How can one prove $e^n$ and $\ln(n)$, modulo 1, are dense in $[0,1]$, for $n=2,3,4...$?

By dense is meant, for any $0<a<b<1$, there is an integer $m$ such that $0<a<f(m) mod 1<b<1$


1 Answer 1


They are not uniformly distributed, which would mean e.g. $$\lim_{n\to\infty} \frac{|\{k<n\colon f(n)<a\}|}n=a\quad\text{for all }a\in[0,1].$$ For example between $e^M$ and $e^{M+1}$ there are about $e^{M+1}-e^{M+\frac12}$ numbers $n$ with $\ln(n)\bmod 1>\frac 12$ and $e^{M+\frac 12}-e^{M}$ numbers $n$ with $\ln(n)\bmod 1<\frac 12$. These counts differ by a factor of $\sqrt e$ and that will be the relative proportion the larger the range of $n$ one checks becomes.

But they are dense in $[0,1]$ and that is the property you are looking for (as reflected by the edit of the question).

For the logarithm: Let $\epsilon>0$ be given. Find $N$ such that $\frac1N<\epsilon$. Then $0<\ln(n+1)-\ln n<\frac1n<\epsilon$ for all $n>N$ (because the derivative of $\ln$ is the reciprocal). Therefore the numbers $\ln n\bmod1$ with $N<n<e N+1$ hit every subinterval of length $\epsilon$.

For the exponential this is a bit more difficult.

  • $\begingroup$ I know that this is an old answer, but I don't understand the last paragraph. How does it prove that it is dense in $[0,1]$? $\endgroup$
    – math.n00b
    Apr 19, 2014 at 11:13
  • $\begingroup$ @math.n00b When computing modulo $1$, we may consier $[0,1]$ wrapped to a circle by identifying $0$ and $1$. On the tour from $N$ to $\approx eN+1$, the logarithm grows by $1$, i.e. we walk around the circle once, and we always walk forward in steps $<\epsilon$. Therefore we cannot "jump" over any interval of length$\epsilon$ without hitting it. - Revisiting this asnwer, however, i have still litttle more to say about the exponential thantahat it is "a bit more difficult", hm. $\endgroup$ Apr 19, 2014 at 12:35
  • $\begingroup$ @math.n00b. For the log's case, only consider $\{log(2^n)\}$,then you will get the density property. $\endgroup$
    – David Lee
    Jun 30, 2016 at 1:05

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