Can there be a line in $\mathbb{R}^n$ whose only rational point is $ {0} $?

Question

This question popped up when I had to prove that a linear transformation whose rational null space points were only ${0}$ was in fact an isomorphism.( in a specific context where the linear transformation could be restricted to $\mathbb{Q}^n$ so it was quite simple and didn't imply the result).
So I tried proving that if the kernel were to contain a line in $\mathbb{R}^n$ then this line would necessarily pass by some rational points other than $0$.
But so far I couldn't prove it, so now I am questioning its validity even though I couldn't construct any counter example.
Any ideas?

Answer 1

A non-constructive argument is also easy.

Let $S$ be the unit sphere in $\Bbb R^n$, and let $L$ be the set of lines through the origin that contain at least one other rational point. Each $\ell\in L$ meets $S$ in a two-element set $P_\ell$. Let $P=\bigcup_{\ell\in L}P_\ell$; $L$ is countable, so $P$ is countable. For each of the uncountably many $x\in S\setminus P$ the line through $x$ and the origin contains no rational point other than the origin.

Answer 2

You can take the line $\left\{t\left(1,1,\ldots,1,\sqrt2\right)\,\middle|\,t\in\Bbb R\right\}$. Its only point such that all of its coordinates are rational is $(0,0,\ldots,0)$.