 So here we'll introduce a new way of representing integers known as the charge field model. And the basic problem is that the number line is actually very limited in what you can do with it. You can easily represent some additions, but the problem is that you can run into difficulties when trying to represent subtractions. And the reason is that we have to view this subtraction A-B as the removal of a set of B objects from a set of A objects. And if we can do this with whole numbers, it's something that's possible with the number line, but if A is smaller than B, if we're trying to take away more objects than what we have, then the ability to use the number line is severely limited. And so a more flexible way of handling integer arithmetic, of representing integer arithmetic, is known as the charge field model. And this is based on two key ideas. Suppose I have a whole number, and what I'm going to do is, if n is a whole number, a set of n-1s, well first of all, by my definition of multiplication, this is n times 1, and that's going to correspond to the positive integer n. We don't ordinarily indicate the sign of a positive number. It's actually helpful to do so in this particular case. So, rather than thinking about this as n-1s, and as the product n times 1, what I'll do is I'll think about this as n plus 1s, and what I have is n times a plus 1. And the reason that that's useful is that if I think about the negative integers, what I can think about here is that I can take a set of n negative 1s, and this is the product, again, by my definition of multiplication, this is n times negative 1, and that's going to correspond to the negative integer minus n. Now, important note for later on, n is assumed in this particular instance to be a whole number. In general, an expression like this might or might not be a positive number, might or might not be a negative number, it depends on what n is. Here we're actually assuming that n is in fact a whole number. And finally, the last key idea in the charge field model goes back to our very definition of what the additive inverse is, if I take plus 1, and I add on the additive inverse, what I get is zero. In other words, positive and negative units cancel each other out. Well, let's take a look at an example. Let's use the charge field model to represent the integers 5 and negative 3, and then represent the sum 5 plus negative 3. So, first thing to notice is that my integer 5, well, that's 5 plus 1s, so I can represent it as a set of 5 plus 1s. So, let's see, so there's my 5 plus 1s, and I will very cleverly indicate each of these with a plus 1 to indicate that it's a plus 1. Now, I also have the integer negative 3, and so again negative 3, well, that's 3 times negative 1, so that's 3 negative 1s, so I can represent that as a set of 3 negative 1s. So there we go. And then finally, I want to put the two sets together, and when I join those two sets, again, addition is the result of combining two sets. Addition is defined as the union of two sets, and so there is a representation of my sum 5 plus negative 3. Now, one more thing we can actually do. So again, this is actually our representation of the sum 5 plus negative 3. It's the union of the two sets, but we can actually simplify this a little bit. Because the definition of additive inverse tells us that whenever I have a plus 1 and a negative 1 together in a set, again set being defined as the addition, then I can replace that. Those things are exactly the same as having nothing. So I have a plus 1, and what I'm going to do is I'm going to locate pairs of positive and negative units, and I'm going to eliminate them. So I look around. Well, here's a pair. Here's plus 1 and a minus 1. Here's a plus 1 minus 1 pair. So I'll go ahead and eliminate that. Throw it away. Here's a plus 1 minus 1 pair. I don't need that anymore. I'll get rid of it. Here's a plus 1 minus 1 pair. I'll get rid of it. And finally, here's what I have left. I have this one and this one. And what I have here is a representation for the number positive 2. And so what I can do at this point is I can conclude that when I take the sum 5 plus negative 3, it's the same as positive 2. Well, what can we do with a difference? What can we do with a subtraction? So again, I can interpret any set of subtraction as the removal of this. I want to remove negative 3 from 5. So I want to start with a representation of 5. There it is. I have 5 plus 1s. And I want to remove 3 negative 1s. So again, this amount that I'm trying to remove is the same as 3 negative 1s. So I go here and I don't see any negative 1s I can remove. So I want to remove 3 negative 1s. There are no negative 1s here. So what can we do? Well, we can throw our hands up and say, I can't solve this problem. I'll go do something else. But if we have a little bit of persistence and give ourselves a little bit of time to think about it, the thing to remember is that anytime I have a plus 1 and a minus 1, I can replace them with 0. But that equality works both ways. Anytime I have nothing, I can put down a plus 1 and a negative 1, as long as I put them down together. In other words, what I can do is I can create positive and negative units. I can create pairs of positive and negative units from nothing at all. So I take this nothing that I have here, and I create a pair of positive and negative units. So I'll take that nothing, and there's a pair of positives and negatives. And I'll take some more nothing, but over here I'll take some more nothing, and I'll create another pair. And I'll take this nothing over here, and I'll create a couple of pairs of positive and negative units. So I have my 5, and then I have a whole bunch of other stuff in here, but all of these are positive and negative unit pairs. All of these are nothing. But that nothing is important, because now I have enough negative ones to be able to remove three of them. So I'm going to remove three of those negative ones. They're gone. And I should do some simplification here. Again, I have this last remaining positive and negative pair. If you think about it, we made a few too many pairs. We didn't need that many. And that last positive and negative pair, well again, a positive and a negative give us zero. I could just drop that entirely. And here's my sum of, sorry, here's my difference of 5 minus negative 3. It's this amount here, which is going to be, if I count it, there's 8 plus 1's there. This sum is equal to 8. How about a different value, negative 3 minus 5? Well again, same sort of idea. I want to remove 5 from negative 3. So I'll start with a representation of negative 3. And I want to be able to remove 5 plus 1's. I'm subtracting 5. I'm removing 5 plus 1's. So I'm going to do the same thing. I don't have any plus 1's here, but that's okay. I can create pairs of positive and negative units. And let's do this at least somewhat efficiently. I'll create those pairs until I have enough plus 1's to be able to remove. So here's a plus 1. That gives me 1 plus 1 I can remove. I now have 2 plus 1's I can remove. 3, 4, 5. I now have 1, 2, 3, 4, 5 plus 1's I can remove. So let's get rid of them. And my representation, my final answer for this, is going to be this set represents that difference, negative 3 minus 5. And I can count here what do I have. I have a total of 8 negative units. This is going to correspond to the value negative 8.