1.3. RATIONAL APPROXIMATION 7 Proof. Suppose there are infinitely many p/q satisfying the inequality. Then by the exercise above, q gets arbitrarily large. Thus there exists a p/q with q b such that |a/b p/q| 1/q2. This implies that |aq bp| b/q 1, which is a contradiction. We next consider rational approximation of irrational numbers. The question is, “If α is irrational, are there any rational numbers p/q satisfying the inequality p/q| 1/q2?” The affirmative answer follows from a theorem of Dirichlet on rational approximation of any real number. Theorem 1.3.7 (Dirichlet). Let α be a real number and n a positive integer. Then there is a rational number p/q with 0 q n satisfying the inequality α p q 1 (n + 1)q . Proof. If n = 1, then p/q = [α] or p/q = + 1] satisfies p/q| 1/2. Suppose that n 2. Consider the n + 2 numbers 0,α [α], [2α],...,nα [nα], 1 in the interval [0, 1]. Assume that the numbers in our list are distinct, which is the case if α is irrational. By the pigeonhole principle, two of the numbers differ in absolute value by at most 1/(n + 1). If one of the numbers is 0 and the other is [iα], then i n, |iα [iα]| 1/(n + 1), and α [iα] i 1 (n + 1)i . After [iα]/i is reduced to lowest terms p/q, the rational number p/q satisfies the required inequality. Similarly, if the two numbers are [jα] and 1, then j n, and reducing ([jα] + 1)/j to lowest terms p/q, we have p/q satisfies the required inequality. Finally, if the two numbers are [iα] and [jα], where i j, then |jα [jα] (iα [iα])| = |(j i)α + ([jα] [iα])| 1 n + 1 . Then, j i n, and α [jα] [iα] j i 1 (n + 1)(j i) . Thus, after ([jα] [iα])/(j i) is reduced to lowest terms p/q, the rational number p/q satisfies the inequality. In the event that the n +2 numbers are not distinct, then α itself is a rational number with denominator at most n. For this case, either there exists 1 i n so that α = [iα] i
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