Getting the members of a set by giving the sum of them

Question

I have a set of numbers of powers of 2, say $\{1, 2, 4, 8, 16\}$. I don't know how can I prove that the sum of some members of the set are unique, but that is not my question (although I appreciate answering this question too). My question is: is it possible to get the members by giving the summation of those members? For example, for number 12, the result should be the set $\{4, 8\}$ and for number 7, it is $\{1, 2, 4\}$.

Answer 1

You are basically considering the binary numeral system (we usually take into account only the decimal numeral system, or radix-$10$, since most of us are born $5$ fingers for each hand and $2$ hands so that $2 \dot 5 = 10$).
Now, $\sum_{j=0}^{n-1} 2^{j} = 1 + 2 + 4 + 8 + \dots + 2^{n-1}$ is what we get by considering the first $n$ elements of your sequence and it is just a matter of showing that $\sum_{j=0}^{n-1} 2^{j} < 2^n$ holds for any $n \in \mathbb{Z}^+$ (which is always true since $\sum_{j=0}^{n-1} 2^{j} = 2^n-1$), so you can consider this as the inductive step of a direct proof by induction, but you can achieve the same goal with a reductio ad absurdum, of course (see Martin's comment above and the corresponding link).