 One way to view numbers is as solutions to equations. Consider an equation like x plus a equals b, where a and b are natural numbers and a is not equal to 0. In some cases, there are whole-number solutions. x plus 3 equals 10. In other cases, there are not. x plus 10 is equal to 3. Or are there? So remember, it doesn't matter what something is. What matters is what it does. So for example, suppose x plus 3 equals 5, and y plus 8 equals 2. And so we can treat x and y as numbers defined by the equation. And the question we might ask is, what equation will x plus y be the solution to? Now for the moment, let's not worry about proving any of the following. This is just an exploratory argument. So we might have x plus 3 equals 5, y plus 8 equals 2. And if I want to get an equation that involves x plus y, well, I could just add the two equations and get x plus y plus 11 is equal to 7. And so here's an equation that x plus y solves. Now let's take another example. Let's suppose z plus 10 equals 4, and y plus 8 equals 2. And it turns out, or we want to show, that z and y solve the same equation. So we know z plus 10 equals 4. And I'd like to say something plus 8. And so here we notice that 10 is 8 plus 2. And likewise I can break 4 into 2 plus 2. And the plus 2 I can cancel and get z plus 8 equals 2, which is the same equation that y solves. And this leads to the following idea. If we view a number as a solution to an equation, then we can identify the number with the terms of the equation. And we can do this in any number of ways, but here's one easy way. The equation x plus 3 equals 5. Well, let's associate that with the ordered pair 5, 3. And likewise, y plus 8 equals 2. We'll associate that with the ordered pair 2, 8. And z plus 10 equals 4 becomes 4, 10. Now, we noted that 2, 8 and 4, 10 solve the same equation. That means they are equivalent. And that means that we'd like to have some way that we can identify when ordered pairs are equivalent. So after some thought, and we'll see why this works, this leads to the following. When a, b, c, d be ordered pairs of natural numbers, then a, b is equivalent to c, d, whenever a plus d is equal to b plus c. If you want, the outside numbers add to the inside numbers. While squiggle is a relation, let's prove that it's an equivalence relation. Well, no, you should do your own homework. But it is an equivalence relation, and because it's an equivalence relation, it partitions the ordered pairs into equivalence classes. And this allows us to define the integers as follows. Let squiggle be the equivalence relation. The integers are the equivalence classes. And again, if you look at that and say, that doesn't look like an integer, remember it doesn't matter what something is, what matters is what it does. And we'll show that these equivalence classes, under the right definitions, act exactly as we'd expect the integers to behave. So let's try to define addition. Now, if this ordered pair corresponds to the equation x plus b equal to a, then we can informally think about the ordered pair a, b as the integer a minus b. So let's think about the addition of these ordered pairs. So if I want to add a, b plus c, d, well a, b, I can think about as the integer a minus b, c, d, I can think about as the integer c minus d, and so when we add the integers, we can simplify this in any number of ways, but we have this connection between the difference a minus b and the ordered pair a, b. So let's try to write this sum as a difference between two quantities. Now, it's okay to think about something informally, but sooner or later you have to be rigorous. And that means setting down a definition which we will agree and abide to. So this suggests the following definition. Let's take two equivalence classes. Then the sum is going to be the equivalence class generated by the component-wise sum. And that's a great definition, but we do need to make sure the result is independent of the representation. And remember this falls under the category of verifying or refuting that addition is well-defined. So let's pull in our definition. By definition, the sum of two integers is this. Being well-defined means that it doesn't matter what representatives we use in our computation. And so we should verify we get the same equivalence class even if we use different class representatives. So remember we can use any element of the equivalence class as the class representative. So if A, B is equivalent to Aprime Bprime and CD is equivalent to Zprime Dprime, then the equivalence class AB is the same as the equivalence called Aprime Bprime. The equivalence class CD is the same as the equivalence class Cprime Dprime, and their sum will be the equivalence class Aprime plus Cprime Bprime plus Dprime. and we want to show that these two equivalence classes are equal. So it's useful to remember that if two equivalence classes intersect, they are identical. So we know that AB is equivalent to A'B'. Definitions are the whole of mathematics, all else is commentary. Our definition of that equivalence tells us, and similarly, since CD is equivalent to C'D', we know, now everything in sight is a natural number, and we've proven, or should have proven, all the properties of the natural numbers. So we can add, there's a lot of different ways we could list the sum, so let's get an idea of our destination. We want to show that the two equivalence classes are equal. We can do that if we can show that the class representative is in the other equivalence class. Because again, if two equivalence classes intersect, then they are identical. By definition, you only get to be in the equivalence class if you are equivalent to the class representative. And they're equivalent if the outer sum A' plus C' plus B plus D is equal to the inner sum A plus C B' plus D'. And so we stare at this a moment and try and figure out how we can get from our starting point to where we want to be. And in this case, we see that if we just add both sides, then since equals is an equivalence relationship, we can switch the sides. That completes our bridge, but let's make sure we can drive forward. If I have these two, I can add them and end up with the sum. We can switch the order. Disentangling the sum gives us the equivalence relation. So we're in the equivalence class. And if the two equivalence classes intersect, then they have to be identical. And so addition is well-defined. It doesn't matter which class representative we use.