How to represent the floor function using mathematical notation?

Question

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?

Answer 1

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.

Answer 2

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

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 $M \leq m$.

Answer 3

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

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

Answer 4

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

$x-(x$ mod $1$)

Answer 5

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

Answer 6

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.

Answer 7

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.

Answer 8

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

Answer 9

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: $\forall x \in \mathbb{R} \ \exists! n \in \mathbb{Z} \ n \le x < n+1 \quad \lfloor x \rfloor := n$.

Answer 10

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

Answer 11

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.

Answer 12

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

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