Tiling Rectangles by Squares

The following is a a special case of a theorem due to Dehn (1903):

---

(Theorem, Dehn 1903) A rectangle $R$ with sidelengths $1$ and $x$, where $x$ is irrationnal cannot be tiled by finitely many squares with disjoint interiors.

---

Proof. Suppose by contradiction that there is such a tiling by squares and consider an enumeration of the squares $S_1,\ldots, S_n$ with sidelengths $s_1,\ldots,s_n$.

Viewing $\mathbb R$ as an infinite dimensional vector space over $\mathbb Q$ we can recognize the set of rational linear combinations of $s_1,\ldots,s_n$ and $x$ as a linear subspace $V \subseteq \mathbb R$.

Since $1$ and $x$ are linearly independent over $\mathbb Q$ we can extend $\{ 1,x\}$ to a basis $\{1,x,e_1,\ldots,e_k\}$ of $V$. A linear functional $\psi:V\rightarrow \mathbb R$ is then defined by its value on the basis elements. We can then choose $\psi$ such that $\psi(x)=-1$ and $\psi(1)=1$ and $\psi$ collaspes to $0$ on all other basis elements.

Define then, for a rectangle $\Lambda$ of sidelengths $\alpha$ and $\beta$ the quantity $\Psi(\Lambda)=\psi(\alpha)\psi(\beta)$.

Now, on the one hand, we clearly have $\Psi(R)=\psi(x)\psi(1)=-1$. But on the other, if we prolong all the horizontal and vertical edges in the tiling of $R$ to partition $R$ into smaller rectangles $R_i$, we see by linearity of $\psi$ that:

We have thus reached a contradiction.