Edited by Richard Walker, Monday 9 March 2026 at 10:22
To recap, Proizvolov's Identity says that
If we split the numbers into any two groups of numbers, arrange the first group in ascending order, the second in descending order, calcule the differences between the corresponding numbers in the two groups, and add up the differences, the total will always be .
In my earlier post I got as far as showing this will be true if taking the maximum of each pair of numbers at corresponding position in the two groups gives the integers in some order and similarly taking the minimum gives the integers in some order.
You can find a formal proof in the Wikipedia article but I found it a bit hard to follow and I think I would find it even harder to explain in a simple way, so I've made up an example that I hope will show why it must be true in a more intuitive way. This uses the same numbers as in my previous post.
The blue columns are the group in descending order, the green shaded columns the group in ascending order. The pair of numbers we going to focus on are those at position 4. You can see they are 5 and 6 but ignore that, another time they might have been different numbers, with a different maximum and minimum, it would not change the argument. I claim that, irrespective, one of them must be in the range and the other in the range . In our example , so that translates to and .
Why must this be? Well first suppose they are both less than or equal to 5. Then travelling along the path marked by circles the green bars at 1, 2 and 3 must be less than 5 (because the green bars are an increasing series); then we have the two values at position4, which we have just supposed are both 5 or less; then the blue bar at 5 must be less than 5 (because the blue bars are a decreasing series). Thus all 6 numbers along the trajectory of the circles would have to be distinct integers in the range , which is impossible because there are only 5 numbers in that range.
Now suppose both the numbers at position 4 are greater than or equal to 6. Then by a parallel argument to the one just given but following the trajectory marked by the squares, it would mean we had 6 distinct integers in the range , which again is impossible because there are only 5 numbers in that range.
So one of the numbers at position 4 must be in the lower range and one in the upper and the same reasoning shows that the same applies to the numbers at every position, and we can extend this to the general case using the appropriate notation and make it into a proper proof of Proizvolov's Identity, but my aim in this post has been to offer an insight into why the identity is true in a simple way that doesn't use anything complicated.
Proizvolov's Identity—How To Complete The Proof
To recap, Proizvolov's Identity says that
If we split the numbers into any two groups of numbers, arrange the first group in ascending order, the second in descending order, calcule the differences between the corresponding numbers in the two groups, and add up the differences, the total will always be .
In my earlier post I got as far as showing this will be true if taking the maximum of each pair of numbers at corresponding position in the two groups gives the integers in some order and similarly taking the minimum gives the integers in some order.
You can find a formal proof in the Wikipedia article but I found it a bit hard to follow and I think I would find it even harder to explain in a simple way, so I've made up an example that I hope will show why it must be true in a more intuitive way. This uses the same numbers as in my previous post.
The blue columns are the group in descending order, the green shaded columns the group in ascending order. The pair of numbers we going to focus on are those at position 4. You can see they are 5 and 6 but ignore that, another time they might have been different numbers, with a different maximum and minimum, it would not change the argument. I claim that, irrespective, one of them must be in the range and the other in the range . In our example , so that translates to and .
Why must this be? Well first suppose they are both less than or equal to 5. Then travelling along the path marked by circles the green bars at 1, 2 and 3 must be less than 5 (because the green bars are an increasing series); then we have the two values at position4, which we have just supposed are both 5 or less; then the blue bar at 5 must be less than 5 (because the blue bars are a decreasing series). Thus all 6 numbers along the trajectory of the circles would have to be distinct integers in the range , which is impossible because there are only 5 numbers in that range.
Now suppose both the numbers at position 4 are greater than or equal to 6. Then by a parallel argument to the one just given but following the trajectory marked by the squares, it would mean we had 6 distinct integers in the range , which again is impossible because there are only 5 numbers in that range.
So one of the numbers at position 4 must be in the lower range and one in the upper and the same reasoning shows that the same applies to the numbers at every position, and we can extend this to the general case using the appropriate notation and make it into a proper proof of Proizvolov's Identity, but my aim in this post has been to offer an insight into why the identity is true in a simple way that doesn't use anything complicated.