Positive-definite over $\mathbb Q$ form is positive definite over $\mathbb R$? I was reading P. Etingof's "Introduction to the representation theory" when I found this problem (and I've trouble with it): we have a quadratic form $Q(x) = \sum x_i^2 - \frac{1}{2}\sum b_{ij}x_ix_j$, $b_{ij} \in \mathbb N$ (if it can be important, $b_{ij}$ is an adjacency matrix of some graph) and $Q$ is positive-definite over $\mathbb Q$.
Now our goal is to prove that $Q$ is positive-definite over $\mathbb R$.
It's obvious that it's at least positive-semidefinite (Etingof's hint is to use that $\mathbb Q$ is dense in $\mathbb R$ and here it works). But why positive-definite?
My idea was to use the Sylwester criterion two times... But is it true for rational numbers?

The map $Q: \mathbb{R}^n \to \mathbb{R}$ is continuous, and for every nonzero rational $v$, we have that $Q(v) > 0$. Hence around every rational nonzero $v$, there is a neighbourhood where $Q > 0$. Every nonzero real $v$ falls into one of these neighbourhoods.
