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Euclid's Elements start with five Postulates, including the fifth one, the famous Parallel Postulate. Less well, known however is the Postulate that forms the basis for motivation behind the fifth: the fourth one, which states that "all right angles are equal." Students who see this for the first time might find this puzzling, because obviously two angles which are equal to a 90 degree angle are equal to each other, since Common Notion 1 says that "things which are equal to the same thing are are also equal to one another". But then they realize that the matter is so straightforward: the definition of a right angle is an angle produced when two lines intersect each other and produce equal adjacent angles, and it's not clear why an angle produced by one such pair of lines should bear any relation to an angle produced by another such pair of lines.

So Euclid's fourth Postulate is not redundant for the reason that beginning students might think. But my question is, is it nevertheless a redundant postulate, although for far less trivial reasons? David Hilbert, in his Foundations of Geometry (Grundlagen der Geometrie in German), claims to prove Euclid's fourth Postulate in theorem 15 (on page 19 of the PDF or page 13 according to the book's internal page numbering), prefacing the proof by saying "it is possible to deduce the following simple theorem, which Euclid held - although it seems to me wrongly - to be an axiom."

Now it's fair to say that Hilbert was working in a different (and more rigorous) system of axioms than Euclid was, but I think Hilbert's proof should be seriously considered for two reasons. First of all, why would he dub Euclid's decision to call "all right angles are equal" a Postulate as "wrong" if it merely reflected a stylistic difference concerning what you choose as starting assumptions and what you consider theorems? But more importantly, by tracing back all the assumptions used in the proof of theorem 15, it seems to me that only four of Hilbert's axioms are ultimately used: IV 3, IV 4, IV 5, and IV 6. And I don't think Euclid would have objected to any of these statements:

IV 3 follows directly from Euclid's Common Notion 2.

IV 4 is partly stated in Euclid's Book I Proposition 23, which doesn't depend on the fourth postulate, and the part of IV 4 which (I think) is not stated is easily provable in Euclid's system.

IV 5 follows from Euclid's Common Notion 1.

IV 6 is just part of Euclid's Book I Proposition 4, which doesn't depend on the fourth postulate at all.

So could Euclid have proven his fourth Postulate as a theorem instead of just assuming it?

Any help would be greatly appreciated.

Thank You in Advance.

EDIT: It seems to me that the key idea driving Hilbert's proof is that Euclid's Book I Proposition 4, i.e. the Side-Angle-Side (SAS) congruence theorem, implies that the supplements of equal angles are equal. Can anyone confirm or deny that this implication is in fact valid, and if it is valid, that the conclusion can be used to show that all right angles are equal?

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I suggest one of the modern treatments, as Hilbert gave an outline but left many parts unfinished. Two that I have are Euclid: Geometry and Beyond, by Robin Hartshorne; Euclidean and Non-Euclidean Geometries, by Marvin Jay Greenberg. –  Will Jagy Nov 2 '13 at 4:25
    
@WillJagy That may be true in general (and those references seem interesting), but the reasoning leading up to theorem 15 seems pretty detailed. But in any case, what really interests me is not Hilbert's system as such, but rather whether Hilbert's proof can be transplanted to Euclid's system, or whether the fact that Euclid's fourth postulate is a theorem is just due to some peculiarity of Hilbert's system. –  Keshav Srinivasan Nov 2 '13 at 4:33
    
I wonder that Euclid did not apply the "method" of proof of prop. I.4 to prove Post. 4. That method is open to objection by modern standards, but if Euclid could use it on I.4, why not on the postulate? –  Michael E2 Nov 2 '13 at 19:13
    
@MichaelE2 Euclid treated the method of superposition (which is what he used to prove Book I Proposition 4) the same way he treated the Parallel Postulate: something that happens to be true, but should be avoided if you have a more conservative method of proof. So he probably wouldn't have used it directly to prove the fourth Postulate. However, he definitely would have used Book I Proposition 4 to prove the fourth Postulate if he could, and Hilbert's proof is simple enough that Euclid could have easily discovered it, so it makes me question the validity and assumptions behind Hilbert's proof. –  Keshav Srinivasan Nov 2 '13 at 19:28
    
Then I wonder that he didn't assume I.4. –  Michael E2 Nov 2 '13 at 19:32

1 Answer 1

Euclid's right-angle postulate excludes the existence of cone points: right angles at the vertex of a cone are smaller than right angles elsewhere on the cone. So this postulate cannot be proved insofar as the other axioms apply on a cone, which one could argue that they do.

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Does the side-angle-side congruence theorem apply to a cone? That is Euclid's Book I Proposition 4, and Hilbert uses that result (which he takes as an axiom) in his proof of Euclid's fourth postulate. –  Keshav Srinivasan Nov 2 '13 at 15:15
    
Actually, Hilbert assumes part of side-angle-side as an axiom and proves the rest of it as a theorem. But regardless, the point is that both Euclid and Hilbert establish the side-angle-side congruence result in a way that doesn't depend on the assumption that all right angles are equal, and yet Hilbert manages to prove that all right angles are equal from the side-angle-side congruence result. So let me put it this way: does Euclid's fourth postulate follow from Euclid's other postulates plus the side-angle-side congruence result? –  Keshav Srinivasan Nov 2 '13 at 17:17
    
@Keshav I guess this will depend on the definition of angle. You can take the base point of the triangle to be the vertex of the cone and compare it to another triangle with its base point elsewhere. Then the hypotheses of SAS may be fulfilled in the sense that the sides are equal and the angles are both right (though actually different), and yet the triangles are not congruent. So in this sense SAS fails on the cone, and, conversely, SAS in this sense rules out the possibility of cone points. –  Viktor Blasjo Nov 2 '13 at 20:05
    
But why would you say that the two angles are equal just because they're both right angles, if you're not assuming that all right angles are equal? –  Keshav Srinivasan Nov 2 '13 at 21:17
    
@Keshav You wouldn't necessarily, but you might. You could for example choose to define the meaning of an angle between two rays as how big a part of a full revolution one ray needs to be rotated to coincide with the other. Then one could argue that the (actually smaller) right angle at the cone point equals the right angle elsewhere. –  Viktor Blasjo Nov 3 '13 at 0:25

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