Rational Approximation by Continued Fractions The convergents of a continued fraction expansion of x give the best rational approximations to x. Specifically, the only way a fraction can approximate x better than a convergent is if the fraction has a bigger denominator than the convergent. The first lemma says that the denominators of convergents of continued fractions increase. Lemma. Let a 0, a, a 2,...be a seuence of integers, where a > 0 for. Define p 0 = a 0, 0 = Then + > for > 0. p = a a 0 +, = a p = a p +p 2, = a + 2, 2. Proof. Let > 0. Note that is a positive integer. So + = a + + > a + =, where a + because the a s are positive integers from a on. The convergents of a continued fraction oscillate around the limiting value, and the convergents are always fractions in lowest terms. In fact, the convergents are the best rational approximations to the value of the continued fraction. I ll state the precise result without proof. Theorem. Let x be irrational, and let c = p be the -th convergent in the continued fraction expansion of x. Suppose p, Z, > 0, and x p < x p. Then +. Here s what the result means. Draw the line through the origin in the t-y plane with slope x. Plot the points (p,) and (p, ). The hypothesis x p < x p says that the vertical distance from (,p) to y = xt is less than the vertical distance from (,p ) to y = xt. y (,p ) y = xt x - p x - p (,p) t
The conclusion says that +. In fact, since + >, > : The denominator of p is bigger than that of p. In other words, the only way the point (p,) can be closer to the line is if its y-coordinate is bigger. I can restate the theorem in the form of a corollary in which you can see the fractions in uestion approximating x. Corollary. Let x be irrational, and let c = p be the -th convergent in the continued fraction expansion of x. Suppose p, Z, > 0, and x p < x p. Then >. Proof. Given the hypotheses of the corollary, suppose on the contrary that. Since x p < x p, I can multiply the two ineualities to get x p < x p. Apply the theorem to obtain +. But then +, which contradicts the fact that the s increase. Therefore, >. This result says that the only way a rational number p can approximate a continued fraction better than a convergent p is if the fraction has a bigger denominator than the convergent. Example. Here are the convergents for the continued fraction expansion for π: a p c 3 3 3 7 22 7 22 7 5 333 06 333 06 3 3 292 03993 3302 03993 3302 3 3.4592920, which is in error in the seventh place. The theorem says that a fraction p closer to π than only if > 3. 3 can be The next result is sort of a converse to the previous two results. It says that if a rational number approximates an irrational number x sufficiently well, then the rational number must be a convergent in the continued fraction expansion for x. 2
Theorem. Let x be irrational, and let p be a rational number in lowest terms with > 0. Suppose that x p < 2 2. Then p is a convergent in the continued fraction expansion for x. Proof. Since for 0, the s form a strictly increasing seuence of positive integers. Therefore, for some, < +. Since < +, the contrapositive of the preceding theorem gives x p x p = x p < 2 2 = 2. Hence, x p <. 2 Now assume toward a contradiction that p is not a convergent in the continued fraction expansion for x. In particular, p p, so p p, and hence p p is a positive integer. Since p p, p p = p p = p x+x p p x + x p < + 2 2 2. (The second ineuality comes from the Triangle Ineuality: a+b a + b.) Subtracting from both sides, I get 2 < 2 2 2, so <. But I assumed, so this is a contradiction. Therefore, p is a convergent in the continued fraction expansion for x. Example. Show that is the best rational approximation to π by a fraction having a denominator less 3 than 000. Suppose that p < 000. Since p is a fraction in lowest terms that is a better approximation to π than, and that 3 is a fraction is a better approximation to π than 3, π p < π 3. Since < 000, 2 2 < 2000000, so 2 2 > 2000000 = 5 0 7. 3
But π 3 = 2.66764... 0 7. Thus, 2 2 > 5 0 7 > The hypotheses of the theorem are satisfied, so p expansion of π. π 3 > π p. But the other convergents with denominators less than 000 3, 22 7, 333 06 than 000 are poorer approximations to π than 3. must be a convergent in the continued fraction with denominators less Hence, is the best rational approximation to π by a fraction having a denominator less than 000. 3 Example. (a) Compute the first 6 convergents c 0,...c 5 of the continued fraction for /3. (b) Show that 278 25 is the best rational approximation to /3 having denominator less than 55. (a) x a p c 2.22398009056935 2 2 2 4.4646825446245 4 9 4 9 4 2.5200682352479 2 20 9 20 9 6.57865204239306 6 29 58 29 58.72854274376962 49 67 49 67.373335342782462 278 25 278 25 2.67855704363653 2 705 37 705 37 (b) Suppose that p is a fraction in lowest terms which is a better approximation to /3 than 278, and also 25 that < 55. Since p is a better approximation to /3 than 278 25, /3 p < /3 278 25 =.99094... 0 5. Since < 55, < 55 2 < 24025 2 2 < 48050 2 2 > 48050 = 2.086... 0 5 4
So I have 2 2 > 2.086... 0 5 >.99094... 0 5 > /3 p. (The ineualitiesareapproximate, but thereisenoughroombetween2.086... 0 5 and.99094... 0 5 that there is no problem.) By the approximation theorem, p is a convergent for /3. But no convergent with < 55 is a better approximation than 278 25. This contradiction shows that 278 25 is the best rational approximation to /3 having denominator less than 55. c 207 by Bruce Ienaga 5