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?
$$\lfloor-0.8\rfloor=-1$$
$$\text{or}$$ $$\lceil-0.8\rceil=-1$$
$\endgroup$ 45 Answers
$\begingroup$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:
$\qquad\qquad$
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 $x$.
$\endgroup$ 2 $\begingroup$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 $$ \lceil x \rceil = \min \{ z \in \Bbb{Z} : x \le z \}. $$
$\endgroup$ 1 $\begingroup$The first one is true: $\lfloor x\rfloor\le x\le\lceil x\rceil$, no matter if $x$ is negative or not.
$\endgroup$ 1 $\begingroup$When $x\in \mathbb{R}\setminus \mathbb{Z}$ ($x$ is not an integer):
$\lfloor-x\rfloor= -\lfloor x\rfloor - 1.$
When $x \in \mathbb{R}$:
$\lceil-x\rceil= -\lfloor x\rfloor.$
$\endgroup$ 1 $\begingroup$The correct answer is it depends how you define floor and ceil. You could define as shown here the more common way with always rounding downward or upward on the number line.
OR
Floor always rounding towards zero. Ceiling always rounding away from zero. E.g floor(x)=-floor(-x) if x<0, floor(x) otherwise
If gravity were reversed, the ceiling would become the floor. So from a physics standpoint the standard mathematical definition might be inadequate.
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