Homepage › Solution manuals › Ivan Niven › An Introduction to the Theory of Numbers › Exercise 7.4.6* ($\xi - h_n/k_n = (-1)^n k_n^{-2} (\xi_{n+1} + \langle 0,a_n,a_{n-1}, \ldots,a_2,a_1\rangle)^{-1}$)

An Introduction to the Theory of Numbers

Exercise 7.4.6* ($\xi - h_n/k_n = (-1)^n k_n^{-2} (\xi_{n+1} + \langle 0,a_n,a_{n-1}, \ldots,a_2,a_1\rangle)^{-1}$)

Prove that for n ≥ 1 ,

ξ − h n k n = ( − 1 ) n k n − 2 ( ξ n + 1 + ⟨ 0 , a n , a n − 1 , … , a 2 , a 1 ⟩ ) − 1 .

Answers

Proof. As in (7.9), we have

ξ − h n k n = ⟨ a 0 , a 1 , a 2 , … , a n , ξ n + 1 ⟩ ( by (7.8) ) = ξ n + 1 h n + h n − 1 ξ n + 1 k n + k n − 1 − h n k n ( by Theorem 7.3 ) = − ( h n k n − 1 − h n − 1 k n ) k n ( ξ n + 1 k n + k n − 1 ) = ( − 1 ) n k n ( ξ n + 1 k n + k n − 1 ) ( by Theorem 7.5 ) = ( − 1 ) n k n 2 ( ξ n + 1 + k n − 1 k n ) .

Moreover, by Problem 7.3.4,

k n k n − 1 = ⟨ a n , a n − 1 , … , a 2 , a 1 ⟩ .

Therefore

⟨ 0 , a n , a n − 1 , … , a 2 , a 1 ⟩ = 0 + 1 ⟨ a n , a n − 1 , … , a 2 , a 1 ⟩ = k n − 1 k n .

Hence

ξ − h n k n = ( − 1 ) n k n 2 ( ξ n + 1 + ⟨ 0 , a n , a n − 1 , … , a 2 , a 1 ⟩ ) .

So ξ − h n k n = ( − 1 ) n k n − 2 ( ξ n + 1 + ⟨ 0 , a n , a n − 1 , … , a 2 , a 1 ⟩ ) − 1 is proven. □

Note: The difficulty is not there, but rather in the Problem 7.3.4.

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2025-07-30 08:12
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