We say a real number $x$ has bounded partial quotients if its continued fraction expansion $[a\_0; a\_1, a\_2 \cdots]$ is bounded by some constant $M=M(x)$. The set $A$ consisting of those numbers whose partial quotients are bounded forms a dense, uncountable subset of $\mathbb{R}$ which includes the algebraic numbers of degree $\leq 2$. It appears that it is an open problem whether or not $A$ contains any algebraic numbers of degree $>2$.

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> Question : Is it known whether or not the set $A$ has measure zero?
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The reason this question is interesting has to do with Diophantine approximation. We say a real number $x$ is *badly-approximable* if there exists a positive constant $C=C(x)$ such that $$\left|x-\frac{p}{q}\right|>\frac{C}{q^2}$$

for all rational $p/q \neq x$. Here, the name fits, since for *any* irrational $x$, $|x-p/q|<\frac{1}{q^2}$ for infinitely many pairs $(p,q)$, and badly-approximable numbers are precisely those for which we cannot do better than this, i.e., merely scaling the numerator by a certain constant $C$ ruins everything.

It turns out that the property defining the set $A$ discussed earlier (i.e., bounded partial quotients) is in fact completely equivalent to this property of being badly-approximable. Thus asking about the size of $A$ is essentially asking about the size of the set of worst-approximable real numbers.