Linear independence in $mathbb{Z}_2$ and $mathbb{R}$

Since these vectors have entries in ${ 0, 1 }$, it follows that any linear dependence of such vectors over $Bbb{R}$ is actually (up to a constant multiple) a linear dependence over $Bbb{Q}$, which can easily be turned into a linear dependence over $Bbb{Z}$ by clearing denominators, and then by reducing mod $2$ can be turned into a linear dependence over $Bbb{Z}_2$. (If this linear dependence over $Bbb{Z}_2$ is initially trivial, that means the $Bbb{Z}$-coefficients of the original linear dependence all had a factor of $2$ in them to begin with, which may be cancelled out; if the resulting linear dependence over $Bbb{Z}_2$ is still trivial, we repeat as necessary until not all the $Bbb{Z}$-coefficients are even, at which point we get the desired nontrivial linear dependence relation over $Bbb{Z}_2$.)

But the other direction isn't true: the vectors $[1, 0, 0, 1], [1, 1, 0, 0], [0, 1, 0, 1]$ are linearly dependent over $Bbb{Z}_2$ (their sum is $[0, 0, 0, 0]$), but linearly independent over $Bbb{R}$.