# 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.

$\blacksquare$

## Sources

- 1971: George E. Andrews:
*Number Theory*... (previous) ... (next): $\text {1-2}$ The Basis Representation Theorem: Exercise $4$