Every publication about the irrationality measure $\mu(\alpha)$ mentions as an obvious fact that $\mu(r) = 1$ for a rational $r$.
Since I am new to this topic, it didn't look quite obvious to me, so I tried to prove.
This is what I've got (please, check that it's allright):
For a rational number $r = \frac{a}{b}$ we need to prove that for any positive $\varepsilon$ the set of fractions $\frac{p}{q}$, satisfying inequality
$$
0 \lt \left| r – \frac{p}{q} \right| \lt \frac{1}{q^{1+\varepsilon}} \qquad (1)
$$
is finite.
Firstly, since $1 \geq \frac{1}{q} \geq \frac{1}{q^{1+\varepsilon}}$, the inequation (1) never holds if $\left| r – \frac{p}{q} \right| \geq 1$, therefore we can safely assume that
$$
0 \lt \left| r – \frac{p}{q} \right| \lt 1,
$$
or
$$
q \left(r – 1\right) \lt p \lt q \left(r + 1\right) \qquad (2)
$$
Then from
$$
\left| r – \frac{p}{q} \right| = \left| \frac{a}{b} – \frac{p}{q} \right| = \frac{|aq-bp|}{bq} \geq \frac{1}{bq}
$$
it follows that (1) is not fulfilled if $q \gt b^{\frac{1}{\varepsilon}}$ (i.e. $b\lt q^\varepsilon$), therefore number of denominators satisfying (1) is finite. Further on, from (2) it follows that the number of numerators
for given denominator is finite. This proves that $\mu(r) \leq 1$.
I've got stuck proving that $\mu(r) \geq 1$. I expect it to be true for any number, not only rational. As far as I can see, that suggests that for any $\varepsilon \gt 0$ (or maybe even $\varepsilon \geq 0$), no matter how big $k$ is, there can be found $q \geq k$, so that
$$
\left| x – \frac{p}{q} \right| \lt \frac{1}{q^{1-\varepsilon}}
$$
has a solution.
Best Answer
Let $x$ be a real number. Then $0 \leq \left| x - \frac{p}{q} \right| \leq \frac{1}{2} \cdot \frac{1}{q}$ for all nonzero integers $q$ because the set of rational numbers with denominator $q$ partitions the reals into intervals of length $1/q$ and so $x$ can never be more than half a $1/q$ from the nearest such rational. This means the set of rational numbers of denominator $q$ within half of $1/q$ of $x$ contains either one or two members and is therefore finite. If we shrink the interval further, then the midpoints of the intervals have no such fractions within that distance. Therefore, $\frac{1}{2} \cdot \frac{1}{q}$ is the lower bound to have a nonempty set.
Then $$ \frac{1}{2} \cdot \frac{1}{q} = \frac{1}{q^{\log_q(2)}} \cdot \frac{1}{q} = \frac{1}{q^{1 + \log_q(2)}} \text{.} $$ As $q$ increases, $\log_q 2$ goes to zero, so $\mu$ takes the value $1$ on the rationals.