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Is $n\lceil x \rceil = \lceil nx \rceil$, for all integers $n$ and real numbers $x$? If not, are there some similar (or less general) rules available? If yes, can this somehow be generalized?

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See here. There are some identities there you might find interesting. For example this one. – Dejan Govc Apr 3 '12 at 21:04
You'll find some inequalities (for $\lfloor \cdot \rfloor$) here. The answer uses the Hermite's identity, mentioned by Dejan. – draks ... Apr 4 '12 at 15:30
up vote 4 down vote accepted

No, for example $2\lceil \frac 14 \rceil \ne \lceil 2\cdot \frac 14 \rceil$

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No, $n\lceil x\rceil$ is not in general equal to $\lceil nx\rceil$, as easy examples will show you; for instance, just take $n=2$ and $x=1/2$, so that $n\lceil x\rceil=2$ and $\lceil nx\rceil=1$.

To see what happens in general, suppose that $n$ is positive and that $\lceil x\rceil=m$; this amounts to saying that $x=m-1+\alpha$, where $0<\alpha\le 1$. Then $nx=(m-1)n+n\alpha$, so $$\lceil nx\rceil=\lceil (m-1)n+n\alpha\rceil=(m-1)n+\lceil n\alpha\rceil\;.$$

Since $0<\alpha\le 1$, there must be a unique integer $k$ such that $0<k\le n$ and $$\frac{k-1}n<\alpha\le\frac{k}n\;.$$ Then $k-1<n\alpha\le k$, so $\lceil n\alpha\rceil=k$, and $$\lceil nx\rceil=(m-1)n+k\;.$$ On the other hand, $n\lceil x\rceil=mn$, which is equal to $n\lceil x\rceil$ if and only if $k=n$, i.e., if and only if $$n-\frac1n<\alpha\le 1\;.$$

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