If you drop the condition that the two sets have equal size then the claim is no longer true. Take the set {1,1,1,3,3}.

Is there a set of naturals, not all equal, that exhibit this behavior when for each division set A doesn’t need to be the same size as set B?

Great to hear from you!

I agree that with the matrix-solution field extensions are a little overkill (although I still like it better). I wrote it like that so as to use the same argument as in the more elegant matrix-less solution. I think that the matrix-less solution with the field extension is definitely more elegant.

And I will indeed start to update again, I have tons of new stuff to post…

A More Elementary Proof that the Rank Does not Change:
The rank of the matrix can be defined as the maximal number K, such that there exists a K x K on minor which is non singular. Clearly, singularity does not change when the field is extended, as it can be defined over any field as “Determinant = 0″, and the determinant does not change. Thus the rank does not change either.

Good to see you updating again!

Good solution. In the last part (the proof for the reals/complex) you use an assertion (without proof) that the rank is the same over Q and over R. The assertion is of course true (I prove it below for other readers). But then the construction of the matrix is not necessary and a more elegant solution exists (again, see below).

The Missing Proof that the Rank Does not Change:
Let T be a set of m vectors from Qn, independent over Q. Then T is also independent over R. Otherwise, let a1,…,am (in R) be the non-trivial coefficients than make the sum 0. Let e1,…,ek be a basis for the extension field Q(a1,…,am). Then in at least one coordinate, we have a non-trivial sum over Q that is 0. Contradiction.

A More Elegant Solution (no need to talk about matrices):
Let S be our set of reals. As S is finite, take a finite-degree extension field of Q that contains all the elements of S. Take a basis, e1,…,en of this extension field. Now, our property holds for S i.f.f. it holds for the projection of the elements of S onto each of the coordinates defined by e1 and en, and it does hold for each coordinate, as it holds for Q.

For Naturals:

First, lets notice that if S has the property, so does tS for every scalar t. Also, S+t has the property, because each partition is into two sets of equal size N.
Second, we notice that for every element x in S, sum(S-x) = sum(A) + sum(B) = 2 * sum(A) and it’s an even number.
Therefore, sum(S) is even iff x is even – which is correct for every element x. Therefore all elements of S has the same parity.
Now, first we take S and subtract min(S) from it. it will still have the property, still be all natural numbers, and now has a 0 element.
Since 0 is even, all elements are even, and so we can divide by 2 – again and again. if not all elements were 0, sometime, one of them will become odd, and this is a contradiction.
Therefore, they were all 0 – so all of S elements were equal.

For Integers:

Simply subtract min(S), and now all of S’s elements are non-negatives and integers.

For Rationals:

Simply multiply S by LCM(denominators of S) – now they’re all integers.

For Reals and Complex:

Since Q is a field, and we know that every set of rational numbers that have the property are all equal, we can write the following:
Let M be a 2N+1 x 2N+1 matrix, such that – the main diagonal is all 0′s. every Row has exactly N 1′s and N (-1)’s.
Then the dimension of the Nullspace of this matrix is 1: only vectors of type (x,x,x…,x)
So Rank(M) = 2N

but if Rank(M) = 2N in Q, it’s also the rank of the matrix in any other superfield of Q, such as R and C.
So the only vectors in the nullspace of this matrix are (x,…,x) in those fields too.