# 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

• I have tried to edit the post. Together with the OPs edits, I do not think the close reason "unclear what you're asking" applies any longer. – Martin Sleziak Jun 19 '16 at 14:45

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:

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