## Really Equal? Naturally!

You have a set *S* of 2N+1 real numbers, with the following property: if you remove any one element, you can partition the remaining 2N elements into two sets, *A* and *B*, each of size N, such that the sum of the N numbers in set *A* equals the sum of the N numbers in set *B*. Prove that all the numbers in the original set are equal.

### Spoiler Alert - Hint Ahead

Start by trying to solve the riddle for the special case where the set contains only *natural numbers* instead of general reals.

## Spoiler Alert - Solution Ahead!

We will solve the riddle for the case of natural numbers, integers, rational numbers, and finally real numbers. Note that everywhere below I use the regular set notation to denote multisets.

### Solving for the Natural Numbers

** Theorem 1**: If the numbers in

*S*are all natural numbers, then they are all equal.

** Lemma 1**: If \(S \subset \mathbb{Z}\), then all the numbers in

*S*must have the same parity, i.e. either all numbers are even, or all numbers are odd.

** Proof of Lemma 1**: Indeed, let’s consider two numbers \(x, y \in S\). Since you can split the remaining numbers into 2 sets of equal sums that must mean that \((\sum_{a \in S}a) - x\) is even and \((\sum_{a \in S}a) - y\) is even. Therefore

*x*and

*y*have the same parity. \(\blacksquare\)

** Lemma 2**:

*S*is a set satisfying the above property i.f.f. the set \(\{x - t \mid x \in S, t \in \mathbb{Z}\}\) is satisfying the above property.

** Lemma 3**: If

*S*contains only even numbers, then

*S*satisfies the above property, i.f.f. the set \(\{\frac{x}{2} \mid x \in S \}\) satisfies the above property.

The proofs of Lemmas 2 and 3 are trivial.

** Proof of Theorem 1**: By

**, all elements of**

*Lemma 1**S*are either even or odd. If all elements are odd, by

**, we can subtract 1 from all elements of**

*Lemma 2**S*, making all elements even. If all the elements are even, by

**, we can divide all elements of by 2. Consider the sequence of sets you get by repeated application of this process, diving by 2 if all elements are even, and removing 1 if all elements are odd, until one of the elements of the set,**

*Lemma 3**x*, becomes 0. Call the set at that point

*S’*. Let

*y*be another element of

*S’*. If

*y*is not 0, y must be an even number from Lemma 1. Keep diving all elements by 2 (from

**) at some point the image of**

*Lemma 3**y*will be an odd number, leading to a contradiction. Therefore for all \(y \in S', y = x\). Therefore all elements of

*S’*are equal. Therefore all elements of

*S*are equal. \(\blacksquare\)

### Solving for the Integers

** Theorem 2**: If the numbers in

*S*are all integers, then they are all equal.

** Proof of Theorem 2**: Let \(m = \text{min}(S)\). The set \(S' = \{x - m \mid x \in S\} \subset \mathbb{N}\)
satisfies the conditions of

**. Therefore all elements of**

*Theorem 1**S’*are equal. Therefore all elements of

*S*are equal. \(\blacksquare\)

### Solving for the Rationals

** Theorem 3**: If the numbers in

*S*are all rational numbers, then they are all equal.

** Proof of Theorem 3**: Let \(S = \{\frac{a_i}{b_i} \mid 0 \leq i \leq N\}\). Let \(g = \prod_{0 \le i \le N}b_i\). Let \(S' = \{x \times g \mid x \in S\}\). Then \(S' \subset \mathbb{Z}\) and

*S’*satisfies the condition of

**. So all the elements of S’ are equal. So all the elements of**

*Theorem 2**S*are equal. \(\blacksquare\)

### Solving for the Reals

** Theorem 4**: If the numbers in

*S*are general real numbers, then they are all equal.

** Proof of Theorem 4**: Look at field extension of \(\mathbb{Q}\) in \(\mathbb{R}\) generated by the elements of

*S*. You can look at its elements as a finite dimensional vector field over \(\mathbb{Q}\). The claim is true i.f.f. it is true separately in each dimension. Each of the dimensions separately satisfies the conditions for

**, therefore the values in each dimension are all equal, therefore the values in all dimensions are equal, therefore the numbers in**

*Theorem 3**S*are all equal. \(\blacksquare\)