13

It is well known that the Cantor set is uncountable. Hence it contains irrationals. What are the 'nice' irrationals in the Cantor set.

Here, I am expecting irrational numbers in the form of square roots of $\frac{1}{n}$, cube roots of $\frac{1}{n}$, or their combinations, or $\pi/n$, $e/n$ ($n\in\mathbb{N}$), or rational powers of $e$, $\pi$, or any such nice form. (In fact we can take a number with ternary expansion with $0$'s and $2$'s, which is not repeating; but I would like to see numbers not in ternary form.)

Bart Michels
  • 26,355
Groups
  • 10,238
  • Here Cantor set is the standard Cantor set, obtained my removing middle $1/3$rd parts, successively, from $[0,1]$, $\cdots$. – Groups Nov 06 '14 at 08:08
  • 1
    I don't think there is a proof that any 'nice' irrational is not absolutely normal, let alone being an element of the Cantor set. – bof Nov 06 '14 at 08:08
  • What is 'absolutely normal'? – Groups Nov 06 '14 at 08:08
  • Wikipedia – bof Nov 06 '14 at 08:11
  • Roughly speaking, a number is absolutely normal if its digits to any base (in particular base $3$) are randomly distributed. For a number to belong to the Cantor set, its base $3$ representation is all $0$s and $2$s, which is highly non-random. Nobody has ever proved that the digits of any 'nice' irrational number are have a non-random distribution. – bof Nov 06 '14 at 08:16
  • @bof: I can't help but nitpick: by "random" you mean "uniform". –  Nov 06 '14 at 08:19
  • 1
    @Hurkyl: Can't help but nitpick the term "random"? You've come to the right website! ;-) – Asaf Karagila Nov 06 '14 at 08:20
  • @Hurkyl: I write A, I say B, I mean C, but it should be D, okay?? I said "roughly speaking" because I was too lazy to think how to say it right. Um, for each k, all possible blocks of length k have the same asymptotic density, have I got it right now?? Thanks for the correction! – bof Nov 06 '14 at 08:28
  • An interesting conjecture is that all irrationals in the Cantor set are transcendental. See here. – Andrés E. Caicedo Feb 14 '16 at 04:50

1 Answers1

8

Define the Jacobi theta function (or whatever this variant is called): $$\theta(q)=\sum_{n=-\infty}^\infty q^{n^2}=1+2\sum_{n=1}^\infty q^{n^2}.$$ Then $$\theta\left(\frac13\right)-1=2\sum_{n=1}^\infty 3^{-n^2}=0.2002000020000002\ldots_3$$ is an irrational number in the middle-thirds Cantor set. You can decide how 'nice' $\theta(1/3)$ is!

Chris Culter
  • 26,806