Let's consider $A$ the subspace of $\mathbb R^4$ given by $A=\mathrm{Span}(Y_1,Y_2)$ where
$$\begin{cases} Y_1=(0,1,\sqrt 2,\sqrt 3), \\ Y_2=(1,0,\sqrt 3,\sqrt 2).\end{cases}.$$
I was wondering about the answer of the following question:
Does there exist a rational subspace $B$ of $\mathbb R^4$ of dimension $2$, such that $\dim(A\cap B)\geqslant 1$?
Some context.
By rational subspace, I mean a subspace of $\mathbb R^4$ which admits a rational basis, in other words a basis $(a,b)$ with $a,b\in\mathbb Q^4$.
I conjecture the answer to be no, but I can't prove it so far.
We can reformulate the problem the following way: does there exist a vector $v\in A\setminus\{0\}$ such that
$$\exists \xi_1,\xi_2\in\mathbb R,\quad \exists r_1,r_2\in\mathbb Q,\quad v=r_1\xi_1+r_2\xi_2\quad ?$$
- We can write the problem in coordinates, which leads us to find $\alpha,\beta,\xi_1,\xi_2\in\mathbb R$ and $a_1,a_2,b_1,b_2,c_1,c_2,d_1,d_2\in\mathbb Q$, such that
$$\alpha Y_1+\beta Y_2=a \xi_1+b\xi_2,$$
which gives
$$\begin{cases} \beta=a_1\xi_1+b_1\xi_2 \\ \alpha=a_2\xi_1+b_2\xi_2 \\ \alpha\sqrt 2+\beta\sqrt 3=a_3\xi_1+b_3\xi_2 \\ \alpha\sqrt 3+\beta\sqrt 2 = a_4\xi_2+b_4\xi_2,\end{cases}$$
but it does not seem to help at all.
Since I am quite stuck with this problem, any help, hints, or new ways to look at the problem would be much appreciated.
$$ (\sqrt2 + \sqrt3) (Y_1-Y_2) = \pmatrix{ 1\\ -1 \\ \sqrt2 + \sqrt3 \\ -(\sqrt2 + \sqrt3)} = \pmatrix{ 1\\ -1 \\0\\0} + (\sqrt2 + \sqrt3)\pmatrix{0\\0\\1\\-1} $$