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Let $\mu (r)>2$ be the irrationality measure of a transcendental number $r$, and consider the following set of points $P \in\mathbb{R}$:

$P=\{r\in \mathbb{R}: \mu(r)=Constant\}$

Is this set a fractal, and If so, then what is it's dimension?

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The known upper bound of e.g. $\mu (\zeta (3))$ is $5.513891$, but it might be improved. So what does it means $\mu (r)=$ Constant $=5.513891$ in $P$? –  Américo Tavares Nov 17 '10 at 23:15
    
I would say that we should distinguish between two cases: $\mu(r)=\mu (\zeta (3))$ and $\mu(r)=5.513891$. I'm slightly confused by this comment, because if $\mu (\zeta (3))$ is an irrational number, then it's impossible to write it in decimal notation in any case. –  Matt Calhoun Nov 18 '10 at 15:00
    
Correction: ... what does it mean ... . $\zeta(3)$ is indeed an irrational number with an irrational measure less or equal to $5.513891...$. And if $\zeta(3)$ is not a transcendental number? (Though I believe one day someone will prove it is). –  Américo Tavares Nov 18 '10 at 15:25
    
If $\zeta (3)$ is algebraic, then $\mu(\zeta(3))=2$. –  Matt Calhoun Nov 18 '10 at 18:12
    
Yes, I know. But what does the condition $\mu(r)=$ Constant mean, when applied to $r=\zeta(3)$? What is the value of the Constant that would make $r=\zeta(3)$ be in $P$? –  Américo Tavares Nov 18 '10 at 20:31

1 Answer 1

up vote 3 down vote accepted

It is a fractal much like the cantor set with dimension 2/r. That is Jarniks theorem. You can find a proof in the Falconer book Fractal Geometry: Mathematical Foundations and Applications.

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Thanks! I own this book, I should have checked it more thoroughly. –  Matt Calhoun Feb 19 '11 at 7:45

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