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 |α − 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 = [α + 1] satisfies |α − p/q| ≤ 1/2. Suppose that n ≥ 2. Consider the n + 2 numbers 0,α − [α], 2α − [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 − [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 − [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 − [iα] and jα − [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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