6.3: Very Good Approximation
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 8854
Continued fractions provide a representation of numbers which is, in a sense, generic and canonical. It does not depend on an arbitrary choice of a base. Such a representation should be the best in a sense. In this section we quantify this naive idea.
A rational number \(a/b\) is referred to as a "good" approximation to a number \(\alpha\) if \[\frac{c}{d} \neq \frac{a}{b} \hspace{5mm} \text{and} \hspace{5mm} 0<d \leq b\] imply \[d\alpha  c > b\alpha a.\]
Remarks. 1. Our "good approximation" is "the best approximation of the second kind" in a more usual terminology.
2. Although we use this definition only for rational \(\alpha\), it may be used for any real \(\alpha\) as well. Neither the results of this section nor the proofs alter.
3. Naively, this definition means that \(a/b\) approximates \(\alpha\) better then any other rational number whose denominator does not exceed \(b\). There is another, more common, definition of "the best approximation". A rational number \(x/y\) is referred to as "the best approximation of the first kind" if \(c/d\neq x/y\) and \(0<d\leq y\) imply \(\alpha  c/d>\alpha  x/y\). In other words, \(x/y\) is closer to \(\alpha\) than any rational number whose denominator does not exceed \(y\). In our definition we consider a slightly different measure of approximation, which takes into the account the denominator, namely \(b\alpha  a/b=b\alpha a\) instead of taking just the distance \(\alpha  a/b\).
[good] Any "good" approximation is a convergent.
Proof. Let \(a/b\) be a "good" approximation to \(\alpha = [a_0;a_1,a_2,\ldots,a_n]\). We have to prove that \(a/b=p_k/q_k\) for some \(k\).
Thus we have \(a/b>p_1/q_1\) or \(a/b\) lies between two consecutive convergents \(p_{k1}/q_{k1}\) and \(p_{k+1}/q_{k+1}\) for some \(k\). Assume the latter. Then \[\left\vert \frac{a}{b}  \frac{p_{k1}}{q_{k1}} \right\vert \geq \frac{1}{bq_{k1}}\] and \[\left\vert \frac{a}{b}  \frac{p_{k1}}{q_{k1}} \right\vert < \left\vert \frac{p_k}{q_k}  \frac{p_{k1}}{q_{k1}} \right\vert = \frac{1}{q_kq_{k1}}.\] It follows that \[\label{l7} b>q_k.\] Also \[\left\vert \alpha  \frac{a}{b} \right\vert \geq \left\vert \frac{p_{k+1}}{q_{k+1}}  \frac{a}{b} \right\vert \geq \frac{1}{bq_{k+1}},\] which implies \[\left\vert b\alpha  a \right\vert \geq \frac{1}{q_{k+1}}.\] At the same time Theorem [inequ] (it right inequality multiplied by \(q_k\)) reads \[\left\vert q_k \alpha  p_k \right\vert \leq \frac{1}{q_{k+1}}.\] It follows that \[\left\vert q_k \alpha  p_k \right\vert \leq \left\vert b\alpha  a \right\vert,\] and the latter inequality together with ([l7]) show that \(a/b\) is not a "good" approximation of \(\alpha\) in this case.
This finishes the proof of Theorem [good].
Exercises

Prove that if \(a/b\) is a "good" approximation then \(a/b \geq a_0\).

Show that if \(a/b>p_1/q_1\) then \(a/b\) is not a "good" approximation to \(\alpha\).
Contributors
Dr. Wissam Raji, Ph.D., of the American University in Beirut. His work was selected by the Saylor Foundation’s Open Textbook Challenge for public release under a Creative Commons Attribution (CC BY) license.