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I saw this problem several years ago, and I discovered a solution to it. I've since learned a somewhat more efficient solution based on the same idea.

Call a rectangle in the $(x,y)$ plane rational if it has sides parallel to the axes and at least one of its sidelengths is rational. Suppose $R$ is a rectangle which can be disected into a finite collection of rational rectangles (i.e., they cover and intersect only on their boundaries). Show $R$ is rational.

The rather elegant solution (not by me, but similar to mine) to this problem is as follows. First, scale up the whole picture by the least common denominator of all the rational sidelengths, so each rectangle in the disection has an integer sidelength. Then integrate $\sin{(2 \pi x)} \sin{(2 \pi y)}$ over $R$. Via the dissection, the integral is clearly $0$, and hence $R$ must be rational. The only things this proof need are the fundamental theorem of calculus and the antiderivative of $\sin$.

For quite a while, I was satisfied with this solution. The machinery of the integral surprisingly has all the necessary tools to prove this. But now I want to understand what it is that makes this proof work.

I've tried to unpack this proof into a sequence of simple statements, but I find that it always seems to get more complicated. Hence, I ask, is there a way to simultaneously simplify this argument and avoid calculus (with an argument similar to this one, not a totally new one)? Is there a way to unpack some of the definitions to make this argument more transparent?

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There was a slight typo before. "Call a rectangle in the (x,y) plane rational if it has sides parallel to the axes and at least one of its sidelengths is rational." –  Logan Maingi Sep 16 '11 at 23:39
    
I should point out that the terminology here is non-standard; usually a "rational rectangle" means something somewhat different. –  Logan Maingi Dec 15 at 7:21

2 Answers 2

up vote 6 down vote accepted

Anyone interested in this problem would enjoy the very nice expository article (fourteen proofs!) by Stan Wagon.

Added: And more than fourteen proofs makes the result even more true.

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This is an excellent reference just from skimming it, and I suspect it will totally answer my question. Give me some time to read it first, but it is very much appreciated. –  Logan Maingi Sep 16 '11 at 23:44

You can replace the need for calculus with just the concept of volume under the curve, by replacing the sine function with a periodic piecewise linear function with average 0 over its period, for example lines connecting the point sequence (0,0),(1/4,1),(3/4,-1),(1,0) and repeating with period 1. You can find the relevant volumes with basic geometry of triangles instead of integral calculus.

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