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For every solution of the equation $x^2+y^2=1$, $x,y\in \mathbb Q$, the matrix $$ \begin{pmatrix} x & y \\ -y & x \end{pmatrix} $$ is a rotation that maps $\mathbb Q^2$ to itself. An example is given by the rotation $$ \begin{pmatrix} \textstyle\frac45 & \textstyle\frac35 \\ \textstyle-\frac35 & \textstyle\frac45 \end{pmatrix} $$ which has infinite order, and thus provides an answer to your second question. |
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For every solution of the equation $x^2+y^2=1$, $x,y\in \mathbb Q$, the matrix $$ \begin{pmatrix} x & y \\ -y & x \end{pmatrix} $$ maps $\mathbb Q^2$ to itself. An example is given by the rotation $$ \begin{pmatrix} \textstyle\frac45 & \textstyle\frac35 \\ \textstyle-\frac35 & \textstyle\frac45 \end{pmatrix} $$ which has infinite order, and thus provides an answer to your second question. |
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