A number $\alpha$ is said to satisfy the Diophantine condition with
exponent $\beta$ iff for some constant $C>0$ the estimate
$$
\left| \alpha - \frac{p}{q} \right| > \frac{C}{q^{2+\beta}}
$$
holds for every rational fraction $p/q \in \mathbb{Q}$.


A question:


1. Is it true that if $\alpha$ satisfies the Diophantine condition
with exponent $\beta$ then the number $1/\alpha$ also satisfies the
Diophantine condition with some exponent.



Inverse to a Diophantine number
Of course, with the same $\beta$. Note that in the definition we may restrict only to the fractions $p/q$ satisfying inequality $|\alpha-p/q|<|\alpha|/2$. Next, $$\left|\frac1{\alpha}-\frac qp\right|=\left|\alpha-\frac pq\right|\cdot|\alpha^{-1}|\cdot \left|\frac qp\right|.$$
If $|\alpha^{-1}-q/p|<|\alpha^{-1}|/2$, the first multiple may be estimated from below as $$\frac C{q^{2+\beta}}=\frac{C(q/p)^{2+\beta}}{p^{2+\beta}}>\frac{C(2\alpha)^{-(2+\beta)}}{p^{2+\beta}},$$
two other multiples are bounded from below.


