Representation of Ternary Expansions

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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.

$\blacksquare$


Sources

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