Given $\varepsilon > 0$ and an irrational $\alpha$, there exists a rational number $m / n$ that is ($\varepsilon / n$)-close to $\alpha$?

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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.