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Find the domain of x

$$\left \lfloor x \right \rfloor + \left \lfloor x+\frac{1}{2} \right \rfloor + \left \lfloor x-\frac{1}{3} \right \rfloor =8$$ My approach

When x is an integer $x+x+x-1=8$ or x=3

But for case when x is not an integer i am not able to substitute, manually x is $\frac{10}{3}$ which i did by trial method.

My final answer is $3\le x <\frac{10}{3} $

Samar Imam Zaidi
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  • I just want to know whether we can do it by substituting and term of x – Samar Imam Zaidi Sep 12 '18 at 20:08
  • Hint: Left hand side is an increasing function. Your approach is good, $x$ can't be less, it can only increase. Next increase is of LHS function is after we add $1/3$ to $x$. – Jakobian Sep 12 '18 at 20:15
  • The equation when $x$ is an integer should be $x + x(x-1) = 8$ with solution $x = \sqrt{8}$ (which however is a non-integer). Or have you missed a plus sign between $\lfloor x+\frac12 \rfloor$ and $\lfloor x-\frac13 \rfloor$? – md2perpe Sep 12 '18 at 21:14
  • I made the modification I added + sign – Samar Imam Zaidi Sep 13 '18 at 02:09

1 Answers1

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Let $x = k + y$, where $k$ is an integer part of $x$ ($k=\left \lfloor x \right \rfloor$) and $y$ is its fractional part. Consider $y$-s belonging to the following intervals $\left[0, \frac{1}{3}\right)$, $\left[\frac{1}{3}, \frac{1}{2}\right)$ and $\left[\frac{1}{2}, 1\right)$.

mwt
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