Dirichlet approximation theorem proof

Question

I want to show that

For all irrational numbers $a \in \mathbb{R}$ there exists an $Q \in \mathbb{N}$ such that $|a-\frac{p}{q}| \lt \frac{1}{qQ}$ for $p,q \in \mathbb{Z}$ and $1 \le q \le Q$.

My attempt:

Let $a \in [0,1)$ and $Q \in \mathbb{N}$. Divide $[0,1)$ in $Q$ disjunct subintervals $I_i := [\frac{i}{Q}, \frac{i+1}{Q})$ for $i \in \{0,1,\ldots, Q-1\}$. Consider the fractional parts $\{na\} \in [0,1)$ of the number $na$ for $n \in \{0,1,\ldots, Q\}$. Since there are in total $Q+1$ such fractional parts but only $Q$ subintervals, by the Pidgeonhole Principle there must be at least one subinterval that contain at least two fractional parts. Let such two fractional parts be $\{na\}, \{ma\} \in I_i$. Wlog let $m \lt n$. Then

$$|\{na\} - \{ma\}| \lt \frac{1}{Q}$$

since those two fractional parts are in the same $I_i$ which has a width of $\frac{1}{Q}$. By the definition of fractional parts

$$\{na\} - \{ma\} = na - \lfloor na \rfloor - (ma - \lfloor ma \rfloor) = na - ma - \lfloor na \rfloor + \lfloor ma \rfloor = aq - p$$

for $q := n-m$ and $p := \lfloor na \rfloor - \lfloor ma \rfloor$. Also $p,q \in \mathbb{N}$, because $m \lt n$. It follows then

$$|aq - p| \lt \frac{1}{Q} \iff |a - \frac{p}{q}| \lt \frac{1}{qQ}$$

If $a \ge 1$ then we can write $a = a' + k$ for some $a' \in [0,1)$ and $k \in \mathbb{N}$. Then with $p' = p+qk \in \mathbb{N}$

$$|a - \frac{p}{q}| = |a' + k - \frac{p}{q}| = |a' - \frac{p+qk}{q}| = |a' - \frac{p'}{q}| \lt \frac{1}{qQ}$$

by the first case.

If $a \lt 0$ then the rational number must be negative too, so let $p \in \mathbb{Z}$ and $p \le 0$. Then with $a' := -a \gt 0$ and $p' := -p \ge 0$

$$|a - \frac{p}{q}| = |-(a' - \frac{p'}{q})| = |a' - \frac{p'}{q}| \lt \frac{1}{qQ}$$

by the first case again.

My question:

Is my proof correct? Is there anything which could be done better?

I'm still not really sure about the dependencies. $Q$ is really $Q_a$ so so only dependend on $a$?! And is $p,q$ really $p_a,q_a$ only dependent on $a$ too or also on $Q$?

Could I instead take numbers $na$ for $n$ from another subset than $\{0,1,\ldots, Q\}$ provided it has the same cardinality?

How can I guarantee that $p' \in \mathbb{N}$ because $k$ could be a negative integer. Also I think the final argument I gave, why considering $a \in [0,1)$ is sufficient, is not really the essence.

Answer 1

Firstly your statement is simply wrong for negative $a$, take $a = - \pi$ for example. Then for all positive integers $p$ and $q$ we always have $$ |a - \frac{p}{q}| \geq \pi, $$ clearly violating the stated inequality. If you want to allow negative $a$ in Dirichlet's approximation theorem, then you must also be willing to accept negative $p$.

Your proof looks correct to me and very similar to the standard proofs of Dirichlet's approximation theorem. Note that $Q, p, q$ all depend on $a$, but this is allowed: look at the order of the quantifiers in the theorem!

A lot of other subsets than just $\{0, 1, \ldots, Q\}$ work, for example $\{10, 11, \ldots, Q + 10\}$ works perfectly fine as well.