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Does the decimal portion of $n\log_3 2$ include all irrational numbers between 0 and 1 repeatedly, where $n$ is all positive integers?

Oddly enough I ran into this question while fooling around with the Collatz Conjecture.

Lacking an answer, directions as to how to find an answer (what to read up on) would be useful.


Decimal Portion: integer portion → 3 | .141592653 ← decimal portion

repeatedly: cycling through again and again


As a follow-up question, and perhaps the one I should have initially asked:

Would the decimal portion of $n\log_3 2$ include all numbers in the decimal portion of $log_3(n + 1/2)$? (again, n representing all positive integers)

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    No. You only obtain countably many numbers this way, but there are uncountably many rationals between 0 and 1.2012-05-24
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    It contains all *finite* strings of digits infinitely often. A consequence of the fact that $\log_3 2$ is irrational.2012-05-24
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    @Casey: Take any nummber $a$ in the unit interval, irrational or rational. Given any $\epsilon>0$, there are infinitely many $n$ such that the fractional part of $\log_3 2$ is at distance less than $\epsilon$ from $a$.2012-05-24
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    @GEdgar Are you certain that the one follows from the other? That seems decidedly non-trivial...2012-05-24
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    The fractional part of $n\log_3 2$ is probably dense in $[0,1)$. Perhaps this suffices for your needs?2012-05-24
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    @Steven: Yes. And each finite string appears asymptotically as often as every other of the same length.2012-05-24
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    @Didier I think that's the definition of a normal number. As far as I know, proving a number is normal is ridiculously hard.2012-05-24
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    @chris Nope. You are confusing (i) the expansion of a single number and (ii) the sequence of iterates of a number by a transformation (here the rotation by $\log_32$ on the unit circle). I agree that finding normal numbers (case (i)) may be difficult (that is, unless one relies on basic probabilistic techniques), but (ii) is a different, easier, story.2012-05-24
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    Difference between density of fractional part of $n\log_3 2$ compared to fractional part of $10^n\log_3 2$2012-05-24
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    For the fact that the fractional part of $n a$ is uniformly distributed in $[0,1]$ when $a$ is irrational, see http://en.wikipedia.org/wiki/Equidistribution_theorem2012-05-24
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    @JyrkiLahtonen: again, note that it's $n a$, not $10^n a$.2012-05-24
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    Oops. I read GEdgar's comment too quickly. Sorry about that. Thanks for pointing it out, Robert.2012-05-24
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    @GEdgar The number 0.101001000100001... is irrational but clearly does not contain all finite strings of digits. Have I misunderstood?2012-05-24
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    @Ronald: call that number $x$. No, $x$ does not contain all finite strings, but the collection of all $n x$ does.2012-05-24
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    @GEdgar Thanks, understood.2012-05-24
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    Mea culpa on my mental slip there - I'd forgotten the equidistribution theorem, which certainly makes it easy.2012-05-24
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    @Didier Ah thank you. I had missed that.2012-05-24

2 Answers 2

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Let $A=\{\mbox{decimal portion of } n\log_3 2\; |\; n \mbox{ is a positive integer}\}$, and $B=\{a\in[0,1]\;|\;a \mbox{ is irrational}\}$. Set $A$ is countable (there is a trivial one to one correspondence between $A$ and set of natural numbers), but $B$ is uncountable. So $B\nsubseteq A$. But obviously $A\subset B$.

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    As for the second question, it doesn't have this problem. And to be clear, I assume you mean if $$A=\{n\log_3 2-\lfloor{n\log_3 2}\rfloor: n\in \mathbb{Z}^{\geq 0}\},$$ and $$B=\{ \log_3(n+\frac12)- \lfloor\log_3(n+\frac12\rfloor: n\in \mathbb{Z}^{\geq 0}\},$$ then is it true that $A\subseteq B$?2012-05-24
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Your follow-up question is unclear. I think you are asking whether every fractional part of $\log_3(n+(1/2))$ is also the fractional part of $m\log_32$ for some $m$. In fact, the fractional part of $\log_3(n+(1/2))$ is never the fractional part of $m\log_32$ for any $m$. If it were, it would mean that $$m\log_32-\log_3(n+(1/2))$$ is an integer. But this is $\log_3(2^{m+1}/(2n+1))$, which can't be an integer, since $2^{m+1}/(2n+1)$ can't be a power of 3.

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    It can be a negative power of 3, can' it be?2012-05-30
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    Only if $m=-1$, and the question referred to positive values.2012-05-31
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    you're right. I didn't notice that.2012-05-31