What is meant by "convergent just preceding ##\frac{a}{b}##" ?

[Adesh]

TL;DR Summary

Convergents in a continued fractions are things that we get when truncate the process at any stage.

Continued Fractions are themselves quite odd to learn and see in modern Algebra courses. The convergents are the numbers that we get when stop the division process at any stage. I know the laws related to convergents, let $p_n$ denote the numerator of $n$th convergent and $q_n$ the denominator. It can be proved easily that $p_n q_{n-1} - p_{n-1} q_{n} = (-1)^{n}$, and I managed to become comfortable with it (after a fierce struggle). But the problem has come when I was reading Higher Algebra by Hall and Knight, Chapter XXVI, Article 347, in it we find

To find the general solution in positive integer of the equation $ax -by =c$.

Let $\frac{a}{b}$ be converted into continued fractions, and let $\frac{p}{q}$ denote the convergent $\color{red}{ just~ preceding~\frac{a}{b}; ~then~ aq -bp= \pm 1}$.

Now, everything that is in red is causing me the problem, first of all what does it mean "just preceding $\frac{a}{b}$", we can talk about first convergent, second convergent ... n th convergent but what "preceding" thing is quite not obvious. In the example after this article, the book says

In converting $\frac{42}{29}$ into a continued fraction the convergent just before $\frac{42}{29}$ is $\frac{13}{9}$.

But if we convert $\frac{42}{29}$ into continued fraction we would find

$$\frac{42}{29} = 1+ \frac{13}{29} $$

$$\frac{42}{29} = 1+ \cfrac{1} {
\frac{29}{13}} = 1 + \cfrac{1} {
2+ \cfrac{3}{13} } $$
$$\frac{42}{29} = 1+ \cfrac{1} {\cfrac{29}{13}} =
1 + \cfrac{1} {
2+ \cfrac{1}{
\cfrac{13}{3} }} = 1+ \cfrac{1} {
2+ \cfrac{1}{
4+ \cfrac{1}{3} } }$$

So, you see I never got anything like $\frac{13}{9}$ in the continued fraction of $\frac{42}{29}$. Please explain me what is the book is doing and what he meant by "convergent just preceding ...".

Thank you.

 

[etotheipi]

If the completed continued fraction is $$x = a_0 + \frac{1}{a_1 + \frac{1}{a_2 + \dots}}$$ then the convergents are $a_0$, $a_0 + \frac{1}{a_1}$, $a_0 + \frac{1}{a_1 + \frac{1}{a_2}}$, and so on. In this case, your continued fraction is $[1:2,4,3]$, so the third convergent is $$1 + \frac{1}{2 + \frac{1}{3}} = \frac{13}{9}$$

 

[Adesh]

If the completed continued fraction is $$x = a_0 + \frac{1}{a_1 + \frac{1}{a_2 + \dots}}$$ then the convergents are $a_0$, $a_0 + \frac{1}{a_1}$, $a_0 + \frac{1}{a_1 + \frac{1}{a_2}}$, and so on. In this case, your continued fraction is $[1:2,4,3]$, so the third convergent is $$1 + \frac{1}{2 + \frac{1}{3}} = \frac{13}{9}$$

Thank you so much.