TITLE: Tiling a rectangle with both rational and irrational sided squares
QUESTION [8 upvotes]: We define a 'tiling of rectangle with squares' as the process of dividing the rectangle into finitely many squares so that they do not overlap and cover up the whole rectangle.
Here is my question:
Is there a rectangle with such a square tiling that contains squares of both rational and irrational side lengths?
The following statement is well-known:

A rectangle can be tiled with squares if and only if the ratio of its side lengths is a rational number.

The proof of this statement can be found in Proofs from THE BOOK. From this we can assume the sides of such a rectangle, if it exists, to be 1 and p/q, p and q being integers. However, I really have no idea how to move further from this point.
Hope you could help me with this problem. Thanks!

REPLY [2 votes]: Since I have a good solution now, I'm coming to answer my own question :)
Using the so-called Smith's diagram, every tiling is equivalent to an electrical circuit, with all resistances equal to 1:
https://en.wikipedia.org/wiki/Squaring_the_square#Perfect_squared_squares
In the Smith's diagram, the squares are considered as resistors, the horizontal lengths as currents and the vertical lengths as voltages.
Without loss of generality, the vertical side length of the original rectangle can be set as 1. Applying Kirchhoff's circuit laws to this problem, we get a set of linear equations. When we consider the currents as unknowns, then the coefficients in the system are all +1 or -1. The right hand side of the system consists of just 0 and 1 (which is the only voltage source). Therefore, from linear algebra, we know that all currents have to be rational numbers. That means the answer to the question is No.