The Floor Function/Box Function

Definition
We define the floor function as : \lfloor x\rfloor = greatest integer less than or equal to x. e.g., \lfloor 0.786\rfloor = 0; \lfloor - 6.987\rfloor = - 7.Graph of the floor function
The function is shown in the image (1st quadrant only)(see the attachment)
Image
This function is discontinuous and indifferentiable at each and every points. :P

Some other definitions
We define “curvy x” as: \{ x\} = x - \lfloor x\rfloor
We define the ceiling function as: the least integer greater than or equal to x; ie \lceil x\rceil = least integer greater than or equal to x. For example; \lceil - 6.598\rceil = - 6; \lceil 8.956\rceil = 9.

Hence it is easy to notice that \lceil x\rceil = 1 + \lfloor x\rfloor for x\not\in\left\{\mathbb Z\cup \mathbb{R}_{<0}\right\}

A challenge
Can you write the function f(x) where x is approximated to x';i.e.:
x = 2.569\sim 3 = x' ; x = 1.326\sim 1 = x'

Common concepts
1.\lfloor x\rfloor + \lfloor y\rfloor \leq \lfloor x + y\rfloor
2.\lfloor x\rfloor = n\Leftrightarrow n\le x\le n + 1
3. \left\lfloor \frac {\lfloor x\rfloor}{n}\right\rfloor = \left\lfloor \frac {x}{n}\right\rfloor

(Proof:
Let x = m + \alpha (0\le \alpha\le 1); m = qn + r.
\lfloor x\rfloor = m\implies \frac {\lfloor x\rfloor}{n} = \frac {m}{n} = q + \frac {r}{n}
Hence\left\lfloor \frac {\lfloor x\rfloor}{n}\right\rfloor = q...................(1)
Also \frac {x}{n} = q + \frac {r + \alpha}{n}\implies \left\lfloor \frac {x}{n}\right\rfloor = q..................(2)
Comparing (1)&(2) yields the desired result. )

Examples
\boxed{1}. (alex2008) Prove that 2^{2004}|1 + \lfloor (\sqrt {13} + 3)^{2004}\rfloor.
(solution by Potla)
See this url for proof. (refer to post # 126, 127) :)
\Box
\boxed{2} For every n\in\mathbb{N}; find the largest k\in\mathbb{N}, such that:
2^k|\lfloor(3 + \sqrt {11})^{2n - 1}\rfloor
Hint: Define a_n = (3 + \sqrt {11})^n + (3 - \sqrt {11})^n. Consider different values of a_n to get a recurrence relation. Then set x = (3 + \sqrt {11})^n; y = (3 - \sqrt {11})^n.

\boxed{3}.(alex2008 – inequation) Solve the following inequation :

\lfloor x\rfloor^3 - (x - 1)\lfloor x\rfloor^2 \le 1 - \{x\}
(solution by omegatheo)

\lfloor x\rfloor^3 - (x - 1)\lfloor x\rfloor^2\e 1 - \{x\}\implies \lfloor x\rfloor ^2\left(\lfloor x\rfloor - x + 1\right)\l....
\Box
\boxed{4} (alex2008; concepts of functions needed) Find all the functions f: \mathbb{R}\rightarrow \mathbb{R} such that f(x) + f(\lfloor x\rfloor) + f(\{x\}) = 2x\ ,\ (\forall)x\in \mathbb{R}
(solution by Farenhajt)
Putting x = 0 we get f(0) = 0\quad(1)

Putting x = n\in\mathbb{Z}, we get 2f(n) + f(0) = 2n\stackrel{\mathrm{with\,(1)}}{\iff}f(n) = n\quad (2)

Putting x = n + \alpha where n\in\mathbb{Z} and 0\leqslant\alpha < 1, we get

f(n + \alpha) + f(n) + f(\alpha) = 2n + 2\alpha\stackrel{\mathrm{with\,(2)}}{\iff}f(n + \alpha) + f(\alpha) = n + 2\alpha\qua...

Putting n = 0 in (3) we get f(\alpha) = \alpha

Now (3) becomes f(n + \alpha) = n + \alpha

Hence \left(\forall x\in\mathbb{R}\right)f(x) = x
\Box
\boxed{5} (alex2008)Let be a,b,c,d\in \mathbb{N}^* . Show that :
\left\lfloor \frac {a^2 + b^2}{2cd + 1} \right\rfloor + \left\lfloor \frac {b^2 + c^2}{2da + 1} \right\rfloor + \left\lfloor ...
(solution by Farenhajt)
Since abcd\geq 1, we get \left\lfloor{2\over 2abcd + 1}\right\rfloor = 0, hence

a^2 + b^2 < 2cd + 1\iff a^2 + b^2\leq 2cd
b^2 + c^2 < 2da + 1\iff b^2 + c^2\leq 2da
c^2 + d^2 < 2ab + 1\iff c^2 + d^2\leq 2ab
d^2 + a^2 < 2bc + 1\iff d^2 + a^2\leq 2bc

Summing the right-hand inequalities up, we get

(a - b)^2 + (b - c)^2 + (c - d)^2 + (d - a)^2\leq 0

and the conclusion follows.
\Box
\boxed{6} Solve the equation \left\lfloor \frac {x + 1}{2} \right\rfloor = \left\lfloor\frac {2x + 1}{3} \right\rfloor, where \lfloor a\rfloor is the integer part of the real number a.
Solution (makar)
Let \ \left\lfloor\frac {x + 1}{2}}\right\rfloor = \left\lfloor{\frac {2x + 1}{3}}\right\rfloor = k,k\in\mathbb N
\implies k\le\frac {x + 1}{2} < k + 1 and \ k\le\frac {2x + 1}{3} < k + 1
\implies 2k - 1\le x < 2k + 1 and \ \frac {3k - 1}{2}\le x < \frac {3k + 2}{2}
Now first analyze the case in which there is no solution,there are two such cases:
2k + 1\le \frac {3k - 1}{2} or \ \frac {3k + 2}{2}\le 2k - 1
\implies For \ k\in ( - \infty, - 3]\cup [4,\infty) there is no solution.
\implies there is a solution for k = - 2,\ - 1,\ 0,\ 1,\ 2,\ 3 only.
Solving for these values of \ k we get
\ \boxed{\boxed{x\in \left[ - \frac 72,\ - 3\right)\bigcup\left[ - 2,\ - 1\right)\bigcup \left[ - \frac 12,\ \frac 52\right)\...
\Box
References
http://www.artofproblemsolving.com/Foru … p?t=273532
http://www.artofproblemsolving.com/Foru … p?t=273536
http://www.artofproblemsolving.com/Foru … p?t=273069
http://www.artofproblemsolving.com/Foru … p?t=281136

MY APOLOGISES

Okay; now : PRACTICE.
EXERCISES

1. Prove that \forall n\geq 3(n\in\mathbb{N} we have: 8| \left\lfloor \left(\sqrt [3]{n} + \sqrt [3]{n + 2}\right)^3\right\rfloor

2. Prove that \forall n\in\mathbb{N} we have the following identity:
\sum_{r = 0}^n \left\lfloor x + \frac {r}{n}\right\rfloor = \lfloor nx\rfloor
Where x\in\mathbb{R}.

\boxed{3.} ( important lemma)The “golden ratio” is defined as \frac {1 + \sqrt {5}}{2}. Prove that : (n\in\mathbb{N})
(a)\lfloor \alpha(\lfloor n\alpha\rfloor - n + 1)\rfloor = n; or n + 1.
(b)\lfloor (n + 1)\alpha\rfloor = 1. \lfloor n\alpha\rfloor + 2 (if \left\lfloor \alpha ( \lfloor n \alpha\rfloor - n + 1 ) \right\rfloor = n) ; and 2.
\lfloor n\alpha\rfloor + 1 otherwise.

\boxed{4}. If a,b,c\in\mathbb{R} and \lfloor na\rfloor + \lfloor nb\rfloor = \lfloor ac\rfloor \forall n\in\mathbb{N} we have a\in\mathbb{Z} or b\in\mathbb{Z}.

\boxed5. (inequality; USAMO 1981) Prove that \forall x\in\mathbb{R}^ + and n\in\mathbb{N} we have: \lfloor nx\rfloor = \sum_{r = 1}^n \frac {\lfloor rx\rfloor}{r}.

I leave you all (mainly beginners) to digest these first, and then I will post the solutions. If any of you have some solutions, please post it as a comment to this post. :)

box_function.JPG
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