 So let's continue to look at equivalence classes and talk about operations on equivalence classes. So here's a very general overview for how mathematical research is conducted. One basic strategy for mathematical research is to start out with something. So for example, some set. We'll define a relation on the elements of this set S. So for example, we might define an equivalence relation. And now we have a defined operation on these elements of S. And for example, given two of the elements of S, what can we do to create something new? And here's an important distinction between a relation which tells us two things are somehow connected to each other and an operation which takes two things and then creates something that is entirely new. We see where this relation and these operations take us. And if they don't take us too far, we define a few more relations and a few more operations. And much mathematical research goes along the same line. So again, a useful component, useful guide for becoming a mathematician is he never stopped doing this process. He never stopped trying out new things. So we defined a thing, the equivalence classes of S induced by some equivalence relation. We know that the equivalence classes are set. So we don't really need to define new relations for them. We can just recycle the standard set relations. But how about the binary operations on those equivalence classes? So I have a thing, equivalence classes. What can I do with a couple of equivalence classes? Well, again, the equivalence classes are set. So we could use the binary operations we've already defined for sets, union, intersect, complement and so on. But we find that these don't do anything interesting or at least they don't do anything really new. If we want to create new mathematics, we have to create a new type of binary operation. So we can try to define a new operation on these equivalence classes. However, there is an important warning. One of the problems we're going to run into when trying to define binary operations on equivalence classes is the following. So here's my equivalence classes A and B. It's very natural to try and define that binary operation in terms of A and B. The problem is that I can have two equivalence classes that are the same but represented by entirely different elements. So this equivalence class A could be the same as the equivalence class C. B could be the same as D. And if I define my binary operation in terms of A and B, I want to make sure it identifies the same equivalence class as the one using C and D. In particular, I want to make sure that the result of the operation does not rely on which element we use to represent the equivalence class. And mathematically, we say that the operation is well defined if the result does not rely on the particular representation of the operands. If I switch the operands around, as long as I'm dealing with the same equivalence classes, I should get the same answer. So again, let's try it out. Our overview here is I have some set S. I have some equivalence relation. That equivalence relation induces a partition of S into a set of equivalence classes. And because these equivalence classes are things, they form a set of elements, a good question to ask is can we define a binary operation on this set of elements? So we can define a binary operation any way we want to. But what we have to make sure of is that when we define that binary operation on the equivalence classes, we have to make sure that it's well defined. So, well, let's continue to work with our ordered pairs in N2, with our equivalence relation, A, D equivalent to C, D, whenever the outer sum A plus D is equal to the inner sum B plus C. And again, it's worth keeping in mind that A, B, C, and D are all elements of the natural numbers. So A plus D, B plus D, B plus C, these are all defined quantities. However, things like A minus C, B minus D might not actually have meaning in our set of natural numbers. So again, that's why we define it in terms of plus, because adding two natural numbers is always possible. We don't want to introduce minus because subtracting two natural numbers might not be possible within the natural numbers. So now this equivalence relation induces a set of equivalence classes. So here's an equivalence class. Here's another equivalence class. How do I want to define the equivalence class sum? Well, because A, B, C, and D are all in the set of natural numbers, let's try this out. We're going to define the sum of the two equivalence classes to be the equivalence classes formed by the, if you want a fancy term here, component-wise sum of the elements that define the equivalence class. And so here we've defined the sum of two equivalence classes in terms of the component-wise sum of the representatives of the two equivalence classes. And we're done. Except we do need to ask the all important question, is this operation well-defined? And again, what that means is if I can represent this equivalence class in a different way, I want to make sure that the sum doesn't change. Because if I add these two, I get this equivalence class. If I add these two, I get what looks to be a different equivalence class. And what I want to make sure of is that these two equivalence classes are really the same thing, as long as these two are the same thing. So let's take a look at that. And so here's a good, useful method of proving things. I want to show that this is a well-defined operation, which means that whenever I have two different ways of representing the same equivalence class, that the sum doesn't depend on that particular representation. So useful proof strategy is to know what our goal is. We want to show that this sum and this sum are the same. Know where we're starting. The assumption that we're making is that these two are equivalent. And we should work our way towards the middle. So here's where I'm starting. Here's where I want to end. If I can meet in the middle, I will form a proof. So, well, I know these two are equivalent. Means that the outer sum is equal to the inner sum a plus b prime has to be b plus a prime. That is our definition of that equivalence relation. And I have no idea what to do next. So let's work the other end of the bridge. So my definition of sum of equivalence classes is going to be component-wise. So when I add these two together, this is a plus c b plus d. This is a prime plus c b prime plus d. Now I know that if two equivalence classes are equal, then the representative has to be in the other equivalence class. So I know that a plus c b plus d has to be an element of this equivalence class. And, well, if I'm going to be in an equivalence class, that means I have to be equivalent to the other one, the representatives. So our representatives are going to be equivalent to each other. And my definition of our equivalence relation is the outer sum a plus c b prime plus d has to equal the inner sum b plus d a prime plus c. So this comes from our definition of the equivalence relation. I can do a little bit of algebra because everything inside is a natural number, so I don't have to worry about anything peculiar happening. And so now what do I have? a plus b prime c plus d b plus a prime c plus d. And actually I can get from this line here to this line here because all I've done is I've taken this line and I've added c plus d to both. I've regrouped things. I'm allowed to do that. I've used my equivalence definition here that if I have the outer sum equal to the inner sum, the two ordered pairs are equivalent. And that gives me my inclusion in the equivalence class. That means the equivalence classes are the same. That means the sum is the same regardless of which representative I use. And my operation is well defined. Now one minor qualifier here, I change the first term. What if I change the second term? And we should check to make sure that it's also going to give us the same result. But we'll leave that as an exercise and we'll continue to talk about equivalence classes next.