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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?

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

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

José Carlos Santos
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Mirrana
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5 Answers5

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

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

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

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 $x$.

pizzapants184
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amWhy
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    With a vertical number line you even have "above" for ceiling and "below" for floor... – koalo Aug 27 '17 at 11:31
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    This is really by convention of definition not intuition. Pivoting towards or away from zero also has its intuition and places where it is useful. floor(x)={-floor(-x) if x<0, floor(x) otherwise. – Gregory Morse Oct 14 '21 at 15:21
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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 $$ \lceil x \rceil = \min \{ z \in \Bbb{Z} : x \le z \}. $$

Andreas Caranti
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    Again this does not have to be so. By convention or common definition. Other definitions define it as rounding towards or away from 0 – Gregory Morse Oct 14 '21 at 15:27
  • @GregoryMorse I see your point, but in that case you aren't really interested in the true floor or ceil of the value itself. Instead, you are then interested in the floor or ceil of the deviation of the value from zero, i.e. it's absolute value. So, you can define such an alternative as the floor or ceil of the absolute value of the given number, with the sign tacked on afterwards. – Gustav Streicher Jan 21 '24 at 00:02
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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.

cthl
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0

When $x\in \mathbb{R}\setminus \mathbb{Z}$ ($x$ is not an integer):

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

When $x \in \mathbb{R}$:

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

Sil
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Inceptio
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    Don't you mean $\lfloor -x \rfloor=-\lceil x\rceil$? For example, what happens when $x$ is an integer, say $3$? – mursalin Oct 11 '15 at 18:16
-1

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.

Gregory Morse
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