# Convergent of continued fractions the best rational approximation of a number? [duplicate]

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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## marked as duplicate by Srivatsan, Asaf Karagila, t.b., Guess who it is., Mike SpiveyDec 6 '11 at 5:20

@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. –  Gerry Myerson Nov 14 '11 at 5:21
It's not true. For example, $3/1$ is a convergent of $x = \frac{15}{4} = 3 + 1/(1 + 1/3)$, but it is not a best approximation since $|4/1 - x| < |3/1 - x|$.
Technically, $x$ is required by the OP to be irrational. I suppose the counter-example can be fixed easily to handle that (e.g., by picking $x$ to be sufficiently close to $15/4$)... –  Srivatsan Nov 14 '11 at 6:42
The correct result is that if $|\alpha−c/d|<|\alpha−p_k/q_k|$, where $k\ge 1$, then $d>q_k$. There is no claim about $k=0$. If we change the question to what are the best approximations to $\alpha$ with denominator $\le s$, the answer changes to convergents or secondary convergents. (The definition of secondary convergent is too long for a comment.) –  André Nicolas Nov 14 '11 at 7:34