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It's clear to me how these functions work on positive real numbers: you round up or down accordingly. But if you have to round a negative real number: to take $\,-0.8\,$ to $\,-1,\,$ then do you take the floor of $\,-0.8,\,$ or the ceiling?

That is, which of the following are true?

$$\lfloor-0.8\rfloor=-1$$

$$\text{or}$$ $$\lceil-0.8\rceil=-1$$

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6  
Look at the number line - Floor: Go to the next integer left of where you are. - Ceiling: Go to the next integer right of where you are. –  k.stm Mar 28 '13 at 15:26
    
Lower limit: Lower than or equal to it. Upper bound: More than or equal to it. So, First one is right.! –  Inceptio Mar 28 '13 at 15:26
    
@K.Stm. +1 beat me to it :) –  Tyler Mar 28 '13 at 15:27
    
To almost any mathematician, it is obvious: $\lfloor -1.8\rfloor=-2$. I remember however bumping into a computer implementation that gave $\lfloor -1.8\rfloor=-1$. –  André Nicolas Mar 28 '13 at 15:44

4 Answers 4

up vote 13 down vote accepted

The first is the correct: you round "down" (i.e. the greatest integer LESS THAN $-0.8$).

In contrast, the ceiling function rounds "up" to the least integer GREATER THAN $-0.8 = 0$.

$$ \begin{align} \lfloor{-0.8}\rfloor & = -1\quad & \text{since}\;\; \color{blue}{\bf -1} \lt -0.8 \lt 0 \\ \\ \lceil {-0.8} \rceil & = 0\quad &\text{since} \;\; -1 \lt -0.8 \lt \color{blue}{\bf 0} \end{align}$$

In general, we must have that $$\lfloor x \rfloor \leq x\leq \lceil x \rceil\quad \forall x \in \mathbb R$$

And so it follows that $$-1 = \lfloor -0.8 \rfloor \leq -0.8 \leq \lceil -0.8 \rceil = 0$$


K.Stm's suggestion is a nice, intuitive way to recall the relation between the floor and the ceiling of a real number $x$, especially when $x\lt 0$. Using the "number line" idea and plotting $-0.8$ with the two closest integers that "sandwich" $-0.8$ gives us:

$\qquad\qquad$enter image description here

We see that the floor of $x= -0.8$ is the first integer immediately to the left of $-0.8,\;$ and the ceiling of $x= -0.8$ is the first integer immediately to the right of $-0.8$, and this strategy can be used, whatever the value of a real number $x$.

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+1 for mentioning my name!! : - ) –  k.stm Mar 28 '13 at 18:02

Keep in mind that $\lfloor x \rfloor \le x \le \lceil x \rceil$, so it's the first one.

More precisely, for $x \in \Bbb{R}$, $$ \lfloor x \rfloor = \max \{ z \in \Bbb{Z} : z \le x \} $$ while $$ \lceil x \rceil = \min \{ z \in \Bbb{Z} : x \le z \}. $$

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The first one is true: $\lfloor x\rfloor\le x\le\lceil x\rceil$, no matter if $x$ is negative or not.

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When $x>0$,

$\lfloor-x\rfloor= -\lfloor x\rfloor - 1$

$\lceil-x\rceil= -\lfloor x\rfloor$

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