Applications of the course to Number Theory

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1 Alications of the course to Number Theory Rational Aroximations Theorem (Dirichlet) If ξ is real and irrational then there are infinitely many distinct rational numbers /q such that ξ q < q. () 2 Proof Let Q be given. For a real number α let [α] be the largest integer no greater than α, called the integer art of α and set {α} = α [α], the fractional art of α. Consider the fractional arts {0ξ}, {ξ}, {2ξ}, {3ξ},...,{Qξ} and the intervals [i/q, (i + )/Q], 0 i Q. There are Q + fractional arts distributed amongst the Q intervals. So there must exist some interval containing at least two of the fractional arts, i.e. {aξ} and {bξ}, 0 a < b Q, say, lying in [j/q, (j + )/Q]. Being in the same interval means that {bξ} {aξ} /Q. Write aξ = m + {aξ} and bξ = n + {bξ} for aroriate integers m and n. Then {bξ} {aξ} = (bξ n) (aξ m) = (b a)ξ (n m). Writing q = b a so 0 q Q and = n m we find that we can solve ξ q < qq for some 0 q Q. Since this is true for all Q this gives the infinity of solutions for (4) (noting that /qq /q 2 when q Q). This can be imroved Theorem 2 (Hurwitz) If ξ is real and irrational then there are infinitely many distinct rational numbers /q such that ξ q <. 5q 2 Proof Not given (The easiest way is to used continued fractions.) The constant 5 is best ossible, in that the result does not hold if it is relaced by a larger value. So if c > 5 then there exist irrational ξ for which there are only finitely many distinct /q satisfying ξ /q < /cq 2. In articular ξ = ( + 5)/2 would be such an excetion. Yet it can be shown that the number of excetions are relatively rare.

2 Theorem 3 If f(q)/q increases with q and q= f(q) is divergent then, for almost all ξ we can find an infinite sequence of distinct rationals /q, q > 0 satisfying ξ q < qf(q). Proof Not given This result shows that orders of aroximation such as < q 2 log q and < q 2 log q log log q are usually ossible, for almost all ξ. I leave it as an exercise to the student to rove that q= q log q and q log q log log q diverge. (The easiest method is to bound above by integrals.) Though we don t give the roof of Theorem 3 we do rove a (artial) converse below. Borel Cantelli Lemma Observation Let A i, i be an infinite collection of sets. An element x will lie in finitely many of these A i, if and only if q= N : n N, x / A n. So the element x will belong to infinitely many of these A i if and only if ( N : n N, x / A n ) N, n N : (x / A n ) N, n N : x A n x A k. N k N 2

3 Theorem 4 (Borel Cantelli Lemma) If A, A 2,... F and i= µ(a i) < then µ{x : xbelongs to infinitely many A i } = 0. Proof By the observation it suffices to rove that ( ) µ A k = 0. N k N Let ε > 0 be given. By the definition of convergence of the series in the assumtions we have that there exists M such that For this M we also have Hence µ(a i ) < ε. i=m A k A k. N k N k M ( ) µ A k N k N ( ) µ A k k M < ε. i=m µ(a i) since µ is monotone, since µ is sub-additive, True for all ε > 0 imlies the required result. Theorem 4 has many alications in Probability Theory but here we give one in Number Theory, concerning rational aroximations. Theorem 5 Let f : N R be given. Define D [0, ] by α D if, and only if, there exist infinitely many /q,, q Z, > 0 such that α q < qf(q). Then if q= f(q) < (2) 3

4 we have that the Lebesgue measure of D is zero. Proof Define A q = 0 q ( q qf(q), q + ). qf(q) Then α D if, and only if, α A q [0, ] for infinitely many q, so it suffices to show, subject to (2), that µ ( N k N (A k [0, ]) ) = 0. Yet Hence µ (A q [0, ]) µ (A q ) 2 0 q qf(q) 2(q + ) qf(q) 4 f(q). µ (A q [0, ]) q= q= 4 f(q) <. So the sets A q [0, ] satisfy the conditions of Theorem 4 and hence µ ( N k N (A k [0, ]) ) = 0, that is, µ(d) = 0. Note This theorem shows that Dirichlet s result cannot be extended to ξ q < q 2 log 2 q, for instance, for many ξ. (I ll leave it to the student to check that q= q 2 log 2 q converges but agin the easiest way is to bound the sum from above by an integral.) It is obvious that this result is a artial converse of Theorem 3, where we also needed that f(q)/q increases with q. For such f we see that there are two cases for the sum in (2), it either diverges as in Theorem 3, when a roerty holds for almost all numbers, or the sum converges as in Theorem 5, when the roerty holds for almost no number. We say that the roerty satisfies a zero-one law (There is never a case in the middle.) As remarked above this shows that Dirichlet s result on aroximations cannot be substantially imroved for all ξ. Yet there are numbers ξ that have excetionally good aroximations. 6 Liouville numbers 4

5 Theorem 6 For any algebraic number α of degree n > there exists M = M(α) > such that α q > Mq n for all integers, q, > 0. Proof If /q is chosen such that α q > then the result is trivial so assume that /q satisfies qα q. Assume α is a root of f(x) = a 0 + a x + a 2 x a n x n, where a i Z. Given any /q we must have f(/q) 0 for if not we would be able to write f(x) = (qx )g(x) for some olynomial g with integer coefficients but with deg g = n. Also, since α is algebraic of degree strictly greater than, we have that g(α) = 0 in which case α is algebraic of degree n. This would be a contradiction. So ( ) 0 f = a 0q n + a q n + a 2 2 q n a n n. q q n Thus a 0 q n + a q n + a 2 2 q n a n n is an integer since all, q, a i Z not equal to zero. Hence (and this is the trick ) we must have a 0 q n + a q n + a 2 2 q n a n n and ( ) f q q. (3) n For a real number x close to α we can use the Mean Value Theorem to get f(x) = f(x) f(α) = f (ζ) x α for some ζ : ζ α α x. Choose x = /q which by assumtion above satisfies /q α and so ζ satisfies ζ α /q α. Define Then, combining with (3) gives M = su(, f (ζ) : ζ α ). 5

6 q n ( ) f = f (ζ) q q α M q α, which is the required result. Examle We can follow the method of roof of the above theorem when α = ( + 5)/2. Then f(x) = x 2 x and f (x) = 2x. As we take better aroximations /q to α then ζ, which lies between α and /q must get closer to α, that is, f (ζ) must get closer to f (α) = 5. So we can take M no smaller than 5, confirming the otimal nature of the Theorem of Hurwitz above. Liouville s Theorem has been imroved such that given any algebraic number (whatever its degree) and any κ > 2 then there exists a constant c = c(α, κ) with α q > c q κ for all rationals /q. From Dirichlet s Theorem this is seen to be best ossible in that we cannot take κ 2. Strangely, there is no known formulae or method for calculating c(α, κ) in general. Only for some articular α and κ is it known. For instance, c( 3 2, 2.955) 0 6, that is, 3 2 q > 0 6 q for all rationals /q. Definition A real number α is a Liouville number if α is irrational and for all n there exists integers, q > 0 such that q α < q. n Examle is a Liouville number. α = k= 0 k! Verification Let α N be the sum of the first N terms so 6

7 α N = N! = 0N! + 0 N! N + 0 N! = 0n + = 0 N! 0, N! for some integer, of the form 0n + and so corime to 0 N!. Then 0 α = N! 0 + (N+)! (N+2)! 0 (N+3)! 2 < < 0 (N+)! (0 N! ). N So for every N we can find a very good rational aroximation to α, so α is a Liouville number. Theorem 7 Every Liouville number is transcendental. Proof Assume not, so there exists a Liouville number α that is algebraic for some degree n. Note that n > since α is irrational. Then Theorem 6 imlies that there exists M such that α q > Mq n for all integers, q > 0. Choose an integer k n such that 2 k > 2 n M. Then since α is Liuoville we can find integers, q > 0 such that q α < q < k Mq n by the choice of k. This is a contradiction so the assumtion is false and every Liouville number is transcendental. Let E be the set of all Liouville numbers. Theorem 8 The set E has zero Lebesgue measure in [0, ]. Proof By definition α E if, and only if, α Q c and for all k there exist integers > 0 and q such that 7

8 So q α < q. k say. Note that E = Q c = Q c + k= = q 2 G k, k= G k [0, ] q 2 q =0 Let µ be the Lebesgue measure on R. Then µ (G k [0, ]) q 2 = q 2 = q 2 To bound this sum observe that 4 q 2 < ( q q, k q + ) ( q q, k q + ). q =0 q =0 ( µ q q, k q + ) 2 2(q + ). q q dt t k since t k in the range of the integral. Adding gives q 2 < dt t = k k 2. Hence µ (G k [0, ]) 4/(k 2). But E [0, ] G k [0, ] for all k so µ (E [0, ]) 4/(k 2) for all k 2. Hence µ (E [0, ]) = 0. 8