Proizvolov's Identity

At the beginning, the applet displays $2N$ natural numbers: $1, 2, 3, \ldots, 2N - 1, 2N.$ 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.


Please try switching to IE 11 (Windows) or Safari (Mac), for no other browser nowadays runs Java applets. If asked whether to allow the applet to load, click Yes - the applet is signed with a security certificate from a trusted company.

Proizvolov's Identity


What if applet does not run?

Thus the set of numbers $\mathbb{N}_{2N} = \{1, 2, 3, \ldots, 2N - 1, 2N\}$ is split into two sequences, one decreasing and one increasing:

$a_{1} \gt a_{2} \gt a_{3} \gt \ldots \gt a_{N}$ and
$b_{1} \lt b_{2} \lt b_{3} \lt \ldots \lt b_{N}.$

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

$|a_{1} - b_{1}| + |a_{2} - b_{2}| + \ldots + |a_{N} - b_{N}| = N^{2}.$

The proof is based on the observation that

For every $i,$ $i = 1, \ldots, N,$ one of the numbers in the pair $\{a_{i}, b_{i}\}$ belongs to $\mathbb{N}_{N} = \{1, 2, \ldots, N\},$ the other to $\mathbb{N}_{N+1, 2N} = \{N+1, N+2, \ldots, 2N\}.$

Proof 1

Assume on the contrary, that, for example, for some $i,$ both numbers in the pair belong to $\mathbb{N}_{N}:$

(1)

$\begin{align} a_{i} &\lt N + 1\\ b_{i} &\lt N + 1. \end{align}$

This would imply that (together with $a_{i})$ there are at 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)

$\begin{align} a_{i} &\gt N\\ b_{i} &\gt N. \end{align}$

leads to a contradiction in a similar manner. Thus indeed, in any pair $a_{i}, b_{i}$ one number belongs to $\mathbb{N}_{N},$ the other to $\mathbb{N}_{N+1, 2N}.$ This implies that

$\begin{align} |a_{1} - b_{1}| &+ |a_{2} - b_{2}| + \ldots + |a_{N} - b_{N}|\\ &= [(N+1) + (N+2) + \ldots + 2N] - [1 + 2 + \ldots + N]\\ &= [1 + 2 + \ldots + 2N] - 2·[1 + 2 + \ldots + N]\\ &= N(2N + 1) - N(N + 1)\\ &= N^{2}. \end{align}$

Proof 2

Let's assign the numbers in $\mathbb{N}_{N}$ one color, and another to the remaining numbers. It's a consequence of a general result that, if there is one monochrome pair, there necessarily is another one but of a different color. This leads to an immediate contradiction with the monotonicity of the sequences in two columns. Say, let one pair be $\{a',a''\},$ the other $\{b',b''\}.$ Then either $a'\lt b'$ and $a''\lt b''$ or $a'\gt b'$ and $a''\gt b'',$ a contradiction in both cases. From here the solution proceeds as before.

(This solution arose in an answer to a question posted by James Tanton on twitter.)

Proof 3

There is an elegant geometric proof by Staurt Anderson. It can be found on a separate page.

Proof 4

Grégoire Nicollier has observed that situation exploited by Proizvolov's identity can be generalized:

Let $S=\{a_1,\dots,\,a_n\}\cup\{b_1,\dots,\,b_n\}=\{c_1,\dots,\,c_{2n}\}$ be the given set of different numbers with $a_1\lt a_2\lt \cdots\lt a_n,$ $b_1\gt b_2\gt \cdots\gt b_n,$ $c_1\lt c_2\lt \cdots\lt c_{2n}.$ Then the sum

$\displaystyle\sum_{k=1}^{n}|a_{k}-b_{k}|$

is independent of the split of $S$ into the two so ordered sets.

Indeed, observe, that there is no pair $(a_k,\,b_k)$ whose both elements are smaller than $c_{n+1}$ or greater than $c_n$. Otherwise, there would be $k+(n+1-k)=n+1$ numbers in $S$ smaller than $c_{n+1}$ or greater than $c_n$, a contradiction. Each term $|a_k-b_k|$ is thus equal to $c_m-c_\ell$ for different $m$, $\ell$ with $\ell\le n\lt m$.

As an example, if $c_{k}=k^{2},$ then $\displaystyle\sum_{k=1}^{n}|a_{k}-b_{k}|=n^{2}(2n+1),$ independent of a particulare split of $S$ into two sets.

Grégoire also observed that the statement of the generalization holds when the inequalities are not required to be strict.

References

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

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