Solving $\lfloor {2x} \rfloor + \lceil {x} \rceil - 4x = 0$

Question

I need to find: $A=\{x \in \mathbb{R}\vert\,\lfloor {2x} \rfloor + \lceil {x} \rceil - 4x = 0\}$

Let $x \in \mathbb{R}$. Then $$\begin{align*} \lceil {x} \rceil =4x- \lfloor {2x} \rfloor &\iff 4x- \lfloor {2x} \rfloor -1 < x \leq 4x -\lfloor {2x} \rfloor\\ &\iff \bigg[ 4x- \lfloor {2x} \rfloor -1 < x \bigg] \land \bigg[ x \leq 4x -\lfloor {2x} \rfloor \bigg]\\ &\iff (3x-1 < \lfloor {2x} \rfloor ) \land (\lfloor {2x} \rfloor \leq 3x) \end{align*} $$ Now define $B=\{x\in \mathbb R\mid(3x-1<\lfloor {2x} \rfloor \}$ and $C=\{x\in\mathbb R\mid\lfloor {2x} \rfloor \leq 3x\}$. Then $A=B\cap C$.

My confusion: $-\dfrac{2}{3} \in (B \cap C)$ but $-\dfrac{2}{3} \notin A$

In which step have I gone wrong?

Answer 1

The first step missed the condition that $\lceil x\rceil$ is an integer

The first step in your equivalence transformation as quoted below is wrong.

$$ \lceil x\rceil =4x- \lfloor 2x\rfloor \iff 4x- \lfloor 2x\rfloor -1<x\le 4x- \lfloor 2x\rfloor $$

The definition of $\lceil x \rceil$ is "the least integer that is not smaller than $x$". Or, what is equivalent, "the integer that is $\ge x$ and $<x+1$". Hence, the first transformation should have been

$$ \lceil x\rceil =4x- \lfloor 2x\rfloor \\ \iff 4x- \lfloor 2x\rfloor \text{ is an integer and } x\le 4x- \lfloor 2x\rfloor < x +1 $$

A full solution

$\lceil x\rceil =4x- \lfloor 2x\rfloor$ implies $4x$ is an integer. There are three cases.

• $x$ is an integer.
  Then $x = 4x-2x$.
  $x=0$
• $x$ is not an integer but $2x$ is an integer.
  Then $\lceil x\rceil = x+\frac12$.
  $x +\frac12 = 4x-2x$.
  $x=\frac12$.
• $2x$ is not an integer but $4x$ is an integer, i.e., $\lfloor 2x\rfloor = 2x-\frac12$. There are two subcases.
  • $x$ is $\frac14$ plus an integer, i.e., $\lceil x\rceil = x+\frac34$.
    $x+\frac34=4x-(2x-\frac12)$
    $x=\frac14$
  • $x$ is $\frac34$ plus an integer, i.e., $\lceil x\rceil = x+\frac14$.
    $x+\frac14=4x-(2x-\frac12)$
    $x=-\frac14$

Hence $x\in\{-\frac14, 0, \frac14, \frac12\}$.

Answer 2

"In which step have I gone wrong?" In the first one: The left hand side implies that $x$ is divisible by $4$ while the right side does not.

Here is how I would approach it.

Notice that $\lfloor 2x \rfloor$ and $\lceil x\rceil$ are always whole numbers so that $\lfloor 2x \rfloor+\lceil x\rceil -4x$ is a whole number if and only if $x=n/4$ for some $n\in\mathbb Z$. Now assume $x=n/4$ is in $A$. Then $$\lfloor 2x \rfloor+\lceil x\rceil -4x=0$$ which for $n$ means $$\lfloor n/2 \rfloor+\lceil n/4\rceil =n.$$ Now notice $$ n/2-1+n/4\le\lfloor n/2 \rfloor+\lceil n/4\rceil \le n/2+ n/4+1 $$ and thus $$ n/2-1+n/4\le n\le n/2+ n/4+1 $$ which by multiplying with $4$ yields $$ (3n-1\le 4n\le 3n+1) \iff (-4 \le n \le +4) $$ Thus the only possible values for $n$ are $\{-4,\dots,4\}$, which are the values $\{-1,-1+1/4,-1+2/4,\dots,1\}$ for $x$. (You still have to check which of these values are in $A$. But after that you are done)

Answer 3

Solving $\lfloor {2x} \rfloor + \lceil {x} \rceil - 4x = 0$

Alternative approach:

First of all, the answer of Alex, which is more elegant than my approach, may well have been intended by the problem composer.

My standard approach to problems like this is to assume that $~x~$ has form $~P + r ~: ~P \in \Bbb{Z}, ~0 \leq r < 1.$

Note that $r = 0 \implies$

• $\lfloor {2x} \rfloor = 2x.$
• $\lceil {x} \rceil = x.$
• Therefore, $~3x - 4x = 0 \implies x = 0.$

So, one of the solutions is $~x = 0,~$ and (besides that solution), you can assume, without loss of generality, that $0 < r < 1.$ This allows you to make the simplifying assumption that $~\lceil {x} \rceil = P + 1.$

It remains to express $~\lfloor {2x} \rfloor~$ based on $~P~$ and $~r~$. Here, you have that :

• Case 1: $~0 < r < (1/2) \implies \lfloor{2x}\rfloor = 2P.$

• Case 2: $~(1/2) \leq r < 1 \implies \lfloor{2x}\rfloor = 2P + 1.$

Since Cases 1 and 2 above are mutually exclusive, the problem can be resolved by taking the union of $~\{x=0\},~$ and all the values of $~x~$ that satisfy either Case 1 or Case 2.

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$\underline{\text{Case 1}}$
$0 < r < (1/2).$

Then $~(2P) + (P+1) - 4(P+r) = 0 \implies 1 = P + 4r.$

Since $~P \in \Bbb{Z},~$ and $~0 < r < (1/2),~$ Case 1 requires that $~r = (1/4),~$ which implies that $P = 0.$

So, the only satisfying value for Case 1 is $x = 0 + (1/4) = (1/4).$

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$\underline{\text{Case 2}}$
$(1/2) \leq r < 1.$

Then $~(2P+1) + (P+1) - 4(P+r) = 0 \implies 2 = P + 4r.$

Since $~P \in \Bbb{Z},~$ you must have that $~r~$ is some value in $~(1/2) \leq r < 1,~$ such that $~4r~$ is an integer. Therefore, $~r~$ must be some element in $~\{ ~1/2, ~3/4 ~\}.$

$~r = (1/2) \implies 4r = 2 \implies P = 0 \implies x = (1/2).$

$~r = (3/4) \implies 4r = 3 \implies P = -1 \implies x = -1 + (3/4) = -1/4.$

Therefore, the satisfying values for Case 2 are $\{ ~1/2, ~-1/4 ~\}.$

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So, the final answer is

$$\{ ~0 ~\} \cup \{ ~1/4 ~\} \cup \{ ~1/2, ~-1/4 ~\}.$$