Show that every nonzero integer has balanced ternary expansion? show that every nonzero integer can be uniquely represented in the form 
$e_k3^k + e_{k-1}3^{k-1}+ … + e_13+e_0$ where $e_j= -1, 0, 1$ for $j = 0,1,2,…k$ and $e_k \neq 0$ 

Proof by Strong Induction.
Let $P(n) : n = 3^m + 3^{m-1}b_{m-1} + \cdots + 3^0b^0 \text{ where each b = -1, 0 or 1 }$.
Base Case: P(1) is true. Since $1 = 3^0$.
Assuming $P(1), P(2), \cdots, P(n-1)$ is true to prove it for $n$.
Considering cases:
Case 1: $n = 3m$
$$\implies n = 3(3^m + 3^{m-1}b_{m-1} + \cdots + 3^0b_0)$$
$$\implies n = 3^{m+1} + 3^{m}b_{m-1} + \cdots + 3^1b_0 + 3^0*0$$
Case 2: $n = 3m+1$
$$\implies n = 3(3^m + 3^{m-1}b_{m-1} + \cdots + 3^0b_0) + 1$$
$$\implies n = 3^{m+1} + 3^{m}b_{m-1} + \cdots + 3^1b_0 + 3^0*1$$
Case 3: $n = 3m-1$
$$\implies n = 3(3^m + 3^{m-1}b_{m-1} + \cdots + 3^0b_0) - 1$$
$$\implies n = 3^{m+1} + 3^{m}b_{m-1} + \cdots + 3^1b_0 + 3^0*(-1)$$
So we find that any number can be expressed like a balanced ternary expansion.
