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Let's say we have axiomatic geometry as defined by Hilbert's axioms. For line segments, angles, triangles, squares, etc. we have the notion of congruency to determine whether two of them are "the same".

But this doesn't seem sufficient to determine whether two figures of different shape have the same area. For example, I don't see how the Pythagorean theorem can be proved using only the notion of congruency.

So basically my question is: how is area defined in axiomatic geometry?

Marc
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  • Can the axiomatization you have in mind prove that the ratio between the circumference of a circle and its diameter is a constant? Then you must use some notion of continuity and limits, and the same could be applied to areas as well. Otherwise, you are only working with shapes that can be triangulated, and in that case you can just assume the area of the rectangle as a definition. – dxiv Oct 16 '16 at 21:55
  • As a separate comment, the Pythagorean theorem has many proofs which don't rely on areas. See for example Proof 6 at Pythagorean Theorem, which is based on similarity alone. – dxiv Oct 16 '16 at 21:59
  • Both your examples include multiplication. I don't see how to define multiplication using Hilbert's axioms either. Maybe I'm missing something really basic here. – Marc Oct 16 '16 at 22:08
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    @Marc: Hilbert's own exposition of how to do algebraic operations on line segments in his system is quite accessible: see http://www.gutenberg.org/files/17384/17384-pdf.pdf?session_id=50ca737ae2a89ae1c50440dc92ef4c27b5723ac8 Chapter III Section 15. Hilbert uses triangular decompositions to reason about areas. See Chapter IV. – Rob Arthan Oct 16 '16 at 22:18
  • @Marc Unless I misunderstand your objection, note that the Hilbert axioms imply that segment lengths form an ordered field isomorphic to $\mathbb{R}$ (see for example this answer to Two congruent segments does have the same length?). At that point, the equivalence between $\frac{a}{b}=\frac{c}{d}$ and $a \cdot d = b \cdot c$ becomes a simple agebraic manipulation, which doesn't require any notion of areas. – dxiv Oct 16 '16 at 22:26
  • @RobArthan: Thanks a lot, this looks promising. It will take me some time to digest it. And of course, Hilbert himself leaves the proof of the Pythagorean theorem as an exercise ;-) – Marc Oct 16 '16 at 22:39
  • @dxiv: It is not the notion of area itself which is important to me. I want to know how we can talk about things like the Pythagorean theorem using the axiomatic method directly. So I'd like to see a proof which uses only the axioms (maybe organized in some lemmas and theorems). Of course, we could use the fact that the Euclidean plane as defined by Hilbert's axioms is essentially isomorphic to $\mathbb{R}^2$ with the Euclidean inner product and use this to prove the theorem. But that's not what I'm after. – Marc Oct 16 '16 at 22:50
  • @Marc In Chapter IV Section 20 of Hilbert's own book linked in Rob Arthan's comment, there is the statement that the similarity $a : h_b = b : h_a$ implies the equality of products $a \cdot h_a = b \cdot h_b$. I don't quite see what's the difference between that and the similarity-based Pythagora's proof quoted earlier. – dxiv Oct 16 '16 at 23:10
  • @dxiv: You may be right and I may be confused, so thanks for bearing with me! I haven't really digested the relevant parts in Hilbert's book yet. Maybe we need to define multiplication of line segments (like Hilbert does) in order to prove the Pythagorean theorem from the axioms. Then I understand why you don't see a difference between what you say and what Hilbert does. But naively, if you say that there are proofs which use "similarity alone", I would expect that such proofs don't use multiplication but decompose the squares into sets of pairwise congruent triangles directly. – Marc Oct 16 '16 at 23:21
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    @Marc No problem, yours are perfectly legitimate questions. Only, they don't necessarily have quick short answers. One thing to remember, as Blue points out in his answer, is that Pythagora only holds in the Euclidian geometry. Once you specialized Hilbert's axioms to the Euclidian case, the isomorphism with $\mathbb{R}^2$ follows (rigorously and) naturally, so there is no "sin" to use the algebraic properties of real numbers when dealing with ratios and products of segment lengths. – dxiv Oct 17 '16 at 00:03

1 Answers1

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From Marvin Jay Greenberg's excellent "Euclidean and Non-Euclidean Geometries" ...

What does "area" mean [...]? We can certainly say intuitively that it is a way of assigning to every triangle a certain positive number called its area, and we want this area function to have the following properties:

  1. Invariance under congruence. Congruent triangle have the same area.

  2. Additivity. If a triangle $T$ is split into two triangles $T_1$ and $T_2$ by a segment joining a vertex to a point on the opposite side, then the area of $T$ is the sum of the areas of $T_1$ and $T_2$.

Having defined area, we then ask how it is calculated. [...]

Basically, any strategy for assigning values that satisfy (1) and (2) above can be reasonably interpreted as "area" in a geometry. The calculations are what make things interesting.

  • In Euclidean geometry, one derives that the "one-half base-times-height" formula satisfies the necessary conditions. (Note: So does "one-half base-times-height-times-an-arbitrary-positive-constant".)

  • In spherical geometry, one can show that angular excess ---that is, "angle sum, minus $\pi$"--- works as a triangle's area function (up to an arbitrary constant multiplier). (We can do a sanity check with a simple example: A triangle with a vertex at a sphere's North Pole, and with opposite side falling on 1/4 of the Equator, covers one-eighth of that surface of the sphere; therefore, it has area $\frac{1}{8}\cdot 4\pi r^2 = \frac{\pi}{2}r^2$. On the other hand, such a triangle's angular excess is $\left(\frac{\pi}{2}+\frac{\pi}{2}+\frac{\pi}{2}\right) - \pi = \frac{\pi}{2}$, which is, in fact, proportional to the calculated area. (If we work on the unit sphere, we get to ignore the constant of proportionality.))

  • In hyperbolic geometry, angular defect ---"$\pi$, minus angle sum"--- is the go-to function. (This is harder to check than in the spherical case, so I'll note a fascinating consequence: a triangle with three infinitely-long sides happens to have three angles of measure $0$; therefore, such a triangle's area is finite ... specifically: $\pi$! (Constant of proportionality ignored, to maximize the impact of that statement.))

So, because area calculations are so very different in these contexts, you can't expect a single formula to fall out of the basic axioms. At some point, you observe a phenomenon that satisfies (1) and (2), you declare "this is area", and you go on from there.

Importantly, there need not be any direct connection between a geometry's notion of area and that geometry's incarnation of the Pythagorean Theorem. The fact that squares erected on the legs of a Euclidean right triangle have total area equal to that of the square erected upon the hypotenuse is a neat "coincidence". This doesn't happen in non-Euclidean (spherical or hyperbolic) geometry: these spaces don't even allow squares!

(See Wikipedia for a discussion of the non-Euclidean counterparts of the Pythagorean Theorem.)

I've been a bit informal here, but hopefully I've shown that your question "How is area defined in axiomatic geometry?" is actually quite deep.

Blue
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