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.



Can a unit square be cut into rectangles that tile a rectangle with irrational sides?
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 $R^\*.$


