Let $x$ be a real number, and let $R$ be the set of positive real numbers $\mu$ for which $$0<|x-\frac{p}{q}|<\frac{1}{q^{\mu}}$$ has (at most) finitely many solutions $p/q$ for $p$ and $q$ integers. Then the irrationality measure for $x$ is defined as $$\mu(x):=\inf_{\mu \in R} \mu$$ How to prove that $$x\in \mathbb{Q} \Longleftrightarrow \mu(x)=1$$
Any hint would be appreciated.
Hint:
To prove $\mu(x) = 1$ for $x$ rational take $x=a/b$ (we will assume wlog that $b,q \geq 0$).
First note that
$|x-p/q| = \frac{|aq-bp|}{bq} \geq \frac{1}{bq}$
Now if $0<|x-p/q| < \frac{1}{q^\mu}$ then $q^{\mu-1} < b$.
If we take $\mu > 1$ then show that this cannot hold for large enough $q$. For a given $q$ show that there are only a finite numbers of $p$'sh that is possible. Conclude that there can only be finitely many pairs $q,p$ that is possible and therefore $\mu(x) \leq 1$.
If $\mu < 1$ show that we can find infinitely many pairs $q,p$ that satisfy
$|aq-bp| < bq^{1-\mu}$
Conclude that $\mu(x) \geq 1 \to \mu(x) = 1$.