density in the special linear group

Fact used. Over a field, $SL_n$ is generated by elementary matrices $T_{ij}(t)=I+tE_{ij}$, $i\ne j$ (Gauss elimination: any matrix of determinant $1$ is reduced to $I$ by elementary row operations of this type, with the last diagonal entry fixed by the determinant being $1$).

Approximation. Let $g\in SL_n(\mathbb{R})$ and write $g=T_1T_2\cdots T_m$ with $T_l=I+t_lE_{i_lj_l}$. Choose rationals $t_l^{(k)}\to t_l$ ($k\to\infty$) and set
$$g^{(k)}=\prod_{l=1}^m\bigl(I+t_l^{(k)}E_{i_lj_l}\bigr).$$
- Each factor has determinant $1$ and rational entries, so $g^{(k)}\in SL_n(\mathbb{Q})$.
- Matrix multiplication is continuous, so $g^{(k)}\to g$.

Therefore $SL_n(\mathbb{Q})$ is dense in $SL_n(\mathbb{R})$.

Remark. The naive attempt (approximate $g$ entrywise by a rational matrix) gives a rational matrix with determinant close to, but not exactly, $1$; fixing the determinant by dividing by an $n$-th root is not possible over $\mathbb{Q}$. The generation argument bypasses this. Lesson. To approximate within a group, use generators.