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Suppose we have a subset $S$ of the natural numbers greater than or equal to $2$. Is there a real number $x$ such that $x$ is $k$-normal for all $k$ in $S$, and not $k$-normal for all $k$ in the complement of $S$, no matter what $S$ we choose?

M47145
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user107952
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2 Answers2

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If $r$ and $s$ are powers of the same integer, then $x$ is normal in base $r$ if and only if it is normal in base $s$. Schmidt proved that if $r$ and $s$ are not powers of the same integer then there is an uncountable infinity of numbers that are normal to base $r$ but not even simply normal to base $s$.

The reference is W Schmidt, On normal numbers, Pacific J. Math. 10 (1960) 661–672, MR0117212 (22 #7994).

A lot of work has been done since then to generalize this result. For example, Brown, Moran, and Pearce proved in 1985 that, given a base $s$, every real number can be expressed as a sum of four numbers, each of which is non-normal to base $s$, but normal to every base $r$ that isn't a power of the same integer as $s$.

A more recent paper is Becher and Slaman, On the normality of numbers to different bases, J. Lond. Math. Soc. (2) 90 (2014), no. 2, 472–494. I haven't seen this paper, nor a review, so I can't report on its contents.

Gerry Myerson
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  • $r=2$ and $s=4$ are powers of the same integer, but $x=1/3$ is normal in base $2$ $$1/3 = 0.\overline{01}_2 =0.01010101\dots_2$$ but NOT in base $4$ $$1/3=0.\overline 1_4=0.1111\dots_4$$ – CiaPan Feb 23 '15 at 12:05
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    $1/3$ is "simply normal" in base $2$, as each bit comes up with equal frequency, but "normal" means each pair of bits, each triple of bits, and so on, and so on, come up with equal frequency. The pair $11$ never comes up in $1/3$ base $2$. – Gerry Myerson Feb 23 '15 at 12:15
  • ...and CiaPan has learned something new today. Thank you and David. I didn't notice the 'normal' is much stronger property than 'simple normal'. – CiaPan Feb 25 '15 at 05:14
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If $x$ is normal in base $2$ it will be normal in base $4$. So if $2$ is in $S$ but $4$ is not, there will be no such $x$. Many similar examples are possible.

David
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    Okay, what if we just look at subsets of prime bases? – Nishant Feb 23 '15 at 05:51
  • $1/3$ is normal in base $2$ $$1/3 = 0.\overline{01}_2 =0.01010101\dots_2$$ but NOT in base $4$ $$1/3=0.\overline 1_4=0.1111\dots_4$$ – CiaPan Feb 23 '15 at 12:02
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    @ciapan For a number to be normal, every block of digits should occur with the "correct" frequency. This is not true for $\frac13$ in binary since, for, example, $00$ does not occur $\frac14$ of the time. – David Feb 23 '15 at 12:31
  • ...and CiaPan has learned something new today. Thank you and Gerry Myerson. I didn't notice the 'normal' is much stronger property than 'simple normal'. – CiaPan Feb 25 '15 at 05:14