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Let $x$ and $y$ be some irrational numbers. If the degree of irrationality of $x$ is the same as that of $y$, is it necessarily the case that $x$ and $y$ are rationally dependent ?

ADDENDUM: What if $x$ and $y$ are transcendental, specifically logarithms of some rational numbers ?

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    $\begingroup$ If true, this would imply that only countably many real numbers can have a specified irrationality measure, and this is certainly false when the irr. mes. is $2$ (all but a Lebesgue measure zero set of real numbers have this irr. mes.) and when the irr. mes. is $\infty$ (all but a meager set of real numbers have this irr. mes.). More generally, I'm pretty sure that for each irr. mes. $r$ there exist continuum many real numbers in every open interval having irr. mes. $r.$ (moments later) Maybe I misunderstand what you mean by "degree of irrationality"? $\endgroup$
    – Dave L Renfro
    Apr 27, 2021 at 17:34
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    $\begingroup$ @DaveLRenfro, by ''degree of irrationality'' of the number $\alpha \in \mathbb{R}$, i mean the supremum of the real numbers $\theta$ such that $|\alpha - p/q| \leq q^{-\theta-\varepsilon}$ for every $\varepsilon > 0$ and infinitely many rational numbers $p/q$ (Liouville-Roth)... $\endgroup$
    – PRIMES is in P.
    Apr 27, 2021 at 18:02
  • $\begingroup$ Isn’t it likely that all logarithms of rationals $>1$ have irrationality measure $2$? $\endgroup$
    – Emil Jeřábek
    Apr 27, 2021 at 18:39

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