Math and Logic Puzzles: Redux

[BNL]

Spoiler: Rationals

Consider a set of and odd number of real numbers good if it satisfies the condition. (for brevity)
Then,
S is good <=> S+r is good for real r (using the same division)
S is good <=> rS is good for nonzero real r (using the same division)

Call S := S+r and S := rS operations. Operations do not change whether a set is good or not

Suppose S contains rational numbers. Multiply them by the lcm of all denominators, so we have only integers, then subtract each element by the smallest element in S.
Now S contains only 0 and positive integers. If there are any odd numbers in S, S is not good because removing either the odd number or 0 will cause the sum of the remaining elements to be odd and they can't be divided into two equal sets.

If S contains only 0 it is clearly good. Otherwise keep dividing S by 2 until there is an odd number (it will happen as positive integers can only be finitely large).

[BNL]

Spoiler: Reals

If we expand to real numbers, we can convert the real numbers into a vector space like this: Let S = {a1,a2,...a13}. Let T0, T1, T2, ... T13 be sets like this: T0 = {}, and Ti = Ti-1 if q1a1+q2a2+...+qi-1ai-1 = ai has a solution in rational q1,q2,...,qi-1, and Ti = Ti-1 U {ai} if there is no solution. Let T = T13 = {r1,r2,...,rk}. Then each ai can be expressed uniquely as q1r1+q2r2+...+qkrk, where q1,q2,...,qk are rational.

Now we have shown that we can equate S to a vector space of k dimensions over the rationals, and we can use the same argument for the proof where S contains rationals.