# Inverse to a Diophantine number [closed]

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.

## closed as off-topic by Alexey Ustinov, js21, David Handelman, Felipe Voloch, Denis SerreDec 8 '17 at 15:16

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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.