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I'm probably overthinking this.

What constraints must you place on $v\in \mathbb R$ : $n \left \lfloor {v} \right \rfloor $ = $\left \lfloor {n v} \right \rfloor $ if $n$ is an arbitrary integer?

I can tell that $v \in ℤ$ works eg. $2 \left \lfloor {3} \right \rfloor = \left \lfloor {2 \times 3} \right \rfloor, $ but I'm wondering if I'm missing a more subtle set of constraints on $v$.

Any help (even instructive comments/hints) would go a long way.

Thanks for your time.

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  • $\begingroup$ What is V? ${}{}{}{}{}$ $\endgroup$ – Namaste Jul 31 '18 at 20:53
  • $\begingroup$ @amWhy V is a number $\endgroup$ – McMath Jul 31 '18 at 20:54
  • $\begingroup$ why not right it as $v$ or $x$....? $\endgroup$ – Namaste Jul 31 '18 at 20:54
  • $\begingroup$ @amWhy sure thing, I'll change it to v. No biggie :) $\endgroup$ – McMath Jul 31 '18 at 20:55
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This is for $n>0$ and $v>0$.

Write $v=x+y$, where $x$ is an integer and $0\le y<1$, so $\lfloor v\rfloor=x$. Next, $$ \lfloor nv\rfloor=\lfloor nx+ny\rfloor=nx+\lfloor ny\rfloor. $$

Can you do the rest?

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  • $\begingroup$ referencing Batominoski's answer, I'm guessing 'the rest' is: $0 ≤ y < \frac{1}{n}$. Thank you for sharing your answer, it's very illuminating. $\endgroup$ – McMath Jul 31 '18 at 21:36
  • $\begingroup$ @McMath There are the cases for $n<0$ and $x<0$… $\endgroup$ – egreg Jul 31 '18 at 22:03
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Here is a powerful result (stronger than needed, though), which can be used to deal with this problem. From the theorem below, you would see that $\{x\}\in\left[0,\dfrac1n\right)$ for $x\in\mathbb{R}$ to satisfy $$\lfloor nx\rfloor=n\,\lfloor x\rfloor\,,$$ where $n\in\mathbb{Z}_{>0}$ is fixed. Here, $\{x\}$ is the fractional part of a real number $x$,

Theorem. For each $x\in\mathbb{R}$ and $n\in\mathbb{Z}_{>0}$, $$\lfloor nx\rfloor=\sum_{k=0}^{n-1}\,\left\lfloor x+\frac{k}{n}\right\rfloor\text{ and }\lceil nx\rceil=\sum_{k=0}^{n-1}\,\left\lceil x-\frac{k}{n}\right\rceil\,.$$

For $n\in\mathbb{Z}_{<0}$, use the fact that $\lfloor x\rfloor =-\big\lceil (-x)\big\rceil$ for all $x\in\mathbb{R}$. The theorem above shows that $$\{x\}\in \{0\}\cup\left(1+\frac{1}{n},1\right)$$ if $\lfloor nx\rfloor=n\,\lfloor x\rfloor$ for a given $n\in\mathbb{Z}_{<0}$. For $n=0$, any $x$ works.

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  • $\begingroup$ thank you, your approach makes perfect sense. I appreciate you taking the time to help $\endgroup$ – McMath Jul 31 '18 at 21:11

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