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Proofreading, while the question is on the front page anyway
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LSpice
LSpice
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Is Monsky's theorem dependingdependent on the axiom of choice?

The extension of the 2-adic valuation to the reals used in the usual proof uses clearly uses AC. But is this really necessary  ? After all, given aan equidissection in $n$ triangles, it is finite, so it should be possible to construct a valuation for only the algebraic numbers, and the coordinates of the summits (with a finite number of "choices"), and then follow the proof to show that $n$ must be even. Or am I badly mistaken  ?

Is Monsky's theorem depending on axiom of choice?

The extension of the 2-adic valuation to the reals used in the usual proof uses clearly AC. But is this really necessary  ? After all, given a equidissection in $n$ triangles, it is finite, so it should be possible to construct a valuation for only the algebraic numbers, and the coordinates of the summits (with a finite number of "choices"), and then follow the proof to show that $n$ must be even. Or am I badly mistaken  ?

Is Monsky's theorem dependent on the axiom of choice?

The extension of the 2-adic valuation to the reals used in the usual proof clearly uses AC. But is this really necessary? After all, given an equidissection in $n$ triangles, it is finite, so it should be possible to construct a valuation for only the algebraic numbers, and the coordinates of the summits (with a finite number of "choices"), and then follow the proof to show that $n$ must be even. Or am I badly mistaken?

add AC tag
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edit approved Aug 14, 2020 at 22:54
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edit approved Aug 14, 2020 at 22:54
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Is Monsky's theorem depending on axiom of choice  ?

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Feldmann Denis
Feldmann Denis
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Is Monsky's theorem depending on axiom of choice ?

The extension of the 2-adic valuation to the reals used in the usual proof uses clearly AC. But is this really necessary ? After all, given a equidissection in $n$ triangles, it is finite, so it should be possible to construct a valuation for only the algebraic numbers, and the coordinates of the summits (with a finite number of "choices"), and then follow the proof to show that $n$ must be even. Or am I badly mistaken ?