I wonder if the various lattice-point theorems, such as 
<a href="http://en.wikipedia.org/wiki/Pick%27s_theorem">Pick's Theorem</a> or 
<a href="http://en.wikipedia.org/wiki/Minkowski%27s_theorem">Minkowski's Lattice Theorem</a>,
have been generalized to the collection of points
with rational coordinates no more than height $h$?

A rational $a/b$ in lowest terms has *height* $\max( |a|,|b| )$.
For example, here are the rationals of height $\le 5$:
$$\left\{\frac{1}{5},\frac{1}{4},
\frac{1}{3},\frac{2}{5},\frac{1}{2},\frac{3}{5},\frac
   {2}{3},\frac{3}{4},\frac{4}
   {5},1,\frac{5}{4},\frac{4}{
   3},\frac{3}{2},\frac{5}{3},
   2,\frac{5}{2},3,4,5 \right\}$$
(Incidentally, the length of this list is given 
by integer sequence <a href="http://oeis.org/A018805">A018805</a>.)

Let us call rationals of height $\le h$, *$h$-rationals*, 
and points with both coordinates $h$-rationals,
*$h$-rational points*
(my own terminology).
Generalizations from lattice points to $h$-rational points
seems natural, and likely
has been explored...? If so, I would appreciate a reference!
That's my primary question.

Here is a specific, Pick's Theorem -like question:

> Can the number $i$ of $h$-rational points inside a polygon $P$
be expressed in a form that is not tantamount to
enumerating each interior point?
Assume we are
given the vertex coordinates of $P$, and perhaps the number
of boundary points on each edge.

If $P$ is an axis-aligned rectangle, then the number of interior points $i$
can be computed from the number of points $b_x$ lying on the
bottom side and the number lying on the left side $b_y$:
$i = (b_x - 2)(b_y - 2)$. So, for the rectangle with lowerleft
corner $(1,1)$ below, $b_x=8$ and $b_y=7$ leads to $i=30$.
But just knowing the total number of boundary points $b=26$,
a la Pick's Theorem, is inadequate to determine $i$.
It seems feasible that there is a nonlinear transformation that
maps the $h$-rational points to the integer lattice, allowing
Pick's Theorem to be applied there.
Answering this question when $P$ is a triangle should lead to a result for arbitrary
$P$ via triangulation.
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![RatsHeight5][1]
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  [1]: https://i.sstatic.net/1PWcB.jpg