Is there any real number except $1$ which is equal to its own irrationality measure? If so, then what is the cardinality of the set of all such numbers? Is the set dense on any interval? Is it measurable?
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If $x$ has irrationality measure $\mu$, and $y$ has irrationality measure $\tau<\mu$, does $x+y$ automatically have irrationality measure $\mu$? Linguistically, I'd want this to be true. It seems like it might be helpful. – 2'5 9'2 Mar 16 '12 at 02:00

See also http://mathoverflow.net/questions/249194/irrationalitymeasureofthenumberisitself – Gerry Myerson Sep 06 '16 at 22:47
2 Answers
Using the property stated in that article:
$$\mu(x)=2 + \limsup \frac{\log a_{n+1}}{\log q_n}$$
where the continued fraction expansion for $x$ is $[a_0,a_1,...]$ and the $n$th convergent is $\frac{p_n}{q_n}$.
Start with $a_0=2$ and $a_1=2$, so $q_0=1$, $q_1=2$.
Now, assume you have a continued fraction $$\frac{p_n}{q_n}=[a_0,...,a_n]$$
Define $a_{n+1}$ to be the least integer such that $2+\frac{\log a_{n+1}}{\log q_n}>\frac{p_n}{q_n}$.
Then $x = [a_0,a_1,...] = \lim \frac{p_n}{q_n}$ will satisfy your requirement.
Just show a bound on $2+\frac{\log a_{n+1}}{\log q_n}\frac{p_n}{q_n}$.
In particular, you can use that $\log (a_{n+1}1)>(\log a_{n+1}) 1$ to show that if $2+\frac{\log a_{n+1}}{\log q_n}\frac{p_n}{q_n}>\frac{1}{\log q_n}$, then $$2+\frac{\log (a_{n+1}1)}{\log q_n}>\frac{p_n}{q_n}$$ which would violate our definition of $a_{n+1}$. So $$\mu(x)=2+\limsup \frac{\log a_{n+1}}{\log q_n} = \lim \frac{p_n}{q_n}= x$$
So there exists such an $x$.
You can easily get uncountably many such $x$ by choosing any values $a_{2n}\in\{1,2\}$ and then choose the $a_{2n+1}$ by the above condition, again making the $\limsup$ equal to the limit of $\frac{p_n}{q_n}$.
I think the same argument can be made to show that the set is dense in $[2,\infty)$. Basically, you can make such a $x$ starting with any finite sequence $[a_0,...,a_n]$ with $a_0\geq 2$. Indeed, it is uncountable in any finite subinterval $[a,b]$ with $b>a\geq 2$.
I don't think this resolves the measurability issue, contrary to my earlier claims.
It feels like $\{x:\mu(x)=x\}$ should be measurable, since it feels fairly constructive. On the other hand, it feels like if the set $\{x:\mu(x)=x\}$ has nonzero measure, then $\{x:\mu(x)=x+\alpha\}$ should have nonzero measure, when $\alpha\in\mathbb R$, and thus we'd have an uncountable set of disjoint positive measures, which I believe is not possible.
So my guess is that the set is measurable with measure $0$.
But that is all gut, no proof.
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Just thinking out loud, this is not my area of expertise at all but I found the question intriguing. This should be a comment but it's too long.
Using $\mu$ for the irrationality measure, and since: $$\begin{align}\mu(x)=1&\text{if}\ x\in\mathbb{Q}\\ \mu(x)=2&\text{if}\ x\in\mathbb{A}\text{ [Roth]}\\ \mu(x)\ \overset{\underset{\mathrm{def}}{}}{=}\ \infty&\text{if}\ x\ \text{is Liouville}\\ \mu(x)\geqslant2&\text{otherwise} \end{align}$$ then $1$ is the only rational or algebraic number satisfying your condition. So any possible solution must be: (1) transcendental, (2) $>2$, and (3) not a Liouville number. Unfortunately we only know (or have upper bounds on) the irrationality measures of a very few such numbers and none of them work, and there are uncountably many transcendental numbers that are not Liouville.
There is a construction method for numbers $x$ that have a given $\mu$, namely $$x = [\lfloor a\rfloor;\lfloor a^b\rfloor,\lfloor a^{b^2}\rfloor,\lfloor a^{b^3}\rfloor,\dots]\ \ a>1,b=\mu1$$ See Brisebarre, 2002. So I suppose one could set $x=\mu$ and work from there. But I don't find that anyone has done so, yet.
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1Nice observation. If the function $x(\mu)$ is continuous, I guess then we might be in business. :) – Srivatsan Dec 31 '11 at 06:10

The question of the continousness of $\mu(x)$ is fascinating. Its value is discontinuous ($\infty$) for all Liouville numbers, but since the Lebesgue measure of the Liouville numbers is $0$, I suppose it's nevertheless possible that it's a continous function. If anybody would want to undertake this question, I understand that Roth won the Fields for that kind of thing. :) – Mark Beadles Jan 01 '12 at 01:49


Oh, I did not mean the continuity of $\mu(x)$; my apologies if that wasn't clear. It is clearly discontinuous because it is $1$ at every rational number and $\geqslant 2$ elsewhere. What I meant was the continuity of the number $x(\mu)$ whose irrationality measure $\mu$ (where $x(\mu)$ is regarded as a function of $\mu$). – Srivatsan Jan 01 '12 at 17:16

Ah, I see. But of course we can construct an infinite number of $x$'s given any particular $\mu$. – Mark Beadles Jan 01 '12 at 17:24

That is a good point, but easily redeemed. Precisely, I am asking whether such a number can be continuously defined; i.e., does there exist a continuous function $x : [2, \infty) \to \mathbb R$ such that for all $\mu \geqslant 2$, $x(\mu)$ has irrationality measure $\mu$. – Srivatsan Jan 01 '12 at 17:43