Let $[a_0,a_1,\cdots,a_{N}]$ be a finite simple continued fraction,and let $p_n$ and $q_n$ be the numbers defined in Exercise 10.Prove that
\begin{equation}
p_nq_{n-2}-p_{n-2}q_n=(-1)^na_n
\end{equation}
for $n=2,\cdots,N$.
Proof:According to 数论概论(Joseph H.Silverman) 定理39.2 连分数相邻收敛项之差定理,
\begin{equation}
\frac{p_{n-1}}{q_{n-1}}-\frac{p_{n}}{q_{n}}=\frac{(-1)^n}{q_nq_{n-1}}
\end{equation}
And
\begin{equation}
\frac{p_{n-2}}{q_{n-2}}-\frac{p_{n-1}}{q_{n-1}}=\frac{(-1)^{n-1}}{q_{n-1}q_{n-2}}
\end{equation}
So
\begin{equation}
\frac{p_{n-2}}{q_{n-2}}-\frac{p_n}{q_n}=\frac{(-1)^n}{q_nq_{n-1}}+\frac{(-1)^{n-1}}{q_{n-1}q_{n-2}}
\end{equation}
So
\begin{equation}
p_{n-2}q_n-p_nq_{n-2}=\frac{(-1)^nq_{n-2}}{q_{n-1}}+\frac{(-1)^{n-1}q_n}{q_{n-1}}
\end{equation}
So
\begin{equation}
p_nq_{n-2}-p_{n-2}q_n=(-1)^n[\frac{q_n}{q_{n-1}}-\frac{q_{n-2}}{q_{n-1}}]
\end{equation}
So
\begin{equation}
p_nq_{n-2}-p_{n-2}q_n=(-1)^na_n
\end{equation}(According to 数论概论(Joseph H.Silverman) 定理39.1 连分数的递归公式).