NonEuclidean Geometry by Herbert Meschkowski (PDF)

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Noneuclidean Geometry focuses on the principles, methodologies, approaches, and importance of noneuclidean geometry in the study of mathematics. The book first offers information on proofs and definitions and Hilbert’s system of axioms, including axioms of connection, order, congruence, and continuity and the axiom of parallels. The publication also ponders on lemmas, as well as pencil of circles, inversion, and cross ratio. The text examines the elementary theorems of hyperbolic geometry, particularly noting the value of hyperbolic geometry in noneuclidian geometry, use of the Poincaré model, and numerical principles in proving hyperparallels. The publication also tackles the issue of construction in the Poincaré model, verifying the relations of sides and angles of a plane through trigonometry, and the principles involved in elliptic geometry. The publication is a valuable source of data for mathematicians interested in the principles and applications of noneuclidean geometry.

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⭐I give some sample highlights. As we all know, Euclid’s parallel postulate implies that the angle sum of a triangle is equal to two right angles. The converse is also true. Proof: Consider a point A and a line BC with AB perpendicular to BC, i.e. A+C=right. Now mark the point D on BC such that AC=CD. This isosceles triangle has the angles D, D, (2*right-C); therefore D=C/2. Continue in the same way, making E=C/4, etc. The angle at A tends to A+C/2+C/4+…=A+C=right so any line going through A not perpendicular to AB will eventually be contained in one of these triangles, so it will cut BC, so it is ruled out as a possible parallel as the parallel postulate says. QED.Thus in hyperbolic geometry the angle sum of a triangle cannot always be equal to two right angles. In fact, it will always be less. To prove this we need to prove first that the angle sum of a triangle can never be greater than 2*right. Proof: Take a triangle ABC, extend the base AB and put copies of the triangle next to each other along this line. Thus the B of the first triangle sits against the A of the second and so on. Let C’ be the angle between them, i.e. B+C’+A=2*right. We claim that C’>=C. If C>C’ then the base of the triangle would be greater than the distance between the tips. But this would mean that, for some n, to walk from A to C, then along the line through the tips, then back down to the n:th A would be shorter than walking the straight line from A to the n:th A, which is absurd. Thus C’>=C and so A+B+C<=B+C'+A=2*right. QED.In hyperbolic geometry the angle sum of a triangle is always less that two right angles. Proof: If the angle sum was always two right angles then the parallel postulate would hold in hyperbolic geometry, which it does not, so there must be at least one triangle with angle sum less than 2*right. We say that this triangle has positive defect 2*right-(A+B+C). Defect is clearly additive, and as we just proved it can never be negative, so if we cut the triangle into pieces at least one of them will have positive defect. By continuing to cut the piece that has positive defect we can obtain an arbitrarily small triangle with positive defect. Now consider an arbitrary triangle ABC. We just showed that we can find a triangle with positive defect that fits inside ABC. Obviously there is some triangular dissection of ABC that has this triangle as one of its pieces. Thus, since defect is additive and nonnegative, ABC must have positive defect as well. QED. ⭐

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