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I am trying to solve the following problem:

For what real numbers $x$ is: $\lfloor2x-3\rfloor-3\lfloor x+2\rfloor=0$?

I have no idea how to do this, please help me

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  • $\begingroup$ Try plotting $x \mapsto \lfloor 2x-3 \rfloor$ and $x \mapsto 3 \lfloor x+2 \rfloor$. Note that $\lfloor x + n\rfloor = \lfloor x \rfloor +n$ for any integer $n$. $\endgroup$ – copper.hat Nov 28 '17 at 5:48
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As for integer $a, \lfloor{y+a}\rfloor=\lfloor y\rfloor+a$

$$6+3=\lfloor{2x}\rfloor-3\lfloor x\rfloor$$

Now let $x=I+f$

If $f<.5,$

$$9=2I-3I\iff I=?$$

Else $f\ge.5,$ $$9=2I+1-3I\iff I=?$$

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  • $\begingroup$ @samjoe, Thanks, please find the updated version $\endgroup$ – lab bhattacharjee Nov 28 '17 at 5:56
  • $\begingroup$ This method is pretty nifty! $\endgroup$ – jonsno Nov 28 '17 at 6:04
  • $\begingroup$ @samjoe, Thanks for your help $\endgroup$ – lab bhattacharjee Nov 28 '17 at 6:05
  • $\begingroup$ You forgot to multiple $\lfloor x\rfloor$ by $3$ $\endgroup$ – ℋolo Nov 28 '17 at 6:19
  • $\begingroup$ @Holo, Thanks for ur input $\endgroup$ – lab bhattacharjee Nov 28 '17 at 6:20
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Hint:

We know:

$x,y\in [z,z+1)$ then $\lfloor x\rfloor=\lfloor y\rfloor$

$\lfloor a+b\rfloor=\lfloor a\rfloor+b$ for integer $b$

With those 2:

$\lfloor2x-3\rfloor-3\lfloor x+2\rfloor=\lfloor2x\rfloor-3-3(\lfloor x\rfloor+2)=\lfloor2x\rfloor-3\lfloor x\rfloor-9=0\\\implies\lfloor2x\rfloor-3\lfloor x\rfloor=9$

Now divide it into 2 cases, when $\lfloor x\rfloor=\lfloor x+0.5\rfloor$ and when $\lfloor x\rfloor+1=\lfloor x+0.5\rfloor$

Can you continue from here?


Explanation about the cases:

I got to the equation $\lfloor2x\rfloor-3\lfloor x\rfloor=9$, let's look only on $\lfloor2x\rfloor$, if $\lfloor x+0.5\rfloor=\lfloor x\rfloor+1$ then $\lfloor2x\rfloor=2\lfloor x\rfloor+1$ elsewhere $\lfloor2x\rfloor=2\lfloor x\rfloor$

Solution to case one:

If $\lfloor x+0.5\rfloor=\lfloor x\rfloor+1$:

$$\lfloor2x\rfloor-3\lfloor x\rfloor=9\\2\lfloor x\rfloor+1-3\lfloor x\rfloor=9\\\lfloor x\rfloor=-8\\x=-8+c,c\in[0,1)\\\text{we know that $\lfloor x+0.5\rfloor=\lfloor x\rfloor+1$ so:}\\\lfloor -8+c+0.5\rfloor=\lfloor-8+c\rfloor+1=\lfloor-8+c+1\rfloor\\c\in[0.5,1)$$

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  • $\begingroup$ I understand what you have done but when you divide it into 2 cases, from where did you get the 0,5? Please help me i don't know how to continue $\endgroup$ – Carlos Toapanta Nov 28 '17 at 10:54
  • $\begingroup$ @CarlosToapanta add edit the solution, see if it is clearer $\endgroup$ – ℋolo Nov 28 '17 at 11:36
  • $\begingroup$ I think that x must be equal to -7+c instead of -8+c $\endgroup$ – Carlos Toapanta Nov 28 '17 at 14:05
  • $\begingroup$ @CarlosToapanta no, you maybe did a mistake because of the $-$, try to put values in the form of $-8+c=k\in(-7,-7.5]$ and you will see. $\endgroup$ – ℋolo Nov 28 '17 at 14:39
  • $\begingroup$ Yes you are right, so to finally solve the case i have to apply x=-8+c yo get x? $\endgroup$ – Carlos Toapanta Nov 28 '17 at 14:52

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