We can't give the structure of linear space on $\mathbb{Q}(\sqrt{2})$ for a set of rational numbers Please help me prove, that we can't give the structure of a linear space over $\mathbb{Q}(\sqrt{2})$ for a set of rational numbers with standard addition operation.
$\mathbb{Q}(\sqrt{2})$ is a field of numbers $(a + b\sqrt{2}), \text{where  } a,b\in \mathbb{Q}$
Edit
We can assume, that in this case "1" is a neutral element for multiplication? The neutral element must be defined along with field definition or operation definition and I have freedom to choose it?

In a vector field sturcture, there are given 'scalar' type entities and 'vector' type entities. Now scalars are elements of the field $\Bbb Q(\sqrt 2)$ and vectors should be elements in some $U\subseteq \Bbb Q$, if I am right.
$1$ is present in the field, and of course acts as unit, but multiplying vector by vector is not defined.
The main point is, that if $v\in U\subseteq\Bbb Q$, and $x\in \Bbb Q\subseteq \Bbb Q(\sqrt 2)$, then
$x\cdot v$ in this vector space structure has to be $x\cdot v$. You can see this by writing $x=n/m$.
And therefore all $\Bbb Q$ is used up with a given $v\ne 0$, and then, "there is no more space for" $\sqrt 2\cdot v$, though this should differ to all $x\cdot v$ for $x\in\Bbb Q$.
