Here is a very clever geometry proof I once saw. I do not recall where it was I encountered it, but It sticks out in my memory for its creative simplicity; I take no credit for it.

Statement: suppose we have a box (right rectangular prism), which is divided completely into a finite number of smaller boxes, not necessarily the same size or shape. If for each of these smaller boxes, at least one of the edges (dimensions) is of integer length, then at least one edge of the overall box must be of integer length.

Let us choose Cartesian coordinates so that the edges of the box are parallel to the axes. Then the smaller boxes must have edges parallel to the axes, as well. Thus, each box may be represented as the region in space with for some real numbers *x*_{1}, *x*_{2}, *y*_{1}, *y*_{2}, *z*_{1}, *z*_{2}.

Now, consider the complex-valued function . Let us take the volume integral of this over one of our boxes. Then

.

Now, the complex exponential is periodic with period 2π, so if and only if

is an integer multiple of 2π, which is true if and only if *x*_{2}–*x*_{1} is an integer. Thus, we see that the product , and thus the integral, is zero if and only if at least one of the side lengths *x*_{2}–*x*_{1}, *y*_{2}–*y*_{1}, or *z*_{2}–*z*_{1} is an integer.

Thus, for all our small sub-boxes, as (at least) one of the lengths is an integer, the integral of is zero over every small box. But since our overall box is formed entirely from these smaller boxes, the integral of over it must be zero, and thus one of the sides must have integer length.

Analogous proofs work for rectangles in the plane, and for rectangular parallelepipeds in any number of dimensions.

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Tags: Box, Complex Exponential, Geometry, Integral, Math, Monday Math

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