MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\mu(x) = 2$?

share|cite|improve this question
$\mu(e)=2$ follows quickly from the continued-fraction expansion (and generalizes to $e^{2/k}$ if I remember right). – – Noam D. Elkies Feb 26 '12 at 19:38
Your definition is garbled. Perhaps $s=r$. – Gerald Edgar Feb 26 '12 at 20:01
Yes you are right, I must flipped the letters around in my head while typing, thanks for pointing that out. – Stanley Yao Xiao Feb 26 '12 at 21:39
You probably mean $\mu(q)=\infty$ for a rational $q$? Everything has irrationality measure at least 2 by the pigeonhole principle – Anthony Quas Feb 26 '12 at 22:49
@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 $\mu(x) = 1$. – Zack Wolske Feb 27 '12 at 0:41
up vote 12 down vote accepted

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.

share|cite|improve this answer
Neat - $\tanh(1/k)$ is equivalent to $e^{2/k}$ by fractional linear transformation, but I didn't know about $\tan (1/k)$. – Noam D. Elkies Feb 27 '12 at 5:53

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!

share|cite|improve this answer
Mea culpa, mea maxima culpa. But, what would Serge Lang say? – Gerry Myerson Mar 18 '12 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.

share|cite|improve this answer
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)? – Noam D. Elkies Feb 27 '12 at 5:28
To the best of my knowledge, no. – Gerry Myerson Feb 27 '12 at 5:59
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. – Kevin O'Bryant Feb 27 '12 at 15:40
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. – Kevin O'Bryant Feb 27 '12 at 15:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.