 How can we define the multiplication of the integers? Again, it's helpful to think about A, B as the integer A minus B, and we note that if we multiply our two ordered pairs, well, if we multiply A minus B times C minus D, we get. And if we want to read this back as an ordered pair, we need to express it as something minus something, which gives us our component. And this suggests a way we can define the multiplication of two equivalence classes when the equivalence classes represent integers. And of course, this is well-defined, and the product of two integers is an integer. Well, maybe you should prove these things. Now remember, our class representative has components that are natural numbers, and this means we can determine whether A is less than B, A is equal to B, or A is greater than B. Definitions are the whole of mathematics, but we know what less than, equal or greater than means, so let's just go straight to and define positive and negative. Now since we're thinking of this equivalence class as the integer A minus B, this suggests the following. The equivalence class AB is going to be called positive if A is greater than B, negative if A is less than B, and if A and B happen to be the same thing, this equivalence class is just the additive identity zero. Remember we showed that a little earlier. And now that we have positive and negative, let's prove one of the great mysteries of mathematics that the product of two negatives is positive. So let's take two negative integers, the equivalence class AB, and the equivalence class CD. Definitions are the whole of mathematics, since we want to talk about the product, let's go ahead and pull in that definition of product. And so the product of the two integers is. Now this will be kind of a complicated proof, so let's go ahead and put in where we want to be. We want to conclude that our product is positive. Definitions are the whole of mathematics, all else is commentary. We have a definition for positive and negative. Since we want our product to be positive, that means our first term has to be greater than the second term, which allows us to construct one step back on our bridge. On the other hand, the integers we're starting with are negative, which means the first term is actually less than the second term. And so we know that. Now since inequalities are directional, it's helpful if they all point in the same direction. So here we have less than, less than, but at the end we have greater than. So let's switch this last inequality around. And remember that a, b, c, and d are natural numbers and so we can assume that we've proven all of their properties because you have done all your homework right. So I want an ad on the left. Well if I multiply this first inequality by d on both sides, I do get ad less than bd. I also want a bc. So if I multiply this inequality by b, I get my bc. Oh and now I can add these two inequalities together and I'm stuck. Now we can try and fix this proof, but let's think about this for a moment. One objection we might have is that this expression we're working with is kind of complicated and so we might ask whether we can simplify this. Well remember, definitions are the whole of mathematics, all else is commentary, and our definition relies on this notion of greater than or less than for the natural numbers. And we didn't define it because everybody knows what it is, but maybe we should define it. Here's one possibility. Let a and b be natural numbers, a is less than b if there is some other natural number not equal to zero where a plus k is equal to b. So since we want a to be less than b and c to be less than d, then we know that a plus k is equal to b and c plus m is equal to d. And since a, k, b, c, m and d are all natural numbers and all of the properties we've proven for the natural numbers still hold. So I can rewrite one side using the additive identity zero. Definitions are the whole of mathematics, all else is commentary. Remember that our definition for our equivalence relation is, and so I can rewrite both of these statements in terms of our equivalence relation. Well what good does that do? One of the things to remember is that since we are working with equivalence classes, we can use any other class representative. And so instead of the equivalence class with class representative a, b, we'll use another class representative, zero, k. And likewise for the equivalence class with class representative c, d, we'll use a different class representative, zero, m. And now we have much easier equivalence classes to work with, and we can find their product. And since k, m is greater than zero, this product will be positive. Now it's worth pointing out a few things about this proof. First of all, the first time we tried it, we tried to use just the definitions that we had, and that didn't work so well because we ended up getting stuck at the midway. And part of the problem is that we didn't have a definition for what it meant for one natural number to be less than another. It was only after we introduced a definition that we were able to complete the proof. In fact, definitions are the whole of mathematics, all else is commentary. If we take out the lines of the proof that rely on the definitions, well, we'd have to remove this line because that came from the definition of when an integer was positive or negative, and then this statement came from our definition of what it means for one natural number to be greater or less than another. So this next statement is gone. Now we can keep the next statement because we actually prove that zero is the additive identity, but recovering the equivalence of the ordered pairs came from our definition of our equivalence relation, and so this line, too, would be gone. Our next line is actually a theorem, so we'll keep it, but after that we use our definition of multiplication, and then our definition of positive and negative again. And as you can see, if we take out everything that relies on the definition, we don't actually have anything left that makes any sense at all, including our conclusion. And so again, definitions are the whole of mathematics, all else is commentary.