## Properties of Convergents of Finite Continued Fractions

Theorem (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. Then
a)
$p_k q_{k-1}-p_{k-1}q_k=(-1)^k \rightarrow \frac{p_k}{q_k}- \frac{p_{k-1}}{q_{k-1}}=\frac{(-1)^k}{q_kq_{k-1}}, \; 2 \lt k \lt n$

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{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}

b)
\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}

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.