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 $\mu(x) = 2$?

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    $\begingroup$ $\mu(e)=2$ follows quickly from the continued-fraction expansion (and generalizes to $e^{2/k}$ if I remember right). – $\endgroup$ Feb 26, 2012 at 19:38
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    $\begingroup$ Your definition is garbled. Perhaps $s=r$. $\endgroup$ Feb 26, 2012 at 20:01
  • $\begingroup$ Yes you are right, I must flipped the letters around in my head while typing, thanks for pointing that out. $\endgroup$ Feb 26, 2012 at 21:39
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    $\begingroup$ @Anthony: Consider $x=0$, so that we want $\frac{p}{q} < \frac{1}{q^r}$. This has no solution for any $r \geq 1$ when $p \neq 0$. The statement of the question should say that $\frac{p}{q} \neq x$, then it follows that all rationals have $\mu(x) = 1$. $\endgroup$ Feb 27, 2012 at 0:41
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    $\begingroup$ The result that for almost all numbers $\mu(x)$ is $2$ is Khinchin's Diophantine approximation theorem. See en.wikipedia.org/wiki/Diophantine_approximation for some details. Schmidt's book on Diophantine approximation has proofs and references. $\endgroup$ Feb 27, 2012 at 8:45

3 Answers 3


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.

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    $\begingroup$ Neat - $\tanh(1/k)$ is equivalent to $e^{2/k}$ by fractional linear transformation, but I didn't know about $\tan (1/k)$. $\endgroup$ Feb 27, 2012 at 5:53
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    $\begingroup$ Also $J_0(1)/J_1(1)$ and $I_0(2)/I_1(2)$ and $J_0(1)/J_1(2)$ seem to have such predictable continued fraction expansions. Where might I learn more about this? $\endgroup$
    – Gro-Tsen
    Apr 8, 2020 at 16:40
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    $\begingroup$ I think you mean $J_0(2)/J_1(2)$. $\endgroup$ Apr 12, 2020 at 4:40
  • $\begingroup$ You might look at Cuyt et al, Handbook of Continued Fractions for Special Functions $\endgroup$ Apr 12, 2020 at 4:44
  • $\begingroup$ I think the result about $\mu(x)=2$ should hold for any irrational $x$ where the simple continued fraction coefficients $a_n$ are bounded by some (finite-degree) polynomial $P(n)$. I get this from looking at the formula: $$\mu(x)=2+\limsup_{n\to\infty}\frac{\ln a_{n+1}}{\ln q_n}$$And the observation that, in the very worst case, $q_n$ is bounded below by the Fibonacci numbers ($q_n$ is the denominator of the $n$th convergent). However I am not sure. Do you agree? $\endgroup$
    – FShrike
    Aug 18, 2022 at 12:20

Yes, there are uncountably many "explicit" real numbers that are (i) badly approximable and (ii) transcendental and (iii) have easy-to-write-down binary expansions. See my paper with van der Poorten, Folded Continued Fractions, J. Number Theory 40 (1992), 237-250. I'm surprised Gerry Myerson didn't remember that!

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    $\begingroup$ Mea culpa, mea maxima culpa. But, what would Serge Lang say? $\endgroup$ Mar 18, 2012 at 10:08

A real irrational number $x$ is said to be "badly approximable" if there is a positive constant $c$ such that $$\left|x-{p\over q}\right|\gt{c\over q^2}$$ for every rational $p/q$. It is known that $x$ is badly approximable if and only if its continued fraction has bounded partial quotients. So these numbers have irrationality measure 2.

  • $\begingroup$ Is any explicit number with this property known to be transcendental, or to be definable by a closed form other than a continued fraction with a given sequence of bounded partial quotients (other than a quadratic irrationality, which is known to happen iff the sequence is eventually periodic)? $\endgroup$ Feb 27, 2012 at 5:28
  • $\begingroup$ To the best of my knowledge, no. $\endgroup$ Feb 27, 2012 at 5:59
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    $\begingroup$ Fix an irrational $\alpha>1$, $A=\\{ \lfloor n \alpha \rfloor: n\geq 1\\}$, and let $a_i=2$ if $i\in A$ and $a_i=1$ if $i\not\in A$. The number $[a_0;a_1,a_2,\dots]$ is known to be transcendental. The idea of the proof is that if the scf of $\alpha$ is almost periodic, then there are quadratic irrationals too close to $\alpha$: by a theorem of Schmidt (analogous to Liousville's Theorem) algebraic numbers cannot be super-well-approximated by other algebraics. $\endgroup$ Feb 27, 2012 at 15:40
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    $\begingroup$ A result of Florian Luca et al is that if the scf of $\alpha$ has exactly 2 values of partial quotients, and the indices where one of the values happens is automatic (in the sense of the Allouche & Shallit) book, then $\alpha$ is transcendental. So, for example, if you take $a_i=3$ if the binary expansion of $i$ has an even number of 1's, and $a_i=7$ otherwise, you get a provably transcendental number. $\endgroup$ Feb 27, 2012 at 15:43

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