8. Dirichlet s Theorem and Farey Fractions

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1 8 Dirichlet s Theorem and Farey Fractions We are concerned here with the approximation of real numbers by rational numbers, generalizations of this concept and various applications to problems in number theory A property of the integers which we freuently use is that if h <, then h = 0, alternatively that if h 0, then h The rationals are dense in R but two rationals with small denominators cannot be too close together Thus when a/ and b/r are two different rational numbers we have a b ar b = r r and since they are uneual the numerator is non-zero Thus a b r r () There is a very simple, and useful, theorem due to Dirichlet which tells us how well a real number can be approximated by a rational number a/ in terms of the denominator Theorem 8 (Dirichlet) For any real number α and any integer Q there exist integers a and with Q such that α a (Q + ) As an immediate conseuence of casting out all common factors of a and in a/ we have Corollary 82 The conclusion holds with the additional condition (a, ) = [ ) n Proof of Theorem 8 Let I n denote the interval Q+, n Q+ and consider the Q numbers {α}, {2α},, {Qα} (Here we use { } = to denote the fractional part) If one of these numbers, say {α}, lies in I, then we are done We take a = α and then 0 α a < Q+ Similarly when one of the numbers lies in I Q+, then Q+ α α <, whence Q+ α ( α + ) < 0 and we can take a = α + When neither of these situations occurs the Q numbers must lie in the Q intervals I 2,, I Q, so there must be at least one interval which contains at least two of the the numbers (the pigeon hole principle, or box argument, or Schubfachprinzip) Thus there are, 2 with < 2 such that (α 2 α 2 ) (α α ) < Q+ We put = ( 2 ), a = ( α 2 α ) This turns out to be a very powerful theorem and in many applications it is all that one needs to know about the approximation of reals by rationals It is obviously best possible Take b =, r = Q + in () above

2 2 8 DIRICHLET S THEOREM AND FAREY FRACTIONS Theorem 83 Suppose that α is irrational Then there exist infinitely many rational numbers a/ with (a, ) = such that α a/ < 2 In particular there are arbitrarily large for which this ineuality holds Proof Choose Q to be an integer > and choose a, in accordance with Corollary Then α a / < (Q +) 2 Now, given a /,, a n / n with (a m, m ) = and α a m / m < m 2 we obtain a n+, n+ as follows Since α is irrational we have α a m / m (m =,, n) Choose Q n+ > max { α a /,, α a n / n } and then choose a n+, n+ in accordance with Corrollary Obviously α a n+ / n+ n+ (Q n+ + ) < 2 n+ and α a n+ / n+ < min { α a /,, α a n / n } so we must have a n+ / n+ distinct from any of a /,, a n / n Moreover it is clear that for any m the a m is uniuely defined by the ineuality α a m / m m (Q m + ) Thus the m are distinct and so there are arbitrarily large m When α is rational, say α = a 0 / 0, then the ineuality α a/ has only a finite number of solutions in a, with (a, ) = since, by (), we have α a/ 0 when a/ a 0 / 0 Indeed the ineuality α a/ < 0 has the uniue solution a/ = a 0 / 0 Theorem 84 The real number α is irrational if and only if for every ε > 0 there are a Z, N such that 0 < α a < ε Proof If α R\Q, then choose Q = /ε Then by Theorem, there are a, such that α a Q+ < ε Moreover, α a If α Q, then there are b Z and r N such that (b, r) = and α = b/r Choose ε = 2r and suppose that there are a Z, N such that α a < ε Then α a/ < 2r and α a/ = b/r a/ = b ar r Thus b ar < 2 Hence b = ar, whence α a = 0 Example e = 0 Then 0 < k=k+ K! k=k+ k! is irrational To prove this let = K!, a = K! K k! = e a and K! k! = K + k=k+ if K is large enough There is a useful generalisation of this (k K )! ( ) e k K + < ε K+ k=0 k!

3 568 NUMBER THEORY II 3 Theorem 85 (Liouville) Suppose that α is an algebraic number of degree n ( ) Then there is a positive constant c = c(α) such that α a > c n whenever a Z, N and a/ α (this latter condition can be omitted when n 2) Proof By algebraic of degree n we mean that α is a root of a non-constant polynomial with integer coefficients and the degree n corresponds to the minimal degree amongst all such polynomials It is not hard to see that we may suppose that there is a uniue polynomial P (λ) = a 0 λ n + a λ n + + a n such that (i) a j Z for 0 j n, (ii) a 0 > 0, (iii) (a 0, a,, a n ) =, (iv) P (α) = 0, (v) n minimal Firstly a polynomial satisfying (i) and (iv) must exist by definition of α Taking one of minimal degree ensures (v) By multiplying through by ± we can ensure (ii) and by taking out common factors we can ensure (iii) Moreover if there were two distinct such polynomials P and P, then by (ii) and (iii) the one cannot be a multiple of the other so we could obtain, by considering a 0P (λ) a 0 P (λ), one of lower degree satisfying (i) and (iv) and then repeat the above process to obtain (ii) and (iii) and so contradict (v) It suffices to show that there is a c(α) such that if α a/, then α a/ > c(α) n for then we can replace c(α) by min (, c(α) ) and so the conclusion follows also when α a/ Since the a j are integers we have n P ( ) a Z Moreover P (a/) 0, for otherwise we could factor out λ a/ and obtain a polynomial of lower degree Q(λ) = P (λ)/(λ a/) which satisfies Q(α) = 0 Although in the first instance this could be guaranteed only to have rational coefficients by multiplying through by a suitably integer we could recover a polynomial Q of degree n with integer coefficients and satisfying Q (α) = 0 Hence n P ( ) a On the other hand, by the mean value theorem of the differential calculus P (a/) = P (α) P (a/) = (α a/)p (β)

4 4 8 DIRICHLET S THEOREM AND FAREY FRACTIONS where β lies between α and a/ Since we are supposing that α a/ it follows that P (β) max{ P (λ) : λ α, α + } = c(α) Hence n P (a/) α a/ c(α) Example The number θ = k=0 is transcendental, ie is not algebraic To see this suppose on the contrary that it is algebraic and let n be its degree Let = K = 2 K!, a = a K = K k=0 2K! k! Then 0 < θ a/ = k=k+ 2 k! 2 (K+)! l=0 = 2, and so if K is sufficiently 2 l K+ large we have 0 < θ a K / K < c(θ) K n which contradicts Liouville s theorem In the uadratic case we know the α which give rise to the largest c(α) 2 k! Theorem 86 Let α = and suppose that c < 5 α a/ < c 2 has only finitely many solutions Then the ineuality Proof The irrational number α is a root of the polynomial P (x) = x 2 x Moreover the other root of P (x) is 5 2 Thus P (a/) is non-zero and 2 P (a/) Z Hence P (a/) 2 The polynomial P (x) has an expansion P (x) = 5(x α) + (x α) 2 about the point α If α a/ < c 2, then the right hand side has absolute value < 5c 2 + c2 4 when x = a/ Hence and this only holds for 5c < c 2 2 < c c 5 The box principle has many useful generalizations and applications, mostly in combinatorics, but there are several famous ones in number theory Of course it is an immediate conseuence of the principle of induction that if n sets partition a set of n + objects then one of them contains two objects More generally when n sets partition a set of kn + objects one of them must contain k + objects, and conseuently if a finite number of sets partition an infinite set, then one of them is also infinite We can use Dirichlet s theorem to treat Pell s euation, x 2 dy 2 = When d is a perfect suare the solubility of the euation is boringly trivial By factorising

5 568 NUMBER THEORY II 5 the left hand side and euating each factor to ± we see that the only solutions are x = ±, y = 0 in that case Therefore we henceforward suppose that d is not a perfect suare In particular d is irrational Let α = d in Dirichlet s theorem Since d is irrational, by the method of proof of Theorem 2 we can obtain an infinite seuence of triples of integers a,, Q ; a 2, 2, Q 2 ; a 3, 3, Q 3 ; with a n d < n (Q n + ), Q n+ > a n d n n Thus a 2 n dn 2 an = a n n d + n d a n n d + 2n d Q n + Q n + < 2 d ( ) Q n + + 2Q n d Thus we have found infinitely many solutions to the ineuality x 2 dy 2 < 2 d Hence, by the box principle, there exists an integer t with 0 < t < 2 d such that there are infinitely many pairs x, y with x 2 dy 2 = t (2) Again by the box principle, there are infinitely many pairs x and y so that not only (2) holds but x is in a fixed residue class modulo t and y is in a fixed residue class modulo t Let x 0, y 0 be a given such pair and let x, y be another with x and y large (obviously if one is, then so is the other) Then Choose u = xx 0 dyy 0 t x y d, v = yx 0 xy 0 t Then v y x o y 0 d t with y since d is irrational Moreover u 2 dv 2 = t 2 ( (xx 0 dyy 0 ) 2 d(yx 0 xy 0 ) 2) = t 2 (x 2 x 2 0 dy 2 x 2 0 dx 2 y d 2 y 2 y 2 0) = t 2 (x 2 dy 2 )(x 2 0 dy 2 0) = Thus we have produced infinitely many solutions to Pell s euation It is, at least theoretically, possible to calculate solutions, for a given d, by this method, but this is

6 6 8 DIRICHLET S THEOREM AND FAREY FRACTIONS very inefficient and there is a much faster way via the theory of continued fractions However, it is now possible to obtain the structure of the complete solution set to Pell s euation Let x n, y n (n = 0,, 2, ) be the solutions of (2) with t = and 0 < x 0 < x < x 2 < and y n > 0 Note that this implies that y 0 < y < y 2 < Given any two solutions x, y; x, y it is readily seen that xx + yy d, xy + x y is another Therefore x n x 0 + y n y 0 d x n+ and x n y 0 + x 0 y n y n+ Thus (x n + y n d ) (x 0 + y o d ) x n+ + y n+ d On the other hand x n+ x n y n+ y n d = ( + y 2 n+ d)( + y2 nd) y n+ y n d > 0, and y 2 n+( + dy 2 n) ( + dy 2 n+)y 2 n = y 2 n+ y 2 n > 0 so that y n+ x n x n+ y n > 0 Moreover x n+ x n y n+ y n d, y n+ x n x n+ y n is also a solution of (2) Thus x n+ x n y n+ y n d x 0 and y n+ x n x n+ y n y 0 Therefore x n+ + y n+ d ) ) = (x n+ + y n+ d (x n y n d x 0 + y 0 d x n + y n d Combining this with the previous displayed formula we see that for each n 0 we have x n+ +y n+ d = (x0 +y 0 d)(xn +y n d) Therefore the complete solution to Pell s euation (2) is given by x + y d = (x 0 + y 0 d) m, m =, 2, where x 0, y 0 is the least solution Of course this argument gives no easy way of finding the least solution The set F n = { a is the Farey series of order n Thus F 5 is } : 0 a n, (a, ) = 0, 5, 4, 3, 2 5, 2, 3 5, 2 3, 3 4, 4 5, An analysis of F n gives an alternative line of approach to uestions of diophantine approximation Theorem 87 (i) If a and b r are successive terms of F n, then b ar = (ii) If a, c s and b r are three successive terms of F n, then c s = a + b + r Proof (i) Since (a, ) = we can solve the congruence ay (mod ) with n < y n Let x = (ay + )/, so that x ay = Clearly (x, y) = Thus x/y F n Also x y = a + y, so that x y comes later in the series If it is not b/r, then x y b xr by = r yr yr

7 and Hence 568 NUMBER THEORY II 7 b r a = b ar r r y = x y a yr + r = y + yr > n yr y which is absurd (ii) By (i), c as = and sb cr = Solving for c and s gives and (ii) follows c(b ar) = b + a, s(b ar) = + r As defined, the elements of F n lie in [0, ], but we could just as well take F n = { a } : n, (a, ) = and uite clearly the above theorem holds for Fn in place of F n Suppose that a, a, a+ + are three successive terms of Fn Clearly a < a + a + < a < a + a < a + + Thus we can partition R into intervals I(a/) of the form Moreover so that (a + a ±, + ± ) = and Since the a+a± + ± I(a/) is of the form [ a + a +, a + a ) (a + a ± ) a( + ± ) = a ± a ± = ±, a + a ± + ± a = ± ( + ± ) are not in F n we must have + ± n + Hence each interval [ a θ (n + ), a ) + θ + (n + ) with 0 θ ± Since the I(a/) partition R we have another proof of Dirichlet s theorem