Difference between Adjacent Convergents But One of Simple Continued Fraction
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Theorem
Let $n \in \N \cup \set \infty$ be an extended natural number.
Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be a continued fraction in $\R$ of length $n$.
Let $p_0, p_1, p_2, \ldots$ and $q_0, q_1, q_2, \ldots$ be its numerators and denominators.
Let $C_0, C_1, C_2, \ldots$ be the convergents of $\sqbrk {a_0, a_1, a_2, \ldots}$.
For $k \ge 2$:
- $p_k q_{k - 2} - p_{k - 2} q_k = \paren {-1}^k a_k$
That is:
- $C_k - C_{k-2} = \dfrac {p_k} {q_k} - \dfrac {p_{k - 2} } {q_{k - 2} } = \dfrac {\paren {-1}^k a_k} {q_k q_{k - 2} }$
Proof
Let $k \ge 2$.
\(\ds p_k q_{k - 2} - p_{k - 2} q_k\) | \(=\) | \(\ds \paren {a_k p_{k - 1} + p_{k - 2} } q_{k - 2} - p_{k - 2} \paren {a_k q_{k - 1} + q_{k - 2} }\) | Definition of Numerators and Denominators of Continued Fraction | |||||||||||
\(\ds \) | \(=\) | \(\ds a_k \paren {p_{k - 1} q_{k - 2} - p_{k - 2} q_{k - 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a_k \paren {-1}^k\) | Difference between Adjacent Convergents of Simple Continued Fraction |
$\blacksquare$