# Is there a surface we know of that fulfills all of Euclid's postulates but has a negation of the first postulate or the fourth postulate? [closed]

The way I view Euclid's postulates are as follows:

1. A line segment can be made between any two points on surface A.

2. A line segment can be continued in its direction infinitely on surface A.

3. Any line segment can form the diameter of a circle on surface A.

4. The result of an isometry upon a figure containing a right angle preserves the right angle as a right angle on surface A.

5. If a straight line falling on two straight lines make the interior angles on the same side less than 180 degrees in total, the two straight lines will eventually intersect on the side where the sum of the angles is less than l80 degrees.

The way I view the five postulates is simple. Each postulate defines some quality a surface has.

For instance, 2 seems to define whether a surface is infinite/looped or finite/bounded, 3 seems to force a surface to be circular (or a union of circular subsets), and 5 I believe change the constant curvature of a surface (wether it is 0 or nonzero).

I want to determine what "quality" 1 and 4 define in the context of the surface itself. 1 seems to imply discontinuity vs continuity, and I think 4 would imply non-constant curvature. However, I am unsure. Ultimately I would like to assign each of these a quality of a surface that they define such that all surfaces can be "categorized" under some combination of postulates, but that is irrelevant.

What two surfaces individually violate the first postulate and violate the fourth postulate?

## closed as primarily opinion-based by The Great Duck, астон вілла олоф мэллбэрг, Daniel W. Farlow, JonMark Perry, user91500Jan 17 '17 at 6:10

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• Violating 4 is quite easy, you just have to mess up the usual notion of the angle. Say, take the standard Euclidean plane and then declare that the lines $x=0$, $y=0$ not to be orthogonal; instead declare that $x=0$ is orthogonal to the line $x=y$. – Moishe Kohan Oct 20 '16 at 14:51
• Note also that Axiom 5 is not Euclid's; maybe you did have in mind the original 5th postulate: "That, if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles." – Moishe Kohan Oct 20 '16 at 14:58
• @MoisheCohen Axiom 5 is equivalent to Euclid's. You're right, it is a different wording but its an equivalent statement. Also, axiom 4 probably trickier than just that. From what I read regarding it, the 4th postulate is more about how isometries affect angles, and the 4th one declares that isometries preserve right angles. It's not what euclid originally wrote but from I understand it is equivalent to his meaning of "equality" of "right angles". Plus, what I'm looking for is more a surface, not just a redefinition. Sure, saying certain lines are orthogonal like that will break it. – The Great Duck Oct 20 '16 at 18:19
• however, that ignores the tricky part: finding a surface that actually has that quality. After all, orthogonality is pretty well defined from what I understand, so then the issue is finding a surface messy enough to break that while not breaking the others. I admit, it might not exist, but it is still a worthy thought to pursue. – The Great Duck Oct 20 '16 at 18:21
• As I said, the surface is $R^2$, but you change the notion of angles. As for Axiom 5, it is equivalent to Euclid's only if you keep other axioms and you asked about violating Axiom 4. If you use Euclid's 5th postulate instead of Playfair's Axiom then violating Axiom 4 becomes more difficult since Euclid's 5th postulate uses angles (unlike Playfair's Axiom). – Moishe Kohan Oct 21 '16 at 0:03