# Representation of Ternary Expansions

## Theorem

Let $x \in \R$ be a real number.

Let $x$ be represented in base $3$ notation.

While it may be possible for $x$ to have two different such representations, for example:

$\dfrac 1 3 = 0.100000 \ldots_3 = 0.022222 \ldots_3$

it is not possible for $x$ be written in more than one way without using the digit $1$.

## Proof

It is sufficient to show that two distinct representations represents two distinct numbers.

Let $a$ and $b$ two real numbers representable as the form above.

Their signs are easy to distinguish, so we consider $\size a$ and $\size b$.

There is some $n$ such that:

$\size a, \size b < 3^n$

In that case, $\dfrac {\size a} {3^n}$ can be represented as:

$0.a_1 a_2 a_3 \ldots$

and $\dfrac {\size b} {3^n}$ can be represented as:

$0.b_1 b_2 b_3 \ldots$

where $a_i, b_i$ are either $0$ or $2$.

Let $N$ be the smallest integer such that $a_N \ne b_N$.

Without loss of generality assume that $a_N = 2$ and $b_N = 0$.

We have:

 $\ds \frac {\size a} {3^n}$ $=$ $\ds \sum_{j \mathop = 1}^\infty \frac {a_j} {3^j}$ $\ds$ $=$ $\ds \sum_{j \mathop = 1}^{N - 1} \frac {a_j} {3^j} + \sum_{j \mathop = N}^\infty \frac {a_j} {3^j}$ $\ds$ $\ge$ $\ds \sum_{j \mathop = 1}^{N - 1} \frac {a_j} {3^j} + \frac 2 {3^N}$ because $a_N = 2$, $a_j \ge 0$ $\ds$ $>$ $\ds \sum_{j \mathop = 1}^{N - 1} \frac {a_j} {3^j} + \frac 1 {3^N}$ $\ds$ $=$ $\ds \sum_{j \mathop = 1}^{N - 1} \frac {b_j} {3^j} + \sum_{j \mathop = N + 1}^\infty \frac 2 {3^j}$ Sum of Infinite Geometric Sequence $\ds$ $\ge$ $\ds \sum_{j \mathop = 1}^{N - 1} \frac {b_j} {3^j} + \sum_{j \mathop = N}^\infty \frac {b_j} {3^j}$ because $b_N = 0$, $b_j \le 2$ $\ds$ $=$ $\ds \frac {\size b} {3^n}$

and thus $\size a$ and $\size b$ are distinct.

$\blacksquare$