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Since $\mathbb{Q}$ is dense in $\mathbb{R}$, given a real $\varepsilon > 0$ and an $\alpha$, there exist $m, n \in \mathbb{Z}$ such that $$\left|\frac{m}{n} - \alpha \right| < \varepsilon \,.$$

However, is it true that, given a real $\varepsilon > 0$ and an $\alpha$, there exist $m, n \in \mathbb{Z}$ such that $$\left|\frac{m}{n} - \alpha \right| < \frac{\varepsilon}{n} \,?$$

I believe that's true. My intuition is that, for big $n$, the multiples of $1 / n$ make up a sufficiently precise "grid" to account for $(\varepsilon /n)$-good approximations of $\alpha$. The problem is that, with bigger $n$, the error $\varepsilon / n$ is also smaller. I'm trying to write an actual argument that justifies this, and any help would be appreciated.

Gonçalo
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hdecristo
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    look at $\alpha n$ modulo $1$ and apply the pigeon principle (also called Dirichlet box principle) when you split $[0,1]$ in boxes of length less than epsilon – Conrad Jan 27 '24 at 20:45
  • Take a look at Thue–Siegel–Roth theorem on diophantine approximation to discern more precisely on this subjet. – Ataulfo Jan 28 '24 at 00:15
  • Related: https://math.stackexchange.com/questions/2557926/dirichlet-approximation-theorem-proof. – Gonçalo May 08 '25 at 02:14

1 Answers1

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If $\varepsilon>1$, the inequality is satisfied with $n=1$ and $m=\lfloor\alpha\rfloor$. If $\varepsilon\leq 1$, use Dirichlet's approximation theorem: for any real numbers $\alpha$ and $N$, with $1\leq N$, there exist integers $m$ and $n$ such that $1\leq n\leq N$ and $$ |n\alpha - m|<\frac{1}{N}. \tag{1} $$ With $N=\frac{1}{\varepsilon}\geq 1$, the inequality $(1)$ is equivalent to $$ \left|\alpha-\frac{m}{n}\right|<\frac{\varepsilon}{n}. \tag{2} $$

Gonçalo
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