floor ceiling proof

Question

Hi I would like to proof without using induction that: $$ \left\lceil\frac{n}{m}\right\rceil \leq \frac{n+m-1}{m} $$

I tried:

$$ \left\lceil\frac{n}{m}\right\rceil \leq \frac{n}{m}+\frac{m}{m}-\frac{1}{m} $$

But dont see any pattern.

does any one have hints?

THX

score 3 · Answer 1 · answered Oct 29 '14 at 16:54

3

$$\left\lceil\frac{n}{m}\right\rceil < \frac{n}{m} + 1$$ $$m\left\lceil\frac{n}{m}\right\rceil < n + m$$ $$m\left\lceil\frac{n}{m}\right\rceil \leq n + m - 1$$ $$\left\lceil\frac{n}{m}\right\rceil \leq \frac{n + m - 1}{m}$$

answered Oct 29 '14 at 16:54

filipos

1,572

score 1 · Answer 2 · answered Oct 29 '14 at 16:43

Note that $$\left\lceil\frac n m\right\rceil<\frac nm+1=\frac{n+m}m.$$ The inequality is sharp and the smallest difference can be $\frac1m$ because only fractions with denominator $m$ can be on the right side. Therefore $$\left\lceil\frac n m\right\rceil\le\frac{n+m}m-\frac1m=\frac{n+m-1}m.$$

score 0 · Accepted Answer · answered Oct 29 '14 at 16:31

0

Write $n=km+r, 0 \le r \lt m$ using the usual division with remainder.

answered Oct 29 '14 at 16:31

Ross Millikan

383,099

ok I tried that and get $$k+\frac{r}{m}+1-\frac{1}{m}$$. Should I see a pattern? – Mamba Oct 29 '14 at 16:37
Now note that $\lceil \frac nm \rceil = k$ if $r=0$ and $k+1$ if $r \gt 0$ – Ross Millikan Oct 29 '14 at 16:39
Based on Ross Millikan's comment...write $ k = \left( \frac{n}{m} - \frac{r}{m} \right) $ – Redsbefall Oct 29 '14 at 16:55

floor ceiling proof

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