First of all, I apologize for the rather silly question. This came up while I was scouring the Internet on a mathematical terminology appropriate for a concept that I need for a paper which I am currently writing.
So here it is:
Is $X + \frac{2}{X}$ a rational quadratic form, where $X \in \mathbb{Q} \setminus \mathbb{Z}$?
Or to put in another way:
Can the rational function $X + \frac{2}{X}$ be represented by a quadratic form, where $X \in \mathbb{Q} \setminus \mathbb{Z}$?
I know that I can just check Cassels' Rational Quadratic Forms, unfortunately the nearest university library is a bit far from my place of residence (not to mention being closed during the holidays).
Cassels will tell you that a quadratic form $Q(x)$ on $\mathbb{F}^n$ is just a function such that
1) $Q(\lambda x) = \lambda^2 Q(x)$,
2) $B(x,y):= (Q(x+y) -Q(x) -Q(y) )/2$ is bilinear.
Given any degree two polynomial in one variable, $p(x)$, we can always turn it into a quadratic form in two variables $x,y$ by setting $P(x,y) = y^2 p(x/y)$.
Maybe the issue here is that if we're looking at the zero locus given by the rational function $X+\frac{2}{X}=0$, then this is the same as the zero locus of $x^2 + 2 =0$. This is not very interesting over $\mathbb{Q}$, however, if we make this polynomial homogeneous then it becomes $X^2 + 2Y^2=0$, where $[X:Y] \in \mathbb{P}^1 \mathbb{Q}$ is a point in the projective space with homogeneous coordinates. Note that the zero locus of a quadratic form is well defined, where as any other level set (i.e. $Q(X,Y) = c \neq 0$) is not.
The rationality simply means that you're taking $X,Y \in \mathbb{Q}$ and your coefficients are rational too.