Category: Aljabar

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.
(Answer D)

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.
(Answer E)

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 $\boxed{3 \le x < 4}.$
(Answer E)

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.
(Answer E)

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 $\boxed{a < b}.$
(Answer C)

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 $\boxed{2 < x < 3}.$
(Answer C)

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 $2.5 \leq x < 3.$
(Answer D)

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 $\boxed{0}.$ 
(Answer A)

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 $\lceil x \rceil = 37.$
(Answer C)

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 $2.012.$
(Answer C)

Answer a)
$\lfloor 1,\!2 \rfloor = 1.$
Answer b)
$\lfloor -0,\!1 \rfloor = -1.$
Answer c)
$\lceil -\dfrac34 \rceil = 0.$
Answer d)
$\left \lfloor \dfrac12 + \left \lceil \dfrac12 \right \rceil \right \rfloor = \left \lfloor \dfrac12 + 1 \right \rfloor = \left \lfloor \dfrac32 \right \rfloor = 1.$
Answer e)
$\lceil 2 + \lceil -\dfrac12 \rceil \rceil = \lceil 2 + 0 \rceil = \lceil 2 \rceil = 2.$
Answer f)
$\left \lfloor \dfrac12 \cdot \left \lfloor \dfrac52 \right \rfloor \right \rfloor = \left \lfloor \dfrac12 \cdot 2 \right \rfloor = \lfloor 1 \rfloor = 1.$

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 $\begin{cases} 0, & \text{if}~x~\text{is an integer} \\ -1, & \text{if}~x~\text{is not an integer}. \end{cases}$

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 $x.$ $\blacksquare$

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.  $\blacksquare$

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

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 $x.$ $\blacksquare$

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 $n = 6, 8, 12, 24.$

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 $y = -1.$ $\blacksquare$

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 $$\displaystyle \sum_{k=1}^n \left \lfloor \sqrt{k} \right \rfloor= \left \lfloor \sqrt{n} \right \rfloor(n+1)-\dfrac{\left \lfloor \sqrt{n} \right \rfloor\left(\left \lfloor \sqrt{n} \right \rfloor + 1\right)\left(2\left \lfloor \sqrt{n} \right \rfloor + 1\right)}{6}.$$