Inspired by this question I ask this. For which $a$ is $n\lfloor a\rfloor+1\le \lfloor na\rfloor$ true for all sufficiently large $n$?

The original question concerned $a=e$, the usual $2.71828\ldots$, where the inequality holds by a power series argument. In the discussion Ross Millikan pointed out that if $a$ is natural, the inequality fails, and he also mentioned that if $a$ is just above a rational, the inequality fails. My followup calculation is below.

Suppose $a$ is irrational. Let $a=\frac{m}{n}+\delta$, where $0<\delta<\frac{1}{n}$. Write $m=nq+r$, where $0\le r<n$. Then the LHS is $n\lfloor a \rfloor+1=n\lfloor \frac{m}{n}+\delta \rfloor+1 = n\lfloor \frac{nq+r}{n}\rfloor + 1 = nq+1$.

The RHS is $\lfloor n (\frac{m}{n}+\delta)\rfloor=m=nq+r$. Hence the desired inequality is false so long as $r=0$.

However, this does not answer the question of which $a$, if any, manage to avoid $r=0$ consistently. To avoid simple counterexamples, I ask for all sufficiently large $n$ rather than all $n$. Perhaps this can be answered in one sentence with a continued fraction result.

Clarification: I intended to ask whether there was some $N_a$, depending on $a$, for which the inequality would hold for all $n>N_a$. The calculation was for a specific $n$, because that was the case I had considered previously, so it was only tangentially relevant. fgp kindly solved both the question I meant to ask and the question it looked like I was asking.

  • $\begingroup$ I'm assuming $a < 1$ doesn't count? ;-) $\endgroup$
    – fgp
    Apr 29, 2013 at 23:14
  • $\begingroup$ I don't see why not, @fgp. Good observation. $\endgroup$
    – vadim123
    Apr 29, 2013 at 23:25
  • $\begingroup$ Why do you use $n$ both for the title inequality and for the denominator of your rational? If they are meant to be the same, which is supposed to be set equal to which (i.e. which was is chosen first)? $\endgroup$
    – Kallus
    Apr 30, 2013 at 0:06

1 Answer 1


The question seems to be a bit confused about whether $n\lfloor a\rfloor + 1 \leq \lfloor an \rfloor$ is supposed to hold for all $n$, for all $n \geq N$ where $N$ is chosen independent from $a$, or for all $n \geq N_a$ (i.e., what is sufficiently large may depend on $a$).

In the latter case, the following shows that the inequality is true for all $a \in \mathbb{R}^+\setminus\mathbb{N}$ if $n \geq N_a$. Assume $a \in \mathbb{R}^+\setminus\mathbb{N}$ and let $$ \langle a \rangle := a - \lfloor a \rfloor $$ then $\langle a\rangle \neq 0$ and $$ na = n\lfloor a\rfloor + n\langle a\rangle\text{.} $$ Since $\lfloor m + x\rfloor = m + \lfloor x \rfloor$ if $m\in\mathbb{N}$ it follows that $$ \lfloor na \rfloor = n\lfloor a\rfloor + \lfloor n\langle a\rangle\rfloor $$ and since $\lfloor n\langle a\rangle\rfloor \geq 1$ for $n \geq \frac{1}{\langle a\rangle} =: N_a$ you have for those $n$ that $$ n\lfloor a\rfloor + 1 \leq \lfloor an \rfloor \text{.} $$

The argument about $a$ being close to a natural only applies if you choose $n$ first (i.e. independent from $a$). Otherwise, for an $a \in \mathbb{R}\setminus\mathbb{N}$, you can always pick an $n$ such that $a$ is more than $\frac{1}{n}$ bigger than any smaller natural. If you do choose $N$ first, the above shows that it's only those $a$ which lie too close to a natural which cause problems, since the inequality $\lfloor n\langle a\rangle\rfloor \geq 1$ is the only one where the value of $n$ enters the picture. Thus, if you pick $N$ first, the inequality holds exactly if $$ \langle a \rangle = a - \lfloor a \rfloor \geq \frac{1}{N} \text{.} $$

For completness' sake, it should be stated that the inequality quite obviously holds for no $a$ if $n \leq 1$, and for $n > 1$ it follows from the above that $n\lfloor a\rfloor + 1 \leq \lfloor an \rfloor$ for all $n > 1$ if and only if $$ \langle a \rangle = a - \lfloor a \rfloor \geq \frac{1}{2} \text{.} $$


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