Can a square be cut parallel to its sides to make a rectangle of non-square-rational proportion? For arbitrary positive integers $m$ and $n$, a unit square can be dissected along a regular grid dividing it into $mn\times mn$ subsquares and reassembled into an $m/n\times n/m$ rectangle. But can it be cut another, nonrational, way into rectangles to form a rectangle whose ratio of sides is not the square of a rational number? (The usual rules apply: finitely many cuts, and no gaps, overlaps, or discarding.)
 A: Here's a quick proof using linear algebra, adapted from my answer to a related question, that this rectangular equidissection is not possible:
Let $z$ be an irrational side length of the target rectangle. Using the axiom of choice, it is possible to choose a $\mathbb Q$-linear function $f : \mathbb R \to \mathbb R$ satisfying $f(1) = 1$ but $f(z) = 0$.
Now for any given axis-aligned rectangle $R := [x_0, x_1] \times [y_0, y_1]$, we define an invariant
$$F(R) := f(x_1 - x_0)\cdot f(y_1 - y_0).$$
Clearly $F(R)$ does not change if $R$ is translated, or rotated by $90^\circ$.
Expanding, we also see $$F(R) := f(x_1)f(y_1) - f(x_1)f(y_0) - f(x_0)f(y_1) + f(x_0)f(y_0).$$
Using this expansion, we immediately see that whenever a rectangle $R$ is cut into rectangular pieces $R_1, \ldots, R_n$, we have
$$F(R) = F(R_1) + \ldots + F(R_n),$$
because all the summands not corresponding to the corners of $R$ cancel out.
Together, this implies that whenever two sets $\mathcal A$ and $\mathcal B$ of rectangles have a common rectangular dissection, it is true that
$$\sum_{R \in \mathcal A}F(R) = \sum_{R \in \mathcal B}F(R).$$
However, $F([0,1]^2) = 1$ and $F([0,z]\times[0,w]) = 0$ for any $w$, so you cannot dissect a unit square (or even any finite set of rectangles with rational side lengths) into finitely many rectangles you can reassemble into a rectangle with one side of length $z$.

Note: Above we used the axiom of choice, this is not actually necessary for the proof to work. For any given finite rectangular dissection of some rational rectangles, simply let $V$ be the $\mathbb Q$-vector subspace of $\mathbb R$ generated by all side lengths of subrectangles in that dissection. Note that $z$ should lie in $V$ if the dissection is intended to fit into the target rectangle. Now simply define $f$ only on $V$ instead of on $\mathbb R$, and you will not require the axiom of choice to ensure the existence of $f$ because $V$ is finite dimensional over $\mathbb Q$. The rest of the proof works as before.

Note also that this proof easily generalizes to higher dimensions. No set of rational cuboids has a finite axis-aligned cuboidal equidissection with a cuboid with some irrational edge length.
