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Suppose we have $a$, which is in real, and $k$ which is a natural number and $k \geq1$.

Which one is larger? $$\lfloor a \rfloor \hspace{5mm} \text{ or } \hspace{5mm} k \, \left\lfloor \frac{a}{k} \right\rfloor $$

I feel that $\lfloor a \rfloor$ is no smaller than $ k \, \left\lfloor \frac{a}{k} \right\rfloor$.

I'm wondering if there is a formal proof or theory for this?

Thank you very much!

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    $\begingroup$ Try writing $ a = kn + r$ for the largest possible integer $n$ for which $r$ is nonnegative. Then divide by $k$ and ... . $\endgroup$ – Ethan Bolker Mar 14 '18 at 18:44
  • $\begingroup$ Welcome to MSE. Please use MathJax. $\endgroup$ – José Carlos Santos Mar 14 '18 at 18:51
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$a = k*n + r$

$\cfrac{a}{k} = n + \cfrac{r}{k}$

$k*floor(\cfrac{a}{k}) = n$

$floor(a) >= k*floor(\cfrac{a}{k})$

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Just consider the definition: $$\biggl\lfloor\frac ak\biggr\rfloor=n\iff n\le\frac ak < n+1\implies nk\le a ,\enspace\text{so}\quad nk\le \lfloor a\rfloor. $$

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  • $\lfloor a \rfloor$ is the largest integer which is less than or equal to $a$.

  • $k\left\lfloor \frac{a}{k} \right\rfloor$ is an integer which is less than or equal to $a$.

Need I say more?

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