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For a real number $\alpha$, let the irrationality measure $\mu(\alpha) \in \mathbb{R}\cup \{\infty\}$ be defined as the supremum of all real numbers $\mu$ such that $$ \left| \alpha-\frac{p}{q}\right| \le \frac{1}{q^\mu} $$ has infinitely many solutions $\frac{p}{q} \in \mathbb{Q}$, where $q \ge 1$. It's known that if $\alpha$ is rational, then $\mu(\alpha)=1$ and if $\alpha$ is algebraic and irrational, then $\mu(\alpha)=2$ by Roth's Theorem. The set of all $\alpha$ such that $\mu(\alpha) >2$ has Lebesgue measure 0 by Khinchin's Theorem.

One can explicitly write down real numbers $\alpha$ such that $\mu(\alpha) = \infty$ (e.g. Liouville's constant) and transcendental real numbers $\alpha$ such that $\mu(\alpha)=2$ (e.g. $\alpha =e$, see also this thread). It's also possible to find upper bounds on $\mu(\alpha)$ for some real numbers such as $\pi$. But I don't know of a single example of a real number $\alpha$ whose irrationality measure is known, finite and greater than 2.

Question: Is there an example of a real number whose irrationality measure $\mu(\alpha)$ is known exactly and satisfies $2<\mu(\alpha)<\infty$?

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    $\begingroup$ I imagine if one "toned down" Liouville's construction and wrote $\alpha=\sum_{n=1}^\infty 10^{-\lfloor\mu^n\rfloor}$, then one could show that $\mu(\alpha)=\mu$. There is something to be shown, though, since the obvious rational approximations by terminating decimals only demonstrate that $\mu(\alpha)\ge\mu$. $\endgroup$ Commented Jun 18, 2013 at 17:13
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    $\begingroup$ The Champernowne constant has irrationality measure $10$ (the base used in its construction). $\endgroup$ Commented May 1, 2017 at 5:20

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The answer is yes - see for example

Yann Bugeaud Diophantine approximation and Cantor sets Math. Ann. (2008) 341:677–684

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    $\begingroup$ I'm pmretty sure $\mu(\alpha)=1$ for rational $\alpha$ is correct. $\endgroup$ Commented Jun 18, 2013 at 17:10
  • $\begingroup$ Sorry - corrected. $\endgroup$ Commented Jun 18, 2013 at 21:57
  • $\begingroup$ Thanks Carl, that's the kind of reference I was looking for. $\endgroup$ Commented Jun 19, 2013 at 17:29
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Irrationality measure is a question about approximation by rationals. The continued fraction expansion gives the best approximations and controls their quality. Irrationality measure is a kind of asymptotic growth of the continued fraction expansion. Asking about the irrationality measure of a particular number is asking properties of its continued fraction expansion. But if you are willing to specify numbers by their continued fraction expansion, it is easy to write down a continued fraction expansion with the desired measure. Inductively define the continued fraction $[a_1,a_2,\ldots]$ by setting $a_{n+1}=\lceil q_n^{\mu-1}\rceil$, where the convergent is $\frac{p_n}{q_n}$. Then $q_{n+1}=a_{n+1}q_n+q_{n-1}$, so the error $(q_nq_{n+1})^{-1}$ is about $q_n^{-\mu}$.

It's a bit of a cop-out, but it's definitely worth mentioning.

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  • $\begingroup$ Thanks, I'm happy to specify numbers by their continued fraction expansion. But don't you mean $\mu-2$ instead of $\mu-1$? (sorry if I'm being stupid here) $\endgroup$ Commented Jun 19, 2013 at 17:39
  • $\begingroup$ Yes, you're right, it should be $a_{n+1}=q_n^{\mu-2}$. Then $q_{n+1}\sim q_n^{\mu-1}$ and the reciprocal error is $q_{n+1}q_n\sim q_n^\mu$. $\endgroup$ Commented Jun 19, 2013 at 18:39
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The irrationality measure of the Champernowne constant $$ C_b= \sum_{n=1}^{\infty} n \cdot b^{-\left(\sum\limits_{k=1}^{n}\left\lceil\log_{b}(k+1)\right\rceil\right)} $$ in base $b>2$ is exactly $b$. For example, $$ C_{10} = 0.123456789101112131415\ldots $$ has irrationality measure 10.

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  • $\begingroup$ What about $b = 2$? $\endgroup$ Commented Jun 19, 2025 at 10:21
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The Rabbit Constant which is defined through the Fibonacci numbers.

Let $\phi=\frac{\sqrt{5}+1}{2}$ and $F_{n}=\operatorname{round}\left(\frac{\phi^{n}}{\sqrt{5}}\right)$.

We let $R$ denote the value of the continued fraction $[0; 2^{F_0}, 2^{F_1}, \dots ]$

Then $R$ has irrationality measure $1+\phi$.

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    $\begingroup$ This is a particular case of Ben Wieland's answer from 2013. $\endgroup$ Commented Jun 15, 2025 at 4:52

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