Exercise 7.4.5 ($\xi_n = \langle a_n,a_{n+1},a_{n+2},\ldots \rangle.$)

Answers

Proof. Let $ξ$ be an irrational number. The sequence ${(ξ_{n})}_{n \in ℕ}$ is defined inductively by (7.7), and by (7.8)

\begin{align} ξ = ⟨ a_{0}, a_{1}, \dots, a_{n - 1}, ξ_{n} ⟩ . & (1) \end{align}

Let $n$ be any fixed integer. For all $m > n$ ,

⟨ a_{0}, a_{1}, \dots, a_{n - 1}, ⟨ a_{n}, a_{n + 1}, a_{n + 2}, \dots, a_{m} ⟩ ⟩ = ⟨ a_{0}, a_{1}, \dots, a_{m} ⟩ .

Taking the limit of both members when $m \to \infty$ , we obtain

\begin{align} ⟨ a_{0}, a_{1}, \dots, a_{n - 1}, ⟨ a_{n}, a_{n + 1}, a_{n + 2}, \dots ⟩ ⟩ = ⟨ a_{0}, a_{1}, a_{2}, \dots ⟩ . & (2) \end{align}

The comparison of (1) and (2) gives

ξ = ⟨ a_{0}, a_{1}, \dots, a_{n - 1}, ξ_{n} ⟩ = ⟨ a_{0}, a_{1}, \dots, a_{n - 1}, ⟨ a_{n}, a_{n + 1}, a_{n + 2}, \dots ⟩ ⟩ .

Put

α = ξ_{n}, β = ⟨ a_{n}, a_{n + 1}, a_{n + 2}, \dots ⟩ .

Since $ξ = ⟨ a_{0}, a_{1}, \dots, a_{n - 1}, α ⟩ = ⟨ a_{0}, a_{1}, \dots, a_{n - 1}, β ⟩$ , by Theorem 7.3

ξ = \frac{α h_{n - 1} + h_{n - 2}}{α k_{n - 1} + k_{n - 2}} = \frac{β h_{n - 1} + h_{n - 2}}{β k_{n - 1} + k_{n - 2}} .

Since the homographic function $x \mapsto \frac{x h_{n - 1} + h_{n - 2}}{x k_{n - 1} + k_{n - 2}}$ is one-to-one, this implies $α = β$ , so

ξ_{n} = ⟨ a_{n}, a_{n + 1}, a_{n + 2}, \dots ⟩ .

(NZM say p. 335 to prove this equality: “(...) it becomes obvious if we apply to $ξ_{n}$ the process described at the opening of this section”.) □

2025-07-29 10:25