By lattice we mean a finitely generated free abelian group $L$ equipped with an integral non-degenerate symmetric bilinear form $L\times L\rightarrow\Bbb{Z}, \ (x,y)\mapsto x\cdot y$. We call $L$ hyperbolic, if it has signature $(1,r-1)$.
Suppose $L$ is hyperbolic. How to prove that if $x^2>0$ and $y^2>0$ then $x\cdot y\neq0$ ?
When the rank $r=2$, it is an easy direct computation: we can write $x^2=2x_1x_2$ and $y^2=2y_1y_2$ so in the expression $xy=x_1y_2+x_2y_1$ we see that the two summands must have the same sign. What about the general case?
Edit: I might have made a mistake for the case $r=2$ above. In fact I realized that I tacitly assumed $L$ to be even and unimodular and hence isometric to the hyperbolic plane $$U=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$ while this might not be the case for any hyperbolic rank-2 lattice… how does one go?
First of all, you do not need $L$ to be a lattice, the same will work in the vector space $W=L\otimes {\mathbb R}$. Consider the span $V$ of $x$ and $y$ in $W$ and restrict the bilinear form to $V$. This restriction has the signature $(1,1)$ and, hence, you can use the argument you already know.
Edit: Here are the details for the 2d case: Consider two vectors $x, y\in W\cong R^{2}$ such that $x\cdot x>0, y\cdot y>0$ and assume that $x\cdot y=0$. Then the Gramm matrix of your bilinear form (restricted to $W$) in terms of the basis $x, y$ is $$ \left[\begin{array}{cc} x^2 & 0\\ 0 & y^2 \end{array}\right] $$ which has two positive diagonal entries. Therefore, the restriction of your bilinear form to $W$ is positive-definite. This contradicts the assumption that the original form has signature $(1,r-1)$.