Almost Binary!?

The correct sequence of answers is "Yes", "No", "Yes". Let us see the proof of any of these answers.

Question 1. The answer is "Yes".

Proof: If $x$ can be represented by the formula (*) then $-1=\sum_{k=1}^\infty \frac{-1}{2^k}\leq x \leq \sum_{k=1}^\infty \frac{1}{2^k}=1,$ then $x\in [-1, 1].$

Now we can prove the converse implication: $x\in [-1, 1] \Rightarrow x\in S.$ If $x=-1$ or $x=1,$ then $x$ can be represented as $\sum_{k=1}^\infty \frac{-1}{2^k}$ or $\sum_{k=1}^\infty \frac{1}{2^k}.$ respectively. Therefore, in this case, $x\in S.$ Let us consider now the case when $x\in [0, 1),$ then using the binary representation $x=\sum_{k=1}^\infty \frac{b_k}{2^k},$ where $b_k$ is 1 or 0 for any natural number k. Then $2x=\sum_{k=1}^\infty \frac{2b_k}{2^k}$ Subtracting 1 from both sides of this equation and using that $1=\sum_{k=1}^\infty \frac{1}{2^k},$ we get that $2x-1= \sum_{k=1}^\infty \frac{2b_k}{2^k}-\sum_{k=1}^\infty \frac{1}{2^k}= \sum_{k=1}^\infty \frac{2b_k-1}{2^k}=\sum_{(2b_k-1)=1}\frac{1}{2^k}+\sum_{(2b_k-1)=-1}\frac{-1}{2^k}$ Now, adding 1 to both sides of the equation, $2x=1+\sum_{(2b_k-1)=1}\frac{1}{2^k}+\sum_{(2b_k-1)=-1}\frac{-1}{2^k}$ Finally, we divide both sides by 2, $x=\frac{1}{2}+\sum_{(2b_k-1)=1}\frac{1}{2^{k+1}}+\sum_{(2b_k-1)=-1}\frac{-1}{2^{k+1}}$ and this is a possible representation of $x$ in the form given by the formula (*).

If $x\in (-1, 0),$ then $-x \in (0, 1).$ Since $-x$ can be expressed in the form (*), then it is easy to see that $x$ is also representable in that form.

Question 2. The answer is "No".

Proof: Notice that $\frac{1}{4}= \sum_{k=3}^\infty \frac{1}{2^k}.$ Then $\frac{1}{2}= \frac{1}{2}+\frac{1}{4}- \sum_{k=3}^\infty \frac{1}{2^k}= \frac{1}{2}-\frac{1}{4}+ \sum_{k=3}^\infty \frac{1}{2^k}.$ So, the number $\frac{1}{2}$ is an example of a number with two distinct representations in the form (*).

Question 3 . The answer is "Yes".

Proof: Let $x= -\frac{21}{32}.$ Then it is easy to see that $-x= \frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}$ Multiplying by 2, $2(-x)= \frac{2}{2}+\frac{2}{2^3}+\frac{2}{2^5},$ Subtracting 1 and using that $1=\sum_{k=1}^\infty \frac{1}{2^k},$ $2(-x)-1= \frac{1}{2}+\frac{-1}{2^2}+\frac{1}{2^3}+\frac{-1}{2^4}+\frac{1}{2^5}-\sum_{k=6}^\infty \frac{1}{2^k}.$ Hence, solving for $-x,$ as we did in the proof in question 1, $-x=\frac{1}{2} +\frac{1}{2^2}+\frac{-1}{2^3}+\frac{1}{2^4}+\frac{-1}{2^5}+\frac{1}{2^6}-\sum_{k=6}^\infty \frac{1}{2^{k+1}},$ and, therefore, $x=\frac{-1}{2} +\frac{-1}{2^2}+\frac{1}{2^3}+\frac{-1}{2^4}+\frac{1}{2^5}+\frac{-1}{2^6}+\sum_{k=6}^\infty \frac{1}{2^{k+1}}.$

Then, for this representation of $x,$ we have obtained that $a_1=-1, a_2=-1, a_3= 1, a_4=-1, a_5=1 .$ Therefore, that the sum $a_1+a_2+a_3+a_4+a_5= -1.$ This completes the proof that the answer to this question is "Yes."