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This is a direct follow-up to Conjecture on irrational algebraic numbers.

Take the decimal expansion for $\sqrt{2},$ but now think of it as the base $11$ expansion of some number $\theta_{11}.$ Is there an easy (or, failing that, hard) proof that $\theta_{11}$ is transcendental? Of course, same question stands for $\theta_k,$ for your favorite $k>10.$

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    $\begingroup$ $\theta_{11}$ is certainly irrational. There is a well-known (but still far from proved) conjecture that all algebraic irrationals are normal in all bases. So (of course) your number is transcendental. But your question (whether there is an easy proof of it) is not answered by this. $\endgroup$ – Gerald Edgar Jul 6 '14 at 22:21
  • $\begingroup$ @GeraldEdgar What is the justification for the well-known conjecture? $\endgroup$ – Igor Rivin Jul 7 '14 at 1:49
  • $\begingroup$ Algebraic irrationality "is independent of" base expansions; there are countably many algebraic irrationals; Therefore "with probability 1", all algebraic irrationals are normal to all bases. $\endgroup$ – Anthony Quas Jul 7 '14 at 4:27
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    $\begingroup$ Since the digits are the same but the base is larger, in a certain sense, $\theta_{11}$ has a quicker rational approximations than $\sqrt 2$. Can this fact affect the irrationality measure of $\theta_{11}$? $\endgroup$ – Pietro Majer Jul 7 '14 at 12:42
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    $\begingroup$ @NikitaSidorov And of course, we all speak for ourselves. $\endgroup$ – Igor Rivin Jul 7 '14 at 16:42

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