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Comment by thaumasiotes | Hacker News Reader

> He asked whether one could accomplish the task by chopping a circle into a finite number of pieces that could be moved within a plane and reassembled into a square of equal area — an approach known as equidecomposition

Huh. The Banach-Tarski theorem ("you can chop a sphere into a finite number of pieces and, by moving them within 3-space, reassemble them into a sphere of double the radius") strongly suggests this is possible. What's so interesting about the revised question?

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Actually, a Banach-Tarski-like result is impossible in 2D space, since there is a Banach measure (= volume definition to all subsets of the plane) that extends the usual volume definition (e.g. for circles).

The crucial idea that makes Banach-Tarski work in 3D is the insight that the set of rotations around an axis through the origin in 3-space has a free subgroup F on 2 generators (finite strings of A's, B's and their inverses). From this fact the proof is quite easy, but this comment is too small for it.

Banach-Tarski doesn't work in 2-dimensional space; there isn't a finite collection of subsets of the plane which can be assembled to make both one disc of radius one and two discs of radius one.

I believe that Banach-Tarski would make it much easier to disect a sphere and make a cube.

Chopping and rearranging something that's the same spherical shape (but different size) is different from chopping a 2D square and rearranging into a circle. Presumably, if it were easy, Tarski himself would have shown it, given that he's the one who posed the question.

> Chopping and rearranging something that's the same spherical shape (but different size) is different from chopping a 2D square and rearranging into a circle.

As the other responses point out, the Banach-Tarski construction uses properties of 3D space that do not occur in 2D space. But I don't think the "spherical shape" of the original volume or the final volume is relevant; the point of the construction is that the intermediate pieces can't really be said to have a shape.

As other people point out, 2-space is not 3-space. Only weakened version of B-T work in 2-space (e.g. infinite number of pieces).

Tarski, of course, was familiar with his work of the year before formulating the B-T paradox when he posed this question.

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