. I will choose k. So let me say k is equal to say 7, just to something not. OK, k7. Now you give me a sequence. So a0 to al is equal to give me a big number. Give me a big number, Marianne. 3, 2, 7, 8, 1, 9, 2, 5. OK, a0 to al. This is a0, a1, a2, a3, a4, a5, and 6. So I will prove to you that I can find an n. So then there exists n such that 7 to the power of n is equal to 3, 2, 7, 8, 1, 9, 2, 5, 1, 2, 3. So this is a0, a1, a2, a3, a4, a5, a6, a7, a8, a9, a9. Up to a something. I don't know how many digits this will have. I'm just saying that the first seven digits will be exactly those ones here. OK, is the statement is clear? Yes. Excuse me? Yes, we exclude the, no, no. We don't exclude k power 10. We just exclude that k is a power of 10. Because, of course, if k is a power of 10, then you never get anything. You only get 10, I mean, you only get powers of 10 when you take powers. So that is the reason, because this is trivially false if k is a power of 10. If k is equal to 10, then all the powers will be 10, 10 squared, 10 cubed, 10 to the 4, and this cannot be 2. So what is this? But this is the only restriction on k. So we can choose any k we want, any sequence we want. I claim that some power of k will start with this sequence. This is the statement. It's not a trivial statement. I mean, it's quite remarkable and surprising, if you want. And I will show now that there's actually, the proof of this is the dynamical systems. In fact, there's a lot of relations between dynamical systems and number theory in many different ways. I will now touch upon them at several moments in the course. And this is the first simple setting in which we can see such a relation. So how are we ever going to prove this using dynamical systems? Excuse me, I did not hear your question. You're sorry? L, not necessarily. You mean if I can choose this is equal to just A0 to AL equals all the same, was this your question? Yes, I can choose them. No, that was not your question. L is not fixed for any L greater than or equal to 0, OK? And any given finite sequence like this, I have this statement. Was that your question? Tell me your question. Yes, not different from what? I didn't understand your question. So here you have the sequence of powers of k. F is what? No, of one of these. They change. Of course, every time you change the power, the initial digits change completely. No, because there is a, oh, if we prove that no two powers can have the same sequence of digits. So you're thinking about a strategy for the proof? Aha, OK. Well, I don't know if you can do that, because we want to allow any possible value of L and any possible finite sequence. Let me show you how we do this. So proof, we want to show that there exists some n in n and some a L plus 1 a L plus m such that k to the n is equal to a 0 a L a L plus 1 a m, right? So this is the sequence that I had fixed, OK? To say that, sorry, there exists some m in n and some sequence such that this is true. This is the sequence that we had fixed. So what we need to show is that this number is of this form. This is exactly the sequence of digits I had wrote here. So this is L plus m. And what does this mean? This means we can write this as a 0 a L times 10 to the m plus a L plus 1 a, sorry, it's just the scientific notation for this number, right? So this number times n to the m, which is exactly, these are m terms here. So if we let p, so let p equals a 0 to a L, then this is the same, this is equivalent to saying that there exists n, m, and q natural numbers with q less than 10 to the m such that k to the n is equal to 10 to the m times p plus q. So I'm just rewriting this. So this here is just equal to this. I just defined p as this times 10 to the m, right? And then this is just a number that is less than 10 to the m, because this has m terms, so it's less than 10 to the m. So to say that there exists, so what we need to show is that there exists such a sequence that this is true, and this is the same thing as saying that there exists a q less than 10 to the m such that this is true. So far I haven't really done much. But now what does this mean? So this is the same thing. This is equivalent to saying that there exists n, n in n such that k to the n is greater than 10 m p and less than, so I think this should probably be an equal sign, less greater than or equal. In my notes it's strict, but I think it should be less than or equal, less than 10 m p plus 1. Oh wait, where should be the equal sign? Should be, yes, it should be equal and it should be strict. OK, because q is less than 10 to the m, so I just like this 10 to the m p plus 10 p. q is less than 10 to the m, so q is less than 10 to the m, so I put this 10 to the m p plus 1 is here. OK, we agree that this is the same thing? Yes. OK, so this is what we need to show. If we can show this, we've proved a lemma. So if we can show that there exists m and n such that this holds, then we will have shown that there exists n, m and q with q less than 10 to the m that this holds, of course if that's true. And if this holds, that means that this number is of this form and we have proved that result. So we need to show this, that there exists m and n such that this holds. So taking logs, this is equivalent. So logs base 10 is equivalent to saying that. So we take logs base 10, so this is m log 10 base 10 plus log base 10 of p less than or equal to n log base 10 of k less than log base 10 of 10 plus log base 10 of p plus 1. But these log base 10 of 10 is just, what did I forget? m here. So log base 10 of 10 is just equal to 1. So this is equivalent to m plus log base 10 of p less than or equal to n log base 10 of k strictly less than m plus log base 10 of p plus 1. So we've got this m on both sides. So this is saying that this number here squeezed between these two numbers because we've got the m on both sides, we can put the m in the middle term and we can write this is log base 10 of p less than or equal to n log base 10 of k minus m less than log base 10 of p plus 1. So notice that we've removed from the two extreme sides, we've removed n and m. n and m are just now in the middle, right? So these are fixed by p because p is fixed from the beginning because p is the sequence that we chose, right? p is the sequence that we chose. So p is fixed from the beginning. So sufficient to show that there exists n and m in n such that log base 10 of p less than or equal to n log base 10 of k minus m less than or equal to log base 10 of p plus 1. We have just repeat what I had before, but just to emphasize the fact that these are the two variables that we have to play with, these are fixed. These are fixed. What does this mean? We have n and m are integers. So what this means is just that we are interested in how close this is to being an integer. All we need is this to be very close to being an integer because then we can put that integer here. We are free. n and m are free. It doesn't matter what we choose. We just need to show that there exists n and m such that this is true. So whatever this value is, it doesn't really matter. What we're saying is that if p is very big, then the difference between these will be very small. So this gives you the gap inside. We can also write it like this. n log base 10 of k minus m should be inside this gap. Log 10 of p log inside this small interval. So this interval might be very small. What we're saying is that this is very small. And what we're saying is that before, for this to be very small, we need this to be very close to being an integer. Is that clear? Because as long as this is very close to being an integer, we can just choose m to be that integer and we're OK. So how do we show that that is very close to being an integer? So there should probably be some absolute value here that I did not put in. So it is sufficient to show that this is in this interval. Somewhere here I can introduce the absolute value. I want the difference between these two terms to be small, to be in this interval. So why is that true? So for this, what this means is sufficient to show that n log base 10 of k, mod 1, is in this interval. Because what does mod 1 mean? Mod 1 means precisely how close it is to the integer 0, basically. So if this is very close to some integer, if this is very close to m, then this will be mod 1. This will be very small. So we just need to show that this mod 1 is inside this interval. And now we're done, because now why does this follow? How do we know that this is true? Exactly. So this is just an irrational n log. So you let theta equals log 10k is irrational. So we want to know the value of this mod 1. This is given. So let fn of x. So the sequence let fn of 0 equals 0. So we start with the point 0. We iterate n log 10 of k, mod 1. So this is just 0 plus n theta, where theta is equal to log base 10 of k. This is just equal to n. So this sequence of numbers here is nothing but the orbit of 0 under the irrational translation or rotation of log 10 to the power k. So you start with 0, and then you rotate. You get log, OK. You get theta, then you get 2 theta, then you get 3 theta, and so on, where theta is precisely log base 10 of power k. And so the n theta, you get n log base 10 of power k. So what we need to show is that this is inside this interval here. This interval is some interval in the circle. Here we have some interval, which is this interval here. And all we're trying to show is that there exists some n such that the orbit of 0 will intersect this interval. And that is true because this is irrational. So since k is not a power of 10, log base 10 of k is irrational. Since log base 10 of k is irrational, then of 0 under fx equals x plus theta, where theta is equal to log base 10 of k, is dense for all x, in particular for x equals 0. And so there exists some n in n such that fn of x belongs to this interval here, log base 10 of p, log base 10 of p plus 1. And this is just equal to fn of 0. And this is just equal to n theta, which is equal to n log base 10 of k. So it belongs to this interval. I guess probably it was not necessary. At some point when I was looking at this, I thought it was not necessary, it was necessary, but it's not actually necessary, you're right. Because I can still show that it comes close. So maybe that was a bit confusing. Maybe it was even incorrect. I'm not sure, but we do not need that, you're right. Yeah, yeah, I don't need it at all. OK, so today we had a little glimpse into the wider world of dynamics, as I said before. So these are some of the more interesting examples of dynamical systems. So even though I have spent a lot of time on defining the concepts of structural stability, anthropological conjugacy, and so on, which we will apply to some more complicated systems, until now we had concentrated on those techniques and looked at very simple systems. I just wanted to hear now today in the next lecture to give you some examples of some systems in which the dynamics is much more interesting. And as you can see also, there's sometimes some quite remarkable applications of two number theories. You see there's no, when you just look at the problem, you do not see any connection with the dynamics at all. But it's just a very simple corollary. We use here in a crucial way, we reduce the problem to one of the rotational on the circle, and we solve it just by using the fact of what we know about the rush. And there's many more interactions like this. So I think we will call it a day for today. Thank you.