# When does $\lfloor (n-1)x \rfloor + \lfloor x \rfloor = \lfloor nx \rfloor$?

I am trying to find the conditions under which $\lfloor(n-1)x\rfloor + \lfloor x \rfloor = \lfloor nx \rfloor$. The trivial case is whenever $x \in \mathbb{Z}$. If $n = 2$, then $x - \lfloor x \rfloor \lt \frac{1}{2}$.

As $n$ increases however, the pattern becomes more complicated. For example, for $n = 3$, if $x - \lfloor x \rfloor \lt \frac{1}{3}$ then $\lfloor 3x \rfloor = 3\lfloor x \rfloor$, and $\lfloor 2x \rfloor = 2\lfloor x \rfloor$ so the identity is satisfied. However, if $\frac{1}{2} \lt x - \lfloor x \rfloor \lt \frac{2}{3}$, then the identity is also satisfied.

I tried seeing whether Hermite's identity can help but I can't see an obvious way of applying it.

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Firstly, I'd try and check when $\lfloor nx \rfloor = n \lfloor x \rfloor$. – Quintofron Sep 3 '12 at 19:13

Let $m=\lfloor x\rfloor$ and $\alpha=x-m$. Then

$$\lfloor (n-1)x \rfloor + \lfloor x \rfloor =\lfloor(n-1)m+(n-1)\alpha\rfloor+m=nm+\lfloor(n-1)\alpha\rfloor\;,$$

and $$\lfloor nx \rfloor=\lfloor nm+n\alpha\rfloor=nm+\lfloor n\alpha\rfloor\;,$$

so $\lfloor (n-1)x \rfloor + \lfloor x \rfloor = \lfloor nx \rfloor$ iff $\lfloor(n-1)\alpha\rfloor=\lfloor n\alpha\rfloor$, i.e., iff there is an integer $k$ such that $$k\le(n-1)\alpha<n\alpha<k+1\;.\tag{1}$$

$(1)$ is equivalent to $$\frac{k}{n-1}\le\alpha<\frac{k+1}n\;,$$

so we want the non-empty intervals of the form $$\left[\frac{k}{n-1},\frac{k+1}n\right),\quad k=0,\dots,n-1\;.$$

Now $\frac{k}{n-1}<\frac{k+1}n$ iff $kn<(k+1)(n-1)=kn+n-k-1$ iff $n>k+1$, so

$$\lfloor (n-1)x \rfloor + \lfloor x \rfloor = \lfloor nx \rfloor\quad\text{iff}\quad x-\lfloor x\rfloor\in\bigcup_{k=0}^{n-2}\left[\frac{k}{n-1},\frac{k+1}n\right)\;.$$

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Thank you! So to extend it a bit, it looks like if I were to compute the intersection of all these intervals I will get that the only case where the identity is verified for all $p \in \mathbb{N}$ up to $n$ is when $x$ itself is an integer (for large $n$). – maruko Sep 3 '12 at 19:53
More precisely: Assume $0\le x <1$. Then $\lfloor (n-1)x\rfloor+\lfloor x\rfloor = \lfloor n x \rfloor$ holds for all $n\in\{1, 2, ..., N\}$ if and only if $x<1/N$. But this is immediately clear because $0=\lfloor 0 x\rfloor = \lfloor 1 x\rfloor =\ldots = \lfloor N x\rfloor$. – Hagen von Eitzen Sep 3 '12 at 20:37