For elliptic curve $ y^2 = x^3 +3x+7$ I found the finite group $ E(\mathbb{F}_{11})= \left\{ \mathcal{O}, (1,0),(5,2),(5,9),(8,2),(8,9),(9,2),(9,9),(10,5),(10,6) \right\}$.
I have to find a generator of a subgroup of $ E(\mathbb{F}_{11})$, but for every point $P$ in $ E(\mathbb{F}_{11})$ when I try to calculate $[2]P$ I get a rational number. Am I doing something wrong?
For example if $P=(5,2)$ then by the formula $x([2]P)={(\frac{3x_1^2+a}{2y_1})}^2-2x_1$, where $a$ is the coefficient next to $x$ in the elliptic curve, and $x_1$ is the first coordinate of $P$ and $y_1$ is the second. But here I get $x([2]P) = (\frac{3*25+3}{2*2})^2-2*5 =(\frac{39}{2})^2-10 $. Should I just ignore the denominator and use $(\frac{39}{2})^2-10 =\frac{1521-10*16}{16}=\frac{1361}{16}$ and the answer for $x$ would be $1361$ mod 11$= 8$?