Welcome back to our lecture series Math 4220, Abstract Algebra 1 for students at Southern Utah University. As usual, I'll be your professor today, Dr. Angela Miseldine. This is our first video for lecture 11, which we're going to talk about the idea of a haze diagram. Although there's a little bit of background we're going to have to get into before we get directly to the haze diagrams. So if you've been following us with this series, you might be a little bit surprised at the following heading. Section 19.1 from Tom Judson's textbook, what just happened, weren't we just like in chapter 3? Well, this is the background I'm referring to. Section 19.1 chapter 19, in fact, is about partially ordered sets, lattices, the algebra of partially ordered sets, and the like. So I want to just do a quick review of partially ordered sets before we talk about the idea of a haze diagram because, again, the haze diagram is going to be a lattice which naturally comes from these objects here. So in summary stack, a partially ordered set is kind of like an equivalence relationship with one very important tweak. So first of all, a partial order is a relation on the set X. So we say that a partial order, say less than or equal to here, is a relationship on the set X itself. So what that means is this is kind of funny looking here, but the partial order, the relation less than or equal to is a subset of the set X cross X right here. Now, if you are curious, this isn't a proper less than, this is actually the symbol backslash p, r, e, q, sorry, e, c, e, q, right here. And the idea is the p, e, the p, r, e, c is short for predecessor. If you drop off the equals there, you always get the predecessor without the line under it. But I always put the equal here. So you have predecessor. If you want to go the other way around, you can kind of like this curly thing. It's hard to write with my pen here. If you want to go the other way around, this would backslash s, u, c, c, e, c. So successor, predecessor versus successors there. So a partial order is a relationship on a set X that has size three conditions and these conditions will look very, very similar. So the first one is the reflexive property. Oh, okay, I see that. So that says that X will be related to X for every element inside the set. The third condition, what happened number two, we'll come back to that one. The third condition is the transitivity property, right? So if X is related to Y and Y is related to Z, then X is related to Z. And partial orders here are trying to capture the idea of a symbol like less than or equal to. That's why we're picking something that kind of looks like less than or equal to, but isn't exactly. So that we feel like it could be different, but they're very similar, right? The fact that it's less than or equal to means it should be reflexive. When it comes to partial orders, they should be transitive, right? If X is less than Y and Y is less than Z, then X should be less than Z, right? If it just keeps on getting bigger, that should be retained, right? So I skipped the second one. So let's come to this one. Because if you look at the first one and the second one, this is exactly what we expect for an equivalence relationship. Now for a partial order, we actually tweak the symmetric property. We don't actually take the symmetric property, we take what's called the anti-symmetric property. So if X is related to Y and Y is related to X, then in fact, it must have been that X is equal to Y. So remember the symmetric property for equivalence relationships. This would say something like if X was related to Y, then Y is related to X, right? So the symmetric property says that whenever a relationship holds, the symmetric relationship holds automatically. We're not saying that. What we're saying is if the symmetric relationship holds, it actually must have been the case that the two things are equal. The only way you can reverse the direction is if with equality. And if you think of the usual less than or equal to symbol, if X is less than or equal to Y and Y is less than or equal to X, which is basically saying that X is less than or equal to Y and Y is less than or equal to X, right? The only way that could happen is this would force equality, right? X equals Y. That's the anti-symmetric property. And so if X is equipped with a partial order, we call it a partially ordered set. Although some people call this a poset for short, partially ordered set. Some people actually take offense to the word poset. I think it's too crude or something. I'm okay with either one. Partially ordered set versus a poset, good with either one. So a set equipped with a partially ordered set, we call it, sorry, a set equipped with a partial order. Sometimes it's called a partial ordering. This is a poset or a partially ordered set. So if two elements belong to the set X and there's a relationship between them, X is less than or equal to Y. That's usually how you read it. X is less than or equal to Y or Y is less than or equal to X. One of those things happens. Then we say that the elements are comparable. And if no such relationship exists between X or Y, there's no relationship between X and Y in either direction. Then we would say that the two elements are incomparable or incomparable. However you want to pronounce that word. And so the archetype of what a partial order is, is in fact the partial order that exists on the real numbers. We say that two real numbers A and B satisfy this relationship. We say that A is less than or equal to B. Whenever the difference A minus B is non-negative. So if the bigger number is subtracted by the smaller number, that actually will give you something positive. Of course, if they're equal, that'll give you zero. So we say that B is bigger than A or equal to when their difference is non-negative. Zero is a possibility. Now you can restrict partial orders from one set to a smaller set. This works out pretty easily. We talked about in the previous lecture how you can restrict a binary operation to talk about a subgroup. That's not such a big deal when it comes to a partial order. If you look at a subset, you just retain whatever relationships existed previously. And so we can make the rational numbers into a partial order set. The integers, the natural numbers using the same partial order that existed on the real numbers. Now I should caution you that this partial order does not extend to the complex numbers. We don't have a well-defined notion of when a complex number is less than or equal to another one. At least not by extending this partial order right here. So the real numbers is an example of a poset. Another archetype of what a poset is all about is the power set. So given a set X, its power set, which is commonly referred to as P of X, but honestly in the literature, it's probably getting more mentioned as two to the X, at least in common torques in set theory. That's how we like to look at it, two to the X here. And the reason I say this is from common torques is because the power set, if X has a cardinality of N, the power set is going to have a cardinality of two to the N. And so we oftentimes use two to the X in that situation to develop the power set. It's also borrowing from the notation that if you have the set Y to the X, this is the set of functions from X to Y, like so. And so this two to the X is thinking that a subset can be identified with a function from two to X, whereby two here, I mean, you have the options of zero and one. These binary sequences there. So we don't need to go into the common torques too much, but you'll often see two to the X used to describe the power set of some set X. I do believe in Judson's textbook, he uses P of X, so we'll probably stick with that when it is commonly used. The power set can be made into a partially ordered set, where the partial order is going to be the set containment symbol. Now be aware that set containment is a partial order. It satisfies the conditions. It's reflexive, right? Because what does the subset symbol mean here? So we say that A is a subset of B if and only if for all little a's and a's, it follows that little a is inside of B as well, right? So that's what it means to be a subset. So if you take an element of X, if you take an element X inside of A, it's an A. Therefore, since X was chosen arbitrarily, A is a subset of A, the reflexive property holds. What about anti-symmetry? Suppose that A is less than or equal to B and B is less than or equal to A, right? So A is a subset of B and B is a subset of A. Well, that implies that the two sets are equal to each other. That's actually how we define set equality, that two sets are equal exactly when A is a subset of B, that is every element of A belongs to B and B is a subset of A. That is every element of B belongs to A. So yeah, that one's kind of like automatic in that regard. And then transitivity, if A is a subset of B and B is a subset of C, we would want that A is a subset of C, right?  But since B is a subset of C, every element of B, which includes X, is an element of C. And as X was an arbitrary element of A, this then shows that A is a subset of C. And so therefore the set containment symbol is an example of a partial order. And now unlike the previous example, this actually gives us a reason why do we call it a partial order? Why partially? What's so partial about it? Well, if you were to take a set like say X equals the set 1, 2, 3, then some of the subsets of X would be like Y could equal 1 and Z could equal 2, 3. So these are both subsets of X, right? They both are subsets of X, so they belong to the power set. But how do these compare? It's not true that Y is a subset of Z. That's actually false. It's also not true that Z is a subset of Y. Y doesn't belong to Z. It's not a subset of Z because one's not inside of Z. But Z is not a subset of Y because 2 doesn't belong to Y. And so this would be an example of two elements which are incomparable. And that's why we call it a partial ordering. That we don't say that Y is bigger than Z or Z is bigger than Y. There's no comparison there. And so because of the possibility of in comparisons, we do call these things partial. Now, kind of basically using the same reasoning as we did before. This is the main reason why we're starting talking about it in this section. If you have a group G, we could take the family, the set of all subgroups of that group, and then using that symbol less than or equal to right. Then by basically the same arguments as before, the set of subgroups of a group is a partially ordered set in terms of the set containment symbol, which we draw it a little bit differently for subgroups. So that way to indicate that a subset is a subgroup. It's a different symbol, but it'll also be a partially ordered set. And this is the post set that we're very much interested in right now. And then just as a last example, sort of like another canonical example, if you have a set of positive integers, then and you take the relationship of divisibility, where X could be any positive. So X is going to be the positive divisors of some positive integer N. Then with respect to divisibility symbol, this also forms a partially ordered set. I'm actually going to leave it up to an exercise to the viewer here to prove that one. I don't want to steal the thunder there, but these give some important examples of partially ordered sets.


I am an engineer, so that means that with my colleagues, I get to invent and make things. I get to invent the way that we make materials, I get to make products, I get to make devices. And what I've learned over the course of the years of inventing and making things, is that I want what I make to impact the world. But the world of production is changing very rapidly too. So as we'll discuss today, production is not about making one amazing thing once. So as cool as this 3D printed material is, that's not production. That's invention, that's innovation, that's what happens at the lab scale. But production, today's production paradigm is making things faster, making things more customized, making things in a more distributed way. And as we'll see, making things that are much more digitally integrated than ever before. And this is itself producing a lot of challenges and a lot of opportunities. So let's take one example. This is my phone, this is my iPhone. I bought it in Boston, near where I work. Many of the people who designed this device, the electronics, the hardware, the materials, they were educated at MIT. But this device wasn't designed at MIT, it was designed in Cupertino, California. And in fact, right now, the largest employer of the mechanical engineers coming out of MIT, go work at Apple to do just that. But as we know, my phone was designed in California, but it wasn't produced in California, it wasn't manufactured in California. Instead, it was manufactured here in Shenzhen, in Southern China. And it was manufactured at such low cost, so quickly, such high quality, that I have owned that phone, as well as other phones that I've since broken their screens. We can't live without this kind of device, but is this the production paradigm of the future, where we design it here and we make it there? So the answer turns out to be no. Several studies have shown that you need to actually be able to innovate, prototype, manufacture where you invent for companies to be successful in the long term. And so I'll show you just one study from MIT. Suzanne Berger and her colleagues did a study called Production in the Innovation Economy. And they looked at 150 production firms that came out of just one university, MIT, over one decade. So of those 150 companies, you see that there's a wide variety of production sectors that are affected, ranging from biopharma to robotics to semiconductors to medical devices. So we touch a lot of sectors with what we invent and then are able to spin off into production firms. But when you look at those companies and say which of these a decade later are still operating, here's what you see in this graph, and there's two things that you see here. Red means the companies are closed, blue means they merged with another company, green means they're still operating. So the first thing you see is that even in an environment as rich as the Boston Cambridge area with a highly skilled workforce, venture capital investment, lots of manufacturing capabilities in that environment, not all companies make it. That's not a surprise. Some companies succeed, some companies fail. But if you look at the companies that succeeded, the ones that are still operating are the ones that had the manufacturing innovation ecosystem right around them. So not a state away, not a globe away, but down the street or a car's ride away, very close to the companies. So this is part of what we've learned is production in today's innovation economy. Why is that? Because those companies could prototype quickly, could fail quickly, could develop their supply chains quickly to build that first production cycle and then move on to the next idea and prototype that. So that very quick turnaround, that ability to fail quickly is a hallmark of those successful companies. So we've always needed this connection between innovation and product, right between our ideas and our heads and then our ability to make it with our hands and we've needed that manufacturing ecosystem. But what's special about those manufacturing ecosystems is now changing. Part of it is the proximity between the small and medium companies and the large companies, the physical proximity. Part of it is that you don't only just need more hands, different kinds of hands, highly skilled hands, but you need to figure out how those hands can work with robotic hands and this human machine interface that is part of more automated production today. That requires a new skill set. But it's that transition that we're all making and that companies are some struggling and some excelling to make that allows us to get from a steel body car with highly automated mass production to a 3D printed car in my lifetime, right? In a single lifetime. And so let's talk about 3D printing as an example. This is also called additive manufacturing and this is thought of as a new manufacturing capability and advanced manufacturing technology, but it's not new. In fact, most of the people who invented this manufacturing technology and the machines they built are retired. This has been around for decades, but when it first came out, when we were first developing this in the basement of university labs where the 3D printers took up a room of this size, there were a lot of basic science questions, a lot of R&D to de-risk that kind of technology. And now we've gone from 3D printing very intricate toys out of polymers to actually using that same capability to make something like this, the replica of a Shelby Cobra. Okay, so this is a prototype vehicle built last year in the U.S. as a collaboration between the Department of Energy and companies and universities and built from design to production in 12 weeks. And this is the U.S. Secretary of Energy Moniz taking a spin in that prototype vehicle so it really does work. You can do more than prototyping with additive manufacturing now. You can actually have companies that can turn a profit by customizing that 3D printed car for you on demand. Okay, so this sounds impressive and it sounds rapid when you see it all on the same slide but it took my entire lifetime to get from here to there. So why is that road of advanced manufacturing innovation so bumpy, so long? We can think about this in terms of a graph. I'm an engineer, I like to graph things. On the vertical axis we'll make this dollars. And this could be profit and loss. This could be investment. And on the horizontal axis we'll make this the production scale. So here we're talking about the very initial idea. This is the basic research and development where we generally fund this through governments and universities. To the far end where we're able to actually make a product at profit and sell it to the customer, the consumer, the defense market, what have you. And not just with one big factory but increasingly with a series of micro factories that are near the customer. Okay, so that's the range of space but in between the idea and the profitable product is this. This valley of debt, this is where most advanced manufacturing technologies and companies die, right in the middle. Because this is where the profits are very low. You're investing but you're not actually seeing a return on investment. So why does this gap exist for advanced manufacturing and what can we do about it? This is the challenge. So the gap exists for a few reasons. One of these reasons is that in this middle this is where we're de-risking the technology. 3D printing for cars didn't work right away. It was only working for a very narrow set of materials and it certainly wasn't working at yield or cost that would ever turn a profit. Manufacturing is capital expenditure intensive. We're not making apps, we're building things, real things that require real machinery and real training of the individuals who are making them. So that CAPEX intensive operation, that requires investment. And there's low or long return on investment for banks, for venture capitalists, for governments. And so it's hard to finance this middle gap. Third, you need a workforce that can actually do this new kind of manufacturing. And that means retraining your existing workforce, training your new kinds of students to do this and that doesn't happen overnight as well. So for all these reasons, there's a big gap. And then you can think about the role of the government. If regulation or tax policy is uncertain, not that it's not there, but that it's uncertain or it's changing rapidly, this prohibits investment further by companies. So we don't have vertically integrated companies that take us across this gap anymore for the most part. So how do we bridge the gap? And so increasingly, the way that we're bridging the gap in advanced manufacturing technology is through public-private partnerships. And this is where governments, and this could be federal government, state government, city government, are working with the companies directly, small, medium, and large, and with academia from community and technical colleges to universities to work together and side by side with economic development organizations to fund and make the sustainable manufacturing innovation ecosystem. So it has a local flavor to it, a regional strength and investment flavor to it, where every stakeholder has some skin in the game. Every stakeholder is putting real money, real time in, to build this manufacturing innovation ecosystem. And so we'll take a look at a couple of examples of these at the national level, but there's actually levels that are below that and also successful. So here's a map of Germany. Germany has a very strong and well-earned reputation for being very industry responsive, very manufacturing forward, and those manufacturing careers are perceived to be of high value to the citizens of the country. The Fraunhofer Institutes, there's 60 of these in Germany. There's billions of dollars of investment by the German government in these institutes are a big reason for the success. Not only because they have the local governments paying into the system, and not only because a third of the money for these institutes has to come from industry or privately funded projects, but because there's also a focus on the workforce training and a strong model of apprenticeships and internships within companies so that researchers, students are paid to learn as part of their training. And this has become so well-received that the Fraunhofer's aren't even contained within Germany anymore. You may have one in your country. In fact, they exist in North and South America, near where I live, near where we are right now in countries like Japan. Now if we look at Japan as a particular model, they have a different approach, a more concentrated approach in terms of the manufacturing technologies they're developing, but still it's a public private partnership. They're between the AIST, a federal agency, and companies like Kuwata, and focusing on manufacturing technologies that are of interest to the consumers and also play to the technical strengths of the country. This is a strength of manufacturing in Japan and has been a focus of investment along with 3D printing, where the government and these private companies are working hand in hand. But in the UK, there's a different model. This is called the Catapult Innovation Centers, and seven of these in the UK are actually focused on manufacturing. Each of these is partnered with industry, ranging from Boeing to automobile companies, very focused on workforce training and growing small companies and making them healthy. In fact, one of the most recent stories that came out of these was that this program of the Catapult Centers helped produce the materials for sample delivery for Mission to Mars. So a lot of really interesting activities coming out, and this type of public-private partnership as we make this EU transition will be very important to keep an eye on for the production economy within the UK. And as we've already talked about in China, you actually see a lot of this activity happening, but a transition going on here as well. Because in China, we don't want to just make things that are invented elsewhere. In China, we also want to invent and innovate on the products, the processes, the materials, and have them made here or elsewhere. So that's a transition. And the China government has just announced 15 manufacturing innovation centers that will be launched in the coming years to just that end. Workforce training in Africa in the manufacturing center targeted towards women made in India campaigns to focus on production-conscious consumers in that part of the world. So trends all over. But in the US, when we tend to rely on market competition rather than industrial policy and planning, we also have a network for manufacturing innovation. And in fact, President Obama in 2013 announced the National Network for Manufacturing Innovation. In the US, this is anchored by 15 different institutes, each on a different topic. The first one was called America Makes, and that was on 3D printing or additive manufacturing. But now we have a whole range, including digital manufacturing in the middle of the country and one that is so new in smart manufacturing for clean tech that it was announced last week and it doesn't even have its official logo yet. And this is a subset of the 15, five more under competition right now in topics ranging from biopharma to robotics manufacturing. So each of these, although on a different topic, is intended to work together so that the 3D printed materials in Ohio support the composites manufacturing in Tennessee so that the integrated photonics that are made in Massachusetts support the flexible electronics that are manufactured in California. So there's a synergy and then there's also this focus on workforce training so that the skills are adaptable and can move from advanced manufacturing industry to industry with everyone having skin in the game. So I'll close with just a couple in which MIT is heavily involved. One of these is called Advanced Institute for Manufacturing of Integrated Photonics or AIM Photonics. Here's an image of what those are. So it looks like a semiconductor wafer but in fact the devices on it use light, use photonics to pass information and represent a new class of lasers, communication devices, kinds of devices we'll use for driverless cars. And so the way that we get there through these institutes includes startup companies such as the one that's represented by this individual here. People who've worked in labs for years to drive the technology and then teach it at the university level shown here but also a real focus on who the manufacturers of the future are going to be. Who's going to do these wafer runs within the companies? Who's going to assemble and test the devices before they go out the door to the consumer? Who's going to figure out how they actually work in production lines to speed productivity of other kinds of manufacturing? That's the focus of these institutes, the technology and the workforce training together. And then as a reminder on the next one that MIT's heavily involved in for advanced fibers and textiles, when we learn how to manufacture objects or materials in a new way, we can open up whole new product segments that we haven't even thought of yet. And that's part of the new production paradigm. So what I'm showing you here, these funny objects that are about the size of your thumb are called preforms. They're made out of polymers but other materials. But when we draw them down to fibers that are about the diameter of your hair, these also can become photonic devices where they can sense and communicate and pass information. But it's in the form of a fiber which we can then make in the form of a textile. And not just a textile that you would wear, but a textile that you would use for highly complex structures. This is the basis of one of the other institutes in the U.S. that'll officially launch in October, advanced functional fabrics of America or Afoa as you see here. And again, the focus is on the workforce training. This sounds great, who's gonna make it? Where's it gonna be made? And so the focus on the workforce training and the skills gap is important to every country, but also in the U.S. as you can see from this poster, some countries like Germany have made great strides in manufacturing being an attractive career, being part of production as attractive. But not all countries, in the U.S. it is still a struggle and it's really a call for a new generation of a Rosie the Riveter, right? And that Rosie is gonna need to have skills in digital design and manufacturing, not only hands-on manufacturing, because what we're making is made in a different way. So because of that, it leads to one final challenge in the workforce training and how companies and governments and universities start to communicate with each other and I call that production integrity. So this represents intellectual property, passing digital data from machine to machine country to country. This represents how we actually adapt to changes in production when data standards are also adapted. So how we address this final phase of this current industrial revolution, this production integrity with a skilled workforce that's adaptable so that we can invent it here and make it here, that's the challenge of the next production paradigm. Thank you. Thank you.


Good morning, everyone, and welcome back to my channel, Apples and Tiaras. Today I am in my third grade classroom. If you are not already subscribed to my channel, go ahead and pause this video, click that subscribe button. And if you're not following me on Instagram, run on over there, follow me on Instagram because I do a lot of updating on Instagram and through stories and things like that. So today I thought it would be fun to start kind of a new series. I've been feeling kind of a slump in my content area for my teacher portion of this channel. If you guys have been with me a while, then you know that where I wanted to go with my channel here on YouTube was more of like a life as a teacher channel showing all aspects, motherhood, balancing work and managing a household, managing work and other parts of life. Just a working mom who happens to be a teacher. So I've been trying to balance teaching content with non-teaching content. So today's video is going to be teaching related. I would like to start a series on how to teach certain subjects and how to teach certain skills. So right now I have begun teaching fractions here in my third grade classroom and I have had an abundance of people asking me, how do you introduce fractions to your third graders? So what I would like to do real quick is just get a show of hands down in the comments below.  So raise your hand down below in the comments. You can use an emoji or say hi or something. If you're a student teacher or an aspiring teacher.  Those are the two grade levels that I have personally taught. So I would love to be able to reach to you individually just because I have taught those grade levels and I feel that I am knowledgeable in the content enough to, first of all, teach the children. Second of all, teach you how to teach it. So I'm gonna show you just basically what I do for teaching fractions, introducing fractions. I'm gonna show you the resources and the materials that I use and then I'm gonna kind of talk through what I say to the kids to make sure that I drill in the most important parts of fractions. So we're gonna take a little walk over to my teaching station and we're gonna just get started. We're gonna dive right in. First things first, think about learning targets. I'm gonna share a resource with you where I find my learning targets because I feel like sometimes with these standards it's hard for me to formulate a learning target because there are so many parts to each piece. So I'm gonna just share with you a website real fast and then we're gonna get started. So the website that I use to find or create my learning targets and this isn't the only one I use. This is just if I'm unsure about which learning targets I want. I use wiki-teacher.com. So this is what the homepage looks like and I know I could do a screen recording but I'm on my work computer so I just want to make this more like quick so that I can get into the actual stuff. But basically up at the top here you can click on unwrapped standards, choose the grade level and the area that you're looking for and then it comes up with all of the standards listed here. So I'm just gonna scroll down to the fraction standards and click on the first one, NFA one. Understand a fraction one B as the quantity formed by one part when a hole is partitioned into B equal parts. Understand a fraction AB as the quantity formed by parts of size one B. So basically that's the kids knowing that the denominator is the amount of equal parts, a hole is made of and the numerator is how many parts are being represented. So I'm just gonna click this and then once you get into the actual standard that you want there are a lot of resources here so there's learning targets and I'm gonna click on that in a minute. Vocabulary, essential questions and then there's also alignments here and ELL consideration. So if you're wanting to see what they did in second grade or the grade before or what they're gonna be doing in the grade after you you could click on vertical alignment and it will share with you all of the standards that are either before or after in the grade level before or after you. So this is kinda nice because you'll kinda know like what they're gonna be expected to do with the information you teach them this school year. So my favorite spot is vocabulary and learning targets. So I look here to see what vocabulary I might need to use. So we have compose, decompose, denominator, numerator, fraction and then partition which is a big one, unit fraction which is humongous especially for this standard. Students need to know that just one part of each hole is called a unit fraction and I actually tie it into little unit squares. These are obviously not units, these are rods but I tie it into unit squares so I show them a unit square and then tie it into a fraction thing. So anyway, then you can go up to learning targets and this is where I like to find all of my learning targets. If I'm not sure of how I wanna word it or create it myself. So most of the time, there will be student friendly learning targets in this list as well but for this specific standard, there are not. Basically this is what I use to find my learning targets. So most of the time the student friendly one will say like students can or I can and then these are the learning targets. So the first learning target, I wouldn't use understand that a fraction is a number. I would use that as part of my teaching and just explaining to the kids that a fraction is a number that represents a part of a whole. So that would be more of like a teaching point but this could easily be a learning target for you. Understand that the denominator of a fraction represents the number of equal sized parts that make a whole. So again, this might be more of like a vocabulary learning target, something that you could have for a long period of time because the kids will not remember this the first day. Hopefully they do but if they don't, this is something that you're gonna be covering most days of your fraction unit and then the same thing for numerator here. Okay, so this one right here is one of the first ones that I used. Compose fractions with unit fractions. This is not how I worded it. I worded it more like I can divide a whole into equal sized parts and label each part as a unit fraction. So I kind of took it from here and I reworded it more for kid friendly and that was one of my learning targets. So each day I have a different learning target or I build upon them. So as you can see yesterday's learning target was I can represent a fraction on a number line and identify them. So not only was I having my students build their own number lines and put fractions on them but I also had them looking at already created number lines and identifying a fraction on the number line. So they had to do two things, not just one. So you wanna be very cognizant of your learning targets for what you actually want your student to be able to do at the end of your lesson. Anyway, so I am still pushing the fraction vocabulary in my classroom. So I have this big thing displayed here because my students don't have this memorized yet and we are in week two of fractions. So I have that labeled there and then another thing that I have is just the names of fractions and what they look like. So that students, if they have a hard time spelling eights or looking at how many equal parts and actually what they're called, this is here just in case they need it as a reference. Now, a lot of what I use for teaching my students is I usually create some sort of like vocabulary page so that kiddos can look. This isn't a fraction one but this is like for properties. I have the vocab word, the definition and then I show an example. This is all through avid. This is three column notes. So I try to use avid as much as possible but for fractions since there wasn't a lot of vocabulary, I just started with an Amy Grossbeck anchor chart and I use these for my students to take notes on. These are on TPT. If you'd like to buy them just look up Amy Grossbeck anchor charts and I'm pretty sure she has them for many grade levels but I use these for vocabulary and for examples. So this was day one of my fraction unit. We started with just vocabulary and then I had the students draw and take notes and then we also talked about what is an equal part and what is not because when students are drawing and creating their own fractions, modeling their own fractions, you wanna make sure that they're not creating unequal parts when they're drawing. So one thing that I also taught them was that when you have a fraction that you're creating that has an even number of parts, you can cut it in half first and then think about how many parts would be on each side. This is really good because when you're getting into equivalent fractions and then putting fractions on a number line when the number lines labeled zero, one half and one whole, students can get confused and not know what to do with the one half that's labeled. So yesterday my kids and I talked about, well, what other fractions could we rename as one half? So we listed all of the ones that we would probably see up on the board so that when we saw a number line labeled one half, if it had a numerator of one of, or a denominator of one of these, we could automatically put one of these on the one half mark so that we already knew it was there. So anyway, we have equal parts here and not equal parts. Again, this is Amy Grossbeck. I use her anchor charts to teach pretty much everything just because she has everything laid out here. All of the vocab is really good. Then we went into what a unit fraction is because my first learning target was they had to, first of all, divide a piece into equal parts. Secondly, they had to label each part. So we went through and we filled in what a unit fraction is. Then we moved into there are ways to represent fractions. We started with just parts of a whole. We focused on these two first and then once I got through parts of a whole, that was a whole day, we did one of these. So I showed the students how to name it in a word form. I showed them how to draw models of each one. So they just practiced. Sometimes if I know my kids are working really well, I will let them, I'll do two of them and then I'll let them do the rest and I'll check. That way it's a quick formative assessment for me to see where they're at. And then again, this is a blank one because I put mine up on the board, but this is that anchor chart that's right over there. So all of the kids have these in their notebooks. And then again, unit fraction, I just stapled these on top because I ended up not gluing them in. But then we have this one and I showed them on a number line how if you are having a hard time with the number line, you can just create a rectangle around it and then the number line just goes right through it. So it's basically the same thing as creating parts of a whole, especially if you're doing it as a strip diagram there. And then we went into fraction bars. So they saw a lot of this on their prodigy game. My students use prodigy to practice their skills for math intervention while I pull small groups. And so this was something they were seeing a lot there. So I wanted to make sure to cover it. This is mostly used for like problem solving with fractions. And then Monday we started fractions on a number line. Now I did have the anchor chart for this but I forgot to make copies for the kids. So I literally just copied the anchor chart onto my notebook and showed the students and they copied it themselves. So some days we just fill in the blanks and some days we create the anchor chart in our notes. So this was just a step by step how to put a fraction on a number line. So this went along with my learning target. I can create a number line and represent a fraction on a number line. So this was the very first day they worked with number lines. Then I went on to draw a number line and I asked them what fraction is being represented by the point. This was the second part of that learning target. I can identify a fraction on a number line. This has already been made for them. Most of the time there'll be a half part label here but I wanted to save that for another day because I just wanted to introduce them counting each line on the number line, not counting zero. This is super important when you get to number lines. Zero does not count. You start counting at the first mark past zero. One, two, three and four. Also super important to make sure that you teach your students that one hole is equal to a fraction that has the same numerator and the same denominator. That way when they start seeing that the same number is on top and bottom, they already know it's one hole and they can label that right away. So that is as far as we've gotten. Another thing that I use are these exit slips and I forget the seller of these but I will try to link them down below. But there are exit slips for pretty much every standard and I just use these as quick checks. Like I'm probably gonna give them this one before I start teaching today to just see what they remember from the day before. So there's three different exit slips that I could use before, after or even during a lesson just for a formative assessment, a quick check and then I just keep track in my little notebook. So hopefully this video is helpful to some of you who are teaching third grade for the first time or you're interested in teaching third grade and you're just wanting to learn a little bit about how to teach something. While I'm teaching I'm very energetic, I'm very bright, I have the kids chanting, I have the kids getting up, I have the kids talking. So a lot of what I do is getting them engaged in the learning but also taking notes and practicing throughout. So I always have their whiteboards out, they're always drawing pictures, they're always doing examples. I rarely use the textbook. Sometimes it's just for practice, sometimes I steal the problems from the textbook and use them on their whiteboards. So instead of them doing it in a book they do it on their whiteboard. So I try to use my district's curriculum as best I can but I also use other things to supplement. So I hope this video was informative. Hopefully it was helpful to those of you who asked me about how do I introduce fractions and absolutely use that resource wikiteacher.com to get your learning targets from and to just unwrap the standards a little bit more if you're not sure about them or you have a hard time understanding them because they are written kind of funny and we do have to really analyze them as teachers. So hopefully you guys enjoyed this video. Again, don't forget to subscribe and follow me on Instagram for more updates and I will see you guys on Sunday for my next video which will not be teaching related. I'm going to try to put one teaching related video and one non-related video on my channel each week so you'll be getting two videos each week. Hopefully if I stay on my schedule but I will see you guys Sunday and I hope you guys have a great week and bye.


Ralph Lauren yeah, you know they got a brand guy Louvet now. Why is this important? You know Ralph Lauren is a great merchandiser and He has picked merchandisers as CEOs and they've clashed This is a guy who is all about that the previous guy was trying to supply chain management to ran into trouble There's a clash here's a pure brand guy coming out of Proctor very very strong manager Maybe that guy Emerging with Ralph Lauren can do line extensions of what's a great brand so now it's coming down it yields 2.75 But a certain point you're just going to say you know what I want to own Ralph Lauren


Most of you are still here. That's good. After Tuesday, so. Well, I'm in the middle of it right now. How is it so far? Okay. Okay. I graded the first problem, and I was surprised at how many issues there were with stating the first and second principle of induction. Oh, you haven't... Oh, what does that mean? It means that I haven't seen it. Oh, right. Okay. I appreciate your honesty. You have a man of great integrity. Okay. Well, now you know something about it. Okay. Okay. I won't say much about it then. Well, you'll get your exam back on Tuesday. It'll be done on Tuesday. All I will say is that some of you need to study your definitions a little bit harder. That's all I'm going to say for now, I suppose. So, what's that? It's a lead job and definitions, Joe. Oh, yeah. Yeah. Well, you already knew that it was probably going to be on the exam anyways. Yeah. Yeah. So, okay. So, the one thing I'll say is... Okay. Can you guys, do you mind if you just stop talking for a second? I appreciate it. The thing I want to emphasize, this isn't going to give anything else away, but, you know, some of you, I think, are...  For all or there exists, you know, for all acts there exists, some of you really don't know what that is saying, and that's why there are a lot of mistakes on these definitions. I'm confident of it because I've taught this kind of stuff before.  Some of you will, instead of saying s an element of with the epsilon, or sorry, instead of saying s is a subset of, right, with the sort of turned u and the bar underneath it, you'll say s is an element of n when you actually mean that s is a subset of n. So, these are the kinds of things. This is just basic mathematical lingo. Some of you still don't quite have that down yet. So, that's important. It's very important to get that down because changing those symbols, you're saying something completely different. And often, by just changing a symbol, what you say doesn't make sense. Yes. Yes, you will. Because the definition is right in your notes. It's right in the book. You guys should have done better on this. I'm sorry, you really should have done better on this. I told you it'd be on there. It's in the notes. It's in your book. All you had to do is memorize it. That's it. So, yes, you are going to get hosed a bit on this. Yes, that's correct. You'll get partial credit. But, you know, there's really no excuse to not get this. And most of you did not get it. Even though I said it many, many, many times. And I even told you, for example, okay, my rant will be over, okay. But I even said, this has nothing to do with properties. I said that in class. The day before, well, it was a week before because of the snow day. But I said, you shouldn't be writing p of n when you're defining the first principle of finite induction. I didn't do it. I didn't do it. I never did this when we talked about it, right? You use this idea when you're actually proving statements using the principle. But the principle, the first principle of finite induction has nothing to do with properties. I saw a lot of p of n prove for p of n plus 1 and all this stuff. No, that's not what it says. It doesn't say that in the notes either. Okay? And I'm just a little disappointed because I told you that specifically. I think I was even yelling, don't say p of n. If I asked you this on the test, no p of ns. I still saw, like, half of you writing p of ns. Which, you know, just... No, that's bad. That's not good. Okay. So, all right, I'm done scolding you for now. I'll scold you more next Tuesday. Okay. So, okay, and this is... Yeah, this is not good. Now, I have no more edge lines to make things neat anymore. Okay. This section is actually, I think, called the Euclidean algorithm. We're not going to do the Euclidean algorithm. We're going to do some other stuff in this section. If you knew... Well, maybe some of you do know this, but you really should be thanking me for not doing the Euclidean algorithm in class. It's kind of messy, and it's just... Yeah. But, I mean, to actually write it out rigorously, you have all these... You have all these remainders and things, and then it just... No. No. No. So, well... Yeah. Well, I... Here's the thing, though. In the homework, there's very little on the Euclidean algorithm in the homework anyway. So, really what I'm going to do is I'm going to actually do the stuff that's going to help you more with the homework problems. So, I think, really, this is the best way to go, honestly. So, basically what we're going to do is we're going to just keep talking about the GCD for a bit, and then we're going to talk about the least common multiple, and we're going to try to relate these things together. So, what I'm going to do is just kind of go straight into a theorem. This is still talking about the GCD, and so we're going to suppose that A and B are integers, right? That's what this means. A and B are in Z, and they're not both zero. If you remember, we talked about the GCD defining it. We don't want both A and B to be zero, because then there is no greatest common divisor, because everything divides zero. So, we kind of imposed this in order for the GCD to exist in a natural way. And the theorem says, for any natural number K, the GCD of K, A, and K, B is equal, so basically this is saying you can factor out the K. This is equal to K times the GCD of A and B. Okay, this isn't that surprising. You have the largest number that divides both A and B. It stands to reason that if you want to find the largest number that divides both K, A and K, B, it's probably going to just be K times that number, probably. The book goes through this and does a proof using the Euclidean algorithm, which there's nothing wrong with it, but it's a little bit overkill in a way. There's kind of a nicer way to do this, which I'm going to show you now. So what we're going to do is we're going to let D, B, the GCD of K, A and K, B. Okay, so again, it would do you some good just to try to follow the argument here. So what we're going to do is we're going to prove that D is equal to K times the GCD of A and B. So I'm going to kind of go through this slowly here. We want to prove this, right? So I'm just calling the left side D. That's all I'm doing. So proving this then is equivalent to proving that D is equal to K times the GCD of A and B. There's nothing deep going on yet.  So I'm just going to call this D. Okay. So here's what we're going to do. We're going to let D prime be equal to the GCD of A and B. Thank you. Thank you for noticing. I appreciate that. I'm really glad about that. I'm going to sleep better tonight knowing that. So D prime is going to be the GCD of A and B. I could have done this all in one statement, but I'm just going to do this gradually. What is it that we have to show then? So we're going to show that D is equal to K times D prime. D is K times the GCD of A and B. And so I'm just letting D prime be the GCD of A and B. So we're trying to prove that D is equal to K times the GCD of A and B. And so I'm just going to call D prime the GCD of A and B. So now we just need to prove that D is equal to KD prime. All I'm doing, if you go back up to what we're trying to prove, we're going to go to the left-hand side and D prime B, this guy. Just for notational convenience. That's all. So now we just have to show that D is equal to KD prime. So here's what we're going to do. So the first thing is that D prime divides A and some of these that just are right from the definition, I'm not going to write out explicitly just for space, but I would really encourage you to try to follow along here. Why does D prime divide A? Well, D prime is the GCD of A and B. I'm just reminding you what the definition is. The GCD of A and B in particular is a divisor of A and B. It's the greatest one. So since D prime is the divisor of A, D prime divides A. It's just from the definition of GCD. Okay. Thus KD prime divides K times A. See if you believe that. I could write all this out, but like I said, I'm not just to save space. How would you get this? If you know that D prime divides A, how do you know that KD prime divides KA? How would you prove that? Well, this just means D prime times X is equal to A. Just multiply through by K. Then you get the second statement. That's all you have to do. Does that make sense? Just multiply it through. If A divides B, then C A divides C B. It's probably part of the theorem, too, somewhere. That would be my guess. I don't remember off the top of my head, but probably. Okay. Similarly, KD prime also divides KB. For the same reason. It's the same argument. It's just with the B replaced. D prime is the GCD of A and B, and you just go through the same argument. It's exactly the same. Just with B in place of the A. What do we know then? KD prime divides KA and KB. There's something I have to think about now for a second. Whatever this is, it's an integer that divides both KA and KB. Notice that D is the GCD of KA and KB. It's the greatest common divisor of KA and KB. Since KD prime is a divisor of KA and KB, therefore KD prime has to be less than or equal to D, because D is the biggest one that divides both. Yes, yes. That's what I'm saying, because KD prime is a common factor of KA and KB. D, by definition, is the greatest common factor of KA and KB. Since KD prime is a factor of KA and KB, it has to be less than or equal to the greatest common factor. That's incorrect. It's just a little bit. It just doesn't really follow the standard. I don't know that you lose points for it, but I prefer that you write it out. Any questions about this last part? KD prime being less than or equal to D? Here's the strategy. We're trying to prove that D is equal to KD prime. We know that KD prime is less than or equal to D. If I can prove that D is also less then they have to be equal to each other. You can't have one thing less than another thing and that thing less than the first thing. They're going to have to be equal. Let's see here. Can I hide this somehow? Yeah, right here. Up here? Yeah, I got it. There we go. Thank you. I'll give you some extra credit for that. I'm on video too. I'm not going to dig myself any deeper than I already am. The second thing we want to observe is let's note that there's not much to say here. K certainly divides KA and K divides KB. You guys by that? How do you know that K divides? Why is this true? It definitely matters. It definitely matters. A. I'm just trying to get you to think about the definitions. Think about what this means. What does this mean? This means that there's some integer you multiply KA to get KA. Well, obvious, A. K times A is KA. And same thing here. That's why K divides KB. I don't want you to just sort of memorize this vertical dash. I want you to, when you see it go, what does that mean? We're saying K divides KB. B. I want you to try to think about this. Some of you really need to have practice, really associating meaning to these things. You're making it overly complicated. You don't even need to introduce this third auxiliary variable. It's just simply this. K times A equals KA. So by definition, K divides KA. You don't need a C. It's sort of like saying, let me call this X. Then X is messy. I don't need to do that. I can just say a rag that's messy. You see what I'm saying? Yeah. That's also, I'm glad that you brought that up though. I still see a lot of you do this. A lot of you are throwing in as many variables as you can. You're doing an induction proof. Let N equal K. And then let's prove it for K plus 1. Let's call it K prime now. There's no reason to do that. Just stick with 1. It just makes it so much easier. I've kind of felt the author for this in part because he does this a lot. He'll change variables 8,000 times and approve for no reason. But you really don't need to do that. Yeah.  There's no reason to do that. It's just confusing for no reason whatsoever. Especially at this level. It's really bad for them to do this. Yeah. Right. Well, it is what it is unfortunately. But okay.  I'm going through this kind of slowly. But I do want you to think about this. By the way, I should point this out. Your next homework will be do a week from this coming Tuesday. Okay. So you don't have any homework due next week. Okay. So why is this true? Can anybody tell me why this is true? Guys, can you try not to do that? I would appreciate that. Okay. Why is this true? It comes from the definition that we talked about. This is in your notes. I'm just going to remind you. The GCD of A and B has the property that number divides the GCD. That's actually in your notes. That's kind of how we define the GCD. And D, remember, D by definition is the GCD of KA and KB. So since K divides KA and K divides KB, it's a common divisor of KA and KB. It has to divide the GCD. And that's where that comes from. Okay. Yeah.  I have no question in asking at the end. Okay. So what do we know? Hence, k times alpha equals D for some integer alpha, right? That's what it means to divide D. Okay. So I'm going to, underneath this, I'm going to put an asterisk down here because we're going to use this here in a second. Okay. And since D is the GCD of we know that D definitely divides KA. I'm doing this. This proof may not be the most streamlined, but I'm doing this because I just want to remind you of all of these things. Okay. Definition, right? It's the greatest common divisor. In particular, it is a divisor of both of them. That's why D divides KA. Okay. Okay. But, remember that by star so k alpha divides KA, right? And that's just coming back from over here. Sorry, I should move this a little bit. Bless you. D divides KA, but remember that D is equal to k alpha, so k alpha divides KA. What can we conclude from this? Thus, if we know that k alpha divides KA, is there some obvious conclusion we can make from this? I mean, this is vague, but what can I say as a result of this? That has to do with alpha and A. Alpha divides A because you can just cancel the k's out. If k alpha divides KA, then that means k alpha x equals KA for some integer x. You can cancel the k's. You know k's not 0, right? Because we're already assuming that for any k and n. Remember what n starts. 1, 2, 3, 4, 5. k's not 0, so we can cancel it out. Okay. So alpha divides A and you can do the same kind of thing. You can play the same game to conclude that alpha also divides B, right? Because D divides both KA and KB. So if we do the same argument with D dividing KB, you're going to get that alpha divides B. So we're almost done. So we're running out of room here, but so if we know that alpha divides A and alpha divides B, what can we say let's see, what do we do? We let D prime be the GCD of A and B. Right? It's the greatest common divisor of A and B. Alpha is a common divisor of A and B. D prime is the greatest common divisor of A and B. So alpha is less than or equal to D prime. Okay? Alright, let's see if you believe this. Hence k times alpha is less than or equal to k times D prime. Do you buy that? Because k's positive. That's true by K, for sure. Okay, and now by this starred equation k alpha is D, right? So we get that D is less than or equal to kD prime. See that? I'm just replacing the k alpha because we know that it's equal to D. So what do we have? We're basically done now. Okay? What do we prove? We prove that kD prime is less than or equal to D and we also prove that D is less than or equal to kD prime. So they have to be equal. And that's exactly what it is that we're trying to show. Okay? If one number is less than or equal to the other and then the reverse, they have to be the same number. Do you guys understand that? Does that make sense? Okay? So that's it. One and two, give it to us. Yeah, let me say something more about that here in a second. Okay, I'm going to put a little box here because this is not part of the proof. Just sort of to address this. You don't even need to write this down. Okay? But suppose you have real numbers A and B and suppose you know that A is less than or equal to B and B is less than or equal to A. Okay? If, suppose they weren't equal. So if A wasn't equal to B, then it's not, so if A is not B then, okay, we know A is less than or equal to B. If A weren't equal to B, then they would have to be less than B or equal. Right? Then you'd have A is less than B and B is less than A. Right? Because of this. Okay? If you're assuming this, I want to show you that A and B have to be equal. Well, if they're not equal because A is less than or equal to B and they're not equal, it has to be less than B. And the same thing here. So then you get this, but by transitivity A is less than itself and that's not possible. No, I mean I'm sort of explaining where it comes from. You know, there's always this question of, well, how much do you have to actually show? You know, how much do you have to prove? And there is no exact answer I can give you. It just doesn't exist. I can't say you have to show, if any of these two million things show up, you have to show these, but these five billion things you don't. I can't do that. So this is something that I wouldn't expect you to show. I also wouldn't expect you to know that you wouldn't have to show it. I'm just letting you know that you don't have to. Okay? Okay, so that's it. So we've proved this theorem. All right, any questions here? No? Everybody's asleep? Okay. All right, that's fine. Okay, so you guys have this down then? Okay. All right. So I'm going to prove now a lemma. This is actually, I think the rest of the proofs are not as kind of tedious as this one. And this is sort of a corollary, but a and b are integers, which are not both zero. Usually, there's not a big difference. Usually lemmas are things that you prove that are not as complicated or important as a theorem, and then you use them to prove a theorem. So it's sort of like, it's sort of like a piece.  And so you might have several lemmas and then you'll put them together to prove the theorem. Instead of having a 10-page proof, you'll break it up into pieces. It just makes it easier to follow that way. Okay, and this, I don't think you're going to have a hard time buying this. The GCD of a and b is equal to the GCD of minus a and minus b. Not really a whole lot to this, really. But what we're going to do is we're just going to do something similar to what we did before. Don't worry, this is not very long. D is going to be the GCD of a and b and we're going to let D prime be the GCD of minus a and minus b. In fact, I'm not even going to do the whole proof. I'm just going to do part of it. D divides a and D divides b thus and I'm not going to write out the details. If somebody wants me to, I will. But if D divides a, that just means D times some other integer is equal to a, then D definitely also divides minus a because if Dx equals a, then D times minus x equals minus a. Oh, yeah, I believe. Okay, yeah. Well, we're going to just incorporate that. Actually, that's good. Okay, because now we can just reinforce this. So you've got experience with this. So I guess I won't need to say much else about this. So D divides minus a and D divides minus b. So what can we say? Think about the idea that we had before when we were trying to prove that the GCDs were the same. What did I do? I showed that one was less than equal to the other and the other one was less than equal to the first thing on the other side. So, because it's the same thing as before. Since D is a divisor of minus a and minus b, and D prime is the largest divisor of minus a and minus b. That means that D has to be less than or equal to D prime because it's the largest one. And D is one of them. So if you do the same thing the other way, you get that D prime is less than or equal to D and you're done. And that's it. Oh, that D was really bad. I'm regressing now. Okay. So that's it. You show that they're both less than or equal to the other one and then you conclude that they have to be equal. All right. Yeah. Oh, no, because that's what I want though. Because I mean it's true that D prime divides minus a and minus b. But what I really want is that D divides minus a and minus b. So, because D divides a, and as I said, whatever you have to multiply x by to get a, multiply minus x by to get minus a. And so D is definitely going to divide minus a and minus b. The reason why we want D to divide minus a and minus b is that then we can conclude that D is less than or equal to D prime. Because D prime is the largest thing that divides minus a and minus b. That allows us to conclude that the first one is less than or equal to the second one. That's why we want the D there. So we can get that it's less than or equal to. And we flip the argument around so we get D prime is less than or equal to D and then we conclude that they're the same. What you're saying is definitely correct. It's just that we want the D there so we can get the less than or equal to and approve. Is this okay? Okay. So now I'm going to introduce another thing that I don't know if I've ever used this before. Maybe I have, maybe I haven't. Okay. Why am I doing this instead of calling a theorem? Well, it doesn't matter really. It's just, usually in mathematics, in, you know, research papers, a proposition is sort of a theorem that's not that great. A lemma is a theorem that's really crappy. A proposition is okay and a theorem should be something that's interesting. Okay. So now we're going to let a and b be a lens of z, z rather, sorry. Again, which are not both zero. And again, the reason why I do this is because otherwise the GCD is not defined. Let's see if I can get this on here. For any non-zero integer k, the GCD of k a and k b is equal to the absolute value of k times the GCD of a and b. Okay. So I'm going to ask you a few questions as we go through this. I'd like you guys to be as involved as possible because I think that this is something that you guys should be able to do. All right. So what we're going to do is we're going to assume that let a and b be integers. We're going to assume that they're not both zero. And now we're going to let k also be non-zero. Okay. So we want to prove this. And what we're going to do is we're just going to consider two cases. k is not zero. It has to be either positive or negative. So we're going to just look at these cases separately. And this will fall out pretty quickly. Case one is that k is positive. And the GCD of k, a and k, b. So how can we proceed with this? We want to show that it's equal to that. And there's hardly anything to do. We've basically already done it. I think back to what we did in the beginning of class today. This is equal to k, right? Right. It's equal to k times the GCD of a and b. This was what? Theorem one, right? Theorem one. Was it theorem one? Yeah. Okay. Because theorem one, remember, the hypothesis was that k is positive in theorem one. And in this case, k is positive. So we can definitely use theorem one. And this is, of course, the same thing just one second. This is the same thing as the absolute value of k times the GCD of a and b because k is positive. So we definitely get this to work out in this case. Yes. Yes. Yes. Absolute value of k. Okay. I'm running out of room here. But if you wanted to write a note beside how I got this, it's just because we're assuming k is positive. Absolute value of k and k are the same thing since k is positive. I'm just running out of room. That's why. Okay. And this is going to get messy probably because I'm going to try to squeeze this all on this one page here. But case two is that k is less than zero. Okay. So this requires just a little bit of work. Hmm. GCD of, we're going to try to do that to prove the same thing. GCD of k and kb is equal to okay. So I did prove this lemma for a reason and it's going to come into play here. Joe. Yeah. Yes. Yeah. And that's exactly what I was going to say. Yeah. It's lemma one. Yeah. We're going to use lemma one here. Okay. And you'll see how this applies in a second. This is equal to the GCD of minus kA and minus kB, right? By lemma one. Which is the same thing as the GCD of minus k times A and minus k times B, right? Same thing. Minus kA, the minus on the outside is the same thing as minus k times A. Same thing. Okay. So now, how can we finish this? So what can we say now? Now what can we do? We can pull out the minus k because k was negative. We're assuming k is negative. So the minus k is now positive. So now we can use theorem one to pull the minus k out. Make sense? Okay. And that's why we have the lemma because now we can flip it to being positive and we can pull it out just like before. But we're not completely done. There's still one other thing we have to know, right? Because we're trying to prove this. So we almost have it, but really what we want though or what we need I should say is that minus k is actually the absolute value of k, right? Why is that true? Because k is negative, right? Remember what the absolute value does to a negative real number. It just erases the negative essentially. It makes it positive.  Is that wrong? Well, but you still need, you still sort of need this, right?  Yeah, yeah. But here's the thing though. We can pull out the k though because k is positive and theorem one, k has to be positive in order to pull it out. So since k is negative, we can't just pull it out. We have to use the lemma to make it positive first and then pull it out and then we say because k is negative, minus k is the absolute value of k. You really, I don't think you can really get around and you have that. Okay. Sorry about things getting all squished up here, but this is the end of the proof. So this is the absolute value of k times the GCD of A and B. And again, I'm running out of room. I apologize. If you want to put something in your notes to remind you why this is true, it's because k is less than zero. That's why minus k is the absolute value of k. Is this okay? Of course, you guys learned this a long time ago. I think you guys are probably fine with this. Okay.  So it allows you to sort of know what the GCDs of other things are without having to go through all the computation. Okay. So, the last thing we're going to talk about, if everybody's got this down, okay, are we okay? You guys okay? And this is not going to be nearly as painful as it was before. Sorry, let me move this around here. I'm going to call this a theorem, although there's not much to it really. A and B are non-zero integers. Then there is a least positive integer M such that A divides M and B divides M. Okay. Because of time here, I'm thinking I may not write the proof down. There's really nothing to this proof. It's just if A and B are non-zero integers, then A and B both divide the absolute value of A times the absolute value of B. And that's certainly positive. That's not hard to show. That's very easy to show. And so, just by the well-ordering property, just take the least positive integer that A and B both divide. That's it. That's all there is to it. So, like I said, I think I'm not going to write this out. I want to spend time on more complicated things. So, you don't have to prove this. Just take this as fact. It's very simple to establish this. So, let me go to the definition then. And you've also seen this before. These come in multiple. So, this is denoted LCM. And this is simply the smallest positive integer M such that A divides M and B divides M. Okay. So, it's just the smallest positive integer that's multiple of both A and B. Okay. So, just to refresh your memory here, I don't mean this to be an 8th grade algebra exercise or anything, but just to make sure that everyone's on board here. What's the least common multiple of 8 and 12? 24, right? Smallest number, positive number that 8 and 12 both divide into, right? What about this? Sorry. The least common multiple of minus 6 and 7. Okay. I'm going to call this grade algebra teacher if you don't get this right. Okay. Apparently, you really like your algebra teacher. Nope. It's 42. You and you are both, they're coming to class next week. Third grade teacher, okay. All right. I'll take care of her for you. It's okay. Keep forgetting about this camera here. I'm going to incriminate myself. Okay. Because by definition, the least common multiple by definition is the smallest positive integer. And that's why I did this. So remember, by definition, the least common multiple and the GCD, they're positive. Positive. Yeah. I understand it has to be positive. It doesn't, but that's not the definition. The definition is just simply the smallest positive number that both of them divide into. Minus 6 divides into 42 because minus 6 times minus 7 is 42. We're not multiplying by to be a positive number. It can be any integer. Okay. Are we good with this? No? Is this okay? All right. So, this is the last page. Okay. So this is the last, we're going to do one more theorem and then we're done. Okay. And if we finish early, then we finish a little bit early today. All right. We go to the next page. Yeah. We good? Yeah, Joe. About this LCM thing. In this case, with the 6 and 7, I noticed that it's just the same thing is just multiplying the two of them. Does that work? Does that always work if one of them happens to be a prime number? No.  If it was 7 and 7, for example, the LCM would be 7. It wouldn't be 49. But actually, you're going to have an answer to that question in the next theorem, I think. So you'll see how this works. So the point is, here's the point. What do you mean to? The least common multiple is equal to the product if and only if the two numbers are relatively prime, as she said. Relatively prime. So they don't share any common. So the GCD is one. GCD is one. So this is going to come from this next theorem. Okay. That's too much work. No, no, no. I'll take care of that myself. It's okay. I appreciate the offer, though. I do appreciate the offer. You would do it and we'd get it on video. That would... I'll think about doing it in the future. I'm technically challenged, so this could be a problem for me. Okay. Here's the last theorem. And we're going to let... Now this time it's even easier. So A and B now are natural numbers. Then the GCD of A and B times the least common multiple of A and B is equal to A and B. So, here's the proof. Like I said, this is the last proof, so muster up some energy to try to pay attention to this and then we'll be done. Okay, so let's just assume A and B are natural numbers. And let's let D be the GCD of A and B. So we know a lot about the GCD. We're going to try to prove something about the least common multiple. So here's what we're going to prove. We're going to show that the least common multiple of A and B equals A, B over D. That is what we're going to...that was bad. Well, that's okay. A, B over D. Least common multiple is A, B over D. So I want you to see that this is the same thing as what we're trying to prove, right? Because we just divide through both sides of the equation by the GCD of A and B. We're calling that D, remember. It's the same thing as asserting that least common multiple of A and B is A and B is A, B over D, right? Just divide through. And then use the fact that we're calling it D. Okay. Here's one thing I want you to note which I'm not going to really justify in detail. But A, B over D is definitely a natural number. It's a positive integer. In order for it to be the least common multiple, by definition it certainly has to be a positive integer, right? That's how we define the least common multiple. In particular, it has to be a positive integer. How do we know that A, B over D is a positive integer? How do we know that? Anybody? Joe? D was the GCD, right? D is the GCD of A and B, yes. Because if D divides A and B divides B and A and B are integers. So if you do A times B, you get an integer and D has to divide that. Yes, that's right. And because they're all positive, right? Remember our assumption is that A and B are natural numbers. Remember the definition of natural number. They're all positive. So A, B over D is also positive because A and B are positive, right? And D is the common divisor of A and B. D is a factor of A. So this is certainly, it's not a proper fraction. It is actually a whole number, right? So we're just going to prove several things here. So the first thing, what do we need, and this is the reason why I'm doing it this way is because I really want you to think about the definition of least common multiple. So we're claiming that A, B over D is the least common multiple of A and B. So one thing that has to be true, there are three things that have to be true for this to be the case, aside from it being a positive integer, which we already know. One is that A, B over D has to be a multiple of A and a multiple of B. And it also has to be the smallest one that's a multiple of A and B. So there are three things that we have to check. One, we need to know that it's a multiple of A. Two, that it's a multiple of B. And three, that it's the smallest multiple of A and B. So that's what we're going to prove. A is a factor of A, B over D. So in other words, A times some integer is A, B over D. And there's not much to this. Note that A times B over D, of course, this is so simple you shouldn't even be thinking very hard about this. You guys buy that? A times B over D is A, B over D? How do we know that? Does that immediately imply that A is a factor or that A divides A, B over D? No. Because what if B over D was a half? It has to be an integer, right? It has to be an integer. Yes. Yes. I'm just trying to get everyone to think this way. But I want to be very clear on this. This is a common mistake that you might make in your homework. Just because you say you want to prove that A divides B. If we know that A times something is B, that does not mean that A divides B. Within the real numbers, you can always do this, right? A times B over A is always B. So just knowing that you can multiply by something to get B does not mean that it divides. Everything has to be an integer. Everything has to be an integer, right? Because you could prove that 2 divides 3, right? 2 times 3 halves is 3. Oh, therefore 2 divides 3? No, it doesn't. Because 3 halves are not an integer. You need it to be an integer. I've never taught number 3, but I'm sure that that's the case that you're going to make is not really realizing that you need to know that what you're multiplying by has to be an integer. That's very important. Yes. Right, sure. We're actually going to do that here in a little bit. But I'm just going to write it out in words just because I want to emphasize this really in words. And B over D is an integer since D divides B. That's the important part. 2. Okay, I want to make sure I get this done. B also divides A, B over D. And the proof is just the same idea. In this case, you just say that B times A over D is A, B over D and D divides A because D is the GCD of A and B. It's the same argument. I'm just flipping the letters around. Okay, and this is where we actually have to do a little bit of work. Let's suppose now that C is a positive integer and that A divides C and B divides C. So in other words, C is a common multiple of A and B. We want to show though that A B over D is the smallest positive common multiple of A and B. So what we have to prove is that if C is any common multiple of A and B, then A B over D has to be less than or equal to C. In other words, A B over D, that shows that it's actually the least one. It means that it's less than or equal to anything that's a common multiple of A and B. It has to be the least. So what we have to prove, we need to show that A B over D is less than or equal to C. And I want you to pause for just a second here. I'm going to kind of reiterate what I said before. C is positive and C is a common multiple of A and B. We are trying to establish that A B over D is the smallest possible one. So since C, we're assuming is one, we need to show that A B over D is less than or equal to it. That shows that it's the smallest possible one. Okay. So, and it's just going to go like this. Since A divides C, A X equals C for some integer X, right? And so this is just coming from our system here. A and B both divide C, right? Since B divides C, B Y equals C for some integer Y, okay? That's just definition. We're assuming A and B both divide C, so this just follows from definition. Okay. And the last thing that we're going to say before we put all this together is as D is equal to the GCD of A and B, what do we know about D? D is equal to A alpha plus B times beta for some integers alpha and beta. You guys remember this? You know why I'm writing this down? The GCD is a linear combination, right? When you prove that, that's actually, that was actually built into the, sort of built into the definition almost, that I gave you. Okay. Let me move this thing out of the way here. See if I can squeeze all this on here. All right. This is going to get a little messy. I apologize for that, but you guys can all fall all these steps though. So I'm going to take C over A, B over D. I'm doing this for a reason. No, no, no. This is the clearest way to do this. You'll see here at the end why I'm doing it this way. So how do you do this? C divided by a fraction, you just flip and multiply, right? You guys know that, right? So you just multiply C by D over A, B, and you get this. Okay. So this is equal to C times A alpha plus B beta, right? Over A, B. You guys see where I'm getting this now? I'm sorry, I'm running out of room here, but D is A alpha plus B beta, so I'm just replacing the D with this. That's all I'm doing. It's equal to that, so it's just going in right there. Okay. So this is, if you distribute, this is C A alpha plus C B times beta simplify this. Hopefully I can do this in well, if anyone's at all lost here, just let me know. This is C alpha over B plus C beta over A. You guys see how I got that? We can just split the fraction up into two pieces. We can cancel the A out for the first piece, so that's where the C alpha over B comes from. In the second piece, we can cancel the B out to get C beta over A. Yes. Yes. We're not, although we still got a little work to do. Well, all we have right now is that we divided by, that's kind of where we're going. That's roughly the idea. We've divided by A B over D. What we're going to show though is that this is actually an integer. Once we've got that, that really gives it to us. Don't you already know that? Right, but we don't know necessarily because we need to show that these are integers too. This is a fraction. Maybe this isn't an integer. Yes. The answer is yes. I'm just going to write the details out. But we do need to, we just need to say a few things to really nip this in the bottom and make it concrete. But your idea is basically right. Yeah. Okay, so let me, I'll tell you what, just to make this part a little bit easier, I just got boxed in here and I'm running out of space. So I'll call this one, this equation one. I'm going to call this equation two. I'm still going to have to do a new page here, I think. Okay, so what I can do now is, so the last part we had before was C alpha over B plus C beta over A. Right? But look over here. C over B is Y. You see that? C over B is Y. So I'm just replacing this part with Y. That's where I got that. So what's going on up here? C over A is X. C over A is X. So I'm replacing the C over A with X to get X B over, X beta over there. Does that make sense? Okay. Okay, so, yeah, now we're almost done now. My, can I go on here? Almost done. Almost done. Okay. So here's what we get if we put all this together. So A B over D times Y alpha plus X beta equals C. If you look, I can go back to the other page if you want to, but remember what we had before. We had C over A B over D is equal to Y alpha plus X beta. Right? That's what we had before. So all I'm doing now is just dividing both sides by A B over D. Go back to the previous, what you just had before I wrote that, multiply both sides by A B over D. That's all we're doing. And you should see where this is coming from. Is this okay? Just A B over D on both sides. So what can we say then? Well, we don't even need to say that we just know that A B over D certainly divides C, right? These are all integers, right? Y alpha, X and B, these are all integers. So this product in sum is certainly an integer as well. So A B over D definitely divides C. What is it that we wanted to prove? We wanted to prove that A B over D is less than or equal to C. That's what we wanted to show. But because it divides it, everything's positive, right? And this is actually a theorem from 2, 2. A B, D and C, these are all positive numbers. If you know that one positive integer divides another one, then that first one has to be less than or equal to the other one, for sure. Right? For sure. And if you want some justification for this, this is also just theorem one of section 2.2. When I say theorem one, I mean according to my labeling. Part F. So we've actually, I mean, this last part we actually already proved this, essentially. Okay. And don't worry. I'm not proving this. I'm just stating this as a corollary. I'm not going to prove it. A and B are natural numbers. Then, and this is what I was just saying just a few minutes ago. The least common multiple of A and B is equal to A times B if and only if the GCD of A and B is 1. Well, yeah. Yeah. Yeah. So the idea here is just that, you know, the GCD is small as possible, of course, when two integers are relatively prime. But if they don't have anything in common, then that makes the least common multiple as big as possible. If you think of prime factorizations, it sort of makes sense. The prime factorization of A and B, they don't have any primes in common. So the only way for, if you have any common multiple of A and B, it's going to have to be all of those primes together because there's nothing in common. So that's sort of the idea. Okay. So I think I will stop there. To start on this, but let me go ahead and give this to you. Okay. So this is 2.4. Okay. So I'm not giving you a ton of problems here. So this is 3, 4A, 5A, 6, 9 and 10A. I should tell you that you will almost certainly have one more assignment that's going to be due aside from this also next Tuesday. Okay. On Tuesday, that will probably be due as well. So you'll still have another week to finish it.  You'll get your exams back on Tuesday. You'll get your previous assignment back on Thursday. I'd rather not pass all that back in one day. So the exam will be Tuesday because I assume you want to get that back probably more than your last homework. And then the homework will be Thursday. And then the next Tuesday will be your next assignment. Okay.