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zmudz
 one year ago
Find all \(x\) for which
\(\left x  \left x1 \right \right = \lfloor x \rfloor\)
The answer is suppose to be in a range, but when I put in that it can be all real numbers, that is wrong.
zmudz
 one year ago
Find all \(x\) for which \(\left x  \left x1 \right \right = \lfloor x \rfloor\) The answer is suppose to be in a range, but when I put in that it can be all real numbers, that is wrong.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I would try using the definition of the absolute value function to rewrite the LHS into something that you can easily match up with the RHS. \[x=\begin{cases}x&\text{for }x>0\\0&\text{for }x=0\\x&\text{for }x<0\end{cases}\] This means \[\begin{align*}xx1&=\begin{cases}xx1&\text{for }xx1>0\\0&\text{for }xx1=0\\x+x1&\text{for }xx1<0\end{cases}\\[2ex] &=\begin{cases}xx1&\text{for }x>\dfrac{1}{2}\\[1ex]0&\text{for }x=\dfrac{1}{2}\\[1ex]x+x1&\text{for }x<\dfrac{1}{2}\end{cases} \end{align*}\] You also have \[\begin{align*} x1&=\begin{cases}x1&\text{for }x>1\\0&\text{for }x=1\\x+1&\text{for }x<1\end{cases} \end{align*}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Putting all this together, you get \[\begin{align*} xx1&= \begin{cases} 1&\text{for }x>1\\x&\text{for }x=1\\[1ex]2x1&\text{for }\dfrac{1}{2}<x<1\\[1ex]0&\text{for }x=\dfrac{1}{2}\\[1ex]2x+1&\text{for }x<\dfrac{1}{2} \end{cases} \end{align*}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now, given that \[\lfloor x\rfloor=\begin{cases}&\vdots\\ 0&\text{for }0\le x<1\\ 1&\text{for }1\le x<2 \\&\vdots\end{cases}\] you have the isolated solution \(x=\dfrac{1}{2}\) (since both LHS and RHS are zero) and the range \([1,2)\).
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