# What is meant by “the convergent just preceding $\frac{a}{b}$” in continued fractions?

I’m reading about continued fractions and integer solutions of a linear equation. In Higher Algebra by Hall and Knight, article 347 we have

To find the general solution in positive integers of $$ax-by =c$$ Let $$\frac{a}{b}$$ be converted into a continued fractions, and let $$\frac{p}{q}$$ denote the convergent just preceding $$\frac{a}{b}$$; then $$aq -bp = \pm1$$.

Now, I’m totally puzzled about what does it mean to say “just preceding $$\frac{a}{b}$$”. Please help me in understanding what he meant by that and how he found out that equality. I know for a continued fraction, if $$\frac{p_n}{q_n}$$ denotes the n th convergent then $$p_n q_{n-1} - p_{n-1}q_{n}= (-1)^n$$ But in the above case he has used a very different equality.

It seems as if you have in fact answered your own question. The "convergent just preceding $$\frac{a}{b}$$" in the sequence of continued fraction convergents to $$\frac{a}{b}$$ is simply the convergent just before $$\frac{a}{b}$$. This exists because $$\frac{a}{b}$$ is rational and has a finite sequence of continued fraction convergents that ends with $$\frac{a}{b}$$, and so there is a convergent $$\frac{p}{q}$$ just before that. By the property you stated, we have that $$aq-pb$$ is either $$1$$ or $$-1$$.