 Euler began the modern study of continued fractions with his 1744 essay on continued fractions. To continue discussion, it helps to introduce some terminology. In modern terms, the continued fraction has partial denominators a—yes, I know it's not really a denominator—b, c, d, and so on. It should be clear that a rational number has a finite continued fraction expansion. It should also be clear that, as long as we make the obvious restriction that our partial denominator should be positive integers, that an irrational number has an infinite continued fraction expansion. But there are irrational numbers, and there are really irrational numbers. Let's take a look at that. So Euler used the approximation for square root of 2 and found the sequence of partial denominators. So remember we can find these by applying the Euclidean algorithm. So we'll divide to get a quotient and remainder, divide to get a quotient and remainder, and lather, rinse, repeat. And a rather remarkable thing happens, our quotients all seem to be two, corresponding to the relationship. Now we shouldn't generalize from one example, and so Euler considered square root of 3, and he didn't say what approximation he used, but if we use this approximation and apply the Euclidean algorithm we get, and so in this case our terms seem to alternate, the remarkable regularity of the terms led Euler to note it does not follow from the division that the quotients will continue in this fashion, but it seems likely, and in fact it can be proven. Which if you read that carefully is a very interesting thing to say, because Euler is essentially saying that we begin with our suspicion of what is true and then go on to approve. So again it helps to introduce some modern terms, if the sequence of partial denominators repeat after some point we say the continued fraction is periodic, and Euler considered the problem of finding the value of a periodic continued fraction. So let's consider the simplest case where our partial denominators are all the same, except for that first denominator which again isn't really a denominator but we call it one. So we find x-a equals, and at this point we'll make an important observation. The repitent, this portion of the continued fraction that repeats over and over again, is the same as the whole thing, it's x-a, and so we can replace this repitent with x-a and get an equation that doesn't extend off to infinity. And once we do that we can solve for x and find, now by assumption a and b are positive so our continued fraction will also have a positive value, so the positive solutions to this quadratic equation will be, and so x can be expressed in terms of a and b. So note that if a is equal to one and b is equal to two, we get x is equal to the square root of two and the continued fraction expansion, which is what we found earlier. For convenience we can let b equal to a, and what that will do is that we'll zero out this portion that isn't under the radical. And then our formula for x gives us, and so in general, and this allows us to find things like a continued fraction expansion for square root of five. So our formula gives us, and if we let a equal to two, then we get square root of five equal to, and a convergence will be, which give us excessively better approximations to the square root of five.