Ternary Expansion of $\frac{10}{27}$

Question

I am reviewing the decimal expansion of real numbers in the context of real analysis.

I am learning the ternary expansion, but I don't understand the following statement:

$$x=\frac{10}{27}=0.101=.100222\dots$$

In a ternary expansion, I should be able to write: where $a_k\in\{0,1,2\},$

$$x=\sum_\limits{k=1}^\infty a_k\frac{1}{3^k}.$$

So I understand that the second inequality means:

$$x=\frac{1}{3}+\frac{0}{3^2}+\frac{1}{3^3}=\frac{10}{27}.$$

But why is the final equality true with $2^s$ keep going? What is the analogy in the decimal expansion?

Consider the sum \begin{eqnarray} 9 \sum_{n=1}^{\infty } \left( \frac{1}{10} \right) ^n. \end{eqnarray} In two different ways. — Donald Splutterwit, Sep 18 '19 at 19:36

score 0 · Answer 1 · answered Sep 18 '19 at 23:06

0

$\dfrac{10}{27}=\dfrac{3^2+1}{3^3}$ Then in base $3$ you have $\dfrac{101}{1000}=0.101$ Now you have to verify that, similarly to $0.00999999...=0.001$ in base $10$ you have $$0.100000+0.002222222...=0.1000000...+0.01=0.101$$

answered Sep 18 '19 at 23:06

Piquito

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score 0 · Accepted Answer · answered Sep 18 '19 at 23:09

Hint $$\sum_{k=4}^\infty \frac{2}{3^k}=\frac{2}{3^4} \sum_{m=0}^\infty \frac{1}{3^m}=\frac{2}{3^4}\frac{1}{1-\frac{1}{3}}=\frac{1}{3^3}$$

Therefore $$\frac{10}{27}=\frac{1}{3}+\frac{0}{3^2}+\frac{1}{3^3} \mbox{ and } \\ \frac{10}{27}=\frac{1}{3}+\frac{0}{3^2}+0+\sum_{k=4}^\infty \frac{2}{3^k} $$

Ternary Expansion of $\frac{10}{27}$

2 Answers2