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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.

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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 $-1$.

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