Proof of the balanced ternary expansion

Understanding this proof would be easier of you read the Basis Representation Theorem proof. Let \(n\) be the positive integer we are trying to expand. According to the division algorithm, if we divide \(n\) by 3:

$$ n = q_0 3 + r_0 $$

where \(r_0\) and \(q_0\) are unique integers and \(r_0\) is less than \(3\), so \(r_0\) can be 0, 1 or 2. Instead of \(q_0\) and \(r_0\), we will use \(c_0\) and \(a_0\), where \(c_0 = q_0\) and \(a_0 = r_0\) when \(a_0\) is 0 or 1. In the case where \(r_0 = 2\), \(c_0 = q_0 + 3\), and therefore \(a_0 = -1\). This means \(a_0\) is either -1, 0 or 1:

$$ n = c_0 3 + a_0 $$

Similarly:

$$ c_0 = c_1 3 + a_1 $$

where \(a_1\) is -1, 0 or 1, and \(c_1\) is either the qoutient or the sum of 3 and the qoutient. Repeating this process until \(c_k\), where \(c_k = 0\):

$$\begin{gather} c_1 = c_2 3 + a_2 \\ c_2 = c_3 3 + a_3 \\ \vdots \\ c_{k-1} = 0 * 3 + a_k \end{gather}$$

Replacing \(c_0\) in the first equation we get:

$$\begin{gather} n = (c_1 3 + a_1) 3 + a_0 \\ n = c_1 3^2 + a_1 3 + a_0\end{gather}$$

Replacing \(c_1\) we get:

$$\begin{gather} n = (c_2 3 + a_2)3^2 + a_1 3 + a_0 \\ n = c_2 3^3 + a_2 3^2 + a_1 3 + a_0 \end{gather}$$

Doing this repeatedly:

$$ n = a_k 3^k + a_{k-1} 3^{k-1} + \cdots + a_2 3^2 + a_1 3 + a_0 $$

In every \(a_j\) for \(j=0,1,\ldots,k\), the value for \(a_j\) is either -1, 0, 1.