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How can I prove the following equation?

$$ \lfloor nx \rfloor = \lfloor x \rfloor + \Big\lfloor x + \frac{1}{n} \Big\rfloor + \Big\lfloor x + \frac{2}{n} \Big\rfloor + \Big\lfloor x + \frac{3}{n} \Big\rfloor + \Big\lfloor x + \frac{4}{n} \Big\rfloor + \Big\lfloor x + \frac{5}{n} \Big\rfloor+ \dotsb + \Big\lfloor x + \frac{n-1}{n} \Big\rfloor $$ $n∈N$ and $x∈R$

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Replacing $x$ by $x':=x+{1\over n}$ adds $1$ on the left hand side. On the right hand side the first term $\lfloor x\rfloor$ disappears, and on the far right the term $\lfloor x+1\rfloor$ appears. It follows that the right hand side increases by $1$ as well.

Therefore it suffices to prove the formula for $0\leq x<{1\over n}$. Inspection immediately reveils that both sides are $=0$ in this case.

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If you take a closer look, you will notice that the second term within each floored term will be less than one.
E.g in $$\left\lfloor x+\frac{4}{n}\right\rfloor$$As you can see $\frac{4}{n}$ is less than one.
So, we can conclude that each term will be reduced to $\lfloor x\rfloor$ if $n$ and $x$ are integers.
We will get
$$\lfloor nx\rfloor = n\lfloor x\rfloor$$
Since $n$ and $x$ are integers, we can remove the floor function.
Hence proved.

This identity will not be true for all values of $n$ and $x$ if they can be any number other than integers. Try putting any fractional value.

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  • $\begingroup$ This is a intuitive conclusion and it's not generally true, for example for $x∈R$ $\endgroup$ – Alireza Ghaffari Feb 14 '15 at 6:50
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Try to rewrite $x$ as $$x=\lfloor x \rfloor + \frac kn + \alpha$$ where $0\le k<n$ and $0\le\alpha<\frac1n$.

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