Are the singular values of matrices in $SL_2(Z)$ dense in $(0,1)$?

Question

Given a $2 \times 2$ real matrix $A$, we denote by $\sigma_1(A) \le \sigma_2(A)$ its singular values. If $\det A=1$, then $0<\sigma_1(A)\le 1$.

Define $X=\{\sigma_1(A) \, | \, A \in SL_2(Z)\}$. Is $X$ dense in $(0,1]$?

I know that $1 \in X$ (since $A=\text{Id} \in SL_2(Z)$), and that $0 \in \overline X$. Indeed, consider $ A_n=\begin{pmatrix} 1 & 0 \\\ n & 1 \end{pmatrix}. $ We have $\lim_{n \to \infty}\sigma_2(A_n)=\infty$, so $\lim_{n \to \infty}\sigma_1(A_n)=0$.

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Here $SL_2(Z)$ is the group of $2 \times 2$ matrices with integer values and determinant $1$.

Answer 1

No, the squared singular values are the roots of $X^2-Tr(AA^\top)X+1$ where $Tr(AA^\top)$ is an integer, and given $r\ne 0$ there are only finitely many values of $n$ such that $|r^2-nr+1|\le 1$, thus only finitely many values of $n$ such that $\exists s, s^2-ns+1=0$ and $|s-r|< \frac{r}{2}$.

Whence for $\sqrt{r}\ne 0$ to be in the closure of the set of singular values of all $SL_2(\Bbb{Z})$ matrices then $r$ must be a root of $X^2-nX+1$ for some $n$.