1 Continued Fractions

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1 Continued Fractions To start off the course, we consider a generalization of the Euclidean Algorithm which has ancient historical roots and yet still has relevance and applications today.. Continued Fraction Representations of Rational Numbers We start with a simple example, messing with the fraction 6 4. First write the fraction as an integer plus a fraction less than : 6 4 = Now we do something strange; flip the second fraction upside down and repeat: 6 4 = = 4 + = = 4 + = (take out integer part of 4 5 ) (flip 4 5 and take out integer part ) We stop here because the last fraction 4 is the reciprocal of an integer. We call this representation of a continued fraction. To save writing the whole thing out, we simply use the notation = [4;,, 4] 4 Now we need the promised tie-in with the Euclidean Algorithm. Suppose we applied the Algorithm to find gcd(6, 4): 6 = = = = 4 The relationship should be obvious! At each stage we are applying the Division Algorithm: if a > b > 0 then there exist unique integers, the quotient q and remainder r which satisfy a = qb + r and 0 r < b The remainders in the Euclidean Algorithm are underlined. However, it is the sequence of quotients 4,,, 4 that appears in the continued fraction representation of 6 4. Here is another example: = = + = = = = [;,, 4]

2 Sure enough, if we apply the Euclidean Algorithm, 05 = = = + 3 = 4 3 we see the same sequence,,,4 of quotients! What s the Point? Why would any sensible person want to write an already simple expression in such a convoluted manner? One answer is that by truncating a continued fraction we can produce approximations to our original fraction. For example: [4] = 4 [4; ] = 4 + = 9 = 4.5 [4;, ] = = 4 3 = 3 3 = [4;,, 4] = = = 6 4 = Observe how these approximations alternate below and above the final answer. Similarly, for our other example, [] = [; ] = + = 3 [;, ] = + + = 3 = 8 3 = [;,, 4] = + + = = Again, the approximations alternate below and above the final answer. We shall later that, in a certain precise sense, is the best rational approximation to 4 whose denominator is less than 4. Finding rational approximations with small denominators was crucial in the days before calculators! We are in fact in a position to prove that the coefficients for a continued fraction are always the quotients from the Euclidean Algorithm. Have a go yourself before seeing the proof in a few pages. Before seeing more fractions and defining things more directly, we make a historical digression to consider one of the oldest known discussions of irrationality, for it is in the realm of irrational numbers that continued fractions reall show their worth.

3 . Incommensurability and the Ancient Greeks: non-examinable The ancient Greek mathematician Theaetetus ( BCE) developed a notion of irrationality which formed the basis of the longest and most difficult book (Book X) of Euclid s Elements (c.300 BCE). The purpose of Book X was the classification of incommensurables, or to us irrationals. It needs to be understood that the Greeks spoke only of whole numbers and of lengths which were in whole number ratios to each other. Thus 7:5 might represent two rods, the second of which was 5 7 the length of the first. The term rational was not used, instead Theaetetus considered whether two lengths were commensurable, meaning that the two were integer multiples of some common sub-length. In the above case one could produce a third small rod, 7 of which would measure the first rod and 5 the second. Strictly speaking, lengths l and l are commensurable if and only if there exists some sub-length m and integers a, b such that l = am and l = bm. Otherwise said, the lengths are in proportion l : l = a : b In modern times, we are happy equating the concepts of length and number and we d simply say that the ratio l l is a rational number. The question Theaetetus considered was twofold:. If two lengths are commensurable, how do we find the common sub-length m?. If two lengths are incommensurable, how do we prove it? That is, how do we show that no such common sub-length m exists? His answer to the first question is essentially the Euclidean Algorithm, for if l and l are integers, then the common sub-length is precisely the greatest common divisor of l and l. Here is an example with two initial lengths l and l : l l l = 3l + l 3 l 3 l = l 3 + l 4 l 4 l 3 = l 4 l = 7l 4 l = 5l 4 In this case, l 4 is the common sub-length. By choosing units so that l 4 = we see that we ve applied the Euclidean Algorithm to calculate gcd(7, 5) =. The point of this digression is that the Euclidean Algorithm works perfectly well when applied to lengths, not just integers. Indeed, to the Greeks, the Division Algorithm (the main ingredient in the Euclidean Algorithm) may be stated as follows; if l, l are rods of positive length, then exists a unique integer q 0 and a length r which satisfy two properties:. l may be measured by q copies of l plus a single copy of r, i.e. l = ql + r.. r is shorter than l, i.e. 0 r < l. We assume that there is a base length of somewhere with respect to which all other lengths may be compared. This is precisely what choosing units does for us. The Greeks had a philosophical difficulty with this idea: how do you know that all lengths may be measured with respect to a fixed length? In the absence of irrational numbers, they can t. 3

4 We can now give Theaetetus definition of commensurability. Definition.. Two lengths a, b are commensurable if and only the Euclidean Algorithm terminates. The common sub-length is then the last length in the sequence. The lengths a, b are incommensurable if and only if the Algorithm never terminates. In modern language, the ratio a b is rational if and only if the Euclidean Algorithm applied to (a, b) terminates. We can now state a simple Theorem in this language. Theorem.. Suppose that lengths a, b are in the golden ratio: that is a + b : a = a : b. Then a and b are incommensurable. Otherwise said, the golden ratio is irrational. Proof. We apply Theaetetus Algorithm. Observe that since a : b is the golden ratio, we have a b = a + b = + b a a This can be rearranged: a = b + b a = a b = b a > 0 and a b = b a It follows that the first line of the Algorithm is a = b + b a = b + (a b) b = b(b a) a whence the first quotient is. If we consider the new ratio b : a b, ( ) b a b ( a ) ( ) b a b = = b b = = a a b we see that it is also golden! We are right back at the beginning; since b : a b is also in the golden ratio Theaetetus Algorithm will continue ad infinitum, at every step producing a new pair of lengths in the golden ratio. We conclude that a and b are incommensurable. < 0 There are simpler proofs of the irrationality of the golden ratio, but our purpose is to see how it fits with the Euclidean Algorithm. You ve probably seen the famous adjoining picture: each step of the algorithm can be visualized as deleting a square from a golden rectangle, leaving a smaller golden rectangle. b The Greek method of working geometrically makes things very difficult for us algebra-hungry modernites. Indeed it is very hard for us to resist writing b a b ϕ = a 5 + b = ϕ = + ϕ = ϕ ϕ = 0 = ϕ = The Greeks, of course, had no notion of 5, so the above was essentially how they proceeded: fun! I.e. a = b + (a b) = q = and r = a b. 4

5 .3 Continued Fractions and Irrational Numbers Motivated by Theaetetus, we try a continued fraction approach for irrational numbers. If the golden ratio produces an infinite sequence of quotients equal to, is it reasonable to write the following? 5 + ϕ = = [;,,,...] = The answer is yes, though to prove everything will take some time: the critical difficulty is that the ellipses are making a claim about the convergence of some sequence. Defining the correct sequence and proving that it converges to ϕ is the challenge. Before all that, we define the concept of a continued fraction in a manner that s more user-friendly for dealing with irrational numbers. Definition.3. Suppose that x is a real number. Define a pair of sequences (c k ), (R k ) inductively c := x R := x c c n := R n := c n R n R n The notation x is the floor of x: the greatest integer less than or equal to x. If any term R n is zero, then the sequences terminate. The continued fraction representation of x is written [c ; c, c 3,...]. Examples. It is worth revisiting our first example in this language. 6 c = = 4 R = = c = = R = = c 3 = = R 3 = = 4 c 4 = 4 = 4 R 4 = 4 4 = 0 Clearly this produces the same sequence and continued fraction representation we saw earlier.. Now we try this for the golden ratio. If ϕ = 5+, then < ϕ <, whence c = and R = 5. Then c = = = = R = 5 Both sequences have already begun to repeat: we therefore obtain the continued fraction as desired. [;,,,,,,...] 5

6 3. Find the continued fraction expansion of 3. Since 3 < 3 < 4 we have c = 3 and R = 3 3. Thus R = Rinse and repeat: 3 3 = = = c = n c n 3 6 R n Since the sequences (R n ), (c n ) have started to repeat, it follows that they will continue to do so. Thus the continued fraction expansion of 3 is the repeating sequence [3;,,,, 6,,,,, 6,,,,, 6,...] 4. The first few terms of the continued fraction expansion for π can be found similarly, though you ll need a calculator! Unlike the two previous examples, there will be no on-going pattern to find. 3 c = π = 3 = R = π 3 c = = 7 = R = π 3 π 3 π 3 c 3 = = 5 = R 3 = π 3 7π 7π 7π c 4 = = = R 06π 4 = 333 We obtain the continued fraction [3; 7, 5,,...]. These examples suggest the following: 7 = 7π π 3 5 = 06π 333 7π Theorem.4.. A real number has a finite continued fraction representation if and only if it is rational.. If x = a b is rational, then it equals its continued fraction representation, which moreover consists of the sequence of quotients from the Euclidean Algorithm applied to the pair (a, b). 3. If x is any real number, its continued fraction representation is the sequence of quotients found by applying the Euclidean Algorithm (à la Theaetetus) to the pair (x, ). Before proving this, it is worth noting a couple of caveats. If c = x, then x is an integer and its (very boring) continued fraction is itself. The first quotient c in a continued fraction is negative if and only if x is negative: all the other quotients must be positive. 3 The beady-eyed should notice the and 7 that appear prominently in the calculation... 6

7 Proof. Suppose first that x = a b Q+. Write the Euclidean Algorithm for finding gcd(a, b) = r n : a = c b + r 0 r < b b = c r + r 0 r < r r = c 3 r + r 3 0 r 3 < r. r n 3 = c n r n + r n 0 r n < r n r n = c n r n In each case we have c k = r k r k = r k r k rk r k since 0 r k < r k and c k is an integer. Now compute: 4 a b = c + r b = c + = c b + c r + r = c + r c + r r = c + = c + c + + r c n + r n c c n = c + c + c 3 + r 3 r = [c ; c,..., c n ] Thus any positive rational number a b has a finite continued fraction expansion whose coefficients are precisely the quotients in the Euclidean Algorithm. Conversely, if a continued fraction terminates, then it is certainly rational. We have therefore proved parts and. For part 3, x x is less than, and so x = x + (x x ) = c + R is precisely the first line of Theaetetus Algorithm applied to the pair (x, ). The next line is then ( ) = R + R = c R + R R R R ( ) with positive remainder R R = R R < R R. And the third: ) R = R R + R ( R = c 3 R R + R R R 3 R R ( ) with positive remainder R R R 3 = R R R < R R R. Without being too formal (prove by induction if you like), we see that the kth line has the form R R k = c k R R k + R R k ( = R R k + R R k R k R k R k This proves 3: indeed the kth remainder in the Algorithm is the product R R k. 4 More formally, let a = r, b = r 0 and observe that r k r k = c k + r k /r and argue by induction. k 7 )

8 .4 Convergents of Continued Fractions We ve already seen that a continued fraction representation of a rational number is equal to that number. To see that the same is true for irrational numbers, we need a notion of convergence. Definition.5. Given the sequence of quotients (c n ), define new sequences (p n ), ( ) via p 0 = p = c p n = c n p n + p n q 0 = 0 q = = c n + The ratio p n is termed the n th convergent of the continued fraction [c, c, c 3,...]. Example For α = 6 4, the sequences are n c n 4 4 p n If you re struggling to remember the formulæ, look at the pattern in the table: the three red numbers combine to produce the blue: 9 = 4 + Observe that the sequence of convergents ( ) ( pn = 4, 9, 3 3, 6 ) 4 is the sequence of approximations we obtained on page by truncating the continued fraction. The same thing can be seen with any of our other examples: indeed we have the following result. Theorem.6. p n = [c ;..., c n ] The Theorem can be broken into a two-part lemma. Lemma.7. Let c,..., c n be integers such that the continued fraction [c ;..., c n ] exists. 5 We can describe the continued fraction in terms of a matrix multiplication. ( ) ( ) ( ) ( ) ( ) c c c3 cn A B. = = [c C D ;..., c n ] = A C. ( ) ( ) A B pn p. In terms of the sequences (p n ) and ( ), we have = n. C D 5 Strictly c,..., c n are positive integers, and c n. Part holds for any real numbers provided we never divide by zero: something like [;, ] = + clearly cannot be allowed! + 8

9 Proof. We prove part by induction; part is left as an exercise. ( ) ( ) For the base case, observe that [c ] = c = c = A C when c A B =. 0 C D Now fix n and assume that any continued fraction of length n satisfies the condition Then [c ;..., c n ] = A C [c ; c,..., c n ] = c + where ( ) ( ) c c3 0 0 [c ;..., c n ] = c + C A = c A + C A However ( ) ( ) ( ) c A B c A + C c = B + D 0 C D A B By induction, part is proved. ( ) cn = 0 ( ) A B C D We now want to see that the convergents of an irrational number x really converge to x. This is a somewhat tedious piece of elementary analysis: it starts by taking determinants of the statement of Lemma.7: since each matrix on the left has determinant, we have ( ) n = p n p n In particular, gcd(p n, ) = = the convergents are fractions in lowest terms. Moreover, p n p n = ( )n We therefore see that [c ; c,..., c n ] = p n = ( )n + p n = ( )n + ( )n + p n = = ( )n + ( )n + + q q + p = c + n k= ( ) k q k q k Since q k is an increasing sequence (q k+ = c k+ q k + q k q k with equality only if k = and c = ) it follows that the sequence ( ) decreases to zero. The sequence of convergents is therefore the n th partial sum of a convergent alternating series. By the standard theory of such series 6 we see that the limit L satisfies p < p 3 < p 5 < < L < < p 6 < p 4 < p and q q 3 q 5 q 6 q 4 q L p n < + 6 I.e. the proof of the alternating series test. q ( pn ) 9

10 The sequence of convergents is indeed convergent! It remains to see that its limit is x. For this, first observe that for all n N x = c + c + + c n + R n In particular, for the first three convergents, x = c + R > c = [c ] = p x = c + < c c + + = [c ; c ] = p R c q x = c + > c + c + c + = [c ; c, c 3 ] = c 3 + R 3 c 3 q The pattern is obvious; the convergents are alternately smaller (if n is odd) or larger (if n is even) than x. We therefore see that p < p 3 < p 5 < p 7 < < x < < p 8 < p 6 < p 4 < p q q 3 q 5 q 7 q 8 q 6 q 4 q and we ve (finally!) proved the first two parts of the following. Theorem.8. Suppose that x is irrational and has continued fraction representation [c ; c, c 3,...].. x is the limit of the sequence of convergents p n = [c ; c,..., c n ]: p n x = lim n = c +. The nth convergent satisfies x p n < + n= ( ) n 3. Each convergent is closer to x than the previous: x p n+ < x p n + Direct proofs of the latter parts of the Theorem are exercises. Indeed with the slight modification (in the second part) of converting < to, they are true even when x is rational. The latter parts were very useful in the days before calculators, where they could be used to guarantee a particular level of accuracy in an approximation. Given the Theorem, it is now legitimate for us to write x = [c ; c, c 3,...] for any real number x, where it is understood that this means the limit of the sequence of convergents when x is irrational. More theoretically, if you recall the discussion in analysis where the real numbers may be defined as the set of limits of sequences of rational numbers, this procedure produces a concrete example of a sequence which converges to a given irrational number. 0 p 3 q 3

11 Examples With x = 3 = to 5dp, we have n c n 3 6 p n p n With π we obtain [3, 7, 5,, 9,,...] whence, to 8dp we have n c n p n p n According to part of the above Theorem, π 7 < 7 06 = 74 = In fact, π 7 = , so the accuracy estimate is very good. Similarly π < 06 3 = 978 = It is worth noting that the somewhat larger values for the quotients in the sequence (c n ) make the denominators of the convergents for π increase faster and produce better approximations than those for 3. In this sense, the golden ratio ϕ = [,,,,,...] has a very slowly converging sequence of convergents..5 Periodicity of Continued Fractions You might have spotted the following pattern: irrational numbers containing a square root eventually have periodic continued fraction expressions. Example Suppose that x = [;, 4,, 4,, 4,...]. The eventual periodic terms can be written as y = [; 4,, 4,...] = [; 4, y] = y Some simple algebra turns this into a quadratic equation for y: 4y 4y = 0 = y = + from which x = + y = + + = Indeed there is a general result here: (y is clearly positive)

12 Theorem.9. A continued fraction is ultimately periodic if and only if it is the continued fraction of an irrational number of the form x = α + β γ where α, β, γ are rational. Proof. We only prove the = direction, attributable to Euler. The converse, courtesy of Lagrange, is not much more difficuly, but takes a lot longer to write. If x has a continued fraction which is eventually periodic with period n, then x = [d ; d, d 3,..., d r, c, c,..., c n, c, c,..., c n,...] Abusing notation slightly, we may then write α as a finite continued fraction x = [d ; d, d 3,..., d r, y], where y = [c ; c,..., c n, y] By a straightforward generalization of Lemma.7 (see homework), we see that y = [c ; c,..., c n, y] = p ny + p n y + = y + ( p n )y p n = 0 where p k q k are the convergents for y. Since y is irrational and satisfies a quadratic equation with rational coefficients, it follows that it has the form 7 y = δ + ɛ γ Q( γ). Since x may be computed from y using only operations keeping us in the field Q( γ), it follows that x also lies in Q( γ)..6 Best/Diophantine Approximations of Irrational Numbers The question of finding good rational approximations to irrational numbers is very old. The approximation π 7 has been known for thousands of years: we ve now seen that it arises naturally from the consideration of continued fractions. In this section, we make two definitions of what it means to be a best rational approximation of an irrational number x, and see that the convergents of the continued fraction representation of x in fact yield these approximations. Definition.0. Let p, q N, and let x be a positive irrational number.. We say that p q is a best approximation to x of the first kind if, a, b N such that a b = p q and b q we have x p q < x a b Of all fractions with denominators less than or equal to q, the fraction p q is closest to x.. We say that p q is a best approximation to x of the second kind if, a, b N such that a b = p q and b q we have qx p < bx a The distance of qx to the nearest integer is smaller than that of all other bx when b < q. Note that the definitions only have strict inequalities because x is irrational. If x has two distinct rational approximations of the same accuracy (of either type), then it easy to see that x Q. In a given situation, it is therefore often enough to prove that x p q x a b, etc. 7 Q( γ) = {δ + ɛ γ : δ, ɛ Q} is an extension field of the rational numbers. It is closed under addition and multiplication, and all non-zero elements have inverses.

13 Example Consider π = We can produce a table indexed by denominators q of the rational number p q closest to π and whether these are best approximations: q closest p q π p q first kind? qπ second kind? Y Y = 3 n/a N = 3 n/a N = Y N = Y N = Y N = Y Y = N N = N N = N N Note: p q is a best approximation of the first kind π p is smaller than all entries above it. p q is a best approximation of the second kind the non-integer part of qπ is smaller than all entries above it. It takes a while to find the next best approximation of the second kind: you shouldn t be surprised to discover that it is the next convergent of π: 333 = π π 333 = Lemma.. Best approximations of the second kind are also of the first kind. Proof. Suppose that p q is a best approximation of the second kind to x. Then qx p < bx a for all b q where a b = p q. But then x p q = q qx p < q bx a = b x a x a q b b q since b q. It follows that p q is a best approximation of the first kind. We now see the link-up with continued fractions. Theorem.. The convergents of x are precisely the best approximations of the second kind, with one caveat: if q = (equivalently c = ) then p q = p = c is not a best approximation of the second kind, for the second convergent p q = p = c = c + will also be an integer, and a better approximating integer than c. Proof. Suppose that a b = p n system of linear equations. with b. Following Lagrange, we consider solutions to a certain 3

14 Recall that + = c n+ + for all n. Since the c n+ are positive integers, it follows that + = + = +. The only way we can get equality is if n = and c =, exactly the caveat in the theorem. We may therefore assume that + > b. Consider the matrix equation ( pn p n+ + ) ( ) y = z ( ) a b Since the determinant of the matrix is ±, there is a unique integer solution ( y z ). We check cases:. If both y, z are zero we have a contradiction, for b = 0.. If y = 0 and z = 0, then b = + z = b +, a contradiction. 3. If z = 0 and y = 0, then yp n = a and y = b forces a b = p n, another contradiction. 4. Finally, if both y, z are non-zero, then 0 < b = y + z+ < + forces y, z to have opposite signs. Since the convergents alternate either side of x, we see that x p n and + x p n+ also have opposite signs. Therefore y( x p n ) and z(+ x p n+ ) have the same sign and thus bx a = (y + z+ )x (yp n + zp n+ ) = y( x p n ) + z(+ x p n+ ) > y( x p n ) x p n In the non-contradictory case, the conclusion is that b a best approximation of the second kind. = bx a > x p n, whence p n If you read the above argument carefully and perform a small modification, we actually get the converse for free! Assume that a b is not a convergent of x: we may assume that b < + for some n. All four parts of the above analysis still hold, as is the conclusion that bx a > x p n. Since b it follows that a b cannot be a best approximation of the second kind. It can also be proved that the best approximations of the first kind are the convergents and some of their intermediate fractions: rational numbers of the form p n+ m + p n + m + where m is an integer. In fact the integers m which work are (roughly) those between c n+ and c n+. We won t pursue this. The primary application however is done. In the days before calculators, if one wanted a rational approximation of a number of desired accuracy, one needed only compute whichever convergent was necessary to achieve this. The same could even be used for rational numbers. For example, suppose we want to find the best approximation to 37 out of all fractions with a denominator at is 4

15 most 5. We would start by computing the continued fraction and the convergents: 37 = [;,,, 4] n c n 4 p n Clearly 3 5 is the answer: indeed 37 being much easier to work with. 3 5 = 5 = 60, so the estimate is very accurate, as well as Hurwitz s Theorem In the next section we shall apply the discussion of convergents to the famous Pell s Equation. To help link this to continued fractions, we shall need the following results. Theorem.3. Suppose that x is a given irrational number.. There are infinitely many rational numbers a b satisfying x a < b b. If a b is a rational number (b N) in lowest terms which satisfies x a < b b then a b is a best approximation of the second kind to x (and thus a convergent of x). Proof. Part is a corollary of Theorem.8: every convergent of x satisfies this inequality. For part, suppose that a b satisfies ( ) but is not a best approximation of the second kind. Then there has to be some n such that b < + Since the convergents are the best approximations of the second kind, it follows that p n (in the second sense) than a b : thus x p n bx a By ( ), we have, x p n bx a < b = x p n < b But then a bp n = a b b b p n x a + b x p n < b + b But this is iff > b, a contradiction. ( ) is closer to x ( a bp n is a positive integer since a b = p n ) ( -inequality) 5

16 Note the distinction between the two parts of the theorem: we d like to have an if and only if, but the result isn t there. a b a convergent of x = x a b < b x a b < = a b b is a convergent of x. A simplified version of Hurwitz s Theorem states that, for a given irrational number x, there are infinitely many rational numbers a b satisfying ( ): by the above, all such must be convergents of x. Indeed, one can prove that at least one out of any two successive convergents satisfies ( ). The result can be improved as far as the following, though we won t prove it. Theorem.4 (Hurwitz). If x is irrational, then there are infinitely many rational numbers a b x a < b 5b which satisfy Indeed all such must be convergents of x, and at least one out of every three consecutive convergents satisfies the inequality. By thinking about the convergents of the golden ratio ϕ = 5+, one can see that 5 is the largest number which can be placed in the denominator without ruling out some examples. Example Here is a table of the first ten convergents of 3 with a check on the accuracy of 3 p n each time: note that the first convergent p q = 3 is not a best approximation, so strictly we should ignore it. n p n p n 3 <? q n In fact all of the convergents checked also satisfy p n 3 <. 5q n 6

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