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Let's say we have continued fractions of irrational numbers of the form $$ [a_0, a_1, a_2,...]: a_0 \in \mathbb{Z}, a_i \in \{1,2\}. $$ Is there any way to determine a number, say $x\in [0,1],$ that cannot be written in the form described above?

EDIT: What about $x\in [0,1],$ that cannot be written as a sum x=a+b, where $a,b$ are of the above form?

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  • $\begingroup$ You only need an irrational number with an entry larger than $2$. Nearly all irrational numbers (like $\pi$ and $e$) have this property. Is this actually the intent of this question ? $\endgroup$
    – Peter
    Commented May 22, 2019 at 10:32
  • $\begingroup$ @Peter Well I'm not sure how to prove that those numbers can't be written in the the form above. $\endgroup$
    – Nom
    Commented May 22, 2019 at 10:39
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    $\begingroup$ Irrational numbers have a unique simple continued fraction. This is well known, but I am not sure whether you can use it. A rational number has two possible representations, for example $[2,4,1]=[2,5]$ , but apart from this, the representation is also unique. Maybe, proofwiki gives a proof. $\endgroup$
    – Peter
    Commented May 22, 2019 at 10:42
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    $\begingroup$ Won't such a continued fraction always be between $[2,1,2,1,2,1,\ldots] = \frac12(\sqrt3-1)$ and $[1,2,1,2,1,2,1,\ldots]=\sqrt3-1$? So for example $\frac13$ cannot possibly be written as a sum of any number of them. $\endgroup$ Commented May 22, 2019 at 10:55
  • $\begingroup$ @hmakholmleftoverMonica $\frac 13$ has a particularly special continued fraction, as well as all the thirds :) $\endgroup$
    – Mr Pie
    Commented Mar 4, 2020 at 11:26

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You can certainly generate irrational numbers in $[0,1]$ that don't have continued fractions satisfying your criterion. Note that irrational numbers have unique continued fractions. A continued fraction $[a_0;a_1,a_2,...]$ with $a_0=0$ lies in $[0,1)$ irrespective of subsequent partial denominators. Thus, any simple infinite continued fraction of the form $[0;a_1,a_2,...]$ where one or more $a_i>2,i>0$ uniquely represents an irrational in $[0,1)$.

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