How do I prove this statement?
$\forall x,y>0\in\mathbb R, \forall\delta >0\exists n_1, n_2\in\mathbb N$ such that $|n_1x-n_2y|<\delta$
I have tried to prove it observing that by the density of rationals numbers wrt real numbers I have $\forall x\in\mathbb R, \forall\varepsilon>0\exists \bar x\in\mathbb Q$ such that $|x-\bar x|<\varepsilon$. Then I can solve the equation above exactly with respect to $\bar x$ and $\bar y$ (namely $n_1\bar x - n_2\bar y = 0$) and by density I get $|n_1x - n_2y|\leq (\varepsilon_xn_1+\varepsilon_yn_2)$, which in my view is not good, since I have an arbitrarily small quantity times an arbitrarily large quantity.
The book I'm reading says that the proof is trivial: can someone help me?
Thanks to the comments I solved the problem.
The crucial result is Dirichlet's approximation theorem, which states that $\forall\alpha\in\mathbb R\forall N\in\mathbb N\ \exists p,q\in \mathbb N$ such that $|p\alpha-q|<\frac{1}{N+1}$. Then one observes (I use the notation of my question) $|n_1x-n_2y| = y\Big|n_1\frac{x}{y}-n_2\Big|<\frac{y}{N+1}$ by the previous theorem.