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For arbitrary positive integers $m$ and $n$, a unit square can be dissected along a regular grid dividing it into $mn\times mn$ subsquares and reassembled into an $m/n\times n/m$ rectangle. But can it be cut another, nonrational, way into rectangles to form a rectangle whose ratio of sides is not the square of a rational number? (The usual rules apply: finitely many cuts, and no gaps, overlaps, or discarding.)

John Bentin
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  • Should the cuts be parallel to the sides of the square? Otherwise the task seems trivial to me: just draw the diagonals and reassemble. – Michael Hoppe Dec 14 '13 at 13:34
  • @MichaelHoppe: The question title says "Can a square be cut parallel to its sides ... ?". Anyway, I will repeat this in the text of the question. – John Bentin Dec 14 '13 at 15:56
  • We're not changing the area, so it's certain that the product of the new height and width will be rational, and both dimenions will need to be irrational. – NovaDenizen Dec 14 '13 at 16:08
  • @NovaDenizen: Indeed, the product of the new height and width is $1$. But I don't see immediately why this implies that both dimensions are irrational, because irrational numbers can have a rational sum: for example, $\sqrt2+(3-\sqrt2)=3.$ – John Bentin Dec 14 '13 at 16:34
  • @JohnBentin You get area by multiplication not by addition. – Džuris Dec 14 '13 at 17:02
  • @JohnBentin An irrational multiplied by a rational is always irrational. – NovaDenizen Dec 14 '13 at 17:05
  • If one side in the new rectangle has length $r$, the other will have length $\frac1r$, resulting in a ratio of $1/r^2$, so the problem is equivalent to finding a reconstruction with irrational side length. – NovaDenizen Dec 14 '13 at 17:08
  • @Juris: The lengths of the sides of the rectangle are built up by adding lengths of the sides of the component rectangles, not by multiplying them. – John Bentin Dec 14 '13 at 17:12
  • @NovaDenizen: Yes, I follow your argument now: the side lengths would have to be irrational. – John Bentin Dec 14 '13 at 18:25
  • Would an infinite number of cuts be acceptable? I guess not. – MvG Dec 15 '13 at 18:06
  • @MvG: No. Allowing that would make the problem trivial. – John Bentin Dec 15 '13 at 19:17
  • @MvG: I have made the exclusion of the infinite case explicit now. – John Bentin Dec 15 '13 at 19:27

1 Answers1

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Here's a quick proof using linear algebra, adapted from my answer to a related question, that this rectangular equidissection is not possible:

Let $z$ be an irrational side length of the target rectangle. 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 $f(z) = 0$.

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, $F([0,1]^2) = 1$ and $F([0,z]\times[0,w]) = 0$ for any $w$, so you cannot dissect a unit square (or even any finite set of rectangles with rational side lengths) into finitely many rectangles you can reassemble into a rectangle with one side of length $z$.

Note: Above we used the axiom of choice, this is not actually necessary for the proof to work. For any given finite rectangular dissection of some rational rectangles, simply let $V$ be the $\mathbb Q$-vector subspace of $\mathbb R$ generated by all side lengths of subrectangles in that dissection. Note that $z$ should lie in $V$ if the dissection is intended to fit into the target rectangle. 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 also that this proof easily generalizes to higher dimensions. No set of rational cuboids has a finite axis-aligned cuboidal equidissection with a cuboid with some irrational edge length.

Magma
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  • Sorry, I don't understand what you mean by “two sets $\mathcal A$ and $\mathcal B$ of rectangles have a common rectangular dissection”. Do you mean that each of the two sets $\mathcal A$ and $\mathcal B$ of rectangles is a rectangular dissection of a common rectangle (in this case $[0,\pmb,, 1]^2$ ? – John Bentin Oct 30 '21 at 06:03
  • The other related question was about dissecting two square and reassemble the parts into one. A and B would be the set of two squares and the set of one. The sentence says both sets can be dissected into the same set of rectangles. For this question you can simplify it to "If two rectangles A and B have a common rectangular dissection then F(A) = F(B)." – Florian F Oct 30 '21 at 06:30
  • @JohnBentin Two sets of rectangles $\mathcal A$ and $\mathcal B$ have a common rectangular dissection if the rectangles in $\mathcal A$ can be dissected into smaller rectangles which can then be reassembled to form the rectangles in $\mathcal B$ (i.e. you can cut $\mathcal A$ and $\mathcal B$ into the same pieces). – Magma Oct 30 '21 at 09:14