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

  • $\begingroup$ 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

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

  • $\begingroup$ @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


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

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.