Cut two squares into rectangles to reassemble a single square. This problem would belong to puzzling SE, except that I suspect it to be impossible.  So I post it here, to see if someone can provide an argument proving the impossibility.
The problem: Is it possible to cut two unit squares into a finite number of rectangles and rearrange these into a solid square of side length $\sqrt 2$?  (without overlap, discarding of parts, holes in the assembly, or such.)
A more general version of the problem has been posted already
Can a square be cut parallel to its sides to make a rectangle of non-square-rational proportion?
but it didn't receive an answer.
 A: Here's a quick proof using linear algebra that this rectangular equidissection is not possible:
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 $f(\sqrt 2) = 0$.
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, $F([0,1]^2) + F([0,1]^2) = 2$ and $F([0,\sqrt{2}]^2) = 0$, so you cannot dissect two unit squares into finitely many rectangles you can reassemble into a square of side length $\sqrt 2$. $~~~\square$

Above we used the axiom of choice, this is not actually necessary for the proof to work. For any given finite rectangular dissection of two unit squares, simply let $V$ be the $\mathbb Q$-vector subspace of $\mathbb R$ generated by all side lengths of rectangles of that dissection. Note that $\sqrt 2$ should be in $V$ if the dissection is intended to fit into the $\sqrt 2$ square. 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 here that the only property of $\sqrt 2$ we required was that it is irrational. This proof trivially generalizes to any other irrational side length of square or rectangle, immediately implying that whenever any finite rectangular dissection of one or more rectangles with rational side lengths can be reassembled into another rectangle, that rectangle must also have rational side lengths.
