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.
Thus the set of numbers N2N = {1, 2, 3, ..., 2N - 1, 2N} is split into two sequences, one decreasing and one increasing:
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a1 > a2 > a3 > ... > aN and
b1 < b2 < b3 < ... < bN.
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The applet is supposed to suggest a theorem due to V. Proizvolov, a well known Russian mathematician:
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|a1 - b1| + |a2 - b2| + ... + |aN - bN| = N2.
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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.
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This would imply that (together with ai) there are in the very least (N - i + 1) 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
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
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| |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. |
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References
- S. Savchev, T. Andreescu, Mathematical Miniatures, MAA, 2003, p. 66.
Copyright © 1996-2008 Alexander Bogomolny
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