It is well known that real numbers are either rational or irrational.
However, one can ask whether some irrational numbers are in some sense harder to approximate by rational numbers than others. One way to make this notion precise is the Irrationality Measure, which assigns a positive number $\mu(x)$ to each real number $x$. Almost all transcendentals, and all (irrational) algebraic numbers have $\mu (x)=2$, including $e$. But some transcendentals can have $\mu(x)>2$. Liouville numbers, for example, have infinite $\mu(x)$. They can be approximated very well by rationals, while algebraic irrationals can not. So, counter-intuitively, the irrationality measure of an irrational is larger when it is better approximated by a rational.
I am confused about the last statement, why the larger is $\mu(x)$, the better $x$ is approximated by rational numbers. I would appreciate if someone could give some explanations and comments. Thanks in advance.
Best Answer
By definition the irrationality measure of $x$ is the infimum of $\mu$ such that $| x - p/q| < 1/q^\mu$ has only finitely many solutions in integers $p,q$.
By a theorem of Dirichlet, for every irrational $x$ there are infinitely many $p,q\in\mathbb{Z}$ such that $|x - p/q| < 1/q^2$, so the irrationality measure $\ge 2$. But typically you can't do better than that, so the irrationality measure is just $2$. If $p/q$ is a very good approximation to $x$ you might have $|x - p/q| < 1/q^\mu$ for some larger $\mu$ ($q$ being an integer $> 1$, larger $\mu$ means smaller $1/q^\mu$ and a better approximation). And if that happens not just once but for infinitely many $p/q$, you have a number of irrationality measure $\ge \mu$, which has better rational approximations than a number whose irrationality measure is just $2$.