Values in Caring for Proof

Abstract

In response to Gowers (Mathematical knowledge. Oxford University Press, Oxford, pp 33–58, 2007) who sought qualities of proofs that make them memorable, we suggest converting this into qualities that make proofs reconstructible. But Re-constructibility is a property of the person, situation and proof. This leads us to consider what values are being displayed when students are offered proofs, and the notion that, as a caring profession, there are tensions between showing care for students and care for mathematics. We offer interpretations of the commonly used words appreciation, comprehension and understanding in relation to proofs, and use these to analyse and develop three proofs of the irrationality of √2, in order to suggest that in addition to immersion in the mechanics of proof such as justifications, warrants, sentence structure and the like, what students need in order to be able to re-construct proofs is to internalise the act of personal narrative or self-explanation concerning not simply the steps and stages of a proof, but their appreciation, comprehension and understanding of the key ideas, including conceptual insights and technical handles.

Appendix: Extending the Carpet Reasoning

The carpet-proof extends to √3 using either a triangle or a square, but what details need to be checked, what narrative has to be supplied to turn either of the diagrams into a proof, and how does it depend on to whom the reasoning is to be presented?

For the triangle version, the uncovered triangle turns out to be integer sided and equal in area to the total area of the three overlap triangles. For the square version, the three lightly shaded squares are each one third of the area of the whole. Let them have edge length b while the overall has edge length a; the uncovered area is made up of six squares each with edge length (a − b)/2 and their total area is equal to the area of the two mid-shaded squares with edge length a − 3w, which overlap in the little dark-shaded square with edge length b − 2w. Although the mid-shaded squares are each equal in area to three of the small white squares, these do not have integral sides unless a and b are of the same parity. By constructing a square which has twice the edge length of one of these, with edge length 2(a − 3w) < a, it can be presented as 3 squares each with edge length a − b < b and so the infinite descent works.

The conceptual insight to construct a double sized square comes from Miller and Montague (2012, p. 110). Instead of requiring that some visible region be decomposed appropriately, they extended or varied the construction to construct a new object which can be suitably decomposed and yet involves integer sides smaller than the initial ones. Miller and Montague also wondered about other surds. They found the triangle version for √3, which they extended to √6 and √10 with greatly increased effort in calculating length, taking into account the multiple overlaps.

But for √15 and larger, the newly constructed triangle turns out not to be smaller than the original and so the reasoning breaks down (Miller and Montague 2012, p. 112). They observe that the method could not in case extend indefinitely, as some triangular numbers are also square!

Aiming for √5 using the corners idea with pentagons turns out to be rather more difficult, because the overlaps are not pentagons and the uncovered floor area is more than a pentagon.

Miller and Montague (2012, p. 112–113) found a clever way of decomposing the overlaps into two pieces, a regular pentagon at the top and a triangle at the bottom. The triangle matches the adjacent uncovered region (as indicated in the right-hand figure by the completion of the bottom overlap pentagon). It takes some facility to relate the edge-lengths of the resulting regular pentagons and the edge length of the central regular pentagon in order to verify that the they all have integer sides and so constitute an infinite descent and so a contradiction. Let a ² = 5b ², with the large pentagon having edge-length a and the smaller ones edge-length b. Then the inner uncovered pentagon has edge-length 5b − 2a, and the short edges of the kite are length a − 2b. Since

$$\frac{5b - 2a}{a - 2b} = \frac{a}{b} \; \text{if} \; \text{and}\; \text{only}\;\text{if} \;a^{2} = 5b^{2}$$

 and since 0 < 5b − 2a < a and 0 < a − 2b < b, the contradiction arises. There seem to be some resonances with the algebraic proof!

However, extending this approach to other numbers is likely to be increasingly difficult, since for d ≥ 6 it is even less clear how to reconfigure the overlaps, and for d > 25 placing regular d-gons in the corners of a regular d-gon leads to extra overlaps (but see Bogomolny webref for alternative geometrical proofs). It is even less clear how cube-roots or others might be handled geometrically.