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

242 michael hallettintuitively inspired ‘purity of method’ question, it issues in a result on theabstract, conceptual level; not just this, but the result also shows why theassumption behind Euclid’s I, 1 is justified in an elementary way.Hilbert begins to consider this metamathematical problem on p. 65 of hislecture notes, pp. 64ff. of the Ausarbeitung. He constructs in effect the smallestPythagorean sub-field of the reals which contains 1 and π, whichyieldsacountable model of the axioms I–III (indeed of I–V, the whole system ofthe 1898/1899 lectures and the first edition of the Grundlagen) when the usualanalytic geometry is constructed from pairs of its elements taken as coordinates.However, the number √ 1 − (π/4) 2 does not exist in this field.⁴³ Since 1, 1,and π are in the field, the model will possess three lines which satisfy the2triangle inequality, but from which no triangle can be constructed. This is wellillustrated by the diagrams in the Ausarbeitung, pp. 66 and 68, hereFigs.8.6and 8.7.One can see from Fig. 8.7 that the upper apex of the triangle depicted inFig. 8.6 ought to have coordinates (π/4, √ 1 − (π/4) 2 ); but this is not a pointin the model. The same example shows that there can be a line partially withinand partially without a circle but which intersects the circle nowhere: take thevertical line x = π 4 in the diagrams; this ought to meet the circle x2 + y 2 = 1atpoints with y-coordinates ± √ 1 − (π/4) 2 , but again these points are missing.1 1π2Fig. 8.6. Model of the Failure of the Triangle Inequality Property.⁴³ Hilbert’s quick sketch of the argument is as follows. Suppose √ 1 − (π/4) 2 were in the Pythagoreanfield constructed, then, since this field is minimal, it would be represented by an expression formedfrom π and 1 by the five operations allowed; Hilbert denotes this expresssion by A(1,π).Butthen,as he points out, A(1, t) must represent the corresponding element √ 1 − (t/4) 2 of the correspondingminimal field constructed from 1 and the real number t,whatevert is. However, while A(1, t) is alwaysreal, it is obvious that √ 1 − (t/4) 2 will be imaginary for t sufficiently large t. Hence,A(1, t) will notalways represent it.