10
$\begingroup$

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!

$\endgroup$
0

2 Answers 2

3
$\begingroup$

No, such a tiling is not possible.

Let's suppose we have some tiling of a rectangle by squares; for now, just look at the incidence structure of the tiling, i.e. the relationships between which edges overlap or are included in one another in which directions. (Informally, imagine that we've drawn a diagram of the tiling but haven't yet checked that it's to scale.)

This incidence structure will give us lots of linear equations constraining our squares; for instance, the sum of the side lengths of all squares on a given horizontal line must add to the corresponding side length of the rectangle.

We'll also get some linear inequalities, corresponding to statements like "these three squares don't extend past the right-hand border of this other square along the $x$-axis". This is just to ensure that the incidence structure we described above is preserved.

Note that at least within the region bounded by our inequalities, satisfying all these linear equations is both necessary and sufficient for a valid tiling (so long as our side lengths are nonnegative); given a solution to these equations, simply fit each new square into a corner of the existing tiling, and any gaps or overlaps would correspond to some violation of the linear constraints. This is because any other square you might not align with as intended has already had its coordinates precisely specified, and there is a constraint requiring the current square to line up with it exactly. (Because we haven't exceeded the bounds of these inequalities, there won't be any unexpected squares to collide with.)

Now, observe that all the coefficients in the above equations and inequalities are rational; thus, if any solution exists, a rational solution exists. So our concern is that there might be some freedom of movement within these constraints - a vector along which we can move some positive distance and preserve the validity of the tiling at all times, thereby generating irrational side lengths. (This will happen if the linear equations specify a subspace with positive dimension, and the inequalities don't prohibit us from moving in one of those directions.)

Let us suppose such a vector $v\neq\vec{0}$ exists. Then for all $0\le t<\epsilon$, the tuple of side lengths

$$(s_1+tv_1,s_2+tv_2,\ldots,s_n+tv_n)$$

will be a valid solution. But now consider another constraint on the tiling: the area of the squares must add up to the area of the rectangle! So the sum

$$\sum_{i=1}^n(s_i+t\cdot v_i)^2$$

must be constant on the range $0\le t<\epsilon$. But we know that at least one of the $v_i$'s is nonzero, so this sum is a quadratic in $t$ with positive $t^2$ coefficient. This is a contradiction! So the tiling can't have any flexibility, and thus will not admit a tiling with irrational side lengths.


As an aside, the above proof goes through with minimal modification for the following more general statement: Given a fixed region $R$ whose sides are axis-aligned with side lengths in the set $S$, and a set of aspect ratios $A$, all tilings of $R$ realizable by a collection of rectangles with aspect ratios from $A$ will have side lengths in field extension $\mathbb{Q}[S\cup A]$. (In fact I believe you can impose stronger restrictions here, but I'd have to think harder before asserting anything with confidence.)

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .