8
$\begingroup$

Let $\alpha$ be an irrational number. Denote by $\mu(\alpha)$ its irrationality measure. Can one say anything about $\mu(\alpha^n)$ for every $n\in\mathbb N$?

Even more, one knows that $\mu(e)=2$. Can one say anything for $\mu(e^{p/q})$ for $\frac pq\in\mathbb Q^*$?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Reguarding your second question: it is known that, for all positive integers $k$, $\mu(e^{2/k})=2$. I do not know if this is also true for other rational powers of $e$. $\endgroup$
    – Manuel Norman
    Commented Oct 14, 2020 at 17:09
  • 1
    $\begingroup$ One can say that if there exist good rational approximations to $\alpha$ then there also exist good rational approximations to $\alpha^n$. But the converse is presumably not true - maybe the square root of Liouville's number already gives a counterexample. $\endgroup$
    – Will Sawin
    Commented Oct 14, 2020 at 17:25

1 Answer 1

Reset to default
3
$\begingroup$

Let $\alpha$ be irrational. There are two cases: $\alpha$ can be either algebraic or transcendental. Products of algebraic numbers are algebraic, while rational powers of transcendental numbers are transcendental. Hence, for all positive integer $n$, $\alpha^n$ is algebraic in the first case, and transcendental in the second one. Roth proved that algebraic irrational numbers all have irrationality measure $2$. Instead, little is known about transcendental numbers. We can only say, in general, that the irrationality measure is $\geq 2$. Thus, by the previous discussion, we can conclude that:

$\alpha$ irrational algebraic $\Rightarrow$ $\mu(\alpha^n)=2$ for all integers $n \geq 1$ (in fact, this also holds for all nonzero rationals $n$, since roots of algebraic numbers are algebraic).

$\alpha$ transcendental $\Rightarrow$ $\mu(\alpha^n) \geq 2$ for all integers $n \geq 1$ (in fact, as before, this also holds for all nonzero rationals $n$).

I think that the actual value of the irrationality measure of a power of a transcendental number highly depends on the particular case. However, it is worth recalling that almost (in the sense of Lebesgue measure) all irrational numbers have irrationality measure $2$.

Improve this answer
$\endgroup$
4
  • 2
    $\begingroup$ Thank you for the case $\alpha$ irationnal. In the transcendental case, at least can one obtain a upper bound of $\mu(\alpha^n)$ depending on $\mu(\alpha)$? $\endgroup$
    – joaopa
    Commented Oct 14, 2020 at 19:03
  • $\begingroup$ That's an interesting question. I think that it might be possible. Maybe an upper bound can be derived using the limit formula involving the convergents of the simple continued fractions (of the powers, in this case) found by Sondow. This is just an idea - I have not checked if it works $\endgroup$
    – Manuel Norman
    Commented Oct 14, 2020 at 19:54
  • 4
    $\begingroup$ I wonder why the downvote. Maybe I'm missing something, but I don't see anything here that would merit downvoting, even if this states what some might consider as relatively obvious and well-known -- both of which are highly dependent on the reader and thus within the scope of one of the original purposes of this site. Regarding your last sentence, it's also worth recalling that almost all (in the sense of Baire category) irrational numbers (or even real numbers) have irrationality measure $\infty,$ so their abundance depends on the notion of largeness one uses. $\endgroup$
    – Dave L Renfro
    Commented Nov 9, 2021 at 7:27
  • 2
    $\begingroup$ FYI, the same dichotomy of Leb. measure and Baire category for irrationality measure holds for normal numbers -- Lebesgue-almost-all real numbers have irrationality measure $2$ (min. value, when the inconsequential countably many alg. numbers are omitted) and are normal (minimum digit frequency variation), and Baire-almost-all real numbers have irrationality measure $\infty$ and are highly non-normal (see Olsen and Stylianou), so this answer might be of interest. $\endgroup$
    – Dave L Renfro
    Commented Nov 9, 2021 at 7:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .