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

34 marcus giaquintoPythagoras’s theorem is used to infer that the base of the right-angled trianglehas length √ ab,wherea and b are the diameters of the larger and smaller of theosculating circles respectively; then visualizing what happens to the trianglewhen the diameter of the smaller circle varies between 0 and the diameter ofthe larger circle, one infers that √ ab (a + b)/2andthat √ ab = (a + b)/2justwhen a = b. Reflecting on this way of reaching the inequality one can retrievethe premisses and the steps, and by expressing them in sentences construct asound non-visual argument. This is quite different from the visual genesis ofour belief that the triangles of a square either side of a diagonal are congruent.In that case reflection on our thinking does not provide us with any argument,and we are reduced to saying ‘It’s obvious!’In some cases visual thinking inclines one to believe something but onlyon the basis of assumptions suggested by the visual representation that remainto be justified given the subject’s current knowledge. In such cases there isalways the danger that the subject takes the visual representation to show thecorrectness of the assumptions and ends up with an unwarranted belief. Insuch a case, even if the belief is true, the subject has not discovered the truth,as the means of belief-acquisition is not reliable and demands of rationalityhave been transgressed. This can be illustrated by means of an example takenfrom Nelsen’s Proof Without Words (Montuchi and Page, 1988) presentedinFig. 1.5.Reflection on this can incline one to think that when the positive realnumber values of x and y are constrained to satisfy the equation x.y = k (wherek is a constant), the positive real number values of x and y for which x + y isminimal are x = √ k = y. (Let ‘#’ denote this claim.) Suppose that a personknows the conventions for representing functions by graphs in a Cartesiancoordinate system, and knows also that the diagonal represents the functiony = x, and that a line segment with gradient −1from(0, b) to (b, 0) represents(0, j)xy = k(0, 2 k)(k,k)(2k, 0)(j, 0)Fig. 1.5.