Prove: $\forall n \in \mathbb{N}, \exists k \in \mathbb{N}, \forall x \in \mathbb{R}, \lfloor nx \rfloor - n \lfloor x \rfloor \leq k$.
So far I have only tried some examples and it works, but I don't have any approaches to the proof. Could someone guide me?
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The archimedian principle says for any $r >0$ we can find $wr \le x < (w+1)r$ so we can do that for $r = \frac 1n$.
We can always find an integer $w$ so that $\frac wn \le x < \frac {w+1}n$.
And division algorithm says we can always find integer $q$ and $r$ so that $w = qn + r; 0\le r < n$.
So $q + \frac rn \le x < q +\frac {r+1}n$.
So $[x] =q$ an $n[x] = qn$.
And $w = nq+r \le xn < w+1 = nq +(r+1)$.
So $[nx]=nq +r$. So $[nx] - n[x] = (nq+r)-nq = r$ and $0 \le r < n$. Or $0 \le r \le n-1$
So just select $k = n-1$ and that's it.
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An alternative way to think of this is to realize there's nothing special about a floor function to the nearest integer.
There are floor functions to the nearest multiple of any positive value.
If $m \le x < m+1$ then $nm \le nx < nm + n$ and therefore $nm \le n[x]< nm+n$ and $n[x]$ is floor function to the nearest multiple of $n$ of $nx$.
So $n[x]$ is a floor function to the nearest multiple of $n$. and $[nx]$ is a floor function to the nearest multiple of $1$. The must those can differ by is $n-1$.
Let $n\in\mathbb{N}$ and $x\in\mathbb{R}$, $\lfloor nx\rfloor\leqslant nx$ and $n\lfloor x\rfloor\geqslant n(x-1)$ so that $\lfloor nx\rfloor-n\lfloor x\rfloor \leqslant n$. It suffices to take $k=n$, which is independant of $x$.