By Evangelos Kranakis (auth.)

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**Example text**

For the given primes p, q, let P = {1,2, ... ,(p-l)/2} and Q = {1,2, ... ,(q-l)/2}. It is clear that for any cEQ there exists be Q such that either pc == b modq or pc == -b modq (a similar property holds for Pl. Define n(p, q) = the number of times that pc is congruent modulo q to an integer in -Q, as c runs through Q. 2 (Gauss's Lemma) (plq) = (-l)n(p,q). Proof of the Lemma: For each cEQ, one can find Be, be such that (10) where be E Q and Be = +1 or -1. The mapping c ---+ be(c E Q, be E Q) is 1 - 1 (and hence also onto).

Vii) AnlBn - A n-2IBn - 2 = an(-I)n-l/(BnBn_2), n> 2. (viii) A2n-t/ B2n-l < A2n+11 B 2n+1 < Q' < A2nl B2n < A2n-21 B 2n- 2· (ix) limn_co Ani Bn = Q'. Proof: The proof of the theorem, although long and detailed, is straightforward by induction on n and is left as an exercise to the reader. Notice that (ix) follows from (v) and the fact that the sequence Bn has exponential growth. In fact, an easy induction on n, using the definitions of An, Bn, will show that An, Bn ~ fn ~ Rn-2, where R is the golden mean.

Aa < . B3 BI At first it will be shown that AlB is either a convergent or else lies between two convergents of a. Indeed, assume on the contrary Al A BI < B" Recall that adl = approximation that Ad B 1 " It follows from the definition of Diophantine lTa - a I< IAB - a I= IA-BaBI ~ IA - aBI < laT l l which is a contradiction. contrary that Hence, Ad BI ~ - I a , AlB" Next, assume on the A2 A B2 > B" Recall that B2 = a2" It follows that > _1 " IBA _ al > IAB _ A21 B2 - B2B Thus, IA - 1 1 1 aBI > - = - ~ - = lal - al = IAI - aB11, B2 a2 a2 which contradicts the definition of Diophantine approximation.