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Since the Cantor set is uncountable, it must contain irrationals. I am aware that they can't be normal, so the irrationals in the Cantor set are transcendental. Are there any explicit constructions of such numbers, or can we only indirectly show their existence?

Could you please provide the construction for the base $10$?

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    An interesting related question is which _rational_ numbers are in the Cantor set. It's not just the endpoints; it's also $1/4$ and $3/10$ and various others. Of course it's anything whose ternary expansion repeats and has only $0$s and $2$s, but that doesn't say what the denominators are. Denominators of rationals in the Cantor set are on a page at OEIS.2012-01-28

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Certainly there are irrationals in the Cantor set that can be described simply and explicitly, such as the number that has base $3$ expansion $0.200200020000200000200000020\dots.$ If the number were rational, its base $3$ representation(s) would be ultimately periodic. But it isn't, because of the increasing number of $0$'s between consecutive $2$'s.

Added: An interesting related question is whether there is a closed form irrational number in the Cantor set. The meaning of that question is not clear since we have not defined closed form. However, let $\alpha=\sum_0^\infty \frac{2}{3^{n(n+1)/2}}.$ Then $\alpha -2$ is an irrational number in the Cantor set, for basically the same reason as the example we gave in the main post. But $\alpha=\sqrt[8]{3}\; \vartheta(0, 1/\sqrt{3}),$ where $\vartheta$ is the Jacobi $\vartheta$-function. Unfortunately, $\vartheta$ is a pretty exotic function. If we define closed form more narrowly, I do not know what the answer would be.

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    @Nate Eldredge: I sort of take it back about structureless. There seems to be an unusual number of repeated digits, probably because $1/9=0.1111\dots$2012-01-28
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As André indicates in the comments, any non-repeating ternary number containing no instances of $1$ is irrational and in the Cantor set. However, it is unknown whether or not all (and as far as I know any) irrational algebraic numbers are normal in any base. So the question of whether or not all irrationals in the Cantor set are transcendental may be open.

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    cantorset, has it been shown that there are no normal numbers in the Cantor set? As of 2010, it has been proven that the set of numbers normal to no base in the Cantor set has Hausdorff dimension $\log 2 / \log 3$ (see http://arxiv.org/pdf/0909.4251v3.pdf by Ryan Broderick, Yann Bugeaud, Lior Fishman, Dmitry Kleinbock, and Barak Weiss). Your claim is far stronger than what they proved. Or by normal do you mean absolutely normal (in which case it would be trivially true)?2014-03-24