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

visualizing in mathematics 35the function x + y = b. Then looking at the diagram may incline the subjectto think that for no positive value of x does the value of y in the functionx.y = k fall below the value of y in x + y = 2 √ k, and that these functionscoincide just at the diagonal. From these beliefs the subject may (correctly)infer the conclusion #. But mere attention to the diagram (or a visual image ofit) cannot warrant believing that the y-value of x.y = k never falls below thatof x + y = 2 √ k and that the functions coincide just at the diagonal; for theconventions of representation do not rule out that the curve of x.y = k meetsthe curve of x + y = 2 √ k at two points extremely close to the diagonal, andthat the former curve falls below the latter in between those two points. Sothe visual thinking is not in this case a means of discovering proposition #.But it is useful because it leads the subject to propositions which, once she hasestablished their truth,⁶ would provide her with what is needed to transforman inclination to believe into knowledge.We can say about the case just discussed that visual thinking is a heuristic aidrather than a means of discovery. When it comes to analysis mathematiciansof the 19th and 20th centuries were right to hold that visual thinking rarelydelivers knowledge. Visualizing cannot reveal what happens in the tail of aninfinite process. So visual thinking is unreliable for situations in which limitsare involved, and so it is not a means of discovery in those situations, let alonea means of proof. A more optimistic view is presented in Brown (1999), whereit is proposed that reflection on a single diagram (Fig. 1.6) suffices to prove theIntermediate Zero Theorem.The idea is that reflection on the diagram warrants the conviction that aperceptually continuous graphical line with endpoints above and below thehorizontal axis must cross the axis, and that in turn justifies the theorem: if af(b)abf(a)Fig. 1.6.⁶ This is easy enough. (i) For each equation it is trivial that if x = y, their common value is √ k.So the functions expressed by those equations meet at the diagonal. (ii) To show that the y-valuesof x.y = k never fall below the y-values of x + y = 2 √ k, we need only show that for positivex, 2 √ k − x k/x. As a geometric mean is less than or equal to the corresponding arithmetic mean,√[x.(k/x)] [x + (k/x)]/2. So 2√k x + (k/x). So2√k − x k/x.