Partition rectangle into finite number of squares For what positive real numbers $x,y$ can an $x\times y$ rectangle be partitioned into a finite number of squares?
When $\dfrac{x}{y}$ is a rational number, it is not hard to see that we can partition it.
Also, we can scale the width to $1$, so, say we're dealing with a $1\times c$ rectangle. For what values of $c$ can we do the partition?
 A: A rectangle with sides $x$, $y$ and $\frac{x}{y} \not \in \mathbb{Q}$ cannot be partitioned into squares. This is a classical result due to Dehn, Sprague and others.
Indeed, if we have a partition of the rectangle into squares of sizes $l_i$ we get the equality of areas
$$x\cdot y = \sum l_i^2$$
This can be proved by further dissecting the rectangle along the sides of the squares and using the distributive law. The same argument in fact can be used to prove the stronger equality in the $\mathbb{Q}$-vector space $\mathbb{R}\otimes_{\mathbb{Q}}\mathbb{R}$:
$$x\otimes y = \sum l_i \otimes l_i$$
Assume now $\frac{x}{y}$ is not rational. Then there exists a $\mathbb{Q}$-linear functional $\phi \colon \mathbb{R}\to \mathbb{Q}$ so that $\phi(x) = 1$ and $\phi(y)= -1$. From the above equality we obtain:
$$\phi(x) \cdot \phi(y) = \sum_i \phi(l_i) \phi(l_i)$$
that is $-1$ is a sum of squares in $\mathbb{Q}$, contradiction.
Note that we do not need to use the existence of a Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$, we could as well work with the $\mathbb{Q}$ span of $x$, $y$ and all of the $l_i$'s.
A: Another proof of this fact uses the following (not very complicated) fact from linear algebra:
If a system of linear equations with rational coefficients (that includes the right-hand sides as well) has a solution then it has a rational solution.
