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Let $A=\{ \sum_{i=1}^\infty \frac{x_i}{3^i}: x_i = 0 \,\text{ or }\, 2 \}$, show that $\frac{3}{4} \in A$

My attempt:

Let $2\bigl(\frac{1}{3^{n_1}}+\frac{1}{3^{n_2}}+\frac{1}{3^{n_3}} + \dotsm\bigr) = \frac{3}{4}$

$\Rightarrow 2(\frac{3p+1}{3^{n_1}}) = \frac{3}{4}$, if $n_1>n_2>n_3>...$

$\Rightarrow 8(3p+1) = 3^{n_1+1}$

$\Rightarrow 8 = 3q$, where q is an integer.

Which is impossible.

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  • $\begingroup$ 3=3/8?? typo i guess! $\endgroup$ – Albus Dumbledore Sep 6 '20 at 10:54
  • $\begingroup$ @Quantum No. The title is (sum of reciprocals of poweres of $3$) = $3/8$. $\endgroup$ – José Carlos Santos Sep 6 '20 at 10:55
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    $\begingroup$ One problem I see with your proof is that you denote $n_1$ to be the biggest power. But you can't do that, since they could be infinitely many and not have a biggest. $\endgroup$ – Todor Markov Sep 6 '20 at 10:57
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    $\begingroup$ math.stackexchange.com/questions/289803/… See John's Answer. $\endgroup$ – Sumanta Sep 6 '20 at 10:59
  • $\begingroup$ If you don't see yet why your proof is incorrect, you can try to exaggerate it a bit: $\pi=3+\frac{1}{10}+\frac{4}{10^2}+\frac{1}{10^3}+\frac{5}{10^4}+\ldots$. If what you've attempted was possible, you would get $\pi=\frac{q}{10^{n}}$ for some $n$, i.e. $\pi$ would be rational. $\endgroup$ – Stinking Bishop Sep 6 '20 at 11:08
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The greedy algorithm works here. Since $2/3 < 3/4$, we pick $x_1 = 2$. Then the remainder is $1/12$, which we compare against $2/9$ and is smaller, so $x_2 = 0$. The next power of $3$ is $27$, and we have $2/27 < 2/24 = 1/12$, so $x_3 = 2$, leaving a remainder of $1/108$. We next observe that $$\frac{1}{108} = \frac{3}{4} \cdot \frac{1}{3^4},$$ so it turns out that $$\begin{align} \frac{3}{4} &= \frac{2}{3^1} + \frac{0}{3^2} + \frac{2}{3^3} + \frac{1}{3^4} \cdot \frac{3}{4} \\ &= \frac{2}{3^1} + \frac{0}{3^2} + \frac{2}{3^3} + \frac{1}{3^4} \left( \frac{2}{3^1} + \frac{0}{3^2} + \frac{2}{3^3} + \frac{1}{3^4} \cdot \frac{3}{4} \right) \\ &= \frac{2}{3^1} + \frac{0}{3^2} + \frac{2}{3^3} + \frac{0}{3^4} + \frac{2}{3^5} + \frac{0}{3^6} + \frac{2}{3^7} + \frac{1}{3^8} \cdot \frac{3}{4} \\ &= \cdots \\ \end{align}$$ and it becomes apparent that the pattern is $$x_{2k+1} = 2, \quad x_{2k+2} = 0.$$ We can formally verify this by noting that our sum is now $$\sum_{k=0}^\infty \frac{2}{3^{2k+1}} + \frac{0}{3^{2k+2}} = \frac{2}{3} \sum_{k=0}^\infty \frac{1}{9^k} = \frac{2/3}{1 - 1/9} = \frac{3}{4}.$$ The astute observer will note that the sequence of $x_i$ corresponds to the base-$3$ or ternary digit expansion of the rational $3/4$; i.e., $$\frac{3}{4} = (0.202020\ldots)_3.$$ We went through two cycles of the expansion instead of just one: we could have noted that $$\frac{3}{4} = \frac{2}{3^1} + \frac{0}{3^2} + \frac{1}{3^2} \cdot \frac{3}{4}.$$

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To answer the question in the body: $$ 9 \cdot \frac 34 = 6 + \frac 34 \implies \frac 34 = 0.202020\cdots \text{ in base } 3 $$

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