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Since 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? Thanks again.

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Sorry, I meant base_10 expansions, forgot to specify. – cantorset Jan 28 '12 at 1:38
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. – Michael Hardy Jan 28 '12 at 4:56
up vote 5 down vote accepted

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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Sorry, I meant base_10 expansions, forgot to specify. – cantorset Jan 28 '12 at 1:38
@cantorset: André has used ternary expansions because of the nice characterization of elements of the usual Cantor set as those elements of $[0,1]$ not containing a $1$ in some ternary expansion. Thus it is natural to give the answer as a ternary expansion. As with any real number, you can convert André's answer to base 10 if you want (or choose a different Cantor set which is better adapted to decimal expansions). But giving an irrational number by its ternary expansion certainly counts as an "explicit construction"! – Pete L. Clark Jan 28 '12 at 1:45
There is an explicit algorithm to convert the ternary number above to decimal. – André Nicolas Jan 28 '12 at 1:48
0.743518234716047631162591395743236508537458617409028492242942029887873113897604‌​290655219506942231313224728388961234176742266448916358824448821277574456351111637‌​601216470868144693850267451136656708073649568782164121285016361248958784444411497‌​727979416399918073812121368613024850582033807547641809074634170536753398707712555‌​589104601090603721049692752883902340565206794595269772044562193657066762746318051‌​109373786108156255957228680173839129475552600524462318811584708238967071857755371‌​846633623065615483281056398149073408191245450049887452116801365348289782465887094‌​1918601963689801969747536097135... – Nate Eldredge Jan 28 '12 at 4:22
@Nate Eldredge: For a number of years, I have tried (with probably total lack of success) to persuade students not to go to the calculator too early. The calculator produces a structureless stream of digits, and mainly we solve problems by using structural information. The digit stream above is a very nice illustration! – André Nicolas Jan 28 '12 at 4:29

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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So, when he said "they can't be normal, so the irrationals in the Cantor set are transcendental" he was using something unknown. – GEdgar Jan 28 '12 at 1:49
Unknown to me at least. – Dan Brumleve Jan 28 '12 at 1:53
I was actually using Wikipedia, missed it was a conjecture not a proven fact: It has been conjectured that all algebraic irrational numbers are normal. Since members of the Cantor set are not normal, this would imply that all members of the Cantor set are either rational or transcendental. thanks a lot by the way! – cantorset Jan 28 '12 at 1:55
There may be some shortcut way of proving all irrationals in the Cantor set are transcendental that doesn't require proving all irrational algebraic numbers are 3-normal. All we would need to show is that every irrational algebraic contains at least one $1$ when written in base $3$. – Dan Brumleve Jan 28 '12 at 2:15
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 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)? – Bill Mance Mar 24 '14 at 5:01

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