# Decartes geometry: real numbers and the plane

In Decartes's geometry, we express every point in the plane (Euclidean goemetry) with a pair of real numbers, so we can transfer the geometry problem to algebra problem, but how can we know that the pairs of real numbers cover all the points in the plane?

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His actual name is Descartes (with an s at the end), just fyi. Though I don't quite understand your question because to me, the plane is nothing more than $\mathbb{R}^2$ by definition. Perhaps other people have a different take on the matter. –  Cameron Williams Jul 14 '14 at 23:02
@CameronWilliams Thanks –  89085731 Jul 14 '14 at 23:05
I edited your post slightly to make it more understandable for others. Please make sure that what I have changed is in-line with what you meant. –  Cameron Williams Jul 14 '14 at 23:08
@CameronWilliams thanks a lot, that's it. –  89085731 Jul 14 '14 at 23:09
We can use the axioms of Hilbert. (Euclid's are quite inadequate.) One can show (Hilbert shows) that any plane that satisfies his axioms is isomorphic to the Cartesian plane over a complete ordered field. There is (up to isomorphism) only one of these. –  André Nicolas Jul 14 '14 at 23:15

## 3 Answers

It's actually not a foolish question.

The first thing you can prove about the Cartesian plane is that it satisfies all of Euclid's posulates. So if you have a Euclidean geometry theorem, it is true in the Cartesian plane.

There might, however, be theorems about the Cartesian plane that are not provable in Euclidean geometry.

Note, in mathematics, we aren't actually concerned with the plane as a "real" object that exists in the world, but rather, axiomatic systems that describe objects.

As one commenter above notes, Euclid's axioms are actually deficient for certain purposes. Hilbert came up with a far more thorough axiom system which is much more closely related to the Cartesian geometry.

Nothing in Euclid's axioms says anything about "how many" points there are. It is entirely possible to do Euclidean planar geometry just in $F^2$ where $F$ is the smallest set of numbers containing all rationals, and is closed under taking sums, products, and square roots of positive values. So that's way fewer points than $\mathbb R^2$.

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It depends what one means by theorem about the Cartesian plane. Tarski's axiomatization is a complete theory. –  André Nicolas Jul 14 '14 at 23:34
@AndréNicolas Not sure what that is referencing in my above. Do you mean Tarski's axiom to replace "Euclidean geometry?" I'm talking about naive actually-Euclid Euclidean geometry. :) –  Thomas Andrews Jul 14 '14 at 23:37
Yes, technically Tarski's axioms. But we don't quite need that, since if we identify the Euclidean plane with $\mathbb{R}^2$, one can use the decision procedure for real-closed fields to settle any geometric question. Geometric question here is somewhat narrow, stuff about lines and circles. In particular one cannot quantify over the natural numbers. –  André Nicolas Jul 14 '14 at 23:42
If we identify the Euclidean plane with... The whole point of the question is, why can we? Sure, if we do, we can do all these things, but that's somewhat off-topic. @AndréNicolas –  Thomas Andrews Jul 14 '14 at 23:45
If we think of it as Cartesian plane over a real-closed field, the same is true, and the Tarski axiomatization, or any careful axiomatization, is sufficient to prove the isomorphism. –  André Nicolas Jul 14 '14 at 23:48

We need to be precise about the axiomatization of the Euclidean plane. One such axiomatization is the one due to Hilbert (Foundations of Geometry, 1899). Note that the partial axiomatization present in Euclid's Elements is quite inadequate.

Hilbert shows that any plane that satisfies his quite geometric axioms must be isomorphic to the coordinate plane over a complete ordered field. And it is a standard result that up to isomorphism there is only one complete ordered field, namely $\mathbb{R}$.

The axiomatization by Hilbert is second order. If we use a first-order axiomatization, such as the one due to Tarski, we must replace complete ordered field by real-closed field. And then there are many non-standard models of the geometry. One could imagine then that the genuine plane, whatever that may mean, is not isomorphic to $\mathbb{R}\times\mathbb{R}$. In other words, one could imagine that the genuine plane is either bigger or smaller than the ordinary Cartesian plane over the reals.

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This question can be simplified to whether or not the points on a line (given a choice of origin and unit) correspond to numbers.

In modern formulations of Euclidean geometry, there are completeness axioms which ensure that you have the "correct" number line. For example, here is Hilbert's version.

(note I've avoided saying "real number", as you might want other things in more general settings -- e.g. in first-order logic all you can really ask for from a number system is to be a real closed field, and there are corresponding completeness axioms that would guarantee that)

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could u please give me a link for that axiom? –  89085731 Jul 14 '14 at 23:25
I've edited it in –  Hurkyl Jul 14 '14 at 23:27