Irrationals

Approximations to a given probability

There are many uses of continued fractions, one such common application is seeking a fraction with a relatively small denominator that well approximates a given probability. For example, a baseball player has a batting average of 0.334, rounded to 3 digits, others may be interested at the possible "at bats" the player could have given this batting average. We have 0.334 = 167/500, now it's quite common to feel quite certain that 500 must be the fewest "at bats" possible, but is this correct?

We have a rational number x that when rounded to 3 digits is 0.334, so we have 0.3335 < x < 0.33345. Let us look at the continued fraction expansion of these two numbers: 0.3335 = [0; 2, 1, 666], 0.3345 = [0; 2, 1, 94, 1, 1, 3]. We now want to choose the continued fraction a such that [0; 2, 1, 666] < a < [0; 2, 1, 94, 1, 1, 3], note that the expansion are the same until the 3rd entry, hence we want the 3rd entry of a, which we denote by a3 to satisfy 94 < a3 < 666. Since we are after a rational number with a small denominator, we shall choose a3 to be 94 + 1 = 95. Then we get a = [0; 2, 1, 95] = 96/287. Thus the fewest at-bats possible is 287.

Irrationals

So far, we have only worked with rationals. In my last post, I explained Euclid's Algorithm which we use to compute the continued fraction expansion of a rational number. Recall that a rational number is a number that can be expressed the quotient or fraction p/q of two integers, where q is non-zero. Now, what about an irrational number? Like π or e, which have never-ending and non-repeating decimals. Firstly, let me introduce the floor function, consider a real number x, then the floor function of x gives the largest integer less than or equal to x, which we denote by ⌊x⌋. For example ⌊2.133⌋ = 2. Now we have the following procedure to compute the continued fraction expansion of an irrational x:

let 

Then all the ak will be the entries in the continued fraction expansion of x. Confused with too many ak and xk? Don't worry, let's do an example together! Consider x = sqrt(15), then following the above procedure, we have

Since x3 = x1, the results will repeat, with x4 = x2 and so on, hence sqrt{15} = [3;1,6,1,6,...] ​

If you use the above procedure to compute the continued fraction expansion of different irrational a few times, you would discover that the expansion is infinite! Whereas the expansion for rational numbers is always finite. 

Apart from finiteness, there is one feature I found particularly interesting for the continued fraction expansion of irrationals. One way to get an irrational number easily is to consider a natural number which is non-square, let's call it d, then if you consider its square root: sqrt(d), which is an irrational. For this kind of numbers, if you compute their continued fraction expansion, you would see that its expansion is always periodic, like sqrt(15) in the example above. Furthermore, if the period consists of p terms a1,a2,··· ,ap , then ap = 2⌊d⌋ and the sequence a1, a2,···, ap is symmetric, in other words, 2a0 =ap, a1 =ap−1, ..., ak =ap−k.

Why are we interested in such number sqrt(d)? You shall see very soon in my next post on Pell's Equation!


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