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I'm looking for an irrational number which does not have all $b$ distinct digits in its base-$b$ representation and which can be expressed in "closed form" for some reasonable definition thereof.

Acceptable would be +, -, *, /, ^, sin, $\pi$, $e$, and anything else 'reasonable'. Unacceptable would be algorithms, unbound $\sum$ or $\prod$, "write in base 2 and interpret in base $b$", etc.

The problem may be hard, in which case I would be willing to accept an answer explaining that this is so (ideally with a reference).

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  • $\begingroup$ Are you certain that such numbers exist? More to the point - and this may sound a bit crazy, but I think there's actually some justification for it - are you certain that the existence of such numbers isn't dependent on which axioms of set theory you take? $\endgroup$ – Steven Stadnicki Jul 31 '13 at 1:53
  • $\begingroup$ @StevenStadnicki: I'm not at all certain! It would be very interesting if someone could show that this was dependent on nonconstructive axioms, though I have no real hope of seeing that. $\endgroup$ – Charles Aug 1 '13 at 7:09
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Regarding hard: it is conjectured that each irrational element of the ring of periods is a normal number. So, they are excluded. This already excludes a lot. $e$ is also conjectured to be normal. Basically, since everything but a set of (Lebesgue) measure zero is normal, you should really have to resort to constructions (that are in some sense artificial) to avoid normal numbers. So, this excludes most notions of closed form.

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  • $\begingroup$ This is not, technically speaking, an answer but I'm going to accept this as the best evidence I have that such a number probably doesn't exist for a reasonable interpretation of "closed form". $\endgroup$ – Charles Aug 1 '13 at 7:12
  • $\begingroup$ Answer on a PSQ $\endgroup$ – user334732 Dec 8 '18 at 21:27
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Consider $b = 3$ where we have the digits $0$, $1$, and $2$. We are looking for an irrational number (in closed form) which has ternary expansion which uses at most two of the three digits. In fact, it has to use exactly two digits, otherwise the result would be rational.

To simplify the problem somewhat, let's try to find such a number in $[0, 1]$. Now we have to consider which two digits will appear in the ternary expansion; let's try for $0$ and $2$. So the question is:

For which $x \in [0, 1]$ does $x$ have a ternary representation consisting only of $0$ and $2$?

The answer is:

Elements of the Cantor set.

As the Cantor set is uncountable and $\mathbb{Q}$ is countable, there are uncountably many irrational elements of the Cantor set. If you can find one which matches your criteria, you're done. Unfortunately, I don't know whether such elements have ever been explicity found

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    $\begingroup$ I believe the OP is looking for numbers that are 'digitally deficient' in all bases - I presume he's already aware of how to do the deed in just one base and either wants an explicit number which does the deed in all bases or a proof that none exists. $\endgroup$ – Steven Stadnicki Jul 31 '13 at 1:52
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    $\begingroup$ It is up to OP to say, but I expect one base would already be interesting. All bases $\gt 2$ seems awfully ambitious. $\endgroup$ – André Nicolas Jul 31 '13 at 2:15
  • $\begingroup$ Especially since artificial numbers in one base typically look random in other bases. I don't think a nonzero number can avoid the digit $b-1$ in base $b$ for all $b>N$, $N \geq 2$. $\endgroup$ – Eric Tressler Jul 31 '13 at 2:19
  • $\begingroup$ I read it as a condition for a particular base rather than all bases, but I can see that the question could have been intended to mean for all bases. I'll leave my answer here anyway. $\endgroup$ – Michael Albanese Jul 31 '13 at 2:27
  • $\begingroup$ @AndréNicolas: Indeed, I want only one base -- that's already hard enough. $\endgroup$ – Charles Aug 1 '13 at 7:10

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