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I would like to know whether the following statement is true:

Let $x \in \Bbb R$ be any real number. Is there an integer $n ≥ 1$ such that $\lfloor nx \rfloor$ has only $0$'s and $1$'s as digits in its decimal expansion?

If $x$ is a rational number, then the statement is true from this question. I know that the statement is wrong if we replace the floor part of $nx$ by the decimal part (for instance see here). I believe that the statement is false, but I failed to build a counter-example.

Any help would be appreciated. Thank you!

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Watson
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  • without thinking on this too hard, it would suggest to me that if it were true we have found a bijection between real numbers on the interval [0,1) to $\mathbb Q$ which we know does not exist. – Doug M Sep 16 '16 at 18:13

1 Answers1

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If we allow $n$ to be an arbitrary integer (and $x$ be positive), then the answer is positive.

Imitating the case where $x$ is a fraction we consider the numbers $a_1=1, a_2=11, a_3=111, ... $. Let $r_1,r_2,r_3, ...$ be the remainders when the $a_i$ are divided by $x$ (i.e. $a_j=n_j*x+r_j$ where $n_j$ are integers and $0 \le r_j<x$). If $i>[x]+1,$ then the distance between some remainders $r_k, r_l$ $(1 \le k<l<=i)$ is smaller than 1. As $|r_l - r_k|<1,$ then $a_k-a_l=(n_k-n_l)*x+r_k- r_l$ with $a_k-a_l$ having only 0's and 1's as digits in its decimal expansion and $-1<r_k-r_l<1.$ Hence $[(n_k-n_l)*x] = a_k-a_l$ if $r_k \le r_l$ and $[(n_k-n_l)*x] = a_k-a_l-1 = -111...10000...0001$ if $r_k > r_l.$

However, this answers your question if $x<0$ as $[(n_k-n_l)*z]=[(n_l-n_k)*(-z)]$ for $z>0$ and $n_l-n_k>0.$

My guess is that you are mainly interested in the case where $n$ is a natural number and $x>0.$

parsimonious
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