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Suppose $S$ is set of numbers such that every number in it expands in decimal digits,every digit is 0 or 1,and $\lim_{n\rightarrow\infty}\frac{C_{n}(0)}{n}=\frac{1}{2}$ where ${C_{n}(0)}$ and ${C_{n}(1)}$ are the counting numbers of 0,and 1 from first digit to n digit respectively.

Now, is there any irrational algebraic number $x\in S$?. Obviously if there is one, there must be other irrational algebraic number

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  • $\begingroup$ @Fan Zheng Thank you for your editing $\endgroup$
    – XL _At_Here_There
    Commented Aug 20, 2017 at 3:38
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    $\begingroup$ It is conjectured that every irrational algebraic number is normal, so even without your limit condition one expects the answer to be no. Of course this normality conjecture is very hard -- your question is clearly easier, though not obviously (to me) within reach of current techniques. $\endgroup$
    – Daniel Litt
    Commented Aug 20, 2017 at 4:11
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    $\begingroup$ I was just about to say what Daniel just said. See en.wikipedia.org/wiki/Normal_number#Properties_and_examples It seems your question would be an open problem (not really in the purview of MO). $\endgroup$
    – Todd Trimble
    Commented Aug 20, 2017 at 4:14
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    $\begingroup$ Again, as I said in my first comment, it's a notoriously unsolved problem whether the conjecture is true, and such problems are generally not considered on-topic for MO. See mathoverflow.net/help/on-topic $\endgroup$
    – Todd Trimble
    Commented Aug 20, 2017 at 4:39
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    $\begingroup$ So there's already a discussion at Math.SE? Started an hour ago? math.stackexchange.com/questions/2399799/… Cross-posting is very much frowned upon. Together with the fact that famous well-known problems are generally off-topic, this problem should be closed here. $\endgroup$
    – Todd Trimble
    Commented Aug 20, 2017 at 4:54

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