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Let $\frac{a}{b}$ and $\frac{c}{d}$ be two reduced fractions with $bc-ad > 1$ (and hence $\frac{a}{b} \lt \frac{c}{d}$) and $a,b,c,d$ positive. It is well known that there are integers $u,v$ such that $0 \leq u \leq a, 0 \leq v \leq b$ and $av-bu=1$. Let $q$ be the largest positive below $\frac{ \max(b,d)+v}{b}$, i.e. the euclidean quotient of $\max(b,d)+v$ by $b$). Then the following inequality holds :

$ cv-du \leq q(bc-ad) \tag{*} $

I know that (*) is true, but the only proof I know relies on the heavy machinery of Farey sequences and the Stern-Brocot tree. Is there a more direct proof ?

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    ....but the "-es" at the end could be considered as explaining it, just as "Archimedes" becomes "Archimedean", with a final "-ean". As often happens in modern Greek, the upsilon is pronounced like an "f" when it's followed by a "k" and in some other contexts. Googling, I was surprised to learn that the immortal geometer's name seems to be a commonplace surname in Greece today. See http://www.psy.auth.gr/efklides, for example.2012-03-05

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Let $bc-ad=r$. Then $bcv-avd=rv,\quad bcv-(1+bu)d=rv,\quad (cv-du)b=rv+d,\quad cv-du=(v/b)r+(d/b),\quad cv-du=(v/b)(bc-ad)+(d/b)$ which is pretty close to what you want.

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    Yes, but you are not my thesis supervisor.2012-03-06