Thank you very much. It's really a pleasure to be back. And I just wanted to thank Alina, Michael, and Philip for keeping this amazing seminar series running. I know it's a lot of work. So thank you very much for doing this for the community. If you have questions, just interrupt me or put them in the chat. And Alina, could I ask you if I might not see the chat? So if it comes in, if you could just let me know. So this is a little fun project that I'd like to share with you on a very basic question on smallest denominators. So here's the problem. So we look at an interval, just a second. We look at an interval of length delta centered at a point x. And we look at all rational points that are in there. Of course, we have infinitely many. And we ask, what is the smallest denominator of all rational points that fall into this interval? And now let's shift this interval across, say, x from 0 to 1 and see what the distribution of those smallest denominators is. So what I've plotted here is just a few examples of interval sizes where we have on the x-axis here just the center point x. And you see how it moves through. And as the size, the length of the interval becomes smaller and smaller, you see a more and more complicated function emerge. And of course, as maybe it's not a surprise to you, it very much depends on where the interval is. And circle method people will recognize pictures like that. So that's the question. What is this function, the smallest denominator function? So recently, Shannon Haynes looked at this problem and said, well, let's work out the expected value for this function. So in other words, just integrate over x. And what they proved is that it's 16 over pi squared times 1 over the square root of the length of the interval plus an error term that you see here. And the constant 16 over pi squared is kind of an interesting one, right? It has pi squared in it and so on. So the question is really where this is coming from. And I'd like in this talk to shed a little more light on that question. But also, of course- So actually, can I ask, isn't it in the case that log of delta as delta becomes small, becomes very big? So how is that an error term? Ah, so the log of delta, yeah, you can put it as log of 1 over delta if you want, right? Delta to the minus 1 half even grows faster than log delta. Are you happy with that answer? So log delta grows slower than delta to the minus 1 half. OK, thanks, yeah. Yeah? OK, so and here is just a little illustration. That's sort of 16 over pi squared here. And you see it sort of makes sense, yeah? So the little numerical illustration that I've given you is in the right ballpark of the Chen Haines result. Now, there has been a conjecture from the 70s that I'm now going to talk about, where instead of looking at x as a continuous variable, we just sample on discrete points of the form j over n. And in order to explain that to you, I'm just going to redefine slightly my smallest denominator. So I'm not going to take it in an interval centered at x, but just as an interval where the left point is x. And I'll do this because what I want to do now is I want to simply divide my unit interval up into these small boxes. And I'm just asking now what is the, so this is 1 over n, 2 over n, and so on. And I'm just going to ask what is the smallest denominator in each of these boxes. And if you could say, well, isn't that the same thing? Because x and j over n are very close to each other. Well, this is a rapidly varying function, the smallest denominator function. And it's sort of illustrated here. You lose a few values. For instance, this one you don't get. And here you see a whole host of values. So by sampling of the order of x being j over n and the interval of size 1 over n, it's actually a more subtle problem. And indeed, following the Chen-Hains paper, Balazar and Martin proved that if we take the expectation of the smallest denominator function at those discrete points, we still get the same answer. So I've written it in this way. So we first have this discrete average. And then remember the size of the interval delta is now 1 over n. So that is again over delta to the minus 1 half. And they prove this result with an error term. And the proof is more involved than Chen-Hains, in both Clustermann sums and some refined estimates. Precisely for the reason that, as I explained, it's a more subtle evaluation now. Right. So we have the expectation value. And so a natural question is, well, do we understand the whole distribution of our smallest denominators? And here is a theorem that says yes, we do. And the limit distribution is given by this nice formula here. So let's just recap. So we have the smallest denominator. The expectation was going like d to the minus 1 half, if you remember. So I'm normalizing by d to the 1 half to make this a quantity that fluctuates around a constant. And we're asking, what is the probability that for random accents, a sub-interval d of 0, 1, what's the probability and the limit of small delta? And that's the answer. Now, here is a little numerical experiment where I've partitioned the unit interval into the 3,000 points that appeared in my abstract. So we have, again, our unit interval. And we're just sampling now the 3,000 points. So that's 1 over 3,000 here. In each point, we choose the smallest denominator, and that's the histogram. And you see it follows pretty nicely the curve. Unfortunately, my little laptop couldn't do more than 3,000 points without breaking down. But I always do like doing experiments that have a sort of, that actually show also the fluctuations around the limit. So now, how can we prove this theorem? And I should have mentioned there's also a nice paper by Albert Artilis. And Albert is in the audience who proved the existence of the limit and expressed the limit in terms of a certain random variable on the space of lattices. And I'll come back to that in a minute. But the nice question here is, where is this formula coming from? And that's the thing I'm going to explain to you next. But before I do this, let me just say this is not the first time I've seen this distribution. In my work with Andreas Stoenbergson 10 years ago, we found exactly the same distribution in the shortest cycle length of a large random circulant directed graph of in and out degree 2. So here are two examples of such graphs, or actually four examples, this one, this one, this one, and that one. And what we do, so a circulant graph is basically a graph where you connect the end vertices you have in a circulant fashion by telling you which neighbors you connect. So for instance, C8 here connects every other neighbor in this way. And the three refers to connecting the third, which is drawn here. And this one here works similarly. And now this is the distribution corresponding to the directed graph. And you see that these guys here have exactly in each vertex two coming in and two coming out. I won't say more about this, just a little advertisement to have a look at this paper that proves such limit laws for all sorts of geometric quantities, such as a diameter, cycle length, et cetera, four circulant graphs. OK, so let me show you the proof of the theorem for the limit distribution and explain to you also what the limit distribution is. Apart from the explicit formula, what I'll show you is that it's connected to the statistics of ferry fractions. So here is the ferry sequence of level Q. So all denominators reduce fractions with denominator at most capital Q. And we can write it in this way. So Z2 hat here denotes the set of primitive lattice points. And you may recall that the number of points in our ferry sequence of level Q grows like Q squared. And here we have exactly 1 over Z to 2, so 3 over pi squared. Now, the key points in this observation is that you can relate the distribution of the smallest denominator to the so-called void statistics of ferry fractions. And that solves the problem in one step, because we can then refer to the large body of work that we have on the distribution of ferry fractions. So the smallest denominator is bigger than L times delta to the minus 1 half if and only if the set of primitive lattice points so that Q is less than L to the minus 1 half. And we have a rational point in this interval is empty. So I hope that that makes sense to everyone. So if I want my smallest denominator to be bigger than this quantity here, I can only have that if I don't have any denominator that's smaller. That's all this says. So that's the key observation to make everything work. And now that you stare at this and you can convince yourself that that condition here is equivalent to the ferry fractions of level Q not intersecting this particular interval here, if we choose the level to be exactly L times delta to the minus 1 half and the length of this interval, well, that was just exactly given by delta. So you just need to tweak things in such a way that this all works. So if you plug this in here, the Q and the S, you'll see that the second equivalence that we have here is indeed true. So in other words, to recap, the smallest denominator in the interval around x is larger than L times d to the minus 1 half if and only if the ferry fraction doesn't intersect this interval here. And now remember that x is random. And when x is random, the probability that this happens is called the void statistics. And as delta goes to 0, Q goes to infinity. Remember, L is fixed here in all of this. The limit of the void statistics for the ferry fraction was proved in a very beautiful paper by Kargayev and Zyglians Zygliansky in 1997. And they showed that the void statistics converges to the so-called Hall distribution. And so this would be the density. The Hall distribution is the density of the void statistics. So the probability to having a void bigger than S, that's exactly what we need here, is therefore given by the integrated density, the distribution function. And we have an explicit formula for the Hall distribution that was calculated by Hall in 1970 already. And so now we just need to remember what S was. S was this thing here. So we plug it in. So that's our limit distribution. And in order to get the density, we just need to take the derivative of this with respect to L. And you will get the formula for eta of S that I had on the previous slide here, not on the previous on this slide, by plugging in the Hall distribution and working it all out. So that's the end of the mystery. So by using this equivalence in essentially one step, you understand that the distribution of the smallest denominators are simply related to the gaps in the Hall sequence through this relation. Great. Now, of course, it is having the convergence in distribution doesn't imply the convergence of moments. And that was precisely what Chen and Haynes proved and also the proof of the Kruswig-Meier conjecture that I had on the previous slide by Balazar and Martin. So we need to do some extra work to get this. And what I'll show you now is that we can use the techniques that I'm going to show you to even get convergence of moments and therefore get an extension of the theorems that were just restricted to the expectation value. So what this theorem E here tells you is that indeed, we don't just get convergence of the expectation, but for all moments, we have a restriction here on the moment. And that has to do with the fact that the limit distribution that I had given by the density eta has a heavy tail. So actually, moments with real part alpha greater equal to 2 or larger will diverge. So the limit will be infinite. So with this normalization here, we need to restrict alpha to real parts strictly less than 2. And then we get convergence of moments, both for the continuous average and also the discrete sampling here that was studied. And there was sort of the challenge of a crucifix-mayer conjecture. And we also have an explicit expression for the limiting moment. And this function here is a close relative to the function computed by Kargayev and Zeljanski for the moments of the void distribution of faring fractions. You've seen that the two are related. For the convergence in law, the two, as I've shown you, are actually equivalent. For the moments, they are not because now you're, if you like, integrating over L, so L is no longer fixed. So the theorems of Chen and Haynes and Kargayev and Zeljanski are not equivalent, even though they're quite close in spirit. Good. So the nice observation is that this limit function here, m alpha, is elementary apart from the beta function here that we have. And that's Euler's integral of the first kind, which has explicit expressions. And you can work it out. So in particular, if you look at the expectation value, which corresponds to alpha equal to 1, you'll get your 16 over pi squared. You don't need to do a big calculation to find that. So everything's consistent. That's great. And in fact, you can do more generalized moments. So for instance, you could say, well, I'm going to combine my alpha moment with a log moment like that, where n can be any integer. And you can prove a limit theorem for this. And it was sort of fun to work out some of these things. So here's a list of the mixed alpha moment and nth log moment. So here, if alpha is 0, that'll give you the first log moment, the 0th log moment, which is of course 1, because it normalizes the probability density. But then here, you get this formula, for instance. So it's all quite pretty, what you see. You get the Catalan constant for certain negative moments. Anyway, that was all fun. But now let's take a gear up and actually ask the question in higher dimensions and explain to you how and can prove these theorems, not using analytic number theory techniques, but techniques from homogeneous dynamics, which to my mind is certainly an easier first step in higher dimensions, because we do not have good continued fractions available. The analytic theory is more complicated. But having said that, all the results, I'm going to present in principle, could also be proved by analytic techniques. I'm not going to use any egotic theory that cannot be made effective. Now, before I go into this, are there any questions about what I've talked about so far? As I said, I'm going to explain now really what's behind the statistics of smallest denominators and also the corresponding distributions of fairy sequences. Everyone happy? Good. OK. Right. So let's think about higher dimensions and how we can set the problem up there. So again, we look now at rational numbers in Qn. And instead of having an interval, we choose an arbitrary test set A that we're going to fix. So think of A as being just a ball. We scale it down by delta and ask, what is the smallest denominator of all rational vectors in that test set shifted by X? So X, you can think again as the center or the endpoint of the intervals we looked at before. And we have this question. Now, I'm just going to get up and switch the light back on because I've been too quiet. This might happen a few times. OK. So that's what we're interested in here. So that's our main object. And the playground will be the space of lattices, which we can realize as this quotient space here, g mod gamma, where g is the group of n plus 1 times n plus 1 matrices with real coefficients and determinant 1 modulo, the modular group, which stabilizes the cubic lattice. So you can think of g mod gamma as parametrizing the space of lattices. g has a harm measure, which Minkowski showed we can normalize in such a way that it becomes a probability measure on g mod gamma. And now here, this guy is a cone, if you like, with cross-section A. So the picture you want to have in mind here is that this sort of looks like this. And this would be A. I've drawn it as a disk, where here we have X. This is the X direction. And here we have the Y direction. And I hope this makes sense. So this is an Rn plus 1. And now I'm defining a certain probability measure, P0A. The 0 here just stands for the fact that I don't want to have any point of this lattice intersecting my cylinder. Now, this is not quite a lattice. It's the primitive lattice points in Zn plus 1. g here is an element in my group g, modulo gamma. So that distorts my cubic lattice. And by taking the measure of g for which this holds, I'm effectively looking at a random lattice here. And I'm asking for the probability that a randomly linearly distorted lattice does not intersect the cylinder. Now, why is this distribution important? Well, it's important because it gives us the limiting distribution for the minimal denominator. And again, there's been a paper by Albert Artilis who explained exactly this using the space of lattices without reference to the fairy fractions, though. And I'm going to just show you again, as I've done previously, and how the fairy fractions explain this in one step. OK, so what does this theorem say? Well, for any test set, think of it as a ball and any d here. Boundary of measure 0, of course. Otherwise, this might not work. And any L greater than 0, we have a convergence in law. So the probability that q min appropriately normalized. And now we don't have the square root of d here, but we need to adjust our normalization according to the dimension we're in by this factor has a limit distribution. And this limit distribution, EL, is given by this formula. In other words, it's given by the probability that a random primitive lattice does not intersect our cone. We know that the tail of this distribution is power-like as to the minus n. That was proved by Andreas Trömbergson in 2011. There's also a beautiful paper by Afrija and Margules, to prove this as an upper bound for very general sets A. And so this means in particular that if we now just remember the relationship between S and L, which is this, this would be S here, we get this kind of decay for our limit distribution. And you see, we can only have finitely many moments. So the L to the n plus first moment would diverge. Now, the proof of this theorem proceeds exactly as before. So remember, this is the definition of our fairy sequence now in higher dimension. So the fairy sequence level q is exactly given by those fractional points with denominator at most capital Q. Now, the growth of the sequence goes like q to the n plus 1. Remember, in one dimension, it was q squared. Now it's q to the n plus 1. And we use exactly the same argument as in the one-dimensional case, observing that the smallest denominator can only be bigger than L normalized in this way if and only if the fairy sequence doesn't intersect with this set. And here is, again, the relation that you need to work out between qS on one hand and L and delta on the other. And now we are done because we know everything about the fairy statistics in higher dimensions as well. I worked this out in about 10 years ago, drawing heavily on my joint work with Andreas Trömböckson, where we looked at much more general questions on distribution of lattice points in randomly sheared and rotated set. So that's that. And now, well, let's look at the moments. And now I'm going to give you the proof on actually how you what you need to do in addition to understand the moments. So this is now in higher dimensions if you realize, again, with this new normalization factor. And we simply take the smallest denominator integrated over x, or raise it to the power alpha, integrate over x, and we get the limit distribution here. And as I said earlier, this guy here decays like 1 over L plus n plus 2, actually, it should be. Let me just see here. There might be a mistake here. I'll double check that. I'll get back to you tomorrow on this. So in order to have a finite moment, you need to restrict alpha here. And that's the correct restriction. So how do you prove such a statement? Well, if the real part of alpha is 0, then you can, you don't need to do anything in addition because you already have convergence in distribution. And the reason for this is, so if you understand the probability of Q min being larger than L, you can use a usual approximation argument of bounded continuous functions by characteristic function to prove a statement like that. So that's the usual weak convergence statement. So for any bounded continuous function f, we have this convergence. And now you simply realize that if the real part of alpha is 0, this will be a bounded continuous function, and you're done. Now, if the real part of alpha is bigger than 0, then you need to do something more. You need to work harder. And the first step is to realize that you can write the alpha's moment and integration by parts here as an integral over your distribution, which is here. So this gives you the alpha's moment. And so in order to prove convergence of this integral, you need to ensure that you don't get too much weighted infinity of your distribution. And that's not just guaranteed if you have convergence in distribution. So you need an estimate that holds for all l and delta uniformly. And now this is the kind of estimate you need to prove here. So you need to exclude that you have accumulation of mass at infinity. So you integrate from r to infinity, the kind of object you want to work out, and then take the limb soup over delta to 0 and then r to infinity. And if this is 0, it guarantees that you don't have that escape of mass. And then you can commute the limits delta to 0 and r to infinity. So then you can just simply replace in the limit the integral over l. And you get that the alpha's moment is just given by the alpha's moment of the limiting distribution. OK, so the way this works is just to recall, we have this equivalence that I've showed you before. So this is equivalent to the ferry fractions not having a void distribution. And now something really quite pretty happens. So you remember that the ferry fraction not having a certain point in this interval can be lifted to a higher dimensional setting. And that's exactly where the argument of the, not the argument, where exactly this observation comes in that the ferry fraction not intersecting this test set is exactly equivalent of the primitive lattice points not intersecting my cylinder. And that, in turn, is now equivalent of, so what you do here is you multiply on both sides with an appropriate alpha s. And then you see that this condition is equivalent to this condition here, where now your cone is a standard cone that's homothetically dilated by l. So if I had a blackboard, I will do that calculation. But I think you can do that calculation yourself to get from here to there. And then what you need to estimate in order to get this escape of mass bound, well, you need to estimate the measure of x for which this condition is true. And the point is that when l is very large, you look at a very large set that is not allowed to intersect this primitive lattice. So our cone contains a ball because we assumed that A has non-empty interior. And we can understand the shape of lattices that avoid a large ball of radius roughly l. If you, these by Mahler's compactness criterion, these balls, these lattices that avoid a ball of large radius need to be high up in the cusp. And so what we need to now do is we need to look at these shared lattices here that do this. And you can estimate the measure of those x for which you are high up in the cusp by a calculation that Vion Kim, who is also here in the audience and I did recently, and you can show that this is exactly bounded above by l to the minus and l plus 1. So the geometric picture here is really that you are now in the space of lattices and you can geometrically interpret this action, the shearing action, by x as a horosphere. So what we're really estimating is the part of the horosphere that's high up in the cusp in the modular space and getting a uniform bound on that. So that's what you need to do if you want to prove this. And that then completes the proof of the upper bound at infinity. So we have no escape of mass provided alpha is less than n plus 1. And if you have a negative moment, you need to do exactly the same thing. Now you do your integration by parts the other way around and you need to estimate now the set of x for which your minimal denominator is less than l. So what changes from the previous consideration is now that we need to avoid escape of mass near 0, because if the moment is negative, this is bad near 0. It explodes. And so we just need to make sure that the measure of this set is very, very small for small l. Now since we look at the minimal or the smallest denominators less than something, this now translates into the condition that our fairy fraction does have a point in that interval, and therefore that our primitive lattice does intersect with the comb. That in turn for small l again means we have to have a short vector in our lattice and we can again, by a different calculation now, estimate the set of x for which our lattice, this lattice here has a short vector of length less than l. And that gives us the corresponding estimate that we have here. So now the measure decays like l to the n plus 1 for very small l. And again, we get the correct estimate and everything's fine. Good. So we need to get the moments to really prove those additional escape of mass estimates. And that's not a surprise, but it is a significant additional input to go from the convergence in distribution to the convergence of moments. So now I want to come to the discrete sampling aspect, which was the recent proof of this longstanding conjecture from the 1970s. And I want to show you that this also works in higher dimensions. So as you've seen before, we had to look at the measure of x for which a shear had a certain distributional property. And we also had this matrix appear, which was related to our small parameter delta. And what's really underlying all the theorems about the convergence in distribution of this fairy statistics that I showed you on the previous slide are equidistribution theorems on the space of lattices of this form where you integrate, there should be a dx here, where you integrate the horosphere that corresponds to this action. And then you sort of translate it by this diagonal action here, the aq. And as q tends to infinity, one can prove, and these are theorems that go back to Margulis' thesis and a very beautiful paper of Eskin and McMullen who built on that, you can prove that these kind of averages equidistribute with respect to the high probability measure on g mod gamma. And so that's an old story. But if we now move to the discrete average case, we need to replace this continuous average over x by a sum over the j over these small elements. And so instead of having now a continuous average over our whole cycle, I just sum, I still transport everything by this diagonal action. And my claim is now that we still get equidistribution. And once you have this theorem i, everything follows exactly in the same way as before in the continuous average. So the theorems we have known already for 10 years can then just be adapted to also the discrete averages. And it's a non-trivial thing to move from here to there. And I'm going to show you now why it is non-trivial and how, nevertheless, ergodic theory really helps us to understand if you like why the two averages have to be the same in a sort of very nice robust fashion. So what I'd like to show you is that theorem i is true, which means that this kind of average here converges to the same average as in the continuous limit. So I'd like to relate the two in some way. So the first thing I do is I define a probability measure nu j, and this is again a typo that should be nu i. So I define a sequence of probability measures nu i by simply fixing a subsequence of integers n i here. These are my scales over which I sum. And so I have a sequence of probability measures. And what I would like to prove is that any such subsequence converges to my limiting high measure mu. Good. Now the key observation now is that we have a very nice commutation relation here where we can commute through the h and the a. So again, that's very simple linear algebra. But what happens here is that the object that I'm interested in, which is the x0 plus n to the minus 1j, I'll take this part here and I'll convert it through and you get exactly this thing. So why is this important? Well, you've seen in my theorem that I had this assumption here that q to the n plus 1 has to be less or equal to n to the power n. And you see here why this is important, because then this thing here will be of constant or decreasing order. What this also means in geometric terms is that actually as you vary the points J, so as you go from J to J plus 1 and so on, they are a finer distance apart. And so that's already a good thing because that means we can use the same escape of mass estimates as in the continuous setting. But how can we use that connection? Because they're finer distance apart generally to actually reduce the problem to the continuous average. So it's easy if we choose the sequence of q's and i's, the q i's and n i's in such a way that this thing goes to 0 because then actually they're not just a finer distance apart, but they're a very small distance apart.  So you can approximate the continuous average. You can use the continuous average to approximate the discrete average. So that's when this thing here converges to 0 along a certain subsequence. So let's assume that doesn't happen, but it converges to a positive limit tau 0. If that happens, then my points are a finer distance apart. So how can I then use this type of Riemann sum argument? I can't, right? Because they are just really a finer distance apart. I need something a little more sophisticated. And the sophisticated argument will use the ergodicity of the action. Now, I won't go through the full details here. I'll put those slides online. And you can look at them at your leisure. But this is a really pretty argument that was shown to be by Manfred Einziedler. And the idea here is that you can use the ergodicity of various actions to show that the limits have to coincide. And the argument is the following. So we're interested in the limit of this sequence of measures, nu j. So the first thing I do is I define a slightly fettant nu j. And I'm sorry, this should be nu i in this way. So I'll just, you remember, I have, this is a discrete measure. I'm just summing over those points at j here. And now I just smooth it slightly by epsilon. So I'm bringing it a little closer to the continuous average, but not quite. And epsilon will be very, very small. This here is my continuous average. I want to show that these two have the same limit. And I'm also defining the complementary probability measure, which is defined in such a way that if I add those up, so the one I'm interested in, the nu epsilon, and the complementary measure, I'll get exactly mu. Now, we use an argument that says that, well, let's suppose, first of all, my fettant measure is converging to a limit along a certain subsequence. I know, of course, by the equidistribution that I already have for the continuous average, that mu i will always converge to mu. So along the same subsequence, we have also this convergence. And by this relation here, I also know now that mu i epsilon bar will converge to the right limit. And because my mu i's and mu i's and mu i bars satisfy this relation, also the limit has to satisfy this relationship. Now we know that mu is ergodic with respect to this so-called horospherical action up here by this subgroup. We also know that mu epsilon is ergodic with respect to this action. And if you have an ergodic measure, you know that it's extremal. So it can't be written as a linear combination of other invariant measures. And this means that the three measures have to be the same and have to be harm measure. And then the last point that you don't just need to check is that the fettant measure, mu i epsilon, is actually close to the original measure you're interested in. And so that's it. So this is the ergodic theoretic argument. If you want to do this with analytic number theory methods, you have to do harmonic analysis. And it's much more complicated. But it can also be done. So it'll involve in the one-dimensional setting clusterman sums to resolve this difference between the discrete and the continuous average. And in higher dimensions, it will involve the higher dimensional analogs of clusterman sums. And there are results in the literature which would give you that. OK. So I think I'm coming to the end of the talk. Just two more slides. The first one is just to say, we can, using this equivalence, prove analogous statements also for the higher dimensional distance function to ferry fraction. So you say, OK, what is the distance of a given point x to the ferry fractions? In other words, to the nearest rational point with denominator at most q. So that's what's written here. This is something that Kargayev and Sigliansky mentioned that result earlier looked in 1997. And the techniques that I showed you prove that you also have the convergence of the beta moment of this distance function to ferry points, provided your moment isn't too large. And another statistics that I really like is what we call the pigeonhole statistics. So the pigeonhole statistics, say, for a sequence of numbers in the unit interval, fractional parts of your favorite sequence, works as follows. So you, again, partition your interval into n boxes. So each box has size 1 over n. And you look at the first, say, n elements of your sequence. You can also look at the first sn element of your sequence. That's actually what we're doing here. Or maybe I'm, yeah, well, let me not do this because I think I'm looking at the first n over s integer part elements of my sequence. So you have n boxes, and you ask yourself, where are the elements of my sequence going to be? So there might be two here, there might be none in the second box, then there might be three here. And you're simply asking, well, how many boxes are there that have no element of my sequence in it? How many elements, how many boxes are there that have one, that have two, et cetera? So that's the pigeonhole statistics. And we can, again, use our discrete average to show that the pigeonhole statistics for the fairy sequence converges has a limit. And this limit is, again, given by the probability that a random primitive lattice intersects this cone that I had on the previous slides in exactly k points. So I think I'm done. Here are references. The main references, this paper highlighted in yellow, where you find everything that I've described here. The main upshot of my lecture was, I hope to illustrate to you, that this beautiful problem of finding the smallest denominator is actually related to some really quite sophisticated mathematics, surprisingly, maybe, but that it also relates to a whole body of work on the fairy statistics that was started with Hall in the 1970s and that many people worked on. I only mention a few here, but there's a lot of work also by Boca Zaresco and their coworkers that can connect it to this beautiful problem. And you see some beautiful limit distributions that, certainly in one dimension, you can analyze explicitly. In higher dimension, it's still a big open challenge. We don't really know much about those lattice distributions, except for the tail estimates that I've showed you, the heavy tails. And so it's rather difficult to get explicit formulas for those. Good. So that's all I had to say. I'm going to go back and switch the light back on. And while I do that, you can maybe think about if you have any questions for me on this. Thank you.


So this one says acrolein, and the structure is shown there, absorbs its maximum wavelength at 217 nanometers with a molar absorptivity of 16,000 mils per meg per centimeter. What is the concentration weight, or the weight concentration required to observe an absorbance of 0.80 when the path length is 1.0 centimeters? So again, we're using Bayer's law for this one here, but we're solving for concentration. So what is concentration? That's what we're looking for. And it says the weight concentration. What that means is milligrams per milliliter. So of course, all we're going to do is rearrange this equation. So c equals a divided by epsilon b like that. So c is going to be a is 0.80. So divided by epsilon, so epsilon is this. So 16,000 milliliters. And this is 1 milligram times 1 centimeter. And we're going to also divide it by path length. So we'll just put it over here. 1.00. So is everybody cool with me doing it that way? So again, I just flipped this over because we're dividing it. So that will allow us to cancel our units out centimeters, centimeters, giving us a mid per mill weight concentration. Does that make sense? So notice the molar absorptivity units are different in this. And the concentration units are different in this one than the last one we did. So molar absorptivity and concentration units kind of mirror each other. So now all we've got to do is pull it in. So 0.8 divided by 16,000. 5.0 times 10 to the negative fifth milligrams per mole. So that would be the concentration. Okay. Are there any questions on this one? Again, pretty straightforward, I think, just rearranging the equation.


What's good, YouTube? This your boy T-World back to y'all again with another art tutorial, man. If you new to the channel, go ahead and hit that like button, comment, subscribe, make sure you click post notifications so you will be notified every time you budge out to the heat. In today's video, I will be giving you guys a full walkthrough on how to make the cartoon head. Start to finish. So without further ado, let's jump right into it. OK, the first thing you want to make sure, you want to make sure the size of your layout is 18 by 18 inches. Before you start your project, go to File, go to New, and then just make sure you got this size. This is a good size for Instagram and stuff like that. That's the size I use. Since I already got it saved, I ain't got to do it.  If y'all are always wondering how I get those lines, I use a digital tablet. It's called UG, where I draw right onto the monitor. But all I do is come right here, and I go to New Brush, then I hit OK, and then I click Pressure. Then I hit 3, and I put 3 right there. And now make sure I got my extreme. And all it is, it's not a certain style of line. It's just that I start out light. It's just me using a pressure technique. So I start out light, and then I press down hard in the center, and then I end my lines with light pressure. So it'll give me that effect, like it's a calligraphy pen or something. But yeah, that's how I get that effect. I know a lot of people be asking me, what kind of brush you use, and I wish I had your brush, and I'm not even using a brush. Now this is the setting for my tablet. So I put my tablet on high pressure. That mean I got to press down firm for it to respond. So with that, it didn't even help me press down even lighter since it's harder to pick up. So when I end my lines, it'll be a little more sharper. But yeah, all you got to do is set your pressure sensitivity if you're using a tablet, and you'll be able to get those exact lines you see me using. Now the first thing you want to do when you're dealing with making your cartoon heads is paying attention to detail with your line work. So what I like to start out with is the nose and the eyes. So what we want to do is we'll do all the line work, make sure our lines are clean. Clean lines play a big part in this piece. Just the more you do it, the better you'll get at it. So if you ain't that good, just stick to it. Mute this picture so I can make sure my lines are looking clean. I'll just clean and set it every now and then.  Make sure you don't leave no openings. So when it's time to do the coloring technique, you won't have no problem. Because it's a lot of people telling me, like, man, every time I try to do the coloring technique, you're showing your videos that are always not colored certain areas and stuff like that. It's because the reason it's not. Because just say if I wanted the color to be a different color with the technique, I'm going to show you all later in the video. If I got a little opening like this, it won't be able to recognize this as an area. It'll just go ahead and color everything outside of this as well. So just keep that in mind. As you're doing your line work, make sure you're closing off areas that you know you're going to want to be a different color later. I'll just shape up this head. I brush back small, 25, trying to make some clean cartoons. You really got to utilize your pressure. You got to learn a certain technique yourself to invest in you a tablet so you can have that option. You get more control when you get to really draw yourself other than using your inner hair. We're going to fill in the hair, and we're going to fill in the facial hair as well.  Make sure you have these settings for your pencil tool. Anyone hit OK. All right here, pull that color to the top, move black to the top, and make sure your bottom color is locked. Now we're going to come back with a pencil tool and we're going to fill in this hair. When I'm filling it in, I go ahead and try to add a little detail. And you'll see what I'm talking about.  I want to go ahead and try to give it like a hairy effect. It's just like the base layer for the hair. It is not like what we're going to do. I finally get to what I was doing. If it went ahead and filled in the area after I close that area, do all that with the hair. I like to fill in my eyes as well with that same technique. And this technique right here helps you cover a lot of ground as much as I can with it. It's good to use for like a base. And then you go on top of it with your fine detail. In the detail, so you use that effect for the base. That'll cover a lot of ground for you. I was going to go back to our brush tool. Other thing I like to do too, I like to copy some of these little pieces of hairs I made. I'll select it. I just drag over with my mouse.  Another thing too, doing all these hairs will eventually, if you don't have a strong computer, it'll start to slow down on your computer. So what I do is like once I got to a good point, I go ahead and I copy my line work. Go to your selection tool right here. You want to drag to copy your line work. Once you got it all selected like this, you want to go to object, expand the pinch, and then you want to click merge on your path finder. If you can't find your path finder, just go to window and then select your path finder, where it's in right there and drag it out here. See what I just did, I made all my lines one. So it saved me some space on my computer if it's not fast enough. You know what I'm saying? If you don't got a fast computer, you want to stay doing that because doing these little tedious pieces of hair, it'll start to slowly slow down your computer because it's too much going on. So that's one tip you need to know. Just like I told you, every now and then we're going to make sure that the object expands a tedious work. Same thing, but we're going to ease our way down. We're going to make also, we're going to place some under our previous lines. It's just how far you want to go into depth. You ain't got to go this deep into it, me personally. And really when you create an art, don't worry about finishing fast. Just take your time because it's all about the finished piece. It ain't really about how fast can you do or how fast can you draw something. Just really take your time and you'll get faster with time. You know what I'm saying? But the more you do something, the better you'll get at it. And then just focus on perfecting your craft. Don't worry about speed. You know what I'm saying? Just see this fade from this side. I'm small as we can give it. Good thing about Adobe Illustrator, you get to zoom and expand the pen. A little minute because I ought to let extra hair. But yeah, there you go. Find you a good base color to start out with. It ain't got to be too perfect because I always, I'm going to show you what I do. I'm going to find my colors, click OK. Right click, your shape. Go to your rectangle tool. And you want to drag that color over the entire picture. Now this is why I say it's important to make sure your lines are closed off because this is going to let us know if our lines are closed off or not. So after you drag your color over your picture, you want to right click the color. Go to arrange and send to back. And you will know if you did this step right. If you look over here at your color layer and you see your line work on top of your color. So after you do that, you want to highlight everything. Then you want to go to merge on your power finder. Still loading, wait up. OK, then you want to right click onto your color, click isolate, select group. Now on this outer color right here, and press delete. My lines are closed off because it didn't delete inside my cartoon. Another thing I like to do to get these, get the color out of the hair, what you want to do. Double click onto just any area right here. Come over here, click the drop down menu. Click it again. And you want to stroll down until you see the last black line work. Leave all this open. Leave the colors open. Just lock all the black layers. So what I do is hold the left trigger on the mouse. And I stroll up to lock everything that's black. So I go down to the bottom to my last black line work. And I stroll up, holding my. And we finally made it to the top. What you want to do, you want to unlock these top two right here. We showed up, and we locked all our black line work. And you want to make sure these two are unlocked.  You see how I got it selected, but it don't got our black line selected because we locked out the line time. That's how you got to hit delete twice. This here, this part, isolate selected. We only get to select that area. Hold shift or any deeper into coloring the base color. I want to edit, edit color, adjust color balance. I just choose whatever I think look good to me. I don't really. The base color laid out. It's time to do the shading. So what you want to do, you want to create your layer in between your line work and your color layer. And when I'm doing my set, my shading on all my cartoon heads, I start with the darkest tones. So we're going to start with a dark shadow. And then I use my I use my pencil tool to do my all my shadows. So we're going to start with the eyes sample the skin tone. Always put your shadows on different layers. Like I put all my dark shadows on one layer and then I put all my light shadows in the lighter I go. I put it on a separate layer just in case I want to go in the just colors at the end when you're doing shadows, you ain't going to find at first. So you want to already have it on a different layer so you can go back and change the color if needed real good at shading. It's going to take a lot of trial and illustration.  So you got to kind of see how it's shading look using these clean lines. So it takes some time to get because it's a different style.  But illustrator, you got to really draw out the shadows. It's a different style. You start make creating your see how we got it on our own layer. We're going to highlight make sure you put this to all under our dark shadow. So every shadow layer we add is going to fall. We're going to make sure we put it under the next shadow. OK, now we're just going to go a little lighter. This is our in between tone shadow. We made above the shadow, the dark shadow because it's going to fall right under it. Anyway, it's good to go ahead and put it on separate layers. And we're going to have to change this color layer, but you already know the color. It's going to be perfect right now. Now we're going to add the details inside the hair. So what you want to do, create your layer above everything. So let your gradient right here to your gradient.  You want the center to be a bright color. Be straight black. You want it to be kind of faded into like a brownish, blueish color. Do a fade for some hair texture. Now you want to know how do I export my artwork when I'm finished? I go to file, export, export as, export as. Make sure you close your art board, export. 300 pixels. And this I say that as a PNG file. Okay. And there you have it, man. Make sure you stay tuned, more heat coming soon. And I'm out this thing.


وَأَقُولُوا فِي القُرْآنِ مَا جَاءَتْ بِهِ أَيَادُوْهُ فَهُوَ الْكَارِيمُ الْمُنْزَالُوَ وَأَقُولُوا قَالَ اللَّهُ جَلَّ جَلَالُوهُ وَالْمُسْطَافَ الْهَاديِ وَلَا أَتْ أَوَّالُوَ الحمد لله الذي شرح صدور أوليائه بالإيمان وفتح لهم أبواب النصوس بقواعد البيان وصلى الله وسلما على من أنزل الله عليه الكتابة والميزان وعلى آله وصحابته ومن تبيعهم بإحسان أما بعد now we're going to go into the third chapter, العلم after the author, رحمه الله الشيخ محمد بن صالح العثيمين he spoke about the definition of فِق and if you remember correctly what was the definition that he gave for فِق the definition that he gave for فِق was معرفة الأحكام الشرعية العملية بأدلتها التخصيلية what was the first thing that he said معرفة الأحكام الشرعية and if you remember correctly he said that معرفة والمراد بقولنا معرفة is العلم والظرن so the first point of definition of الفِق was معرفة is to know and he said two things fall under معرفة العلم والنالج and a ضن high speculation and we spoke about that we spoke about that in our first first lesson so because the first point of the definition of فِق علم falls under it the author, رحمه الله الشيخ مهم to speak about knowledge and also to speak about also a ضن which is high speculation so that's why he's not going to go into العلم knowledge the author, رحمه الله he says العلم تعريفه العلم يدراك الشيء على ما هو عليه يدراك جازمة كي يدراك أن كل أكبر من الجزء وأنني يتشرط في العبادة فخرج بقولنا يدراك الشيء عدم الإدراك بالكلية ويسمى الجهل ويسمى الجهل البصيط مثل أن يسأل متى كانت غزوة بدر فيقول لا أدري وخرج بقولنا على ما هو عليه يدراك على وجش يخالف ما هو عليه ويسمى الجهل المركم مثل أن يسأل متى كانت غزوة بدر فيقول في السنة الثالثة من الهدراء وخرج بقولنا يدراك جازمة يدراك الشيء بحيث يحتمل عنده أن يكون على غير الوجه الذي أدركه فلا يسمى ذلك علمى ثم إنتر ثم إنترجح عنده ثم إنترجح عنده أحد الاحتمالين فالرجح ظن والمرجوح وهم وإن تساول أمران فهو شق وبهذا تبين أن يعشق الأتي الأول علم وإدراك الشيء على ما هو عليه إدراك جازمة الثانية جهل بصير وعدم الإدراك بالكلية الثالث جهل مركب وإدراك الشيء على وجه يخالف ما هو عليه الرابع ظن مع احتمال ضد راجع شك وهو ادراك الشيء مع احتمال ضد مساول ادراك شيء مع احتمال ضد مساول عندما كانت بطلاف بدر حيث أنه يبقى في الثلاثة سنة بعد مجموعة حيث أنه كان في الثلاثة سنة بعد مجموعة وما أنه يوجد من our words أكثر أكثر متأكد من ذلك أنه يوجد في ذلك أنه يوجد في ذلك أنه يوجد في ذلك أنه يوجد في ذلك لذا يجب أن يكون مختلف من the way it was perceived فهذا ليست مجموعة لكن، إذا كنت أمامتت أن يفعل في ذلك وإذا كنت أمامت أن تائدك الإنت does not matter إنه لن يجب أن أتصبح أكثر من الجميع وإذا كانت نفسا سافة تقول أن الان Nicholas وحاجة لتعرفة أنه يوجد في ذلك ربما عدد أن نتخلل ونحن أنه مهمة توطيل ونحن أنه رائع ونحن أنه تعرف أن врائع and حيث يجب أن يكون هناك مجموعة مجموعة في مجموعة مجموعة مجموعة ثانية مجموعة with the prospect of it being against a less correct view مجموعة fifth مجموعة with the possibility of it being against the most correct view مجموعة sixth مجموعة with the possibility of it having something equally correct opposing it now insha'Allah we're going to speak about what the scholars refer to and they call it مراتب الإدراك the levels of perception the levels of perception the highest of them is knowledge and the second one is ignorance and ignorance is two types and then we're going to go into ظن I will see what ظن is speculation وهم which is suspicion and shak which is doubt and what each one the difference between it we will see insha'Allah my translation may not be accurate but my explanation will be insha'Allah the ظهر رحمه الله he started with علم and knowledge what does knowledge mean it means إدراك with شي it's to perceive something what إدراك it means in the Arabic language التصور perception so pay attention here but a person is perceiving something perception in his head something is in his head he perceives it he sees it in that way now what's in your mind when it comes to the perception and what it is externally have to be the same knowledge is what what's in your mind and what you think and what you've perceived and what it's out there have to be together they have to be the same and there's a third thing that's needed as well which is you have to have certainty that is what it's called knowledge so it's three things knowledge stands on what is it إدراك with شي there is a perception number one علامة who are as it is as it is and last but not least إدراك and جازمن the person is certain about it so for example somebody asks you and they said to you they asked you what is this straight away there is a perception in your head something you've perceived something you looked and you've perceived something second is you speak out and you say oh this is a cup a plastic cup is your perception and what it is are they the same yes and then you're certain you are what you are certain about it so when we say perception let's start how many points did we say that knowledge stands on three the first one is perception what did we push out of the definition what did we dismiss to get out of the definition what we're going to see is ignorance or one type of ignorance which is called جهل بسط we're going to see it جهل بسط the simple type of ignorance the person doesn't have a perception you say to this person what is this say it to you I don't know in his head there's no perception that's called what جهل بسط we got rid of جهل بسط okay and the chef gave another example which is asking the person متى كانت غزوة بدر when was the battle of بدر and he said لا أدري I don't know so there's no perception in his head at all we got rid of that in the first point علامة هو as it is what did we get rid of what did we push out of the definition what did we get rid of we got rid of the second type of ignorance which is جهل مركب a compounded ignorance because the second type of ignorance which is known as the compounded ignorance is what the person has perception but the problem is the reality and the perception are they are going against each other so what he's perceived and what it really is out out there or what it externally is is not in correlation they're not together so you ask the person and you say to them what is this and he says to you this is a mobile phone so is there a perception in that person's mind yeah there is the person has perceived something they perceive the mobile phone but is this a mobile phone I'm holding up it's not a mobile phone it is a what it's a plastic cup and the Shaykh gave the example of the same question when was the battle of bedr and instead of saying the second year of the hijriah after the hijriah he said third year so he has a perception but his perception is incorrect we got rid of that in the second point of the of the way the علم we got rid of it and just as a side point why is it called compounded ignorance we said the ignorance of how many types is two types it's an easy ignorance and a compounded ignorance the reason why it's called a compounded ignorance is because number one the person does not know عدم العلم there's no knowledge there and the second thing is in there is اعتقاد العلم the person believes they know so it's compounded of those two things that's why it's called compounded ignorance he actually thinks he knows and that type of ignorance is hard to remove when you don't know when you think you know are you with me it's when you don't know and you think you when you think you know and that type of person is in a sad situation the poet he said وفي الجهلي قبل الموت موت اللي أهلي he said that ignorance is a death ignorance he said وفي الجهلي inside ignorance there is قبل الموت موت اللي أهلي it's a death for the individual before the real death it comes to them if you're ignorant you're dead way before the actual death comes to you so pay attention he said وفي الجهلي in ignorance there's a death the person who's ignorant is dead he's dying he's on his deathbed or he's even a dead person he's a dead walking person before the actual death comes to him and look what he said فأجسامهم قبل القبور and their bodies are the graveyard before they're actually going to the actual graveyard you see and then he said وإني مرأا if an individual لم يحيا بالعلمي ميتون if a person is not revived with knowledge then he dies he's gonna be dead فليس لو حتى النشور نشور and then guess what anything that tries to awaken him and anyone who tries to bring him out won't be able to because ignorance is the biggest enemy ودهالك and the Prophet ﷺ he referred to ignorance as what the Prophet ﷺ he referred to it as a disease that needs to be cured you know the famous حليطة of the man who asked his companions his friends his colleagues he asked them and he said to them after he had a wound on his head and he said to them I can't use water I am scared that it will go into my head and I might die from it and etc can you guys find me a ruling out of it and they said to him we can't as long as the water is there you have to use it because they only knew that the only time that you can do تيامم is when there is no water there so he said okay so he used the water and he died he took their فتوة so then they went to the Prophet and they told the Prophet they said the man he asked us this question and we gave him this ruling the Prophet ﷺ he said they killed him may Allah kill them then the Prophet ﷺ he said why did they not ask if they didn't know then the Prophet ﷺ said فإنما شفا أولعي سؤال the cure to ignorance was asking the cure to ignorance is what asking وداركة فوتي he said وليس العما طولس وليس العما blindness and a person being blind is not طولس طولس سؤالي you ask questions too much you're not blind if you're questioning and you're interrogating and you're investigating and you're asking he said that's not blindness like he said إنما تمام العما طولس سؤال it's to be silent on ignorance for the rest of your life you're too shy to ask you don't want to ask you don't want to verify you don't want to know وليس العما طولس سؤالي وإنما تمام العما طولس طولس سؤال طولس سؤال a person if he doesn't know he should say but there is a point I want to mention here as a side benefit and I think it will benefit the people to know which is the difference between knowledge and intellects okay there is a difference between علم and عقل صح knowledge is something and intellect is another thing and there's nothing better than a line of poetry a poet he said علم العليمي وعقل العقل اختلفة knowledge and intellect had a عقل and علم had a discussion they talk to each other they had a dispute argument they said to each other منذ الذي فيهما قل أحرز الشرفة which of us has reached a pinnacle of nobility who from amongst us me or you is the best knowledge do you think you're better than me عقل is do you think you're better than me which of you me and you is better فال علم قال knowledge said أنا أحرصتوا غايةهم knowledge said I reached a pinnacle I reached a highest level I'm at the highest level و العقل قال لأن عقل he said عقل said through me Allah is known in the Quran what does Allah تعالى say in some places لعلكم تعقلون you know the eye what did Allah say أم خلكم من غير شيء أم هم القالقون أم خالق السماوات ولا ربا بلا يوقينون these verses are making you think عقل who created this who brought this عقل is saying I'm better than you knowledge you know I'm better than you because it's through me Allah was known فأفصح العلم وقال له knowledge thought for a bit and then it brought our eloquency from its mouth and it spoke word of God and it said بأينا الله which of us who are you في قرآن التصفح which of you who are you does Allah refer to himself as does Allah call himself عقل or does he call himself علي فأقن العقل then عقل realized أن العلم سيده and knowledge is his master فقبل العقل رأس العلم وصرافة and he kissed him on the forehead and they went out together so this teaches us what what does this teach us that it's really knowledge that is greater than العقل here I want to point out something very important and I really want to use this opportunity to debunk and dismantle a group of people the point that I want to mention إن شاء الله is the relationship this is a big topic itself a topic that deserves series not one or two or three series to be discussed which is the concept of the عقل and the textual evidence there are some people who exceeded their limits when it comes to this issue meaning that you've come with exaggeration in it and there's a group of people who are negligent when it comes to the place that the عقل has so it's fair إن شاء الله to bring everything into its perspective what we have to understand is the عقل what is this relationship with the text شيخو الإسلام ابن تيمي said something very powerful and I think this itself should be written in control صلاحة I believe that Muslim lives by this justice statement alone they will benefit والله فلاح في الدنيا والآخرة in this world and hereafter which is he said شيخو الإسلام ابن تيمي he said الشرع وال العقل the شريعة and the religion it gave governorship and it made intellect a governor okay ثم عزله and then it chose to take its position away from it what does شيخو الإسلام ابن تيمي I mean are you with me when you came into Islam are you with me you came to Islam not because you knew the Quran of the sunnah you came into it from the ways that brought you to Islam it was your عقل you used your brain so what happened is the عقل brought you into Islam and then after you came into Islam the شريعة I said to you no longer do we need it the text is enough for you and that is why he said ما لا يعرفه البدريون that which the people of bedr did not know فليس من الدين it's not from the religion and what the Sahab is in the know but people didn't know they are not from the religion there's another statement that شيخو الإسلام ابن تيمي he said which I liked he said الشرع قاضن the الشرع the Quran and the sunnah are the judge والعقل شاهد and the عقل is what it's the witness in a court room there's a judge and his witness is right the judge here is the شرع and the شاهد the witness is who the عقل ويجوز للقاضي it is permissible for the judge طرد الشاهدي to dismiss the witness متشع whatever he wants you can take him up we don't need your testimony does that make sense so the شرع when he wants to it can dismiss the opinions of people's logic because it has a higher position it's تقاضي it's what judges are you with me and if somebody doesn't really become aware of that he will fall into and your intellect is only permitted to what to look into to research to object to agree all in accordance to what the شرع permits it the عقل is only can look into and observe and research and agree or disagree if the شرع lets it if it doesn't it can't do that عبدالله عباس رحمه الله he said something he said that the same way pay attention here the same way that your eyesight are limited in how far you can see your عقل is also limited in how far you can think and that's the problem with many people they've given the عقل an unrestricted rights that the عقل can do what it wants and how it wants to the extent that they believe it can govern the book of Allah and the sunnah are you with me so that's very important coming back to what the author said what was the third thing that he said جازمان when we're talking about knowledge we said the three points the first one is perceiving something the second one was as it is and the third one was with certainty there's no doubt in your heart when we said certainty three things are drops three things are what we push out of the definition let me just repeat because I did go on a a tangent I did go off topic when we said perception what was it that got out of the definition when we made sure it was was not in the definition is the جهل البصير the simple ignorance there's no perception صح when we said as it is what did we push out of the definition the compounded ignorance okay and the reason we explain why it's called compounded ignorance صح because we said number one there's no knowledge there and the second problem is the person thinks they they believe that they know they think they know but they don't know so that's why it's compounded it's two things on top of each other that's why it's called جهل مرقب the third thing in the definition of knowledge I said is what learning it with certainty here what drops is three things we get rid of how many things three things the first one is وان speculation is not in there because speculation وان it's better to learn the Arabic word because in English the translation is not definitely it's not definite and it's not accurate okay it's better to learn it in Arabic and then learn what the the word means and it should be explained to you so what are the three things that we pushed out of the definition it is وان what does وان mean وان means you've perceived something as it is but there is in your heart a doubt that it's not so the persons asked what is this okay he goes this is a plastic cup we all together is a plastic cup so he has perception and his perception and his answer is in accordance to what the reality it's in accordance to the that which is in front of him but in his heart he hasn't got that certainty that comes with the knowledge right but here the certainty the percentage is very high how many percent is it it's 70 that's called what وان what is that called that 70% that he's got in his heart it's called what it is called وان what about the remaining 30% the remaining 30% is called وان if he let's turn the table if he said this is a cup a plastic cup there is a perception it is in it is in accordance to the reality in front of him but the percentage of what he thinks of it is how much 30% it's not 70% this is called وان are we all together does that make sense it's very low what about if it's the same 50-50 this is called شك does that make sense those are the six that the author mentioned you learnt all of them in a very simple way all of those are called what all of those six are called they are called it's called the level of perception صحرح and this is very good for you to learn it it's very good for you to learn it because atheists they try to make you what they try to push you on these concepts on these terms it's very important that you learn it because these are knowledge that are talked in علم المنطق علم المنطق you learn two things that's what it really stands on تستيق تصور that's really what you learn علم المنطق and that's another topic for itself but there's a seventh one that some scholars add to it there's a seventh one that some scholars add to it which is called what is it called ال الاعتقاد الاعتقاد means a level between علم and ظن it's a level between what it's higher than it's higher than ظن and it's lower than علم they say اتقاد is what يقبل التشكيك it can't accept speculation to be opened on it okay because knowledge doesn't accept speculation to be opened on it somebody told you now this is a mobile phone however 50,000 times that he says it it won't change right like in belief يقبل التشكيك it can't accept can't accept can't accept why do you see a Christian becoming a Muslim you believed in something and now he moved away from it because speculation came into it doubt that's اعتقاد belief now the وطرحم الله he goes into the the types of knowledge we've learned the types of ignorance and now he's going to talk about the types of knowledge he says اقصام العلم ينقسم العلم إلى قسمين ضروري ونظري فالضروري ما يكون ادراك المعلوم فيه ضروريا بحيث يضطر إليه من غير نظر ولا استدلال كالعلم بأن كل أكبر من الجزء وأن النار حاره وأن محمد رسول الله والنظري ما يحتاج إلى نظر واستدلال كالعلم بوجوب النية في الصلاح كتب الأعلم سيطه العلم إلى انها سواء الاجتماعي وقلوب and المدلل سيطه العلم هو أن سيطه العلم بأن لن يساعد على أحد ولكن لا يحتاج لن يظهر أو تحاول such as the knowledge that the hole is greater than the part and that the fire is hot and that محمد صلى الله عليه وسلم is the messenger of Allah number 2 سيطه العلم is what needs to be يجب أن يجب أن ينظر إلى كثير من المعلم that the intention is necessary in the prayer. Here the author, he speaks about knowledge, but from more angle, because the categorization of knowledge is from many different ways. You can categorize knowledge from many different perspectives. And you should understand that things can be categorized from many different perspectives. For example, I can categorize a person from his height. You can say he's either taller, sure. Or I can say the person's weight, the athletic, the average, the chubby, whatever. Or the person's skin color. So one person you can categorize them in many different ways. The same is when it comes to knowledge. This is one of its types of categorization, which is how did the information come about? باعتبار طريق إدراك البعلوني. How did you perceive this information? So it's categorized into how many? Two. The first one is علم ضروريون. What does علم ضروريون mean? It means essential knowledge or necessary knowledge. It is knowledge that doesn't require any observation. And it doesn't require for one to look for it. Or even ask evidence for it. It's what philosophers and others might say self-evident. It doesn't require evidence. It's called necessary knowledge. You don't ask evidence for it. And from them is that the fire is hot. That the fire is hot. For example that the house is bigger than the door. That's called what? علم ضروريون. And from that enters Allah's existence. Allah's existence is ضروريون. It doesn't require any evidence. Trying to prove Allah's existence and trying to bring evidence for Allah's existence is a logical absurdity. It's a what? It's a logical absurdity. To try to say that I'm going to prove God's existence. God Almighty is self-evident. You talking this minute and saying what you're saying is an evidence of Allah's existence. Because Allah says to us in the Quran. سنريهم آياتنا في الأفاق وفي أنفسهم حتى يتبين لهم. حتى يتبين لهم أنه الحق. Allah says we're going to show our signs in the sky and the new universe. And we're also going to show them our signs in them.  Allah says in another ayah. وفي أنفسكم أفلا تبصيرون.  So nowadays when you see Muslims saying that we're going to prove God's existence. It kind of gives validity to the opposite. The person you're debating. And you're interlocking. You're accepting from him that his argument has premise. And some things they are not accepted even if there's a person opposing you. For example if somebody wanted to debate for the rape for instance. You can't say we'll prove his theories wrong. He should be even given a platform. Or should he be given any consideration in his argument. The same is when it comes to Allah's existence. It doesn't require evidence should not be looked for. Rather the person who says Allah doesn't exist. The burden of proof is on him. He needs to provide that. Why doesn't God exist? Until now 80s haven't proven why God doesn't exist. Make sense? So that's It's not something that is known out of necessity. The second type is called It's called And that is empirical evidence. Empirical knowledge. Empirical knowledge is. That which is known from looking. Observing. Researching. Travelling. Asking. Demanding. Requesting. It comes through those methods. For example. Is an intention required. For the salah or is it not. For example. That the zakat is not done from foods. This is. Empirical evidence. It does require research. It does require. Looking into it. It's not something everybody would know. This one the scholars they say it comes from You have to stimulate your mind and think. Okay. And. This one comes about with a correct premise. Then with the correct premise produces and it brings about. A correct result. The premise is right. And from there comes a. Upright. Justifiable. Result. But there are things. That can be debated. Whether they are. علم ضروري. Or علم النظري. Whether they ask. Knowledge that is known out of certainty. Or whether this is knowledge that is known. By observation. By looking. It can sometimes change. Somebody can say to someone. I see this to be. علم ضروري. And another one could say no. علم النظري. It can happen. And sometimes it is subjective. صحيح. But some things are not subjective. And we mentioned some of them. شيخو الإسلام مبنوتيميا. Has a kitab called. الردو على المنطخيين. شيخو الإسلام. The first volume paid 40. He goes in great details with this issue. The issue of. Knowledge. That which is known out of necessity. And that which needs. Observation. And looking. و لذلك what's very sad is. Those atheists today. Who reject Allah's existence. When it's knowledge that is known out of necessity. It's علم ضروري. Allah's existence. سبحانه وتعالى صح. A poet he said. و ليس يصحف الأدهاني شئ. إذا احتاج النهار إلا دليل. إلا دليل. It's not befitting for the mind. إذا شخص يساعدك. To prove. The sun. When the sun is out. You see. It's not befitting. Shows how it did with it. If you start arguing that person. Then. I have the right. Is the third person to say. Both of you are crazy. For even trying to argue with him. Makes you crazy. It's sad that. Today. They dismiss. That in fact. The existence is ضروري. And they accept as evidence. Something that they themselves admit. Which is علم النظري. Which is science. Science is based upon what? Science is based upon empirical evidence. It's based upon deduction. They don't reach a conclusion in something. Unless they what? They have to do. And that's a discussion. For another day. I'm going to conclude that. Insha'Allah. Anything which I have said. That was wrong. Incorrect this from me. شيطان. اللهم يزمسني. سبحانك اللهم بحمدك. أشد ولا إلا الله. أستغفرك. أتوب إليه.


We would love to pay you by PayPal. Click here in the dashboard, then enter your PayPal info. And now if you don't notice that message, you can go directly from the dashboard, click Profile, then Edit Profile, and then you can enter your PayPal info, click Save, and then to verify, click Edit Profile again, make sure that's your PayPal. Now that's my PayPal, so you shouldn't use George at NE.TV, you should use your PayPal, unless you want to fund dinners for George. That's always appreciated. Alright guys, I hope you enjoyed this short FAQ. My name is George, head of Freedom, CEO and founder, and we are building some awesome tools for you. I hope you enjoy what we are building for you, and we'll see you next time. Bye!


Hey guys, it's Nate, AKA The Foot Accountant. Welcome back to the channel. I'm gonna talk about some insane icon flips that I've had today. I have my biggest ever icon flip in FIFA history. I have all the FIFA slides that I've played that I've traded. I have my biggest one ever today. So I wanna talk about that. But I also wanna talk about this Bruno Fernandez and today's content because this is causing some market movements. This Arturo Vidal SBC is another one that is causing a lot of market movements today. This SBC, you know, we'll talk about it in a bit. But I think it's good value. Sick dynamic image.  A lot of people are after. But I wanna talk about this icon stuff first. If you need to trade right now in this game, if you have the ability to trade with icons right now in FIFA 20, and that means if you have anywhere over like 300,000 coins, you can trade with icons. There are people on this game. Of course, icon swaps right now is what everybody is grinding towards. They're grinding towards the icon swap tokens which has people thinking about icons which I think is causing more people to use icons, try them out. And it's just creating a market where you can trade with these icons because they fluctuate a decent amount. All these cars today, these five right here, after the Maldini, all of these add up to somewhere around like 150,000 coins a profit. I made 70K on this card. I made 50K on this card. And then these are like some 10, 20K flips right here with these cards right here, the Blanc Pirlo and Ronaldinho. So those are really solid flips. You know, making 20, 30K a card on most of them with a couple other big ones, that's really solid. But this one today was the massive, massive flip that I had and I just wanna talk about this for a second because sometimes you just find these gems on the 59th minute. And I'm gonna talk about some filters and stuff. If you guys have some curiosity with icon trading, you can kind of talk about that. But I'll show you the pictures. These are taken from my companion app today where I sniped this, four million coins flat for this Maldini, sold it for 6.7 million, which is 2.3 million coins after tax. My biggest ever flip on FIFA. Now I had it flipped like a month ago where I bought a prime route or a moments rule for a million coins less than what it went for on the market. And I thought that was gonna be one of my biggest flips ever. But this card right here, 2.3 million. Honestly, boys, it was just kind of lucky, right? I just got lucky with somebody listing this Maldini up on the market for this price. And right now there is one up on the market, 6.4 million. There was one up on the market right now. Is this fresh? This is not fresh, but this guy was just nowhere to be found earlier today on the market. And now he is there and I just kind of got lucky with the quick sale. I was able to sell that card in 20 minutes, which usually when you're trying to flip one of these upper tier, upper echelon icons, it doesn't happen that way. And today, fortunately, it did. So I got insanely, insanely lucky with that flip. But you can find those slips just as easy as I can. It's all about being quick and fast to the 59th minute. This is how I find those deals. I set a filter. This is the exact filter that was on today when I signed Maldini. 950,000 minimum buy it now. And a 4 million max buy it now. And basically what I'm doing is I'm filtering out a lot of those lower icons. And I'm trying to just get into this budget where I'm comfortable with. And I know the prices of a lot of cards in the 1 million to like 2 to 3 million coin range. I know a lot of the icons in that range. So that's why I search here.  That's how you refresh the market. So I search with this filter and I want to get, I want to use right bumper or R1 and get to the 59th minute as fast as I possibly can. So as you can see there, I need to lower my search filter. It's nighttime. There's less icons out there. If you get to the 59th minute in two pages, like I did, like right here, this is not showing all the listings. You can tell because it's like one hour range. There's more cards getting listed up other than this.  So I'm going to go down to 850 with 4 million and see if I can get to the 59th pretty quickly. Dude, there's another one. This icon filter stuff is a bit glitchy is what you'll find out. You really have to find a filter and once you find one, you can work it for, you know, 10, 20, 30, 40 searches. That was close. But once you find a good filter, you can actually, you can absolutely just change your refresh, your minimum or your maximum bid price, which is the one that I set at 15 million and you keep searching. Now this filter looks a bit too open, right? Because dang, look at that. Something is messing up with these icon filters. I might have to lower this top range down a little bit too. Knock it down to three mil because your entire goal with getting to the 59th minute is trying to get there in less than, I like to say less than 15 searches, less than 15 pages, which right now for me is seeming impossible with the filters that I'm trying to work with and it's not working. But I want to show you guys a filter to get on. Of course, the filters during the day are different. I sniped that Maldini during the daytime so there were more cards on the market. So that's why my filter could be wider. But right now, since there's less cards on the market in the evening time, that's why I have to widen up my filter in the nighttime and you tighten it up during the day. That's what I'm trying to say because there's more cards on the market during the day. I'm going to widen this up just a little bit more. Sometimes changing one of the minimum bid or the minimum buy it now or the maximum buy it now by even like 10,000 coins can make your filter work so much better or worse. So this is basically what I did today though, boys. I get to the 59th minute and you look to see which cards were just listed and this is why it's so fun because you never know what's going to be there at the 59th minute.  You're like, oh my goodness, that's a huge undercut. What a deal I just found. Okay, this is not going to work for me right now. I don't know what the deal is but this filter is not working. This is bad for the video because I'm supposed to be showing you how I snuck this card today, right? But this filter is not working at night time like it does during the day, which is unfortunate. I think we have a good one here though. I think we found a good one here. But again, your whole goal for this is to set a good filter, get to the 59th minute as fast as you can and look for undercuts on cards and you'll be amazed. It's hard to get to the 59th minute for icons because you can't filter them really specifically unless you go by like midfield or defender which you can sort by midfielder or defender. But I like just going back just like this because I feel like I can find the best deals back here when I don't have super restrictive searches. 1.1 mil for Stoikov isn't bad. 1.3 mil is not terrible for that. But again, the biggest thing again with these, 1.05. That's not bad either. It's not terrible but it's not good right now. The thing you have to look for with a lot of these cards is you have to remember the tax. That's the second biggest thing. The first biggest thing is knowing your prices. The second biggest thing is the tax. Get into the 59th minute looking for deals, knowing prices to know which card okay. I can sell this card for X amount of coins during the day at this price and then knowing how much tax there is because if you buy a card for a million and you need to sell it at 1.1, that's decent profit. You're making 45,000 coins profit right there because there's 55,000 coins of tax at 1.1 million. 5% tax on every single card sale in this game. So you got to make sure that you leave yourself some leverage for actual coin profits after you sell the card and after there's the tax calculation is put in there. 815 for blunt. That seems pretty cheap. I'm going to keep searching down the line though. 1.5 for Stevie G is not the best. 1.8 is not the best. Anything else back here? 960, nope, nope. I'm not seeing any big undercuts at the moment. So nothing big here but this is, you get the gist of it right? And if you're on a lower budget you can set a different filter right? Let's say you have like, you have 500,000 coins and you want to go on this really short filter right here. You can just kind of memorize the cards that are in this price range. It might now work as well for this. If you're going to do like a lower range if you're like 300,000 coins or anybody can do this budget. I like this little method right here. It just searches for icons on a minimum bid. You basically can get to the 59th minute in one page and you just have to kind of know prices. 423 for this Nedved. I can see that, oh, there's another one at 423. That's not a deal. So basically what you're doing here is you're refreshing this min price by you move it once every time you search and you get to that 59th minute as fast as possible and look for, these are icons that have less than 70,000 start price or open bid icons basically. That's what we're looking for with this filter. So if you have any more questions about the filters or anything drop them down in the comments below. I'll try to help out the best I can. I have another video in icon trading on my channel. If you go down to the FIFA 20 trading methods playlist you will see more videos about icon trading where I go more in depth on some of that stuff. But this is my favorite way to trade in the game just because the uncertainty you never know what somebody is going to list a card for on the game. And that's what happened to me with this Maldini today. This got listed up at 4 million flat. I knew it was too cheap. I instantly bought it and then relisted at 6.7 after kind of looking at foot bin and seeing where he possibly could have sold that over the past couple of days. Foot bin is your friend with all of these cards but realize that foot bin does not tell you the whole story. So learning a card's price takes time. Watch that card over a day, over two days. Add them to your watch list, your transfer targets. You can tell that I'm learning cards every single day. I'm looking at these prices. Are these viable prices? Some things that I see that might be an undercut but I don't think I can actually make profit on. And then I'm adding open bids like this Viara. If I can snag that for a mil, I can know I can probably sell that for like 1.07 so I can make 20K-ish maybe on that kind of card. So just adding cards to your transfer targets is huge for memorizing prices and knowing where you can get a card to sell at. So that's enough talk about the icons. Enough of me talking about my big flip on the day. Let's talk about this other content. Bruno Fernandez Player of the Month overpriced in my opinion for the upgrade that he got. Now, I want to kind of change your minds and how you're looking at this card because a lot of people are looking at this 89 and they're saying this guy barely got an upgrade. He got like a, he got plus zero pace and he got like plus ones all the way around. Technically, the Bruno Fernandez was upgraded from his inform. The Player of the Month got a plus two from the inform and it got a decent stat boost all around. But this winter refresh card, they don't take promo cards into account when they're doing Player of the Month boost or like inform boosts. They don't usually, usually. Sometimes they, you know, they can do whatever they want. It's their game. It's EA's game. But they don't usually skip over a promo card which is what they did not do here. They went from the 88 rated inform. They gave him a plus two and that's his Player of the Month. But if you compare it to the 89, there's really not that much of an upgrade. Again, it's plus ones all around and plus zero pace. So there's a lot of people that are not happy with this SBC. But I will tell you that there are a lot of people doing this SBC because it is Bruno Fernandez. He's gonna get a dynamic image at some point and that's gonna make the demand for this SBC go even up more.  If you're a man with the possibility that this could be his highest rated card of the year, I don't know if he gets in a team of the season, especially because the second half of the season which was cut short due to the virus stuff and the canceling of the games, at least for now. So I don't know if he's gonna get a team in the season card, right? So if you do wanna do this SBC, if you're a Manchester United fan, you want this guy in your squad, if I can persuade you to go get the tradable version, the 89 rated winter refresh, I would love to persuade you to get this card instead because it's gonna be way better for you in the long run because yes, this card's gonna drop in value, but you're not then sinking your coins into an untradable SBC that you're not going to be able to recover from, right? If you get this card, which is 644,000 coins at the moment, 643K, what's this one, 650? So if you wanna get the best rating Bruno Fernandez, I would go for this one. He's down a little bit today because of the SBC. I don't think he's gonna rebound that much higher because the SBC is like 750K, but if you see this guy around 600K flat, that's a decent buy price in my opinion. Even if you can go at the 88 rated for your team, you know he's at 330K, that's not a terrible price. These cards dropped a little bit today because of the player of the month, but this player of the month and the Vidal SBC as well coming out on the same day, this was calculated by EA for sure, absolutely calculated. And again, why wouldn't high-rated's rise? Boys, if you look at my transfer list, apart from the icons that I have flipped today, hopefully you guys are swimming in the money, swimming in the money. I've been listing these Kool-A-Ballies and these Keelinis at 45,000 coins for a week now after I bought them last week at the beginning of foot player days and the Konami Bowl promotion where these cards dropped massively after the Ramos SBC came out. They dropped on that Friday with two for one packs. They dropped in the weekend and I told you guys, buy, buy, buy because I knew there was gonna be more SBCs coming, the packs were gonna be done and that is exactly what happened and now we're getting SBC after SBC, after SBC and these cards are rising. If you did buy, I think it's a perfect exit time on your high-rated fodder today into tomorrow in the next couple of days because there are a lot of SBCs in here that a lot of people are doing. Bellerin, a lot of people did this. People might even be chipping away at Neymar still or at Messi, Bruna Fernandez, people are interested in now. This is an end game at CDM center mid, Vidal, that people are doing. Not really sure if people are doing Banega. People are doing Boatang though from the Players tab. I can't tell you how many, wait, is this one? Wow, this Vidal doesn't expire. I didn't even know that. Wow, that's a GG. Yo, I'm gonna end up doing that SBC at some point, probably during TOTS, during upgrade packs. If that does happen this year, that's when I'm gonna end up doing that SBC for sure. But we got, where's he at? Where is he at? It is Drew and Boatang. This is another big-time card that a lot of people are doing and again, these kind of act like Icon SBCs that are sitting out here in the Players tab that don't have any expiration date. I did not know that Vidal didn't have an expiration date. That's a GG. That's a GG e-sports. I like that because Vidal is actually a, I think Vidal is actually a decent value card and this again is really, really pushing up high-rated today. Taking a look at some of the high-rated, Illasich, 41,000 coins. I bought him on my video yesterday for 37.  So if it's anything that is 86 and above, I would say take the money before rewards. This week, take the money before rewards because you know what? Don't even chance it with the pack supply. Take the coins now if you want to because so many people are doing the Vidal, the Boatang, the Bruno Fernandez. This is SBC time. This is when SBC cards are the highest.  But you're making 10,000 coins a card on some of these high-rateds. So is that worth it? You're waiting a week for 10K a card? I think it's absolutely worth it. That was a great investment. I'm glad I made that call. And hopefully you guys got in on that because they are flying, absolutely flying right now. So again, Vidal right now, it looks like costs around 400,000 coins. But a lot of people are doing this card because of the links that you get with them. Let's check them out for a second. If you look at his in-game stats too, this is the definition of an end-game box-to-box center mid. He's high, high work rates, which is perfect. Four-star weak foot, three-star skills. He's got the four-star weak foot. That's all that matters. He's got 80... He got a plus 11 pace boost from his footmas card. And this was the biggest complaint people had today. They said, hey, I don't like re-released players getting an SBC. I don't like it either. But when I see a plus 11 pace boost on a card that, you know, Barcelona links is 454,000 coins a little bit expensive right now. Yeah, I think it is. Will he drop down in the future? Yes, I think he will. High-rateds are inflated as we just looked at. This card will drop down. If you want to wait two, three weeks to do this, we're probably going to get a promo. And this SBC price will probably be in like the 375,000 coin range, which I'm a big fan of that then because I think this card is worth every single bit of 375K, even as an untradable card because this is insane, right? This is going to be a nuts card. And again, I hear the people that are saying they're upset because of the re-release content. But the stat boost is so massive. At this point in the game, this is a very usable card. And if, you know, if you're going to re-release a card like they did here with Vidal, make sure the stat boost is massive. And they did that here with the pace, which is the biggest improvement that I've seen on pace on a card like this for a while. And that's what this card needed to go to the foot miss to this, it needed the pace boost to make him Hullet Gang, club 80, whatever you call it. So that's a GG. Huge fan of this again. I've probably talked about that too much.  Last thing I got to talk about, you know I'm from America, right boys? I am a fan of the MLS, not a huge fan. I don't follow it that much, but this card that dropped today as well, this center back right here, the MLS League SBC. I'm going to check on this right now because I would, I would almost bet you that the MLS League SBC is up a lot. I'm not a League SBC guy. I don't do a ton of them. How much is the MLS League SBC? It's 300,000 coins right now, according to Footbin. But that 88 O'Para card is one of the best center backs in terms of statistics that we have seen on FIFA. Look at this SBC, the last 72 hours. This SBC cost you 150K two days ago. And now it's 300K almost, not quite 300K, 238, but I bet a lot of these guys are extinct. So it actually does cost 300K. I bet some of the guys on these MLS teams are extinct. You guys might know better than me, so put in the comments down below which one of these SBCs are harder to do. But I know a ton of people are going after this card because look at the stats. 92 pace with 86 acceleration, 96 sprint speed. And this is what does it for me. 97 agility. This guy's six foot two with medium high work rates, which is perfect. He's got 97 agility, 95 strength, 92 aggression. You throw something on there that boosts his tackling, which is already high enough. You boost his tackling and maybe you can boost in like dribbling stats or whatever. Look at his passing, 99 short passing and 90 long passing. If you're doing any league SBC, it's got to be the MLS. Maybe wait a couple of weeks until this hype dies down a little bit, but there's going to be people that are talking about this for the rest of FIFA because that Opara card is nuts. So that's a big time GG from EA with this card right here. You know, it's going to be hard to link him. You might need some icons. If you do the Western McKinney SBC, that would help. I'm trying to think of some other players that you can link him to some other Americans in this game, but regardless, if you can link them with icons, go for it because this card is crazy, crazy good. I'm really curious to see how he plays in game. There's probably some YouTube reviews out there on his card, but I'm sure that he plays insane. So again, I thought it was a decent day of content from EA Sports. And of course, we're feeling good as traders and investors with these cards selling for the prices that they are selling if you did get in them, if you bought them earlier. Because again, it's a big time of SBCs. I expect a promo on Friday. I do fully expect a promo on Friday with, maybe for birthday, maybe carnival, maybe something else related to just FIFA content in general, but I do expect a promo on Friday. So I would get rid of your high rated gold sooner rather than later. All right, boys, if you enjoyed the video today, smash the thumbs up on it, comment down below if you have any questions, and of course, subscribe to the channel. If you're new, it's been Nate, the Foot Accountant. Catch you guys later. Peace out.