This week I've been attending the Conference on Meaningfulness and Learning Spaces at UC Irvine, in honor of the 80th birthday of my co-author Jean-Claude Falmagne. I believe videos of the talks will eventually be online and I'll put up a link to mine when that happens.

Jean-Claude himself gave a talk, in which the following cute geometric fact came up as an example: If $$f(x,y)$$ is the function that computes the length of the hypotenuse of a right triangle with side lengths $$x$$ and $$y,$$ then $$f$$ obeys the associative law. It's not difficult to see this from the algebraic form of the function, but Jean-Claude instead presented a geometric demonstration of this fact, based on the properties of an orthoscheme, a tetrahedron with four right triangles as its sides.

Three of the edges of the orthoscheme, forming a path (labeled $$a,$$ $$b,$$ and $$c$$ in the figure) are at right angles to each other and can be aligned to lie parallel to the coordinate axes; these three edges, together with the opposite long edge of the orthoscheme, form a cycle that divides the surface of the orthoscheme into two topological disks, each formed by two right triangles. One of these two disks forms a geometric representation of the formula $$f(a,f(b,c))$$ for using the Pythagorean formula to derive the length of the long edge as a function of the two short edges. The other forms a geometric representation of the different formula $$f(f(a,b),c).$$ Since these two formulas compute the same value (the length of the long edge) they must be equal to each other, and this equality is the associative law.

This result turns out to be prototypical of all associativity laws for strictly-monotonic functions of the real numbers. In the Pythagorean theorem, there is a monotonic invertible function $$g$$ (squaring) such that $$f(x,y)=g^{-1}(g(x)+g(y)).$$ Every monotonic invertible one-variable function $$g$$ gives rise to a monotonic associative two-variable function $$f$$ in this way, and (by a result of János Aczél) every monotonic associative function can be decomposed in this way. They all can be interpreted geometrically via the orthoscheme, as monotonic transformations of the Euclidean distance.

### Comments:

None:
2014-02-28T22:21:51Z

In the last paragraph, instead of $$g(x), g(y)$$, I think you want $$g(x) + g(y)$$.
-- Ravi Boppana

11011110:
2014-03-01T01:22:33Z

Yes — thanks for catching this.