Proizvolov's identity

In mathematics, Proizvolov's identity is an identity concerning sums of differences of positive integers. The identity was posed by Vyacheslav Proizvolov as a problem in the 1985 All-Union Soviet Student Olympiads.Template:Sfnp

To state the identity, take the first 2N positive integers,

1, 2, 3, ..., 2N − 1, 2N,

and partition them into two subsets of N numbers each. Arrange one subset in increasing order:

$A_1<A_2<\cdots <A_N.$

Arrange the other subset in decreasing order:

$B_1>B_2>\cdots >B_N.$

Then the sum

$|A_1-B_1|+|A_2-B_2|+\cdots +|A_N-B_N|$

is always equal to N².

Example

Take for example N = 3. The set of numbers is then {1, 2, 3, 4, 5, 6}. Select three numbers of this set, say 2, 3 and 5. Then the sequences A and B are:

A₁ = 2, A₂ = 3, and A₃ = 5;

B₁ = 6, B₂ = 4, and B₃ = 1.

The sum is

$|A_1-B_1|+|A_2-B_2|+|A_3-B_3|=|2-6|+|3-4|+|5-1|=4+1+4=9,$

which indeed equals 3².

Proof

For any $a,b$, we have: $|a-b|=\max \{a,b\}-\min \{a,b\}$. For this reason, it suffices to establish that the sets $\{\max \{a_i,b_i\}:1\le i\le n\}$ and :$\{n+1,n+2,\dots ,2n\}$ coincide. Since the numbers $a_i,b_i$ are all distinct, it suffices to show that for any $1\le k\le n$, $\max \{a_k,b_k\}>n$. Assume the contrary that this is false for some $k$, and consider $n+1$ positive integers $a_1,a_2,\dots ,a_k,b_k,b_{k+1},\dots ,b_n$. Clearly, these numbers are all distinct (due to the construction), but there are at most $n$ of them, which is a contradiction.

References

• .

External links

• Proizvolov's identity at cut-the-knot.org
• A video illustration (and proof outline) of Proizvolov's identity by Dr. James Grime