# Representation of Integers in Balanced Ternary

## Theorem

Let $n \in \Z$ be an integer.

$n$ can be represented uniquely in balanced ternary:

$\displaystyle n = \sum_{j \mathop = 0}^m r_j 3^j$
$\sqbrk {r_m r_{m - 1} \ldots r_2 r_1 r_0}$

such that:

where:

$m \in \Z_{>0}$ is a strictly positive integer such that $3^m < \size {2 n} < 3^{m + 1}$
all the $r_j$ are such that $r_j \in \set {\underline 1, 0, 1}$, where $\underline 1 := -1$.

## Proof

Let $n \in \Z$.

Let $m \in \Z_{\ge 0}$ be such that:

$3^m + 1 \le \size {2 n} \le 3^{m + 1} - 1$

where $\size {2 n}$ denotes the absolute value of $2 n$.

As $2 n$ is even, this is always possible, because $3^r$ is always an odd integer for non-negative $r$.

Let $d = \dfrac {3^{m + 1} - 1} 2$.

Let $k = n + d$.

We have that:

 $\displaystyle \size {2 n}$ $\le$ $\displaystyle 3^{m + 1} - 1$ $\displaystyle \leadsto \ \$ $\displaystyle \size n$ $\le$ $\displaystyle d$ Definition of $d$ $\displaystyle \leadsto \ \$ $\displaystyle -d$ $\le$ $\displaystyle n \le d$ $\displaystyle \leadsto \ \$ $\displaystyle 0$ $\le$ $\displaystyle n + d \le 3^{m + 1} - 1$

Let $k = n + d \in \Z$ be represented in ternary notation:

$k = \displaystyle \sum_{j \mathop = 0}^m s_j 3^j$

where $s_j \in \set {0, 1, 2}$.

By the Basis Representation Theorem, this expression for $k$ is unique.

Now we have:

 $\displaystyle d$ $=$ $\displaystyle \dfrac {3^{m + 1} - 1} {3 - 1}$ by definition $\displaystyle$ $=$ $\displaystyle \sum_{j \mathop = 0}^m 3^j$ Sum of Geometric Progression

Hence we see:

 $\displaystyle n$ $=$ $\displaystyle k - d$ by definition $\displaystyle$ $=$ $\displaystyle \sum_{j \mathop = 0}^m s_j 3^j - \sum_{j \mathop = 0}^m 3^j$ $\displaystyle$ $=$ $\displaystyle \sum_{j \mathop = 0}^m \paren {s_j - 1} 3^j$ $\displaystyle$ $=$ $\displaystyle \sum_{j \mathop = 0}^m r_j 3^j$ where $r_j \in \set {-1, 0, 1}$

Hence $n$ has a representation in balanced ternary.

The representation for $k$ in ternary notation is unique, as established.

Hence the representation in balanced ternary for $n$ is also unique.

$\blacksquare$