Mancosu - Philosophy of Mathematical Practice (Oxford, 2008).pdf

236 michael hallettClearly, AB = BA ′ and OB = OB; both angles ̸ OBA and ̸ OBA ′ are π 2 .Thequestion is now to compare OA and OA ′ . What Hilbert does is to rotate OA ′positively through the angle t. This results in new coordinates (x ′′ , y ′′ ) for theend-point, wherex ′′ + iy ′′ = e (1+i)t · (e t cos t − ie t sin t)= e (1+i)t · e t (cos t − isint)= e (1+i)t+t · (cos t − isint)= e (1+i)t+t−it= e 2tTherefore, x ′′ = e 2t ,andy ′′ = 0. (See Fig. 8.4.) Thus, the rotation takes thepoint A ′ to the x-axis, namely to the point (e 2t , 0). But this point cannotcoincide with (1, 0), since e 2t > 1. (This is because the power series expansionof e 2t is 1.t 0 + 2t + 4t2 +···; since the expansion of 1 is 1.t 0 + 0 + 0 +···,2!the series for e 2t − 1is2t + 4t2 +···. The first coefficient of this is 2, hence2!e 2t − 1 > 0, and so e 2t > 1.) The length of OA ′ is therefore greater than 1, andso different from that of OA. The base angles in the extended triangle OAA ′are the same; the Angle-Sum Theorem holds, since the Parallel Axiom does.In consequence, we have an example of a triangle for which the base anglesare equal, but in which the sides opposite these angles are not, violating theITT (in its converse version).Simple calculation in Hilbert’s model also shows that OB + BA ′ < OA ′ ,violating the Euclidean Triangle Property. As for the theory of triangle area,Hilbert shows that the Pythagorean Theorem for right-angled triangles holdsin the form ‘The sum of the square surface areas over the two legs of theright angle equals the square over the hypothenuse’, since this depends onlyon the weaker Triangle Congruence Axiom. But the squares over BA ′ and BAare the same; since the square over OB equals itself, those over OA and OA ′must be the same, too. But OA < OA ′ ; hence the usual analytic conclusionfrom the Pythagorean Theorem (‘The length of the hypothenuse is the squareroot of the sum of the squares of the lengths on the other two sides’) fails,because we will have here sums which are the same, but squared hypothenuselengths which are different.³⁹ For this reason, the geometry constructed is³⁹ As Hilbert says in Hilbert ( ∗ 1902, 125a): ‘...one can no longer conclude that the sides are equal fromthefactthatthesquaresare’. See Hallett and Majer (2004, 597).