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An Introduction to the Theory of Numbers

Exercise 6.2.2 (Rational numbers $h/k$ such that $|\xi - h/k| < 1/(\lambda k^\alpha),\ (\alpha >2)$)

Take ξ = ( 1 + 5 ) ∕ 2 . Let λ > 0 and α > 2 be real numbers. Prove that there are only finitely many rational h ∕ k satisfying

| ξ − h k | < 1 λ k α ) .

Answers

Proof. By the proof of Theorem 6.12 applied to 3 > 5 , there exists only finitely many different rational numbers h ∕ k such that

| ξ − h k | < 1 3 k 2 .

Morevover, for all k ∈ ℕ ∗ ,

1 λ k α ≤ 1 3 k 2 ⟺ k α − 2 ≥ 3 λ ⟺ log ⁡ k ≥ 1 α − 2 log ⁡ ( 3 ∕ λ ) ( because  α > 2 ) ⟺ k ≥ k 0 = ⌊ e [ 1 ∕ ( α − 2 ) ] log ⁡ ( 3 ∕ λ ) ⌋ + 1 .

Hence there are only finitely many rational numbers h ∕ k with k ≥ k 0 such that

| ξ − h k | < 1 λ k α .

Note that for each fixed positive integer k < k 0 ,

| ξ − h k | < 1 λ k α ⇒ | h k | − | ξ | ≤ | ξ − h k | < 1 λ k α ⇒ | h | < k | ξ | + 1 λ k α − 1 .

Therefore, for each positive integer k < k 0 , there are only finitely many integers h such that

| ξ − h k | < 1 λ k α .

This shows that there are only finitely many rational h ∕ k satisfying

| ξ − h k | < 1 λ k α .

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2025-07-12 08:36
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