Axiomatic geometry pdf

Authors Authors and affiliations Audun Holme. This process is experimental and the keywords may be updated as the learning algorithm improves.

This is a preview of subscription content, log in to check access. There exist other variants of this notation as well, so it may be prudent of a reader to check out the notation in use before drawing any conclusions. Google Scholar. This statement has to be qualified somewhat. Suppose that the model is constructed in terms of the real numbers. This will be the case for all the models we are going to consider here. Then we may conclude from the existence of the model that if there is no contradiction among the axioms leading up to the construction of the reals, then there is no contradiction in our axiomatic system under consideration.

As the former statement is taken for granted, we may say that our new system is free of contradictions. This is the fine print of math. We should realize it is there, but not spend our lives worrying about it. Corollary 3. For any ray, there exists at least one opposite ray. This is part of [G] Betweenness Exercise 3. The ordering of the pair of rays does not matter. Given points A, B, C, the following are equivalent: 1.

AB and AC form an angle. BC and BA form an angle. CA and CB form an angle. So, the rays are distinct and not opposite, and they form an angle. Then, by Corollary 3. Notation 3. The edges of a triangle meet each other only at the vertices: Lemma 3. The other intersections are proved similarly. It follows from Lemma 3. The conclusion follows immediately. The converse of Theorem 3. However, if we exclude this case, then there is the following: Theorem 3. The following fact is going to be used several times, both in these Notes, and in the HW solution set.

Step 1 is to show B, C, D are distinct. However, by Theorem 3. Notation 4. Theorem 4. By Lemma 3. This Theorem is part of [G] Proposition 3. Under certain conditions, the point E is unique. E is on at least one of the segments CB or AC, so to prove uniqueness, there are two cases.

By Corollary 3. Since A, D, and F are not on a by a Lemma 3. Theorems 4. The above Proof of Pasch 2 using Pasch 1 is based on an argument appearing in Theorem 5. It is also [G] Betweenness Exercise 3. This is part of [G] Proposition 3. By Lemmas 3. Similarly, B is not on m. Now, we need to show A and B are on the same side of m. Supposing A and B are not on the same side of m, Theorem 3. Corollary 4.

This is the rest of [G] Proposition 3. This is the converse to [G] Proposition 3. This is part of the Corollary on p. This is the rest of Corollary on p. To summarize the above results Theorem 4. This is the rest of [G] Betweenness Exercise 3. Cases 1. For the last part of the Theorem, all that remains is to consider Case 2.

Lemma 4. This part of [G] Proposition 3. For any ray, there exists exactly one opposite ray. Note that Theorem 4. The above Theorem is Proposition 3. There are three cases. A, Z, and C are distinct by Theorem 3. Exercise 4. By the construction from the Proof of Theorem 4. Similarly, there is a line m so that A and C are on m, B is not on m, and there is a set H m, B , the half plane containing B, bounded by m. Lemma 5. Theorem 5. This Theorem is stated on [G] p. Definition 5.

Similarly using Corollary 3. Further, if the ray does not contain a vertex, it meets exactly one side. The set of all points in a given model of order geometry is a convex set. The convexity of the empty set is trivially true, for lack of a counterexample.

A ray is a convex set. A half plane is a convex set. Let S be the intersection of all the sets Si. If every Si is convex, then S is also convex.

This follows immediately from Theorems 3. The interior of an angle is a convex set. The interior of a triangle is a convex set. Given any set S, there exists a unique convex hull ch S. So, there is at least one convex set containing S; let ch S be the intersection of all convex sets which contain S. Such an intersection is convex by Theorem 5. It is contained in any convex set T which contains S, because it is the intersection of T with all the other convex sets containing S.

We can conclude that the two sets are equal. This is [G] Exercise 4. Note that there are 24 ways to order the four points, eight of which were already named, and the rest falling into two other sets of eight. By Theorem 5. The interior of a convex quadrilateral is a convex set, and the diagonals of a convex quadrilateral intersect at exactly one point.

This point is in the interior of the quadrilat- eral. Exercise 6. Use the trigonometric function identities, for example on [G] p. Theorem 7. The set of all L. Exercise 7.

This means: calculate F F z , and simplify. First, simplify the quantity by adding fractions. Then compute the limit. See Exercise 7.