# Binary Numeration System with Alternating Signed Digits and Its Graph Theoretical Relationship
## 2 The Proof of Theorem 2  
The proof will be a simple series of simple claims and their proof. 1. Suppose that  
\[s_{\alpha}2^{\alpha}-s_{\alpha}2^{\beta}+s_{\alpha}2^{\gamma}\mp\ldots\]  
is a non-empty alternative binary representation of the number \(n\). Then, \(n\) is not \(0\) and its sign is equal to the sign of \(s_{\alpha}\). Indeed,  
\[n=s_{\alpha}(2^{\alpha}-2^{\beta})+s_{\alpha}(2^{\gamma}-2^{\delta})+\ldots,\]  
where the \"brackets\" are positive. (Brackets contain two terms, with the possible exception of the last one which might have only one term.) 2. 0 has a single representation, the empty sum. 3. It is enough to prove the theorem for positive integers, assuming that the most significant digit is \(+1\). 4. Let \(n\) be a positive integer and take an alternative binary representation of it with \(\ell(>0)\) terms. Consider the representation first \(k(<\ell)\) of the representation and denote their sum by \(S_{k}\). If \(k\) is even, then \(S_{k}<n\). If \(k\) is odd, then \(S_{k}>n\). Indeed,  
\[n=(2^{\alpha}-2^{\beta})+(2^{\gamma}-2^{\delta})+\ldots=2^{\alpha}-(2^{\beta} -2^{\gamma})-\ldots,\]  
where the \"brackets\" have positive values. 5. Each positive integer \(n\) has an alternating binary representation. Let us write down \(n\) in the binary number system, i.e., as the sum of different powers of two (with strictly decreasing order of exponents). Substitute \(2^{k+1}-2^{k}\) for each power of two: \(2^{k}\). We obtain a new representation of our initial number as the sum of signed powers of two; furthermore, the signs alternate. The exponents will be decreasing but not strictly decreasing. There can be only two powers of the same exponent, they must follow each other, and they must have different signs. Thus, they cancel. By performing the cancellations, we obtain a representation of our starting number in the alternating binary number system. 6. Let \(n\) be a power of two, i.e. \(n=2^{k}\) for a suitable natural number \(k\). Then, \(n\) has exactly two alternating binary binary representations: \(2^{k}\) and \(2^{k+1}-2^{k}\). We know (4.) that \(2^{\alpha-1}\leq 2^{\alpha}-2^{\beta}\leq n\leq 2^{\alpha}\). 7. If \(n\) is between two consecutive powers of two (for a suitable natural number \(k\) we have \(2^{k}<n<2^{k+1}\)), then the first term of each representation of \(n\) is necessarily \(2^{k+1}\). Again, we know (4.) that \(2^{\alpha-1}\leq 2^{\alpha}-2^{\beta}\leq n<2^{\alpha}\). 8. If \(2^{k}<n<2^{k+1}\) (\(k\in\mathbb{N}\)), then every representation of \(n\) is \(2^{k+1}-\rho\) where \(\rho\) is the representation of the number \(2^{k+1}-n(>0)\)  
9. If \(2^{k}<n<2^{k+1}\) (\(k\in\mathbb{N}\)), then \(2^{k+1}-n\) is strictly smaller than \(n\). Then, the proof is completed by a simple induction. In the case of a non-zero number in one of the two options for the representation, the last is non-zero, and the preceding digit is \(0d\), or \(d\tilde{d}\), where \(d\in\{-1,1\}\). There are several simple ways to make the representation unique. Theorem 2.3: _The alternating binary representation will be unique if any of the following additional conditions are satisfied:_  
1. _The last non-zero digit must be_ \(1\)_The last non-zero digit must be_ \(-1\)_._ _The last non-zero digit must be immediately preceded by a_ \(0\) _digit (or we have only one non-zero digit)._ _The last non-zero digit must be immediately preceded by a non-zero digit._ _The number of non-zero digits is odd._ _The number of non-zero digits is even._  
The proofs are straightforward. If we choose any of the rules above, we obtain a numeral system. For example, we can fix (c) as the basic rule of binary alternating coding of integers. We will denote the \(-1\) digit as \(\bar{1}\) and we use subscript \(2\)-\(Alt\) when representing \(n\) using place-value notation. We list a few integers with small absolute value and their representation.