Okay, today, I'm just going to give you a little update on 2023 try to center myself in the camera here with still being near my microphone. Anyway, 2023, this is my plans I've been doing a lot of stuff with Android lately, and I want to share what I'm learning with you So that you can learn also so I don't forget I could always look up my own videos when I forget how to do something Don't freak out if you have no cares about Android my plan is for right now is normally Release two videos a week. I release one on Wednesday and one on Saturday. And so my plan is I think every Wednesday I'm gonna release an Android video and every Saturday I'm going to release another topic probably shell scripts or some sort of general Linux Information, but I just wanted to let you know that that's my plan for at least the next couple months Because I have a lot of things I want to show you from again I love, you know messing with Android devices messing with the boot loaders boot animations Logo screens. We're going to look at doing GSI's which are generic system images Which are Android images that run on pretty much any device that you can unlock the bootloader on that was made in the last four years or so Three years So yeah, I'm gonna go over that sort of stuff, but also I'm finally Getting the hang of some Android development. It's something I have tried to do a number of times in the past I really had have very little interest in Android development because to me Developing something specifically for one system is just kind of crazy when I write programs I want to run on everything and on my devices I can already write shell scripts bash scripts I can write stuff in C or Python or Perl or JavaScript and all the stuff runs great on my phones and my desktops and our cross-platform with other operating systems and I can even do goo is I could Because with Turmux you can do pop-up notifications toaster notifications You know notification notifications at top. I said pop-up notifications pop up dialogue boxes And I can write those so that they're compatible with all systems just doing a check on what system you're running on And and whatever programming language I'm doing I can always output a gooey as HTML and Make buttons and stuff without having to worry about you know SDKs and stuff like that But I really really want to learn how to do some Android development besides I've made stuff in Godot And you can export as an APK for Android, which has been useful in the past But there is still some limitation there if you're really wanting to interact with Android but mainly I want to be able to one Some of these open source Android projects I would love to modify them and I've never really gotten that very far in the development So I haven't really messed with it and part of the reasons I haven't I've tried so many times over the years to just Compile a hello world script or APK for Android and I never never never Completed it and the reason is I hate Java. No, I mean I have nothing against Java itself I just hate developing in it. It's to me. It's a sloppy language. No offense if you program in Java I'm not saying it's bad language. I just I don't like it Especially when it comes to Android development and here's the issue. This is what happened every single time I wanted to just create a hello world for for Android. So so like again, I could do bash scripts I could do C programs C++ I could pile those right on the phone and they run like that and I don't have to do very much You know, but you want you read these tutorials for compiling hello world applications for Android And it's like well first you need to download Android studio and the SDK, which is like four gigs worth of files And then and then you have to download the SDKs for different platforms of Android And then if you were which which that right there just having to download gigs worth of stuff Is ridiculous to make a hello world program and then and then to create an APK You've got to create like five different folders a manifest manifest File then you had to create an XML for the layout and then a JavaScript file It's like all this just to print hello world to the screen is ridiculous And again, I would start doing this and just halfway through the download of like Android studio or even just trying to download Just the SDK, which is still huge. I would get I would lose interest and I would just be like forget it I'm moving on to something else. I finally have it down to where I Can download just the SDK I even I even I'm jumping back here for a second I even found Examples online of writing stuff in C compiling as an APK, but you still had to download the SDK and every time I did it It was at least, you know, hun a couple of gigs worth of files The ways the tutorials all did it. I finally figured out how to do it and to get it under a gig I want to say it's somewhere between the two numbers come to mind as 500 megabytes and 800 megabytes Which is still huge and ridiculous But I'm going to go over tutorials in that I have notes up online I have a few different projects that I've created all with the notes to where you can down basically I'm downloading Just the SDK command line tools which ends up being about a gig and a half But half of that is an emulator to emulate Android on your desktop and I so I can delete that and I end up with Yeah, probably about 800 megabytes worth of files, which is Acceptable if I really really want to go down this route which I I just want to know more about the Android development and I want to share that with you and These tutorials out there again They're a lot of them use the Android studio which I have no interest in using and yet They all tell you to download gigs and gigs worth of files I want to show you how to do it from the shell without some unnecessary gooey And without unnecessary files. So try to getting it down as minimal as possible using Grail, which I finally have started to get the hang of and So yeah, I've been doing that and I even One of the other things I want to do years ago If I want to take one of my websites or any website and package it as an APK there was the option of App builder It was by Adobe and it was part of the Cordova project or something like that Anyway, I figured out looking at some example code and minimizing it how to Create an APK that it's just it's very very small and you can package up a website like that And I've played around with different options to where you can you know run scripts locally or Through the web browser anyway I'll get into all that and I wrote a bash script that basically you run it and it pulls down the template It asks you for an app name an app ID the website It tries to pull the fav icon for the app and I've been doing that a lot and the question is why I've always I've always been like Why would you install an application when you can just put a link to a website? and I've created a lot of applications on my website for myself for my wife and a lot of stuff for Co-workers and an issue a lot of people have is during updates And I'm not sure if it's when they're updating their browser or updating their system the icons get lost So I'm hoping packaging them as applications will allow them to not be lost when they're doing system updates And that's part of I'm talking about I mentioned in another video How stupid it is how web browsers save links now because it used to be and you can still do it on on Like desktops if you do it the proper way the way I'm saying your proper way Where you create an icon say open up this website either with a default browser or if you want to specify a browser Nowadays you create especially on mobile devices. Let's say I have brave browser or Firefox or whatever I say add link to home screen or shortcut to home screen and it doesn't just make an icon that links that web page It basically adds that web page to a database with an ID and then as an icon for that specific browser and that ID so when you click on it it goes to the browser and says okay, what website is this ID and then it loads it Which is ridiculous because if you ever clear your cache or Apparently doing some updates or move to a new system and you try to copy all their information and certain browsers like brave browser Don't let you do a standard backup of the application. It gets lost And I think they're just trying to force you to use their browsers and not flip-flop between browsers anyway I'm rambling a bit, but I'm just explaining why I'm going down this route And I still hate it in some ways, but at least I'm learning stuff and I'm learning more about Permissions in the applications and how the applications are built I don't plan on building you know these big applications because again I wouldn't waste my time designing something as an APK that can't run on other things But I can wrap other programs. I've written in other languages so that I can take it that's for You know that runs on any system and just basically package it up as an APK to run an Android devices So people install it that way so I'm gonna be doing a lot of that stuff So again a lot of bootloader stuff partition stuff rooting stuff changing boot animations Creating applications like hello world stuff from script packaging up websites as APKs Change modifying APKs creating something with Godot and then modifying the APK to give it other permissions and do other things So, you know a lot of hacky stuff. So I hope I hope even if you're not interested in Android development Maybe maybe this will help because even if you're not interested in doing it It's nice to know it. So, you know, you can understand these other programs out there Yeah, in case you wanted to you know add to or modify just just modifying an APK It's someone else created or an application for Android someone else created So yeah, that's what's coming up Over the next couple months. So I'm not leaving, you know, like I said every Saturday I'm gonna be doing whatever I normally do a standard Linux videos on different topics But then Wednesday at least for the next couple of months Is going to be Android stuff. So don't hate me if you're not interested in Android Just skip those videos or watch them and maybe learn something Thanks for watching and as always I hope that you have a great day films by Chris comm be sure to subscribe So you don't miss any of what I'm talking about and have a great day


They, okay, but they, they weren't in second second.  Yeah, but Willie D was hard. Man listen, I seen this.   Willie D's verse, go always step up the play. We ain't talking about verse, we talkin' about super stuff. It don't matter, superstar, nah, he made up ghetto bars too. That's like three. He wrote all of his verse. That's like three. God dang, give my boy a shot. You shouldn't have left it, man. That's like. Yeah, we on boss talk, one on one, one on one. Yeah, we gon' talk anywhere around the world. I gotta ask you this, you think Sqauffa's was wrong for not taking Willie D up there with you?  Sqauffa, Sqauffa, I seen a little white boy. So you saying he shouldn't have took Willie D?  Why? What? Why? They didn't call Willie D, they called Sqauffa. Well, he sung a song that Willie D was on, the ghetto boy, man, they didn't ask for Willie D. Right since second half. They, okay, but they, they weren't in second, Sqauffa was a, Sqauffa's was a fucking, a motherfucker, beast by itself. Yeah, but Willie D was hard. Man, listen, I seen it in the ball head hoes. I seen it. We in the alcohols and you know how to do this. Y'all, hey, listen, here's some songs, man, that you go back and listen to on the ghetto boys. Willie D's verse, go all the way to step up to play. We ain't talking about verse, we talking about super stuff. The superstar, now he made up ghetto boys too. That's like three. He wrote all of his verse. That's like three. God dang, get my boy inside. That's like, that's like, that's like three six, my fam bring a PJ Pat. Project Pat? Yeah. Nigga, you crazy?  Project Pat can rap. Man, you saying about some, you don't know what you're, you don't know what you're, I'm a white boy, I'm a white boy, came to me the other day, right? Yeah.  He said, cool, I looked him in his eye, I said, it's some stuff you gonna see that will make it hard to smile. You know what I'm saying? That one, look at me, see it. What type of stuff I'm finna see? You know what I'm saying? You see it, man.  Look at nigga say.  Nigga quoting the song, nigga.   Scarface? Scarface? Hey, man. Listen, how about West said Scarface is the hardest nigga out of Texas legend. And nobody have a bigger legend style. I think I agree with him. Legend wise. The Scarface? Yeah. Legend wise. There's nobody living that's a bigger legend than Scarface in Texas. A lot of people say that. I agree with that. A lot of people. Living. Erika. Erika Baidu? Yeah. That's a different genre though. No, it's still, it's still cold. Do you know what I'm saying? It's still culture, but I still say when it comes to legend status, man, it's hard to get around Scarface, man. You got a lot going. But hey, man. Listen, man. Top five legends. Are you talking about legend? In the world? In the world? No, no, no. In Texas, top five legends. Oh, this is what we're gonna do. Scarface. Bon. Pimp. Mothry. And it's one more. The boy bag. Hey, Big Moth. Yeah, he started, he started the same.


It is ironic, I think, that type 2 diabetes is, in essence, too much sugar than that which is healthy, and the Enneagram type 2 can be much the same. At our best, we are consumed with compassion. At worst, we brew a perfect poison, looking to meet our own needs by neglecting ourselves and choosing instead to be needed by others. Until we learn to regulate the health of our souls and receive as we give, we must wait and carefully, cautiously taste the sweetness of life.


Thank you very much for the invitation. This talk is joint work or based on joint work with Nunufleitas, Alan Klaus and Haluk Sengun. And the title is the Fermat equation and the unit equation. So to start us off, everyone knows the statement of Fermat's last theorem proved by Andrew Wiles says that this equation here has no non-trivial solution. So every solution has to satisfy ABC is equal to zero, where if you take your ABC to be a rational numbers and the exponent L to be at least three. You may also know this other theorem by Jarvis and Meakin. They proved Fermat's last theorem over Q2. And you notice that there's a slight difference apart from the field. There's a slight difference in that their exponent starts with L equals five. So they don't allow L equals three. And the reason for that is that this equation has solutions for L equal to three when you take your field to be Q2. So you can ask what should the statement be for a general number field? What should the conjecture be? And here's an attempt at formulating a conjecture. So this is, I've called this the asymptotic Fermat conjecture and it says that if K is a number field and it does not contain the cube roots of unity. So Zeta three is a primitive cube root of unity. Then there's a constant BK such that if your exponent L is bigger than BK, so you take a prime bigger than BK, the only solutions to the Fermat equation with exponent L and unknowns belonging to K are the trivial ones that satisfy ABC equal to zero. Okay, so I'm going to call that this, the asymptotic Fermat conjecture or AFC for short. And you might be wondering why we have this restriction on the field, cube root of one is not in K. And so as you know, if you add the cube roots of one, you get zero. So as a consequence of that, for every L bigger than three, you're going to have, if your cube root of one is in K, you're going to have a non-trivial solution to the Fermat equation for all large, for all exponents bigger than three. And so maybe the statement of the conjecture should be, let K be a number field, then there's a constant L bigger than, there's a constant BK such that for exponent L bigger than BK, the only solutions to the Fermat equation are either the trivial ones or permutations of this. Okay, so the talk is going to be about the relationship between this conjecture and the unit equation or the S unit equations. I'm going to get back, get to that eventually and attempts at proving not this conjecture and complete generality because it's unrealistic, but proving this for some interesting infinite families of number fields. Okay, so to start us off, we're going to have a crash course on what you might call the modular approach to the Dierfantian equation or the proof of Fermat's last theorem, but in a slightly broader context. So we look at not exactly the Fermat equation, but the Fermat equation where we've put in an extra coefficient, so P is going to be some fixed prime, some fixed odd prime and yeah, I'm going to look at this equation and what I'm trying to do is to apply the strategy of the proof of Fermat's last theorem to make deductions about the solutions. Okay, so it's following Sehr and Mazer. Now, the first step you do and you do this because it makes things slightly easier later is you take the three terms and one of them we're going to call capital A and one of them we're going to call capital B and one of them we're going to call capital C, but we permute them in a way so that the middle one, capital B is the even one and because everything is co-prime, the other two will have to be odd and one of them will have to be minus one mod four, so that's my A and the one that's one mod four will be my C. So we do this permutation and then we write down this elliptic curve, it's called the fly curve that yeah, okay, so there is a mysterious step that happens after this or several mysterious steps. So theorems of Mazer, Wiles and Libit tell you that somehow this elliptic curve that we've built from an unknown solution to this equation is related to a modular form, so an U form of way two and level two P. So there's no point for the purpose of this talk, it is to sort of delve too deeply into what is going on here. So we'll take this as a sort of a starting point and see what it tells us. Now, here we have a congruence of two Galois representations, but I want to get rid of this relationship straight away and write down a more elementary consequence of this. So by the way, this is the hardest part of the talk after a couple of slides will be reduced to doing very easy manipulations, okay. So this relationship, whatever it means, it implies a congruence between the traces of Frobenius of this fry elliptic curve and the coefficients, these are the coefficients of the cusp expansion of F. So I get a relationship for every prime Q that I choose, I'm going to get something like this. And what's lambda? Lambda is some prime ideal that divides my own energy L and it's a prime ideal of the ring of integers of the field of coefficients of this uniform. Okay, so to summarize, if I pick a P and I look at this equation, somehow a solution is associated to a new form of weight two and level two P and this is the relationship and there's only finitely many new forms of weight two and level two P. So if I didn't have the P, so I'm just looking at the Fermat equation, there would be no new forms of weight two and level two and I would say contradiction and the proof of Fermat's last theorem, that's the end of it. But in other cases, in some other cases, you're going to have finitely many new forms, you can compute them by algorithms of Climona and Stein and you can say, well, now that I know the new forms and I know their coefficients, what do they tell me about the solutions? So here's what happens next. This prime ideal lambda, which I said is a prime ideal that is above L, it's going to divide some explicit constant that I build from F. For every prime Q, I get this constant here. So because, well, you're in one of two cases. So if you're in the second case, then it divides either this one or that one. And if you're in the first case, you'd say, okay, AQF, I know it because I have the new form. But I don't know what this is, this integer is, but this is something in, I know that this is something in Z and it lives in the Hasse interval. So it's an integer whose absolute value is less than twice the square root of Q. So lambda must divide, divides this difference, so it divides one of these differences. So I've got this number for every prime Q I choose. I can calculate it and I know that lambda divides it by for some mysterious reason. And because lambda is also divides the prime L, what's L? L is the exponent in my original equation. I know that L divides the norm of this. So this is living in a number field that's generated by the coefficients of L. L divides the norm of this thing. So this is something I can calculate and you might think about, you look at this and you say, oh, great. I've bounded the exponent L in this equation. I have a bound now because it divides some norm. And that would be true if this number here is non-zero. So you can say, well, when should it be non-zero? So one situation it should be non-zero is if this AQF, if I have a prime Q for which AQF is an irrational number, okay? It's not in Z, it's an irrational number because here I'm subtracting something in Z from something not in Z. I will get something that's non-zero and likewise here. So this would be non-zero. So the conclusion is if F is irrational meaning that for some of its coefficients are irrational numbers, I'm going to get a bound, an explicit bound on L which is the exponent in the equation that I'm working with. And so I'm reduced now. So I say, good. Well, maybe I no longer want to solve this equation. I want to do something realistic. I just want to bound the exponent. So if I'm bounding the exponent, I'm now happy to delete all the irrational new forms from my list. And I'm left with the rational ones. And rational new forms, so by something called the Eichler-Shumor theorem. So this is the converse of the modularity theorem. Modularity theorem tells you you start off with elliptic curve, there's an associated eigen form with rational coefficients. Okay, Eichler-Shumor, it's the other way around. You start off with this modular form with rational coefficients. It's associated to an elliptic curve. So there is an elliptic curve E. It's not the same as our fly curve, something else, but it's something that you get from this rational new form and it will, its conductor is equal to the level of the new form. So it's going to be 2P. And because I've started with an explicit, with a fixed P as in my coefficients of this twisted pharma equation, I can go away and look in Cormoranus tables and list, let's say, all elliptic curves of conductor 2P and ask, what does that tell me? But you can take the argument a little bit further, this kind of argument here. And you can show that there's an even bigger constant, so it doesn't matter what we call it, an even bigger constant such that if L, if the exponent L is bigger than this even bigger constant, then you can assume that the elliptic curve E not only has conductor 2P, but full two torsion. So if I don't have full two torsion, then I'm going to find some prime Q for which this constant is non-zero. And I'm going to get a bound on the exponent L. That's how the argument works. Okay, so now you're sort of reduced to saying, if I want to get a bound on the exponent L, I have to rule out elliptic curves of conductor 2P and full two torsion. And well, let's try to write them down. So here's how you do it. You say elliptic curve with full two torsion, it has a model like this where U and V are integers and the discriminant, because we have conductor 2P is going to be a power of two times a power of the prime P and yeah, but the discriminant has this form here. It's 16 times the discriminant of this cubic polynomial. And so you say, well, U must be plus or minus a power of two times the power of P, V must be a power of two times the power of P, U plus V must be a power of two times the power of P, up to side. We take these two and we substitute them in here and we get an equation. And now you can go away and solve this. Okay, it's very easy. And it's an easy exercise to show that when you solve this equation, the prime P has to be either a mercenary, or you can substitute them in here. And the prime P has to be either a mercen prime or a Fermat prime. Okay, because, you know, mercen and primes and Fermat primes are one away from a power of two. So you can see that they're going to give you a solution to this equation. Let me just say that this equation has other solutions. For example, one plus one equals two. But if you work backwards, if you take the solution one plus one equals two, you put it in here, you're going to get something like conductor 32, not conductor 2P. So although we've reduced to solving this equation, we are not interested in all solutions. Not all solutions when we go back will give us conductor 2P. Okay. And so here's a theorem of Sarah and Mazer, which when this was proved, it was assuming Sarah's modularity conjecture, but it's now unconditional. If P, you know, thanks to theorems of Wiles and the Bit, or thanks to the proof of Sarah's modularity conjecture by Cary and Vandenberg. So if P is neither Mersenne nor Fermat, then in this equation here, the prime exponent is bounded by some explicit constant, which is going to be something like, I mean, this was worked out by Alan Klaus and something like P to the 12P, if I remember correctly. Okay. So as a summary, if I have a Fermat type equation and I want to use this Fry curves and Galois representations of elliptic curves and modular forms to say something about it, then you end up having to do one of three things, either computing all new forms of a certain level or computing all elliptic curves of a given conductor and full two torsion, or computing all solutions to an S unit equation. But well, maybe the word all is wrong here for the last one, because as we saw, okay, well, what's an S unit equation? We'll come back to that. So here is an example of an S unit equation. So I'm going to define it in a minute, but notice that here we weren't interested in all solutions. We were only interested in some solutions that satisfy some further conditions, because when we work backwards, not all the solutions to the S unit equation are going to give us elliptic curve with the correct conductor. Okay. So what's an S unit equation? Let's talk about that. So here's the general setup. You have a number field K, S is a set of prime, I should say finite set. S is a finite set of prime ideals. And I have this thing called the group of S units. So this is the notation, OS star. It's elements in the multiplicative group of the field that are only supported on the prime ideals appearing in S. So if I take any, if I, it's alpha, such that the valuation at any prime, not in S is zero. Okay, so if I take alpha and I think of it as a fractional ideal, I factor it, I'll only see things in S. And the S unit equation is this equation. So epsilon plus delta equals one, where epsilon and delta are S units. And so here's a very simple example. If you take your K to be Q and S to be two, then the S units, well, they're the numbers whose, the rational numbers whose valuation is zero at all odd primes, the primes not in S. So you get powers of two, up to, up to side. And if you look at the solutions to this equation in powers of two up to side, well, you get three solutions, half, half, two minus one, minus one, two. And if you take, okay, if you throw in another prime and odd prime P, then your S units are these, and the solutions to the S unit equation will be these three, plus if P is a firmer or mercent prime, the things that you get from that. So firmer prime minus a power of two is one. All right. So more of a sort of a little bit more background on S unit equations. So by the way, we finished the hard part of the talk. So everything is going to get much easier now. But, okay, so we do know, thanks to a classical theorem of Siegel, that S unit equation has only finitely many solutions. And there's this really beautiful theorem of Iversa. This is I think the most famous result in the subject, which is the number of solutions to the S unit equation is bounded uniformly in terms of three things. So it's the R1 and R2, that's the signature. So the number of real and complex embeddings. And S is the number of prime ideals in the finite set S. Okay, so you have this bound. And there's other bounds, there's bounds on the heights of solutions. For example, you to be Joe and Giori, but they take more notation to write down. So I've missed them out. That's a famous, famous result. And there's algorithms. So if you start off with an explicit field K and an explicit set S, there are algorithms that find all the solutions to the S unit equation. Starting with D-Vegas. So these, you know, D-Vegas algorithm. And I think everything that follows it uses essentially Baker's bounds plus LLL to compute the solutions. And this is still, you know, a lot of progress with these algorithms have been made in the last few years. So, but things are still happening. So here's a theorem which really surprised me when I saw it. And it surprised me because it's, the statement is very simple, but in a subject that's 100 years old, you don't expect a new discoveries of this generality and elegance. So this is a theorem due to the Antifolu. And it says if you have a number field K, N is going to be the degree. And there's two assumptions. One is that three doesn't divide the degree. And the second one is that the prime, the three splits completely in K. Then the unit equation has no solution. So here the unit equations, S is the empty set. Okay. So my OS star that I had before is the units of the field K. It has no solution. And notice that somehow to the Antifolu theorem is saying I have a restriction on the degree. I have a local condition on the field K, not on the unit equation, just on the field K. And these two together, they tell me that the unit equation has no solutions. So one of the things that we're going to be doing a little bit later is looking at similar theorems. Okay. That has been discovered since of this type. Okay. Restriction on the degree plus local condition imply that the unit equation has no solutions. I'll also say that I believe that, and maybe some people might correct me, but if you fix a number field and let's say, you know, of degree five, so the first can, well, let's go a bit higher. Let's say degree seven. So the first condition is satisfied. And you look at the set of number fields of degree seven where three splits completely, I believe that set should have positive density somehow. But I think we do not know how to prove that yet. So potentially to the antifluous theorem, saying for every degree other than the one, okay, for every degree, not divisible by three, we have this set of positive density of number fields, conjecturally, for which we can rule out solutions to the unit equation. And so one of the natural questions, now that we've said that there's a connection between the Fermat equation and the S-Unit equation is can we prove results of the type, restriction on the degree of the number field, local conditions on the number field implies, instead of unit equation has no solutions, the asymptotic Fermat equation, Fermat conjecture. So a statement about the Fermat equation, not the unit equation. So that's what we're trying to answer. And so here's the main tool. So we spoke a little bit, we spoke about this example of cell and laser. So here's some kind of generalization to number fields. It says that if you take K to be a totally real number field, and you, so we're going to have these two sets of prime ideals. S is going to be the prime ideals dividing two. And T is going to be the prime ideals dividing two of residual or inertial degree one. So what this means, the second condition is, I'm taking the prime ideals where the residue field is F2, they go in T, everything else, all of the prime ideals above to go in S and the ones with residue field F2 go in T and I'm assuming that T is non-empty and now I suppose that every solution to the S unit equation, this S unit equation here with this S satisfies some extra condition, which is if I take a solution here, I can find a prime Q in this set T where the valuation of epsilon and delta is not big. It's bounded by something explicit. So that's the condition. Then the asymptotic firmac injection holds for K. So that means that the exponent L in my Fermat equation over K, that's bounded, is bounded. And so this theorem is one of several theorems of the form. Local conditions plus assumption on the solutions, okay, so assumptions on the S unit equation, assumptions on the solution of the S, local assumptions on the solutions of the S unit equation implies the asymptotic firmac injection. There's other theorems like that, but we're just seeing a specimen. So how is this proved? Well, there's nothing actually very clever. It's just following exactly the same strategy as SEAR and MESA, but of course, because you're dealing with elliptic curves over number fields and module forms over number fields, you need different, you need, yeah, you need contemporary versions of the theorems of Wiles and MESA and Dribbit. And so here's some of the, so one of the things you need is Merrell's uniform boundedness theorem. Another, so that's sort of, you can think of this as a, something that replaces MESA's Isogony theorem in the proof of Fermat's last theorem. And you need, so these are, you could think of them as some vast generalizations of the theorems of Wiles and Dribbit. So modularity, lifting and optimization theorems due to many people. I've mentioned some Kissen, Barnett, Lam, G, Galati, Broi Diamond, and so on. So, and there's also, so the theorem that I've mentioned on the previous slide, it assumes that the number field is totally real. And the reason for that is that these modularity and level optimization theorem, they're sort of, yeah. There are, there are the sort of, so we do know, for example, thanks to recent theorems of Alan, Harry and Thorne, that a positive proportion of elliptic curves over each imaginary quadratic field are modular, okay. But that's, but we're still missing something called level optimization. That's the analog of Ribbit's theorem over general number fields. So there's a verge, but if you assume some standard conjecture in the Langlands program, then you can just write down almost exactly the same theorem for general number fields. Okay, so here's a, here's a sort of a very explicit example. You take this quartic field, this is totally real and to ramifies in this quartic field. So S and T are going to be just this set. And the great thing about this is that Nigel Smart figured out the solutions to this S unit equation, which is exactly the ones you need in the thing that you need to apply the theorem mentioned a few slides ago. And you find, so there's 585 solutions, but you find that they all satisfy this condition, that their valuations are sufficiently small and the theorem tells you that the asymptotic firma conjecture is true for K. Okay, so what I'm going to do from now on is go through some elementary calculations with S unit equations and to try and sort of come up with a theorem like the Antifolu's theorem for number fields where some prime ramifies. So Antifolu's theorem we had three totally splits. Here we've got a prime P that isn't necessarily three and P, the assumption is that P is totally ramified. So when I factor P in the ring of integers, I get this Gothic P to the N where N is the degree. And so I'm going to do some very explicit algebraic number theory. So the first observation is that if I have something in the ring of integers, epsilon, then its norm is going to be congruent to epsilon to the N modulo P. Okay, so let's see how this proof goes. Well, totally ramified, one of the things it tells me is that the residue field of Gothic P is the same as the residue field of the prime P. So it's a field Fp. And so that means when I look at epsilon modulo P, it must equal something in here. So it must equal something that's in Z. So epsilon is equal to A, it's congruent to A for some A in Z. And now, I mean, I'm going to, I need a definition for norm. So we will know, we used, well, I often think of norms as products of conjugates, but that is only convenient when you're dealing with a Galois extension. So here's one way to define norms is you look at this linear transformation, okay? Which takes an element of okay and multiplies it, okay, and multiplies it by epsilon. And so this, if you think of okay as a Z module, so it's something like Z to the N once you fixed a basis, then you can represent this T epsilon by a matrix and N by N matrix. And the norm is defined as the determinant of this matrix. But because of this here, epsilon is the same as A mod A mod P where A is something in Z, our matrix is going to be, when I think of it mod P is going to be A times the identity matrix because both of these on an integral basis, reduced mod P, they just do the same thing. And now when I take norm of this, that I'm going to get the norm of this one, AIN, A times the identity, which is A to the N and A is the same as epsilon. So I've proved this, what I was trying to prove. Okay, now let's take this result and apply it not to arbitrary elements but to units, okay? So here's what I get, is if I have a number field, same assumption as before, okay? I've added P is odd, that's not important. P is totally elaborate and the GCD of the exponent N with P minus one over two is one. So I'm assuming that these two things are co-prime. The degree N is co-prime to P minus one over two. I'm taking now not an arbitrary algebraic integer but a unit in okay. And the statement is that under these assumptions, the every unit is plus or minus one, modulo gothic P, the unique prime ideal above my small P. Okay, so how do we prove this? Well, the previous labor tells me the epsilon to the N is congruent to the norm of epsilon but epsilon is a unit so it has norm plus or minus one. And now I'm almost there. What I do is I think of epsilon to the N, I think of the image of epsilon to the N in this group here, okay? So this is, I'm taking the multiplicative group of the residue field of P and I'm courting out by plus or minus one, the subgroup plus or minus one. So epsilon to the N maps to the identity element in this group because it's congruent to plus or minus one mod P. But if you think about this group, it's the same as this one here because the residue field of a totally lamified prime is the residue field of the prime underneath it. And this thing has ordered P minus one over two and I've conveniently assumed that the N, the exponent N here which is the degree of the number field, I've assumed that it's coprime to the order of this group. So because epsilon to the N maps to the identity element, so the order of the image of epsilon divides N and divides P minus one over two and these two are coprime. So epsilon is the identity element, okay? In this group. So the identity element. So it has to be congruent to plus or minus one mod P. Okay, so I think this is, you're probably thinking this is all ridiculous for a web seminar, but here we get a conclusion, which is that I impose these two conditions I've seen before. Okay, so I'm going to have P's, I'm going to assume that P is not three, P is at least five. It's totally a lamified and the degrees coprime to P minus one over two and the statement is the unit equation has no solution. So assumption on the degree, local assumption on K, and we get the same conclusion as in trying to lose a theorem. And again, I mean, it should be true. It should hold for positive proportions of number fields where this condition on the exponent is satisfied. So how do you prove it? Previous lemma tells me epsilon has to be plus or minus one because it's the unit. Delta has to be plus or minus one mod P. You take these and you substitute them into this equation and you get plus or minus one plus or minus one is congruent to one. And that doesn't work unless P is three. So we get a contradiction for P not equal to three. Okay, I'm almost embarrassed to call this a theorem after I've shown you the proof anyway. So, well, can we do something like this for the Fermat equation? And this is what the next theorem is about. And it says that if you have a totally real number field of degree N and you take P as a totally alamified prime, any totally alamified prime that's at least five, and you have an assumption on the restriction on the degree, which is the GCD of N and P minus one is one. And you suppose two is totally alamified or inert in K, then the asymptotic Fermat injection holds for K. So here we have what we asked for before. We have to get the asymptotic Fermat injection, it's enough to have only restrictions on the degree and the local behavior of some primes in K. So there's, I mean, obviously infinite families of number fields of every odd degree that satisfy these conditions. And this is, again, one of several theorems. So there's theorems for even degrees that do similar things. So let's have a look at the proof of this. Well, I'm assuming that the prime P is totally alamified. So it looks like this. If I factor it, I have a gothic P to the N and two, it's either inert or totally alamified. So there's a unique prime above two, which I called Q. So I have a prime gothic P that's above P and a prime gothic Q above two. And the exponent here is one or N, because it's either inert or totally alamified. And I'm thinking not about the unit equation, but about the S-unit equation, because that's what we need to deduce the asymptotic Fermat injection. And recall that we're not trying to prove anymore that there are no solutions, because there are solutions. I can take two minus one equal one. That's a solution to this equation. But what I'm trying to do is to control the valuations of the solutions, because that's the theorem that we saw earlier, is if the valuations aren't big, then you're fine. And then asymptotic Fermat injection hold. Okay, so here's a typical case of one of the several cases you're going to have to look at where epsilon is a unit and delta, so if epsilon is a unit, delta is one minus epsilon is going to be an integer. Delta has high valuation, okay? So it's divisible by four. That's already big valuation in the case where the exponent here is N. Okay, so let's try to, let's see how we will out this case. This is the typical case of the proof. So you say, ah, so epsilon is one minus delta. Delta is divisible by four. So epsilon is one mod four. And okay, so that tells me that the norm of epsilon has to be one mod four. But epsilon's a unit. So it's norm is plus or minus one. So I get plus or minus one is congruent to one mod four. So I see that the norm will have to be one. So using this condition, I've deduced that the norm of epsilon is equal to one. And now we switch primes. We're not thinking about the prime of two. We're thinking about the prime above P. And we remember that the norm of epsilon is congruent to epsilon to the N mod P. But we've already said that the norm of epsilon is one. So we get epsilon to the N is congruent to one mod P. And now I use this condition, condition three. Condition three is saying that N is coprime to the order to the, to the, to the, okay, the order of the multiplicative group of the residue field of P. Okay, which is P minus one is prime to P minus one. So I deduce the epsilon is one mod P. Okay, so that's my second deduction. And now I look at Delta, what's Delta Delta is one minus epsilon. So Delta is one minus epsilon. That's one minus one modular this prime ideal. And I've written contradiction. Why is this a contradiction, because this prime has to divide Delta, and Delta is an S unit. It's only allowed to be divisible by Q. It's not allowed to be divisible by the other prime P. Okay, so this is a typical case also all the proof is just elementary algebraic number theory like this. And you deduce this, this theorem, which gives local sufficient local conditions for the asymptotic firm architecture to hold for a number field K. Okay, so let me, in the last few minutes, let me give another interest. So an interesting family of Dumber fields for which we can say something about the unit equation and the asymptotic firm architecture. Okay. And this is the ZP extensions of Q. So these are very popular and you are sour theory. And basically, if you take any, well, I said odd prime, that's the important. Okay, so let's take an odd prime P. And you look at the field of the Pn plus one lutes of unity. It's Galois group is isomorphic to this. And you can take the fixed field of this subgroup here. Now, you're going to get something whose Galois group is this said more P to the NZ. So this is this we denoted by Q and P has nothing to do with periodic numbers. It's, it's a number field that is totally a real it's Galois it has degree P to the end. This is it's Galois group. So if you fix your P, okay, P totally ramifies in here, because it totally ramifies in this one.  And this thing Q infinity P is called the cyclotomic ZP extension of Q. So in EOS our theory. People study things like the ranks of elliptic so you fix an elliptic curve let's say over Q, and you go up this tower, and you want to know how the ranks behave. But you can ask sort of similar questions for other diaphragm teen problems. And as a corollary of the theorem that we wrote before, because in the in the theorem for for unity equations. We had that the prime P is the prime P totally ramifies well that's true for this thing. And P had to be at least five so you have to rule out three, but it seems that turns out that two is allowed. Okay, and the degree had the degree had to be co-prime to P minus one or the degree here is P to the end, it's co-prime to P minus one. So you did use the unit equation has no solutions for this family of number fields. So point out that this doesn't work for P equals three, and that there are solutions so it's an open problem. You know, can can you determine the solutions for the unit equation for Q and three as n varies but but you know if you fix if you fix and you can you can just run a program and you get you get the solutions. And I'll just say that this corollary is in the spirit of this conjecture. So it's a conjecture of solid and passion that says that if you if you have any curve of genius at least to over rationales. Then the, as you're going up the tower, the set of rational points stabilizes. So X Q infinity is finite for all primes P. So if you, it might look like these two are not connected, but there's a common generalization pointed out to me by Min Yong Kim, where instead of curves you look at hyperbolic curves, and then this one becomes a special case of this conjecture. Okay, so let's look, let's let's go to further and finish off with this slide. So for these ZP extensions or the layers of the ZP extensions. It turns out that the what the theorem I wrote before immediately immediately gives you this theorem or this corollary I should have called a corollary, which is, if you take a prime P. That's not three. Okay, maybe you can include two. Okay, P is at least five. And for odd primes you need an extra condition, which is P is non V fridge. Non V fridge means that two to the P minus one is not one more P squared. Then the asymptotic firma conjecture holds for each one of these number fields. Okay, so maybe I should point out a few, a few things. So why, why do we need this non V fridge assumption. Well, it's equivalent to two being inert as you go up in this tower. So, so in the theorem that we had before for the firma equation over number fields with local condition wanted to to be either ramified or totally ramified or inert. So this gives you this condition. And so this, I mean, so I think a very vexing open problem is to show that there are infinitely many non V fridge primes. Okay, it's, it's really surprising that we don't know how to do that. But it seems that almost all primes are, are non V fridge. The only ones that we know to be V fridge are these two 1093 and 3511. And this condition also pops up in, in, in, in the classical theory of firm as last theorem. V fridge showed the first case. So P, okay, the exponent P does not divide any of the variables, like in the first case of the firm as last theorem holes for exponent P. And we sort of notice the word effective here. So, for the constants I'm talking about the, the, the bound on the exponent in the firma equation for that to be effective, what you need is modularity of elliptic curves over the particular number field that you're talking about. That, so for, so if activity makes use of the following beautiful theorem of jack thorn, which is the elliptic curves over these ZP cyclotomic extensions of q are modular. Okay, that's an excellent place to stop. Thank you very much for your attention.


You can help a child learn about adding by using any small objects. You can place two crayons in front of the child and ask how many crayons are there? The child can count one, two, and say, two. Then you could add three more crayons and say, if we get three more crayons, how many are there now? The child can count one, two, three, four, five. And they can say, five. You can say, that's right. When we have two crayons and we add three more crayons, we get five. You can do this activity with any other numbers. Take for instance, start with one crayon. You could add two more crayons and have them count, or you could start with four crayons and then add even two more crayons and have them count. You can do this with any small objects around the house, such as rocks, candy, or even buttons. This activity will help the child start to understand basic addition.


Large-scale manufacturing of high-energy Li-ion cells is of paramount importance for developing efficient rechargeable battery systems. Here, the authors report in-depth discussions and evaluations on the use of silicon-containing anodes together with insertion-based cathodes. This article was authored by Gebrekid and Gebra Selassie-Eshetu, Hongjong, Zabier-Judez, and others.


Now Nigerians have been advised to ask to keep faith in God in order to overcome the current security and economic challenges but definitely in the country. These was the message of the Muslim community in Bios' state as they joined the rest of the world to mark these years Idel Kabir details in this report. Idel Adai celebrated yearly by Muslims to remember Ibrahim's faith in God when he was asked to sacrifice his son. Muslims also use the period to showcase their love for one another and also pray for the betterment of the country. This year, they ask Nigerians to return to God if they must overcome the current challenges facing the nation. Having the child passes a trial period so we believe our own time is now by God's grace it will soon be over. There is no problem that lasts so anything that is hard will definitely come down. It's just matter of time but most importantly we need to be prayerful in our life so people should go back to God in order to pick it from the business. We urge everybody to keep faith because we know that one day Nigeria will be better, Bios'a will be better, that peace all of us are looking for will come and inshallah things will be good again. They also thank the governor of Bios'a state for always identifying with them during their festive periods. He has shown that he's an egalitarian governor. Governor who does not know any tribe, boundary, nation, and race and the likes of it. So hence is the governor he recognizes everybody. That is why when the Muslims are celebrating he's always recognizing us. As you can see the entire Muslim community in Bios'a state we are here collecting our rams, rice, everything from his excellency that sense it through the essay on religious matter rebrand gospel asinba. The Bios'a state government enjoins Muslims and the state to continue to be peaceful and law-abiding in the spirit of the salah. Let them continue to conduct themselves peacefully in Bios'a state because there must be peaceful coexistence among religious organizations or religious groups in this nation. So the way they have conducted themselves peacefully I wish them happy salah on behalf of his excellency the miracle governor of Bios'a state. Citizens are also advised to be toleranced towards one another irrespective of religion ethnicity just see a see plus TV news.


Section 21 of the Family Kitchen Gardener. This is a LibriVox recording. All LibriVox recordings are in the public domain. For more information or to volunteer, please visit LibriVox.org. Recording by Linda Ferguson. The Family Kitchen Gardener by Robert Buist. Section 21. The Quince is supposed to be the golden apples of the ancients. It is a native of Austria and is believed to have been cultivated in Britain for 300 years. Pliny writes in his time of their growing wild in hedged rows, so large as to weigh down the boughs to the ground. Moderns use it only after being stewed, baked or preserved. Quince marmalade is a favourite conserve and Quince wine has been known to cure obstinate asthmatic complaints. There are only two varieties and species that are worth notice. Apple or Orange Quince. The fruit large of a round oval shape, skin very smooth, colour when ripe a bright golden yellow. A clean growing tree and a great bearer, ripe in September and October. Portugal Quince. Fruit round, large size, colour bright yellow. A strong growing tree and bears a fair crop. This variety is used as stocks on which to bud or graft pears. They fruit earlier upon it and are much dwarfed by the process. It is a very general practice with the French and for small gardens may be done to advantage in this country but will not do for orchards. Pyrocynosis or Chinese Quince. Shrub of upright growth with pink flowers. Fruit very large, long oval, smooth and regular formed, colour greenish yellow. Flesh firm, rather dry. Ripe about the end of October. A beautiful preserve of a bright pink colour can be made from the fruit. A specimen before me is really beautiful. Propagation. This is readily accomplished by layers or cuttings as they root in either way very freely. Lay down the shoots early in spring or during the mild winter months and they will be rooted by the following November when they can be planted out in rows till they are strong enough to be removed to the orchard. Cuttings taken off the old plants of the past year's wood or even wood of two years old cut into lengths of about eight inches and planted into moist ground will root the first year and soon attain to be good plants. Soil. A heavy loamy soil is said to be the best for the quince. This is not borne out by results. The finest fruit I have ever seen is grown on deep sandy loam, manured every season. If they are not well cultivated they get knotty and deformed, producing fruit of light character. Pruning. Very little assistance is required from the knife and less to give a direction to the formation of the tree and for shortening any shoots that extend beyond the regular bounds. Raspberry. Rubus idios. Framprosi. French. Himber istroge. German. Some species of this plant are natives of all temperate countries and have been much improved by cultivation. Its fruit is extensively used for making syrups, wines, jams and jellies. It also forms an excellent dessert fruit considered healthful, refreshing and cooling. Of late much has been said and written on the raspberry, but as yet we may say there are about two or three sorts worth general culture. Red Antwerp. Canes dark brown, long, short, jointed, fruit fairly thimble shaped, flesh firm, rich, juicy with a fine, sweet flavour, ripe about the fourth of July. There is a variety called red antwerp generally cultivated with small fruit, readily broken into pieces and wood of a reddish brown colour. Franconia is a hardier variety than the former and does better in colder latitudes. Fruit large, conical, of a bright red colour. Flesh firm, flavour sharp, rich and abundant, ripe about the middle of July. White or yellow antwerp. Fruit nearly as large as the red antwerp of the same shape. Flesh yellow, very tender, rich and very sweet. Wood yellow, a great bearer. Fastorf. Within the past few years this variety, though an old one with a new name, has created quite an excitement in England and not little in this country. We fruited it two years ago and considered it one of the best reds, though we do not think it's the very best. Fruit very large, of an oval conical form. Flesh very rich, juicy abundant and makes a beautiful dessert fruit. It will never be a popular market fruit, being so soft that it will not bear carriage, but will hold its place for home consumption. Ripe 4th July. Ohio ever-bearing. Fruit conical, colour black, large size, produced in clusters on the points of the shoot. Flesh dark red, juice not very abundant, produces through the whole season till frost and quite indispensable on this account. Wood strong, of a dark purple colour. There are several very astonishing and superior raspberries raised from seed by an amateur gentleman of this city, some of them of a beautiful orange and others of a bright amber colour, whose true characters will be known in another year. Propagation. This is of the easiest character. Give the plants rich, deep sandy loam soil and they will send up in abundance of suckers every season, each of which will form a plant and produce fruit the year following. Planting. They should be put out in rows three feet apart and four feet from row to row. Two hundred plants is not too many for a family. Give them plenty of manure every year. Dig deep, but not close to the bottom of the plant. A situation partially shaded or naturally moist, though not wet, is the best locality. A plantation will last twenty years if properly attended to by enriching every year. Pruning. The first fruit I have pruned was the raspberry, and it is the only one that can be reduced to a simple rule. In the autumn cut out all the old wood that produced fruit the past summer, close to the ground. Tie up the new shoots to a stake or trellis, about five feet high, then cut off about a foot of the tops of the shoots, and the work is done. In cold situations the plants, after having been deprived of their old wood, have to be laid down all winter and covered with earth, spruce or pine branches till spring, when they are lifted and tied up as above. The Ohio and Franconia varieties do not require this protection. Strawberry. Fragaria variant. Frésille French. Erdbeerpflanzer, German. The strawberry, so called from the ancient practice, and still continued, of laying straw between the rows to keep the fruit clean. It is not properly a berry, but considered a fleshy receptacle studded with seeds. It is a wholesome and most luscious fruit and wisely distributed by a bountiful hand over nearly every part of the world. Its cultivation has been little regarded till within the past thirty years, and even at the present period is very imperfectly understood. Its healthful influence upon nearly all constitutions, when taken in moderate quantities, is admitted by medical men. The demand for it in a commercial point of view is rapidly on the increase, which has created a desire to know its character and improve its culture. Its uses are generally known. A certain species of beauty is compared to strawberry smothered in cream. A portion of the dessert palatable to all, though the beneficial effects of the fruit, is most certain when fresh from the vine, unmixed and unadulterated. The immense number of varieties now cultivated renders it rather difficult to select from them a few sorts that will continue the season to the longest possible period. In attempting to do so, however, we give preference to varieties obtained from seed in this country which resists the vicissitudes of our climate and give more general satisfaction than any imported variety. Our selection comprises two seedlings of Pennsylvania origin, one of New York and one of Massachusetts. Early May is a pistolet female variety. Fruit above medium size, colour bright red, shape conical, flavour very rich with a delightful aroma. Ripened last year on the 14th of May, season two weeks later than usual, this variety requires to have a row of the Hudson strawberry planted with it to produce the very great crops of which it is capable. It will be a general market fruit. Hobbies seedling. This pistolet female variety is now universally cultivated in every part of the United States and greatly admired. Fruit very large, heart shaped, colour dark red when fully ripe, flavour good with a fine aroma. Ripened last year about the 22nd of May, this variety requires a few of the Hudson or some other stamina sort to be planted near it when it produces extraordinary crops is a general market fruit. Prize seedling. One of the finest flavoured strawberries in cultivation, a staminate variety and produces a crop of fruit when planted alone of a very large size of a rounded heart shaped form. Colour dark crimson when fully ripe with a polished surface, seeds prominent, ripened last year on the 20th of May and continues fully three weeks as bearing an unusual length of time with us. Ross One Phoenix. A staminate plant producing a crop when planted by itself. Fruit very large and frequently of a coxcomb shape of a dark red colour with a smooth polished surface. Flesh firm and of a very rich flavour with a delightful aroma. Ripened last year about the 26th of May. Monthly Kopai. This variety of the alpine strawberry is an improvement on the old sort. The fruit is larger, of a finer flavour and produces copiously the whole year and is veritably a monthly strawberry. There are several varieties recently produced from seeds in this vicinity and other parts of the country which will entirely supplant every foreign sort and we doubt not may even display some of those we have named though they are not yet fully driven. Propagation of Strawberries from Seed. Every person who has any partiality for this fruit that can devote a few hours to their culture in their season should sow the seeds the very best kinds any time from August to April in pots of light earth, water them regularly and they will be above ground in four or five weeks. After they have attained a few leaves to each, plant them into a piece of rich ground in the garden about twenty inches apart. The second season they will produce fruit to prove their merits. The seed is obtained by drying the ripe fruit and washing the flesh from the seed which are all on the outside of the berry. These seeds, when perfectly dry, will keep three years. Flowers that have an entirely green centre are called female or pistolet. Those that have a great many yellow stamens are called male or barren plants. Those that have only a portion of stamens around the base of the green conical centre of the flower are called staminate or perfect blossoms. Soil. All admit that the best soil for this fruit is a deep, light, rich loam if not naturally deep to be made so by trenching. Rich it must be if large and good fruit is required. Therefore prepare the ground of the season before planting and incorporating it with an abundance of manure to the depth of eighteen inches. The exposure must be entirely free from the shade of trees or buildings. For early crops plant on an aspect that has an inclination to the south or southeast. For late crops choose the north or west. By this method the strawberry season is greatly prolonged. Planting. The periods for performing this is in March and April or August and September. In either of these months we have been equally successful. Beds four feet wide each containing three rows and plants fifteen inches apart in the row leaving alleys two and a half feet wide between the beds for the operations of gathering, weeding, hoeing etc. Never take any other crop from among them except a few radishes or lettuce the first season. Destroy the runners after the middle of July unless they are wanted for plants. Hoe them freely and keep the ground in an open condition. Some light rough litter should be sprinkled over the plants during winter in cold localities. In light soils dig in between the rows every autumn a few inches of well rotted dung but in strong and deep alluvial soils it may be dispensed with. In dry seasons give the plants a few waterings after they have done blooming with any liquid manure or other rich water which will greatly promote the swelling of the fruit. A plantation will last three or four years and to have this fruit in perfection plant out a portion each year. For this purpose we advise to plant those sorts that are called pistolet or female plants allowing every sixth row to be a variety that is called stamina or male plant. This latter sort keep within bounds to prevent the runners into mingling with the bearing kinds. It is a prudent precaution to lay straw or other clean material between the rows of the fruiting plants before they come into bloom to prevent the fruit being injured by heavy rains, sand or dust. Gathering the fruit. The common practice of picking the fruit with foot stalks attached is one of the very worst systems causing them to be handled and re-picked before they go to the table. Early in the morning take a vessel, basket or box of convenient size and pick the fruit before it is softened by the sun. Lay hold of the calyx or cup at the base of the fruit with the nail of the first finger and thumb of the left hand and with the first finger of the right give the fruit a gentle but quick draw and it will come off into the hand without the least bruise or damage of any kind and thus proceed till your vessel is full. Strawberries should go to the table without being turned or handled in any way when the full rich aroma of the fruit will be preserved. Those that are carried to market to be retailed for family use should all be in portable boxes in the same way as raspberries. The present mode is disgusting in the extreme. Large tubfuls, bruised and crushed, spooned into quart measures from vessels of very questionable character in both colour and appearance. The denizens carrying home their quantum of mashed matter under the name of strawberry from such a mixture can know little of the delicious aroma and rich flavour of the pure fruit called strawberry. Peppa Capsicum annum Pimoe, French Spanisher Pfeffer, German. There are several varieties of the pepper cultivated for pickling and kitchen purposes. Its natural locality is very generally diffused over all tropic countries requiring in artificial culture a very warm locality, rich light soil and careful cultivation. The green pods or small berries of all the varieties are used for pickling. The ripe fruit is dried and used in small portions as a seasoning of the hottest quality. Bell or sweet Large bell shaped and most esteemed for pickles, the skin being thick and more pulpy than any of the others. Tomato or flat About the size and shape of the tomato is also very generally used for the same purpose. It is of a hotter nature than the former. Cayenne Fruit small, round, tapering, long or curved and of the very hottest quality. We have seen about twenty varieties of the pepper. They're fruit when ripe from about the size of peas to the size of melons and all of a bright red or bright yellow colour. Culture Small portion of seed thinly half an inch deep on a hot bed or in a pot in a warm window any time in March or April and transplant in May or June on good ground one foot apart and 18 inches from row to row. In a mild climate so at the same period in the open ground in a small bed of light soil and transplant when three to four inches high during moist weather or water freely in time of planting. As they grow hoe frequently and earth up the stems similar to cabbage. End of section 21 Recording by Linda Ferguson End of the Family Kitchen Gardener by Robert Buist