Question 1209891: Find all x for which
\Big| x - |x - 2| \Big| = \lfloor x \rfloor + \lfloor 2x \rfloor
Express your answer in interval notation.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's analyze the given equation:
$$|x - |x - 2|| = \lfloor x \rfloor + \lfloor 2x \rfloor$$
We need to consider cases based on the value of $x$.
**Case 1: $x \ge 2$**
* $|x - 2| = x - 2$
* $|x - |x - 2|| = |x - (x - 2)| = |2| = 2$
* $\lfloor x \rfloor + \lfloor 2x \rfloor = 2$
Let $x = n + f$, where $n = \lfloor x \rfloor$ is an integer and $0 \le f < 1$.
Then $2 = n + \lfloor 2(n+f) \rfloor = n + \lfloor 2n + 2f \rfloor = n + 2n + \lfloor 2f \rfloor = 3n + \lfloor 2f \rfloor$.
* If $0 \le f < \frac{1}{2}$, then $\lfloor 2f \rfloor = 0$, so $2 = 3n$, which gives $n = \frac{2}{3}$, but we need $n \ge 2$, so no solution.
* If $\frac{1}{2} \le f < 1$, then $\lfloor 2f \rfloor = 1$, so $2 = 3n + 1$, which gives $3n = 1$ and $n = \frac{1}{3}$, but we need $n \ge 2$, so no solution.
Therefore, there are no solutions for $x \ge 2$.
**Case 2: $0 \le x < 2$**
* $|x - 2| = 2 - x$
* $|x - |x - 2|| = |x - (2 - x)| = |2x - 2|$
* $\lfloor x \rfloor + \lfloor 2x \rfloor = |2x - 2|$
Let $x = n + f$, where $n = \lfloor x \rfloor$ and $0 \le f < 1$.
Since $0 \le x < 2$, we have $n = 0$ or $n = 1$.
* **Subcase 2.1: $0 \le x < 1$**
* $n = 0$
* $|2x - 2| = \lfloor x \rfloor + \lfloor 2x \rfloor = 0 + \lfloor 2x \rfloor = \lfloor 2x \rfloor$
* Since $0 \le x < 1$, we have $0 \le 2x < 2$, so $\lfloor 2x \rfloor = 0$ or $1$.
* If $\lfloor 2x \rfloor = 0$, then $|2x - 2| = 0$, so $2x - 2 = 0$, which means $x = 1$. But $x < 1$, so this is a contradiction.
* If $\lfloor 2x \rfloor = 1$, then $|2x - 2| = 1$.
* $2x - 2 = 1$, so $2x = 3$ and $x = \frac{3}{2}$. But $x < 1$, so this is a contradiction.
* $2x - 2 = -1$, so $2x = 1$ and $x = \frac{1}{2}$.
* Check: $|\frac{1}{2} - |\frac{1}{2} - 2|| = |\frac{1}{2} - \frac{3}{2}| = |-1| = 1$
* $\lfloor \frac{1}{2} \rfloor + \lfloor 1 \rfloor = 0 + 1 = 1$.
* Thus, $x = \frac{1}{2}$ is a solution.
* **Subcase 2.2: $1 \le x < 2$**
* $n = 1$
* $|2x - 2| = \lfloor x \rfloor + \lfloor 2x \rfloor = 1 + \lfloor 2x \rfloor$
* Since $1 \le x < 2$, we have $2 \le 2x < 4$, so $\lfloor 2x \rfloor = 2$ or $3$.
* If $\lfloor 2x \rfloor = 2$, then $|2x - 2| = 1 + 2 = 3$.
* $2x - 2 = 3$, so $2x = 5$ and $x = \frac{5}{2}$. But $x < 2$, so this is a contradiction.
* $2x - 2 = -3$, so $2x = -1$ and $x = -\frac{1}{2}$. But $x \ge 1$, so this is a contradiction.
* If $\lfloor 2x \rfloor = 3$, then $|2x - 2| = 1 + 3 = 4$.
* $2x - 2 = 4$, so $2x = 6$ and $x = 3$. But $x < 2$, so this is a contradiction.
* $2x - 2 = -4$, so $2x = -2$ and $x = -1$. But $x \ge 1$, so this is a contradiction.
**Case 3: $x < 0$**
* $|x - 2| = 2 - x$
* $|x - |x - 2|| = |x - (2 - x)| = |2x - 2| = 2 - 2x$ (since $x < 0$, $2x - 2 < 0$)
* $\lfloor x \rfloor + \lfloor 2x \rfloor = 2 - 2x$
Let $x = -y$, where $y > 0$.
$\lfloor -y \rfloor + \lfloor -2y \rfloor = 2 + 2y$
$-(\lceil y \rceil + \lceil 2y \rceil) = 2 + 2y$
$\lceil y \rceil + \lceil 2y \rceil = -2 - 2y$
Since $\lceil y \rceil$ and $\lceil 2y \rceil$ are positive, the left side is positive, but the right side is negative. So there are no solutions.
**Final Answer:** The final answer is $\boxed{\{\frac{1}{2}\}}$
Answer by ikleyn(52834)  (Show Source):
You can put this solution on YOUR website! .
What is written in this post ?
To the managers of this project !
It looks like that the person who submitted this post to the forum,
is unfamiliar with Mathematical grammar and with the rules of writing Math texts.
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