I’m reading “Diophantine approximation” by W.M.Schmidt.
At the Chapter 2, Theorem 1E which is Dirichlet’s approximation Theorem of simultaneous version, He proved the theorem using pigeonhole principle.
And at the remark, he note that it’s not possible to prove in case $Q \not \in \mathbb{N}$.
When I tried to prove the case with the same manner, I couldn’t find the problem.
Here is the proof. Correct me if I did something wrong.
Let $Q > 1$ be irrational number.
We will use $\lceil r \rceil$ ($r \in \mathbb{R}$) which means the minimum integer greater than $r$.
If we take $0 \leq x_{j} < Q^{\frac{n}{m}} \ (1 \leq j \leq m)$, each $x_{j}$ has $\left\lceil Q^{\frac{n}{m}} \right\rceil$ choices.
Therefore $(x_{j})_{1\leq j\leq m}$ has $\left\lceil Q^{\frac{n}{m}} \right\rceil^{m}+1$ candidates including $(1)_{1\leq j \leq m}$.
We can fill $[0,1]^n$ with $\left\lceil Q^{\frac{n}{m}} \right\rceil^{m}$ number of $\prod\limits^{n}_{k=1}I_{k}$ , where $I_{k}$ is length $\left\lceil Q^{\frac{n}{m}} \right\rceil^{\frac{m}{n}}$ partition of $[0,1]$.
So we can get same formula bounded with $\frac{1}{\left\lceil Q^{\frac{n}{m}} \right\rceil^{\frac{m}{n}}}$
Because of $Q < \left\lceil Q^{\frac{n}{m}} \right\rceil^{\frac{m}{n}}$, $\frac{1}{\left\lceil Q^{\frac{n}{m}} \right\rceil^{\frac{m}{n}}} < \frac{1}{Q}$.
So we got what we wanted.