Problem 1. Call a rectangle in the euclidean plane good if
at least one of its length or width
is a rational number . For example, the 2 ×π and 3/4 × 7/9 rectangles are both
good, while
a
rectangle is not good. Suppose you are given
a big rectangle R, and you are told
that it is somehow tiled by a finite number of good rectangles. Prove that R is
also good.
Note that this tiling can have a small 2 × π rectangle
(good in the horizontal direction)
adjacent to a
rectangle (which is good in the
vertical direction). You have to
somehow conclude that, since everything fits together in a nice finite tiling,
the sum of
horizontal lengths
is rational, or that the sum of vertical lengths
is rational (or possibly that both are rational).
Problem 2. Prove the fol lowing result that we saw
in Calc III.
Ideas for problem 1.
(1) Call a rectangle really good if one of its lengths is
an integer. Given our big rectangle
R tiled by finitely many small good rectangles, we can take L to be the least
common
multiple of the denominators of all the rational lengths of the small good
rectangles.
Scale everything by L, so now we get a huge rectangle, H, which is tiled by
really
good rectangles. We’ll argue that this huge rectangle is also really good (and
hence
we can conclude that R is good — why?).
(2) Consider the function
What is the value of
over a really good rectangle? Why?
(3) What is value of the double integral of f over H? Why?
(4) Place H so that its bottom left corner is at the origin, so it has two other
corners at
(A, 0) and (0,B) say. What can you conclude about A or B? Why?
(5) Why did we have to be careful about the placement of the bottom left corner
of H?
What other points would work? What would be a bad placement for H?
Ideas for problem 2. This is basically an
evaluation of the double integral of 
over the
unit square
[0, 1] × [0, 1] in two different ways .
(1) First method . Use the geometric series
to write out the integrand above as a series in x and y,
and then integrate this series
term by term , first with respect to x and then with respect to y. You can just
do
this formally if you like . However, if you have issues with the fact that the
integral is
not proper (f(1, 1) is not defined), and/or that a geometric series only makes
sense
for |r| < 1, then you should work all this for the rectangle [0, 1]×[0, a] for a
< 1, and
then consider what happens as a -> 1.
(2) Second Method. (Rotate coordinates by π/4!) Use the
substitution
(3) Show that 
(4) Show that the integral becomes the sum of two
integrals
(5) The v-integrals will give you inverse tangents. To do
the resulting horrible u-integrals,
make the substitution 
Have fun.
