For arbitrary positive integers $m$ and $n$, if we dissect a unit square into an $m\times n$ rectangular grid of $1/m\times 1/n$ rectangles, we can reassemble these $mn$ rectangles into an $n/m\times m/n$ rectangle, which is of square-rational proportion $m^2/n^2$. Is there any essentially different way to rectangularly dissect a unit square to a rectangle----that is, to a $1/r\times r$ rectangle, of proportion $r^2$, where $r$ is irrational?

Notes: The usual rules for dissection problems apply (finitely many pieces, with no gaps or overlaps of positive planar measure, and no discarded or additional pieces). This question was posted, unanswered, on MathStackExchange for a month. For what it's worth, my failed attempts to find such a dissection lead me to think it impossible.

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    $\begingroup$ If a unit square is partitioned in congruent rectangles, they must have edges of rational length. Maybe you also allow different size rectangles? $\endgroup$ Commented Jan 15, 2014 at 14:56
  • $\begingroup$ @PietroMajer: Yes, the rectangles cannot all be congruent. $\endgroup$ Commented Jan 15, 2014 at 15:38

3 Answers 3


Suppose that the unit square has been cut into $n$ rectangles: $S=\cup_{i=1}^nR_i,$ where each rectangle $R_i$ has dimensions $s_{2i-1}\times s_{2i}$. Also let $r$ be an irrational number and let $R^*$ be a $1/r \times r$ rectangle. We will define an additive "weight" (which can take negative values) on certain plane sets such that $w(S)=1$ but $w(R^*)=0$. Let $L=\{{1,r,1/r,s_1,s_2,\dots,s_{2n}\}}.$ Clearly any rectangle which can be assembled using the $R_i$ has sides which can be expressed as $\sum_{i=1}^{2n}c_is_i,$ where each $c_i \in \{{0,1\}}$. As usual, let $\operatorname{span}L$ be the set of all linear combinations $\sum_{j=1}^{2n}c_js_j+c'+c''r+c'''\!\!\ 1/r$ with the coefficients $c \in \mathbb{Q}.$ There is a subset $B \subseteq L$ which is a basis in that $\operatorname{span}B=\operatorname{span}L$ and each $r \in \operatorname{span}L$ can be uniquely expressed as a rational linear combination of elements of $B$. Furthermore, we can arrange to have $1 \in B$ and $r \in B.$ Also we may have $1/r \in B$ unless $r$ is the root of an integer quadratic equation $ar^2+br+c=0,$ in which case $1/r=\frac{-a}{c}r -\frac{b}{c}.$

Define the function $f:\operatorname{span}L \to \mathbb{Q}$ by taking $f(x)$ to be the coefficient of $1$ in the unique rational linear combination from $B$ which equals $x$. Then $f$ is a linear transformation: $f(x_1+x_2)=f(x_1)+f(x_2).$ Also, $f(x)=x$ for $x$ rational, and $f(r)=0$ for our special chosen irrational number.

Now, as in the other proofs, use $f$ to give a weight $w_f(R)=f(u)f(v)$ to any rectangle $R$ with sides $u \times v$ with $u,\!v \in \operatorname{span}L,$ and extend the weight additively to any plane set which is a (finite) union of such rectangles with disjoint interiors. As before, $w_f(S)=1 \ne 0=w_f(R^*)$. Hence no dissection of $S$ can be used to assemble $R^*.$


No. It is a famous problem. Suppose it were possible to cut the unit square into finitely many rectangles of sizes $a_i \times b_i$. This means that we have a decomposition $1 \otimes 1 = \sum_i a_i \otimes b_i$. If these rectangle could be reassembled into an $r \times 1/r$ rectangle, we would likewise have $r \otimes 1/r = \sum_i a_i \otimes b_i$. But $r \otimes_{\mathbb{Z}} 1/r = 1 \otimes_{\mathbb{Z}} 1$ if and only if $r \in \mathbb{Q}$.

More concretely, let $f : \mathbb{R} \to \mathbb{R}$ be an additive group homomorphism such that $f(q) = 0$ iff $q \in \mathbb{Q}$; it exists by the Axiom of Choice. By additivity (refining the dissection into an elementary one, given by a complete intersection of lines parallel to the axes), we have $f(1)^2 = \sum_i f(a_i)f(b_i)$, and likewise, $f(r)f(1/r) = \sum_i f(a_i)f(b_i)$. But by construction, $f(1) = 0$ while $f(r)f(1/r) \neq 0$ unless $r \in \mathbb{Q}$.

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    $\begingroup$ Very nice. I'm happy with $\mathrm {AC}$; but, in consideration of the minority who are uncomfortable about it, is it essential---namely, is the theorem unprovable in $\mathrm {ZF}$? $\endgroup$ Commented Jan 15, 2014 at 16:20
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    $\begingroup$ Just a note to anyone who was initially confused in the way I was: this argument should be interpreted in the symmetric square $S^2(\mathbb{R})$ over $\mathbb{Z}$ rather than in the full tensor product $\mathbb{R}\otimes_{\mathbb{Z}}\mathbb{R}$. Otherwise 90 degree rotations of rectangles are not accounted for. $\endgroup$
    – Noah Stein
    Commented Jan 15, 2014 at 20:22
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    $\begingroup$ It is not necessary to define the group homomorphism $f$ on the whole of $\mathbb{R}$. It suffices to define it on a well-chosen finite-dimensional $\mathbb{Q}$-vector space. Thus, AC is not needed here. $\endgroup$ Commented Jan 15, 2014 at 20:25
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    $\begingroup$ Any time you use AC to prove that some object can't exist, this is removable since you can't use such a construction to disprove AC. $\endgroup$ Commented Jan 16, 2014 at 0:38
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    $\begingroup$ @DouglasZare : $\:$ Try applying that argument to "There does not exist a counter-example to AC." $\hspace{.59 in}$ $\endgroup$
    – user5810
    Commented Jan 16, 2014 at 1:47

Using the main idea of one of the proofs of the famous Dehn's tiling problem (see for example here), assume that such a cutting of a square is possible.

Let $S$ the unit square, and $S=\cup_{j=1}^n R_j$ and $$[0,r]\times[0,1/r]=T=\cup_{j=1}^n R_j',$$ were $R'_j$ is produced be translating and/or rotating by $\pi/2$ the $R_j$, and $r$ irrational - we assume that there are no overlaps among the $R_j$'s and no overlaps among the $R_j'$'s, meaning that $R_j\cap R_k$ could only be part of an edge.

As $r$ is irrational, it is possible to extend $\{1,r\}$ to a Hamel basis $\mathcal B$ of $\mathbb R$ over $\mathbb Q$. Define a linear functional $\mathbb f :\mathbb R\to\mathbb R$ as: $$ f(1)=1, \,\, f(r)=0, \,\, \text{and}\,\, f(b)=0, \,\,\text{for all}\,\, b\in\mathcal B \smallsetminus\{1,r\}. $$ Then for any rectangle $R=[a,b]\times[c,d]$ define its $f-$area as $$ A_f(R)=f(b-a)f(d-c). $$ It is not hard to check that $A_f$ is finitely additive, and it is invariant to rotations of $k\pi/2$, $k\in\mathbb Z$ and translations. In particular, $A_f(R_j)=A_f(R_j')$, for all $j$. Then $$ 0=f(r)f(1/r)=A_f(T)=\sum_{j=1}^n A_f(R_j')=\sum_{j=1}^n A_f(R_j)=A_f(S)=f(1)f(1)=1, $$ which is a contradiction. Hence such cutting is not possible.

Note. The use of the Axiom of Choice (i.e., the fact that the linear space $\mathbb R$ over $\mathbb Q$ possesses a Hamel basis) can be avoided, as $f$ does not have to be defined on the whole of $\mathbb R$, but on the subspace of $\mathbb R$ over $\mathbb Q$, which is spanned by the sizes of the rectangles $R_j$. It is also noteworthy, that the additivity of $A_f$ is used in the proof of Dehn's Theorem.


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