# Example Problems and Solutions – Floor Function and Ceiling Function

In mathematics, a function (mapping) is a rule that associates one element of a set with exactly one element of another set. Nowadays, functions can be categorized into various types. Each type of function has its unique characteristics that can be analyzed mathematically, making them applicable in the advancement of knowledge for the benefit of many.

In mathematics, there are two fundamental functions that serve as the basis for many concepts and applications across various disciplines. These are the floor function and the ceiling function. Both of these functions are discrete functions and play a crucial role in modeling, calculations, and data analysis. In this article, we will discuss the basic concepts related to the floor function and the ceiling function, starting with their definitions, properties, and then followed by a series of example problems and solutions related to these two functions.

### Definition: Floor Function

A function $f$ is called the floor function, or sometimes referred to as the greatest integer function, denoted by $f(x) = \lfloor x \rfloor = \text{floor}(x)$ if it maps a real number $x$ to the largest integer that is less than or equal to $x$. Mathematically, it can be written as:
$$\lfloor x \rfloor = \text{max}\{m \in \mathbb{Z} \mid m \le x\}.$$

For example, $\lfloor 3.14 \rfloor = 3$ because $3$ is the largest integer that is less than or equal to $3.14.$ Meanwhile, $\left \lfloor -\dfrac{12}{5} \right \rfloor = 2$ because $-3$ is the largest integer that is less than or equal to $-\dfrac{12}{5}.$ Furthermore, $\lfloor -3 \rfloor = -3$ because $-3$ itself is an integer.

### Theorem: Properties of Floor Function

Let $x$ be a real number and $n$ be an integer.

1. $\lfloor x \rfloor = x$ if $x$ is an integer.
2. $x-1 < \lfloor x \rfloor \le x.$
3. $\lfloor x \rfloor = n \Leftrightarrow n \le x < n+1.$
4. $\lfloor x \rfloor < n \Leftrightarrow x < n.$
5. $n \le \lfloor x \rfloor \Leftrightarrow n \le x.$
6. $\lfloor x + n \rfloor = \lfloor x \rfloor + n.$
7. $\lfloor x \rfloor + \lfloor y \rfloor \le \lfloor x+y \rfloor.$
8. $\lfloor xy \rfloor \le \lfloor x \rfloor \cdot \lfloor y \rfloor.$

In simple terms, the role of the floor function can be explained as follows:

1. Rounding: The floor function is used to round real numbers down to the nearest integer. This is particularly useful in situations where integers are required, such as in modeling or discrete calculations that frequently arise in combinatorics.
2. Piecewise functions: The floor function is used in defining piecewise functions (functions that consist of several different parts based on interval differences).
3. Graph theory: The floor function can be employed in graph theory to calculate the number of specific nodes or vertices in a graph based on a given parameter.

### Definition: Ceiling Function

A function $f$ is called the ceiling function, or sometimes referred to as the least integer function, denoted by $f(x) = \lceil x \rceil = \text{ceil}(x)$ if it maps a real number $x$ to the smallest integer that is greater than or equal to $x$. Mathematically, it can be written as:
$$\lceil x \rceil = \text{min}\{m \in \mathbb{Z} \mid m \ge x\}.$$

For example, $\lceil 5.59 \rceil = 6$ because $6$ is the smallest integer that is greater than or equal to $5.59$. Meanwhile, $\left \lceil -\dfrac{14}{3} \right \rceil = -4$ because $-4$ is the smallest integer that is greater than or equal to $-\dfrac{14}{3}.$ Furthermore, $\lceil 5 \rceil = 5$ because $5$ itself is an integer.

### Theorem: Properties of Ceiling Function

Let $x$ be a real number and $n$ be an integer.

1. $\lceil x \rceil = x$ if $x$ is an integer.
2. $x \le \lceil x \rceil < x+1.$
3. $\lceil x \rceil = n \Leftrightarrow n-1 < x \le n.$
4. $n < \lceil x \rceil \Leftrightarrow n < x.$
5. $\lceil x \rceil \le n \Leftrightarrow x \le n.$
6. $\lceil x + n \rceil = \lceil x \rceil + n.$
7. $\lceil x \rceil + \lceil y \rceil \ge \lceil x+y \rceil.$
8. $\lceil xy \rceil \ge \lceil x \rceil \cdot \lceil y \rceil.$

In simple terms, the role of the ceiling function can be explained as follows:

1. Rounding: The ceiling function is used to round real numbers up to the nearest integer. Similar to the floor function, rounding up is often required in discrete calculations in combinatorics.
2. Piecewise functions: The ceiling function is also used in defining piecewise functions to ensure that the defined function approaches a certain value from above.
3. Discrete modeling: In mathematical modeling and computer science, the ceiling function is frequently employed to approximate data that can only take on discrete values.

The definitions and properties of the floor function and the ceiling function outlined above clearly demonstrate that these two functions have complementary meanings in mathematics. However, despite their apparent opposition, these two functions are interconnected. In other words, there are some important properties that relate the floor function and the ceiling function to each other, which will be explained as follows.

### The Relationship Between Floor and Ceiling Function

Assume that $x$ is a real number. Thus, the following relationships hold.

1. $x-1 < \lfloor x \rfloor \le \lceil x \rceil < x + 1.$
2. $-\lfloor x \rfloor = \lceil -x \rceil.$
3. $\lfloor -x \rfloor = -\lceil x \rceil.$
4. $\lceil x \rceil-\lfloor x \rfloor = 0$ if $x$ is an integer.
5. $\lceil x \rceil-\lfloor x \rfloor = 1$ if $x$ is not an integer.

Furthermore, when discussing the floor and ceiling functions, we will encounter new terminology, namely the fractional part and the integral part. The fractional part of a real number $x$, denoted by $\{x\}$, is the difference between $x$ and the largest integer less than or equal to $x$, which is $\lfloor x \rfloor.$ Mathematically, it can be written as $\{x\} = x – \lfloor x \rfloor.$

Since $\lfloor x \rfloor \le x < \lfloor x \rfloor + 1$, it must hold that $0 \le \{x\} = x – \lfloor x \rfloor < 1$ for every real number $x$. The largest integer that does not exceed $x$ is called the integral part of $x$ because $x = \lfloor x \rfloor + \{x\}.$

For example, the fractional part of $\dfrac54$ is $$\dfrac54 – \left\lfloor \dfrac54 \right\rfloor = \dfrac54 – 1 = \dfrac14.$$In other words, it can be written as $\left\{\dfrac54\right\} = \dfrac14.$ In this case, $1$ is the integral part of $\dfrac54$ because $1$ is the largest integer that does not exceed $\dfrac54.$

### Today Quote

The goal of life is living in agreement with nature.

Multiple Choice Questions

## Problem Number 1

Consider the functions $f(x) = 3x + \lfloor x-0,\!5 \rfloor$ and $g(x) = -3x + \lceil x + 0,\!3 \rceil.$ The correct statements regarding these two functions are $\cdots \cdot$
A. $f(2) < g(2)$
B. $f(0) = g(0)$
C. $f(-1) > g(-1)$
D. $f(3) + g(3) = 6$
E. $f(-1)-g(4) < 0$

Solution

Given $f(x) = 3x + \lfloor x-0,\!5 \rfloor$ and $g(x) = -3x + \lceil x + 0,\!3 \rceil.$
Checking Option A:
Note that $$f(2) = 3(2) + \lfloor 2-0.5 \rfloor = 6 + 1 = 7,$$while $$g(2) = -3(2) + \lceil 2 + 0.3 \rceil = -6 + 3 = -3.$$So, it is clear that $f(2) > g(2).$ This means that the statement in option A is not correct.
Checking Option B:
Notice that $$f(0) = 3(0) + \lfloor 0-0,\!5 \rfloor = 0 + (-1) = -1,$$while $$g(0) = -3(0) + \lceil 0 + 0,\!3 \rceil = 0 + 1 = 1.$$So, it is clear that $f(0) \ne g(0).$ This means that the statement in option B is not correct.
Checking Option C:
Notice that $$f(-1) = 3(-1) + \lfloor -1-0,\!5 \rfloor = -3 + (-2) = -5,$$while $$g(-1) = -3(-1) + \lceil -1 + 0,\!3 \rceil = 3 + 0 = 3.$$So, it is clear that $f(-1) < g(-1).$ This means that the statement in option C is not correct.
Checking Option D:
Notice that $$f(3) = 3(3) + \lfloor 3-0,\!5 \rfloor = 9 + 2 = 11,$$while $$g(3) = -3(3) + \lceil 3 + 0,\!3 \rceil = -9 + 4 = -5.$$So, $f(3) + g(3) = 11+(-5)=6.$ This means that the statement in option D is correct.
Checking Option E:
Notice that $$f(-1) = 3(-1) + \lfloor -1-0,\!5 \rfloor = -3 + (-2) = -5,$$while $$g(4) = -3(4) + \lceil 4 + 0,\!3 \rceil = -12 + 5 = -7.$$So, $f(-1)-g(4) = -5-(-7)=2 > 0.$ This means that the statement in option E is not correct.

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## Problem Number 2

The statement that is not correct regarding the fractional part of real numbers is $\cdots \cdot$
A. $\left\{\dfrac85\right\} = \dfrac35$
B. $\left\{\dfrac17\right\} = \dfrac17$
C. $\left\{-\dfrac{11}{4}\right\} = \dfrac14$
D. $\left\{7\right\} = 0$
E. $\left\{\dfrac{22}{7}\right\} = \dfrac37$

Solution

Checking option A:
$$\left\{\dfrac85\right\} = \dfrac85-\left \lfloor \dfrac85 \right \rfloor = \dfrac85-1 = \dfrac35.$$So, the statement in option A is correct.
Checking option B:
$$\left\{\dfrac17\right\} = \dfrac17-\left \lfloor \dfrac17 \right \rfloor = \dfrac17-0 = \dfrac17.$$So, the statement in option B is correct.
Checking option C:
$$\left\{-\dfrac{11}{4}\right\} = -\dfrac{11}{4}-\left \lfloor -\dfrac{11}{4} \right \rfloor = -\dfrac{11}{4}+3 = \dfrac14.$$So, the statement in option C is correct.
Checking option D:
$$\left\{7\right\} = 7-\left \lfloor 7 \right \rfloor = 7-7 = 0.$$So, the statement in option D is correct.
Checking option E:
$$\left\{\dfrac{22}{7}\right\} = \dfrac{22}{7}-\left \lfloor \dfrac{22}{7} \right \rfloor = \dfrac{22}{7}-3 = \dfrac17.$$So, the statement in option E is false.

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## Problem Number 3

What is the solution of $\lfloor x \rfloor-3 = 0?$
A. $0 \le x \le 3.$
B. $2 \le x < 3.$
C. $2 \le x \le 3.$
D. $3 < x \le 4.$
E. $3 \le x < 4.$

Solution

The equation $\lfloor x \rfloor-3 = 0$ is equivalent to $\lfloor x \rfloor = 3.$ By considering the definition of floor function, we have $3 \le x < 4.$ So, the solution of $\lfloor x \rfloor-3 = 0$ is $\boxed{3 \le x < 4}.$

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## Problem Number 4

What is the solution of $3\lfloor x \rfloor+7 = -3?$
A. $-3 \le x < 7.$
B. $3 \le x < 7.$
C. $-10 \le x < -3.$
D. $-10 < x < 7.$

Solution

The equation $3\lfloor x \rfloor+7 = -3$ is equivalent to $\lfloor x \rfloor = -\dfrac{10}{3}.$ Because $\lfloor x \rfloor$ is an integer, the equation $\lfloor x \rfloor = -\dfrac{10}{3}$ can’t be satisfied by any value of $x.$ Consequently, $3\lfloor x \rfloor+7 = -3$ has no solution.

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## Problem Number 5

Let $a$ be a solution of $0 < 3 \lfloor 3x + 3 \rfloor < 7.$ If $b = 0,$ which of the following statements is definitely true?
A. $a > b.$                          D. $-a = b.$
B. $a \ge b.$                          E. $a = b.$
C. $a < b.$

Solution

Given $0 < 3 \lfloor 3x + 3 \rfloor < 7.$ Since $\lfloor 3x + 3 \rfloor$ is an integer, the possible values for the expression are either $\lfloor 3x + 3 \rfloor = 1$ or $\lfloor 3x + 3 \rfloor = 2.$
First case:
For $\lfloor 3x + 3 \rfloor = 1,$ we have $1 \le 3x + 3 < 2$ which implies $-\dfrac23 \le x < -\dfrac13.$
Second case:
For $\lfloor 3x + 3 \rfloor = 2,$ we have $2 \le 3x + 3 < 3$ which implies $-\dfrac13 \le x < 0.$

This shows that the values of $x$ that satisfy the inequality are $-\dfrac23 \le x < 0.$ So, in this case, $-\dfrac23 \le a < 0.$ Since $b = 0,$ it is clear that $a < b$ for any value of $a$ in that interval. Therefore, the statement that is definitely true is $\boxed{a < b}.$

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## Problem Number 6

Let $x$ be a real number. The values of $x$ that satisfy the equation $\lfloor x \rfloor + \lceil x \rceil = 5$ are $\cdots \cdot$
A. $0 < x < 1$
B. $1 < x < 2$
C. $2 < x < 3$
D. $1 < x < 3$
E. $2 \le x \le 3$

Solution

Given the equation $\lfloor x \rfloor + \lceil x \rceil = 5.$ Consider the following two cases.
Kasus 1: $x$ is an integer.
If $x$ is an integer, then $\lfloor x \rfloor = \lceil x \rceil = x,$ leading to $x + x = 5,$ which implies $x = \frac{5}{2}.$ However, this result contradicts the fact that $x$ is an integer. Therefore, this case does not provide a solution.
Kasus 2: $x$ is not an integer.
If $x$ is a non-integer real number, then it can be written as $x = n + r$ for some integer $n$ and a real number $0 < r < 1.$ Meanwhile, $\lfloor x \rfloor = n$ and $\lceil x \rceil = n+1$. Thus, we have
\begin{aligned} n + (n+1) & = 5 \\ 2n & = 4 \\ n & = 2. \end{aligned}Since $0 < r < 1$, it must be the case that $2 < 2 + r < 3$. Furthermore, because $x = n + r,$ we obtain $2 < x < 3.$ Therefore, the values of $x$ that satisfy the equation $\lfloor x \rfloor + \lceil x \rceil = 5$ are $\boxed{2 < x < 3}.$

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## Problem Number 7

What is the solution of $\lfloor 2x \rfloor + \lfloor x \rfloor = 7?$
A. $2 \le x \le 3.$
B. $2 \le x < 3.$
C. $2,\!5 < x \le 3.$
D. $2,\!5 \le x < 3.$

Solution

Let $x = n + r$ for some integer $n$ and real number $r$ with $0 \leq r < 1,$ so $\lfloor x \rfloor = n.$ This also means $2x = 2n + 2r.$ Thus,
$$\lfloor 2x \rfloor = \begin{cases} 2n, & 0 \leq r < \frac{1}{2} \\ 2n + 1, & \frac{1}{2} \leq r < 1. \end{cases}$$If $0 \leq r < \frac{1}{2},$ the equation $\lfloor 2x \rfloor + \lfloor x \rfloor = 7$ becomes $2n + n = 7,$ so $n = \frac{7}{3}$, which contradicts the fact that $n$ is an integer. Meanwhile, if $\frac{1}{2} \leq r < 1$, the equation $\lfloor 2x \rfloor + \lfloor x \rfloor = 7$ becomes $(2n+1) + n = 7,$ so $n = 2,$ which is consistent with the fact that $n$ is an integer.
So, the solution to the equation is $2 + \frac{1}{2} \leq x < 2 + 1,$ which means $2.5 \leq x < 3.$

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## Problem Number 8

Let $x$ and $y$ be positive real numbers such that $\left\lfloor \dfrac{x}{y} \right\rfloor = 3$ and $\left\lfloor \dfrac{x}{2y} \right\rfloor = 1.$ The value of $\left\lfloor \dfrac{x}{4y} \right\rfloor$ is $\cdots \cdot$
A. $0$                       C. $2$                    E. $5$
B. $1$                       D. $3$

Solution

Given $\left\lfloor \dfrac{x}{y} \right\rfloor = 3$ and $\left\lfloor \dfrac{x}{2y} \right\rfloor = 1.$ By the definition of the floor function, these two equations sequentially imply $3 \leq \dfrac{x}{y} < 4$ and $1 \leq \dfrac{x}{2y} < 2.$ Divide the inequality $3 \leq \dfrac{x}{y} < 4$ by $4$ to obtain
$$\dfrac{3}{4} \leq \dfrac{x}{4y} < 1.$$Similarly, divide the inequality $1 \leq \dfrac{x}{2y} < 2$ by $2$ to obtain
$$\dfrac{1}{2} \leq \dfrac{x}{4y} < 1.$$From this, it can be concluded that the interval of values for $\dfrac{x}{4y}$ is $\dfrac{3}{4} \leq \dfrac{x}{4y} < 1.$ As a result, the value of $\left\lfloor \dfrac{x}{4y} \right\rfloor$ is $\boxed{0}.$

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## Problem Number 9

Let $$x = \dfrac{2}{\dfrac{1}{1.001} + \dfrac{2}{1.002} + \dfrac{3}{1.003} + \cdots + \dfrac{10}{1.010}}.$$The value of $\lceil x \rceil = \cdots \cdot$
A. $35$                 C. $37$               E. $39$
B. $36$                 D. $38$

Solution

Let $$x = \dfrac{2}{\dfrac{1}{1.001} + \dfrac{2}{1.002} + \dfrac{3}{1.003} + \cdots + \dfrac{10}{1.010}}.$$By considering the lower and upper bounds of $x$ when the denominators are the same, we have
$$\begin{array}{ccccc} \dfrac{2}{\dfrac{1}{1.001} + \dfrac{2}{1.001} + \dfrac{3}{1.001} + \cdots + \dfrac{10}{1.001}} & < & x & < &\dfrac{2}{\dfrac{1}{1.010} + \dfrac{2}{1.010} + \dfrac{3}{1.010} + \cdots + \dfrac{10}{1.010}} \\ \dfrac{2}{55/1.001} & < & x & < & \dfrac{2}{55/1.010} \\ \dfrac{2.002}{55} & < & x & < & \dfrac{2.020}{55} \\ 36.4 & < & x & < & 36.7. \\ \end{array}$$Therefore, the value of $\lceil x \rceil = 37.$

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## Problem Number 10

Let $k$ be an integer that satisfies the equation
$$\left\lfloor \sqrt{\lfloor \sqrt{2.012} \rfloor} \right \rfloor = \left \lfloor \sqrt{\sqrt{2.012}} \right \rfloor + \left \lfloor \dfrac{k}{2.012} \right \rfloor.$$The number of values of $k$ that satisfy this equation is $\cdots \cdot$

A. $2.010$                   D. $2.013$
B. $2.011$                    E. $2.020$
C. $2.012$

Solution

Given $$\left\lfloor \sqrt{\lfloor \sqrt{2.012} \rfloor} \right \rfloor = \left \lfloor \sqrt{\sqrt{2.012}} \right \rfloor + \left \lfloor \dfrac{k}{2.012} \right \rfloor.$$From the given equation, we obtain
\begin{aligned} \left \lfloor \sqrt{\lfloor 44,\!\cdots \rfloor } \right \rfloor & = \lfloor \sqrt{44,\!\cdots} \rfloor + \left \lfloor \dfrac{k}{2.012} \right \rfloor \\ \lfloor \sqrt{44} \rfloor & = \lfloor 6,\!\cdots \rfloor + \lfloor \dfrac{k}{2.012} \rfloor \\ \lfloor 6,\!\cdots \rfloor & = 6 + \left \lfloor \dfrac{k}{2.012} \right \rfloor \\ 6 & = 6 + \left \lfloor \dfrac{k}{2.012} \right \rfloor \\ \left \lfloor \dfrac{k}{2.012} \right \rfloor & = 0. \end{aligned}The last equation implies that $0 \leq k < 2.012.$ Since $k$ is an integer, the number of values of $k$ that satisfy the condition $0 \leq k < 2.012$ is $2.012.$

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Bagian Uraian

## Problem Number 1

Evaluate the following.
a. $\lfloor 1,\!2 \rfloor$
b. $\lfloor -0,\!1 \rfloor$
c. $\left \lfloor -\dfrac34 \right \rfloor$
d. $\left \lfloor \dfrac12 + \left \lceil \dfrac12 \right \rceil \right \rfloor$
e. $\left \lceil 2 + \left \lceil -\dfrac12 \right \rceil \right \rceil$
f. $\left \lfloor \dfrac12 \cdot \left \lfloor \dfrac52 \right \rfloor \right \rfloor$

Solution

$\lfloor 1,\!2 \rfloor = 1.$
$\lfloor -0,\!1 \rfloor = -1.$
$\lceil -\dfrac34 \rceil = 0.$
$\left \lfloor \dfrac12 + \left \lceil \dfrac12 \right \rceil \right \rfloor = \left \lfloor \dfrac12 + 1 \right \rfloor = \left \lfloor \dfrac32 \right \rfloor = 1.$
$\lceil 2 + \lceil -\dfrac12 \rceil \rceil = \lceil 2 + 0 \rceil = \lceil 2 \rceil = 2.$
$\left \lfloor \dfrac12 \cdot \left \lfloor \dfrac52 \right \rfloor \right \rfloor = \left \lfloor \dfrac12 \cdot 2 \right \rfloor = \lfloor 1 \rfloor = 1.$

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## Problem Number 2

Let $x$ be a real number. Evaluate the value of $\lfloor x \rfloor + \lfloor -x \rfloor.$

Solution

Let $x$ be a real number. Consider the following two cases.
Case 1: $x$ is an integer.
If $x$ is an integer, then $\lfloor x \rfloor + \lfloor -x \rfloor = x + (-x) = 0.$
Case 2: $x$ is not an integer.
If $x$ is not an integer, it can be written as $x = n + r$ for some integer $n$ and a real number $0 < r < 1.$ Consequently, $-x = -n-r.$ Thus, we have
\begin{aligned} \lfloor x \rfloor + \lfloor -x \rfloor & = \lfloor n+r \rfloor + \lfloor -n-r \rfloor \\ & = n + (-n-1) \\ & = -1. \end{aligned}So, the value of $\lfloor x \rfloor + \lfloor -x \rfloor$ is $\begin{cases} 0, & \text{if}~x~\text{is an integer} \\ -1, & \text{if}~x~\text{is not an integer}. \end{cases}$

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## Problem Number 3

Let $x$ be a real number. Prove that $-\lfloor -x \rfloor$ is the smallest integer greater than or equal to $x.$

Solution

Let $x$ be a real number. Assume that $x$ is an integer. Consequently, $-\lfloor -x \rfloor = -(-x) = x,$ and it is certainly true that $x \geq x.$ Next, assume that $x$ is not an integer. This means $x = n + r$ for some integer $n$ and a real number $0 < r < 1.$ Therefore,
\begin{aligned} -\lfloor -x \rfloor & = -\lfloor -n-r \rfloor \\ & = -(-n + \lfloor -r \rfloor) \\ & = n-\lfloor -r \rfloor \\ & = n-(-1) \\ & = n+1. \end{aligned}
It is clear that $-\lfloor -x \rfloor = n+1 \geq n+r = x,$ and there are no other integers between $n+1$ and $n+r.$
So, it is concluded that $-\lfloor -x \rfloor$ is the smallest integer greater than or equal to $x$ for any real number $x.$ $\blacksquare$

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## Problem Number 4

Prove that $\lfloor x + y \rfloor \ge \lfloor x \rfloor + \lfloor y \rfloor$ for every real number $x$ and $y.$

Solution

Take any real numbers $x$ and $y.$ According to the definition of the floor function, it is known that $\lfloor x \rfloor \leq x$ and $\lfloor y \rfloor \leq y.$ By adding these two inequalities together on their respective sides, we obtain $\lfloor x \rfloor + \lfloor y \rfloor \leq x + y.$ As a result, $\lfloor x+y \rfloor \geq \lfloor \lfloor x \rfloor + \lfloor y \rfloor \rfloor = \lfloor x \rfloor + \lfloor y \rfloor$ because $\lfloor x \rfloor$ and $\lfloor y \rfloor$ are integers. Hence, $\lfloor x \rfloor + \lfloor y \rfloor$ must be an integer.  $\blacksquare$

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## Problem Number 5

Let $x$ be a real number and $n$ be an integer. Prove that $\lfloor x + n \rfloor = \lfloor x \rfloor + n.$

Solution

Let’s assume $\lfloor x \rfloor = m$ for some integer $m$ such that $m \le x < m+1.$ By adding $n$ to all three parts of the inequality, we get $$(m + n) \le (x + n) < (m+n) + 1.$$ As a result, $\lfloor x+n \rfloor = m + n.$ Since $\lfloor x \rfloor = m,$ it follows that $\lfloor x+n \rfloor = \lfloor x \rfloor + n.$
Therefore, it is proven that $\lfloor x + n \rfloor = \lfloor x \rfloor + n.$ $\blacksquare$

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## Problem Number 6

Show that $\left \lfloor \sqrt{\lfloor x \rfloor} \right \rfloor = \left \lfloor \sqrt{x} \right \rfloor$ for every nonnegative real number $x.$

Solution

Take any nonnegative real number $x.$ Let $x = n + r$ for some integer $n$ and a real number $0 \le r < 1.$ Also, let $n = a^2 + b$ where $a$ is the largest integer such that $a^2 \le n,$ and $b$ is a real number. Thus,
$$a^2 \le n = a^2 + b \le x = a^2 + b + r < (a + 1)^2.$$ Consequently, $x < (a + 1)^2$ implies $\sqrt{x} < a + 1$, and finally, $\left \lfloor \sqrt{x} \right \rfloor = a.$ Additionally, $\left \lfloor \sqrt{\lfloor x \rfloor} \right \rfloor = \left \lfloor \sqrt{n} \right \rfloor = a.$ Thus, it is proven that $\left \lfloor \sqrt{\lfloor x \rfloor} \right \rfloor = \left \lfloor \sqrt{x} \right \rfloor$ for any nonnegative real number $x.$ $\blacksquare$

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## Problem Number 7

Evaluate all positive integers $n$ that satisfies the equation $\left \lfloor \dfrac{n}{5} \right \rfloor = \dfrac{n}{6}.$

Solution

Given $\left \lfloor \dfrac{n}{5} \right \rfloor = \dfrac{n}{6}.$ Consider the following two cases.
Case 1: $\dfrac{n}{5}$ is an integer.
If $\dfrac{n}{5}$ is an integer, then $\dfrac{n}{5} = \dfrac{n}{6},$ which implies $5 = 6$ (contradictory).
Case 2: $\dfrac{n}{5}$ is not an integer.
If $\dfrac{n}{5}$ is not an integer, then it can be written as $\dfrac{n}{5} = k + r$ for some integer $n$ and a real number $0 < r < 1.$ Therefore, we have
\begin{aligned} \left \lfloor \dfrac{n}{5} \right \rfloor & = \dfrac{n}{6} \\ \left \lfloor \dfrac{n}{5} \right \rfloor & = \dfrac{n}{5} \cdot \dfrac{5}{6} \\ k & = (k+r) \cdot \dfrac{5}{6} \\ \dfrac{1}{6}k & = \dfrac{5}{6}r \\ \dfrac{k}{5} & = r. \end{aligned}Since $0 < r < 1,$ the last equation implies that the positive integer values of $k$ that satisfy it are $1 \le k \le 4.$

1. For $k = 1,$ we have $r = \dfrac{1}{5}$, so $n = 6.$
2. UnFor $k = 2,$ we have $r = \dfrac{2}{5}$, so $n = 12.$
3. For $k = 3,$ we have $r = \dfrac{3}{5}$, so $n = 18.$
4. For $k = 4,$ we have $r = \dfrac{4}{5}$, so $n = 24.$

So, there are $4$ positive integer values of $n$ that satisfy the equation, namely $n = 6, 8, 12, 24.$

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## Problem Number 8

Show that if $x$ and $y$ are positive real numbers, then $\lfloor xy \rfloor \ge \lfloor x \rfloor \lfloor y \rfloor.$ What are the implications if $x$ and $y$ are both negative or exactly one of them is negative?

Solution

Let $x$ and $y$ be positive real numbers. Also, let $x = a + r$ and $y = b + s$ for some integer $a$ and $b$ and real numbers $r$ and $s$ with $0 \le r, s < 1.$ This implies $\lfloor x \rfloor = a$ and $\lfloor y \rfloor = b.$ Therefore,
\begin{aligned} \lfloor xy \rfloor & = \lfloor (a+r)(b+s) \rfloor \\ & = \lfloor ab + as + br + rs \rfloor \\ & = ab + \lfloor as + br + rs \rfloor. \end{aligned}Since $\lfloor as + br + rs \rfloor$ cannot be negative and $\lfloor x \rfloor \lfloor y \rfloor = ab,$ we obtain $\lfloor xy \rfloor \ge \lfloor x \rfloor \lfloor y \rfloor.$ So, it is proven that if $x$ and $y$ are positive real numbers, then $\lfloor xy \rfloor \ge \lfloor x \rfloor \lfloor y \rfloor.$
Now, let’s consider the case when both $x$ and $y$ are negative real numbers. Using a similar approach, it can be shown that $\lfloor as + br + rs \rfloor$ cannot be positive. Consequently, $\lfloor xy \rfloor \le \lfloor x \rfloor \lfloor y \rfloor.$
Without loss of generality, suppose $x$ is a positive real number, and $y$ is negative. In this case, we cannot determine the direction of the inequality. For example, choose $x = 2$ and $y = -5.5,$ then $\lfloor 2 \cdot (-5.5) \rfloor = \lfloor -11 \rfloor = -11$ and $\lfloor 2 \rfloor \lfloor -5.5 \rfloor = 2(-6) = -12.$ This shows that $\lfloor xy \rfloor > \lfloor x \rfloor \lfloor y \rfloor$ for $x = 2$ and $y = -5.5.$
Next, choose $x = 5.5$ and $y = -1,$ then $\lfloor 5.5 \cdot (-1) \rfloor = \lfloor -5.5 \rfloor = -6$ and $\lfloor 5.5 \rfloor \lfloor -1 \rfloor = 5(-1) = -5.$ This shows that $\lfloor xy \rfloor < \lfloor x \rfloor \lfloor y \rfloor$ for $x = 5.5$ and $y = -1.$ $\blacksquare$

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Read: Problems and Solutions – Fibonacci Number and Fibonacci Sequence

## Problem Number 9

Determine and prove a formula for $\displaystyle \sum_{k=1}^n \left \lfloor \sqrt{k} \right \rfloor$ dalam $n$ and $\left \lfloor \sqrt{n} \right \rfloor.$
Hint: Recall that $\displaystyle \sum_{k=1}^n k^2 = \dfrac{n(n+1)(2n+1)}{6}.$

Solution

Claim that $$\displaystyle \sum_{k=1}^n \left \lfloor \sqrt{k} \right \rfloor = \left \lfloor \sqrt{n} \right \rfloor(n+1)-\dfrac{\left \lfloor \sqrt{n} \right \rfloor\left(\left \lfloor \sqrt{n} \right \rfloor + 1\right)\left(2\left \lfloor \sqrt{n} \right \rfloor + 1\right)}{6}.$$The proof is provided as follows.
We can start with some initial observations by expanding the sigma notation:
\begin{aligned} \displaystyle \sum_{k=1}^n \left \lfloor \sqrt{k} \right \rfloor & = \left \lfloor \sqrt{1} \right \rfloor + \left \lfloor \sqrt{2} \right \rfloor + \cdots + \left \lfloor \sqrt{n} \right \rfloor \\ & = 1 + 1 + 1 + \underbrace{2 + 2 + \cdots + 2}_{\text{5 suku}} + \underbrace{3 + 3 + \cdots + 3}_{\text{8 suku}} + \cdots + \left \lfloor \sqrt{n} \right \rfloor. \end{aligned}These observations lead to the following facts:

1. The sum $\displaystyle \sum_{k=1}^n \left \lfloor \sqrt{k} \right \rfloor$ contributes $1$ for every value of $k$ that satisfies $\sqrt{k} \ge 1.$ There are a total of $n$ such values, namely $k = 1, 2, 3, \ldots, n.$
2. The sum $\displaystyle \sum_{k=1}^n \left \lfloor \sqrt{k} \right \rfloor$ contributes $1$ for every value of $k$ that satisfies $\sqrt{k} \ge 2.$ There are a total of $n-3$ such values, namely $k = 4, 5, 6, \ldots, n.$
3. The sum $\displaystyle \sum_{k=1}^n \left \lfloor \sqrt{k} \right \rfloor$ contributes $1$ for every value of $k$ that satisfies $\sqrt{k} \ge 3.$ There are a total of $n-8$ such values, namely $k = 9, 10, 11, \ldots, n.$
4. In general, for $m = 1, 2, 3, \ldots, \left \lfloor \sqrt{n} \right \rfloor,$ the sum $\displaystyle \sum_{k=1}^n \left \lfloor \sqrt{k} \right \rfloor$ contributes $1$ for every value of $k$ that satisfies $\sqrt{k} \ge m,$ and there are $n-(m^2-1)$ such values of $k.$

From these observations, we can derive the following:
\begin{aligned} \displaystyle \sum_{k=1}^n \left \lfloor \sqrt{k} \right \rfloor & = \displaystyle \sum_{m=1}^{\left \lfloor \sqrt{n} \right \rfloor} (n-(m^2-1)) \\ & = \left \lfloor \sqrt{n} \right \rfloor(n+1)-\sum_{m=1}^{\left \lfloor \sqrt{n} \right \rfloor} m^2 \\ & = \left \lfloor \sqrt{n} \right \rfloor(n+1)-\dfrac{\left \lfloor \sqrt{n} \right \rfloor\left(\left \lfloor \sqrt{n} \right \rfloor + 1\right)\left(2\left \lfloor \sqrt{n} \right \rfloor + 1\right)}{6}. \end{aligned}So, it is proven that $$\displaystyle \sum_{k=1}^n \left \lfloor \sqrt{k} \right \rfloor= \left \lfloor \sqrt{n} \right \rfloor(n+1)-\dfrac{\left \lfloor \sqrt{n} \right \rfloor\left(\left \lfloor \sqrt{n} \right \rfloor + 1\right)\left(2\left \lfloor \sqrt{n} \right \rfloor + 1\right)}{6}.$$

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