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Gerry Myerson
Gerry Myerson
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There is an exercise in "Modern Graph Theory" by Bollobas section II.4 pg 63 that is essentially the same argument, but eliminates the trigonometric functions. For each rectangle $U = [x_1, x_2] \times [y_1, y_2]$ let    $\psi(U) = (x_2 - x_1) \otimes (y_2 - y_1)$ in $\mathbb{Z}(\mathbb{R}/\mathbb{Z}) \otimes \mathbb{Z}(\mathbb{R}/\mathbb{Z})$ (viewed as a $\mathbb{Z}$ module). Then $\sum_{U} \psi(U) = 0$ so the original rectangle must have an integer side.

[Edited after Reid's comment. Here $\mathbb{Z}(\mathbb{R}/\mathbb{Z})$ is the free $\mathbb{Z}$ module with basis $\mathbb{R}/\mathbb{Z}$]

There is an exercise in "Modern Graph Theory" by Bollobas section II.4 pg 63 that is essentially the same argument, but eliminates the trigonometric functions. For each rectangle $U = [x_1, x_2] \times [y_1, y_2]$ let  $\psi(U) = (x_2 - x_1) \otimes (y_2 - y_1)$ in $\mathbb{Z}(\mathbb{R}/\mathbb{Z}) \otimes \mathbb{Z}(\mathbb{R}/\mathbb{Z})$ (viewed as a $\mathbb{Z}$ module). Then $\sum_{U} \psi(U) = 0$ so the original rectangle must have an integer side.

[Edited after Reid's comment. Here $\mathbb{Z}(\mathbb{R}/\mathbb{Z})$ is the free $\mathbb{Z}$ module with basis $\mathbb{R}/\mathbb{Z}$]

There is an exercise in "Modern Graph Theory" by Bollobas section II.4 pg 63 that is essentially the same argument, but eliminates the trigonometric functions. For each rectangle $U = [x_1, x_2] \times [y_1, y_2]$ let  $\psi(U) = (x_2 - x_1) \otimes (y_2 - y_1)$ in $\mathbb{Z}(\mathbb{R}/\mathbb{Z}) \otimes \mathbb{Z}(\mathbb{R}/\mathbb{Z})$ (viewed as a $\mathbb{Z}$ module). Then $\sum_{U} \psi(U) = 0$ so the original rectangle must have an integer side.

[Edited after Reid's comment. Here $\mathbb{Z}(\mathbb{R}/\mathbb{Z})$ is the free $\mathbb{Z}$ module with basis $\mathbb{R}/\mathbb{Z}$]

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Richard Eager
Richard Eager
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There is an exercise in "Modern Graph Theory" by Bollobas section II.4 pg 63 that is essentially the same argument, but eliminates the trigonometric functions. For each rectangle $U = [x_1, x_2] \times [y_1, y_2]$ let $\psi(U) = (x_2 - x_1) \otimes (y_2 - y_1)$ in $\mathbb{R}/\mathbb{Z} \otimes \mathbb{R}/\mathbb{Z}$$\mathbb{Z}(\mathbb{R}/\mathbb{Z}) \otimes \mathbb{Z}(\mathbb{R}/\mathbb{Z})$ (viewed as a $\mathbb{Z}$ module). Then $\sum_{U} \psi(U) = 0$ so the original rectangle must have an integer side.

[Edited after Reid's comment. Here $\mathbb{Z}(\mathbb{R}/\mathbb{Z})$ is the free $\mathbb{Z}$ module with basis $\mathbb{R}/\mathbb{Z}$]

There is an exercise in "Modern Graph Theory" by Bollobas section II.4 pg 63 that is essentially the same argument, but eliminates the trigonometric functions. For each rectangle $U = [x_1, x_2] \times [y_1, y_2]$ let $\psi(U) = (x_2 - x_1) \otimes (y_2 - y_1)$ in $\mathbb{R}/\mathbb{Z} \otimes \mathbb{R}/\mathbb{Z}$ (viewed as a $\mathbb{Z}$ module). Then $\sum_{U} \psi(U) = 0$ so the original rectangle must have an integer side.

There is an exercise in "Modern Graph Theory" by Bollobas section II.4 pg 63 that is essentially the same argument, but eliminates the trigonometric functions. For each rectangle $U = [x_1, x_2] \times [y_1, y_2]$ let $\psi(U) = (x_2 - x_1) \otimes (y_2 - y_1)$ in $\mathbb{Z}(\mathbb{R}/\mathbb{Z}) \otimes \mathbb{Z}(\mathbb{R}/\mathbb{Z})$ (viewed as a $\mathbb{Z}$ module). Then $\sum_{U} \psi(U) = 0$ so the original rectangle must have an integer side.

[Edited after Reid's comment. Here $\mathbb{Z}(\mathbb{R}/\mathbb{Z})$ is the free $\mathbb{Z}$ module with basis $\mathbb{R}/\mathbb{Z}$]

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Richard Eager
Richard Eager
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There is an exercise in "Modern Graph Theory" by Bollobas section II.4 pg 63 that is essentially the same argument, but eliminates the trigonometric functions. For each rectangle $U = [x_1, x_2] \times [y_1, y_2]$ let $\psi(U) = (x_2 - x_1) \otimes (y_2 - y_1)$ in $\mathbb{R}/\mathbb{Z} \otimes \mathbb{R}/\mathbb{Z}$ (viewed as a $\mathbb{Z}$ module). Then $\sum_{U} \psi(U) = 0$ so the original rectangle must have an integer side.