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I have a basic question about the Euclidean metric. I am assuming that I have a Euclidean plane represented by $\mathbb{R}^2$, and that measuring distances on lines parallel to the 'axes' is defined by making it same as distance on $\mathbb{R}$. I am also assuming that the 'axes' are 'perpendicular', otherwise we cannot even start. What I want to define is the distance between two points $(x_1,y_1)$ and $(x_2,y_2)$, such that it corresponds to the distance 'in real life' or in a rough sense, taking a scale and putting it end to end between the two points. It is said that the pythagorean formula provides a way to define a distance that corresponds to this. But I have two questions regarding the validity of the proofs of the pythagorean theorem in this minimalistic setup:

  1. The first one is the rearrangement type of proof, which relies on taking a right angled triangle and embedding it in a bigger square and then moving around pieces. There the question occurs to me is that from where are we getting the fact that the 'rotated' piece remaining after moving these figures around is a 'square' with all sides 'same length c', and that its area is $c^2$? That seems circular (in the logic sense) to me because a square cannot even be defined without having the property that all its 'side lengths' are same, and we don't even have a notion of length for the plane yet! Am I missing something? I know we can set up notions of area for arbitrary sets using measure for instance but what I fail to understand is how do we get the fact that the rotated piece is indeed 'congruent' as a square whose sides are parallel to the axes, without defining congruence through the pythagorean theorem.

  2. The second type of proof I have seen is the one which uses similarity. It uses the notion of angle, which seems to me there is no way of even defining without already having a notion of distance in place? Is there a way to understand it that avoids this circularity? Or even have congruent transformations without having the notions of distance in place?

Rohan Didmishe
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    See Metric foundations of geometry by Birkhoff. – lhf Apr 26 '23 at 11:53
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    Related: Critiques of Euclid's Book 1, Proposition 4 (side-angle-side triangle congruence). – aschepler Apr 26 '23 at 12:37
  • @lhf very fascinating! – Rohan Didmishe Apr 26 '23 at 13:29
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    This is a good question. I think it's answered here: Is Pythagoras' Theorem a theorem? – Ethan Bolker Apr 26 '23 at 14:02
  • @EthanBolker thanks for the answer you supplied in that link, I can see how the parallel postulate is relevant here. What is confusing me is this reliance on the fact that there is a 'square' on the remaining third side of the right angled triangle; how do we even define a square not parallel to the axes without knowing what distance is in a sense? – Rohan Didmishe Apr 26 '23 at 14:52
  • @RohanDidmishe If you're starting with the plane as the set of pairs of real numbers then the usual formula for distance is the definition of distance. Then you define rotations and translations in terms of coordinates and prove that those operations preserve distances. – Ethan Bolker Apr 26 '23 at 15:00
  • @EthanBolker I thank you again for your patience but I still can't convey what I want to ask....it is essentially that why is the formula for distance the one that corresponds to picking up a scale and placing it between two points....I understand once we define it that way we can proceed with everything in an axiomatic fashion....but do I have to accept the pythagorean theorem as an empirical fact to start with? – Rohan Didmishe Apr 26 '23 at 17:49
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    @RohanDidmishe Here's another way to look at your problem. To do geometry starting with coordinates you must first define the lines. Then you use ordinary algebraic operations on the coordinates to prove Euclid's axioms, including the parallel postulate. Then the Pythagorean theorem (stated geometrically) and hence the distance formula follow. (Note that Euclid never actually talks about distance as a number. He only manipulates line segments.) – Ethan Bolker Apr 26 '23 at 18:12
  • @EthanBolker Thank you once more for your answer, I shall think about it more deeply.. – Rohan Didmishe Apr 26 '23 at 18:20

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