By David Hilbert

The fabric inside the following translation used to be given in substance by means of Professor Hilbert as a process lectures on euclidean geometry on the college of Göttingen through the iciness semester of 1898–1899. the result of his research have been re-arranged and placed into the shape during which they seem right here as a memorial deal with released in reference to the social gathering on the unveiling of the Gauss-Weber monument at Göttingen, in June, 1899. within the French version, which seemed quickly after, Professor Hilbert made a few additions, rather within the concluding comments, the place he gave an account of the result of a contemporary research made by means of Dr. Dehn. those additions were included within the following translation.

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However, the rest of l'upffer's development of the theory of proportion should be regarded as inadequate. 54 III. THEORY OF PROPORTION Layoff the segments a = OA and b = OB from the vertex 0 on one of the sides of a right angle and on the other side the unit segment 1 = ~C . Let the circle through the points A, B, C intersect the latter side at the point D . ,-------~,-perpendicular to OC and reflecting in it the point C. In view of the equality of the angles 4 OCA and 4 OBD it follows by the definition of the product of two segments (p.

53 (1900). § 12. INDEPENDENCE OF CONTINUITY AXIOM V 43 urging led to a complete clarification of this problem. The investigations of M. Dehn are basic to Axioms I-III. Only at the end of Dehn's work, were axioms II of order formulated, in a more general way than in the present treatment in order also to encompass Riemannian (elliptic) geometry in the study. These can be formulated, perhaps, as follows: Four points A, B, C, D on a line can always be decomposed into two pairs A, C and B, D so that A, C and B, D are "separated" and conversely.

In order to prove the associative law § 15. ARITHMETIC OF SEGMENTS BASED ON PASCAL'S THEOREM 53 a(be) = (ab)c for segment multiplication, construct from the vertex 0 the segments 1 and b on one side of the right angle and also from 0 the segments a and e on the other side. Then construct the segments d = ab and e = eb and from 0 layoff these segments on the first side. , ae = cd ,, d·ab c or a(eb) = e(ab), and hence it also follows with the aid of the , commutative law that a(be) = (ab)c. 1 As can be seen in the above proof of the e commutative as well as of the associative laws of multiplication use was made only of the special case of Pascal's Theorem whose proof on pp.