I have a question about the proof of the Bolyai-Gerwien theorem presented in the book "Conjecture and proof" from Miklós Laczkovich. In particular, he claims the following key statement:

Any rectangles of the same area are equidecomposible using translations alone

To show this statement he claims that any rectangle of size $a\times b$ is equidecomposible to a rectangle of size $\frac{a}{2}\times{2b}$ using translations alone. This is rather clear, at least geometrically as illustrated in the following picture. enter image description here

It is also quite clear, that this argument shows that any rectangles of size $\frac{a}{n}\times{2n}, n\in\mathbb{N}$ are equidecomposible. Less obviously, any rectangles of size $r\times s, rs=ab,r,s\in\mathbb{Q}$ are equidecomposible.

But then he follows up with the following claim: Repeating this process, we can see that any rectangle R is equidecomsible into a rectangle who's size is at most twice the width While the statement is factually correct(since any two rectangles are equidecomposible), in my opinion the arguments is incomplete.

For example, how does this prove that a rectangle of size $1\times1$ is equidecomposible to a rectangle of size $\frac{1}{\sqrt{2}}\times\sqrt{2}$? For the suggested decomposition to work, you would need an infinite number of steps(which is not allowed in the definition of equidecomposible, which is provided at then end of the question)

I hope this question is not too long, any help is appreciated :)

PS: This is a picture of the whole proof as presented in the afore mentioned book Proof of the theorem by Laczkovitch

PS: This is the definition of equidecomposable: We say that A,B polygons are equidecomposable iff $\exists A_1,\ldots A_n, B_1,\ldots B_n$ such that $A ={\displaystyle\cup_{i=1}^{n}A_i}$, $B ={\displaystyle\cup_{i=1}^{n}B_i}$ and $A_i\cong B_i,\forall i\in\{1,\ldots,n\}$ (that is, there is a rigid movement that transforms $A_i$ into $B_i$) Equidecomposable is an equivalence relation


You misunderstood the statement that you're objecting to, "any rectangle R is equidecomsible into a rectangle who's size is at most twice the width." It doesn't say that $R$ is equidecomposable with every rectangle $R'$ of the sort desctibed (length at most twice the width) but rather with some such rectangle. (It is, as you noted, in fact equidecomposable with every such $R'$, but that's still going to be proved in the rest of the argument.)

  • $\begingroup$ Oh wow! What a stupid mistake I made? $\endgroup$ – miraunpajaro Jul 11 at 18:03

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