# Can one prove existence of incommensurables without the Pythagorean theorem?

Euclid's proof that the side and the diagonal of a square have no common measure, probably going back to Pythagoreans, reduces it to proving the irrationality of $\sqrt{2}$. This reduction uses the Pythagorean theorem, and therefore the axiom of parallels. However, there isn't a common measure for all segments in the hyperbolic plane either, so existence of incommensurables must be independent of the axiom of parallels. Since Cantor's axiom of continuity allows to construct real numbers geometrically it is also enough to produce incommensurables, but what if we leave out these two.

If we take lines with rational slopes only in $\mathbb{Q}^2$ the axiom of congruence for segments isn't satisfied, there is no segment along the diagonal of a rational square congruent to its side and with endpoint at its vertex.

Can one prove existence of incommensurables in elementary absolute geometry, i.e. without the axioms of parallels and continuity? Is commensurability at least consistent with the rest of Hilbert's axioms?

This question is a follow up to What are some examples of proofs using the Pythagorean assumption that all segments are commensurable?.

EDIT: Without the axiom of parallels metric notions, and therefore correspondence between segments and numbers, can not be established. So variations on $\sqrt{2}$ and the golden ratio do not work. Proofs that the Euclidean construction cuts a segment in the golden ratio for example also use some equivalent of the axiom of parallels (angle sum $\pi$ or Euclidean trigonometry). Without the axiom of continuity cardinality arguments do not apply either. On the other hand, there is no obvious model of geometry with infinitely many points where all segments are commensurable, $\mathbb{Q}^2$ with rational slopes does not work exactly because of the unit square and its diagonal.

• You wrote "However, there isn't a common measure for all segments in the hyperbolic plane either, so existence of incommensurables must be independent of the axiom of parallels." Can you amplify on this a little? My initial thought was the exact opposite: to prove the existence of incommensurables, I'd definitely want to be allowed to use the parallel postulate. – John Hughes Aug 28 '14 at 19:25
• There are many possible arguments. The first one only relies on cardinality: $|\mathbb{R}|=2^{\aleph_0}>\aleph_0=|\mathbb{Q}|$. Another possibility is to show that there exist real numbers with an infinite continued fraction, like $\frac{1+\sqrt{5}}{2}=[1;1,1,1,\ldots]$. – Jack D'Aurizio Aug 28 '14 at 19:29
• @JackD'Aurizio You wouldn't be able to construct $\mathbb{R}$ without continuity, but the golden ratio might work using geometric version of the Euclidean algorithm, so-called anthyphairesis. What gives me pause is that I am not sure that the usual Euclidean construction of the golden cut en.wikipedia.org/wiki/Golden_ratio#Geometry still gives the golden cut if performed in the hyperbolic plane for example. – Conifold Aug 28 '14 at 20:10
• @John I meant that incommensurables exist (in the hyperbolic plane) even if the axiom of parallels is dropped, so it isn't logically necessary to produce them as in Euclid's proof. – Conifold Aug 28 '14 at 20:13
• I don't buy that argument: it's possible that with the parallel postulate, incommensurables MUST exists, but in geometries that lack the PP, some contain incommensurables and some do not. – John Hughes Aug 29 '14 at 10:57

## 2 Answers

Irrationals involving the golden ratio do appear in every hyperbolic plane. See https://en.wikipedia.org/wiki/Ideal_triangle. Your edit points out that Euclid's construction of the golden ratio will not work, but your question does not specify just what kinds of constructions are allowed. I think a full answer would depend on making the question more precise.

• @ Colin McLarty Any constructions that follow from Hilbert's axioms without parallels and continuity are allowed. The problem with irrationals you pointed out is that they are specific to the hyperbolic plane. But you have to construct incommensurables, and prove that they are incommensurable, in a general way that doesn't rely on any special properties of Euclidean or hyperbolic planes. Incommensurables certainly exist there since they all satisfy continuity. A model, like hyperbolic plane, can be used to show that some construction does not work in general, but not that it does. – Conifold Aug 30 '14 at 3:12
• Euclidean construction works in that plane but not in the hyperbolic one, ideal triangles work in the hyperbolic plane but not in the Euclidean. Therefore, neither works in general. Relationships between segments and numbers are different in those two planes because metrics are different. Therefore, there can be no such definitive relationship in absolute geometry (even with specified measure unit). – Conifold Aug 30 '14 at 3:26
• @Conifold Does your version of Hilbert's axioms without parallels prove the sum of angles in a triangle differs from two right angles by some (possibly 0) multiple of the area? I expect so, since you even include the continuity axiom. Then they prove the plane is either Euclidean or hyperbolic. And each case proves existence of incommensurables. – Colin McLarty Aug 30 '14 at 10:22
• @ Colin McLarty Axiom of continuity is not included (if it was incommensurables trivially exist by cardinality argument), without it Euclidean or hyperbolic planes are not the only models. That angle sum is $\leq\pi$ can be proved without continuity (I think). The idea to treat $\pi$ and $<\pi$ cases separately is promising but we need a construction that does not rely on continuity in the second case. Unfortunately, construction of lines "intersecting at $\infty$", and hence of ideal triangles, explicitly uses Dedekind cut on angles with intersecting and non-intersecting lines. – Conifold Aug 30 '14 at 21:14
• @Conifold Sorry, I thought your earlier note ruled parallels out and continuity in, but you meant it to rule continuity out too. So one strategy that might work is to look for a construction which in the Euclidean case gives the golden ratio (Wikipedia gives a very simple one) and such that even without parallels (and so without the golden ratio property) you can still prove the two segments cannot be commensurable. – Colin McLarty Aug 30 '14 at 21:56

There is a proof that any square root of a natural number is either a natural number or irrational that uses the unique prime number factorization, not Pythagoras. I don't actually know if there is some subtle dependency on the parallel postulate - hopefully someone else will point that out if that is the case.

If a rational square root of n exists, then (similar set-up to a proof of the irrationality of $\sqrt{2}$) $a^2 = b^2n$ for natural a and b. Since any prime occurring in the factorization of a occurs an even number of times in the factorization of $a^2$, and since the factorization is unique, this same prime occurs an even number of times in $b^2n$. But, in the factorization of $b^2$ any prime also occurs an even number of times. Since the difference of two even numbers is even, this means that any prime occurring in the factorization of n also occurs an even number of times, e.g. n is a perfect square (this is the contradiction).

Of course, we know that not every square root of a natural number is a natural number, so this shows the existence of irrationals (or incommensurables).

• The problem is in relating segments to numbers, to talk about $\sqrt{2}$ in geometry you need a way of producing a segment of length $\sqrt{2}$ given a segment of length $1$. Drawing the diagonal of a unit square gives you that in the Euclidean plane, but proving that the diagonal really has length $\sqrt{2}$ requires the Pythagorean theorem. And it really does because in the hyperbolic plane, where all axioms except the axiom of parallels hold, the diagonal does $not$ have the length $\sqrt{2}$. – Conifold Aug 30 '14 at 0:22
• Here is a simple algebraic proof that the square root of every integer that is not a perfect square is irrational: math.stackexchange.com/questions/471648/…. This proof does not use unique factorization. – marty cohen Sep 17 '14 at 1:40