2

The irrationality measure can be defined as:

Let $x$ be a real number, and let $R$ be the set of positive real numbers $\mu$ for which

$$0<|x-\frac{p}q|<\frac1{q^\mu}$$

has (at most) finitely many solutions for $p$ and $q$ integers. Then, $$\mu(x)=\text{inf}_{\mu\in R}\, \mu$$

My thoughts are:

Suppose for some $x$, $(\mu_0,p_0,q_0)$ is a solution to the inequality, then because $$\frac1{q_0^{\mu_0}}<\frac1{q_0^{\mu_0-\epsilon}}$$ for every $\epsilon>0$, thus $(\mu_0-\epsilon,p_0, q_0)$ is also a solution to the inequality. Thus, we can reduce the irrationality measure as small as possible.

What’s the flaw in my thought?

Szeto
  • 11,159
  • The solutions for $\mu_0$ are indeed also solutions for $\mu_0-\varepsilon$. But maybe there are many others solutions for $\mu_0-\varepsilon$ (which are not solutions for $\mu_0$) – charmd Jun 22 '18 at 05:43

1 Answers1

1

The point of the definition is that there are at most finitely many solutions. You've only ensured that all the solutions are still solutions; you didn't consider whether you've created more solutions. In fact, $\mu$ is defined precisely such that if you reduce it, you get infinitely many solutions.

joriki
  • 238,052