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Possible Duplicate:
A nicer proof of Lagrange's 'best approximations' law?

I was reading through the wikipedia article on continued fractions, and they state, essentially, that for any convergent $\frac{a}{b}$, it is the best approximation you can have. More formally, for an irrational number $x$ with a convergent $\frac{a}{b}$,

$\forall c\forall d \quad |\frac{c}{d}-x| < |\frac{a}{b}-x| \implies d > b$.

However they give no proof of it. Is there a nice one, or did they not give one because it's messy to show?

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    @Srivatsan, it isn't quite a duplicate, since that question asks about $|a-bx|$, and this one asks about $|(a/b)-x|$. But it's certainly true that OP may find something of interest in that other thread.2011-11-14
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    (Cf. my vote to close and @Gerry's comment.) Apologies to james for the hasty vote to close as duplicate.2011-11-14
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    I don't know of a particularly short proof of this fact. The expositions I've seen in textbooks (such as chapter 7 of Niven, Zuckerman, & Montgomery's book) take a few lemmas to establish this.2011-11-14

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