 Well, thank you very much to the organisers for inviting me and for accommodating my change of date. So actually, so today I want to talk about stuff which I sort of looked at. I think the date on the paper was 1999, so those of mine should remind people of a great Prince song 1999. But anyway, so it was a question that actually Bing Lee asked me about this work and it's led to this new stuff. So basically it's going back to this problem of shrinking targets for matrix transformations of the Torah. So let me see if I can get this to work. Yeah. Okay, so the general setup is that you have a D by D integer matrix. So this is going to determine a map, a self map of the D dimensional Taurus and throughout MD is going to be the D dimensional Lebesgue measure on the Taurus. So straight away, I'm going to restrict my attention in this talk to a very special integer matrix, basically this matrix two zero zero three. Okay, because I just there's no point in doing anything more general at this moment, because there's not much that we can do in higher dimensions anyway. Okay, so, so what's the problem then? So if so with T, an integer matrix with non zero determinant, then it preserves, so it's measure preserving an enogotic with respect to two dimensional Lebesgue measure on the Taurus. So in particularly what it means for us is that if you take a ball or with positive measure in the Taurus, then if you take any, but if you take a point X in the Taurus, then the orbit under T, i.e. T to the end of X will line this ball infinitely often. Okay, and this is a statement which is true for almost all points X in the Taurus. Okay, so what you're thinking about here is let me see if I can do this. So you've got your Taurus here, right? If you take some ball, which I'm going to be, it's going to be a square, you take some point X, I need to get a bigger fat pen, take some point X, and you look at the action of X under T, and what it's saying is that almost surely you will land inside this ball infinitely often. Okay, so in other words, trajectory of almost all points will hit the ball infinitely often. So in view of this, it's natural to ask, well, what happens if we allow the ball to shrink with time? Okay, so as the as n increases, what we want to do is decrease in time. So more precisely, what we're going to do is look at this formulation here. Okay, so now instead of the ball being fixed, so I'm going to fix my ball, right around the origin, and that's just for convenience. And the ball with n, so as we look at, as T acts on X, so we're looking at T to the end of X, then the ball is around the origin is going to shrink at this rate where the rate is determined by this function fpsi. Okay, so we've got some decreasing function fpsi. Right, so the first thing to note, let's go through this bit reasonably quickly and then I'll rephrase the setup anyway, right, and then describe it a bit more. So what it is, is this is, so it's a set of points which lie infinitely many pre-images of this ball. Okay, so we can write it as the lymph soup of the pre-images of the ball around the origin. Okay, so since T preserves the measure, what that means by definition is that the measure of the pre-image, so this T to the minus n of the ball is equal to the ball that we started off with. Okay, so we're looking at here is a lymph soup set, so now what we've got, we've already, so this tells us what the measure of this, of these building blocks of the lymph soup set are. So whoops, that's going in the wrong direction. So, sorry I should just say here then, so what's the measure of this, the pre-images, whereas the measure of the ball that we started off with, and the measure, if I'm using the maximum norm, the measure four times fpsi is what we should be thinking about here, is that we've got the ball, so here's the torus, this ball is just this, this is b0 fpsi of n. Yeah, so can, am I frozen, nice, okay, okay, so this this width here is fpsi of n. Right, so now we just apply the Borel-Kenteli lemma, right, so if we apply the Borel-Kenteli lemma, and if the sum of these measures, which is the four fpsi n squares, if these converge, then the set of points which lie in infinitely many of these, i.e. this w tor fpsi is zero. Okay, so that's just the straightforward application of Borel-Kenteli, so of course we're going to be interested then in what happens so when the sum diverges, right, so we've just come, we've just done a very simple proof of the convergence case here, so if it converges, then the measure is zero. So what actually happens is that if the sum diverges, so the sum of these fpsi n squares diverges, then we've got full measure, okay, so this is a statement which potentially I think goes back to the 90s, okay, so if we, now I want to look at this, the case where things converge, so I want to take this specialist again just to keep things explicit, right, so for any tor, I'm going to look at the function x goes to three to the minus x tor, okay, and for this function I'm going to write instead of wt of fpsi, okay, so what we have with this function is pretty clear that for any tor greater than zero, right, so we're looking at the sum of three to the minus n tor squared, right, so this is going to converge, right, so we're in the, so we're in the convergence case for any tor positive, right, now this is, this is zero measure for any tor positive, so we're not going to get any more information about the size of this set, if we stick with Lebesgue measure, so what we're going to do is use Hausdorff measure and dimension to measure its size, okay, so I'm not going to define measure, Hausdorff measure or dimension here, it's just enough to note that when I'm in the situation of Lebesgue measure zero, there's a whole array of sets here of measure zero and I want to see how I distinguish between them, so I'm going to use Hausdorff dimension, okay, so it says then that if tor is in this range between zero and one, then I have this expression tor, so the point is that as tor increases, so as the ball shrinks fast from faster, the size of the set decreases, right, because this is a decreasing function in terms of tor, okay, so this is the result from 1999 which I proved and that's all I really want to say, okay, so we can say two things now, right, so why am I telling you this, right, you may be wondering, it seems like everything known in terms of the size, the Lebesgue known, Hausdorff dimension is known, but what I want to do eventually is look at the situation in which the points on tor are restricted to some curve, okay, so this is going to take us into the world of sort of diaphanetone approximations, but where is the diaphanetone approximation here, right, so let's, let's, it's pretty straightforward of course, right, that, so here's, so I just thank you to F. Ganey-Zorin for providing the pictures, because I was completely lost how to do this, so here's the, here's tor as the unit square and as I said, so this is the, this is going to be the, what's, let's see, I think it's medium, let's go for medium, so this is the origin and so this is the ball around the origin, right, okay, so of course the measure of that ball is just, so this is the side length, what's psi of n here, so the measure of this is four times psi of n squared, right, that's what we used already before, okay, so now, so this is this ball, so now we're going to look at pre-images of this ball, right, so first of all let's have a look at the pre-images of the origin, right, so for this I've got some, I was going to write it down, but then it was, I was really bad with the pen writing, so I decided to type it, so we've got the matrix 2003, then it's pretty easy to see that if I look at rationals pairs where the x-coordinate rational denominator is 2 to the n and the y-rational denominator is 3 to the n, so I'm looking at 2 over 3 to the n and I look at t to the n of this, so this will, so what does this do, we're going for it's multiplying the x-axis points on the x-coordinate by 2 and on the y-coordinate by 3, right, so then here I'm just going to end up with s of origin, right, modulo z squared, so basically the point here is that the rational points of the 2 to the n, 2 over 3 to the n, in the unit square are the pre-images of the origin, right, so let's just let's just draw them, so here they are, so here's, so along the x-axis we're looking at rationals of the denominator 2 to the n, y-axis denominates 3 to the n and so what happened to this actual ball now, well going backwards in the horizontal direction which we're shrinking by a half in the vertical direction which is shrinking by a third, right, so what we're going to end up with is these rectangles, so this is centered around these rational points and so here's a, this should be this one here is this rectangle here, right, so the center here, what's, yep, there you go, so the center is this point s over 2 to the n, 3 over, oops sorry, t over 3 to the n, okay and actually this, this is, there's a mistake here, this, there should be a 2 here as well, because that's the side length, okay, right, so if I call this guy, so I want to call this rectangle, right, so it depends on n, the center, the numerator and the side length is depend, depend the function of psi, right, so let's, here is then the union of these angles to the points, so 2 to the n, t over 3 to the n and so I want to call this union just this r, n, of psi, right, so this is the union over all s and t, where s is between 0 and 2 to the n and t is from 0 up to the n, okay, right, so, so this is just the summary of what I've just drawn, right, so what we've got then is that the set, the pre-image of the ball here is just the union over these rectangles and the side lengths of these rectangles, right, is given by this, but let me, let me just make one thing, sorry I just wanted to make, emphasize one thing, so look, so in the horizontal direction, right, in this, along the x-axis the separation, right, between the rationals is obviously 1 over 2 to the n, right, in the vertical direction the separation is 1 over 3 to the n, yeah, okay, so there's just something to keep in mind, okay, so this means that we can write then our set, our shrinking target set as a limb soup over these rectangles, right, and what is this, well this is nothing more than points x y in the unit square for which this, this holds here infinitely often, right, so in other words this maximum is less than epsilon psi q for infinitely many n, right, so this is obviously now really a question or this, this set very much like a set up of a diaphragm time approximation where, okay, you're not approximating by the same denominators, but you're approximating on one side with denominators of 2 to the n, the other by 3 to the n, okay, what's that mean, so, so I'm just here just recalling the results that we have, so this is why this is a kinship type theorem, right, it's just, you know, if here we add q instead of 2 to the n and 3 to the n we add q, this would be a simultaneous form of a kinship's theorem in classical a diaphragm time approximation and here's the result for the W Torr set, right, so I'm just rewriting it now so we can see the result in terms of the shrinking target set, this set here and the number theoretic set, this set here right, so, sorry, I just, I was trying to get rid of the one second, sorry, it's also psi of n, not psi of q, where is it, oh second part, I hear, yeah, here, sorry, this is f psi of n, thank you, okay, so I can't really see the bottom of my, I have this toolbar which is coming up with the, maybe I can just close it, someone's, oh, I see someone's got a question, oh, okay, thank you, ah, so you get the thing bar coming up when someone asks the question, okay, that's useful, right, okay, so, so what's next, right, so, okay, so what's next is that we want to take the set that we already studied and we know the size in terms of the dimension and the measure and we want to restrict the points of interest to lie on some curve, okay, so we're in two dimensions here, so it's, in general it would be a manifold but here it's going to be a curve, okay, so what we want to do is study the set of points which are this psi approximable but restricted to the curve, right, so this is the set that we're going to be interested in, um, so what the aim would be is to, is to, is to establish the, the, the Lebeser statement and the Hausdorff dimension statement for this set of points restricted to the curve, okay, so this is really very much in line with the classical theory of die counter approximation of manifolds, and so one thing, so if we were looking at the Lebesgue theory, right, there's no point in looking at the two-dimensional Lebesgue measure, right, because the curve is of dimension one, right, so this measure of the curve intersect whatever this set is, right, with, with any at psi it's just going to be zero, okay, so what we want to work with then is of course the measure on the curve, right, so this is just induced one-dimensional Lebesgue measure, okay, so to develop a Lebesgue theory for the curve intersect this set we're going to work with induced measure, and I'm just going to write m for this, okay, oh I've got another, again, q in place of n, yes, thank you, Efkin, yeah this is going to be everywhere because I've just copied it I guess from the previous slide, okay, so, so I'm going to take a very simple case, okay, where the curve c is the diagonal line, right, y equals x, and the point is that it is already interesting and seems to be quite difficult to prove a full result anyway, so let me just take this special case of the diagonal line, okay, so what we prove is the following statement, so this is all as you would expect if you're familiar with the classical diaphragm approximation manifold theory, then what you find here is that, so remember this is the L is just the diagonal, so it's one-dimensional, so this is just one-dimensional Lebesgue measure, this m, right, so what we find is that the measure of this is zero, one, depending on whether the sum that we had before, i.e. the two-dimensional, the two-dimensional volume sum diverges or converges, okay, so and what we also have is that for the dimension, so remember we're going to take this special just, this is just to bring things out explicitly, right, that for tool greater than equal to zero, we consider this function which is 3 to the minus x tool, right, and then what we have is that the dimension of the diagonal with the set W tool is this one minus tour over one plus tour, where tour is between zero and one, okay, the point is that actually it's really quite straightforward to see that if tour is, if we put tour bigger than one in this case, then actually all we get here for early intersect, the set W tool is just the origin, okay, because this is the diagonal line, and remember here what we're approximating by are rationals, all right, where the rational points, on one side it's of s over two to the n, and then t over three to the n, right, so apart from the origin, there aren't any rational points of that form on the actual diagonal, right, so the origin is always going to be there, so what I'm saying here is that when tour is equal to zero, tour is bigger than one, sorry, then the set is actually just the the origin, now a funny thing here is I laugh because so for tour bigger than this one minus, so this I'm going to start calling gamma, right, it's log two over log three, we're not really able to prove the lower bound unless we use the ABC conjecture, so either I'm being stupid and not able to see what one should be doing, and this is the thing that I want to really talk about, or I'm being stupid because I haven't seen that what we've got here, what we've got in front of us is something which is really very difficult, so anyway it's a lose situation, right, so okay so let me describe the main ideas, okay, about how you would go about proving these statements, and it's really about the distribution of these rational points, and in this case it's the distribution of rational points which are close to the diagonal, so the, I should just say of course that, so the result is actually true for a large class of planar curves, basically curves which are locally by leaflets in some sense, but dealing with the diagonal is really the key, right, so trying to understand this set, okay, so remember the picture that we had before, so the set is this limb soup set of rectangles, where the rectangles are centered at these rational points, and the side length is given by essentially up side and over two to the end, and so this is the horizontal, the long side of the rectangle, and this is the short side of the rectangle, right, epsilon three to the end, so I'm going to assume that epsilon n is less than half, this is not a big assumption, and so here we, so it just means that the rectangles, so in the grid that I drew before, so these rectangles in this union are actually disjoint, okay, so it's clear, right, that we're only going to be interested in those rectangles here in this union, right, which actually intersect L, right, because what we're, or we're interested in the end is an intersect this limb soup set, right, so if I have, try this, yeah, blue, blue, nice, okay, so we have a, we've got the diagonal, and we want our rectangles to intersect, right, so this is some rectangle at some point, at some rational point of this form, and what we're interested in then is this intersection here, right, for this limb soup set, and it's pretty easy to see that having this intersection puts a condition on the, on the closeness of the centers of the rectangles to the diagonal line L, right, so this was L, okay, so what is this condition, well if we think of the worst case scenario and when it's still intersects through a rectangle like this, they should be the same size, okay, so here's the center, okay, so what we have then is, so this is some point S over two to the, two to the N, right, and this is some point T over three to the N, right, so what it tells us then is that, so if we, so this is the, the Y equals X line, right, so this is also down here on the X axis, we've got the same point T over three to the N, right, so what it tells us is that the distance, that the distance from this horizontal distance right here, right, is no bigger than the sum of the long side plus the short side, I mean half of it, right, okay, so in other words what we have is that we're going to have this rectangle intersecting the diagonal if and only if the distance between the coordinates in the sense is less than epsilon over two to the N, so this is just the long side, right, the radius of the long side if you want to put it that way, and this is the radius of the short side, okay, so what that means then is that the set that we're interested in can be written as this limb soup of, so this is just the union over the rectangles in this, in this set here, right, which intersect the diagonal, okay, so, and then we take L intersect RN, right, so this was, maybe it's still there, so I hope that's clear what I'm trying, what I'm trying to get at here, right, so we're going to take 20 fixed N, right, we take the, we take all the rectangles which intersect and we're just going to consider this, this union of intervals of the intersections, right, so this is, this is precisely what this is, right, so where to go from here, okay, so now I'd like to get some sort of count, right, for this, for this set of points, for this collection of rectangles, so counting the number of rectangles which intersect the diagonal L is equivalent to counting the number of centres which satisfy this condition here, right, so clearly here I can, this is less than or equal to two times this term here because this dominates the one with a denominator three to the N and so we can obtain an upper bound, so for the number of rectangles by looking at this inequality here, so all I've done is we're multiplying through by two to the N times three to the N, so this becomes three to the Ns, minus two to the Nt is less than or equal to three to the Ns of N and I'm just bounding this by the same factor here, right, so what I get is this here, okay, so this will be an upper bound and throughout now I'm just going to assume that S and T, they can't both be zero, although of course in this, in this collection of rectangles the origin is there, but let's just ignore that for the moment, okay, it's just needed to think of it like this and then, so do we want to count this, well this is actually quite easy to count because three to the N and two to the N are co-prime, right, so this, whoa, that was a big slide, okay, so this is, so we can count this easily and get, this is the number of terms, there's actually the integer part of this, the bound and the upper bound, right, so it's two times three to the Ns of N and for the lower bound we look at pairs S, T, so this is going back to here where we'll, well this is this term here, if we just count those for which it's just less than or equal to Psi N over two to the N, this will certainly be a lower bound, right, so this is, again, the integer part of three to the N of Psi of N, so I've taken the integer part consideration here, so I'll get this three to the N minus one, okay, but it really doesn't matter, everything's going to be up to a constant anyway, so all we're using here to count these terms here, right, is that three to the N and two to the N are co-prime, so there's these unique integers S1 modulo two to the N, T1 modulo three to the N satisfying this linear congruence, okay, and K, right, so we want to then look at solutions of this equal to K, right, and what we have is that there's going to be a unique integer pair, SK TK, where SK just K times S1 and modulo two to the N, and TK is just K times T1 modulo three to the N, right, and these satisfy this congruence, and K is going to be less than two times three to the N of Psi of N here, which is less than three to the N, okay, so there's all these solutions to this, to these congruence, to this congruence, well, linear equation, I guess, right, it's they're all going to be unique, right, because we where the the upper bound is less than three to the N, right, so what that means is that these of course the, so SK and TK are just the centers of our rectangles, okay, well, SK over two to the N and TK over three to the N are the centers of the rectangles associated with the set RN, Psi of N, okay, so I've just summarized here what we've just found that that we've got upper and lower bounds, so this is, I mean, you can obviously do much better than this, but this is good enough for us, that which basically tells us that the number of rectangles that intersect the diagonal is three to the N of Psi of N up to constants, and the point is that this upper bound is enough to prove the the measure statement and the upper bound dimension part of of both theorems, because so this is the easy, these are the easy parts of the of the theorems, and also this was the comment that I made earlier that if we look at the special function, you know, which was of, so it was three to the minus N tool, right, then what we get here is that if two was bigger than one, then this count here is actually less than one, so it's actually zero, okay, but we took out, if you remember the origin, right, but the origin is always there, so so what we get in this case is that the the set L intersect with the the tool well approximate point, this W tool is actually just the origin, right, so it's not particularly interesting, right, so what about the other parts, right, so maybe I should just say, I don't have time, so how long do I have 50 minutes, 50, you have 15 minutes, yes, another 50 minutes, as long as you need, well, 10, 15, whatever you need, no, no, no, I'm sorry, all right, so look, so you know, what I want to do, let me just say one thing here, so this convergent measure part, right, it really is just coming from this count, because remember what we said was that we had, we got the diagonal line, right, and we've got these precisely the rectangles, so we're counting rectangles which intersect, right, and what we have, we're then measuring, whoa, we're then measuring this intersection here, well this intersection is is roughly, well, actually we can assume that all of our rectangles actually intersect properly in this way, and so in other words, none of them are doing this, okay, so anyway, an upper bound is going to be at least two times this shorter length here, right, so two times epsilon psi of n over three to the n, all right, so that's for one of these rectangles, well how many of them are there, well it's bounded above by this three to the n epsilon psi of n, right, so what do we get is that the measure then of the, for a fixed n that of the rectangle, the rectangles intersecting the line is bounded above by this epsilon psi of n squared, right, and again we just use Borel-Cantelli, so if this sum converges then set of points which line infinitely many of these intervals on the diagonal is of measure zero, right, so that's actually how you prove the convergent part, and the convergent and the sort of upper bound for the dimension is pretty similar, but instead of using the Lebesgue volume sum here, you take the diameter here of the intervals and raise it to some power s, okay, so that's, and then you ask for the convergence, another question, who's this? Sam, can you please go ahead and ask your question? By Sam, hello Sam, how are you doing? Good, how are you? Very well, yeah, enjoying your talk, so my question is just, can you replace two to the n and three to the n by pretty arbitrary functions for the convergence theory, because it doesn't seem like you used the fact that they're two to the n and three to the n yet. No, so yeah, so here, basically lacking sequences will do. Okay, all right, right, but see that there's a problem, so on the number theory side it's fine, right, but it doesn't then match necessarily with the matrix problem. Sure, that's okay. I guess this problem you could ignore matrices and just try to say that you're approximating, yeah, you're approximating x by fractions with denominator two to the n and three to the n at the same time and into the arbitrary functions here, or darkenery functions. Yeah, of course you can do that, but see, the point was that we started off with the matrix problem, right, which is, no, but this is fine, right, I mean, so when we're writing this stuff up, we're going to do it in two different problems in some ways, right, but we don't know how, well, we haven't really thought about it very much yet, how to go about trying to do this in the matrix case, okay, so the point is, you know, because these are integers, then two to the n of x, I mean, two to the n of x is taking the fractional part is the same as, you know, looking at two times x fractional part multiplying by two, et cetera, et cetera, right. Yeah, I can see why these come from the matrix problem. Yeah, I was just wondering about something more general, because it seems like you do actually get something much more general here, like maybe you need some, like an early condition, but it, yeah, it seems like it's much more general in the matrix problem, at least on the convergence side. Yeah, okay, great, thanks. Under lacunary, it is finer either. The growth should be about the growth of smooth functions, of smooth numbers. Yep, okay, so for the lower bound, the count's not enough, and so those who, you know, worked in this area know that you also need, so you need the optimal count, but you also need something about the distribution of the actual, okay, so we actually need to have some good information about the distribution of the centres of the rectangle. So remember, what we've shown is that these, these are of the form sk over two to the n, tk over two to the n, where sk is just, remember s1, t1, this is just the solution to the linear congruence, I have such bad memory, so this is, where's it? Two to the n, three to the n, s minus two to the n, t equals one, okay, so this, so s1, t1 satisfies this, and then the sk's and tk's are just k times these reduced modulo two to the n and three to the n respectively, okay, and because we're not going any further than three to the n, these points are unique, okay, so, so let me now explain what sort of information we have, or rather we don't have, so let me just, so let me call, so let me just go back here, let me call this, so this is the number of rectangles we have, right, is three to the n, s i events, so this one's going to call delta n, okay, so we're going to split this, so the distribution of these centres really depends on whether that we have at least two to the n points or less than two to the n points, okay, so what happens, of course, is that if we have more than two to the n points, then all the denomin, all the rationals here, right, so if k is bigger than, if we're running k up to something which is bigger than two to the n, then all the denominators with rationals with them of two to the n, numerators, the fractions with denominated two to the n will actually occur, right, as points on the centres in the rectangles, right, so this is case one, so this is actually the easy case, right, so when delta n, the number of points is bigger than two to the n, okay, so what's happening there, then in this case is, I will be finishing on time, I hope, is that we've got our diagonal line, right, we've got, we've got our, so along the x-axis here, right, whoops, right, what we've got is our rationals with denominator two to the n, right, and what we get is that for each of these rationals with denominator two to the n, so if this is, let me draw this one here, so this is, so this is called, so this is a s over two to the n, right, what we're going to get is some rectangle up here which will intersect it, okay, and then because, so this is the same, this condition here is the same as saying one over three to the n is less than epsilon n over two to the n, right, so not only do all these rationals with denominator two to the n, we will get another rational here, right, so if we were to draw a square around this, right, then in this direction, remember in the vertical direction, the rationals are separated by one over three to the n, right, and this separation, this vertical separation here, right, is less than the side length, the long side length of the rectangles, right, so we're actually going to get these guys turning up as well, right, so for any, so for each s, for each s, right, we're going to end up with how many rectangles here, well it's going to be the side length which is essentially epsilon over two to the n, right, time divided by the separation of rationals vertically which is one over three, divided by the separation which is one over three to the n, so this is three to the n times that, right, okay, so we have this really quite, it's good enough picture that at every point, at every rational with denominator two to the n, we get this sort of blocks, right, where we have a good understanding of the distribution of the, of the points, okay, but this is true then for every rational with denominator two to the n, that's the point, okay, so when this doesn't happen, so when psi of n is, when this, sorry, delta n is less than two to the n, then not all rational points turn up with denominator two to the n and we need to know something about their distribution and that's where things get a bit, oh it seems so elementary but it's quite frustrating, okay, so, okay, so, so this is the case, the different, the case that we really deal with, apart from using a sledgehammer it seems, that when this delta n is less than two to the n and not all the rationals turn up and what we need is to understand the distribution of the rationals that do turn up, right, so remember that these rationals are just, what we're doing, we've got our s one over two to the n, right, and then we're just multiplying this on reducing modulo one, right, and we're going up to this delta of n, okay, so time, time, time, okay, so what we need is some information, so this is to prove the dimension result and what we need is that these rational points that we actually get that turn up in this collection here, right, is that if we put an interval around them of size, so this is just one over the cardinality but give yourself a little bit of space, this epsilon, right, and this should cover the unit interval, right, so actually we don't need such a strong statement but this is what the is asked, right, so what we could actually get away with, so remember of course this is for a particular n, right, so this s one and sk they really actually depend on, I should really put here so this s one depends on n, sk depends on n, right, so this is a statement, right, really so this is a statement which depends on n and what we really want is that the measure of the balls centered around these rational points of donate to two to the end of this size, they cover the unit interval, okay, and for infinitely many n because they don't need this for all n, okay, so a little something about the proof then of what's going on here, so what we've got is we've got delta n points, right, distinct points in the unit intervals, now by the three gaps theorem we know that there are three gaps, right, g1, g2 and g3 where the largest gap g3 is just the sum of g1 and g2, okay, so of course if g1 the smaller gap and g2 are roughly of the same size, right, then we'd have a, then we're in good shape, right, because then we'd have a good distribution of these, of these, of these rational points, right, and so up and so in that case up to a constant, so if the, then all the gaps were the same up to constants what we'd be able to do is actually prove this statement here, it's one, right, with epsilon equal to zero but we might have to put some, some constants in here but who cares, okay, so what we're able to do and this is really actually, this is really actually F. Ganey who figured this out was that if you assume the abc then you can prove that the minimum gap, right, to the power 1 minus epsilon is bigger than g2, okay, so with that information then that establishes this, this, this statement 1, okay, so I think I should, but one last thing then let me just say that in this, this condition this del trend being less than 2 to n, right, in the case of this function psi of tau being 3 to the minus n tau, right, gives you precisely the statement that tau is bigger than 1 minus gamma which is the condition in the theorem that I stated in the Hausdorff dimensions theorem, okay I think I should probably, let me see what I've got, I don't even remember what I've got here, ah okay, so plain, so we could do plain curves but okay let's just, let's call it a day here, that's fine, okay, thank you.