 A really useful idea that it has, for whatever reason, fallen by the mathematical wayside is that of the continued fraction. And this emerges as follows. Suppose I have some positive real number, alpha 1. I can form a sequence of positive integers beginning with alpha 1 as follows. What I'm going to do is I'm going to make alpha 1 equal to a1 plus 1 over alpha 2, where a1 is the greatest integer that is less than alpha 1. One rather neat mathematical concept that has, for whatever reason, fallen by the wayside is that of the continued fraction. And this emerges as follows. Suppose I have some positive real number, alpha 1. What I'm going to do is I'm going to form a sequence of positive integers from alpha 1 in the following way. So first off, I'm going to set alpha 1. I'm going to find it equal to a1 plus 1 over alpha 2, where a1 is the greatest integer less than alpha 1. If you think about alpha 1 as being some decimal number, what I'm going to do is I'm going to split off the integer part, and then whatever's left over, I'm going to express as a fraction of the form 1 over something. Well, anything you've done once, you could do any number of times. So what I can do is I can take this alpha 2, which is in general going to be some other positive real number, and I can write alpha 2 as integer a2 plus 1 over some other real number, alpha 3. And again, a2 is the greatest integer less than alpha 2. And I can continue to do this as far as I want to. And the sequence a1, a2, a3, and so on, again, these are all integers because they are the greatest integer less than some real number. This sequence of integers forms the continued fraction expansion of our real number alpha 1. Now, describing the continued fraction expansion as a sequence of integers is good mathematics, but it's kind of hard to visualize what that actually represents. So here's a slightly more conventional way of looking at it. So again, we have our alpha 1. It's equal to a1 plus 1 over alpha 2. So there's my a1 plus 1 over. Now, alpha 2, I know what that is. Alpha 2 is a2 plus 1 over alpha 3. So I'll fill that in. And again, this is a2 plus 1 over alpha 3. Well, I know what alpha 3 is. It's a3 plus 1 over alpha 4. So I can fill in my a3 plus 1 over, and then alpha 4 is going to be some expression and so on. So when I get my continued fraction expansion in formal mathematics, it's a sequence of integers. But it's convenient to think about it as a fraction of this form, hence the term continued fraction. Well, let's try that out. So let's find the first four terms of the continued fraction expansion of pi equals 3.141, et cetera. And so let's take a look at that. So at each step, what we're going to do is we're going to take our working number and we're going to break it into the greatest integer part and then some leftover part that will express as a reciprocal. So pi is 3 plus 0.14159, et cetera. And I want to express this in the form 1 over. Well, finding that is easy. That's just going to be the reciprocal of 0.14159, which is going to be 1 over 7. something or other. And so my next step, I'm going to take this real number, 7.062, et cetera, and I'm going to express that as an integer plus a leftover part, which I'll express as a reciprocal. And again, I'm going to take this real number and I'm going to express it as an integer plus a leftover portion, which I'll express as a reciprocal. And I'll continue to do that as far as I care to. Now one caveat here, these values are only going to be as good as our original decimal approximation of pi. If I add pi to an infinite number of decimal places, I can recover the indefinite sequence of the continued fraction expansion. But because I have pi to a limited number of decimal places, I'm pretty confident about the first couple of values here. But I'd hesitate to go any farther, starting with this as my decimal approximation for pi. So I'll stop there. And to get the first four terms, my continued fraction expansion are going to be 3, 7, 15, 1, and then some other terms. And again, slightly more conventionally, what we can do is we can express this as a fraction 3 plus 1 over 7 plus 1 over 15 plus 1 over 1 plus and so on. So here's our somewhat expanded form of our continued fraction expansion. Now we'll introduce some other term. Consider our continued fraction, and again, we'll express this in a somewhat more conventional form so we can get a sense of what we're actually talking about here. I have my continued fraction expansion. And if I truncate the continued fraction by using the first k terms of the corresponding sequence, I obtain a convergent. What I mean by that is this. What I'm going to do is I'm going to select however many terms here, and then I'm going to ignore everything after that. So maybe I'll just take that first term or I'll take the first two terms, a1 plus 1 over a2 and ignore everything after that. Or maybe I'll take the first three terms, a1 plus 1 over a2 plus 1 over a3 and ignore everything else and so on. And these are going to be the convergence of the continued fraction expansion. Now because our a1, a2, a3, all of our a values are integers, all the convergence are rational numbers. And we can reduce them through some not too tedious arithmetic. And these convergence have a number of remarkable properties, two of which are of particular importance. First off, they're a sequence of rational numbers, but they always have a limit. The continued fraction expansion of a real number always tends to a limit. And the limit of that convergence is going to be whatever our original real number is. And if I look at the successive convergence, so here's one convergent, there's the next, there's the next, and so on, the successive convergence are alternately greater than or less than the original real number that we started with. So not only do these convergence provide us a way with getting rational approximations to a real number, but we've also managed to trap that real number between two successive convergence. And that's a really useful property that we'll go into a little bit later on.