Sums, Products, and 1-1 Functions

Here is a quote by H. R. Hamley in the Ninth Yearbook of the National Council of Teachers of Mathematics (NCTM, 1934, p 37)

Our purpose in quoting these authorities [Spearman, Dewey, Rignano, Piaget] is partly to uphold the thesis that reasoning is ideal experiment, and partly to support the view that school mathematics should be taught as the symbolic expression of actual and potential activity; in other words, that school mathematics should be presented as a concrete and dynamic, rather than an abstract and static, science. We do not think that the function concept can be grasped by the average student in any other way. Mathematics is the projection of life upon the plane of human imagination.

Below I follow James Tanton's beautiful generalization of the Sums and Products puzzle. James showed that the puzzle admits numerous modification and pointed the way in which modifications can be produced. The function concept plays an important role in the generalization and the derivation requires if not imagination than an ability to follow simple rules. I am afraid, though, that not all of it is a projection of life experiences.

To remind, let there be sequence \(S\) of numbers, \(S = \{A_{1}, A_{2}, \ldots , A_{n}\}\) and a commutative and associative operation "\(\circ\)" applicable to any pair of numbers such that, for numbers \(A\) and \(B\), \(A\circ B\) is another number. The puzzle consists in guessing the result of a step-by-step process such that on every step a randomly selected pair of terms \(a\) and \(b\) is replaced with \(A\circ B\). It was shown that the final number does not depend on the order in which the terms of the sequence are selected and replaced; it is always equal to \(A_{1}\circ A_{2}, \ldots\circ A_{n}\). The question posed and answered by James is how to generate such operations "\(\circ\)".

From the arithmetic and algebra course we know that addition "\(+\)" and multiplication "\(\times\)" both possess the properties of commutativity and associativity. That is a start. We may assume that we know at least one such operation "\(\circ\)". Find any function f, 1-1 between a pair of sets of numbers. For example, \(f(x)=\frac{1}{x}\) is 1-1 between two copies of an open half-line \((0,\infty)\); function \(\mbox{log}(x)\) is 1-1 between \([1, \infty)\) and \([0,\infty)\). All 1-1 functions are invertible, i.e., for any 1-1 function \(f\) there exists the inverse function \(f^{-1}\) such that \(f^{-1}(f(x)) = x\) and \(f(f^{-1}(x)) = x\). In what follows \(f\) will always stand for a 1-1 function.

Given a commutative and associative operation "\(\circ\)" and a function \(f\), we can form another operation "\(\odot\)" as

\(a\odot b=f^{-1}(f(a)\circ f(b))\).

For example, if "\(\circ\)" is just the common addition and \(f=\frac{1}{x}\), \(a\odot b=1/(\frac{1}{a}+\frac{1}{b})=\frac{ab}{a+b}\) which we considered earlier.

That the new operation is commutative is rather obvious; this is a direct consequence of the commutativity of "\(\circ\)". It takes a little more effort to prove associativity:

\( \begin{align} (a\odot b)\odot c &= f^{-1}(f(a\odot b)\circ f(c)) \\ &= f^{-1}(f(f^{-1}(f(a)\circ f(b)))\circ f(c)) \\ &= f^{-1}(f(a)\circ f(b) \circ f(c)) \\ \end{align} \).

And similarly,

\( \begin{align} a\odot (b\odot c) &= f^{-1}(f(a)\circ f(b\odot c)) \\ &= f^{-1}(f(a)\circ f(f^{-1}(f(b)\circ f(c)))) \\ &= f^{-1}(f(a)\circ f(b) \circ f(c)) \\ \end{align} \).

Which proves the associativity of "\(\odot\)".


Related material
Read more...

  • 3-Term Arithmetic Progression
  • Aliquot game (An Interactive Gizmo)
  • Euclid's Game (An Interactive Gizmo)
  • Euclid's Game on a Square Grid
  • Sums and Products
  • Sums and Products, a Generalization
  • Zeros and Nines
  • A Candy Game: Integer Iterations
  • A Candy Game (Change Discharged)
  • Heads and Tails
  • Loop or Halt - An Interactive Gizmo
  • Breaking Chocolate Bars (An Interactive Gizmo)
  • |Contact| |Front page| |Contents| |Algebra|

    Copyright © 1996-2017 Alexander Bogomolny

     62599483

    Search by google: