Every nonzero vector in $\mathbb{R}^n$ can be scaled to be arbitrarily close to $\mathbb{Z}^n-\{0\}$. Problem. Show that for all nonzero $v\in\mathbb{R}^n$ and all $\epsilon&gt;0$, there exists $r&gt;0$ such that $\operatorname{dist}(rv,\mathbb{Z}^n-
\{0\})&lt;\epsilon.$


In other words, we must show that for all $\epsilon&gt;0$ there exists $w\in\mathbb{Z}^n-\{0\}\subseteq\mathbb{R}^n$ such that $|v-w|&lt;\epsilon$.


  Bonus question. Can we take $w$ to be arbitrarily large?


Attempts. The case $n=1$ is trivial: we can take $r=1/v$ so that $rv=1$. But that approach obviously doesn't generalize to $n&gt;1$. Any idea?

More generally, if $v$ is rational, say $v_i=a_i/b_i$, then we can take $r=b_1\cdots b_n$ so that $rv\in\mathbb{Z}^n-\{0\}$.
What you refer to as the 'dist()' function is more commonly referred to as the vector norm, a function which takes a vector and produces the 'length' of a vector.  When you scale a vector by a constant, the norm of the vector scales by that constant (at least for the usual norm in this case).  It follows that the argument given for n=1 can be used nearly verbatim, all you need to do is translate from scaling the vector to scaling the norm.
What you refer to as the 'dist()' function is more commonly referred to as the vector norm, a function which takes a vector and produces the 'length' of a vector.  When you scale a vector by a constant, the norm of the vector scales by that constant (at least for the usual norm in this case).  It follows that the argument given for n=1 can be used nearly verbatim, all you need to do is translate from scaling the vector to scaling the norm.