Properties of Convergents of Finite Continued Fractions
Let
\[[ a_1,a_2,...,a_{n-1},a_n ]\]
be any finite continued fraction and let \[[ p_1,p_2,...,p_{n-1},p_n ], \; [ q_1,q_2,...,q_{n-1},q_n ]\]
/jatex} be the numerators and denominators respectively of the Convergents of Finite Continued Fractions. Thena)
\[p_k q_{k-1}-p_{k-1}q_k=(-1)^k \rightarrow \frac{p_k}{q_k}- \frac{p_{k-1}}{q_{k-1}}=\frac{(-1)^k}{q_kq_{k-1}}, \; 2 \lt k \lt n\]
b)
\[p_k q_{k-2}-p_{k-2}q_k=(-1)^{k-1}a_k \rightarrow \frac{p_k}{q_k}- \frac{p_{k-2}}{q_{k-2}}=\frac{(-1)^{k-1}a_k}{q_kq_{k-2}}, \; 3 \lt k \lt n\]
c)
\[p_k, \; q_k \]
are relatively prime for \[1 \le k \le n\]
.Proof
a) Proof is by induction. Let
\[P(k)\]
be \[p_k q_{k-1}-p_{k-1}q_k=(-1)^k\]
as in a).If
\[k=2\]
then \[p_1=a_1, \; p_2=a_1a_2+1, \; q_1=1, \; q_2=a_2\]
so \[p_k q_{k-1}-p_{k-1}q_k=(a_1a_2+1)(1)-(a_1)(a_2)(-1)^k=1=(-1)^2\]
so \[P(2)\]
is true.Assume
\[P(k)\]
is true.\[p_{k+1}=a_{k+1}p_k+p_{k-1}, \; q_{k+1}=a_{k+1}q_k+q_{k-1}\]
Then
\[\begin{equation} \begin{aligned} p_{k+1}q_k-p_kq_{k+1} &= (a_{k+1}p_k+p_{k-1})q_k-p_k(a_{k+1}q_k+q_{k-1}) \\ &= p_{k-1}q_k-p_kq_{k-1} \\ &= (-1)(p_kq_{k-1}-p_{k-1}q_k) \\ &= (-1)(-1)=(-1)^{k+1}\end{aligned} \end{equation}\]
b)
\[\begin{equation} \begin{aligned} p_kq_{k-2}-p_{k-2}q_k &= (a_kp_{k-1}+p_{k-2})q_{k-2}-p_{k-2}(a_kq_{k-1}+q_{k-2}) \\ &= a_k ()p_{k-1}q_{k-2}-p_{k-2}q_{k-1}) \\ &=a_k(-1)^{k-1} \end{aligned} \end{equation}\]
c) If
\[d=gcd(p_k,q_k)\]
then \[p_kq_{k-1}-p_{k-1}q_k=(-1)^k\]
is a linear combination of \[p_k, \; q_k\]
, is divisible by \[d\]
, which is impossible. 
