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I recently learned of so-called 'far' numbers at a talk. In the talk, it was proven that there is a dense subset of the interval $[0,1]$ of far numbers (however, far numbers were only a minor point of the talk, so we quickly went on to bigger and better things).

A number $x$ might be called far if there is a positive constant $c$ s.t. $\displaystyle \left\| \;x - \frac{k}{2^n} \right \| \geq \frac{c}{2^n}$ for all $k, n \in \mathbb{N}$.

It is not so bad to see that for any odd prime $p$, $1/p$ is a far number. For that matter, for most rationals, it's not so hard to see that they are far. It happens to be that the set of far numbers has measure $0$. But I began to wonder:

Is there a 'far' irrational number?

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  • $\begingroup$ I think that any number which is not Liouville number is a far number. Since Liouville numbers have measure zero, there exists irational numbers which are not Liouville, and those are far.... $\endgroup$
    – N. S.
    Commented Feb 14, 2012 at 18:08
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    $\begingroup$ @N.S. "any number which is not Liouville number is a far number"+"Liouville numbers have measure zero"+"It happens to be that the set of far numbers has measure zero"... $\endgroup$
    – SBF
    Commented Feb 14, 2012 at 18:16
  • $\begingroup$ Shouldn't this have a number-theory tag too? $\endgroup$
    – Aryabhata
    Commented Feb 14, 2012 at 18:45

1 Answer 1

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Consider $x = \sum_{j=1}^\infty a_j/4^j$ where each $a_j$ is either $1$ or $2$. Thus the binary expansion of $x$ consists of two-digit blocks which are either $10$ or $01$. Then $x$ is far. But there are uncountably many choices, so all but countably many of them are irrational.

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    $\begingroup$ I think this is a great solution! It's one of things that I should have thought of first, so that I could claim it as my own. $\endgroup$
    – davidlowryduda
    Commented Feb 14, 2012 at 23:07
  • $\begingroup$ A specific example is $\vartheta_3(0,1/4)/2 - 1/6$ where $\vartheta_3$ is a Jacobi theta function: in Maple's notation $\text{JacobiTheta3}(z, q) = 1+2 \sum_{n=1}^\infty q^{n^2} \cos(2 n z)$ $\endgroup$ Commented Feb 15, 2012 at 23:45

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