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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.

cyanide
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2 Answers2

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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.

Eric Towers
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Here is another solution to this old problem which uses elementary number theory results.

Suppose $x=\frac{a}{b}$ with $a,b\in\mathbb{Z}$, $b>0$ and $\operatorname{g.c.d}(a,b)=1$.

For any $(x,y)\in\mathbb{Z}\times\mathbb{N}$ such that $\frac{x}{y}\neq\frac{a}{b}$ \begin{align} \frac{1}{bx}\leq \frac{|ax-by|}{bx}=\Big|\frac{a}{b}-\frac{x}{y}\Big|\tag{0}\label{zero} \end{align}

Hence, for no $\varepsilon>0$ is there no constant $c>0$ such that $$\Big|\frac{a}{b}-\frac{x}{y}\Big|\leq \frac{c}{y^{1+\varepsilon}}$$ has infinitely many solutions $(x,y)\in\mathbb{Z}\times\mathbb{Z}$, since $y^\varepsilon\xrightarrow{y\rightarrow\infty}\infty$. This shows that \begin{align} \mu(a/b)\leq 1\tag{1}\label{one} \end{align}

To show that $\mu(a/b)\geq1$, notice that there are infinitely integer solutions to the equation \begin{align} qa-pb=1\tag{2}\label{two} \end{align} Indeed, one such pair $(p_0,q_0)$ exists by virtue of $\operatorname{g.c.d}(a,b)=1$. Other solutions $(p,q)$ can be obtained by setting $p=p_0+na$ and $q=q_0+nb$, $n\in\mathbb{Z}$. In particular, there are infinitely many solutions $(p,q)\in\mathbb{Z}\times\mathbb{N}$ exists to \eqref{two}. Hence, infinitely may solutions $(p,q)\in\mathbb{Z}\times\mathbb{N}$ exists for \begin{align} \Big|\frac{a}{b}-\frac{p}{q}\Big|=\frac{1}{bq}\leq\frac1{q}\tag{3}\label{three} \end{align} From \eqref{three}, it follows that \begin{align} \mu(a/b)\geq 1\tag{4}\label{four} \end{align}

\eqref{one} and \eqref{two} implies the desired result.

Mittens
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