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?

Write scalar multiplication as #. Let $\sqrt2\#1=p/q$. Then $\sqrt2\#q=p$, so $\sqrt2\#(\sqrt2\#q)=\sqrt2\#p=p^2/q$. But $\sqrt2\#(\sqrt2\#q)=(\sqrt2\sqrt2)\#q=2\#q=(1+1)\#q=2q$. So $p^2/q=2q$, contradiction. 
