For a given real number $x$, let $R_x$ be the set of real numbers $r$ such that the inequality
$$\displaystyle \left| x – \frac{p}{q} \right| < \frac{1}{q^r}$$
has at most finitely many solutions with integers $p,q$. Define the irrationality measure of $x$, say $\mu(x)$, to be the infimum of $R_x$.
It is known that if $x$ is algebraic and not rational, then $\mu(x)$ is 2, by Roth's Theorem. It is trivial that if $x$ is rational, then $\mu(x) = 1$. I believe it is also known that all real numbers except a set of measure 0 has irrationality measure of 2, but I am unsure of the reference.
For some known transcendental numbers, upper bounds for $\mu$ are known. For example, we know that $\mu(\pi) < 7.6063$ (Salikhov, V. Kh. "On the Irrationality Measure of ." Usp. Mat. Nauk 63, 163-164, 2008. English transl. in Russ. Math. Surv 63, 570-572, 2008.)
Are there any general results concerning a set of transcendental numbers $x$ with $\mu(x) = 2$? Are there any known, 'interesting' numbers (expressible in well-known functions or constants) $x$ with $\mu(x) = 2$?
Best Answer
If the elements $a_n$ of the simple continued fraction of the irrational number $x$ satisfy $a_n < c n + d$ for some positive constants $c$ and $d$, then $\mu(x) = 2$. Besides $e^{2/k}$ for positive integers $k$, interesting examples of such numbers include $\tanh(1/k)$, $\tan(1/k)$, and $I_0(1)/I_1(1)$ where $I_0$ and $I_1$ are modified Bessel functions.