I have trouble showing that $\frac{10}{31}$ and $\frac{45}{69}$ belong/do not belong to the Cantor set. I tried to do the following $$\sum_{n = x}^\infty \frac{2}{3^n} = \frac{2}{3^x} + \frac{2}{3^{(x + 1)}} + \frac{2}{3^{(x + 2)}} + \dots = \frac{2}{3^x} \left[ 1 + \frac{1}{3^1} + \frac{1}{3^2} + \dots \right] = \frac{2}{3^x} \sum_{k = 0}^\infty \frac{1}{x^k} = \frac{2}{3^x} \times \frac{1}{1- \frac{1}{3}} = \frac{1}{3^{x- 1}}$$ and then plug different values of $x = 1, 2, \dots$, but I could not seem to get these numbers. Does it mean they do not belong to the Cantor set? Please help if you can. Thank you so much.
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2$\begingroup$ Well you could just do long division in base 3 to find their ternary expansions. $\endgroup$– Eric WofseyMar 21, 2021 at 21:55
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1$\begingroup$ Cantor set is precisely the subset of $[0,1]$ where the elements have no 1's in their base 3 expansion. $\endgroup$– BenMar 21, 2021 at 22:05
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1$\begingroup$ A number is in the cantor set if in base $3$ it can be written with only the digits $0$ or $2.$ @Anon 1/3 is in the cantor set, even though it is $0.1$ base three because it can be written $0.022\dots$ in base $3.$ So you need to be careful about your statement. $\endgroup$– Thomas AndrewsMar 21, 2021 at 22:15
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1$\begingroup$ $45/69$ is better known as $15/23$. $\endgroup$– Gerry MyersonMar 21, 2021 at 22:18
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2$\begingroup$ $1/3<15/23<2/3$,so $45/69$ gets kicked out immediately. $\endgroup$– uniquesolutionMar 21, 2021 at 22:27
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2 Answers
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Here is the start of a long division base $3$ for the first problem:
0.0222010
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1011)101.0000000
20 22
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10 010
2 022
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2110
2022
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1100
1011
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120
At this point it is clear that $\frac{10}{31}$ is not in the Cantor set, since no later digits can change that $1$ into a $2$. One step earlier it wasn’t clear from the quotient digits alone, since $(0.022201)_3$ could conceivably have continued $$(0.02220122222\ldots)_3\,,$$ which is equal to $(0.022202)_3$, which is in the Cantor set. However, if you realize that the only numbers that have two ternary expansions are rational numbers that can be written with a denominator that is a power of $3$, you can stop as soon as you get that $1$.
What we have at this point says that $(0.022201)_3<\frac{10}{31}<(0.022202)_3$, so it is in one of the deleted open intervals.
The second one is even easier.
answered Mar 21, 2021 at 22:19
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1$\begingroup$ There is no real need to do long division for the second fraction, as $1/3<15/23<2/3$, so it gets kicked out on the first step of the Cantor-set construction. $\endgroup$ Mar 21, 2021 at 22:24
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$\begingroup$ "It wasn't clear one step earlier", but it could have been clear one step earlier because for a fraction to have two base-$3$ expansions, its reduced form would have to have a power of $3$ in the denominator. $\endgroup$ Mar 22, 2021 at 0:30
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$\begingroup$ @AndreasBlass: Good point; I’ve revised the answer to include it. Thanks! $\endgroup$ Mar 22, 2021 at 0:34
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Neither one is in the Cantor set. $\dfrac{10}{31} = 0.02220101110012..._{3}$, and $\dfrac{45}{69} = \dfrac{15}{23} = 0.12212110220122..._{3}$. Both has $1$ in their ternary presentation.
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$\begingroup$ This is correct, but I doubt if it is the point of the exercise. $\endgroup$ Mar 21, 2021 at 22:33
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1$\begingroup$ Thank you. But could you show the contradiction proof? I really want to learn how to show it rigorously. Thanks! $\endgroup$ Mar 21, 2021 at 22:36
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$\begingroup$ @little_sky: Your numbers are specific and all you need is to express it in base $3$ with few minutes of calculations. If you have a general fraction with certain property, then I think a pure proof that you want might be worth the time as you can learn more about the Cantor set. $\endgroup$– user899577Mar 21, 2021 at 22:42
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