Invertible rational matrices are dense in the general linear group

General fact. Let $D$ be dense in $X$ and $U\subseteq X$ open. Take $x\in U$ and a neighbourhood $V$ of $x$. Then $V\cap U$ is a nonempty open set, so it meets $D$; hence $x\in\overline{U\cap D}$.

Application. $X=M_n(\mathbb{R})$, $D=M_n(\mathbb{Q})$ is dense (each entry is approximated by rationals), and $U=GL_n(\mathbb{R})=\{\det\ne0\}$ is open. So $M_n(\mathbb{Q})\cap GL_n(\mathbb{R})$ is dense in $GL_n(\mathbb{R})$.

Subtlety. The elements of this set are the rational matrices that are invertible over $\mathbb{R}$; their determinant is a nonzero rational number, so they are also invertible over $\mathbb{Q}$, with inverse in $M_n(\mathbb{Q})$ (by the adjugate formula). So the set is exactly the group $GL_n(\mathbb{Q})$.

Remark. The same argument shows $\{A\in M_n(\mathbb{Q})\}$ is dense in any open subset of $M_n(\mathbb{R})$. It fails for closed subsets such as $SL_n$ (see the next problems), where rationality interacts with the constraint $\det=1$. Lesson. Open $\cap$ dense is dense in the open set.