Show that the points P = (3.5, 9.5) and Q = (2.5, 8.5) lie on the elliptic curve ($y^2 = x^3 – 36x$) over the real numbers

I am studying elliptic curve cryptography and know basics of elliptic curve(just as much I require in solving these sort of problems) I am doing this problem and totally confused as when I try to do this for P=(3.5,9.5), I get 90.25 on L.H.S and -83.125 on R.H.S, as much as I can understand from the question I can conclude that modulus operator is not required here as we are doing operations over real number and not some $Z_P$ where p is prime.

• $x^3-36x=x(x^2-36)$ is negative, when $x<-6$ or $0<x<6$. So if $x\in(0,6)$ then $y\notin\Bbb{R}$. Also, if $x=n+1/2$ for some integer $n$, then $x^3-36x$ has denominator equal to $8$ (together with an odd numerator), so it cannot be the square of a rational number. – Jyrki Lahtonen Feb 10 '17 at 14:20
• And the denominators are wrong too. When $x=7/2$ or $5/2$, the denominator on the right is $8$, while that is not possible for the square of a rational number (from the left). This thing is totally bollixed up. – Lubin Feb 13 '17 at 1:43

Something is amiss. Neither of those points lie on the elliptic curve $y^2 = x^2 - 36x$. This can be seen by directly plugging in the points, but notice that, more generally, $x^2 - 36x < 0$ if $x \in (0,36).$ And both $3.5$ and $2.5$ are in $(0,36)$. But the LHS is nonnegative for all real $y$.
EDIT: Regarding your changing it to $x^3 - 36x$, the exact same argument applies except now the interval is $(0,6)$ instead of $(0,36)$. Note that $x^3 - 36x < 0$ if (but not only if, but that's not really relevant here) $x \in (0,6)$. And both $2.5$ and $3.5$ are in $(0,6)$. Same issue occurs.