# Irrationality measure of powers

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 $$\frac pq\in\mathbb Q^*$$?

• 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 $e$. – Manuel Norman Oct 14 at 17:09
• 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. – Will Sawin Oct 14 at 17:25

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 transcendantal. 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 $$2$$.
• 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 $\mu(\alpha)$? – joaopa Oct 14 at 19:03