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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. – Andrés E. Caicedo May 24 '12 at 18:02
It contains all finite strings of digits infinitely often. A consequence of the fact that $\log_3 2$ is irrational. – GEdgar May 24 '12 at 18:06
Difference between density of fractional part of $n\log_3 2$ compared to fractional part of $10^n\log_3 2$ – GEdgar May 24 '12 at 19:18
For the fact that the fractional part of $n a$ is uniformly distributed in $[0,1]$ when $a$ is irrational, see – Robert Israel May 24 '12 at 20:40
@JyrkiLahtonen: again, note that it's $n a$, not $10^n a$. – Robert Israel May 24 '12 at 20:42

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 $A\subseteq B$? – Keivan May 24 '12 at 20:32

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? – Keivan May 30 '12 at 19:18
Only if $m=-1$, and the question referred to positive values. – Gerry Myerson May 31 '12 at 11:31
you're right. I didn't notice that. – Keivan May 31 '12 at 16:02

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