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My algebra professor gave me this puzzle a while back. I'm pretty sure I've found the right solution; nonetheless, I wanted to share it and see if you come up with anything really nice or unexpected.

Prove that if you take a unit square and cut it into a finite number of smaller squares (in any way you can think of), the side lengths of the smaller squares are all rational.

P.S. The first tag was my professor's hint.

[Edit] Just to be clear, every piece must actually be a square (e.g. no gluing two triangles into a square).

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I can make one with 2 squares each of sidelength $1/\sqrt{2}$. –  Raskolnikov Dec 19 '10 at 17:32
    
Sorry, I didn't make it clear: it's not that the smaller squares add up to unit area, it's that you can literally assemble them into a unit square. To put it another way, you're only allowed to use scissors on the unit square, and every piece you're left with at the end must be a square. –  Elliott Dec 19 '10 at 17:38
    
So we can't use 'any way we can think of'. Bummer... –  Raskolnikov Dec 19 '10 at 17:39
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But seriously, not only would this be a pretty sadistic (and unrelated, considering the professor taught a course on rings and modules) homework question, it's Christmas break! How could I possibly have homework right now? –  Elliott Dec 19 '10 at 18:53
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@Elliott: I believe you that this is not homework, but I thought I would mention that I have had homework over Christmas break at two different U.S. universities. One was on the quarter system, where the second quarter started after Thanksgiving. (It was a horrible way to schedule the terms.) The second was on the semester system, where classes ended before Christmas but finals were after New Year's. Occasionally the profs would make the last homework assignment due after New Year's as well. –  Mike Spivey Dec 20 '10 at 18:25

3 Answers 3

up vote 11 down vote accepted

This theorem was proved by Max Dehn in early 20th century.

Suppose there is some square whose side $s$ is not rational. Pick some $\mathbb{Q}$-homomorphism $h$ such that $h(1) = 0$ and $h(s) = 1$. Define the area $\sigma(R)$ of a rectangle $R$ with sides $a,b$ by $h(a)h(b)$. We show below that if a rectangle $R$ is partitioned into rectangles $R_i$ then $$\sigma(R) = \sum \sigma(R_i).$$ Take $R$ to be the unit square, and $R_i$ the squares into which it is partitioned. If $R_i$ has side $s_i$ then we get $$0 = h(1)^2 = \sigma(R) = \sum \sigma(R_i) = \sum h(s_i)^2 \geq h(s)^2 = 1.$$ This contradiction shows that all sides must be rational.

It remains to show that $\sigma$ is additive. Extend all sides in the partition of $R$ to get a grid $G_j$. From additivity of $h$ we easily get $$\sigma(R) = \sum\sigma(G_j) = \sum\sigma(R_i).$$

Using very similar methods, you can show the following slightly stronger theorem: Suppose that a rectangle is partitioned into squares. Then all edge lengths are rationally related.

Indeed, if the outside rectangle had sides $a,b$ not rationally dependent then we could pick a homomorphism $h$ such that $h(a)=1,h(b)=-1$, and then get that $-1$ is a sum of squares. Wlog, $a,b$ are both rational, and the previous proof completes the argument.

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+1: Would be nice if you could add a reference too. I believe this book proof appeared much later and Dehn didn't exactly prove it that way. –  Aryabhata Dec 20 '10 at 3:31
    
This is really interesting! I wanted to clarify a few things: (1) is it correct that $h$ gives a map $\mathbb{R} \rightarrow \mathbb{R}$ that's a group homomorphism but not a ring homomorphism, (2) can you give an example (trivial is ok) of a valid $h$, (3) it doesn't feel like we did any actual "work"; is the key step that such an $h$ exists, and (4) why does this reasoning fail if we allow the partitions to be rectangles? –  Elliott Dec 20 '10 at 18:05
    
(1) $h$ really only need be additive, (2) given a basis of the relevant lengths over $\mathbb{Q}$, just define $h(1)=0$, $h(s)=1$, and extent $h$ linearly, giving arbitrary values for the rest of the basis elements, (3) the key is the additivity of $\sigma$, (4) if $R$ is a rectangle then you don't know a-priori that $\sigma(R) \geq 0$ –  Yuval Filmus Dec 20 '10 at 21:09

I believe there is also a proof by associating the tiling with an electric network!

This is a section in the book "Modern Graph Theory" by Béla Bollobás.

It is titled "Squaring the Square" and is numbered section II.2. See this: http://books.google.com/books?id=SbZKSZ-1qrwC&pg=PA46.

A snapshot of the page which talks about Dehn's theorem and it's proof (also found here: http://books.google.com/books?id=SbZKSZ-1qrwC&pg=PA48):



alt text

Note, even though the section starts off talking about the tiling squares being of distinct lengths, the proof of Dehn's theorem (and the problem you seek) is essentially there.

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Since your professor's hint was to use linear algebra, the intended proof is probably that of Hadwiger - employing a Hamel basis of $\:\mathbb R\:$ over $\:\mathbb Q\:$ to construct additive "area" functions in order to deduce a contradiction (as in Yuval's answer). Related results were proved by Dehn circa 1900 by different complicated methods. Although Hamel's work on Hamel bases occurred shortly later in 1905, it was not until much later in the 1950's that Hadwiger and his students noticed that Hamel bases could be employed to greatly simplify Dehn's proofs. For a nice exposition see Freiling; Rinne: Tiling a square with similar rectangles. It's worth remarking that use of the axiom of choice in the construction of a Hamel basis can easily be eliminated for applications of this type, e.g. see the discussion in section 2 of Feshchenko et. al.: Dissecting a brick into bars.

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Nice historical overview! –  Alex B. Dec 20 '10 at 14:42

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