Is it true that ${O}_n({\mathbb R}) \cap {\mathbb Q}^n$ is dense in ${\cal O}_n({\mathbb R})$ for any $n\geq 2$ ? It obviously suffices to consider the density of ${SO}_n({\mathbb R}) \cap {\mathbb Q}^n$ in $SO_n({\mathbb R})$.
This is easy for $n=2$ : consider the matrices of the form
$$ \frac{1}{p^2+q^2}\left(\begin{array}{cc} p & -q \\ q & p \end{array}\right) $$
where $p,q$ are integers.
By the Cartan–Dieudonné theorem, it suffices to prove this for reflection (given by Householder matrices).
Let $H= I_n - 2\frac{vv^T}{\|v\|^2}$. Now let $v\in \mathbb R^n\setminus\{0\}$, $\tilde v \in \mathbb Q^n$ such that $\|v-\tilde v\|_2\le \|v\|/2$. Let $x\in \mathbb R^n$. Then $$ \begin{split} \left((I_n-2\frac{vv^T}{\|v\|^2}) - (I_n -2\frac{\tilde v\tilde v^T}{\|\tilde v\|^2})\right)x &= 2\frac{v^Tx}{\|v\|^2} v - 2\frac{\tilde v^Tx}{\|\tilde v\|^2}\tilde v\\ & = 2\frac{v^Tx}{\|v\|^2}(v-\tilde v) + 2\left(\frac{v^Tx}{\|v\|^2} -\frac{\tilde v^Tx}{\|\tilde v\|^2} \right)\tilde v\\ & = 2\frac{v^Tx}{\|v\|^2}(v-\tilde v) + 2\frac{(v-\tilde v)^Tx}{\|v\|^2} +2\left(\frac1{\|v\|^2}-\frac1{\|\tilde v\|^2}\right){\tilde v^Tx}\cdot\tilde v\\ \end{split}. $$ Since $v\mapsto \frac1{\|v\|^2}$ is continuous from $\mathbb R^n\setminus\{0\}$ to $\mathbb R$, for all $\epsilon>0$ there is $\delta >0$ such that the 2-norm of the right-hand side can be made smaller than $\epsilon \|x\|_2$ provided $\|v-\tilde v\|\le \delta$.
This proves that the set of rational Householder matrices is dense in the set of all real Householder matrices. By Cartan–Dieudonné, the desired result follows.
