28
$\begingroup$

I'm curious as to how the floor function can be defined using mathematical notation.

What I mean by this, is, instead of a word-based explanation (i.e. "The closest integer that is not greater than x"), I'm curious to see the mathematical equivalent of the definition, if that is even possible.

For example, a word-based explanation of the factorial function would be "multiply the number by all integers below it", whereas a mathematical equivalent of the definition would be $n! = \prod^n_{i=1}i$.

So, in summary: how can I show what $\lfloor x \rfloor$ means without words?

$\endgroup$
5
  • 8
    $\begingroup$ "instead of a word-based explanation" - it can happen that insisting on formal notation often hurts clarity more than helping it. In this case, since the textual definition is straightforward, why bother? $\endgroup$
    – J. M. ain't a mathematician
    May 12, 2013 at 3:53
  • 10
    $\begingroup$ For curiosity's sake; consider it a puzzle $\endgroup$
    – Cisplatin
    May 12, 2013 at 3:55
  • 7
    $\begingroup$ What do you mean by that big capital pi with all those other symbols around it? Can you describe that "without words" please? In your accepted answer, what does arctan mean? My point is, all notation was invented to describe some intellectual phenomenon, which may in turn have been based on some earlier idea, etc. until the whole thing ultimately boils down to cavemen telling how many mastadon had been successfully hunted by holding up fingers. $\endgroup$
    – cobaltduck
    May 12, 2013 at 18:06
  • $\begingroup$ @cobaltduck "hire or otherwise persuade a caveman to walk clockwise a circle of radius 0.5/pi until he has covered a distance of x. then have him walk counter-clockwise back to his starting point and subtract from x the distance he walked back" it helps to mark the starting point with a mastodon. ;p $\endgroup$
    – agks mehx
    May 12, 2013 at 20:46
  • $\begingroup$ see alsso: math.stackexchange.com/q/1810859/51028 $\endgroup$
    – Stephan
    Jul 11, 2018 at 22:24

12 Answers 12

Reset to default
86
$\begingroup$

For a real number $x$, $$\lfloor x\rfloor=\max\{n\in\mathbb{Z}\mid n\leq x\}.$$ I'd like to add though, that "$\lfloor x\rfloor$" is mathematical notation, just as much as the right side of the above equation is; the right side might use more "basic" constructions, but you can then ask about $$\max,\qquad\in,\qquad \mathbb{Z},\qquad {}\mathbin{\mid}{},\qquad \leq$$ and so on. At some point you just have to start writing notation and explaining it in words and hope your readers understand. So I disagree with your phrasing of the question.

$\endgroup$
3
  • 4
    $\begingroup$ I've always thought floor/ceiling to be basic themselves, so I suppose "basic" is in the eye of the beholder. (+1, of course.) $\endgroup$
    – J. M. ain't a mathematician
    May 12, 2013 at 3:54
  • 1
    $\begingroup$ that's one step closer to the set-theoretic definition. $\endgroup$
    – Hal Canary
    May 12, 2013 at 20:51
  • 1
    $\begingroup$ With the note that the $\max$ is well-defined because the set is bounded from above (by $x$), and every set of integers bounded above has a maximum element by the well-ordering principle. $\endgroup$
    – asmeurer
    May 12, 2013 at 22:34
31
$\begingroup$

$\lfloor x \rfloor = x - \arctan(\tan(\pi x))/\pi$ ?...

$\endgroup$
29
  • 19
    $\begingroup$ Almost, $\displaystyle\lfloor x \rfloor = (x - 0.5) - \frac{arctan(tan(\pi (x-0.5)))}{\pi}$ $\endgroup$
    – Justin
    May 12, 2013 at 5:22
  • 4
    $\begingroup$ @agksmehx: Zev's answer should be the accepted one - no chickening out there, just the plain professional definition. (Of course this answer is a nice idea, too.) $\endgroup$
    – Hendrik Vogt
    May 12, 2013 at 12:18
  • 7
    $\begingroup$ One tiny quibble is that $\arctan$ is a multi-valued function, and this relies on taking the principal value, otherwise the answer could be any integer. It's somewhat similar to defining $|x|$ as $\sqrt{x^2}$, which could technically be $|x|$ or $-|x|$. $\endgroup$
    – N. Virgo
    May 12, 2013 at 14:35
  • 5
    $\begingroup$ @Clayton: But which answer is clearer as to what $\lfloor x\rfloor$ means (which is usually the goal for things intended to be read by humans)? Besides, a computer seems far more likely to make round-off errors or mistakes trying to mimick $\pi$, an arbitrary real number $x$, subtraction, division, and the functions $\tan$ and $\arctan$, than in trying to compare numbers with $\leq$ (though I don't know anything about programming languages, so maybe I'm wrong about that). $\endgroup$
    – Zev Chonoles
    May 12, 2013 at 19:49
  • 5
    $\begingroup$ But there is a problem, should you use the function with an integer, the result would be its preceding integer. I.e., f(x) = (x - 0.5) - atan(tan(pi*(x - 0.5)))/pi, f(1) = 0. $\endgroup$
    – Severo Raz
    May 12, 2013 at 20:52
20
$\begingroup$

I am an engineer, perhaps this helps if you do not expect too much.

$x-(x$ mod $1$)

$\endgroup$
11
  • 25
    $\begingroup$ Follow-up: "How to represent $\bmod$ using mathematical notation?" $\endgroup$
    – J. M. ain't a mathematician
    May 12, 2013 at 3:53
  • 12
    $\begingroup$ This is a notation that I think most mathematicians would not consider standard... $\endgroup$
    – Mariano Suárez-Álvarez
    May 12, 2013 at 5:05
  • 10
    $\begingroup$ @newzad: The notation $\lfloor x\rfloor$ also has no words, and is much more widely accepted by mathematicians. $\endgroup$
    – Zev Chonoles
    May 12, 2013 at 5:09
  • 1
    $\begingroup$ I would apply mod to the whole expression which would yield zero ... $\endgroup$
    – Dominic Michaelis
    May 12, 2013 at 5:43
  • 2
    $\begingroup$ did nobody notice the directly computable answer below? ⌊x⌋=(x−0.5)−arctan(tan(π(x−0.5)))π $\endgroup$
    – agks mehx
    May 12, 2013 at 8:36
13
$\begingroup$

Recursive way: $\lfloor x\rfloor \equiv \begin{cases} 1+\lfloor x-1\rfloor& x\ge1 \\ -1+\lfloor x+1\rfloor& x<-1 \\ 0&0\le x<1\\ -1&-1\le x<0 \end{cases}$

Also directly(sort of) computes the answer.

$\endgroup$
9
$\begingroup$

$\lfloor x\rfloor=n\leftrightarrow \{n\}=\{y\in\mathbb{Z}:\forall z\in\mathbb{Z}~ z\le x \rightarrow z\le y\}$

This game is fun! Now let's prove 1+1=2.

$\endgroup$
1
  • $\begingroup$ The right side can be radically simplified to just: $\{n\} = \{y \in \mathbb{Z} : x \leq y\}$ ;-) $\endgroup$
    – Musa Al-hassy
    Nov 1, 2017 at 17:43
5
$\begingroup$

You have raised a most-interesting question which has a beautiful solution.

The floor-function is characterized by the following formula,

$\boxed{ \;\; \forall n \in \mathbb{Z},x \in \mathbb{R} :: n \leq \lfloor x \rfloor \equiv n \leq x \;\; }$

Indeed this states that ``$\lfloor x \rfloor$ is the greatest integer that is at-most $x$'':

  1. $\lfloor x \rfloor$ is an ingeter at-most $x$ ---ie $x \leq \lfloor x \rfloor$. We obtain this my taking $n = \lfloor x \rfloor$ in the above formula.

  2. It is also the largest such integer : $\forall n \in \mathbb{Z},x \in \mathbb{R} :: n \leq x \implies n \leq \lfloor x \rfloor$. We obtain this from the characterization by weakening the equivalence $\equiv$ into an implication $\implies$.

Some immediate results from the characterization, for $n \in \mathbb{Z}, x \in \mathbb{R}$, are

  1. $\lfloor x \rfloor < n \equiv x < n$, by negating the equivalence.

  2. $\lfloor x \rfloor \leq x$, by taking $n = \lfloor x \rfloor$. [Contracting]

  3. $\lfloor x \rfloor \leq \lfloor y \rfloor \Leftarrow x \leq y$, proven by the characterization and properties of $\leq$. [Order preserving]

  4. $\lfloor \lfloor x \rfloor \rfloor = \lfloor x \rfloor$, characterization again. [Idempotent]

Other nifty theorems can be easily proven from this characterization, such as

  1. $n = \lfloor x \rfloor \, \equiv \, ( n \leq x \text{ and } x<n+1)$

  2. $\lfloor x+n \rfloor = \lfloor x \rfloor + n$

  3. $\lfloor x \rfloor + \lfloor y \rfloor \leq \lfloor x+y \rfloor $

  4. $0\leq x \implies \lfloor \sqrt{\lfloor x \rfloor} \rfloor = \lfloor \sqrt{x} \rfloor$

Unlike the other answers presented earlier, this characterization shows its power in reasoning. For example, try proving one of the above theorems using one of the definitions presented by the others and then compare that with a proof using the characterization presented here.

These theorems and the characterization are explored in Feijen's ``The Joy of Formula Manipulation'', http://www.mathmeth.com/wf/files/wf2xx/wf268t.pdf.

Best regards,

Moses

$\endgroup$
4
$\begingroup$

I suspect that this question can be better articulated as: how can we compute the floor of a given number using real number field operations, rather than by exploiting the printed notation, which separates the real and fractional part, making nearby integers instantly identifiable.

How about as Fourier series? If we subtract the function $y = x$ from $\lfloor x\rfloor$ then we get a periodic sawtooth wave. Hey look, that Wikipedia page even mentions this relationship.

So if you obtain the Fourier series (a sum of sinusoidal functions of various frequencies and amplitudes) for the sawtooth wave $\lfloor x\rfloor - x$ and add $x$, then you have an approximation for $\lfloor x\rfloor$ which is as good to whatever accuracy you care to apply in evaluating the series. Obtaining the Fourier series for $\lfloor x\rfloor - x$ is a basically a matter of scaling the function to fit: finding the parameters to substitute into the generic equation to produce the right amplitude, frequency and offset.

But another answer is that the printed notation which approximates a number is computable. When a computer evaluates floor(x), it's all math. The printed (or electronically stored) notation for a rational number which approximates a real number is a mathematical object.

$\endgroup$
3
$\begingroup$

$$\lfloor x \rfloor = \text{supremum} \{n \in \mathbb{Z} : n \leq x\}$$

For a subset $S \subset X$, where we have an order $\leq$ on $X$, a supremum or least upper bound of $S$ is an element $M \in X$ such that $$x \leq M, \,\,\,\,\, \forall x \in S$$ and for any $m \in X$ such that $x \leq m$ for all $x \in S$, we have $M \leq m$.

$\endgroup$
4
  • $\begingroup$ Can "max" then be put into mathematical notation as well? Or is this as mathematical as it can get? $\endgroup$
    – Cisplatin
    May 12, 2013 at 3:40
  • 3
    $\begingroup$ @Sim You can make it more and more "formal" as you want. But it is useless beyond a point. $\endgroup$
    – user17762
    May 12, 2013 at 3:46
  • 1
    $\begingroup$ I first read this as "supermum" $\endgroup$
    – Ell
    May 12, 2013 at 10:13
  • 1
    $\begingroup$ @Sim $x = \sup(S) \iff \forall_{y\in U(S)}\left(x \leq y\right) \land x\in U(S)$, $U(S) = \left\{x \mid \forall_{y\in S} \left(y \leq x\right)\right\}$. $\endgroup$
    – Matthias Benkard
    May 12, 2013 at 14:58
3
$\begingroup$

There exists exactly one integer $n$ such that $n \le x < n+1$. We define $\lfloor x \rfloor$ to be this $n$.

Or, if you really like symbols: $\forall x \in \mathbb{R} \ \exists! n \in \mathbb{Z} \ n \le x < n+1 \quad \lfloor x \rfloor := n$.

$\endgroup$
2
$\begingroup$

$\left \lfloor x \right \rfloor = a \in \mathbb{Z}, a \le x \ni a\ge b \;\forall \;b \in \mathbb{Z}, b\le x$

$\endgroup$
1
$\begingroup$

Another expression is \begin{equation} \lfloor x \rfloor = \lim_{n \rightarrow \infty} (\sum_{k=-n}^n \mu(x-k)) - n - 1 \ , \end{equation} where $\mu$ is the step function. The function $\mu$ has the expression \begin{equation} \mu(x) = \lim_{n \rightarrow \infty} f(nx) \ , \end{equation} where $f(x) = e^{-x^2} + \frac{2}{\pi} \textrm{Si}(x)$. We have \begin{eqnarray} f(x) & = & 1 + \frac{2}{\pi}x + \sum_{k=1}^\infty \frac{(-1)^k}{\prod_{j=1}^k j} x^{2k} + \frac{2}{\pi} \frac{(\prod_{j=1}^{2k} j) (-1)^k}{(\prod_{j=1}^{2k+1} j)^2} x^{2k+1} . \end{eqnarray} The Taylor series converges for every $x \in \mathbb{R}$. Hence there is a representation for $\lfloor x \rfloor$ that applies multiplication, addition and limit.

$\endgroup$
1
  • 1
    $\begingroup$ Im not convinced. For starter the term exp(-x^2) can be omitted i think. Because exp( - n x^2) Goes to 0 as n Goes to +oo ... $\endgroup$
    – mick
    May 9, 2016 at 1:12
0
$\begingroup$

$$\large \left\lfloor x\right\rfloor = \sum_{n = -\infty}^{\infty}n\Theta\left(x - n\right) \Theta\left(n + 1 - x\right)\,, \qquad x \not\in {\mathbb Z} $$

$$\large \left\lfloor x\right\rfloor = \lim_{z \to x^{+}}\left\lfloor z\right\rfloor\,, \qquad x \in {\mathbb Z} $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .