The question is to find the solutions to this inequation $$\lfloor x\rfloor^2 - \lfloor x\rfloor - 6 > 0$$

On factorising I got $$(\lfloor x\rfloor-3)(\lfloor x\rfloor+2)>0$$ so $\lfloor x\rfloor>3$ or $\lfloor x\rfloor>-2$ and so on

But in the text book the solution goes like this

$\lfloor x\rfloor>3$ so $x\ge 4$ and $\lfloor x\rfloor<-2$ so $x <-2$ therefore domain is...

Notice that the textbook has reversed the second inequality and it turns out to be correct as I checked it in wolfram alpha.

Can anyone please explain why the book reversed $\lfloor x\rfloor>-2$ into $\lfloor x\rfloor<-2$?


"On factorising I got $(\lfloor x\rfloor-3)(\lfloor x\rfloor +2)>0$, so $\lfloor x\rfloor>3$ or $\lfloor x\rfloor >-2$ and so on"

The bold "so" is wrong: for $a\le b\in\Bbb R$,

$$(t-a)(t-b)>0\iff t< a\vee t>b$$

You applied it incorrectly to the case $a=-2$, $b=3$

  • $\begingroup$ can i say a <t v b>t $\endgroup$ – user453135 Jul 26 '17 at 6:44
  • $\begingroup$ by the way thanks for answering and correcting my mistake. $\endgroup$ – user453135 Jul 26 '17 at 6:45
  • $\begingroup$ @user453135 No, it is not the same thing. For instance, when $a<b$, all $t\in\Bbb R$ verify that condition. $\endgroup$ – user228113 Jul 26 '17 at 6:48

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