# Rational approximation bound for real numbers in (0,1)

I am working on a problem that related to rational approximation of real numbers.

I am looking for a bound of the form:

Given a positive real number, $\alpha \in (0,1)$, there exist positive integers, $p$ and $q$ such that $0< p \leq q$ and,

$$\left|\alpha - \frac{p}{q}\right|<f(q)$$

where $f(q)$ is some function of $q$.

I am almost certain that there are some good bounds given that there are lots of similar statements in the field of Diophantine approximation. Unfortunately, these bounds are for $\alpha \in \mathbb{R}$ and therefore do not necessarily that $p \leq q$. Intuitively speaking, however, it should follow that $\alpha \leq (0,1)$ implies that the rational approximation, $\frac{p}{q} \leq (0,1)$ which would then imply that $p \leq q$.

• Whether $\alpha$ is in $\mathbb{R}$ or in $[0,1]$ doesn't matter. – André Nicolas May 20 '14 at 0:58
• The best rational approximations are all in $[0,1]$, so you can have $f(q) = \frac{1}{\sqrt{5} \, q^2}$. – Daniel Fischer May 20 '14 at 0:59
• @DanielFischer That is also what I suspect... but couldn't that bound imply that $p > q$ for some pathological value of $\alpha$? – Elements May 20 '14 at 1:10
• No, the best approximation of $\alpha$ always has the numerator not greater than the denominator. If $p > q$, then $\left\lvert\frac{q}{q}-\alpha\right\rvert < \left\lvert \frac{p}{q}-\alpha\right\rvert$. – Daniel Fischer May 20 '14 at 1:14
• For rational $\alpha$, you can have equality. For the convergents before the last, you have the same bound as for irrationals. I don't know how good a bound you can get for simultaneous approximations, I'm convinced it must be larger than that for a single approximation, but I don't know how much. – Daniel Fischer May 20 '14 at 9:17