# Approximation of irrationals by fractions

If $\alpha$ is an irrational, and I'm trying to judge the suitability of of a rational $p/q$ as its approximation by the error $\Delta = |\alpha - p/q|$. For a given denominator $q$, I am finding a $p$ for which the error $\Delta$ is minimized. Why is this error "certainly less than $1/2q$ " ?

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If you use convergents of the continued fraction you get particularly good approximations. You can see en.wikipedia.org/wiki/Continued_fraction for information, especially the section "Best rational approximations". If you quit just before a large value in the continued fraction, it is even better. – Ross Millikan Aug 7 '11 at 1:44
@Ross that was precisely the motivation for this question. I encountered the problem when trying to prove the result. – kuch nahi Aug 7 '11 at 1:52

Fix denominator $q$. If $\alpha -p/q$ is greater than $\frac{1}{2q}$ than $\frac{(p+1)}{q}$ is a better approximation.

[EDIT] Equality is impossible since $\alpha$ is assumed irrational.

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Excellent. Thanks! – kuch nahi Aug 7 '11 at 1:39

You can find approximations with error less than $\frac{1}{q^2}$ with continued fractions. The following can be proven with a simple pigeonhole proof.

Approximation Lemma: Let $x$ be any real number and $N$ be a positive integer. Then there are integers $p$ and $q$ with $0 < q \le N$ so that $|p - qx| < \frac{1}{N}$.

Proof: Break up the unit interval $[0,1)$ into $N$ equal subintervals. Notice which subinterval contains the residues $\pmod{1}$ of $kx$ for $0\le k\le N$. There are $N+1$ residues distributed among $N$ subintervals; thus, there must be at least one subinterval with at least two residues, $(k_1 x - j_1)$ and $(k_2 x - j_2)$. We may assume that $k_2 > k_1$. Let $q = k_2 - k_1$ and $p = j_2 - j_1$, then $0 < q \le N$ and, because both residues are in the same subinterval, $|p - qx| = |(k_1 x - j_1) - (k_2 x - j_2)| < \frac{1}{N}$. QED

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