Proizvolov's Identity

At the beginning, the applet displays 2N natural numbers: 1, 2, 3, ..., 2N - 1, 2N. Select any N numbers by clicking on N numbers in turn. The selected numbers will be arranged in a column on the left in the decreasing order. The remaining N numbers will be arranged in the increasing order on the right.


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


Buy this applet
What if applet does not run?

Thus the set of numbers N2N = {1, 2, 3, ..., 2N - 1, 2N} is split into two sequences, one decreasing and one increasing:

a1 > a2 > a3 > ... > aN and
b1 < b2 < b3 < ... < bN.

The applet is supposed to suggest a theorem due to V. Proizvolov, a well known Russian mathematician:

  |a1 - b1| + |a2 - b2| + ... + |aN - bN| = N2.

The proof is based on the observation that for every i, i = 1, ..., N, one of the numbers in the pair ai, bi belongs to NN = {1, 2, ..., N}, the other to NN+1, 2N = {N+1, N+2, ..., 2N}.

Assume on the contrary, that, for example, that, for some i, both numbers in the pair belong to NN:

(1) ai < N + 1
bi < N + 1.

This would imply that (together with ai) there are in the very least (N - i + 1) terms of the sequence {a} that are below N + 1. There also would be at least i terms from the second sequence {b} below N + 1. Together this would give N + 1 distinct natural numbers less than N + 1. But there are only N of them; thus our assumption (1) led to a contradiction.

The case

(2) ai > N
bi > N.

leads to a contradiction in a similar manner. Thus indeed, in any pair ai, bi one number belongs to NN, the other to NN+1, 2N. This implies that

|a1 - b1| + |a2 - b2| + ... + |aN - bN|= [(N+1) + (N+2) + ... + 2N] - [1 + 2 + ... + N]
 = [1 + 2 + ... + 2N] - 2·[1 + 2 + ... + N]
 = N(2N + 1) - N(N + 1)
 = N2.

References

  1. S. Savchev, T. Andreescu, Mathematical Miniatures, MAA, 2003, p. 66.

Related material
Read more...

  • Six integers out of 10: Pigeonhole Principle

  • Pigeonhole Principle (Same sum)

  • Pigeonhole in a Matrix

  • Pigeonhole in Calendar

  • A nice puzzle modeled on the Petersen graph

  • Pigeonhole with Disjoint Intervals

  • All antichains

  • Euclid via Pigeonhole

  • Light Bulbs in a Circle (an Interactive Gizmo)

  • |Contact| |Front page| |Contents| |Games| |Algebra| |Activities| |Store|

    Copyright © 1996-2012 Alexander Bogomolny

     40585929

    Search:
    Keywords: