Proof of $\frac{q_n}{q_{n-1}} = [a_n,a_{n-1},a_{n-2},...,a_2,a_1]$?

Question

Proof of continued fractions axiom.

Let $c=[a_0,a_1,a_2,\dots,a_n,\dots] = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \ddots}}$ be a continued fraction which could be finite or infinite.

By $\frac{p_n}{q_n}$ we denote the $n$-th convergent of the continued fraction, i.e. $$\frac{p_n}{q_n}=[a_0,a_1,a_2,\dots,a_n].$$

How to prove the formula formula: $$\frac{q_n}{q_{n-1}} = [a_n,a_{n-1},a_{n-2},...,a_2,a_1]?$$

This formula is stated in the book Olds C.D. Continued fractions (Math.Assoc.Am., Yale, 1963) as a part of Problem 7 on page 26, together with a related formula $$\frac{p_n}{p_{n-1}} = [a_n,a_{n-1},\dots,a_1,a_0].$$

It is also stated as formula (31) in Wolfram Mathworld article on continued fractions.

visual formula - https://s31.postimg.org/6l4ehkd8r/pic.png

Answer 1

Ok, I guess it is by far simple answer, sorry for posting silly questions :(.

$\frac{q_n}{q_{n-1}}$ = $\frac{a_n*q_{n-1}+q_{n-2}}{q_{n-1}}$ = $\frac{a_n*q_{n-1}}{q_{n-1}}$ + $\frac{q_{n-2}}{q_{n-1}}$ = $a_n$ + $\frac{q_{n-2}}{q_{n-1}}$ = $a_n$ + $\frac{1}{\frac{q_{n-1}}{q_{n-2}}}$

So $\frac{q_n}{q_{n-1}}$ = $a_n$ + $\frac{1}{\frac{q_{n-1}}{q_{n-2}}}$

and $\frac{q_{n-1}}{q_{n-2}}$ = $a_{n-1}$ + $\frac{1}{\frac{q_{n-2}}{q_{n-3}}}$

and $\frac{q_{n-2}}{q_{n-3}}$ = $a_{n-2}$ + $\frac{1}{\frac{q_{n-3}}{q_{n-4}}}$

$\vdots$

So by induction you will have:

$\frac{q_n}{q_{n-1}}$ = $a_n$ + $\frac{1} {a_{n-1} + \frac{1} {a_{n-2} + \frac{1} {\frac{q_{n-3}}{ \ddots}}}}$ = $[a_{n},a_{n-1},a_{n-2},\dots,a_{1}]$